137Chemical Bonding
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7
CHEMICAL BONDING
In lesson 5, you have read about the electronic configuration of atoms of various
elements and variation in the periodic properties of elements. We see various
substances around us which are either elements or compounds. You also know thatatoms of  the same or different elements may combine. When atoms of  the sameelements combine, we get molecules of the elements. But we get compounds whenatoms of different elements combine. Have you ever thought why atoms combineat all?
In this lesson, we will find an answer to this question. We will first explain what a
chemical bond is and then discuss various types of chemical bonds which join theatoms together to give various types of substances. The discussion would alsohighlight how these bonds are formed.
The properties of substances depend on the nature of bonds present between their
atoms. In this lesson you will learn that sodium chloride, the common salt and washing
soda dissolve in water whereas methane gas or napthalene do not. This is because
the type of bonds present between them are different. In addition to the differencein solubility, these two types of compounds differ in other properties as well aboutwhich  you will study in this lesson.
OBJECTIVES
After completing this lesson you will be able to :
/circle6recognize the stability of noble gas configuration and tendency of other
elements to attain this configuration through formation of chemical bonds;
/circle6explain the attainment of stable noble gas electronic configuration throughtransfer of electrons resulting in the formation of ionic bonds;Chemical Bonding
SCIENCE AND TECHNOLOGY  138Notes
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/circle6describe and justify some of the common properties of ionic compounds;
/circle6explain the alternate mode of attainment of stable noble gas configuration
through sharing of electrons resulting in the formation of covalent bonds;
/circle6describe the formation of single, double and triple bonds and depict these
with the help of Lewis-dot method;
/circle6describe and justify some of the common properties of covalent substances.
7.1 WHY DO ATOMS COMBINE?
The answer to this question is hidden in the electronic configurations of the noblegases. It was found that noble gases namely helium, neon, argon, krypton, xenon
and radon did not react with other elements to form compounds i.e. they were non
-reactive. In the initial stages they were also called inert gases due to their non-reactivenature. Thus it was, thought that these noble gases lacked reactivity because of theirspecific electronic arrangements which were quite stable. When we write theelectronic configurations of the noble gases (see table below), we find that except
helium all of them have 8 electrons in their outermost shell.
Table 7.1 : Electronic configuration of Noble gases
Name Symbol Atomic Electronic No. of electrons in the
Number Configuration outermost shell
Helium He 2 2 2
Neon Ne 10 2,8 8Argon Ar 18 2,8,8 8
Krypton Kr 36 2,8,18,8 8
Xenon Xe 54 2,8,18,18,8 8
Radon Ra 86 2,8,18,32,18,8 8
It was concluded that atoms having 8 electrons in their outermost shell are very stable
and they did not form compounds. It was also observed that other atoms such ashydrogen, sodium, chlorine etc. which do not have 8 electrons in their outermost shellundergo chemical reactions. They can stabilize by combining with each other andattain the above configurations of noble gases i.e. 8 electrons (or 2 electrons in case
of helium) in their outermost shells. Thus, atoms tend to attain a configuration in which
they have 8 electrons in their outermost shells. This is the basic cause of chemicalbonding. This attainment of eight electrons for stable structure is called the octet rule .
The octet rule explains the chemical bonding in many compounds.
Atoms are held together in compounds by the forces of attraction which result in
formation of chemical bonds . The formation of chemical bonds results in the lowering139Chemical Bonding
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of energy which is less than  the energy the individual atoms. The resulting compound
is lower in energy as compared to sum of energies of the reacting atom/molecule
and hence is more stable. Thus stability of the compound formed is an importantfactor in the formation of chemical bonds. In rest of the lesson you will study about
the nature of bonds present in various substances. We would explain ionic bonding
and covalent bonding in this lesson . Before you start learning about ionic bonding
in the next section you can answer the following questions to check your
understanding.
 INTEXT QUESTIONS 7.1
1. State octet rule
2. Why noble gases are non-reactive?
3. In the table given below three elements and their atomic numbers are given. Which
of them are stable and will not form compound?
Element At. No. Stable/Unstable
A1 0
B3 6
C3 7
7.2 IONIC BONDING
The chemical bond formed by transfer of electron from a metal to a  non- metal
is known as ionic  or electrovalent bond .
 For example, when sodium metal and chlorine gas are brought into contact, they
react violently and we obtain sodium chloride. This reaction is shown below:
2Na(s) + Cl2(g) ⎯⎯→ 2NaCl(s)
The bonding in sodium chloride can be understood as follows:
Sodium (Na) has the atomic number 11 and we can write its electronics configuration
as 2,8,1 i.e. it has one electron in its outermost (M) shell. If it loses this electron,
it is left with 10 electrons and becomes positively charged. Such a positively chargedion is called a cation. The cation in this case is called sodium cation, Na
+. This is
shown below in Fig. 7.1.Chemical Bonding
SCIENCE AND TECHNOLOGY  140Notes
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Fig. 7.1 Formation of NaCl
Note that the sodium cation has 11 protons but 10 electrons only. It has 8 electrons
in the outermost (L) shell. Thus, sodium atom has attained the noble gas configurationby losing an electron present in its outermost shell. Loss of electron results into
formation of an ion and this process is called ionization . Thus, according to octet
rule, sodium atom can acquire stability by changing to sodium ion (Na
+).
The ionization of sodium atom to give sodium ion requires an energy of
496 kJ mol–1.
Now, chlorine atom having the atomic number 17, has the electronic configuration
2,8,7. It completes its octet by gaining one electron from sodium atom (at. no. 11)
with electronic configuration 2, 8, 1.
Both sodium ion (Na+) and chloride ion (Cl–) combine together by ionic bond and
become solid sodium chloride (NaCl).
Note that in the above process, the chlorine atom has gained an additional electron
hence it has become a negatively charged ion (Cl–). Such, a negatively charged ion
is called an anion . Chloride ion has 8 electrons in its outermost shell and it therefore,
has a stable electronic configuration according to the octet rule. The formation of
chloride ion from the chlorine atom releases 349 kJ mol-1 of energy.
Since the cation  (Na+) and the anion  (Cl–) formed above are electrically charged
species, they are held together by Coulombic force or electrostatic force of attraction.
This electrostatic force of attraction which holds the cation and anion together
is known as electrovalent bond or ionic bond . This is represented as follows:
Na+(g)  +  Cl–(g) ⎯⎯→ Na+Cl–  or NaCl(s)
Note that only outermost electrons are shown above. Such structures are also called
Lewis Structures .
If we compare the energy required for the formation of sodium ion and that
released in the formation of chloride ion, we note that there is a net difference of147 kJ mol
–1 of energy. If only these two steps are involved, the formation of sodium
chloride is not  favourable energetically. But sodium chloride exists as a crystalline
solid. This is because the energy is released when the sodium ions and the chlorideions come together to form the crystalline structure. The energy so released
compensates for the above deficiency of energy.141Chemical Bonding
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You can see that each sodium ion is surrounded by six chloride ions and each
chloride ion is surrounded by six sodium ions  in its solid state structure. The force
of attraction between sodium and chloride ions is uniformly felt in all directions. Thus,
no particular sodium ion is bonded to a particular chloride ion. Hence, there is no species
such as NaCl. Here NaCl is empirical formula and shows that there is one Na+ for
every Cl– Fig. 7.2.
Chloride Ion Sodium Ion
Fig. 7.2 Structure of sodium chloride
Similarly, we can explain the formation of cations resulting from lithium and potassium
atoms and the formation of anions resulting from fluorine, oxygen and sulphur atoms.
Let us now study the formation of another ionic compound namely magnesium
chloride. Mg has atomic number 12. Thus, it has 12 protons. The number of electronspresent in it is also 12. Hence the electronic configuration of Mg atom is 2, 8, 2.
Let us consider the formation of magnesium ion from a magnesium atom. We see
that it has 2 electrons in its outermost shell. If it loses these two electrons, then wecan achieve the stable configuration of 2, 8 (that of noble gas neon). This can berepresented in Fig. 7.3.
Mg
 ⎯⎯→ Mg2+ + 2e–
2, 8, 2 2, 8
Fig. 7.3 Formation of magnesium ion
You can see that the resulting magnesium ion has only 10 electrons and hence it has
2+ charge. It is a dipositive ion and can be represented as Mg2+ ion.
The two electrons lost by the magnesium are gained -one each by two chlorine atoms
to give two chloride ions.
2[Cl(g) + e–
 ⎯⎯→ Cl–(g)]
or 2Cl(g) + 2e–
 ⎯⎯→ 2Cl–(g)
Thus, one magnesium ion and two chloride ion join together to give magnesium
chloride, MgCl2. Hence we can write as in Fig. 7.4.Chemical Bonding
SCIENCE AND TECHNOLOGY  142Notes
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Fig. 7.4 Formation of magnesium chloride
Let us now see what would happen if instead of chloride ion, the magnesium ion
combines with another anion say oxide anion. The oxygen atom having atomic numbereight has 8 electrons. Its electronic configuration is 2,6. It can attain a stable electronic
arrangement (2,8) of the noble gas neon if it gains two more electrons. The two
electrons, which are lost by the magnesium atom, are gained by the oxygen atom.On gaining these two electrons, the oxygen atom gets converted into the oxide anion.
This is shown below in Fig. 7.5.
O + 2e
–
 ⎯⎯→ O2–
2, 6 2, 8
Fig. 7.5 Formation of oxide ion
The oxide has 2 more electrons as compared to the oxygen atom. Hence, it has 2
negative charges on it. Therefore, it can be represented as O2– ion
The magnesium ion (Mg2+) and the oxide ion (O2–) are held together by electrostatic
force of attraction. This leads to the formation of magnesium oxide Fig. 7.6.
Fig. 7.6 Formation of magnesium oxide
Thus, magnesium oxide is an ionic compound in which a dipositive cation (Mg2+)
and a dinegative anion (O2–) are held together by electrostatic force.
Similar to the case of sodium chloride, the formation of magnesium oxide is also
accompanied by lowering of energy which leads to the stability of magnesium oxide
as compared to individual magnesium and oxygen atoms.
Similarly, the ionic bonding present in many other ionic compounds can be explained.
The ionic compounds show many characteristic properties which are discussed
below.143Chemical Bonding
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7.2.1 Properties of Ionic Compounds
Since the ionic compounds contain ions (cations and anions) which are held together
by the strong electrostatic forces of attraction, they show the following generalcharacteristic properties:
(a) Physical State
Ionic compounds are crystalline solids. In the crystal, the ions are arranged in a regularfashion. The ionic compounds are hard and brittle in nature.
(b) Melting and boiling points
Ionic compounds have high melting and boiling points. The melting point of sodiumchloride is 1074 K (801°C) and its boiling point is 1686K (1413°C). The meltingand boiling points of ionic compounds are high because of the strong electrostaticforces of attraction present between the ions. Thus, it requires a lot of thermal energyto overcome these forces of attraction. The thermal energy given to the ioniccompounds is used to overcome the interionic attractions present between the cationsand anions in an ionic crystal. Remember that the crystal has a three dimensionalregular arrangement of cations and anions which is called crystal lattice . On heating,
the breaking of this crystal lattice leads to the molten state of the ionic compoundin which the cations and anions are free to move.
(c) Electrical Conductivity
Ionic compounds conduct electricity in their molten state and in aqueous solutions.Since ions are free to move in the molten state, they can carry current from oneelectrode  to another in a cell. Thus ions can conduct electricity in molten state.
However, in solid state, such a movement of ions is not possible as they occupy fixed
positions in the crystal lattice. Hence in solid state, ionic compounds do not conductelectricity.
In aqueous solution, water is used as a solvent to dissolve ionic compounds. It
weakens the electrostatic forces of attraction present among the ions. When theseforces are weakened, the ions become free to move, hence they can conductelectricity.
ACTIVITY 7.1
Prepare a solution of NaCl by dissolving 1 tablespoon of it in 100 mL water. Take
this solution in a 200 mL beaker and introduce two graphite electrode (obtained fromused dry cell battery), Now connect the electrode with a 3 V dry cell and a bulbin a circuit as shown in Fig. 7.7. Initially take plane water in a beaker (200 mL)and see the glow of bulb. Now replace the plane water by the solution of NaCl,what difference in glow of the bulb is observed? Interpret the result on the basis ofionic bond you have just studied.Chemical Bonding
SCIENCE AND TECHNOLOGY  144Notes
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- +Bulb
Switch
Electrodes
Sodium chloride
solution3V
Fig. 7.7 Aqueous solution of sodium chloride conducts electricity
(d) Solubility
Ionic compounds are generally soluble in water but are insoluble in organic solvents
such as ether, alcohol, carbon tetrachloride etc. However, a few ionic compoundsare insoluble in water due to strong electrostatic force between cation and anion.For example barium sulphate, silver chloride and calcium fluoride.
ACTIVITY 7.2
Take nearly 10 g of NaCl, and two boiling tubes. In boiling tube (1) take 10 mLof water and add nearly 4 g of powdered NaCl. In test tube (2) take nearly 10 mLof ethyl alcohol and add nearly 4 g of powered NaCl. Shake both the test tube vigorouslyand see change in the amount of NaCl added in each case Fig. 7.8. Write your observation
Water1
Ethyl alcohol2Sodium chloride Sodium chloride
Fig. 7.8 Showing solubility of NaCl in water and ethyl alcohol
Before proceeding to the next section in which covalent bonding is discussed, why
don’t you answer the following questions to test your understanding about the ionicbonding?145Chemical Bonding
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INTEXT QUESTIONS 7.2
1. Name the two types of ions present in NaCl.
2. How many shells are present in Na+ ion?
3. What is the number of electrons present in Cl– ion?
4. Name the type of force of attraction present in ionic compounds.
5. In sodium chloride lattice, how many Cl– ions surround each Na+ ion?
6. Show the formation of Na2O, CaCl2 and MgO.
7. Why NaCl is bad conductor of electricity in solid state?
7.3 COVALENT BONDING
In this section, we will study about another kind of bonding called covalent bonding .
Covalent bonding is helpful in understanding the formation of molecules. In lesson2, you studied that molecules having similar atoms such as H
2, Cl2, O2, N2 etc. are
molecules of elements whereas those containing different atom like HCl, NH3, CH4,
CO2 etc. are molecule of compounds. Let us now see how are these molecules
formed?
Let us consider the formation of hydrogen molecule (H2). The hydrogen atom has
one electron. It can attain the electronic configuration of the noble gas helium bysharing one electron of another hydrogen atom. When the two hydrogen atoms comecloser, there is an attraction between the electrons of one atom and the proton of
another and there are repulsions between the electrons as well as the protons of the
two hydrogen atoms. In the beginning, when the two hydrogen atoms approach eachother, the potential energy of the system decreases due to the force of attraction.(Fig. 7.9) The value of potential energy reaches a minimum at some particular distance
Fig. 7.9  Potential energy diagram for formation of a hydrogen moleculeBond length Internuclear distanceattractivePotential Energy (kJ/mol)repulsive
HHHH
HHHHChemical Bonding
SCIENCE AND TECHNOLOGY  146Notes
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between the two atoms. If the distance between the two atoms further decreases,
the potential energy increases because of the forces of repulsion. The covalent bond
forms when the forces of attraction and repulsion balance each other and the
potential energy is minimum.  It is this lowering of energy which leads to the
formation of the covalent bond.
Formation of covalent bond in H2 can be shown as
H.  +  .H  ⎯⎯→  H : H  ⎯⎯→  H2
We will next consider the formation of chlorine molecule (Cl2). A molecule of chlorine
contains two atoms of chlorine. Now how are these two chlorine atoms held together
in a chlorine molecule?
You know that the electronic configuration of Cl atom is 2,8,7. Each chlorine atom
needs one more electron to complete its octet. If the two chlorine atoms share oneof their electrons as shown below, then both of them can attain the stable noble gasconfiguration of argon.
Cl +C lsharing of
one electronClCl
chlorine atom chlorine atom
2, 8, 7 2, 8, 7shared electrons
Note that the sharing pair of electrons is shown to be present between the two
chlorine atoms. Each chlorine atom thus acquires 8 electrons. The shared pair ofelectrons keeps the two chlorine atoms bonded together. Such a bond, which is
formed by sharing of electrons between the atoms is called a covalent bond .
Thus, we can say that a covalent bond is present between two chlorine atoms. This
bond is represented by drawing a line between the two chlorine atoms as follows:
Cl–Cl
covalent bond
Sometimes the electrons shown above on the chlorine atoms are omitted and the
chlorine-chlorine bond is shown as follows:
Cl — Cl
Similarly, we can understand the formation of oxygen molecule (O2) from the oxygen
atoms. The oxygen atom has atomic number 8. It has 8 protons and also 8 electrons.The electronic configuration of oxygen atoms is 2,6. Now each oxygen atom needstwo electrons to complete its octet. The two oxygen atoms share two electrons andcomplete their octet as is shown below:
O +O O
oxygen atom oxygen atom
2, 6 2, 6sharing of 4 electrons
or 2 pairs of electronsO147Chemical Bonding
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The 4 electrons (or 2 pairs of electrons) which are shared between two atoms of
oxygen are present between them. Hence these two pairs of shared electrons canbe represented by two bonds between the oxygen atoms. Thus, an oxygen moleculecan be represented as follows:
OO
The two oxygen atoms are said to be bonded together by two covalent bonds. Sucha bond consisting of two covalent bonds is also known as a double bond.
Let us next take the example of nitrogen molecule (N
2) and understand how the two
nitrogen atoms are bonded together. The atomic number of nitrogen is 7. Thus it has7 protons and 7 electrons present in its atom. The electronic configuration can bewritten as 2,5. To have 8 electrons in the outermost shell, each nitrogen atom requires3 more electrons. Thus, a sharing of 3 electrons each between the two nitrogen atomsis required. This is shown below:
N +
nitrogen atom
(2, 5)N
electronic configurationnitrogen atom
(2, 5)electronic configurationN N
sharing of 6 electrons
or 3 pairs of electrons(2, 8) (2, 8)
Each nitrogen atom provides 3 electrons for sharing. Thus, 6 electrons or 3 pairs
of electrons are shared between the two nitrogen atoms. Hence, each nitrogen atomis able to complete its octet.
Since 6 electrons (or 3 pairs of electrons) are shared between the nitrogen atoms,
we say that three covalent bonds are formed between them. These three bonds arerepresented by drawing three lines between the two nitrogen atoms as shown below:
NN
Such a bond which consists of  three covalent bonds is known as a triple bond .
So far, we were discussing covalent bonds formation between atoms of the sameelements. But covalent bonds can be formed by sharing of electrons between atomsof different elements also. Let us take the example of HCl to understand it.
A hydrogen atom has one electron in its outermost shell and a chlorine atom  has
seven electrons in its outermost shell. Each of these atoms has one electron less thanthe electronic configuration of the nearest noble gas. If they share one electron pair,then hydrogen can acquire two electrons in its outer most shell whereas chlorine willhave eight electrons in its outermost shell. The formation of HCl molecule by sharingof one electron pair is shown below:
H+
hydrogen atom
(1)electronic configurationchlorine atom
(2, 8, 7)electronic configurationH
shared electron pairCl ClChemical Bonding
SCIENCE AND TECHNOLOGY  148Notes
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Similarly, we can explain bond formation in other covalent compounds.
After knowing the nature of bonding present in covalent compounds, let us now study
what type of properties these covalent compounds have.
7.3.1 Properties of Covalent Substances
The covalent compounds consist of molecules which are electrically neutral in nature.The forces of attraction present between the molecules are less strong as comparedto the forces present in ionic compounds. Therefore, the properties of the covalentcompounds are different from those of the ionic compounds. The characteristicproperties of covalent compounds are given below:
(a) Physical State
Because of the weak forces of attraction present between discrete molecules, calledintermolecular forces, the covalent compounds exist as a gas or a liquid or a solid.For example O
2, N2, CO2 are gases; water and CCl4 are liquids and iodine is a solid.
(b) Melting and Boiling Points
As the forces of attraction between the molecules are weak in nature, a small amountof energy is sufficient to overcome them. Hence, the melting points and boiling pointsof covalent compounds are lower than those of ionic compounds. For example,melting point of nephthalene which is a covalent compound is 353 K (80°C).Similarly, the boiling point of carbon tetrachloride which is another covalent liquidcompound is 350 K (77°C).
(c) Electrical Conductivity
The covalent compounds contain neutral molecules and do not have charged speciessuch as ions or electrons which can carry charge. Therefore, these compounds donot conduct electricity and are called poor conductors of electricity Fig. 7.10.
- +Bulb
Switch
Beaker
Rubber
corkEthyl alcohol3V
Fig. 7.10 Ethyl alcohol (a covalent compound) is non-conductor of electricity149Chemical Bonding
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(d) Solubility
Covalent compounds are generally not soluble in water but are soluble in organic
solvents such as alcohol, chloroform, benzene, ether etc.
ACTIVITY 7.3
Take about 5 mL of ethyl alcohol in a test tube. Add few crystal of iodine. Shakethe test tube well. What do you find. The colour of the ethyl alcohol becomes darkbrown. What inference you draw from this. Iodine is soluble in ethyl alcohol. Writeyour observation. Dissolve the same amount of iodine in the same volume of water.(Soluton of iodine in ethyl alcohol is popularly known as tincture iodine and is usedas a antiseptic solution.)
After understanding the nature of covalent bond and properties of covalent
compounds. Why don’t you answer the following questions to test your understaningabout the covalent bonding.
Ethyl alcoholIodine
Before After
Fig. 7.11 Showing solubility of iodine in ethyl alcohol
INTEXT QUESTIONS 7.3
1. How covalent bonds are formed?
2. Show the formation of O2, HCl, Cl2 and N2.
3. How many covalent bond(s) is/are present in following compounds:
(i) H2O (ii) HCl (iii) O2 (iv) N2
4. State loss or gain of elactrons (giving their number) in the following changes :
(i) N ⎯→  N3–(ii) Cl ⎯→  Cl–
(iii) Cu ⎯→  Cu2+(iv) Cr ⎯→  Cr3+
5. Why ethyl alcohol is bad conductor of electricity in its aqueous solutions?Chemical Bonding
SCIENCE AND TECHNOLOGY  150Notes
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WHAT YOU HAVE LEARNT
/circle6The basic cause of chemical bonding is to attain noble gas configuration either
by transfer of electron from a metal to non- metal or by sharing of electrons
between two non-metal atoms.
/circle6Atoms of elements don’t exist freely in the nature. In all, the atoms of all theelements except of noble gases , have less than eight electrons in the valence
shell. Normally gases do not react  with other elements in normal conditions asthey have stable electronic configuration i.e. they have eight electrons in the
valence shell  or outer most shell.
/circle6All the atoms have a tendency to acquire stable  state or noble gas configuration.
Therefore, they combine with atoms of other elements to acquire 08 electronsin the valence shell by giving , taking or sharing of electrons. This is the basic
cause of Chemical bonding and is called Octet Rule.
/circle6Atoms of elements in a molecule are held  together by Chemical Bonding. The
formation of chemical bonds result in the lowering of energy which is less than
the energy of the individual atoms. The resulting compound  is lower in energyand hence more stable.
/circle6There are two types of chemical bonding : ionic bonding and covalent bonding.
/circle6Ionic Bonding: The chemical bond formed by transfer of electrons from a metalto a non- metal is known as Ionic Bond or Electrovalent bond.
/circle6The ionic bond formation takes place in three steps.
(i) Formation of Cations by metals with loss of electrons.
(ii) Formation of Anions by non- metal with gain of electrons.
(iii) Combination of Cations and Anions by electrostatic force of attraction to
form Ionic bond
/circle6Ionic compounds are solid, hard, have high melting and boiling points. They aresoluble in water but insoluble in organic solvents .They are good conductor of
electricity in molten state and in aqueous solution.
/circle6Covalent Bonding: The chemical bond formed by mutual sharing of  equal no.of electrons between two atoms. Covalent bonding is helpful in understanding
the formation of the molecules.H
2, Cl2, O2 and N2 are such molecules formed
by sharing of electrons between similar atoms, while H2O and HCl compounds
formed by sharing of electrons between  dissimilar atoms.
/circle6On the basis of sharing of number of electrons by each atom, covalentcompounds are classified as single bonded, double bonded and triple bonded.
When sharing of one electron  takes place from both  the atoms , single bond
is formed. Like Cl-Cl or Cl
2 and H-H or H2.151Chemical Bonding
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/circle6Double bond is formed when two similar atoms share two pair of electrons e.g.
O=O or O2 and triple bond is formed when there is sharing of three electrons
from each atom. e.g. N ≡N or N2.
/circle6The dissimilar atoms also share electrons but shared pair of electrons shift
towards more reactive atom as in HCl and H2O.
/circle6Covalent compounds mostly have liquid or gaseous state. Some are solid also.
They have low melting point, low boiling point. They are insoluble in water but
soluble in organic compounds. They are non- conductor  of electricity.
