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Classification of Elements and Periodicity in Properties Important Questions & Answers

Get Classification of Elements and Periodicity in Properties chapter’s important questions & answers. View the Important Question bank for Class 11 Chemistry. These important questions will play significant role in clearing concepts of Chemistry. This question bank is designed keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 11 chemistry chapters.  You can access free study material for all three subject’s Physics, Chemistry and Mathematics. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. We have transformed classroom in such a way that a student can study anytime anywhere. With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. Visit eSaral Website to download or view free study material for JEE & NEET. Also get to know about the strategies to Crack Exam in limited time period.

Q.  State modern periodic law.

Q.  Why do elements in the same group have similar physical and chemical properties ?

Q.  In terms of period and group where would you locate the element with _Z_ = 14 ?

Q.  In the modern periodic table, the period indicates the value of (i) Atomic number (ii) Mass number (iii) Principal quantum number (iv) Azimuthal quantum number.

Q.  What would be the IUPAC name and symbol of the element with atomic number 120 ?

$\therefore$ Name of element: Unbinilium Symbol : Ubn.

Q.  Which element do you think would have been named by (i) Lawrence Berkalay Laboratory (ii) Seaborg’s group?

(ii) Seaborgium (_Z_ =106)

Q.  How would you react to the statement that electronegativity of on Pauling scale is 3.0 in all its compounds ?

Q.  Arrange the following ions in the order of increasing size: $B e^{2+}, C l^{-}, S^{2}, N a^{+}, M g^{2+}, B r^{-}$

Q.  Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different ? Justify your answer.

Q.  What do you understand by isoelectronic species ? Give the formula of a species that will be isoelectronic with the following atoms or ions :

(i) $F^{-}\left(10 e^{-}\right): O^{2-}$ (ii) $A r\left(18 e^{-}\right): C l^{-}$ (iii) $M g^{2+}\left(10 e^{-}\right): N a^{+}$ (iv) $R b^{+}\left(36 e^{-}\right): K r$

Q.  Are the oxidation state and covalency of $A l$ in $\left[A I C I\left(H_{2} O\right)_{5}\right]^{2+}$ same ?

Q.  Show by a chemical reaction with water that $N a_{2} O$ is basic oxide and $C l_{2} O_{7}$ is an acidic oxide.

Q.  What is the basic theme of organization of elements in the periodic table.

Q.  Write the atomic number of the element in the third period and seventeenth group of the periodic table.

Q.  In terms of period and group where would you locate the element with  = 114 ? So, it belongs to $7^{\text {th }}$ period and $14^{\text {th }}$ group.

So, it belongs to $7^{\text {th }}$ period and $14^{\text {th }}$ group.

Q.  How would you justify that there are only 18 elements in the fifth period of the periodic table.

Q.  On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.

Q.  Consider the following species : $N^{3-}, O^{2-}, F^{-}, N a^{+}, M g^{2+}$ and $A l^{3+}$ (i) What is common in them ? (ii) Arrange them in the order of increasing ionic radii.

(ii) Since the number of electrons are same, the ionic size decrease with increase in nuclear charge. Therefore, the ions can be arranged in increasing order of ionic radii as $A l^{3+}<M g^{2+}<N a^{+}<F^{-}<O^{2-}<N^{3-}$

Q.  Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell (i) Valence principal quantum number (ii) Nuclear charge (_Z_) (iii) Nuclear mass (iv) Number of core electrons.

Q.  Which of the following statements related to the modern periodic table is incorrect ? (i) The _p_ -block has six columns, because a maximum of 6 electrons can occupy all the orbitals in a _p_ -subshell. (ii) The _d_ -block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a _d_ -subshell.

Q.  What is the basic difference in approach between Mendeleev’s Periodic Law and the Modern Periodic Law?

Q.  How do atomic radii vary in a period and in a group ? How do you explain the variation ?

In contract, the atomic size decreases as we move from left to right in a period. This is because that within a period the outer electrons remains in the same shell but the nuclear charge increases by one unit at each succeeding element. Due to this increased nuclear charge, the attraction of the nucleus for the outer electrons increases and hence the atomic size decreases.

Q.  Considering the atomic number and position in the periodic table, arrange the following elements in the increasing order of metallic character. : Si, Be, Mg, Na, P.

We know that the metallic character increases down a group and decreases along a period as we move from left to right. Therefore, _Na_ is the most metallic element, followed by _Mg_ and _Si_ , while _P_ is the least metallic element. Among and is more metallic than _Be_. Therefore, the overall increasing order of metallic character is $P<S i<B e<M g<N a$

Q.  Write the general electronic configuration of _s_ -, _p_ -, _d_ – and _f_ -block elements :

(ii) $\quad p$ -Block elements: $n s^{2} n p^{1-6}$ where $n=2-6$ (iii) $d$ -Block elements: $(n-1) d^{1-10} n s^{0-2}$ where $n=4-7$ (iv) $f$ -Block elements: $(n-2) f^{0-14}(n-1) d^{0-1} n s^{2}$ where $n=6-7$

Q.  Assign the position of the element having outer electronic configuration, (i) $n s^{2} n p^{4}$ for $n=3 \quad$ (ii) $(n-1) d^{2} n s^{2}$ for $n=4$ and (iii) $(n-2) f^{7}(n-1) d^{1} n s^{2}$ for $n=6$ in the periodic table.

The complete electronic configuration of the element is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4}$ and the element is $S($ sulphur). (ii) $\quad n=4$ suggest that the element lies in the $4^{\text {th }}$ Period. since the $d$ -orbitals are incomplete, therefore, it is $d$ -block element. The group number of the element $=$ no. of $d$ -electrons $+$ no. of $s$ electrons $=2+2=4 .$ Thus the elements lies in group 4 and $4^{\text {th }}$ period. The complete electronic configuration of the element is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2}$ and the element in Ti (Titanium). (iii) $\quad n=6$ means that the element lies in the sixth period. since the last electron goes to the $f$ -orbital, therefore, the element is a $f$ -block element. All $f$ -block elements lie in group 3. The complete electronic configuration of the element is $[\mathrm{Xe}\\} 4 f^{7} 5 d^{1} 6 s^{2} .$ The atomic number of the element $54+7+1+2=64$ and the element $G d($ gadolinium).

Q.  The elements _Z_ = 117 and 120 have not yet been discovered. In which family group would you place these elements and also give electronic configuration in each case

Q.  What are major differences between metals and non-metals?

Non-metals, on the other hand, are strong oxidising agents, they have high ionization enthalpies, have high negative electron gain enthalpies, high electronegativity, form acidic oxides and covalent compounds.

Q.  The elements _Z_ =107 and _Z_ =109 have been made recently; element _Z_ =108 has not yet been made. Indicate the group in which you will place the above elements.

$Z=107 \quad[R n] 5 f^{14} 6 d^{5} 7 s^{2}$ $Z=108 \quad[R n] 5 f^{14} 6 d^{6} 7 s^{2}$ $Z=109 \quad[R b] 5 f^{14} 6 d^{7} 7 s^{2}$ These elements will be placed in $d$ -block in group $7^{\text {th }}, 8^{\text {th }}$ and $9^{\text {th }}$ respectively.

Q.  Which of the elements $N a, M g, S i$ and p would have the greater difference between the first and the second ionization enthalpies ? Briefly explain your answer.

Q.  Energy of an electron in the ground state of the hydrogen atom is $-2.18 \times 10^{-18} \mathrm{J}$. Calculate the ionization enthalpy of atomic hydrogen in terms of $J$ mol $^{-1}$

Energy required for removal of the electron $=0-\left(-2.18 \times 10^{-18} \mathrm{J}\right)=2.18 \times 10^{-18} \mathrm{J}$ This is the energy required to remove the electron from one atom of hydrogen in ground state. The energy required to remove electron from one mole $\left(6.022 \times 10^{23} \text { atoms }\right)$ will be $=2.18 \times 10^{-18} \times 6.022 \times 10^{23}=13.12 \times 10^{5} \mathrm{J} \mathrm{mol}^{-1}$

Q.  Arrange the following elements in the increasing order of metallic character : Si, Be, Mg, Na, P.

Q.  (i) Why do Group-I metals have lower ionisation enthalpy than corresponding Group-II metals?

(ii) Anions are largest than neutral atom because electrons are more than the protons, therefore, effective nuclear charge is less, therefore, distance between centre of nucleus and valence electrons is more.