TERMINAL EXERCISE
1. Why ionic compounds conduct electricity in aqueous solution?
2. Covalent compounds have low melting point than an ionic compound why?
3. Explain the formation of Na+ ion from Na atom.
4. How would you explain the bonding in MgCl2?
5. Which of the following statements are correct for ionic compounds:
(i) They are insoluble in water.
(ii) They are neutral in nature.
(iii) They have high melting points.
6. State three characteristic properties of ionic compounds.
7. How does a covalent bond form?8. What is the number of solvent bonds present in the following molecules?
(i) Cl
2 (ii) N2 (iii) O2 (iv) H2
9. Classify the following statements as true or false:
(i) Ionic compounds contain ions which are held together by weak electro-
static forces.
(ii) Ionic compounds have high melting and boiling points.
(iii) Covalent compounds are good conductors of electricity.
(iv) Solid sodium chloride is a good conductor of electricity.
10. Classify the following compounds as ionic or covalent:
(i) sodium chloride (ii)calcium chloride
(iii) oxygen (iv)hydrogen chloride
(v) magnesium oxide (vi)nitrogenChemical Bonding
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11. An element ‘X’ has atomic no. 11 and ‘Y’ has atomic no. 8. What type of
bond they will form? Write the formula of the compound formed by reacting Xand Y .
12. Name the type of bonds present in H
2O molecule.
ANSWER TO INTEXT QUESTIONS
7.1
1. Every atom has tendency to attain 2 or 8 e– in their outermost shell to get stability
like noble gases.
2. Because they have inert gas configuration which makes it very stable.
3. A and B
7.2
1. Sodium ion Na+ and chloride ion Cl–.
2. Two (2)3. 184. Electrostatic force of attraction5. Six
6.153Chemical Bonding
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7. Due to absence of free Na+ and Cl– ion.
7.3
1. A covalent bond is formed by sharing of equal no. of electrons between two
atoms.
2. O O /c108/c108/c108/c108 /c108/c108 /c108/c108
/c108/c108 /c108/c108
OO/c108/c108 /c108/c108
/c108/c108 /c108/c108
=
Cl H/c108/c108/c108/c108
/c108/c108/c108/c108 ClH– /c108/c108
/c108/c108/c108/c108
Cl/c108/c108/c108/c108
/c108/c108/c108/c108 Cl/c108/c108
/c108/c108
/c108/c108
Cl /c108/c108
/c108/c108
/c108/c108
Cl –/c108/c108
/c108/c108
/c108/c108
/c108/c108
NN/c108
/c108
/c108/c108/c108/c108
/c108
/c108/c108
NN/c108/c108 /c108
3. (i) 2 (ii) 1 (iii) 2 (iv) 3
4. (i) Gain of 3e–
(ii) Gain of 1e–
(iii) Loss of 2e–
(iv) Loss of 3e–
5. Ethyl alcohol do not produce H+ ion in its aqueous solution, hence does not
conduct electricity.Notes
MODULE - 3
Moving Things
 191Motion and its Description
SCIENCE AND TECHNOLOGY
9
MOTION AND ITS DESCRIPTION
You must have seen number of things in motion. For example car, bicycle, bus moving
on a road, train moving on rails, aeroplane flying in the sky, blades of an electricfan and a child on a swing. What makes things move? Are all the motions similar?
You might have seen that some move along straight line, some along curved path
and some to and fro from a fixed position. How and why these motions are different?You will find answers to all such questions in this lesson. Besides studying aboutvarious types of motions, you will learn how to describe a motion. For this we willtry to understand the concepts of distance, displacement, velocity and acceleration.
We will also learn how these concepts are related with each other as well as with
time. How a body moving with constant speed can acquire acceleration will also bediscussed in this lesson.
OBJECTIVES
After completing this lesson you will be able to:
/circle6explain the concept of motion and distinguish between rest and motion;
/circle6describe various types of motion – rectilinear, circular, rotational andoscillatory;
/circle6define distance, displacement, speed, average speed, velocity andacceleration;
/circle6describe uniform and uniformly accelerated motion in one dimension;
/circle6draw and interpret the distance time graphs and velocity time graphs;
/circle6establish relationship among displacement, speed, average speed, velocityand acceleration;
/circle6apply these equations to make daily life situation convenient and
/circle6explain the circular motion.Notes
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 1929.1 MOTION AND REST
If you observe a moving bus you will notice that the position of bus is changing with
time. What does this mean? This means that the bus is in motion. Now suppose youare sitting in a bus moving parallel to another bus moving in the same direction withsame speed. You will observe that the position of the other bus with respect to your
bus is not changing with time. In this case the other bus seems to be at rest with
respect to your bus. However, both the buses are moving with respect tosurroundings. Thus, an object in motion can be at rest with respect to one observerwhereas for another observer, the same object may be in motion. Thus we can saythat the motion is relative.
Let us understand the concept of relative motion. Suppose you are sitting in a vehicle
waiting for traffic signal and the vehicle beside you just starts moving, you will feelthat your vehicle is moving backward.
Suppose Chintu and Golu are going to the market. Golu is running and Chintu is
walking behind him. The distance between the two will go on increasing, though both
are moving in the same direction. To Golu it will appear that Chintu is moving away
from him. To Chintu also, it will appear that Golu is moving ahead and away fromhim. This is also an example of relative motion. See Fig. 9.1.
Fig. 9.1  An example of relative motion
Think and Do
One day, Nimish while standing on the bank of a river in the evening observed
boats were approaching the bank, vehicles passing on the bridge, cattle goingaway from the bank of the river towards the village, moon rising in the sky, birdsflying and going back to their nests, etc. Can you list some thoughts that couldbe emerging in the mind of the Nimish. What type of world Nimish has around
him?
Chintu GoluNotes
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AC
BDEWe can conclude that motion is a continuous change in the position of the object
with respect to the observer. Suppose you are moving towards your friend standingin a field. In what way are you in motion? Are you in motion if you are observing
yourself? Is your friend in motion with respect to you? Are you in motion with respect
to your friend? Now you may have understood that observer with respect to itselfcan not be in motion. Thus, you are moving towards the object with respect to yourfriend and your friend is moving towards you with respect to you in opposite direction.In other words the change in position of the object with respect to observer decideswhether object is in motion. This change should also be continuous. Let us take an
interesting example to understand the concept of motion. There are five players
participating in 200 metre race event. They are running in their lanes as shown inthe Fig. 9.2. The players A, B, C, D and E runs 2, 3, 4, 3, 2 metre respectivelyin one second. Can you help the player to understand that which player is in motionwith respect to which player and which player is at rest with respect to which player?Fill your responses in the table given below.
Fig. 9.2
Table 9.1
Observer player Player in motion Player at rest Remark
A B, C, D E E is in rest with respect to
A because change inposition of A and E in1second is zero while inother cases is not.
B
C
D
Now you will be able to help Nimish to answer some of his questions.Notes
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 194
9.1.1 Types of Motion
In our daily life we see many objects moving. Some objects moving in straight line
and some are not. For example, a ball rolls on a horizontal surface, a stone falling
from a building, and a runner on 100 m race track. In all these examples, you maynotice that the position of moving objects is changing with respect to time along astraight line. This type of motion is called motion in a straight line or rectilinear
motion .
Fig. 9.3  Example of rectilinear motion
Can you think at least two more other example of such motions. You might have
observed the motion of time hands of a clock, motion of child sitting on a merry-
go-round, motion of the blades of an electric fan. In such a motion, an object follows
a circular path during motion. This type of motion is called circular motion .
ACTIVITY 9.1
(A) Suspend a small stone with a string (of length less than your height) with the
help of your hand. Displace the stone aside from the position of rest and release.
(B) Let the stone comes to rest and bring it to the point of suspension with the help
of your hand and release it.(a)  Ball rolling on horizontal surface
(b) Stone falling by hand
(c) A runner on a 100 m race trackNotes
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SCIENCE AND TECHNOLOGY(A)(B)
(C)
(C) Now hold the stone firmly in your hand and whirl it over your head.
Write in table given below, what type of motion of stone you have observed in all
the above three cases with justification.
Table 9.2
Case Type of motion Justification
A
B
C
(A) A person suspend the stone attached to a string, (B) A person oscillate the stone
attached to a string, (C) A person whirling the stone attached to a string
Fig. 9.4 (A), (B) (C)
Have you ever noticed that the motion of the branches of a tree? They move to and
fro from their central positions (position of rest). Such type of motion is calledoscillatory motion . In such a motion, an object oscillates about a point often called
position of rest or equilibrium position. The motion of swing and pendulum of wall
clock are also oscillating motions. Can you think about the motion of the needle of
a sewing machine? What type of motion is it? Now you can distinguish some of themotions viewed by Nimish.Notes
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 196
9.2 DISTANCE AND DISPLACEMENT
For a moving object two points are significant. One is the point of start or origin
where from the object starts its motion and the other is the point where it reaches
after certain interval of time. Points of start and destination are connected by a path
taken by the object during its motion. The length of the path followed by object iscalled distance. There may be a number of paths between the point of start and thepoint of destination. Hence the object may cover different distances between samepoint of start and destination. The unit of distance is metre (m) or kilometre (km).
ACTIVITY 9.2
An object moves from point A to B along three different paths. Measure the distancetravelled by object along these three paths.
Fig. 9.5
In any motion, you will notice that object gets displaced while it changes its positioncontinuously. The change in position of the object is called displacement.
Basically, it is the shortest distance between initial and final position of the object.
The path followed by the object between initial and final positions may or may notbe straight line. Hence, the length of the path does not always represent thedisplacement.
ACTIVITY 9.3
In the following cases measure the distance and displacement and write their valuesin the table given below:
(a) A body moves from A to B
(b) A body moves from A to B then comes to CNotes
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SCIENCE AND TECHNOLOGY(c) A body goes from position A to B and comes back to position A
(d) A body goes from posion A to B and then C
(e) A body moves from position A to B along a circular arc
Fig. 9.6
Table 9.3
Case Distance Displacement
(i)
(ii)
(iii)
(iv)
(v)
Now you can conclude that:
(a) displacement is smaller or equal to the distance.
(b) displacement is equal to distance, if body moves along a straight line path and
does not change its direction.
(c) if a body does not move along a straight line path its displacement is less than
the distance.
(d) displacement can be zero but distance can not be zero.(e) magnitude of displacement is the minimum distance between final position and
initial position.
(f) distance is the length of the path followed by the body.
(g) distance is path dependent while displacement is position dependent.Can you now, suggest a situation in which the distance is twice the displacement?Notes
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 1989.2.1 Graphical Representation of Distance and Displacement
Distance and displacement can also be shown by graphical representation. To draw
a graph, follow the following steps:
(i) Analyse the range of variables (maximum and minimum values).
(ii) Select the suitable scale to represent the data on the graph line adequately.
(iii) T ake independent quantity on x-axis and dependent quantity on y-axis.
Take distance on x-axis and displacement on y-axis. You know that for a motion
along a straight line without changing its direction the distance is always equal to thedisplacement. If you draw the graph, you will find that the graph line is a straightline passing through origin making an angle of 45° with distance axis as shown inFig 9.7.
Fig. 9.7
Let us take another situation where an object moving from one position to anotherand coming back to the same position. In this case the graph line will be a straightline making an angle of 45° with distance axis upto its maximum value and then comesto zero as shown in Fig. 9.8.
Fig. 9.8Notes
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Now you can infer that:
/circle6If graph line is a straight line making an angle of 45° with x-axis or y-axis, the
motion is straight line motion and distance is equal to the displacement.
/circle6For same value of displacement, the distance travelled can be different.
/circle6If graph line does not make an angle of 45° with x-axis or y-axis, the motion
will not be straight line motion.
When an object moves along a circular path, the maximum displacement is equal
to the diameter of the circular path and the distance travelled by object keeps onincreasing with time as shown in Fig. 9.9.
Fig. 9.9
INTEXT QUESTIONS 9.1
Choose the correct answer in the followings:
1. For an object moving along a straight line without changing its direction the
(a) distance travelled > displacement
(b) distance travelled < displacement
(c) distance travelled = displacement
(d) distance is not zero but displacement is zero
2. In a circular motion the distance travelled is
(a) always > displacement
(b) always < displacements
(c) always = displacement
(d) zero when displacement is zeroNotes
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 2003. Two persons start from position A and reach to
position B by two different paths ACB and AB
respectively as shown in Fig. 9.10.
(a) Their distances travelled are same
(b) Their displacement are same
(c) The displacement of I > the displacement of II
(d) The distance travelled by I < distance travelled by II
4. In respect of the top point of the bicycle wheel of radius R moving along a
straight road, which of the following holds good during half of the wheel rotation.
(a) distance = displacement
(b) distance < displacement
(c) displacement = 2 R
(d) displacement = πR
5. An object thrown vertically upward to the height of 20 m comes to the hands
of the thrower in 10 second. The displacement of the object is
(a) 20 m (b) 40 m (c) Zero (d) 60 m
6. Draw a distance-displacement graph for an object in uniform circular motion on
a track of radius 14 m.
9.3 UNIFORM AND NON-UNIFORM MOTION
Let us analyze the data of the motion of two objects A and B given in the table 9.4.
Table 9.4
Time in seconds ( t ) 0 1 02 0 3 0 4 05 0
Position of A (x1 in metre) 0 4 8 12 16 20
Position of object B (x2 in metre) 0 4 12 12 12 20
Do you find any difference between the motion of object A and B? Obviously objects
A and B start moving at the same time from rest and both objects travel equal distance
in equal time. However, the object A has same rate of change in its position and
object  B has different rate of change in position. The motion in which an object covers
equal distance in equal interval of time is called uniform motion whereas the motionin which distance covered by object is not equal in equal interval of time is callednon-uniform motion. Thus, the motion of object A is uniform and of object B is non-
uniform. You can draw the position-time graph for the motion of object A and B and
observe the nature of the graph for both types of motion.
For the uniform motion of object A the graph is a straight line graph and for non-
uniform motion of object B the graph is not a straight line as shown in the Fig. 9.11.Fig. 9.10Notes
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Fig. 9.11  Graph representing uniform and non-uniform motion
9.3.1 Speed
While you plan your journey to visit a place of your interest you intend to think about
time of journey so that you can arrange needful things like eatables etc. for that periodof time. How will you do it? For this you would like to know how far you haveto reach and how fast you can cover the destination. The measure of how fast motioncan take place is the speed. Speed can be defined as the distance travelled by
a body in unit time.
Thus speed =
Distance travelled
time taken
Its SI unit is metre per second which is written as ms–1. The other commonly used
unit is km h–1.
i.e., 1 kmh–1=–1 1000 m 5ms60 60 s 18=×
ACTIVITY 9.4
Here position of four bodies A, B, C and D are given after equal interval of time
i.e. 2 s. Identify the nature of the motion of the bodies as uniform and non-uniformmotion.
Table 9.5
Time (s) → Bodies ↓ 02468
positions (m) → A 0 4 8 12 16
B 0 8 8 10 12
C48 1 2 1 6 2 0D0 6 1 2 1 6 2 0Notes
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 202To identify the nature of the motion you can make a table as given below
Table 9.6
Time taken by body (s) → 2 – 0 = 2 4 – 2 = 2 6 – 4 = 2 8 – 6 = 2
Distance covered by
body (m) ↓
A 4 – 0 = 4 8 – 4 = 4 12 – 8 = 4 16 – 12 = 4
B 8 – 0 = 8 8 – 8 = 0 10 – 8 = 2 12 – 10 = 2C 8 – 4 = 4 12 – 8 = 4 16 – 12 = 4 20 – 16 = 4D 8 – 4 = 4 12 – 6 = 6 16 – 12 = 4 20 – 16 = 4
From the above table you can conclude that body A and C travel equal distances
in equal interval of time so their motion is uniform. But the distances travelled bybody B and D for equal intervals of time are not equal, hence their motion is non-
uniform motion.
To analyze the motion as uniform motion or non-uniform motion, the motion can be
represented by graph. The position-time graph of all the four bodies A, B, C and
D is shown in Fig. 9.12.
Fig. 9.12
Now you can see that the bodies having uniform motion e.g. A and C have their
graph line straight and the bodies having non-uniform motion do not have their positiontime graph line straight. In this graphical representation on axis 1 div = 1s and on
y-axis 1 div = 2m.Notes
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 203Motion and its Description
SCIENCE AND TECHNOLOGYA graph drawn for different distances travelled by object with respect to time is called
distance-time graph as shown in Fig. 9.13.
Fig. 9.13
In Fig. 9.13 distance travelled in 10 s is 22 m. Therefore, the speed of the object
=()
()22 m
10 s = 2.2 ms–1
This motion can be represented by another way i.e., speed = AB
OB. This ratio is also
known as slope of the graph line. Thus the speed is the slope of position-time graph.
Example 9.1  An object moves along a rectangular path of sides 20 m and 40 m
respectively. It takes 30 minutes to complete two rounds. What is the speed of theobject?
Solution:() 22 2 04 0  m Distance travelled
time taken 30 60 s×+=×
           –1 4ms30=
9.3.2 Velocity
If you are asked to reach a destination and you are provided three, four paths of
different lengths, which of the path would you prefer? Obviously, the path of shortestlength but not always. This is also called displacement. In the previous section you
have learnt about distance. When motion is along the shortest path, it is directed from
the point of start to the point of finish. How fast this motion is determines the velocity.The velocity is the ratio of length of the shortest path i.e. displacement to the timetakenNotes
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 204
velocity =Displacement
Time taken
Velocity has same unit as the unit of speed i.e., ms–1 (S.I. unit) or kmh–1.
The shortest path or the displacement is directed from initial position of the object
to the final position of the object. Hence, the velocity is also directed from initialposition of the object to the final position of the object. Thus we can say that thevelocity has direction. Speed does not have direction because it depends upon thetotal distance travelled by the object irrespective of the direction. The quantities which
have direction are called vector and which do not have direction are called scalar
quantity. Thus, velocity can also be expressed as
velocity =
Chan ge in position
Time taken
ACTIVITY 9.5
Observe the motion of an object in the following situations. Find speed and velocity
in each situation and comment over the situation which you find different from other.
Ohject moves from A to B in time 10 s on the scale 1 cm = 10 m
Object moves from A to B than to C in 10 s on the scale 1 cm = 10 m
Object moves from A to B than to C in 20 s on the scale 1 cm = 10 m
Object completes a round of radius 7 m in 10 s
Fig. 9.14Notes
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 205Motion and its Description
SCIENCE AND TECHNOLOGYNow you will be able to distinguish the speed and velocity. Magnitude of
instantaneous velocity is the speed. Now you can understand the importance ofpreplanning your journey to save time, effort and fuel etc.
Example 9.2  In a rectangular field of sides 60 m and 80 m respectively two formers
start moving from the same point and takes same time i.e. 30 minutes to reachdiagonally opposite point along two different paths as shown in Fig. 9.15. Find thevelocity and speed of both the formers.
Fig. 9.15
Solution:  The displacement of both the former in same i.e.,
2260 80+ =3600 6400 10000 +=  = 100 m
∴Velocity A and B, v=displacement
time taken =100 m 1
30 60s 18=× ms–1
speed of A=()80+60 m Distance travelled
time taken 30×60s=
=140
3800 ms–1 =14
18 ms–1
and speed of B=Distance travelled 100s
time taken 30×60s=  =1
18 ms–1
Note:  In this example you can appreciate that the velocity of both the formers is
same but not the speed.
9.3.3 Average speed and average velocity
Speed during a certain interval of time can not be used to determine total distancecovered in given time of the journey and also the time taken to cover the total distanceNotes
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 206of journey. It is because a body does not always travel equal distance in equal interval
of time. In most of the cases the body travels non-uniformly. Thus, in case of non-uniform motion to determine average speed is quite useful. The average speed can
be determined by the ratio of total distance covered to the total time taken.
Average speed =total distance covered
total time taken
Similarly in case of average velocity in place of total distance covered you can take
total displacement.
∴ Average speed =total displacement
total time taken
Let us take few examples to understand the average speed and average velocity.
Example 9.3  If a body covers 50 m distance in 30 s and next 100 m in 45 s then
total distance covered
= 50 + 100 = 150 m
and total time taken = 30 + 45 = 75 s
∴ Average speed =150 m
75 s = 2ms–1
Example 9.4  If an object moves with the speed of 10 ms–1 for 10 s and with
8 ms–1 for 20 s, then total distance covered will be the sum of distance covered
in 10 s and the distance covered in 20 s = 10 × 10 + 8 × 20 = 260 m
∴ The average speed =total distance covered
total time taken
=()260 m 260 m
10+20 s 30 s=
= 8.66 ms–1
Example 9.5  If a body moves 50 m with the speed of 5 ms–1 and then 60 m with
speed of 6 ms-1, then total distance covered
= 50 + 60 = 110 m
and total time taken will be the sum of time taken for 50 m and 60 m = 20 s
Thus, average speed =total distance covered
total time taken
=110 m
20 s = 5.5 ms–1Notes
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Example 9.6  If an object moves 30 m toward north in 10 s and then 40 m eastward
in next 10s, The displacement of the object will be OB
= 2230 40+  =900 1600+  = 2500
= 50 m
∴ The average velocity =total displacement covered
total time taken
=
()50 m 50 m= =  2 . 510+10 s 20 s ms–1
Fig 9.16
Example 9.7  If an object moves along a circular track of radius 14 m and complete
one round in 20 s then for one complete round total displacement is zero and theaverage velocity will also be zero.
From these examples you can conclude that:
(i) Instantaneous speed is the magnitude of instantaneous velocity but average
speed is not the magnitude of average velocity.
(ii) A verage velocity is less than or equal to the average speed.
(iii) A verage velocity can be zero but not average speed.
INTEXT QUESTIONS 9.2
1. Some of the quantities are given in column I. Their corresponding values are
written in column II but not in same order. You have to match these valuescorresponding to the values given in column I:
Column I Column II
(a) 1 kmh–1(i) 20 ms–1
(b) 18 kmh–1(ii) 10 ms–1
(c) 72 kmh–1(iii) 5/18 ms–1
(d) 36 kmh–1(iv) 5 ms–1Notes
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 2082. A cyclist moves along the path shown in the diagram and takes 20 minutes from
point A to point B. Find the distance, displacement and speed of the cyclist.
Fig. 9.17
3. Identify the situation for which speed and average speed of the objects are equal.
(i) Freely falling ball
(ii) Second or minute needle of a clock
(iii) Motion of a ball on inclined plane
(iv) T rain going from Delhi to Mumbai
(v) When object moves with uniform speed
4. The distance-time graph of the motion of an object is given. Find the average
speed and maximum speed of the object during the motion.
Fig. 9.18
5. The distance travelled by an object at different times is given in the table below.
Draw a distance-time graph and calculate the average speed of the object. State
whether the motion of the object is uniform or non-uniform.
Table 9.7
Time (s) → 0 1 02 03 04 05 0
Distance (m) → 02468 1 0Notes
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 209Motion and its Description
SCIENCE AND TECHNOLOGY6. A player completes his half of the race in 60 minutes and next half of the race
in 40 minutes. If he covers a total distance of 1200 m, find his average speed.
7. A train has to cover a distance of 1200 km in 16 h. The first 800 km are covered
by the train in 10 h. What should be the speed of the train to cover the restof the distance? Also find the average speed of the train.
8. A bird flies from a tree A to the tree B with the speed of 40 km h
–1 and returns
to tree A from tree B with the speed of 60 km h–1. What is the average speed
of the bird during this journey?
9. Three players P, Q and R reach from point A to B in same time by following
three paths shown in the Fig. 9.19. Which of the player has more speed, whichhas covered more distance?
Fig. 9.19
9.4. GRAPHICAL REPRESENTATION OF MOTION
It shows the change in one quantity corresponding to another quantity in the graphicalrepresentation.
9.4.1 Position-time Graph
It is easy to analyze and understand motion of an object if it is represented graphically.To draw graph of the motion of an object, its position at different times are shown
on y-axis and time on x-axis. For example, positions of an object at different times
are given in Table 9.8.
Table 9.8  Position of different objects at different time
Time (s) 0 1 2 3 4 5 6 7 8 9 10
Position (m) 0 10 20 30 40 50 60 70 80 90 100
In order to plot position-time graph for data given in Table 9.8, we represent time
on horizontal axis and position on vertical axis drawn on a graph paper. Next, wechoose a suitable scale for this.Notes
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Moving Things
 210For example, in Fig. 9.20 one division on horizontal axis represents 1 s of time interval
and one division on vertical axis represents in 10 m, respectively. If we join differentpoints representing corresponding position time data, we get straight line as shown
in Fig. 9.20. This line represents the position-time graph of the motion corresponding
to data given in Table 9.8.
Fig. 9.20  Position-time graph for the motion of a particle on the basis of
data given in table
We note from the data that displacement of the object in 1st second, 2nd second,.....,
10th second is the same i.e., 10 m. In 10 second, the displacement is 100 m.
Therefore, velocity is 100 m
10 s= 10 ms–1 for the whole course of motion. Velocity
during 1st second = 10 ms–1 and so on.
Thus, velocity is constant i.e., equal to 10 ms–1 throughout the motion. The motion
of an object in which velocity is constant, is called uniform motion.
As you see Fig. 9.20, for uniform motion position-time graph is a straight line.
Like position-time graph, you can also plot displacement-time graph. Displacement
is represented on the vertical axis and time interval on the horizontal axis. Sincedisplacement in each second is 10 m for data in table, the same graph (Fig. 9.20)also represents the displacement-time graph if vertical axis is labeled as displacement.
For good understanding you can observe the following graphs.Notes
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 211Motion and its Description
SCIENCE AND TECHNOLOGY(A) Uniform motion (B) Object is at rest
(C) Non-uniform motion, rate of change (D) Non-uniform motion, rate of change
in position is increasing in position is decreasing
Fig. 9.21  Graph (A), (B), (C), (D)
9.4.2 Velocity-Time Graph
Take time on the horizontal axis and velocity on the vertical axis on a graph paper.
Let one division on horizontal axis represent 1 s and one division on vertical axisrepresent 10 ms
–1. Plotting the data in Table 9.9 gives us the graph as shown in
Fig. 9.22.
Table 9.9  Velocity-time data of objects A and B
Time (s) 012345678
Velocity of A (ms-1) 0 10 20 30 40 50 60 70 80
Velocity of B (ms-1) 0 10 10 10 10 10 10 10 10
Fig. 9.22  Velocity-time graph for the motion of object A and B on the
basis of data given in tableNotes
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Moving Things
 212Lines OR and PQ represent the motion of object A and B respectively. Thus, we
see that the velocity-time graph of motion represented in Table 9.9 is a straight lineand parallel to time axis for object B. This is so because the velocity is constant
throughout the motion. The motion is uniform. Consider the area under the graphin Fig. 9.22 for object B.
Area = (8s) × (10 ms
–1) = 80 m. This is equal to the displacement of the object
B in 8 s.
Area under velocity-time graph = Displacement of the object during that timeinterval
Similarly for object A area under the graph in Fig. 9.22.
=() ( )18 s 80 02×−  ms–1
=() ( )188 0  m2×  = 320 m
This is equal to the displacement of object A in 8 s.
Though, we obtained this result for object B for a simple case of uniform motion,
it is general result.
Let x be displacement of an object in time t, moving with uniform velocity v, then
x = vt (for uniform motion)
You may have seen the motion of objects moving differently. Can you think what
make this difference? Observe the motion of a ball on a floor. The ball slows downand finally comes to rest. This means that the velocity during different time intervalsof motion is different. In other words velocity is not constant. Such a motion is calledaccelerated motion.
9.5 ACCELERATION
In the previous section we have learnt about the non-uniform motion in which the
change in velocity in different intervals of motion is different. This change in velocity
with time is called acceleration. Thus, the acceleration of an object is defined as
the change in velocity divided by the time interval during which this occurs.
Acceleration =Chan ge in velocit y
Time interval
Its unit is ms–2. It is specified by direction. Its direction is along the direction of change
in velocity. Suppose the velocity of an object changes from 10 ms–1 to 30 ms–1 in
a time interval of 2 s.
Fig. 9.23 Changing velocityNotes
MODULE - 3
Moving Things
 213Motion and its Description
SCIENCE AND TECHNOLOGYThe acceleration, a=–1 –130 ms – 10 ms
2.0s = 10 ms–2
This means that the object accelerates in +x direction and its velocity increases at
a rate of 10 ms–1 in every second.
If the acceleration of an object during its motion is constant, we say that object is
moving with uniform acceleration . The velocity-time graph of such a motion is
straight line inclined to the time axis as shown in Fig. 9.24.
For a given time interval, if the final velocity is more than the initial velocity, then
according to Fig. 9.24, the acceleration will be positive. However, if the final velocityis less than the initial velocity, the acceleration will be negative.
Fig. 9.24  velocity-time graph of an object moving with uniform acceleration
When velocity of the object is constant, acceleration will be zero. Thus, for uniform
motion, the acceleration is zero and for non-uniform motion, the acceleration is non-
zero.
Example 9.8  Find the distance and displacement from the given velocity-time graph
in Fig. 9.25.
Fig. 9.25Notes
SCIENCE AND TECHNOLOGYMODULE - 3 Motion and its Description
Moving Things
 214Solution:
Distance travelled = Area of ΔOAB  + Area of ΔBCD
=1
2(25) × (20) + 1
2(10) × (20)
= 250 + 100 = 350 m
Displacement = Area of ΔOAB  – Area of ΔBCD
=1
2(25) × (20) –1
2 (10) × (20)
= 250 – 100 = 150 m
Example 9.9  From the given velocity-time graph obtain the acceleration-time graph.
Fig. 9.26
Solution:  From the given graph acceleration for 0 – 10 s time interval
=15 0
10 0−
− = 1.5 ms–2
acceleration for 10 – 20s time interval in same as for 20 – 30s time interval
=20 15 5
30 10 20−=− = 0.25 ms–2
acceleration for 30 – 40s time interval =20 20
40 30−
− = 0Notes
MODULE - 3
Moving Things
 215Motion and its Description
SCIENCE AND TECHNOLOGY
acceleration for 40 – 50 and 50 – 60s interval =30 20 10
60 40 20−=− = 0.5 ms–2
For all the above time intervals the acceleration-time graph can be drawn as shown
in Fig. 9.27.
Fig. 9.27
INTEXT QUESTIONS 9.3
1. Describe the motion of an object shown in Fig. 9.28.
Fig. 9.28  Position-time graph of an objectNotes
SCIENCE AND TECHNOLOGYMODULE - 3 Motion and its Description
Moving Things
 2162. Compare the velocity of two objects where motion is shown in Fig. 9.29.
Fig. 9.29 Position-time graph for object A and B.
3. Draw the graph for the motion of object A and B on the basis of data given in
Table 9.10.
Table 9.10
Time (s) 0 10 20 30 40 50
Position (m) for A 055555
Position (m) for B 02468 1 0
4. A car accelerates from rest uniformly and attains a maximum velocity of
2 ms–1 in 5 seconds. In next 10 seconds it slows down uniformly and comes
to rest at the end of 10th second. Draw a velocity-time graph for the motion.
Calculate from the graph (i) acceleration, (ii) retardation, and (iii) distancetravelled.
5. A body moving with a constant speed of 10 ms
–1 suddenly reverses its direction
of motion at the 5th second and comes to rest in next 5 second. Draw a position-
time graph of the motion to represent this situation.
9.6 EQUATIONS OF MOTION
Consider an object moving with uniform acceleration, a. Let u be the initial velocity
(at time t = 0), v, velocity after time t and S, displacement during this time interval.
There are certain relationships between these quantities. Let us find out.
We know that
Acceleration =Chna ge in velocit y
Time interval
∴ a=vu
t−
or v=u + at ...(9.1)Notes
MODULE - 3
Moving Things
 217Motion and its Description
SCIENCE AND TECHNOLOGY
This is called as the first equation of motion.
Also, we know that
Displacement = (average velocity) × (time interval)
or s=2vut+⎛⎞
⎜⎟⎝⎠ = 2ua tut++⎛⎞
⎜⎟⎝⎠()vua t=+∵
or s=21
2ut at+ ...(9.2)
This is called the second equation of motion.
If object starts from rest, u = 0 and
s=2102ta t×+
or s=21
2at
Thus, we see that the displacement of an object undergoing a constant acceleration
is proportional to t2, while the displacement of an object with constant velocity (zero
acceleration) is proportional to t.
Now, if we take vuat−=  and .2vust+⎛⎞=⎜⎟⎝⎠ and multiply them, we find that
a.s=()
2vu vutt− +⎛⎞
⎜⎟⎝⎠ = 22
2vu−
or 2 a.s=v2 – u2
or v2=u2 + 2as ...(9.3)
This is called as third equation of motion. In case of motion under gravity ‘ a’ can
be replaced by ‘ g’.
INTEXT QUESTIONS 9.4
1. A ball is thrown straight upwards with an initial velocity 19.6 ms–1. It was caught
at the same distance above the ground from which it was thrown:
(i) How high does the ball rise?
(ii) How long does the ball remain in air? ( g = 9.8 ms–2)Notes
SCIENCE AND TECHNOLOGYMODULE - 3 Motion and its Description
Moving Things
 2182. A brick is thrown vertically upwards with the velocity of 192.08 ms–1to the
labourer at the height of 9.8 m. What are its velocity and acceleration when itreaches the labourer?
3. A body starts its motion with a speed of 10 ms
–1 and accelerates for 10 s with
10 ms–2. What will be the distance covered by the body in 10 s?
4. A car starts from rest and covers a distance of 50 m in 10 s and 100 m in next
10 s. What is the average speed of the car?
9.7 UNIFORM CIRCULAR MOTION
You may have seen the motion of the bicycle on a straight level road. Do all movableparts of the bicycle move alike? If not, then how are they moving differently? Doesthe peddling make a difference in these motions? Like Nimish, number of questionsyou may have in your mind. Let us try to answer these questions. Bicycle is movingon a straight road so its motion is rectilinear motion.
Fig. 9.30  Bicycle moving on a road
Now look at the wheels of the bicycle. Any point on the wheel of the bicycle always
remains at a constant distance from the axis of the wheel and moves around the fixed
point i.e., axis of the wheel. On the basis of this description of motion of the wheelyou can decide very obviously that this motion is circular motion.
Similarly, can you think about the motion of the flywheel of the bicycle? During non-
peddling, there is no circular motion of flywheel and it moves in a straight line thus,
its motion is rectilinear motion. But during the peddling its motion is circular motion
can you think about the motion of any part of the bicycle which has two types ofmotion at the same time? Yes, during the circular motion of the wheel or flywheel,they are also advancing in forward direction on a straight road. Thus, there motionis circular motion as well as rectilinear motion at the same time.
Now consider the motion of an object along a circular track of radius R through
four points A, B, C and D on the track as shown in Fig. 9.31. If object completes
each round of motion in same time, than it covers equal distance in equal intervalof time and its motion will be uniform motion. Since during this uniform motion equaldistance is being covered in equal interval of time, therefore, the ratio of distanceNotes
MODULE - 3
Moving Things
 219Motion and its Description
SCIENCE AND TECHNOLOGYFig. 9.31  Circular motioncovered to the time taken i.e., speed will remain constant. It means in uniform
circular motion speed remains constant.
Now think about velocity, velocity remains
along the direction of motion. In Fig. 9.31you can see the direction of motion changesat every point as shown at point A, B, C and
D. Since there is a change in direction of
motion, therefore, the direction of velocity
also changes. We can say that in uniformcircular motion, velocity changes due tochange in direction of motion and the motionof the object is accelerated motion. Thisacceleration is due to change in the direction
of motion. But in this motion speed remains constant. How interested this motion
is because a body moving with constant speed acquires acceleration.
Think and Do
KI L O M E T R E T O
SP E E D T O N C N E
O ND I S TAANO E
P DI SPLA CDI AAN S V E L O C I TY
TA P P E E R C S A N
KA L U D I N E T RATE A M Y O Y L A E D
MA C H I N E E N L L
EP E P T A D R C E KT OMFTREAEC DRN E N G I N T G CQ
EE N K L O M E T A R
In the above word grid identify the meaningful words, related to description of motion,
in horizontal or vertical columns in sequence and define them (at least three).Notes
SCIENCE AND TECHNOLOGYMODULE - 3 Motion and its Description
Moving Things
 220
INTEXT QUESTIONS 9.5
1. In circular motion the point around which body moves
(a) always remain in rest
(b) always remain in motion
(c) may or may not be in motion
(d) remain in oscillatory motion
2. In uniform circular motion
(a) speed remain constant
(b) velocity remain constant
(c) speed and velocity both remain constant
(d) neither speed nor velocity remain constant
3. A point on a blade of a ceiling fan has
(a) always uniform circular motion
(b) always uniformly accelerated circular motion
(c) may be uniform or non-uniform circular motion
(d) variable accelerated circular motion
WHAT YOU HAVE LEARNT
/circle6If a body stays at the same position with time, it is at rest.
/circle6If the body changes its position with time, it is in motion.
/circle6Motion is said to be rectilinear if the body moves in the same straight line all
the time. e.g., a car moving in straight line on a level road.
/circle6The motion is said to be circular if the body moves on a circular path; e.g. themotion of the tip of the second needle of a watch.
/circle6The total path length covered by a moving body is the distance travelled by it.
/circle6The distance between the final and initial position of a body is called itsdisplacement.
/circle6Distance travelled in unit time is called speed, whereas, displacement per unittime is called velocity.Notes
MODULE - 3
Moving Things
 221Motion and its Description
SCIENCE AND TECHNOLOGY
/circle6Position-time graph of a body moving in a straight line with constant speed is
a straight line sloping with time axis. The slope of the line gives the velocity ofthe object in motion.
/circle6Velocity-time graph of a body in straight line with constant speed is a straightline parallel to time axis. Area under the graph gives distance travelled.
/circle6Velocity-time graph of a body in straight line with constant acceleration is astraight line sloping with the time axis. The slop of the line gives acceleration.
/circle6For uniformly accelerated motion
v=u + at
s=ut + 1
2at2
and v2=u2 + 2as
where u = initial velocity, v = final velocity, and s = distance travelled in t seconds
TERMINAL EXERCISE
1. An object initially at rest moves for t seconds with a constant acceleration a.
The average speed of the object during this time interval is
(a)2at⋅; (b) 2at⋅; (c)2 1
2at⋅; (d)21
2at⋅
2. A car starts from rest with a uniform acceleration of 4 ms–2. The distance travelled
in metres at the ends of 1s, 2s, 3s and 4s are respectively,
(a) 4, 8, 16, 32 (b) 2, 8, 18, 32(c) 2, 6, 10, 14 (d) 4, 16, 32, 64
3. Does the direction of velocity decide the direction of acceleration?
4. Establish the relation between acceleration and distance travelled by the body
5. Explain whether or not the following particles have acceleration:
(i) a particle moving in a straight line with constant speed, and
(ii) a particle moving on a curve with constant speed.
6. Consider the following combination of signs for velocity and acceleration of an
object with respect to a one dimensional motion along x-axis and give examplefrom real life situation for each case:Notes
SCIENCE AND TECHNOLOGYMODULE - 3 Motion and its Description
Moving Things
 222Table 9.11
Velocity Acceleration Example
(a) Positive Positive Ball rolling down on a slope
like slide or ramp
(b) Positive Negative
(c) Positive Zero
(d) Negative Positive
(e) Negative Negative
(f) Negative Zero(g) zero Positive
(h) Zero Negative
7. A car travelling initially at 7 ms
–1 accelerates at the rate of 8.0 ms–2 for an interval
of 2.0 s. What is its velocity at the end of the 2 s?
8. A car travelling in a straight line has a velocity of 5.0 ms–1 at some instant. After
4.0 s, its velocity is 8.0 ms–1. What is its average acceleration in this time interval?
9. The velocity-time graph for an object moving along a straight line has shown in
Fig. 3.32. Find the average acceleration of this object during the time interval
0 to 5.0 s, 5.0 s to 15.0 s and 0 to 20.0 s.
Fig. 9.32
10. The velocity of an automobile changes over a period of 8 s as shown in the table
given below:Notes
MODULE - 3
Moving Things
 223Motion and its Description
SCIENCE AND TECHNOLOGYTable 9.12
Time (s) Velocity (ms–1) Time (s) Velocity (ms–1)
0.0 0.0 5.0 20.0
1.0 4.0 6.0 20.02.0 8.0 7.0 20.0
3.0 12.0 8.0 20.0
4.0 16.0
(i) Plot the velocity-time graph of motion.
(ii) Determine the distance the car travels during the first 2 s.
(iii) What distance does the car travel during the first 4 s?
(iv) What distance does the car travel during the entire 8 s?
(v) Find the slope of the line between t = 5.0 s and t = 7.0 s. What does
the slope indicate?
(vi) Find the slope of the line between t = 0 s to t = 4 s. What does this slope
represent?
11. The position-time data of a car is given in the table given below:
Table 9.13
Time (s) Position (m) Time (s) Position (m)
0 0 25 150
5 100 30 112.5
10 200 35 7515 200 40 37.5
20 200 45 0
(i) Plot the position-time graph of the car.
(ii) Calculate average velocity of the car during first 10 seconds.
(iii) Calculate the average velocity between t = 10 s to t = 20 s.
(iv) Calculate the average velocity between t = 20 s and t = 25 s. What can
you say about the direction of the motion of car?
12. An object is dropped from the height of 19.6 m. Draw the displacement-time
graph for time when object reach the ground. Also find velocity of the objectwhen it touches the ground.Notes
SCIENCE AND TECHNOLOGYMODULE - 3 Motion and its Description
Moving Things
 22413. An object is dropped from the height of 19.6 m. Find the distance travelled by
object in last second of its journey.
14. Show that for a uniformly accelerated motion starting from velocity u and
acquiring velocity v has average velocity equal to arithmetic mean of the initial
(u) and final velocity ( v).
15. Find the distance, average speed, displacement, average velocity and acceleration
of the object whose motion is shown in the graph (Fig. 9.33).
Fig. 9.33
16. A body accelerates from rest and attains a velocity of 10 ms-1 in 5 s. What is
its acceleration?
ANSWERS TO INTEXT QUESTIONS
9.1
1. (c) 2. (a) 3. (b) 4. (a) 5. (c)
6.
Fig. 9.34Notes
MODULE - 3
Moving Things
 225Motion and its Description
SCIENCE AND TECHNOLOGY9.2
1. (a) (iii) (b) (iv) (c) (i) (d) (ii)
2. Distance = 140 m, Displacement = 100 m, Speed = 7 ms–1
3. When object moves with uniform speed4. 2ms
–1, 5 ms–1
5. Average speed = 0.2 ms–1, motion is uniform motion
Fig. 9.35
6. 0.2 ms–17. 63 km h–18. 48 km h–19. R, R
9.3
1. For first five seconds object moves with constant speed i.e. 2ms–1. From 5 to
15 second it remains at rest and then from 15 to 20 seconds it moves with
constant speed 2 ms–1.
The motion of the object is not uniform.
2. Velocity of object A is 4 times the velocity of B.
3.
Fig. 9.36Notes
SCIENCE AND TECHNOLOGYMODULE - 3 Motion and its Description
Moving Things
 2264.
Fig. 9.37
(i) a = 0.4 ms–2, (ii) – a = 0.4 ms–2, (iii) 10 m
5.
Fig. 9.38
9.4
1. (i) 19.6 m, (ii) 4 s 2. Zero and 9.8 ms–2
3. 600 m 4. 7.5 ms–1
9.5
1. (a) 2. (a) 3. (b)Acids, Bases and Salts
SCIENCE AND TECHNOLOGY  154Notes
MODULE - 2
Matter in our Surroundings
8
ACIDS, BASES AND SALTS
From generations, our parents have been using tamarind or lemon juice to give shiny
look to the copper vessels. Our mothers never store pickles in metal containers.
Common salt and sugar has often been used as an effective preservative. How didour ancestors know that tamarind, lemon, vinegar, sugar etc. works effectively? Thiswas common collective wisdom which was passed from generation to generation.These days, bleaching powder, baking soda etc. are commonly used in our homes.You must have used various cleaners to open drains and pipes and window pane
cleaners for sparkling glass. How do these chemicals work? In this lesson we will
try to find answers to these questions. Most of these examples can be classified asacids, bases or salts. In this unit we shall categorize these substances. We shall studyabout their characteristic properties. We will also be learning about pH – a measureof acidity and its importance in our life.
OBJECTIVES
After completing this lesson you will be able to:
/circle6define the terms acid, base, salt and indicator;
/circle6give examples of some common household acids, bases, salts and suggest
suitable indicators;
/circle6describe the properties of acids and bases;
/circle6differentiate between strong and weak acids and bases;
/circle6explain the role of water in dissociation of acids and bases;
/circle6explain the term ionic product constant of water;
/circle6define pH;
/circle6correlate the concentration of hydrogen ions and pH with neutral, acidic
and basic nature of aqueous solutions;155Acids, Bases and Salts
SCIENCE AND TECHNOLOGYNotes
MODULE - 2
Matter in our Surroundings
H+
ACIDS/circle6recognize the importance of pH in everyday life,;
/circle6define salts and describe their methods of preparation;
/circle6correlate the nature of salt and the pH of its aqueous solution;
/circle6describe the manufacture and use of baking soda, washing soda, plaster
of paris and bleaching powder.
8.1 ACIDS AND BASES
For thousands of years, people have known that vinegar, lemon juice, Amla, tamarindand many other food items taste sour. However, only a few hundred years ago itwas proposed that these things taste sour because they contain ‘acids’. The term
acid comes from Latin term ‘accre’ which means sour. It was first used in the
seventeenth century by Robert Boyle to label substances as acids and basesaccording to the following characteristics:
Acids Bases
(i) taste sour (i) taste bitter
(ii) are corrosive to metals (ii) feel slippery or soapy
(iii) change blue litmus red (iii) change red litmus blue
(iv) become less acidic on mixing (iv) become less basic on mixing with
with bases acids
While Robert Boyle was successful in characterising acids and bases he could not
explain their behaviour on the basis of their chemical structure. This was accomplishedby Swedish scientist  Svante Arrhenius in the late nineteenth century. He proposedthat on dissolving in water, many compounds dissociate and form ions and theirproperties are mainly the properties of the ions they form. Governed by this, he
identified the ions furnished by acids and bases responsible for their characteristic
behaviour and gave their definitions.
8.1.1 Acids
An acid is a substance which furnishes hydrogen ions (H+) when dissolved in water.
For example, in its aqueous solution hydrochloric HCl (aq) dissociates as:
HCl (aq) ⎯⎯→  H+(aq) + Cl–(aq)
Some examples of acids are:(i) Hydrochloric acid (HCl) in gastric juice
(ii) Carbonic acid (H
2CO3) in soft drinks
(iii) Ascorbic acid (vitamin C) in lemon and many fruitsAcids, Bases and Salts
SCIENCE AND TECHNOLOGY  156Notes
MODULE - 2
Matter in our Surroundings
(iv) Citric acid in oranges and lemons
(v) Acetic acid in vinegar(vi) T annic acid in tea
(vii) Nitric acid (HNO
3) used in laboratories
(viii) Sulphuric acid (H2SO4) used in laboratories
8.1.2 Bases
A base is a substance which furnishes hydroxide ions (OH–) when dissolved in water.
For example, sodium hydroxide NaOH (aq), in its aqueous solutions, dissociates as:
NaOH (aq) ⎯⎯→  Na+(aq) + OH–(aq)
The term ‘ alkali ’ is often used for water soluble bases.
Some examples of bases are:(i) Sodium hydroxide (NaOH) or caustic soda used in washing soaps.
(ii) Potassium hydroxide (KOH) or potash used in bathing soaps.
(iii) Calcium hydroxide (Ca(OH)
2)  or lime water used in white wash.
(iv) Magnesium hydroxide (Mg(OH)2) or milk of magnesia used to control acidity.
(v) Ammonium hydroxide (NH4OH) used in hair dyes.
8.1.3 Indicators
You might have seen that the spot of turmeric or gravy on cloth becomes red when
soap is applied on it. What do you think has happened? Turmeric has acted as anindicator of base present in soap. There are many substances that show one colourin an acidic medium and another colour in a basic medium. Such substances are calledacid-base indicators.
Litmus is a natural dye found in certain lichens. It was the earliest indicator to be
used. It shows red colour in acidic solutions and blue colour in basic solutions.Phenolphthalein and methyl orange are some other indicators. The colours of theseindicators in acidic, neutral and basic solutions are given below in table 8.1.
Table 8.1 Colours of some indicators in acidic and basic solutionsOH–
BASE
Indicator Colour in acidic
solutionsColour in neutral
solutionsColour in basic
solutions
Litmus red purple blue
Phenolphthalein
Methyl orange red orange yellowcolourless colourless pink157Acids, Bases and Salts
SCIENCE AND TECHNOLOGYNotes
MODULE - 2
Matter in our Surroundings
Acid BaseINTEXT QUESTION 8.1
1. Put the following substances in acid or base bottle.
(a) Milk of magnesia
(b) gastric juice in humans
(c) soft drinks
(d) lime water
(e) vinegar
(f) soap
2. What will happen if you add a drop of the following on a cut unripe apple, curd,
causting soda solution and soap soluton.
(i) phenolphthalein
(ii) litmus
8.2 PROPERTIES OF ACIDS AND BASES
Each substance shows some typical or characteristics properties. We can categorize
a substance as an acid or a base according to the properties displayed. Let us learnthe characteristic properties of acids and bases.
8.2.1 Properties of Acids
The following are the characteristic properties of acids:
1. Taste
You must have noticed that some of the food items we eat have sour taste. The sourtaste of many unripe fruits, lemon, vinegar and sour milk is caused by the acids presentin them. Hence, we can say that acids have a sour taste. This is particularly true ofdilute acids (see table 8.2).