Q.  The first ionization enthalpy $\left(\Delta_{i} H\right)$ values of the third period elements, $N a, M g,$ and $S i$ are respectively $496,737$ and $786 \mathrm{kJ} \mathrm{mol}^{-1} .$ Predict whether the first $\Delta_{i} H$ value for $A l$ will be more close to 575 or $760 \mathrm{kJ} \mathrm{mol}^{-1}$ respectively. Justify your answer.

Al should be lower than that of $M g .$ This is due to effective shielding of $3 p$ electrons from the nucleus by $3 s$ -electrons.

Q.  Explain why cations are smaller and anions are larger in radii than their parent atoms.

Q.  What is the basic difference between the terms electron gain enthalpy and electronegativity?

Q.  (i) Predict the position of the element with atomic number 26 in the periodic table.

electronic configuration is $[\mathrm{Ar}] 4 \mathrm{s}^{2} 3 d^{6}$ (ii) It is because nitrogen has stable electronic configuration i.e., half-filled $p$ -orbitals $\therefore$ its I.E. is more than that of oxygen.

Q.  The first ionization enthalpy $\left(\Delta_{i} H\right)$ values of the third period elements, $\mathrm{Na}, \mathrm{Mg}$ and $\mathrm{Si}$ are respectively $496,737$ and 786 kJ mol $^{-1}$ predict whether the $\Delta i H$ value for $A l$ will be more close to 575 or $760 \mathrm{kJ} \mathrm{mol}^{-1}$. Justify your answer.

since in case of $A l,$ a $3 p$ -electron is to be lost while in $M g,$ a $2 s$ electron is to be lost, therefore, value for $A l$ will be lower than that of $M g\left(737 k J \mathrm{mol}^{-1}\right)$ because of the effective shielding of the $3 p-$ electron from the nucleus by 3 s-electrons. Therefore, $\Delta_{i} H$ for $A l$ will be more close to $575 \mathrm{kJ} \mathrm{mol}^{-1}$

Q.  The size of isoelectronic species $-F^{-}, N e$ and $N a^{+}$ is affected by (i) Nuclear charge $(Z)$ (ii) Valence principal quantum number $(n)$ (iii) Electron-electron interaction in the outer orbital

Q.  The increasing order of reactivity among group 1 elements is $L i<N a<K<R b<C s$ whereas that of group 17 is $F>C D>B r>$

$L i<N a<K<R b<C s .$ In contrast, the elements of group 17. have seven electrons in their respective valence shells and thus have a strong tendency to accept one more electron. The tendency to accept electrons, in turn depends upon their electrode potentials. since the electrode potential of group 17 elements decreases in the order $F(+2.87 V)>C l(+1.36 V), B r(1.08 V)$ and $I(+0.53),$ therefore, their reactivities also decrease in the same order: $F>C l>B r>I$

Q.  Which of the following species has the largest and the smallest size. $M g, M g^{2+}, A l, A l^{3+}$

(ii) Further, cations are smaller than their parent atoms. Therefore, $M g^{2+}$ is smaller than $M g$ and $A l^{3+}$ is smaller than $A l$ (iii) and are isoelectronic ions. Among isoelectronic ions, higher the tye charge, smaller the size. Therefore, ionic radius of $A P^{3+}$ is smaller than that of From the above discussion, it follows that $M g$ has the largest while has the smallest size.

Q.  What is the significance of the terms ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy ?

(ii) Electron gain enthalpy is the energy released when an isolated gaseous atom in the ground state accepts an extra electron to form the gaseous negative ion. The term ground state here means that the atom must be present in the most stable state, _i.e._ , the ground state. The reason being that when the isolated gaseous atoms is in the excited state, lesser amount of energy will be released when it gets converted into gaseous anions after the accepting an electron. Therefore, for comparison purposes, the electron gains enthalpy of gaseous atoms be determined in their respective most stable state, 

Q.  (i) Write the electronic configurations of the elements given below: $A(\text { At. } \mathrm{No} .=9), B(\text { At. } \mathrm{No} .=12), C \quad(\text { At. } \mathrm{No} .=29), D(\mathrm{At}$ $\mathrm{No} .=54), E(\mathrm{At.} \mathrm{No} .=58)$ (ii) Also predict the period, group number and block to which they belong. (iii) Classify them as representative elements, noble gases, transition and inner transition elements.

(ii) Element $A$ receives the last electron in $2 p-$ orbital, therefore, it belongs to $p$ -block elements and its group number $=10+\mathrm{No.}$ of electrons in the valence shell $=10+7=17 .$ Further the period of the element $=$ No. of the principal quantum number of the valence shell $=2^{\mathrm{nd}}$ Element $B$ receives the last electron in $3 s$ -orbital, therefore, it belongs to the $s$ -block elements and its group number $=$ No. of electrons in the valence shell $=2 .$ Further the period of the element $=$ No. of the principal quantum number of the valence shell $=3^{\text {rd }}$ Element $C$ receives the last electron in the $3 d$ -orbital, therefore, it belongs to $d$ block elements and its group number $=$ No. of electrons in the penultimate shell and valence shell $=10+1$ $=11 .$ Further, the period of the element $=$ No. of principal quantum number of the valence shell $=4^{\text {th }}$ Element D receives its last electron in the 5 p-orbital, therefore, it belongs to $p$ -block elements and its group number $=10+\mathrm{No.}$ of electrons in the valence shell $=10+8=18 .$ Further, the period of the element $=$ No. of the principal quantum number of valence shell $=5^{\text {th }}$ Element $E$ receives its last electron in the $4 f$ -orbital, therefore, it belongs to $f$ -block elements. It may be noted here that the filling of 4 f-orbital occurs only when one electron has already entered $5 d$ -orbital. Therefore, element $E-$ belongs to $f-$ block elements and not to $d$ -block elements. since it belongs to lanthanide series therefore, as such it does not have any group number of its own but is usually considered to lie in group $3 .$ However, its period $=$ No. of the principal quantum number of the valence shell $=6^{\text {th }}$ (iii) Elements $A$ and $B$ are representative elements since their last electron enters $p-$ and $s$ -orbital respectively. Elements $C$ is a transition element since it receives its last electron in the $d$ – orbital. Element $D$ is a $p$ -block element with completely filled $s$ – and $p$ -orbitals of the valence shell. Such a type of $p-$ block element is called a noble gas. Element $E$ is an inner transition element since it receives its last electron in the f-orbital.

Q.  Predict the formula of the stable binary compounds that would be formed by the following pairs of elements : (i) Lithium and Oxygen (ii) Magnesium and Nitrogen (iii) Aluminium and Iodine (iv) Silicon and Oxygen (v) Phosphorus and Fluorine (vi) Element 71 and Fluorine.

(ii) Magnesium belongs to group 2 with a valence of 2 while nitrogen belongs to group 5 with a valence of 3. Hence, the formula of compound is $M g_{3} N_{2}$ (iii) Aluminium belongs to group 13 with a valence of 3 and iodine belongs to group 17 with a valence of $1 .$ Hence the formula of the compound is $A l l_{3}$ (iv) Silicon belongs to group 14 with valence of 4 and oxygen belongs to group 16 with a valence of $2 .$ Hence, the formula of the compound is $S i O_{2}$ (v) Phosphorus belongs to group 15 and it has valence 3 or $5,$ while fluorine belongs to group 17 with a valence of 1 Hence the formula of the compound may be $P F_{3}$ or $P F_{5} .$ (vi) Element with atomic number 71 is a lanthanoid called Lutetium $(L u),$ its common value is $3 .$ Fluorine is a group 17 (halogen) element with a valence of $1 .$ Therefore, the formula of the compound formed would be $L u F_{3}$ (Lutetium fluoride).

Q.  Describe the theory associated with the radius of an atom as it (i) gains electron (ii) loses electron.

(ii) **Loss of electrons.** When a neutral atom loses one electron to form a cation, its atomic radius decreases. The reason being that the number of electrons in the cation decreases while its nuclear charge remains the same as the parent atom. Since the same nuclear charge now attracts lesser number of electrons, therefore, the force of attraction of the nucleus on the electrons of all the shells-increases ( effective nuclear charges increases) and hence the size of cation decreases. Thus,

Q.  The first ionization enthalpy values (in $k J$ mol $^{-1}$ ) of group 13 elements are : How will you explain this deviation from the general trend?