Table 8.2 Acids present in some common substances
Substance Acid present
1. Lemon juice Citric acid and ascorbic acid (vitamin C)
2. Vinegar Ethanoic acid (commonly called acetic acid)
3. Tamarind T artaric acid
4. Sour milk Lactic acidAcids, Bases and Salts
SCIENCE AND TECHNOLOGY  158Notes
MODULE - 2
Matter in our Surroundings
Points to ponder
All hydrogen containing
compounds are not acids
Although Ethyl alcohol
(C2H5OH) and glucose
(C6H12O6) contain
hydrogen but do not
produce H+ ion on
dissolving in water. Theirsolutions do not conductelectricity and are not acidic.ACTIVITY 8.1
Go to your neighbourhood shop and procure.
1. Packaged Curd
2. Juices in tetra packs
Test these with a litmus paper to find out if these are acidic in nature.
2. Action on Indicators
We have learnt earlier (section 8.1.3) that indicators show different colours in
presence of acids and bases. Let us recall the colours of the three commonlyused indicators in presence of acids.
Table 8.3 Colours of some indicators in presence of acids.
Indicator Colour in acidic medium
1. Litmus Red
2. Phenolphthalein Colourless
3. Methyl orange Red
3. Conduction of electricity and dissociation of acids
Do you know that solutions of acids in water (aqueous solutions) conduct electricity?
Such solutions are commonly used in car and inverter batteries. When acids are
dissolved in water they produce ions which help in conducting the electricity. This
process is known as dissociation . More specifically, acids produce hydrogen ions
(H+)  which are responsible for all their characteristic properties. These ions do not
exist as H+ in the solution but combine with water molecules as shown below:
H+    +     H2O  ⎯→    H3O+
hydrogen ion hydronium ion
 The H3O+ ions are called hydronium ions . These
ions are also represented as H+(aq).
On the basis of the extent of dissociation occurring in
their aqueous solutions, acids are classified as strongand weak acids.
A. Strong and Weak acids
Acids are classified as strong and weak acids and their
characteristics are as follow :159Acids, Bases and Salts
SCIENCE AND TECHNOLOGYNotes
MODULE - 2
Matter in our Surroundings
The acids which dissociate  partially in
water are called weak acids. All organicacids like acetic acid and some inorganicacids are weak acids. Since theirdissociation is only partial, it is depicted
by double half arrows.
HF(aq) 
/harpoonrightleft H+(aq) + F–(aq)
The double arrows indicates here that
(i) the aqueous solution of hydrofluoric
acid not only contains H+ (aq) and
F–(aq) ions but also the
undissociated acid HF(aq).
(ii)there is an equilibrium between the
undissociated acid HF(aq) and theions furnished by it, H
+(aq) and F–
(aq)
Examples:
(a) CH3COOH Ethanoic (acetic)
acid,
(b) HF Hydrofluoric acid(c) HCN Hydrocynic acid
(d) C
6H5COOH Benzoic acidStrong Acids Weak Acids
The acids which completely dissociate
in water are called strong acids
Nitric acid completely dissociates in
water
HNO3(aq) ⎯→  H+(aq) + NO3–(aq)
There are only seven strong acids
1. HCl Hydrochloric Acid2. HBr Hydrobromic Acid
3. HI Hydroiodic Acid
4. HClO
4Perchloric Acid
5. HClO3Chloric Acid
6. H2SO4Sulphuric Acid
7. HNO3 Nitric Acid
4. Reaction of Acids with Metals
The reaction of acids with metals can be studied with the help of the following acitivity.
ACTIVITY 8.2
This activity may be carried out in the chemistry laboratory of your study centre.
Aim:  To study the reaction of acids with metals.Acids, Bases and Salts
SCIENCE AND TECHNOLOGY  160Notes
MODULE - 2
Matter in our Surroundings
What is required?
A test tube, zinc granules, dilute H2SO4, match box and a test tube holder.
What to do?
/circle6Add a few zinc granules in a test tube.
/circle6Add dil. sulphuric acid carefully along the sides of the test tube.
/circle6Set the apparatus as shown in the Fig. 8.1.
/circle6Bring a burning match stick near the mouth of the test tube, (Fig. 8.1.
Stand
Dilute Sulphuric
acid
Zinc granulesHydrogen
gas bubblesBurning of hydrogen
gas with a pop sound
Burning Candle
Fig. 8.1:  Experiment to study the reaction of dil. H2SO4 with zinc. The gas
burns with a ‘pop’ sound  when a burning match stick is brought
near the mouth of the test tube.
What to observe?
/circle6When dilute sulphuric acid is added to zinc granules, hydrogen gas is formed.
The gas bubbles rise through the solution.
/circle6When the burning match stick is brought near the mouth of the test tube the gasin the test tube burns with a ‘pop’ sound. This confirms that the gas evolved ishydrogen gas.
From this experiment it can be said that dilute sulphuric acid reacts with zinc to
produce hydrogen gas. A similar reaction is observed when we use other metals likeiron. In general, it can be said that in such reactions metal displaces hydrogen fromacids and hydrogen gas is released. The metal combines with the remaining part ofthe acid and forms a compound called a salt, thus,
Acid + Metal ⎯→  Salt + Hydrogen gas161Acids, Bases and Salts
SCIENCE AND TECHNOLOGYNotes
MODULE - 2
Matter in our Surroundings
For example, the reaction between zinc and dil. sulphuric acid can be written as:
Z n      +      H2SO4     ⎯→     ZnSO4     +      H2 ↑
zinc dil sulphuric acid zinc sulphate hydrogen gas
metal acid salt
5. Reaction of acids with metal carbonates and hydrogen carbonates
Reaction of acids with metal carbonates and hydrogen carbonates can be studied
with the help of activity 8.2.
ACTIVITY 8.3
This experiment may be carried out in the chemistry laboratory of your study centre.
Aim:  To study the reaction of acids with metal carbonates and hydrogen carbonates.
What is required?
One test tube, one boiling tube fitted with a cork, thistle funnel and delivery tube, sodiumcarbonate, sodium hydrogen carbonate, dilute HCl and freshly prepared lime water.
What to do?
/circle6Take the boiling tube and add about 0.5 g sodium carbonate to it.
/circle6Take about 2 mL of freshly prepared lime water in a test tube.
Fig. 8.2:  Experimental set up to study the reaction of acids with metal
carbonates and hydrogen carbonates
Thistle funnel
Stand
ClampDelivery tube
Lime water
Na CO23Bubbles of CO2Dilute HClDilute HCl
Boiling tubeCorkAcids, Bases and Salts
SCIENCE AND TECHNOLOGY  162Notes
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/circle6Add about 3 mL dilute HCl to the boiling tube containing sodium carbonate and
immediately fix the cork filled with a delivery tube and set the apparatus as shownin the Fig. 8.2.
/circle6Dip the other end of the delivery tube in the lime water as shown in Fig. 8.2.
/circle6Observe the lime water carefully.
/circle6Repeat the activity with sodium hydrogen carbonate.
What to observe?
/circle6When dilute HCl is added to sodium carbonate or sodium hydrogen carbonate,carbon dioxide gas is evolved.
/circle6On passing CO2 gas, lime water turns milky.
/circle6On passing the excess of CO2 gas, lime water becomes clear again.
From the above activity it can be concluded that if sodium carbonate or sodium
hydrogen carbonate react with dilute hydrochloric acid, carbon dioxide gas isevolved. The respective reactions are:
Na
2CO3(s)  +  2HCl(aq)  ⎯→   2NaCl(aq) + H2O(l) + CO2(g)↑
sodium dil. hydrochloric sodium chloride water carbon
carbonate acid dioxide
NaHCO3(s)  +  HCl(aq)  ⎯→   NaCl(aq)  +  H2O(l)  +  CO2 (g) ↑
sodium dil. hydrochloric sodium chloride water carbon
hydrogen carbonate acid dioxide
On passing the evolved carbon dioxide gas through lime water, Ca(OH)2, the later
turns milky due to the formation of white precipitate of calcium carbonate
Ca(OH)2(aq)     +     CO2(g)  ⎯→    CaCO3(s)    +    H2O(l)
lime water carbon dioxide calcium carbonate water
(white ppt.)
If excess of carbon dioxide gas is passed through lime water, the white precipitate
of calcium carbonate disappears due to the formation of water soluble calcium
hydrogen carbonate.
CaCO3(s)   +     H2O(l)  +  CO2(g)    ⎯→      Ca(HCO3)2(aq)
calcium carbonate water Carbon calcium hydrogen carbonate
(white ppt.) dioxide (soluble in water)
Thus, we can summarize that,
Metal carbonate + Acid ⎯→  Salt + Water + Carbon dioxide
and Metal hydrogen carbonate + Acid ⎯→  Salt + Water + Carbon dioxide
6. Reaction of Acids with metal oxides
We can study the reaction of acids with metal oxides with the help of activity 8.4.163Acids, Bases and Salts
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ACTIVITY 8.4
This activity may be carried out in the chemistry laboratory of your study centre.
Aim :  To study the reaction of acids with metal oxides.
What is required?
A beaker, glass rod, copper oxide and dilute hydrochloric acid.
What to do?
/circle6Take a small amount of black copper oxide in a beaker.
/circle6Add about 10 mL of dilute hydrochloric acid and stir the solution gently with
the help of a glass rod. [Fig. 8.3(a)].
/circle6Observe the beaker as the reaction occurs. [Fig. 8.3(b)].
Black copper oxide
(a) Before the ReactionBeaker
Dilute HCl
(b) After the ReactionBeaker
Bluish green
solutionGlassRod
Fig. 8.3  Reaction between dilute hydrochloric acid and copper oxide (a) before
reaction black particles of copper oxide in transparent dilute hydrochloric
acid and (b) after reaction bluish green solution.
What to Observe?
/circle6When a mixture of dilute HCl and copper oxide is mixed, the black particles
of copper oxide can be seen suspended in colourless dilute hydrochloric acid.
/circle6As the reaction proceeds, the black particles slowly dissolve and the colour of
the solution becomes bluish green due to the formation of copper (II) chloride
(cupric chloride) – a salt.Acids, Bases and Salts
SCIENCE AND TECHNOLOGY  164Notes
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From this activity, we can conclude that the reaction between copper oxide and dilute
hydrochloric acid results in the formation of copper (II) chloride (cupric chloride)
which is a salt of copper. This salt forms bluish green solution. The reaction is:
CuO(s)  +   2HCl(aq)  ⎯→   CuCl2(aq)  +  H2O(l)
copper dil. hydrochloric copper (II) water
oxide acid chloride
Many other metal oxides like magnesium oxide (MgO) and calcium oxide (CaO)
or quick lime also react with acid in a similar way. For example,
CaO(s)    +    2HCl(aq)      ⎯→      CaCl2(aq)  +  H2O(l)
calcium oxide dil. hydrochloric calcium chloride water
(quick lime) acid
So, we can summarize with a general reaction between metal oxides and acids as:
Metal oxide + Acid   ⎯→   Salt + Water
7. Reaction of acids with bases
Let  us study the reaction of acids with bases with the help of the following activity.
ACTIVITY 8.5
This activity may be carried out in the chemistry laboratory of your study centre.
Aim : To study the reaction between acids and bases.
What is required?
A test tube, dropper, phenolphthalein indicator, solution of sodium hydroxide and dil.
hydrochloric acid.
What to do?
/circle6Take about 2 mL solution of sodium hydroxide in a test tube.
/circle6Add a drop of phenolphthalein indicator to it and observe the colour.
/circle6With the help of a dropper add dil. HCl dropwise and stir the solution constantlytill the colour disappears.
/circle6Now add a few drops of NaOH solution. The colour of the solution is restored.165Acids, Bases and Salts
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Dilute NaOHPhenolphthalein
drops
Pink colourTest tube
Dilute HCl
(a) (b)
Fig. 8.4:  Reaction between NaOH and HCl (a) Pink colour solution containing
NaOH solution and a drop of phenolphthalein (b) The solution becomes
colourless on addition of dil HCl
What to Observe?
/circle6When a drop of phenolphthalein is added to a solution of NaOH the solution
becomes pink in colour.
/circle6On adding HCl, the colour of the solution fades due to the reaction between
HCl and NaOH.
/circle6When whole of NaOH has reacted with HCl, the solution becomes colourless.
/circle6On adding NaOH, the solution becomes pink again.
From this activity, we can see that when dilute HCl is added to NaOH solution, the
two react with each other. When sufficient HCl is added, the basic properties ofNaOH and acidic properties of HCl disappear. The process is therefore calledneutralization . It results in the formation of salt and water. The reaction between
hydrochloric acid and sodium hydroxide forms sodium chloride and water.
HCl(aq)   +   NaOH(aq)     ⎯→      NaCl(aq)  +  H
2O(l)
hydrochloric sodium sodium chloride water
acid hydroxide
Similar reactions occur with other acids and bases. For example ,sulphuric acid and
potassium hydroxide react to form potassium sulphate and water.
H2SO4(aq)  +  2KOH(aq)   ⎯→   K2SO4(aq)  +  2H2O(l)
sulphuric potassium potassium water
acid hydroxide sulphate
In general, the reaction between and acid and a base can be written as:
Acid  +  Base   ⎯→   Salt  +  WaterAcids, Bases and Salts
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Warning
Although we talk of ‘taste’ of
acids and bases, it is notadvisable to taste any acid orbase. Most of them are harmful.Similarly touching the solutionsof strong acids and  basesshould be avoided. They mayharm the skin.8. Corrosive Nature
The ability of acids to attack various substances like metals, metal oxides and
hydroxides is referred to as their corrosive nature. (It may be noted here that the
term ‘corrosion’ is used with reference to metals and refers to various deteriorationprocesses (oxidation) they undergo due to their exposure to environment). Acids arecorrosive in nature as they can attack variety of substances.
‘Strong’ is different from ‘corrosive’
Corrosive action of acids is not related to their strength. It is related to thenegatively charged part of the acid. For example, hydrofluoric acid, (HF )is aweak acid. Yet, it is so corrosive that it attacks and dissolves even glass. The
fluoride ion attacks the silicon atom in silica glass while the hydrogen ion attacks
the oxygen of silica (SiO
2) in the glass.
SiO2    +    4HF   ⎯→     S i F4   +    2 H2O
silica hydrofluoric silicon water
(in glass) acid tetra fluoride
8.2.2 Properties of Bases
The following are the characteristic properties of bases:
1. Taste and touch
Bases have a bitter taste and their solutions are
soapy to touch.
2.Action on Indicators
As seen earlier (section 8.1.3) each indicator
shows characteristic colour in presence of bases.The colours shown by three commonly usedindicators in presence of bases are listed below for
easy recall.
Table 8.3 Colours of some common indicators in basic solution
Indicator Colour in basic medium
1. Litmus Blue
2. Phenolphthalein Pink
3. Methyl orange Y ellow
3. Conduction of electricity and dissociation of bases
Aqueous solutions (solution in water) of bases conduct electricity which is due to
the formation of ions. Like acids, bases also dissociate on dissolving in water. Basesproduce hydroxyl ions (OH
–) which are responsible for their characteristic properties.167Acids, Bases and Salts
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Weak bases do not furnish OH– ions by
dissociation. They react with water to
furnish OH– ions.
NH3(g) + H2O(l) ⎯→  NH4OH
NH4OH(aq) /horizontalharpoonextender/horizontalharpoonextender/horizontalharpoonextender/horizontalharpoonextender/arrowrighttophalf/arrowleftbothalf/horizontalharpoonextender/horizontalharpoonextender/horizontalharpoonextender/horizontalharpoonextenderNH4+(aq) +
OH–(aq)
or
NH3(g) + H2O(l) /horizontalharpoonextender/horizontalharpoonextender/horizontalharpoonextender/horizontalharpoonextender/arrowrighttophalf/arrowleftbothalf/horizontalharpoonextender/horizontalharpoonextender/horizontalharpoonextender/horizontalharpoonextender NH4+(aq)
+ OH–(aq)
The reaction resulting in the formation of
OH– ions does not go to completion and
the solution contains relatively lowconcentration of OH
– ions. The two half
arrows are used in the equation to indicate
that equilibrium is reached before the
reaction is completed. Examples of weakbases (i) NH
4OH, (ii) Cu(OH)2 (iv)
Cr(OH)3 (v) Zn(OH)2 etc.The bases which are soluble in water and give OH– ions in their aqueous solution
are called alkalies . All alkalies are bases but all bases are not alkalies . On the
basis of the extent of dissociation occurring in their solution, bases are classified asstrong and weak bases.
A. Strong and Weak Bases
Bases are classified as strong and weak bases and their characteristics are as follow :
Strong Bases Weak Bases
These bases are completely dissociated
in water to form the cation andhydroxide ion (OH
–). For example,
potassium hydroxide dissociates as
KOH(aq) ⎯→ K+(aq) + OH–(aq)
There are only eight strong bases. These
are the hydroxides of the elements ofthe Groups 1 and 2 of the periodic table
1. LiOH Lithium hydroxide
2. NaOH Sodium hydroxide
3. KOH Potassium hydroxide
4. RbOH Rubidium hydroxide
5. CsOH Caesium hydroxide
6. Ca(OH)
2Calcium hydroxide
7. Sr(OH)2Strontium hydroxide
8. Ba(OH)2Barium hydroxide
4. Reaction of bases with metals
Like acids, bases also react with active metals liberating hydrogen gas. Such reactions
can also be studied with the help of activity 8.2 given earlier. For example, sodiumhydroxide reacts with zinc as shown below:
Zn(s)  +  2NaOH(aq)   ⎯→    Na
2ZnO2(aq)  +  H2(g) ↑
zinc sodium sodium hydrogen
metal hydroxide zincate
5. Reaction of Bases with non-metal oxides
Bases react with oxides of non-metals like CO2, SO2, SO3, P2O5 etc. to form salt
and water.Acids, Bases and Salts
SCIENCE AND TECHNOLOGY  168Notes
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For example,
Ca(OH)2(aq)   +   CO2(g)   ⎯→     CaCO3(s)   +  H2O(l)
calcium hydroxide carbon calcium water
(lime water) dioxide carbonate
The reaction can be written in a general form as:
Base + Non-metal oxide  ⎯→   Salt + Water
6.  Reaction of bases with acids
We have learnt the mutual reaction between acids and bases in previous section. Such
reactions are called neutralization  reactions and result in the formation of salt and
water. The following are some more examples of neutralization reactions:
HCl(aq)  + KOH(aq) ⎯→ KCl(aq) + H2O(l)
H2SO4(aq)  + 2NaOH(aq) ⎯→ Na2SO4(aq) + 2H2O(l)
Caustic nature
Strong bases like sodium hydroxide and potassium hydroxide are corrosive
towards organic matter and break down the proteins of the skin and flesh to a
pasty mass. This action is called caustic action and it is due to this property that
sodium hydroxide is called ‘caustic soda’ and potassium hydroxide is called‘caustic potash’. The term ‘caustic’ is not used for corrosive action of acids.
INTEXT QUESTIONS 8.2
1. Name the substances in which the following acids are present:
(a) Ethanoic acid (b) Tartaric acid
2. Which of these acids would be partially dissociated in their aqueous solution?
(a) HBr (b) HCN(c) HNO
3 (d) C2H5COOH
3. An acid reacts with a substance X with liberation of a gas which burns with a
‘pop’ sound when a burning match stick is brought near it. What is the nature
of X?
4. An acid reacts with a substance Z with the liberation of CO2 gas. What can be
the nature of Z?
5. Which of the following oxides will react with a base?
(a) CaO (b) SO2169Acids, Bases and Salts
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8.3 WATER AND DISSOCIATION OF ACIDS AND BASES
In the previous sections, we have learnt that a substance is an acid if it furnishes H+
ions in its aqueous solution and a base if it furnishes OH– ions. Water plays very
important role in these processes, we shall learnt about it in this section.
8.3.1 Role of water in dissociation of acids and bases
If a dry strip of blue litmus paper is brought near the mouth of the test tube containingdry HCl gas , its colour does not changes. When it is moistened with a drop of waterand again brought near the mouth of the test tube , its colour turns red. It showsthat there are no H
+ ion in dry HCl gas. Only when it dissolves in water, H+ ions
are formed and it shows its acidic nature by turning the colour of the blue litmus
paper to red.
A similar behavior is exhibited by bases. If we take a pallet of dry NaOH in dry
atmosphere and quickly bring a dry strip of red litmus paper in its contact, no colourchange is observed. NaOH is a hygroscopic
compound and soon absorbs moisture from
air and becomes wet. When this happens, thecolour of the red litmus paper immediatelychanges to blue. Thus in dry solid NaOHalthough OH
– ions are present but they are not
free and do not show basic nature on coming
in contact with water, OH– ions becomes free
and show the basic nature by changing redlitmus blue. From the above discussion, it isclear that acidic and basic characters ofdifferent substances can be observed onlywhen they are dissolved in water.
(i) When an acid like sulphuric acid or a base like sodium hydroxide is dissolved
in water, the solution that is formed is hotter. It shows that the dissolution processis exothermic . A part of the thermal energy which is released during the
dissolution process is used up in overcoming the forces holding the hydrogenatom or hydroxyl group in the molecule of the acid or the base in breaking thechemical bond holding them and results in the formation of free H
+(aq) and OH–
(aq) ions.
(ii) Many bases are ionic compounds and consist of ions even in the solid state.
For example sodium hydroxide consists of Na+ and OH– ion. These ions are
held very tightly due to the strong electrostatic forces between the oppositelycharged ions. Presence of water as a medium (solvent) weakens these forces
greatly and the ions become free to dissolve in water.Warning
Dissolution of H2SO4 in water is
highly exothermic process.Therefore, to prepare an aqueoussolution, conc. sulphuric acid isadded slowly to water withconstants stirring. Water is never
added to con. sulphuric acid ashuge amount of heat is liberated.Due to that spattering occurs andthe acid can cause serious burnson skins or damage the items onwhich it falls.Acids, Bases and Salts
SCIENCE AND TECHNOLOGY  170Notes
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8.3.2 Self dissociation of water
Water plays an important role in acid base chemistry. We have seen that it helps
in the dissociation of acids and bases resulting in the formation of H+(aq) and OH–
(aq) ions respectively. Water itself undergoes dissociation process which is called‘self-dissociation of water’.Let us learn about it.
Self-dissociation of water
Water dissociates into H+(aq) and OH–(aq) ions as:
H2O(l) /horizontalharpoonextender/horizontalharpoonextender/horizontalharpoonextender/horizontalharpoonextender/arrowrighttophalf/arrowleftbothalf/horizontalharpoonextender/horizontalharpoonextender/horizontalharpoonextender/horizontalharpoonextender H+(aq) + OH–(aq)
The dissociation of water is extremely small  and only about two out of every billion
(109) water molecules are dissociated at 25°C. As a result, the concentrations of
H+(aq) and OH–(aq) ions formed is also extremely low. At 25°C (298K),
[H+] = [OH–] = 1.0 × 10–7 mol L–1
Here, square brackets denote the molar concentration of the species enclosed
within. Thus, [H+] denotes the concentration of H+(aq) ions in moles per litre and
[OH–] the concentration of OH–(aq) ions in moles per litre.
It must be noted here that in pure water and in all aqueous neutral solutions ,
[H+] = [OH–]
Also, in pure water as well as in all aqueous solutions at a given temperature, product
of concentrations of H+(aq) and OH–(aq) always remains constant. This product is
called ‘ionic product of water’ and is given the symbol Kw. It is also called ionic
product constant of water . Thus,
Kw = [H+] [OH–]
At 25°C (298 K), in pure water, Kw can be calculated as:
Kw= (1.0 × 10–7) × (1.0 × 10–7)
= 1.0 × 10–14
8.3.3 Neutral, acidic and basic solutions
We have seen that in pure water H+(aq) and OH–(aq) ions are produced in equal
numbers as a result of dissociation of water and therefore, their concentrations arealso equal i.e.
[H
+] = [OH–]
(i) Neutral solutions
In all neutral aqueous solutions, the concentrations of H+(aq) and OH–(aq) ions
remains equal i.e.
[H+] = [OH–]
In other words the neutral solution is the one in which the concentrations of H+ and
OH– ions are equal.
(ii) Acidic solutions
Acids furnish H+(aq) ions in their solutions resulting in increase in their concentration.
Thus, in acidic solutions171Acids, Bases and Salts
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[H+] > [OH–]
and [H+] > 1.0 × 10–7 mol L–1
In other words the acidic solution is the one in which the concentration of H+(aq)
is greater than that of OH–(aq) ions.
We have seen earlier that the ionic product of water Kw is constant at a given
temperature. It can remain so only if the concentration of OH–(aq) ions decreases.
[OH–] < 10–7 mol L–1
(iii) Basic solutions
Bases furnish OH–(aq) ions in their solutions. This results in an increase in their
concentration. Therefore, in basic solution
[OH–] > [H+]
and [OH–] > 1.0 × 10–7 mol L–1
In other words, the basic solution is the one in which the concentration of H+(aq)
ions is smaller than that of OH–1(aq) ions.
Here also, because of constancy of ionic product of water Kw, the concentration
of H+(aq) decreases. Thus
and [H+] < 1.0 × 10–7 mol L–1
We may summarize the nature of aqueous solution in terms of concentration of
hydrogen ions H+(aq) as shown in table 8.3.
Table 8.3 Concentration of H+(aq) ions in different
types of aqueous solutions
Nature of solution Concentration of H+ ions
at 25°C (298 K)
Neutral [H+] = 1.0 × 10–7 mol L–1
Acidic [H+] > 1.0 × 10–7 mol L–1
Basic [H+] < 1.0 × 10–7 mol L–1
INTEXT QUESTIONS 8.3
1. Why does the colour of dry blue litmus paper remains unchanged even when
it is brought in contact with HCl gas?
2. How does water help in dissociation of acids and bases?3. Identify the nature of the following aqueous solutions (whether acidic, basic or
neutral)
(a) Solution A: [H
+] < [OH–]
(b) Solution B: [H+] > [OH–]
(c) Solution C: [H+] = [OH–]Acids, Bases and Salts
SCIENCE AND TECHNOLOGY  172Notes
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8.4 pH AND ITS IMPORTANCE
When dealing with range of concentrations (such as these of H+(aq) ions) that spans
many powers of ten, it is convenient to represent them on a more compressedlogarithmic scale. By convention, we use the  pH scale  for denoting the concentration
of hydrogen ions. pH notation was devised by the Danish biochemist Soren Sorensen
in 1909. The term pH means “power of hydrogen”.
The pH is the logarithm (see box) of the reciprocal of the hydrogen ion concentration.
It is written as:
pH =
+1log
H⎡⎤⎣⎦
Alternately, the pH is the negative logarithm of the hydrogen ion concentration i.e
pH = –log [H+].
Because of the negative sign in the
expression, if [H+] increases, pH would
decrease and if it decreases, pH wouldincrease.
In pure water at 25° (298 K)
[H
+] = 1.0 × 10–7 mol L–1
log[H+] = log(10–7) = –7
and pH = –log[H+] = –(–7)
pH = 7
Since in pure water at 25°C (298 K)
[OH–] = 1.0 × 10–7 mol L–1
Also, pOH = 7
Since, Kw = 1.0 × 10–14
pKw = 14
The relationship between pKw, pH and pOH is
pKw = pH + pOH
at 25°C (298 K)
14 = pH + pOHLogarithm
Logarithm is a mathematical function
If x = 1 0y
then y = log10x
Here log10x mean log of x to the base
10. Usually, the base 10 is omitted in thenotation thus, y = log x.
e.g. log10
3= 3 × log10
= 3 × 1 = 3
log10–5= –5 × log10= –5 × 1=– 5
Note :  log 10 = 1173Acids, Bases and Salts
SCIENCE AND TECHNOLOGYNotes
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8.4.1 Calculations based on pH concept
In the last section, we learned the concept of pH and its relationship with hydrogen
ion or hydroxyl ion concentration. In this section, we shall use these relations to
perform some calculations.
The method of calculation of pH used in this unit are valid for (i) solutions of
strong  acids and bases only and (ii) the solutions of acids or bases should not be
extremely dilute and the concentrations of acids and bases should not be less than
10–6 mol L–1.
Example 8.1:  Calculate the pH of 0.001 molar solution of HCl.
Solution:  HCl is a strong acid and is completely dissociated in its solutions according
to the process:
HCl(aq) ⎯→ H+(aq) + Cl–(aq)
From this process it is clear that one mole of HCl would give one mole of H+ ions.
Therefore, the concentration of H+ ions would be equal to that of HCl i.e. 0.001
molar or 1.0 × 10–3 mol L–1.
Thus, [H+] = 1 × 10–3 mol L–1
pH = –log[H+] = –(log 10–3)
= –(–3 × log10) = –(3 × 1) = 3
Thus, pH = 3
Example 8.2:  What would be the pH of an aqueous solution of sulphuric acid which
is 5 × 10–5 mol L–1 in concentration.
Solution:  Sulphuric acid dissociates in water as:
H2SO4(aq) ⎯→ 2H+(aq) + SO42–(aq)
Each mole of sulphuric acid gives two mole of H+ ions in the solution. One litre of
5 × 10–5 mol L–1 solution contains 5 × 10–5 moles of H2SO4 which would give
2 × 5 × 10–5 = 10 × 10–5 or 1.0 × 10–4 moles of H+ ion in one litre solution.
Therefore,
[H+] = 1.0 × 10–4 mol L–1
pH = –log[H+] = –log10–4 = –(–4 × log10)
= –(–4 × 1) = 4
Example 8.3:  Calculate the pH of 1 × 10–4 molar solution of NaOH.
Solution:  NaOH is a strong base and dissociate in its solution as:
NaOH(aq) ⎯→ Na+(aq) + OH–(aq)Acids, Bases and Salts
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One mole of NaOH would give one mole of OH– ions. Therefore,
[OH–] = 1 × 10–4 mol L–1
pOH = –log[OH–] = –log × 10–4 = –(–4)
=4
Since pH + pOH = 14
pH = 14 – pOH = 14 – 4
=1 0
Example 8.4:  Calculate the pH of a solution in which the concentration of hydrogen
ions is 1.0 × 10–8 mol L–1.
Solution:  Here, although the solution is extremely dilute, the concentration given is
not of acid or base  but that of H+ ions. Hence, the pH can be calculated from the
relation:
pH = –log[H+]
given [H+] = 1.0 × 10–8 mol L–1
∴ pH = –log10–8 = –(–8 × log10)
= –(–8 × 1) = 8
8.4.2 pH Scale
The pH scale ranges from 0 to 14 on this scale. pH 7 is considered neutral, below
7 acidic and above 7 basic. Farther from 7, more acidic or basic the solution is.The scale is shown below in Fig. 8.5.
Fig. 8.5:  The pH scale
We have learnt earlier that the sum of pH and pOH of any aqueous solution remains
constant. Therefore, when one increases the other decreases. This relationship is
shown in Fig. 8.6
Fig. 8.6:  Relationship between pH and pOH at 25°C.p H+p O H=1 412345678 910 11 12 13 14 0pH
Acidity increases Basicity increases175Acids, Bases and Salts
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1 2 3 4 5 6 7 8 9 1 01 11 21 31 4 0pH of some common substances is shown in table 8.5.
Table 8.5: pH of some common acids and bases
Common Acids pH Common Bases pH
HCl (4%) 0 Blood plasma 7.4
Stomach acid 1 Egg white 8
Lemon juice 2 Sea water 8
Vinegar 3 Baking soda 9
Oranges 3.5 Antacids 10
Soda, grapes 4 Ammonia water 11
Sour milk 4.5 Lime water 12
Fresh milk 5 Drain cleaner 13
Human saliva 6-8 Caustic soda 4% (NaOH) 14
Pure water 7
8.4.3 Determination of pH
pH of a solution can be determined by using proper indicator or with the help of
a pH meter. The latter is a device which gives accurate value of pH. You will studymore about it in higher classes. We shall discuss here the use of indicators for findingout the pH of a solution.
Universal Indicator/pH paper .
It is a mixture of a number of indicators. It shows a specific colour at a given pH.
A colour guides is provided with the bottle of the indicator or the strips of paperimpregnated with it which are called pH paper strips. The test solution is tested witha drop of the universal indicator, or a drop of the test solution is put on pH paper.The colour of the solution on the pH paper is compared with the colour chart/guardand pH is read from it. The pH values thus obtained are only approximate values.
Fig. 8.7:  Colour chart/guide of universal indicator/pH paper.
8.4.2 Importance of pH in everyday life
pH plays a very important role in our everyday life. Some such examples are
described here.
(a) pH in humans and animals
Most of the biochemical reactions taking place in our body are in a narrow pH rangeof 7.0 to 7.8. Even a small change in pH disturbs these processes.Acids, Bases and Salts
SCIENCE AND TECHNOLOGY  176Notes
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(b) Acid Rain
When the pH of rain water falls below 5.6, it is called acid rain . When acid rain
flows into rivers, the pH of the river water also falls and it become acidic. As a result,the survival of aquatic life become difficult.
(c) pH in plants
Plants have a healthy growth only when the soil has a specific pH range which shouldbe neither highly alkaline nor highly acidic.
(d) In digestive system
Our stomach produce hydrochloric acid which helps in  digestion of food. Whenwe eat spicy food, stomach produces too much of acid which causes ‘acidity’ i.e.irritation and sometimes pain too. To get rid of this we use ‘antacids’ which are baseslike ‘milk of magnesia’ (suspension of magnesium hydroxide in water).
(e) Self defence of animals and plants
Bee sting causes severe pain and burning sensation.It is due to the presence of methanoic acid in it. Useof a mild base like baking soda can provides relieffrom pain.
Some plants like ‘nettle plant’ have fine stinging hair
which inject methanoic acid into the body of anyanimal or human being that comes in its contact. Thiscauses severe pain and buring sensation. The leavesof dock plant that grows near the nettle plant whenrubbed on the affected area provides relief.
(f) Tooth decay
Tooth enamel is made of calcium phosphate whichis the hardest substance in our body and can withstand the effect of various foodarticles that we eat. If mouth is not washed properly after every meal, the foodparticles and sugar remaining in the mouth undergoes degradation due to the bacterialpresent in the mouth. This process produces acids and the pH goes below 5.5. Theacidic condition thus created corrode the tooth enamel and in the long run can resultin tooth decay.
INTEXT QUESTIONS 8.4
1. pOH of a solution is 5.2. What is its pH. Comment on the nature (acidic, basic
or neutral) of this solution.
2. pH of a solution is 9. What is the concentration of H+ ions in it.
Fig. 8.8  Nettle plant177Acids, Bases and Salts
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3. What is the nature (whether acidic, basic or neutral) of the following solutions?
(a) Solution A: pH = pOH
(b) Solution B: pH > pOH
(c) Solution C: pH < pOH
8.5 SALTS
Salts are ionic compounds made of a cation other than H+ ion and an anion other
than OH– ion.
8.5.1 Formation of salts
Salts are formed in many reactions involving acids and bases.
1. By Neutralization of acids and bases
Salts are the product (besides water) of a neutralization reaction.
For example,
Base Acid Salt W ater
NaOH + HCl ⎯→ NaCl + H2O
KOH + HNO3⎯→ KNO3+H2O
In general, MOH + HX ⎯→ MX + H2O
In all the above cases we can see that the positively charged cation of the salt comes
from the base. Therefore, it is called the ‘basic radical’. The negatively charged anionof the salt comes from the acid. It is therefore, called the ‘acid radical’ of the salt.For example, in the salt NaCl, the cation Na
+ comes from the base NaOH and is
its basic radical and the anion Cl– comes from the acid HCl and is its ‘acid radical’.
2. By action of acids on metals
In a reaction between an acid and a metal, salt is produced along with hydrogen,
Metal Acid Salt Hydrogen
Zn + H2SO4⎯→ ZnSO4+H2
3. By action of acids on metal carbonates and hydrogen carbonates
Salts are produced in reactions between acids and metal carbonates and hydrogen
carbonates (bicarbonates) along with water and carbon dioxide.
Metal carbonate
or hydrogen Acid Salt W ater Carbon
carbonate dioxide
CaCO3 + 2HCl ⎯→ CaCl2+H2O+C O2
NaHCO3 + HCl ⎯→ NaCl + H2O+C O2Acids, Bases and Salts
SCIENCE AND TECHNOLOGY  178Notes
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Fig. 8.9 Solvey’s process for manufacturing of Baking sodaSaturating tank
Cooling pipesAmmoniacal brineCarbonating tower
CO2CO2Brine
Filter
FilterNH + (traces)32CO
NaHCO (for ignition)3Slaked limeLime kiln
Ammonia recovery tower
NH + a little NH44 3Cl HCO
CaO
HO2Type of salt and the nature of its aqueous solution:
Salt of Nature of Salt
Acid Base Solution pH (at 25°C)
1. Strong Strong Neutral pH = 7
2. Weak Strong Basic pH > 73. Strong Weak Acidic pH < 7
4. Weak Weak More information required –
8.6 SOME COMMONLY USED SALTS
A large number of salts are used in our homes and industry for various purposes.
In this section we would learn about some such salts.
8.6.1 Baking soda
You must have seen your mother using baking soda while cooking some ‘dals’. Ifyou ask her why does she use it, she would tell that it helps in cooking some items
fasters which otherwise would take must longer time. Chemically baking soda is
sodium hydrogen carbonate, NaHCO
3.
(a) Manufacture
Baking soda is manufactured by Solvey’s process. It is mainly used for manufacturingwashing soda but baking soda is obtained as an intermediate.179Acids, Bases and Salts
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Raw materials required
The raw materials required to manufacture washing soda are:
/circle6Lime stone which is calcium carbonate, CaCO3
/circle6Sodium chloride (NaCl) in the form of brine( Conc. NaCl Solution)
/circle6Ammonia (NH3)
Process
In Solvey’s process, carbon dioxide is obtained by heating limestone strongly,
CaCO3(s)   ⎯→   CaO(s)   +    CO2(g)↑
lime stone quick lime carbon dioxide
It is then passed through cold brine (a concentrated solution of NaCl in water) which
has previously been saturated with ammonia,
NaCl(aq) + CO2(g) + NH3(g) + H2O(l) ⎯→  NaHCO3(s)↓ + NH4Cl(aq)
sodium chloride ammonia sodium hydrogen ammonium
in brine carbonate chloride
NaHCO3 is sparingly soluble in water and crystallises out as white crystals. Its
solution in water is basic in nature. It is a mild and non-corrosive base.
Action of heat:  On heating, sodium hydrogen carbonate is converted into sodium
carbonate and carbon dioxide is given off,
2NaHCO3 heat⎯⎯⎯→ Na2CO3 + H2O + CO2↑
sodium carbonate
(b) Use
1. Used for cooking of certain foods.
2. For making baking power (a mixture of sodium hydrogen carbonate and tartaric
acid). On heating during baking, baking soda gives off carbon dioxide. It is thiscarbon dioxide which raises the dough. The sodium carbonate produced onheating the baking soda gives a bitter taste. Therefore, instead of using the bakingsoda alone, baking powder is used. The tartaric acid present in it neutralises the
sodium carbonate to avoid its bitter taste. Cakes and pastries are made flufly
and soft by using baking powder.
3. In medicines
Being a mild and non-corrosive base, baking soda is used in medicines to
neutralise the excessive acid in the stomach and provide relief. Mixed with solid
edible acids such as citric or tartaric acid, it is used in effervescent drinks to cureindigestion.
4. In soda acid fire extinguishersAcids, Bases and Salts
SCIENCE AND TECHNOLOGY  180Notes
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8.6.2 Washing soda
Washing soda is used for washing of clothes. It is mainly because of this chemical
that the clothes washed by a washerman appear so white. Chemically, washing soda
is sodium carbonate decahydrate, Na2CO3.10H2O.
(a) Manufacture
Washing soda is manufacturing by Solvey’s process. We have already learnt about
the raw materials required and part of the process in the manufacture of baking soda.
Sodium carbonate is obtained by calcination (strong heating in a furnace) of sodiumhydrogen carbonate and then recrystallising from water:
2NaHCO
3 heat⎯⎯⎯→ Na2CO3 + H2O + CO2
Na2CO3 + 10H2O ⎯→  Na2CO3.10H2O
sodium carbonate washing soda
(b) Uses
1. It is used in the manufacture of caustic soda, glass, soap powders, borex and
in paper industry.
2. For removing permanent hardness of water.
3. As a cleansing agent for domestic purpose.
8.6.3 Plaster of Paris
You must have seen some beautiful designs made on the ceiling and walls of rooms
in many houses. These are made of plaster of paris, also called POP. Chemically,
it is 2CaSO4.H2O or CaSO4.1
2H2O (calcium sulphate hemi hydrate)
(a) Manufacture
Raw material
Gypsum, (CaSO4.2H2O) is used as the raw material.
Process
The only difference between gypsum (CaSO4.2H2O) and plaster of paris
(CaSO4.1/2H2O ) is in the less amount of  water of crystallization.
When gypsum is heated at about 100° (373 K) temperature, it loses a part of its
water of crystallization to form:
CaSO4.2H2O  ()heat
100°C 373 K⎯⎯⎯⎯⎯⎯ →   CaSO4.1/2H2O + 3/2H2O
gypsum plaster of paris181Acids, Bases and Salts
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The temperature is not allowed to rise beyond 100°C otherwise whole of water of
crystallization is lost and anhydrous calcium sulphate is produced which is called‘dead burnt’ as it does not have the property to set after mixing with water.
(b) Uses
1. In making casts for manufacture of toys and statues.
2. In medicine for making plaster casts to hold fractured bones in place while they
set. It is also used for making casts in dentistry.
3. For making the surface of walls and ceiling smooth.
4. For making decorative designs on ceilings, walls and pillars.
5. For making‘ chalk’ for writing on blackboard.
6. For making fire proof materials.
8.6.4 Bleaching Powder
Have you ever wondered at the whiteness of a new white cloth? How is it made
so white? It is done by bleaching of the cloth at the time of its manufacture. Bleaching
is a process of removing colour from a cloth to make it whiter. Bleaching powder
has been used for this purpose since long. Chemically, it is calcium oxychloride,CaOCl
2.
(a) Manufacture
1.Raw material required : The raw material required for the manufacture of
bleaching powder are:
/circle6Slaked lime, Ca(OH)2
/circle6Chlorine gas, Cl2
Waste gasesSlaked lime
Chlorine
Bleaching powder
Fig. 8.10 Hasen-Clever plant for manufacturing of bleaching powderAcids, Bases and Salts
SCIENCE AND TECHNOLOGY  182Notes
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2.Process : It is manufactured by Hasen-Clever Method. The plant consists of four
cylinders made of cast iron with inlet for chlorine near the base. The dry slaked
lime, calcium hydroxide is fed into the chlorinating cylinders from the top. It moves
down slowly and meets the upcoming current of chlorine. As a result of thereaction between them, it is converted into bleaching power which collects at
the bottom.
Ca(OH)
2   +   C l2    ⎯→     CaOCl2  +   H2O
slaked lime chlorine bleaching
powder
(b) Uses
1. In textile industry for bleaching of cotton and linen.
2. In paper industry for bleaching of wood pulp.
3. In making wool unshrinkable.
4. Used as disinfactant and germicide for sterilization of water.5. For the manufacture of chloroform.
6. Used as an oxidizing agent in chemical industry.
INTEX QUESTIONS 8.5
1. Identify acid radical and basic radical in CaSO4.
2. CuSO4 was prepared by reacting an acid and a base. Identify the acid and the
base that must have been used in this reaction.
3. Which one of the following is the correct formula of plaster of paris?
CaSO4.H2O or 2CaSO4.H2O
WHAT YOU HAVE LEARNT
/circle6Acids are the substances which taste sour, change blue litmus red, are corrosive
to metals and furnish H+ ions in their aqueous solutions.
/circle6Bases are the substances which taste bitter, change red litmus blue, feel slippery
and furnish OH– ions in their aqueous solutions.
/circle6Indicators are the substances that show one colour in an acidic medium andanother colour in a basic medium. Litmus, phenolphthalein and methyl orange
are commonly used indicators.183Acids, Bases and Salts
SCIENCE AND TECHNOLOGYNotes
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/circle6Acids are presents in many unripe fruits, vinegar, lemon, sour milk etc., while
bases are present in lime water, window pane cleaners, many drain cleaners etc.
/circle6Aqueous solutions of acids and bases both conduct electricity as they dissociate
on dissolving in water and liberate cations and anions which help in conductingelectricity.
/circle6Strong acids and bases dissociate completely in water. HCl, HBr, HI, H2SO4,
HNO3, HClO4 and HClO3 are strong acids and LiOH, NaOH, KOH, RbOH,
CsOH, Ca(OH)2, Sr(OH)2 and Ba(OH)2 are strong bases.
/circle6Weak acids and bases dissociate  partially in water. For example, HF, HCN,CH
3COOH etc. are some weak acid and NH4OH, Cu(OH)2, Al(OH)3 etc. are
some weak bases.
/circle6Acids and bases react with metals to produce salt and hydrogen gas.
/circle6Acids react with metal carbonates and metal hydrogen carbonates to produce
salt, water and CO2.
/circle6Acids react with metal oxides to produce salt and water.
/circle6Bases react with non-metal oxides to produce salt and water.
/circle6Acids and bases react with each other to produce salt and water. Such reactionsare called neutralization reactions.
/circle6Acids and bases dissociate only on dissolving in water.
/circle6Water itself undergoes dissociation and furnishes H+ and OH– ions in equal
numbers. This is called self dissociation of water. The extent of dissociation isvery small.
/circle6Concentrations of H+ and OH– ion formed by the self dissociation of water are
1.0 × 10–7 molar each at 25°C.
/circle6Product of concentrations of hydrogen and hydroxyl ions is called the ‘ionic
product’ or ionic product constant’ of water, Kw. It remains unchanged even
when some substance (acid, base or salt etc.) is dissolved in it.
/circle6pH is defined as +1log
H⎡⎤⎣⎦ or –log[H+], likewise pOH = –log[OH–] and pKw
= –log Kw
/circle6In pure water or in any aqueous solution pH + pOH = pKw = 14 at 25°C.
/circle6In pure water [H+] = [OH–]. It is also true in any neutral aqueous solution. In
terms of pH, pH = pOH = 7 in water and any neutral solution.
/circle6In acidic solution [H+] > [OH–] and pH < pOH. Also pH < 7 at 25°C.
/circle6In basic solutions [H+] < [OH–] and pH > pOH. Also pH > 7 at 25°C.
/circle6Universal indicator is prepared by mixing a number of indicators. It shows adifferent but characteristic colour at each pH.Acids, Bases and Salts
SCIENCE AND TECHNOLOGY  184Notes
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/circle6Maintenance of correct pH is very important for biochemical process occuring
in humans and animals.
/circle6If pH of rain water falls below 5.6, it is called acid rain and is quite harmful.
/circle6pH plays an important role in proper growth of plants and also for proper
digestion in our bodies.
/circle6Salts are ionic compounds made of a cation other than H+ ion and an anion other
than OH– ion. They are formed in neutralization reaction.
/circle6Salts are also formed in reaction of acids and bases with metals, of acid withmetal carbonates, hydrogen carbonates and oxides and in reaction of bases withnon-metal oxides.
TERMINAL EXERCISE
A. Objective Type Questions
I. Mark the correct choice
1. Lemon juice contains
(a) tartaric acid (b) ascorbic acid
(c) acetic acid (d) lactic acid
2. Aqueous solutions of acids conduct electricity. This shows that
(a) They contain H+ ions
(b) They contain OH– ion
(c) They contain cations and anions
(d) They contain both H+ and OH– ions
3. Which of the following is not a strong acid?
(a) HCl (b) HBr
(c) HI (d) HF
4. Self dissociation of water produces
(a) a large number of H+ ions
(b) a large number of OH– ions
(c) H+ and OH– ions in equal numbers
(d) H+ and OH– ions in unequal numbers
5. In any aqueous basic solution
(a) [H+] > [OH–] (b) [H+] < [OH–]
(c) [H+] = [OH–] (d) [H+] = 0185Acids, Bases and Salts
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6. In an aqueous solution of HCl which of the following species is not present?
(a) H+(b) OH–
(c) HCl (d) Cl–
7. Which of the following is not a raw material for  manufacturing  washing soda?
(a) Lime stone (b) Ammonia
(c) Slaked lime (d) Sodium chloride
II. Mark the following statements as true (T) or false (F):
1. Acids furnish H+ ions only in the presence of water.
2. Lime water turns blue litmus red.
3. HF is a strong acid.
4. H2 gas is produced when acids react with metal oxides.
5. Corrosive action of acids is due to H+ ions present in them.
6. When the pH of the rain water become more than 5.6 it is called acid rain.
7. Aqueous solutions of all the salts are neutral in nature i.e. neither acidic nor basic
in nature.
III. Fill in the blanks
1. Acids taste ...................... while bases taste ......................
2. Milk of magnesia turns ...................... litmus ......................3. One mole of sulphuric acid would furnish ...................... mole/s of H
+ ions and
...................... moles of SO42– ions.
4. ...................... gas is produced when acids react with metal hydrogen carbonates.5. Lime water turns milky on passing CO
2 gas due to the formation of
......................
6. The reaction between an acid and a base is known as ......................7. Bee sting injects ...................... acid which causes severe pain and burning
sensation.
8. In NH
4NO3 the acid radical is ...................... and the basic radical is
......................
9. Chemically baking soda is ......................
B. Descriptive Questions
1. What is an acid?
2. Give two examples of acids found in food articles.
3. What is a base?
4. Give two examples of bases.Acids, Bases and Salts
SCIENCE AND TECHNOLOGY  186Notes
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5. What are indicators?
6. What is the colour of methyl orange indicator in (i) acidic medium and (ii) basic
medium.
7. Why do solutions of acids and bases conduct electricity?8. Differentiate between strong and weak acids and give one example of each.
9. Write down the reaction between zinc and sulphuric acid.
10. Which gas is evolved when an acid reacts with metal carbonates? Which other
category of compounds would produce the same gas on reacting with acids?
11. What type of oxides react with acids? Give one examples of this type of oxide
and write down the balanced equation for the reaction.
12. What is the name given to the reaction between an acid and a base? What are
the products formed in such reactions?
13. “Corrosive action of acids is not related to their strength”. Justify this statement.14. Give one example each of the following (i) a strong base (ii) a weak base15. List three categories of substances that can react with a base. Give one example
of each and write the chemical reaction involved in each case.
16. What happens when a dry strip of each of red litmus paper and blue litmus paper
is brought in contact with HCl gas? In which case a change would be observedif the strips are moistened and then brought in contact with HCl gas and whatwould be the change?
17. A small palette of NaOH is kept on dry red litmus paper. Initially, no change
is observed but after some time its colour starts changing to blue around the
place where the palette of NaOH is kept. Explain these observations.
18. How does water help in dissociation of acids and bases? Explain.19. What is ‘self dissociation of water’? Name the resulting species and give their
concentrations at 25°C.
20. What is ionic product constant of water? Give its value at 25°C. Will the value
change if an acid, base or a salt is dissolved in water?
21. Give the relationships between the concentrations of hydrogen ions and hydroxyl
ions in (i) pure water (ii) a neutral solution (iii) an acidic solution and (iv) a basic
solution.
22. What is pH? What happens to the pH if the hydroxyl ion concentration in the
solution increases?
23. Predict whether a given aqueous solution is acidic, basic or neutral if its pH is
(a) 7.0, (b) 11.9 and (c) 3.2.
24. Calculate the pH of 1.0 ×10
–4 molar solution of HNO3.187Acids, Bases and Salts
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25. What is the pH of 1.0 × 10–5 molar solution of KOH?
26. What is the pH of 1.0 × 10–2 mol L–1 solution of NaCl?
27. What do you understand by the term ‘universal indicator’?28. What is acid rain?
29. What is the importance of pH for humans and animals, and our digestive system?
30. Which chemical causes pain and burning sensation when somebody accidentally
touches ‘nettle plant’?
31. What is a salt? Give two examples.32. How are salts obtained from an acid? Mention four types of substances that
can be used for it.
33. Give chemical formula of (i) baking soda and (ii) washing soda.34. List the raw materials required for the manufacture of baking soda and describe
the process with the help of suitable chemical equations.
35. Distinguish between baking powder and baking soda. Why is baking powder
preferred for making cakes?
36. Give any two uses of baking soda.
37. What is washing soda? Give its chemical formula. How is it manufactured by
Solvey’s method?
38. Give two uses of washing soda.39. What is the chemical formula of ‘plaster of paris’? How is it manufactured? What
precaution is taken during its manufacture?
40. List any four uses of ‘plaster of paris’.41. What is bleaching? Chemically, what is bleaching powder? Give its any four uses.42. List the raw materials required and the method of manufacture of bleaching
powder. Write the equation for the reaction involved.
ANSWERSTO INTEXT QUESTIONS
8.1
1. Acidic : (b), (c) and (e)
Basic : (a), (d)  and (f)
2. Phenolphthalein: Colourless on unripe apple and pink in solutions of caustic soda
and soap.
Litmus: Red on unripe apple and curd, and blue in solutions of caustic soda and
soap solution.Acids, Bases and Salts
SCIENCE AND TECHNOLOGY  188Notes
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8.2
1. (a) Vinegar (b) tamarind
2. (b) and (d)3. It must be a metal.4. It may be either a metal carbonate or hydrogen carbonate.5. SO
2
8.3
1. It is because HCl gas does not contain H+(aq) ions and is non acidic
2. (i) The heat released in dissolution process help in the dissociation process
by overcoming the forces that hold the hydrogen atom or the hydroxyl groupin the  molecules of the acid or the base, or in breaking the chemical bondholding them.
(ii) Presence of water weaken the electrostatic forces between anion and
cations.
3. (a) Solution A – basic
(b) Solution B – acidic(c) Solution C – neutral
8.4
1. Since pH + pOH = 14
pH = 14 – pOH = 14 – 5.2
= 8.8
Since pH > 7.0, it is basic in nature
2. pH = –log[H+] = 9
∴ log[H+] = –9
or [H+] = 10–9 mol L–1
3. (a) Solution A — neutral
(b) Solution B — basic (since [H+] < [OH–] in it)
(c) Solution C — acidic (since [H+] > [OH–] in it)
8.5
1. Acid radical SO42–
Basic radical Ca2+
2. Acid: H2SO4 (corresponding to the acid radical SO42–)
Base: Cu(OH)2 (corresponding to the basic radical Cu2+)
3. (a) Carbonates (b) potassium salts4. 2CaSO
4.H2OMODULE - 3
MOVING THINGS
9. Motion and its Description
10. Force and motion
11. Gravitation93Atomic Structure
SCIENCE AND TECHNOLOGYNotes
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5
ATOMIC STRUCTURE
In lesson 3, you have studied about atoms and molecules as the constituents of matter.
You have also learnt that the atoms are the smallest constituents of matter. In lesson 4you studied about the chemical reactions, their types and the ways to represent them.You know that according to Dalton’s atomic theory, the atoms of different elementsare different and in chemical reactions the atoms are rearranged between differentreacting substances. However, today we know that the atom is not indivisible as itwas thought by Dalton. The atom  has a structure and contains smaller constituentsin it. In this unit, we would attempt to find out the answers to some of the questionslike, “What is the structure of an atom?”, “What are the constituents of atoms?”,“Why the atoms of different elements are different?” and so on.
We will begin this unit with the study of the discoveries of sub-atomic particles such
as electron, proton etc. Then, we will take up various atomic models proposed onthe basis of these discoveries. We will discuss how various models for the structureof atom were developed and also explain the success as well as the shortcomingsof these models. This will be followed by the description of the arrangement or thedistribution of electrons in the atom. This arrangement is known as electronic
configuration . These electronic configurations are useful in explaining various
properties of the elements. These also determine the nature of chemical bonds formedby it. This aspect is dealt with in lesson  7 on chemical bonding.
OBJECTIVES
After completing this lesson, you will be able to:
/circle6recall the evidences showing the presence of charged particles in matter;
/circle6describe the discovery of electron and proton;
/circle6explain Dalton’ s atomic theory and its failure;
/circle6discuss Thomson’s and Rutherford’s models of atom and explain their
limitations;Atomic Structure
SCIENCE AND TECHNOLOGY  94Notes
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/circle6explain the Bohr’ s model of atom (in brief);
/circle6describe the discovery of neutron;
/circle6compare the characteristic properties of proton, electron and neutron;
/circle6explain various rules for filling of  electrons and write the distribution of
electrons in different shells upto atomic number 20;
/circle6define valency and correlate the electronic configuration of an atom withits valency;
/circle6define atomic number and mass number of an atom;
/circle6describe isotopes and isobars;
/circle6define and compute average atomic mass and explain its fractional value.
5.1 CHARGED PARTICLES IN ATOM
You have read about Dalton’s atomic theory in lesson 3. The theory proposed in
the year 1803 considered the atom to be the smallest indivisible constituent of allmatter. The Dalton’s theory could explain the law of conservation of mass, law ofconstant composition and law of multiple proportions known at that time. However,towards the end of nineteenth century, certain experiments showed that an atom isneither the smallest nor indivisible particle of matter as stated by Dalton. It was shownto be made up of even smaller particles. These particles were called electrons,protons and neutrons. The electrons are negatively charged whereas the protons arepositively charged. The neutrons on the other hand are uncharged in nature. You willnow learn about the discovery of the charged subatomic particles.
5.1.1 Discovery of Electron
In 1885, Sir William Crookes carried out a series of experiments to study thebehaviour of metals heated in a vacuum using cathode ray tubes. A cathode ray tube
Fig. 5.1:  A cathode ray tube; cathode rays are obtained on applying high
voltage across the electrodes in an evacuated glass tube
High voltage95Atomic Structure
SCIENCE AND TECHNOLOGYNotes
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consists of two metal electrodes in a partially evacuated glass tube. An evacuated
tube is the one from which most of the air has been removed. The negatively chargedelectrode is called cathode whereas the positively charged electrode is called anode.These electrodes are connected to a high voltage source. Such a cathode ray tubehas been shown in Fig. 5.1.
It was observed that when very high voltage was passed across the electrodes in
evacuated tube, the cathode produced a stream of particles. These particles were
shown to travel from cathode to anode and were called cathode rays .  In the
absence of external magnetic or electric field these rays travel in straight line. In 1897,
an English physicist Sir J.J. Thomson showed that the rays were made up of a stream
of negatively charged particles. This conclusion was drawn from the experimentalobservations when the experiment was done in the presence of an external electric
field. Following are the important properties of cathode rays:
/circle6Cathode rays travel in straight line
/circle6The particles constituting cathode rays carry mass and possess kinetic energy
/circle6The particles constituting cathode rays have negligible mass but travel very fast
/circle6Cathode ray particles carry negative charge and are attracted towards positively
charged plate when an external electric field is applied (Fig. 5.2)
/circle6The nature of cathode rays generated was independent of the nature of the gas
filled in the cathode ray tube as well as the nature of metal used for making
cathode and anode. In all the cases the charge to mass ratio (e/m) was foundto be the same.
+
B
ACathode AnodeFluorescent screen
Fig. 5.2:  The cathode rays are negatively charged; these travel in straight line
from cathode to the anode (A), however in the presence of an external
electrical field these bend towards the positive plate (B)
These particles constituting the cathode rays were later called electrons . Since it
was observed that the nature of cathode rays was the same irrespective of the metalused for the cathode or the gas filled in the cathode ray tube. This led Thomson toAtomic Structure
SCIENCE AND TECHNOLOGY  96Notes
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conclude that all atoms must contain electrons. This meant that the atom is not
indivisible as was believed by Dalton and others . In other words, we can say
that the Dalton’s theory of atomic structure failed partially.
This conclusion raised a question, “If the atom was divisible, then what were its
constituents?”.  Today a number of smaller particles are found to constitute atoms.
These particles constituting the atom are called subatomic particles .  You have learnt
above that electron is one of the constituents of the atom, let us study the next section
to learn about another constituent particle present in an atom. As the atom is neutral,
we expect the presence of positively charged particles in the atom so as to neutralisethe negative charge of the electrons.
5.1.2 Discovery of Proton
Much before the discovery of electron, Eugen Goldstein  (in 1886) performed an
experiment using a perforated cathode (a cathode having holes in it) in the discharge
tube filled with air at a very low pressure. When a high voltage was applied acrossthe electrodes in the discharge tube, a faint red glow was observed behind the
perforated cathode.  Fig. 5.3
Fig. 5.3  Goldstein’ s cathode ray tube with perforated cathode
This glow was due to another kind of rays flowing in a direction opposite to that
of the cathode rays. These rays were called as anode rays or positive rays. These
were positively charged and were also called canal rays  because they passed
through the holes or the canals present in the perforated cathode. The followingobservations were made about anode rays (canal rays):
/circle6Like cathode rays, the anode rays also travel in straight lines.
/circle6The particles constituting anode rays carry mass and have kinetic energy.
/circle6The particles constituting canal rays are much heavier than electrons and carry
positive charges
+++++
++
+Red glow
Gas at LOW
pressurePositive rays
from Anode
High voltage sourceAnodePerforated cathode
+97Atomic Structure
SCIENCE AND TECHNOLOGYNotes
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/circle6The positive charge on the particles was whole number multiples of the amount
of charge present on the electron.
/circle6The nature and the type of the particles constituting the anode rays weredependent on the gas present in the discharge tube.
The origin of anode rays can be explained in terms of interaction of the cathode rays
with the gas present in the vacuum tube. It can be explained as given below:
The electrons emitted from the cathode collide with the neutral atoms of the gas
present in the tube and remove one or more electrons present in them. This leavesbehind positive charged particles which travel towards the cathode. When thecathode ray tube contained hydrogen gas, the particles of the canal rays obtainedwere the lightest and their charge to mass ratio (e/m ratio) was the highest. Rutherfordshowed that these particles were identical to the hydrogen ion (hydrogen atom from
which one electron has been removed). These particles were named as protons  and
were shown to be present in all matter. Thus, we see that the experiments by Thomsonand Goldstein had shown that an atom contains two types of particle which areoppositely charged and an atom is electrically neutral. What do you think is therelationship between the numbers of these particles in a given atom?
In addition to the two charged particles namely the electron and the proton, a neutral
particle called neutron was also discovered about which you would learn later in this
lesson. Now, it is the time to check your understanding. For this, take a pause andsolve the following intext questions:
INTEXT QUESTIONS 5.1
1. Name two charged particles which constitute all matter.
2. Describe a cathode ray tube.3. Name the negatively charged particles emitted from the cathode in the cathode
ray tube?
4. Why do the canal rays obtained by using different gases have different e/m values?
In addition to the discovery of electrons and protons as the constituents of
atom, the phenomenon of radioactivity  that is the spontaneous emission of
rays from atoms of certain elements also proved that the atom was divisible.
5.2 EARLIER MODELS OF ATOM
In section 5.1 you have learnt that the atom is divisible and contains three smallerparticles in it. The question that arises is, “In what way are the subatomic particlesAtomic Structure
SCIENCE AND TECHNOLOGY  98Notes
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Matter in our Surroundings
arranged in the atom?”. On the basis of experimental observations, different models
have been proposed for the structure of an atom. In this section, we will discuss
two such models namely Thomson model and Rutherford model.
5.2.1 Thomson Model
In lesson 3 you have learnt that all matter is made of atoms and all the atoms are
electrically neutral. Having discovered electron as a constituent of atom, Thomson
concluded that there must be an equal amount of positive charge present in an atom.
On this basis he proposed a model for the structure of atom. According to his model,atoms can be considered as a large sphere of uniform positive charge with a numberof small negatively charged electrons scattered throughout it, Fig. 5.4. This modelwas called as plum pudding  model. The electrons represent the plums in the pudding
made of negative charge. This model is similar to a water-melon in which the pulp
represents the positive charge and the seeds denote the electrons. However, you may
note that a water melon has a large number of seeds whereas an atom may not haveas many electrons.
Electron
Sphere of
positive charge
Fig. 5.4: Thomson’ s plum-pudding model
5.2.2 Rutherford’s model
Ernest Rutherford and his co-workers were working in the area of radioactivity. They
were studying the effect of alpha ( α) particles on matter. The alpha particles are helium
nuclei, which can be obtained by the removal of two electrons from the helium atom.In 1910, Hans Geiger (Rutherford’s technician) and Ernest Marsden (Rutherford’s
student) performed the famous α-ray scattering experiment. This led to the failure
of Thomson’s model of atom. Let us learn about this experiment.
ααααα-Ray scattering experiment
In this experiment a stream of α particles from a radioactive source was directed
on a thin (about 0.00004 cm thick) piece of gold foil. On the basis of Thomson’s
model it was expected that the alpha particles would just pass straight through the99Atomic Structure
SCIENCE AND TECHNOLOGYNotes
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Matter in our Surroundings
gold foil and could be detected by a photographic plate placed behind the foil.
However, the actual results of the experiment, Fig. 5.5,  were quite surprising. It was
observed that:
(i) Most of the α-particles passed straight through the gold foil.
(ii) Some of the α-particles were deflected by small angles.
(iii) A few particles were deflected by large angles.
(iv) About 1 in every 12000 particles experienced a rebound.
Beam of
-particles/c97
Thin gold foilMost particles
are unreflectedSeattered of particles
Circular
fluorescent
screen
Fig. 5.5:  The experimental set-up and observations in the α- ray scattering
experiment   performed by Geiger and Marsden
The results of α-ray scattering experiment were explained by Rutherford in 1911 and
another model of the atom was proposed. According to Rutherford’s model, Fig. 5.6(a).
/circle6An atom contains a dense and positively charged region located at its centre;it was called as nucleus ,
/circle6All the positive charge of an atom and most of its mass was contained in the
nucleus,
/circle6The rest of an atom must be empty space which contains the much smaller andnegatively charged electrons,
(a)( b)
Fig 5.6:  (a) Rutherford’ s model of atom (b) Explanation of the results of scattering
experiment by Rutherford’s model.-
+Electron
Nucleus--
-
-
-
-
---
--
-
-+Atomic Structure
SCIENCE AND TECHNOLOGY  100Notes
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On the basis of the proposed model, the experimental observations in the scattering
experiment could be explained. This is illustrated in Fig. 5.6(b). The α particles
passing through the atom in the region of the electrons would pass straight withoutany deflection. Only those particles that come in close vicinity of the positively chargednucleus get deviated from their path. Very few α-particles, those that collide with
the nucleus, would face a rebound.
On the basis of his model, Rutherford was able to predict the size of the nucleus.
He estimated that the radius of the nucleus was at least 1/10000 times smallerthan that of the radius of the atom. We can imagine the size of the nucleus withthe following analogy. If the size of the atom is that of a cricket stadium then thenucleus would have the size of a fly at the centre of the stadium.
INTEXT QUESTIONS 5.2
1. Describe Thomson’s model of atom. What is it called?
2. What would have been observed in the α-ray scattering experiment if the
Thomson’s model was correct?
3. Who performed the α-ray scattering experiment and what were the observations?
4. Describe the model of atom proposed by Rutherford.
5.3 DRAWBACKS OF RUTHERFORD’S MODEL
According to Rutherford’s model the negatively charged electrons revolve in circular
orbits around the positively charged nucleus. However, according to Maxwell’selectromagnetic theory (about which you may learn in higher classes), if a chargedparticle accelerates around another charged particle then it would continuously lose
energy in the form of radiation. The loss of energy would slow down the speed of
the electron. Therefore, the electron isexpected to move in a spiral fashionaround the nucleus and eventually fall intoit as shown in Fig. 5.7. In other words,the atom will not be stable. However, we
know that the atom is stable and such a
collapse does not occur. Thus,Rutherford's model is unable to explainthe stability of the atom. You know thatan atom may contain a number ofelectrons. The Rutherford’s model also
does not say anything about the way the electrons are distributed around the nucleus.
Another drawback of the Rutherford’s model was its inability to explain the
Fig. 5.7:  The electron in the Rutherford’s
model is expected to spiral into the nucleus101Atomic Structure
SCIENCE AND TECHNOLOGYNotes
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relationship between the atomic mass and atomic number (the number of protons).
This problem was solved later by Chadwick by discovering neutron, the third particle
constituting the atom. You would learn about it in section 5.5.
The problem of the stability of the atom and the distribution of electrons in the atom
was solved by Neils Bohr by proposing yet another model of the atom. This isdiscussed in the next section.
5.4 BOHR’S MODEL OF ATOM
In 1913, Niels Bohr,a student of Rutherford proposed a model to account for theshortcomings of Rutherford’s model. Bohr’s model can be understood in terms oftwo postulates proposed by him. The postulates are:
Postulate 1:  The electrons move in definite circular paths of fixed energy around
a central nucleus ; just like our solar system in which different planets revolve around
the Sun in definite trajectory. Similar to the planets, only certain circular paths aroundthe nucleus are allowed for the electrons to move. These paths are called orbits ,
or energy levels . The electron moving in the orbit does not radiate. In other words,
it does not lose energy; therefore, these orbits are called stationary orbits or
stationary states . The bold concept of stationary state could answer the problem
of stability of atom faced by Rutherford’s model.
32
K18
L8
M2
N+Nucleus(n=4) shellN
(n=3) shellM
(n=2) shellL
(n=1) shellK
Fig. 5.8:   Illustration showing different orbits or the energy levels of fixed
energy in an atom according to Bohr’ s model
It was later realised that the concept of circular orbit as proposed by Bohr was not
adequate and it was modified to energy shells with definite energy. While a circular
orbit is two dimensional, a shell is a three dimensional region. The shells of definite
energy are represented by letters (K, L, M, N etc.) or by positive integers (1, 2,3, …. etc.) Fig. 5.8. The energies of the shells increase with the number n; n = 1,Atomic Structure
SCIENCE AND TECHNOLOGY  102Notes
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level is of the lowest energy. Further, the maximum number of electrons that can be
accommodated in each shell is given by 2n2, where n is the number of the level. Thus,
the first shell (n=1) can have a maximum of two electrons whereas the second shell
can have 8 electrons and so on. Each shell is further divided into various sublevels
called subshells  about which you would study in your higher classes.
Postulate 2: The electron can change its shells or energy level by absorbing or
releasing energy. An electron at a lower state of energy Ei can go to a final higher
state of energy Ef by absorbing  a single photon of energy given by:
E = hν = Ef – Ei
Similarly, when electron changes its shell from a higher initial level of energy Ei to
a lower final level of energy Ef, a single photon of energy hν is released (Fig. 5.9).
n=3
n=2
n=1
A photon is emitted
with energy E=h/c110Increasing energy
of orbits
Fig. 5.9:  The electrons in an atom can change their energy level by absorbing
suitable amounts of energy or by emitting energy.
 INTEXT QUESTIONS 5.3
1. Give any two drawbacks of Rutherford’s model of atom.
2. State the postulates of Bohr’s model.3. How does Bohr model of an atom explain the stability of the atom?
Thus, the Bohr’s model of atom removes two of the limitations of Rutherford’s model.
These are related to the stability of atom and the distribution of electrons aroundthe nucleus. You would recall that the third limitation of Rutherford’s model was its103Atomic Structure
SCIENCE AND TECHNOLOGYNotes
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Matter in our Surroundings
inability to explain the relationship between the atomic mass, and the atomic number
(the number of protons) of an atom. Let us learn how this problem was solved withthe discovery of neutron.
5.5 DISCOVERY OF NEUTRON
You would recall that when we discussed about the failure of Rutherford’s model
we mentioned that it was unable to explain the relationship between the atomic massand the atomic number (the number of protons). According to the Rutherford’s model,the mass of helium atom (containing 2 protons) should be double that of a hydrogenatom (with only one proton). [Ignoring the mass of electron as it is very light].However, the actual ratio of the masses of helium atom to hydrogen atom is 4:1.It was suggested that there must be one more type of subatomic particle presentin the nucleus which may be neutral but have mass.
Such a particle was discovered by James Chadwick in 1932. This was found to be
electrically neutral and was named neutron . Neutrons are present in the nucleus of
all atoms, except hydrogen. A neutron is represented as ‘n’ and is found to havea mass slightly higher than that of a proton. Thus, if the helium atom contained 2protons and 2 neutrons in the nucleus, the mass ratio of helium to hydrogen (4:1)could be explained. The characteristics of the three fundamental particles constitutingthe atom are given in Table 5.1.
Table 5:1 Characteristics of the fundamental subatomic particles
Particle Symbol Mass (in kg) Actual Charge Relative charge
(in Coulombs)
Electron e 9.109 389 × 10–311.602 177 × 10–19–1
Proton p 1.672 623 × 10–271.602 177 × 10–191
Neutron n 1.674 928 × 10–2700
INTEXT QUESTIONS 5.4
1. What is a neutron and where is it located in the atom?
2. How many neutrons are present in the α-particle?
3. How will you distinguish between an electron and a proton?
5.6 ATOMIC NUMBER AND MASS NUMBER
You have learnt that the nucleus of atom contains positively charged particles called
protons and neutral particles called neutrons. The number of protons in an atom
is called the atomic number and is denoted by the symbol ‘Z’ .  All atoms of anAtomic Structure
SCIENCE AND TECHNOLOGY  104Notes
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Matter in our Surroundings
element have the same atomic number. The electrons occupy the space outside the
nucleus. In order to account for the electrically neutral nature of the atom, the numberof protons in the nucleus is exactly equal to the number of electrons. Thus,
Atomic number = number of protons = number of electrons
You would recall that according to Dalton’s theory, the atoms of different elementsare different from each other. We can now say that this difference is due to differencein the numbers of protons present in the nucleus of the element. In other words,different elements differ in terms of their atomic number . For example, the atoms
of hydrogen and helium are different because hydrogen has one proton in its nucleuswhereas the nucleus of helium atom contains two protons. Their atomic numbers are1 and 2, respectively. You have learnt in the Rutherford’s model that the mass ofthe atom is concentrated in its nucleus. This is due to the presence of two heavyparticles namely protons and neutrons in the nucleus. These particles are callednucleons . The number of nucleons in the nucleus of an atom is called its mass
number . It is denoted by ‘A’ and is equal to the total number of protons and neutrons
present in the nucleus of an element.  Thus,
Mass number (A) = number of protons(Z) + number of neutrons(n)
Atomic number and mass number are represented on the symbol of an element. Anelement, X with an atomic number, Z and the mass number, A  is denoted as follows:
AXZ
For example, 12C6 means that the carbon has an atomic number of 6 and the mass
number of 12. This can be used to compute the number of different fundamental
particles in the atom. Let us calculate it for carbon.
As the atomic number is 6 this means:
Number of protons = number of electrons = 6As Mass number = number of protons + number of neutrons
⇒ 12 = 6 + number of neutrons
⇒ number of neutrons = 12 – 6 = 6
Thus, an atom of 
12C6 has 6 protons, 6 electrons and 6 neutrons.
INTEXT QUESTIONS 5.5
1. A sodium atom has an atomic number of 11 and a mass number of 23. Calculate
the number of protons, electrons and neutrons in a sodium atom.105Atomic Structure
SCIENCE AND TECHNOLOGYNotes
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Matter in our Surroundings
2. What is the mass number of an atom which has 7 protons and 8 neutrons?
3. Calculate the number of electrons, protons and neutrons in 40Ar18 and 49K19 .
5.7 ELECTRONIC CONFIGURATION: DISTRIBUTION
OF ELECTRONS IN DIFFERENT ORBITS
As discussed in section 5.4, the electrons move in definite paths called orbits or shells
around a central nucleus. These orbits or shells have different energies and canaccommodate different number of electrons in them.  The question arises that howare the electrons distributed amongst these shells?  The answer to this question wasprovided by Bohr and Bury. According to their scheme, the electron distribution isgoverned by the following rules:
I. These orbits or shells in an atom are represented by the letters K, L, M, N,…
or the positive integral numbers, n = 1,2,3,4,….
II. The orbits are arranged in the order of increasing energy. The energy of M shell
is more than that of the L shell which in turn is more than that of the K shell.
III. The maximum number of electrons present in a shell is given by the formula 2n
2,
where ‘n’ is the number of the orbit or the shell. Thus, the maximum numberof electrons that can be accommodated in different shells are as follows:
Maximum number of electrons in K shell (or n = 1 level) = 2n
2 =   2 × (1)2
= 2
Maximum number of electrons in L shell (or n = 2 level)  = 2n2 =   2 × (2)2
= 8
Maximum number of electrons in M shell (or n = 3 level) = 2n2 =   2 × (3)2
= 18  and so on. See table 5.2
Table 5.2: Electron accommodation capacity of different shells
Value of n Shell name Maximum capacity
1 K-Shell 2
2 L- Shell 8
3 M- Shell 18
4 N- Shell 32
IV . The shells are occupied in the increasing order of their energies.
V . Electrons are not accommodated in a given shell, unless the inner shells are
completely filled.
The arrangement of electrons in the various shells or orbits of an atom of the element
is known as electronic configuration. Keeping these points in mind, let us now studythe filling of electrons in various shells of atoms of different elements.Atomic Structure
SCIENCE AND TECHNOLOGY  106Notes
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Matter in our Surroundings
/circle6Hydrogen (H) atom has only one electron. It would occupy the first shell and
electronic configuration of hydrogen can be represented as 1.
/circle6The next element helium (He) has two electrons in its atom. Since the first shellcan accommodate two electrons; hence, this second electron will also be placedin the first shell. The electronic configuration of helium is written as 2.
/circle6The third element, Lithium (Li) has three electrons. Now the two electrons occupy
the first shell whereas the third electron goes to the next shell of higher energy
level, i.e. second shell. Thus, the electronic configuration of Li is 2, 1.
Similarly, the electronic configurations of other elements can be written. The structures
of the atoms of elements with atomic number 1 to 18 are given in Fig. 5.10.
H
Li
NaBe
MgB
AlC
SiN
PO
SF
ClNe
ArHe
Fig. 5.10:  The structures, according to Bohr’ s model of atoms, of elements
with atomic number 1 to 18.
5.7.1 Concept of Valence or Valency
We have just discussed the electronic configuration of first 18 elements. You can see
from the Fig. 5.10 that different elements have different number of electrons in theoutermost or the valence shell. These electrons in the outermost shell are known asvalence electrons. The number of valence electrons determines the combining
capacity of an atom in an element.  Valence is the number of chemical bonds that
an atom can form with univalent atoms. Since hydrogen is a univalent atom, thevalence of an element can be taken by the number of atoms of hydrogen with whichone atom of the element can combine. For example, in H
2O, NH3, and CH4 the
valencies of oxygen, nitrogen and carbon are 2, 3 and 4 respectively.
The elements having a completely filled outermost shell in their atoms show little or
no chemical activity. In other words, their combining capacity or valency is zero. Theelements with completely filled valence shells are said to have stable electronic107Atomic Structure
SCIENCE AND TECHNOLOGYNotes
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Matter in our Surroundings
configuration. The main group elements can have a maximum of eight electrons in
their valence shell. This is called octet rule ; you will learn more about it in lesson 7.
You will learn that the combining capacity or the tendency of an atom to react with
other atoms to form molecules depends on the ease with which it can achieve octet
in its outermost shell. The valencies of the elements can be calculated from theelectronic configuration by applying the octet rule. It can be seen as follows:
/circle6If the number of valence electrons is four or less then the valency is equal to
the number of the valence electrons.
/circle6In cases when the number of valence electrons is more than four then generally
the valency is equal to 8 minus the number of valence electrons.
Thus,
Valency = Number of valence electrons (for 4 or lesser valence electrons)Valency = 8 - Number of valence electrons (for more than 4 valence electrons)
The composition and electronic configuration of the elements having the atomic
numbers from 1 to 18, along with their valencies is given in Table 5.3.
Table 5.3: The composition, electron distribution and common valency
of the elements with atomic number from 1 to 18
Hydrogen
Helium
LithiumBeryllium
BoronCarbon
Nitrogen
Oxygen
Neon
SodiumFluorine
Magnesium
Aluminium
Silicon
Phosphorus
SulphurChlorine
ArgonH
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
181
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
181
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18–
2
4
5
6
6
7
8
10
10
12
12
14
14
16
16
18
221
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2–
–
1
2
3
4
5
6
7
8
8
8
8
8
8
8
8
887654321–
–
–
–
–
–
––
–
––
–
–
–
–
–
––
–
–
––
–
–
–
–
–
–1
1
1
1
12222
3, 53
3
340
0
04Name of
ElementSymbol
KAtomic
NumberNumber
of
ProtonsNumber
of
NeutronsNumber
of
ElectronsDistribution of
Electrons
L MNValencyAtomic Structure
SCIENCE AND TECHNOLOGY  108Notes
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Matter in our Surroundings
In next lesson, you will study about the importance of electronic configurations in
understanding the periodic arrangement of elements. These electronic configurationsare also helpful in studying the nature of bonding between various elements whichwill be dealt with in lesson 7.
INTEXT QUESTIONS 5.6
1. How many shells are occupied in the nitrogen (atomic number =7) atom?
2. Name the element which has completely filled first shell.
3. Write the electronic configuration of an element having atomic number equal to 11.
WHAT HAVE YOU LEARNT
/circle6According to Dalton’s atomic theory, the atom is considered to be the smallest
indivisible constituent of all matter. This theory could explain the law ofconservation of mass, law of constant composition and law of multiple proportions.However, certain experiments  towards the end of nineteenth century showedthat the atom is neither the smallest nor indivisible particle of matter. It was shownto be made up of even smaller particles called electrons, protons and neutrons.
/circle6 Sir J.J.Thomson discovered that when very high voltage was passed across theelectrodes in the cathode ray tube, the cathode produced rays that travel fromcathode to anode and were called cathode rays .  It  showed that the rays were
made up of a stream of negatively charged particles called electrons.  Thediscovery of electrons meant that the atom is not indivisible as was believed
by Dalton and others .
/circle6 Eugen Goldstein  discovered anode rays by using a perforated cathode (a
cathode having holes in it) in the discharge tube filled with air at a very lowpressure. The discovery of anode rays established the presence of positivelycharged proton in the atom.
/circle6 According to Thomson’s plum-pudding model, atoms can be considered as alarge sphere of uniform positive charge with a number of small negatively chargedelectrons scattered throughout it.
/circle6 The α-ray scattering experiment performed by Geiger and Marsden led to the
failure of Thomson’s model of atom. In this experiment, a stream of α-particles
from a radioactive source was directed on a thin piece of gold foil. Most of theα-particles passed straight through the gold foil, some α-particles were deflected
by small angles, a few particles by large angles and very few experienced arebound.109Atomic Structure
SCIENCE AND TECHNOLOGYNotes
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Matter in our Surroundings
/circle6The results of α-ray scattering experiment were explained in terms of Rutherford’s
model. According to which the atom contains a dense and positively chargedregion called nucleus  at its centre and the negatively charged electrons move
around it. All the positive charge and most of the mass of atom is contained inthe nucleus.
/circle6 The Rutherford’s model however failed as it could not explain the stability ofthe atom, the distribution of electrons and the relationship between the atomicmass and atomic number (the number of protons).
/circle6The problem of the stability of the atom and the distribution of electrons in theatom was solved by Neils Bohr in terms of Bohr’s model of the atom. Bohr’smodel can be understood in terms of two postulates, the first being, ‘  The
electrons move in definite circular paths of fixed energy around a centralnucleus ’ and the second, ‘ The electron can change its orbit or energy level
by absorbing or releasing energy .’
/circle6 In 1932, James Chadwick discovered an electrically neutral particle in atom and
named it as neutron .
/circle6The number of protons in an atom is called the atomic number and is denoted
as ‘Z’. On the other hand the number of nucleons( protons plus neutrons)  inthe nucleus of an atom is called its mass number and is denoted as ‘A’
/circle6The electrons are distributed in different shells in the order of increasing energy.The distribution is called electronic configuration. The maximum number ofelectrons present in a shell is given by the formula 2n
2, where ‘n’ is the number
of the orbit or the shell.
/circle6The valence is the number of chemical bonds that an atom can form with univalentatoms. If the number of valence electrons is four or less, then the valency is equalto the number of the valence electrons.  On the other hand, if the number ofvalence electrons is more than four, then generally the valency is equal to 8 minusthe number of valence electrons.
TERMINAL EXERCISE
1. How did J.J.Thomson discover the electron? Explain his “plum pudding” model
of the atom.
2. What made Thomson conclude that all atoms must contain electrons?
3. Identify the following subatomic particles:
(a) The number of these in the nucleus is equal to the atomic number
(b) The particle that is not found in the nucleus
(c) The particle that has no electrical charge
(d) The particle that has a much lower mass than the others subatomic particlesAtomic Structure
SCIENCE AND TECHNOLOGY  110Notes
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4. Which of the following are usually found in the nucleus of an atom?
(a) Protons and neutrons only
(b) Protons, neutrons and electrons
(c) Neutrons only
(d) Electrons and neutrons only
5. Describe Ernest Rutherford’s experiment with alpha particles and gold foil. How
did this lead to the discovery of the nucleus?
6. What does the atomic number tell us about an atom?
7. What is the relationship between the numbers of electrons and protons in an
atom?
8. How did Neils Bohr revise Rutherford’s atomic model?
9. What is understood by a stationary state?10. What is a shell? How many electrons can be accomodate in L-shell?
11. State the rules for writing the electronic configuration of elements.
ANSWERS TO INTEXT QUESTIONS
5.1
1. Electrons and protons
2. A cathode ray tube consists of two metal electrodes in a partially evacuated glass
tube. The negatively charged electrode is called cathode while the positivelycharged electrode is called anode. These electrodes are connected to a high
voltage source.
3. Electron4. When the electrons emitted from the cathode collide with the neutral atoms of
the gas present in the tube, these remove one or more electrons present in them.This leaves behind positive charged particles which travel towards the cathode.As the atoms of different gases have different number of protons present in them,these give positively charged ions with different e/m values.
5.2
1. According to Thomson’s model, atoms can be considered as a large sphere of
uniform positive charge with a number of small negatively charged electrons
scattered throughout it. This model was called as plum pudding  model.111Atomic Structure
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2. If the Thomson’s model was correct, then most of the α-particles in the α-ray
scattering experiment would have passed straight through the atom
3. The α-ray scattering experiment was performed by Geiger and Marsden.  When
a stream of α-particles from a radioactive source was directed on a thin piece
of gold foil, most of the α-particles passed straight through the gold foil, some
α-particles were deflected by small angles, a few particles by large angles and
very few experienced a rebound.
4. According to Rutherford’s model, the atom contains a dense and positively
charged region called nucleus  at its centre and the negatively charged electrons
move around it. All the positive charge and most of the mass of atom is containedin the nucleus.
5.3
1. The Rutherford’s model could not explain the stability of the atom, the distribution
of electrons and the relationship between the atomic mass and atomic number(the number of protons).
2. The two postulates of Bohr’s model are :
I. The electrons move in definite circular paths of fixed energy around a
central nucleus.
II. The electron can change its orbit or energy level by absorbing or releasing
energy.
3. The Bohr’s model explains the stability of atom by proposing that the electron
does not lose energy when present in a given energy level.
5.4
1.  It is a neutral subatomic particle present in the nucleus of the atom.
2.  An α-particle contains two neutrons.
3.  The electron and proton can be distinguished in terms of their charge and mass.
While the electron is negatively charged, the proton is positively charged.
Secondly, the proton is much heavier than the electron; it is about 1840 timesheavier.
5.5
1. No of protons = 11
No. of electrons = 11
No. of neutrons = 12Atomic Structure
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2. Mass number = number of protons + number of neutrons
Therefore, mass number = 7 + 8 = 15
3.40Ar18 : Number of protons = atomic number = 18
 Number of electrons = number of protons = 18
 Number of neutrons = mass number – number of protons = 40 – 18 = 22
40K19 Number of protons = atomic number = 19
Number of electrons = number of protons = 19
Number of neutrons = mass number – number of protons = 40 – 19 = 21
5.6
1. The electronic configuration of nitrogen is 2, 5. Thus, two shells are occupied.
The first shell (capacity = 2) is completely filled while the second shell (capacity= 8) is partially filled.
2. Helium3. The electronic configuration of an element having atomic number 11 is 2, 8, 1.113Periodic Classification of Elements
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6
PERIODIC CLASSIFICATION OF
ELEMENTS
In the last lesson, you have studied about the structure of atoms and their electronicconfigurations. You have also learnt that the elements with similar electronicconfigurations show similar chemical properties. By the middle of the nineteenthcentury quite a large number of elements (nearly 60) were known. In order to studythese elements systematically, it was considered necessary to classify them. In thislesson, you will undertake the journey through the development of classification ofelements from ancient to modern. You will also study how some properties ofelements vary in the modern periodic table.
OBJECTIVES
After studying this lesson you will be able to:
/circle6describe briefly the development of classification of elements;
/circle6state main features of Mendeleev’s periodic table;
/circle6explain the defects of Mendeleev’ s periodic table;
/circle6state modern periodic law;
/circle6describe the features of the long form of periodic table;
/circle6explain modern periodic classification and
/circle6describe the tr ends in variation of atomic size and metallic character in the
periodic table.
6.1 CLASSIFICATION OF ELEMENTS
6.1.1 Need for Classification of Elements
You must have visited a chemist’s shop. Several hundred medicines are stored init. In spite of this, when you ask for a particular medicine, the chemist is able to locateit easily. How is it possible? It is because the medicines have been classified  into
various categories and sub categories and arranged accordingly. This makes theirlocation an easy task.Periodic Classification of Elements
SCIENCE AND TECHNOLOGY  114Notes
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Before the beginning of the eighteenth century, only a few elements were known,
so it was quite easy to study and remember the properties of those elements and
their compounds individually.  However, by the middle of the nineteenth century, more
the than sixty elements had been discovered. The number of compounds formed bythem was also enormous. With the increasing number of elements, it was becomingmore and more difficult to study their properties individually. Therefore, the need for
their classification was felt. This led to the classifications of various elements into
groups which helped in the systematic study of elements.
6.1.2 Development of Classification
Scientists after many attempts were successful in arranging various elements intogroups. They realised that even though every element is different from others, yetthere are a few similarities among some elements. Accordingly, similar elements were
arranged into groups which led to classification. Various types of classification were
proposed by different scientists. The first classification of elements was into 2 groups-metals  and non-metals .  This classification served only limited purpose mainly
because some elements like germanium and antimony showed the properties of both
– metals and non-metals. They could not be placed in any of the two classes.
Scientists were in search of such characteristics of an element which would never
change. After the work of William Prout in 1815,  it was found that the atomic mass
of an element remains constant, so it could form the basis for a satisfactory
classification. Now, you will learn about the four major attempts made for
classification of elements. They are as follows :
1. Dobereiner’s Triads
2. Newlands’ Law of Octaves3. Mendeleev’s Periodic Law & Periodic Tables
4. Modern Periodic Table
6.1.3 Dobereiner’s Triads
In 1829, J.W. Dobereiner, a German chemist made groups
of three elements each and called them triads  (Table 6.1).
All three elements of a triad were similar in their physical andchemical properties. He proposed a law known asDobereiner’s law of triads . According to this law, when
elements are arranged in order of increasing atomic
mass, the atomic mass of the middle element was nearlyequal to the arithmetic mean of the other two and
its properties were intermediate between those of the
other two.
J.W. Dobereiner
(1780-1849)115Periodic Classification of Elements
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73 9232+=
40 13788.52+=
35.5 12781.252+=Table 6.1: Dobereiner’s triads of elements
S. No. Element Atomic Mass Mean of I and III
1. I. Lithium 7
II. Sodium 23III. Potassium 39
2. I. Calcium 40
II. Strontium 88
III. Barium 137
3. I. Chlorine 35.5
II. Bromine 80III. Iodine 127
This classification did not receive wide acceptance since only a few elements could
be arranged into triads.
6.1.4 Newlands’ Law of Octaves
In 1864, an English chemist John Alexander Newlands arranged the elements in theincreasing order of their atomic masses (then called atomic weight ). He observed
that every eighth element had properties similar to the first element . Newlands
called it the Law of Octaves . It was due to its similarity with musical notes where
every eighth note is the repetition of the first one as shown below :
12345678
lk js xk ek ik èkk uh lk
The arrangement of elements given by Newlands is given in Table 6.2.
Starting from lithium  (Li), the eighth element is sodium  (Na) and its properties are
similar to those of the lithium. Similarly, beryllium  (Be), magnesium  (Mg) and
calcium  (Ca) show similar properties. Fluorine  (F) and chlorine  (Cl) are also similar
chemically.
Table 6.2 : Arrangement of some elements with their atomic masses
according to the Law of Octaves.
Li Be B C N O F
(7) (9) (11) (12) (14) (16) (19)
Na Mg A l Si P S Cl
(23) (24) (27) (28) (31) (32) (35.5)
KC a
(39) (40)
The merits of Newlands ’ Law of Octaves classification are:
(i) Atomic mass was made the basis of classification.Periodic Classification of Elements
SCIENCE AND TECHNOLOGY  116Notes
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(ii) Periodicity of properties (the repetition of properties after a certain interval) was
recognised for the first time.
The demerits of Newlands’ law of Octaves are:(i) It was not applicable to elements of atomic masses higher than 40 u. Hence,
all the 60 elements known at that time, could not be classified according to thiscriterion.
(ii) W ith the discovery of noble gases, it was found that it was the ninth  element
which had the properties similar to the first one and not the  eighth  element.
This resulted in the rejection of the very idea of octaves.
The basic idea of Newlands for using the atomic mass as the fundamental property
for classification of elements was pursued further by two scientists Lother Meyer andD. Mendeleev. Their main achievement was that they both included almost all theknown elements in their work. We shall, however, discuss the classification proposedby Mendeleev which was accepted more widely and is the basis of the modernclassification.
6.1.5 Mendeleev’s Periodic Law and Periodic Table
D’mitri Mendeleev (also spelled as Mendeleef or Mandeleyev ) , a Russian chemiststudied the properties of all the 63 elements known at that time and their compounds.On arranging the elements in the increasing order of atomic masses, he observedthat the elements with similar properties occur periodically. In 1869, he stated thisobservation in the form of the following statement which is known as the Mendeleev’s
Periodic Law .
The chemical and physical properties of elements are a periodic function of their
atomic masses.
A periodic function  is the one which repeats itself after a certain interval. Mendeleev
arranged the elements in the form of a table which is known as the Mendeleev’s
Periodic Table.
Mendeleev’s Periodic Table
Mendeleev arranged the elements in the increasing order of their atomic masses inhorizontal rows till he came across an element whose properties were similar to thoseof the first element. Then he placed this element below thefirst element and thus started the second row of elements.
The success of Mendeleev’s classification was due to the fact
that he laid more emphasis on the properties of elementsrather than on atomic masses. Occasionally, he could not findan element that would fit in a particular position. He left suchpositions vacant for the elements that were yet to bediscovered. He even predicted the properties of such elementsand of some of their compounds fairly accurately. In somecases, he even reversed the order of some elements, if it better
D. Mendeleev
(1834-1907)117Periodic Classification of Elements
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matched their properties. Proceeding in this manner, he could arrange all the known
elements in his periodic table.
When more elements were discovered, this periodic table was modified and updated
to include them. One more group (zero group) had to be added when noble gaseswere discovered.
Table 6.3: Mendeleev’s updated periodic table
Main Features of Mendeleev’s Periodic Table
The following are the main features of this periodic table :
1. The elements are arranged in rows  and columns  in the periodic table.
2. The horizontal rows are called periods. There are six periods in the periodic table.
These are numbered from 1 to 6 (Arabic numerals). Each one of  the 4th, 5thand 6th periods have two series of elements.
3. Properties of elements in a given period show regular gradation ( i.e. increase
or decrease) from left to right.
4. The vertical columns present in it are called groups . There are eight groups
numbered from I to VIII (Roman numerals).
5. Groups I to VII are further divided into A and B subgroups . However, group
VIII contains three elements in each of the three periods.
6. All the elements present in a particular group are chemically similar in nature.
They also show a regular gradation in their physical and chemical properties fromtop to bottom.
Merits of Mendeleev’s Periodic Classification
1. Classification of all elements
Mendeleev’s classification included all the 63 elements  known at that time on the
basis of their atomic mass and facilitated systematic study of elements.Periodic Classification of Elements
SCIENCE AND TECHNOLOGY  118Notes
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2. Correction of atomic masses
Atomic masses of some elements like Be ( beryllium ), Au ( gold), In ( indium ) were
corrected based on their positions in the table. (See box 1)
3. Prediction of new elements
Mendeleev arranged the elements in the periodic table in increasing order of atomic
mass but whenever he could not
find out an element with expectedproperties, he left a blank space.
He left this space blank for an
element yet to be discovered. Heeven predicted the properties of
such elements and also of
some of their compounds.For example, he predicted
the existence of unknown
element for the vacantspace below silicon and thusbelonging to the same group
IV B, of the periodic table.
He called it eka-silicon
(meaning, one position
below silicon ). Later, in
1886, C.A. Winkler  of
Germany discovered this
element and named it as
germanium . The predicted
and the actual properties of
this element were
remarkably similar (see Box2). Ekaboron (scandium)
and eka-aluminium
(gallium) are two moreexamples of unknown
elements predicted by Mendeleev.
4. Valency of elements
Mendeleev’s classificaiton helped in understanding the valency of elements. The
valency of elements is given by the group number. For example, all the elements in
group 1 i.e. lithium, hydrogen, sodium, potassium, rubidium, caesium have valency 1.Box 1
Indium had been assigned an atomic mass of 76
and valency of two. On the basis of its position in
the periodic table, Mendeleef predicted its atomic
mass to be 113.1 and its valency to be three . The
accepted atomic mass today is 114.82 and valencyis three.
Box 2
Predictions for eka-silicon by  Mendeleef
Property Predicted Actual
eka-silicon Germanium
Atomic Mass 72 72·6
Density/g cm–35·5 5·36
Melting point High 1231K
Action of acid Likely to be No action with
slightly HCl, reacts with
attacked hot nitric acid
Action of No reaction No action
alkali  with dil. NaOH
Oxide MO2 GeO2
Sulphide MS2 GeS2
Chloride MCl4 GeCl4
Boiling point 373 K 356 K
of chloride119Periodic Classification of Elements
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Defects of Mendeleev’s Periodic Table
Mendeleev’s periodic table was a great success, yet it had the following defects :
1. Position of Hydrogen
The position of hydrogen which is placed in group IA along with alkali metals
is ambiguous as it resembles alkali metals as well as halogens (group VII A).
2. Position of Isotopes
All the isotopes of an element have different atomic masses therefore, each oneof them should have been assigned a separate position. On the other hand, theyare all chemically similar; hence they should all be placed at the same position.In fact, Mendeleev’s periodic table did not provide any space for differentisotopes. For example, two isotopes of carbon are represented as 
6C12, 6C14
but placed at the same position.
3. Anomalous* Pairs of Elements
At some places, an element with greater atomic mass had been placed beforean element with lower atomic mass due to their properties. For example, cobaltwith higher atomic mass (58.9) was placed before nickel with lower atomic mass(58.7). Other such pairs are :
(i) Tellurium (127.6) is placed before iodine (126.9) and
(ii) Argon (39.9) is placed before potassium (39.1).
4. Grouping of chemically dissimilar elements
Elements such as copper and silver have no resemblance with alkali metals
(lithium, sodium etc.), but  have been grouped together in the first group.
5. Separation of chemically similar elements
Elements which are chemically similar such as gold and platinum have been placedin separate groups.
INTEXT QUESTIONS 6.1
1. Elements A, B and C constitute a Dobereiner’s triad. The atomic mass of A is
20 and that of C is 40. Predict the atomic mass of B.
2. Which property of atoms was used by Mendeleev to classify the elements?3. In Mendeleev’s periodic classification, whether chemically similar elements are
placed in a group or in a period?
*Anomaly means deviation from common rule, irregularity, abnormal, exceptionPeriodic Classification of Elements
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4. Mendeleev’s periodic table had some blank spaces. What did they signify?
5. Explain any three defects of Mendeleev’s periodic table.
6.2 MODERN PERIODIC LAW
Though Mendeleev’s periodic table included all the elements, yet at many places a
heavier element had to be placed before a lighter one. Such pairs of elements ( called
anomalous pairs ) violated the periodic law. Also, there was no place for different
isotopes of an element in the periodic table. Due to these reasons, it was felt that
the arrangement of elements in the periodic table should be based on some other
property which is more fundamental than the atomic mass.
In 1913, Henry Moseley, an English physicist discovered that the atomic number
and not the atomic mass is the most fundamental property of an element.
Atomic number  (Z) of an element is the number of protons in the nucleus of
its atom.
Since atom is as electrically neutral entity, the number of electrons is also equal to
its atomic number i.e.the  number of protons. After this development, it was felt
necessary to change the periodic law and modify the periodic table.
6.2.1 Modern Periodic Law
The Modern Periodic Law  states that the chemical and physical properties of
elements are  periodic functions of their atomic numbers i.e. if elements arearranged in the order of their increasing atomic number, the elements with
similar properties are repeated after certain regular intervals.
Fortunately, even with the revised periodic law, the Mendeleev’s classification did
not require any major revision as it was based on properties of the elements. In fact,
taking atomic number as the basis for classification,  removed major defects from
it such as a nomalous pairs  and position of isotopes.
After changes in the periodic law, many modifications were suggested in the periodic
table. Now, we shall learn about the modern periodic table in its final shape that isbeing used now..
Cause of  Periodicity
Let us now understand the cause of periodicity in the properties of elements. Considerthe electronic configuration of alkali metals i.e., the first group elements with atomic
numbers 3, 11, 19, 37, 55 and 87 ( i.e., lithium, sodium, potassium, rubidium, caesium
and francium) in the table given below:121Periodic Classification of Elements
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Table 6.4 : Electronic configuration of group 1 elements
Element Electronic configuration
3Li 2, 1
11Na 2, 8, 1
19K 2, 8, 8, 1
37Rb 2, 8, 18, 8, 1
55Cs 2, 8, 18, 18, 8, 1
87Fr 2, 8, 18, 32, 18, 8, 1
All these elements have one electron in the outer most shell and so they have similar
properties which are as follows :
(i) They are good reducing agents.
(ii) They form monovalent  cations.
(iii) T hey are soft metals.
(iv) They are very reactive and, therefore, found in nature in combined state.
(v) They impart colour to the flame.
(vi) They form hydrides with hydrogen.
(vii) They form basic oxides with oxygen.
(viii) They react with water to form metal hydroxides and liberate hydrogen.
It is noticed that all the elements having similar electronic configuration have similar
properties. Thus,  the re-occurrence of similar electronic configuration is the
cause of periodicity in properties of elements.
6.3 MODERN PERIODIC TABLE
The periodic table based on the modern periodic law is called the Modern Periodic
Table . Presently, the accepted modern periodic table is the Long Form of Periodic
Table .
It may be regarded as an extended form of Mendeleev’s table in which the sub-
groups A and B have been separated.
Now, you will learn the main features of the long form of periodic table which is
shown in Table 6.5 .Periodic Classification of Elements
SCIENCE AND TECHNOLOGY  122Notes
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Table 6.5 : Modern Periodic Table
6.3.1 Features of Long Form of Periodic Table
The long form of periodic table helps us to understand the reason why certain
elements resemble one another and why they differ from other elements in theirproperties. The arrangement of elements in this table is also in keeping with theirelectronic structures (configuration). In table 6.5, you must have noticed that it is
divided into columns and rows. The columns represent the groups  or family and the
rows represent the periods .
Lanthanoids
Actinoids123Periodic Classification of Elements
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* However, it should be noted here that more and more electrons are added to valence shell
only in case of normal elements. In transition elements, the electrons are added to incompleteinner shells.1. Groups:  There are 18 vertical columns in the periodic table. Each vertical column
is called a group . The groups have been numbered from 1 to 18 (in Arabic
numerals).
All elements present in a group have similar electronic configurations and have
same number of valence electrons. You can see in case of group 1 (alkali metals)and group 17 elements (halogens) that as one moves down a group, more andmore shells are added as shown in Table 6.6.
Table 6.6
Group 1 Group 17
Element Electronic configuration Element Electronic configuration
L i 2,1 F 2,7
Na 2,8,1 Cl 2,8,7K 2,8,8,1 Br 2,8,8,7
Rb 2,8,18,8,1 I 2,8,18,18,7
All elements of group 1 have only one valence electron. Li has electrons in two
shells, Na in three, K in four and Rb has electrons in five shells. Similarly all theelements of group 17 have seven valence electrons however the number of shellsis increasing from two in fluorine to five in iodine.
2. Periods:  There are seven horizontal rows in the periodic table. Each row is called
a period.  The elements in a period have consecutive atomic numbers. The
periods have been numbered from 1 to 7 (in Arabic numerals).
In each period a new shell starts filling up. The period number is also the number
of the shell which starts filling up as we move from left to right across thatparticular period. For example, in elements of 3
rd period (N = 3), the third shell
(M shell) starts filling up as we move from left to right*. The first element of this
period, sodium (Na 2,8,1) has only one electron in its valence shell (third shell)while the last element of this period, argon (Ar 2,8,8) has eight electrons in itsvalence shell. The gradual filling of the third shell can be seen below.
Element Na Mg Al Si P S Cl Ar
Period →
Electronic 2,8,1 2,8,2 2,8,3 2,8,4 2,8,5 2,8,6 2,8,7 2,8,8
configurationPeriodic Classification of Elements
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(a) The first period is the shortest period  of all. It contains only two elements; H
and He.
(b) The second and third periods are called short periods  containing 8 elements
each.
(c) The fourth and fifth periods are long periods  containing 18 elements each.
(d) The sixth and seventh periods are very long  periods  containing 32 elements
each.
6.3.2 Types of Elements
1.Main Group Elements:  The elements present in groups 1 and 2 on left side
and groups 13 to 17 on the right side of the periodic table are calledrepresentative or main group  elements . Their outermost shells are
incomplete,which means their outermost shell has less than eight electrons.
2.Noble Gases:  Group 18 on the extreme right side of the periodic table contains
noble gases . Their outermost shells contain 8 electrons  except He which contains
only 2 electrons.
Their main characteristics are :
(a) They have 8 electrons in their outermost shell (except He which has 2
electrons).
(b) Their combining capacity or valency is zero.
(c) They do not react and so are almost inert.
(d) All the members are gases.
3.Transition Elements:  The middle block of periodic table (groups 3 to 12)
contains transition elements.  Their two outermost shells are incomplete.
Since these elements represent a transition (change) from the most electropositive
element to the most electronegative element, they are named as transition
elements .
Their important characteristics are as follows:
(a) All these elements are metals and have high melting and boiling points.
(b) They are good conductors of heat and electricity.
(c) Some of these elements get attracted towards magnet.
(d) Most of these elements are used as catalyst.
(e) They exhibit variable valencies.
4.Inner Transition Elements:  These elements, also called rare-earth elements,
are shown separately below the main periodic table. These are two series of125Periodic Classification of Elements
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14 elements each . The first series called lanthanoids consists of elements 58
to 71 (Ce to Lu). They all are placed along with the element 57, lanthanum
(La) in the same position (group 3, period 6) because of very close resemblancebetween them. It is only for the sake of  convenience that they are shown
separately below the main periodic table.
The second series of 14 rare-earth elements is called actinoids . It consists of
elements 90 to 103 (Th to Lr) and they are all placed along with the element
89, actinium (Ac) in the same position (group 3, period 7) but for convenience
they are shown below the main periodic table.
In all rare-earths (lanthanoids and actinoids), three outermost  shells are
incomplete. They are therefore called inner transition  elements .
It is interesting to note that the element lanthanum is not  a lanthanoid and the
element actinium is not  an actinoid.
5.Metals:  Metals  are present in the left hand portion of the periodic table. The
strong metallic elements; alkali metals  (Li, Na, K, Rb, Cs, Fr) and alkaline
earth metals  (Be, Mg, Ca, Sr, Ba, Ra) occupy groups 1 and 2 respectively.
6.Non-metals:  Non-metals  occupy the right hand portion of the periodic table.
Strong non-metallic elements i.e., halogens  (F, Cl, Br, I, At) and chalkogens
(O, S, Se, Te, Po) occupy groups 17 and 16 respectively.
7.Metalloids:  Metalloids  are the elements that show mixed properties of both
metals and non-metals. They are present along the diagonal line starting from
group 13 (Boron) and going down to group 16 (Polonium).
ACTIVITY 6.1
Rearrange the alphabets to get the correct name of the element in the space providedand mention its position in the modern periodic table
(a) RGANO ..................... is a noble gas which is placed in group .....................
and third period of the modern periodic table.
(b) HULIMIT ..................... is an alkali metal which is placed in group 1 and
..................... period of the modern periodic table.
(c) MILCUAC ..................... is an alkaline earth metal which is placed in group
..................... and fourth period of the modern periodic table.
(d) POHSROSUHP ..................... is a metalloid which is placed in group 15 and
..................... period of the modern periodic table.Periodic Classification of Elements
SCIENCE AND TECHNOLOGY  126Notes
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6.3.3 Merits of the Modern Periodic Table
The following points overcame the defects of Mendeleev’s periodic table, that is why,
it was accepted by scientists across the world
1.Position of isotopes:  All isotopes of an element have the same atomic number
and therefore, occupy the same position in the modern periodic table.
2.Anomalous pairs:  The anomaly regarding all these pairs disappears when
atomic number  is taken as the basis for classification. For example, cobalt
(at. no. 27) would naturally come before nickel (at. no. 28) even though its atomicmass is little more than that of nickel.
3.Electronic configuration:  This classification is according to the electronic
configuration of elements, i.e., the elements having a certain pattern of electronic
configuration are placed in the same group of the periodic table. It relates theproperties of elements to their electronic configurations. This point will be furtherelaborated in the next section.
4.Separation of metals and non-metals:  The position of metals, non-metals and
metalloids are clearly established in the modern periodic table.
5.Position of transition metals:  It makes the position of the transition elements
quite clear.
6.Properties of elements:  It reflects the differences, the trends and the variations
in the properties of the elements in the periodic table.
7.This table is simple, systematic and easy way of remembering the properties ofdfifferent metals.
INTEXT QUESTIONS 6.2
1. Give any two defects of Mendeleev’s periodic table which has been removed
in modern periodic table. How were they removed?
2. Metalloids are present along the diagonal line starting from group 13 and going
down to group 16. Do they justify their position in the modern periodic table?
6.4 PERIODIC TRENDS IN PROPERTIES
You have learnt about the main features of the long form of the periodic table in theprevious section.and you know that it consists of groups and periods. Let us recalltheir two important features:
1. In a given group, the number of filled shells increases. The number of valence
electrons is the same in all the elements of a given group. However, these valence127Periodic Classification of Elements
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electrons but they are present in higher shells which are farther away from the
nucleus. In view of this, decreases the force of attraction between the outermost
shell and the nucleus as we move downwards in a group.
2. In a given period, the nuclear charge and the number of valence electrons in a
particular shell increase from left to right. This increases the force of attractionbetween the valence electron and nucleus as we move across a period from left
to right.
The above given changes affect various properties which show gradual variations in
groups and periods, and they repeat themselves after certain intervals of atomic
number. They are called periodic properties . Now you are going to learn the
variations of two of such properties in the periodic table.
A. Atomic Size
Atomic size is the distance between the centre of nucleus and the outermostshell of an isolated atom . It is also known as atomic radius . It is measured in
picometre, pm (1 pm = 10
–12 m). Atomic size is a very important property of atoms
because it is related to many other properties.
Variation of atomic size in periodic table.
The size of atoms decreases from left to right  in a period  but increases from top
to bottom in a group . For example, the atomic radii of the elements of the second
period and of group 1 are given below in the tables 6.7 and 6.8 respectively.
Table 6.7 : Atomic radii of period 2 elements
Atomic Number 3 456789
Elements : (in second Li Be B C N O F
period)
Atom radius/pm : 134 90 82 77 75 73 72
Atomic Size
In a period the atomic number and therefore the positive charge on the nucleus
increases gradually. As a result, the electrons are attracted more strongly and they
come closer to the nucleus. This decreases the atomic size in a period from left to
right.
In a group as one goes down, a new shell is added to the atom which is farther
away from the nucleus. Hence electrons move away from the nucleus. This increasesthe atomic size in a group from top to bottom.Periodic Classification of Elements
SCIENCE AND TECHNOLOGY  128Notes
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Table 6.8 : Atomic radii of group 1 elements
Atomic Number Elements Atom radius/pm Atomic Size
(in groups I)
3 L i 134
11 Na 154
19 K 196
37 Rb 21155 Cs 225
B. Metallic and Non-metallic Character
The tendency of an element to lose electrons to form cations is called electropositive
or metallic character  of an element. Alkali metals are most electropositive. The
tendency of an element to accept electrons to form anions is called electronegative
or non-metallic character  of an element.
(a) Variation of Metallic Character in a Group
Metallic character increases from top to bottom in a group as tendency  to lose
electrons increases. This increases the electropositive character and metallic nature.The variation can best be seen in group 14 as shown below.
Table 6.9: Metallic character of groups 14 elements
Element Nature
C Non-metal
Si MetalloidGe Metalloid
Sn Metal
Pb Metal
(b) Variation of Metallic Character in a Period
Metallic character decreases in a period from left to right . It is because the
ionization energy increases in a period. This decreases the electropositive characterand metallic nature. The variation of metallic character in the elements of 3rd periodis shown below.129Periodic Classification of Elements
SCIENCE AND TECHNOLOGYNotes
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Table 6.10 : Metallic character of 3rd period elements
Element Na Mg A l Si P S Cl
Nature Metal Metal Metal Metalloid Non- Non- Non-
Metal Metal Metal
In this section, you have learnt about variation of some properties in periodic table.
Some important trends in periodic table may be summarized in a general way as givenbelow :
Table 6.11 : Variation of various periodic properties in
periods and groups
Property In a Period In a Group
(From left to right) (From top to Bottom)
Atomic number increases increases
Atomic size decreases increases
Metallic character decreases increases
Non-metallic character increases decreases
INTEXT QUESTIONS 6.3
1. Fill in the blanks with appropriate words
(a) The force of attraction between nucleus and valence electrons .................
in a period from left to right.
(b) Atomic radii of elements .................  in a period from left to right.
(c) Atomic radii of elements .................  in a group from top to bottom.
(d) Metallic character of elements .................  from top to bottom in a group.
2. In the following crossword puzzle, elements are present horizontally, vertically
downwards and diagonally downwards. Let us find out how many elements youare able to get within 5 minutes.Periodic Classification of Elements
SCIENCE AND TECHNOLOGY  130Notes
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ZNHYDROGEN
MB I CARBONO
ADETBAR I UMGXYHRMUSAS
NADEOOAOO I
EIUJP X GISLSODI U M YEL I
ID M U X AIG N C
UIO M O G E Y E O
MN D P S BO RON
AECHLOR INE
Please check in the intext answers to find if you missed out any.
3. Let us find how many riddles you can solve.
(i) I am the only noble gas whose outermost shell has 2 electrons. Who am
I?
(ii) I am placed in group 16 of the modern periodic table and essential for
your respiration. Who am I?
(iii) I combine with chlorine to form your table salt. Who am I?
(Hint:  Answers are present in the grid]
WHAT YOU HAVE LEARNT
/circle6The first classification of elements was as metals and non-metals.
/circle6After the discovery of atomic mass (old term, atomic weight) it was thought to
be the fundamental property of elements and attempts were made to correlateit to their other properties.
/circle6John Dobereiner grouped elements into triads. The atomic mass and propertiesof the middle element were mean of  the other two. He could group only a fewelements into triads. For example (i) Li, Na and K (ii) Ca, Sr and Ba (iii) Cl,Br and I.131Periodic Classification of Elements
SCIENCE AND TECHNOLOGYNotes
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/circle6Newlands tried to see the periodicity of properties and stated his law of octaves
as  “ When elements are arranged in the increasing order of their atomic
weights every eighth element has properties similar to the first ”. He could
arrange elements up to calcium only out of more than sixty elements then known.
/circle6Mendeleev observed the correlation between atomic weight and other properties
and stated his periodic law as, “ The chemical and physical properties of
elements are a periodic function of their atomic weights ”.
/circle6Mendeleev gave the first periodic table which is named after him which included
all the known elements. It consists of seven horizontal rows called periods and
numbered them from 1 to 7. It has eight vertical columns called groups  and
numbered them from I to VIII.
/circle6Main achievements of Mendeleev’s periodic table were (i) inclusion of all the
known elements and (ii) prediction of new elements.
/circle6Main defects of Mendeleev’s periodic table were (i) position of isotopes, (ii)
anomalous pairs of elements like Ar and K and (iii) grouping of dissimilar elementsand separation of similar elements.
/circle6Moseley discovered that atomic number and not atomic mass is the fundamental
property of elements. In the light of this the periodic law was modified to “The
chemical and physical properties of elements are periodic functions of theiratomic numbers ”. This is the Modern Periodic Law.
/circle6Modern Periodic Table is based upon atomic number. Its long form has been
accepted by IUPAC. It has seven periods (1 to 7) and 18 groups (1 to 18).
It is free of main defects of Mendeleev’s periodic table. Elements belonging to
same group have same number of valence electrons and thus show same valencyand similar chemical properties.
/circle6Arrangement of elements in the periodic table shows periodicity. Atomic radiiand metallic character increase in a group from top to bottom and in a period
decrease from left to right.
TERMINAL EXERCISE
A.Objective questions
I. Mark the correct choice:
1.Which one of the following was the earliest attempt of classification of
elements?
(a) Classification of elements into metals and non-metals
(b) Newlands’ Law of OctavesPeriodic Classification of Elements
SCIENCE AND TECHNOLOGY  132Notes
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(c) Dobereiner’s Triads
(d) Mendeleef’s Periodic Table
2. The ‘law of octaves’ was given by
(a) Mendeleev (b) Newlands
(c) Lother Meyer (d) Dob ereiner
3. According to the periodic law given by Mendeleev, the properties of an
element are a periodic function of its
(i) atomic volume (ii)atomic size
(iii) atomic number (iv) atomic mass
4. The particle which is universally present in the nuclei of all elements is
(a) neutron (b) proton
(c) electron (d) α-particle
5. Potassium is more metallic than sodium because
(a) both have 1 electron in their outermost shell.
(b) both are highly electropositive.
(c) sodium is larger in size than potassium.
(d) potassium is larger in size than sodium.
7. Which one of the following elements in its chloride does not show the valence
equal to its valence electrons?
(a) NaCl (b) MgCl2
(c) AlCl3 (d) PCl3
8. Which one of the following elements has the least tendency to form cation?
(a) Na (b) Ca(c) B (d) Al
9. Which one of the following does not belong to the family of the alkali metals?
(a) Li (b) Na
(c) Be (d) K
10. The number of elements in the 5
th period of the periodic table is
(a) 2 (b) 8(c) 32 (d) 18
11. The elements with atomic number 9 resembles with the element having atomic
number
(a) 35 (b) 27
(c) 17 (d) 8133Periodic Classification of Elements
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12. In which period of  the periodic table, an element with atomic number 20 is
placed?
(a) 4 (b) 3
(c) 2 (d) 1
II. Mark the following statements True  (T) or False  (F) :
1. The properties of the middle element in a Dobereiner’s triads are intermediate
between those of the other two.
2. The vertical columns in the periodic table are called periods.
3. Mendeleev depended only on the atomic mass of elements for his classification.
4. All elements present in a group are chemically similar.
5. The modern periodic law is based upon atomic mass.
6. The importance of atomic number as the fundamental property was realised by
Henry Mosely.
7. There are 18 groups in the modern periodic table.
8. Non-metals are present in the middle portion of the periodic table.
9. Each period in modern periodic classification begins with filling of electrons in
a new shell.
III. Fill in the blanks:
1. According to the modern periodic law, the properties of elements are periodic
function of their .....................
2. The .....................  number is same as the number of shell which in gradually
filled up in the elements of this period.
3. In normal elements of a particular period the electrons are gradually filled in
.....................  shell.
4. All elements of a particular group have .....................  electronic configurations.
5. In the modern periodic table, groups are numbered from .....................  to
.....................
6. The second and third periods of the periodic table are called .....................
periods.
7. The main group elements are present in group 1 and 2 on the left side and
.....................  to .....................  on the right side of the periodic table.Periodic Classification of Elements
SCIENCE AND TECHNOLOGY  134Notes
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8. All the group eighteen elements (except the first one) contain .....................
valence electrons.
9. All transition elements are metals with .....................  melting and boiling points.10. The group of 14 rare-earth elements belonging to the group 3 and 7
th period
are called .....................
11. All elements present in a given .....................  have the same valency.
12. Atomic size .....................  in a period from left to right.13. Magnesium is .....................  metallic than calcium.
14. Carbon belongs to group .....................  of the Periodic table.
15. All the elements of group 15 have .....................  valence electrons.
B.Subjective Questions
I. Very short Answer Questions (Answer in one word or one sentence).
1. What was the earliest classification of elements?
2. State Newlands’ law of octaves.
3. Which classification of elements failed after the discovery of noble gases?4. State Mendeleev’s Periodic Law.
5. How were the groups numbered in the Mendeleev’s periodic table?
6. Name the fundamental properties of element on which the modern periodic law
is based.
7. How many groups are there in the modern periodic table?
8. How have groups been numbered in the modern periodic table?9. What are normal elements?
10. What are the elements present in the middle portion of the modern periodic table
called?
11. What is atomic size?
12. How does atomic size vary in a period and in a group?
13. Where would the element with largest atomic size be placed in any group?14. Give the number of a group in which metallic, metalloid and non-metallic, all
three types of elements, are present.
II. Short Answer Questions (Answer in 30-40 words).
1. State Dobereiner’s law of triads.
2. Show that chlorine, bromine and iodine (atomic masses 35·5, 80 and 127
respectively) constitute a triad.135Periodic Classification of Elements
SCIENCE AND TECHNOLOGYNotes
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3. What were the reasons for the failure of Newlands’ law of octaves ?
4. Describe Mendeleev’s periodic table briefly in terms of rows and columns and
their raw being.
5. Give any two achievements of the Mendeleev’s Periodic classification.6. What were the defects in Mendeleev’s periodic classification.7. State modern periodic law.8. Briefly describe the modern periodic table in term of groups and period.9. Give names of four classes into which the elements have been classified and
mention to which groups of the modern period table they belong.
10. List the merits of the long form of the modern periodic table and explain any
two of them.
11. How are the electronic configurations of all the elements belonging to a particular
group related? Explain with the help of group 17 elements.
12. How does the electronic configuration of elements belonging to a particular
period vary? Explain with the example of second period elements.
13. Define atomic radius.14. How and why does metallic character vary in a group from top to bottom?
III. Long Answer Questions (Answer in 60–70 words).
1. State Mendeleev’s Periodic Law and describe the periodic table constructed
on this basis.
2. What are the merits and demerits of the Mendeleev’s Periodic classification?3. Describe the modern periodic table in terms of groups and periods.4. What are the following types of elements and where are they located in the
periodic table?
(a) Main group elements (b) Noble gases
(c) Transition elements (d) Inner transition elements.
5. Discuss the merits of the modern periodic table.6. What is the relationship between the electronic configuration and the modern
periodic table?
8. Explain the variation of atomic size in a group and in a period.9. How is metallic character related to ionization energy ? Explain the variation
of metallic character in the periodic table.Periodic Classification of Elements
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ANSWERS TO INTEXT QUESTIONS
6.1
1. Atomic mass of B = 20 + 40302=
2. Atomic mass
3. Group4. These were the positions of elements which were yet to be discovered.5. Any three of the following: (i) position of hydrogen (ii) position of isotopes (iii)
anomalous pairs of elements (iv) grouping of chemically dissimilar element (v)
separation of chemically similar element (vi) no explanation for electronicconfiguration
6.2
1. Anomalous pairs when elements are arranged in the order of their increasing
atomic numbers, these anomalies are automatically removed, since the atomic
number of the first element is less than that of the second although their atomic
masses show revrse trends.
2. Position of isotopes. Since all the isotopes of an element have the same atomic
number, they all will occupy the same position in the periodic table.
6.3
1. (a) increases (b) decreases
(c) increases (d) increases
2. Hydrogen, Carbon, Barium, Sodium, Boron, Chlorine (horizontally)
Magnesium, Iodine, Helium, Neon, Silicon, (vertically downwards)Nitrogen, Oxygen(diagonally downwards)
3. (i) Helium (ii) Oxygen (iii) Sodium
Activity 6.1
(a) Argon (b) Lithium (c) Calcium (d) Phosphorous