$\left(2 k J m o l^{-1}\right)$ than that of $A l$ while that of $T i$ is much higher than those of $A l, G a$ and $I n .$ These deviations can be explained as follows: Al follows immediately after $s$ -block elements while $G a$ and In follows after $d$ -block elements and $T i$ after $d$ – and $f$ – block elements. These extra $d-$ and $f-$ electrons do not shield (or screen) the outer shell-electrons from the nucleus very effectively. As a result, the valence electrons remain more tightly held by the nucleus and hence larger amount of energy is needed for their removal. This explains why $G a$ has higher ionization enthalpy than $A l .$ Further on moving down the group from $G a$ to $I n,$ the increased shielding effect (due to presence of additional $4 d$ -electrons ) outweighs the effect of increased nuclear charge $(49-31=18$ units) and hence the $\Delta_{i} H_{1}$ of $\operatorname{In}$ is lower than that of the $G a .$ Thereafter, the effect of increased nuclear charge $(81-49=32$ units ) outweigh the shielding effect due to the presence of additional 4 $f$ and $5 d$ electrons and hence the of $T l$ is higher than that of In.

Q.  The first $\left(\Delta_{i} H_{1}\right)$ and the $\left(\Delta_{i} H_{2}\right)$ ionization enthalpies $\left(\text { in } k J m o l^{-1}\right)$ and the $\left(\Delta_{e g} H\right)$ electron gain enthalpy (in $k J$ mol $^{-1}$ ) of a few elements are given below: Which of the above element is likely to be : (i) The least reactive metal (ii) The most reactive metal (iii) The most reactive non-metal (iv) The least reactive non-metal (v) The metal which can form a stable binary halide of the formula $M X_{2}(X=\text { halogen })$ (vi) The metal which can form predominantly stable covalent halide of the formula $M X(X=\text { halogen }) ?$

$\left(\Delta_{i} H_{1}\right)$ and positive electron gain enthalpy $\left(\Delta_{e g} H\right)$ and hence it is the least reactive element. since inert gases have positive $\Delta_{e g} H,$ therefore, the element-V must be inert gas. The values of $\Delta_{i} H_{1}, \Delta_{i} H_{2}$ and $\Delta_{e g} H,$ match that of $H e .$ (ii) The element II which has the least first ionization enthalpy and a low negative electron gain enthalpy is the most reactive metal. The values of and match that of $K$ (potassium). (iii) The element III which has high first ionization enthalpy and a very high negative electrons gain enthalpy is the most reactive non-metal. The values of and match that of (fluorine). (iv) The element IV has a high negative electron gain enthalpy but not so high first ionization enthalpy. Therefore, it is the least reactive non-metal. the values of and match that of $I$ (Iodine). (v) The element VIhas low first ionization enthalpy but higher than that of alkalimetals. Therefore, it appears that the element is an alkaline each metal and hence will form binary halide of the formula (where $X=$ halogen). The values of and match that of (magnesium). $M g$ (vi) The element I has low first ionization but a very high second ionization enthalpy, therefore, it must be an alkali metal. since the metal forms a predominary stable covalent halide of the formula $(X=$ halogen), therefore, the alkali metal must be least reactive. athe values of and match that of Li (lithium).

Q.  What do you understand by isoelectronic species? Name the species that will be isoelectronic with each of the following atoms or ions. $\begin{array}{llll}{\text { (i) } F^{-}} & {\text {(ii) } A r} & {\text { (iii) } M g^{2+}} & {\text { (iv) } R b^{+}}\end{array}$

(i) $\quad F^{-}$ has $10(9+1)$ electrons. Therefore, the species nitride ion, $N^{3-}(7+3) ;$ oxide ion; $O^{2-}(8+2),$ neon, $N e(10+0)$ sodium ion, $N a^{+}(11-1) ;$ magnesium ion, $M g^{2+}(12-2)$ aluminium ion, $A l^{3+}(13-3)$ etc. each one of which contain 10 electrons, are isoelectronic with it. (ii) $\quad$ Ar has 18 electrons. Therefore, the species phosphate ion, $P^{3-}(15+3),$ sulphide ion; $S^{-2}(16+2) ;$ chloride ion; $\mathrm{Cl}^{-}(17+1),$ potassium ion, $K^{+}(19-1),$ Calcium ion, $\mathrm{Ca}^{2+}(20-2),$ etc. each one of which contain 18 electrons, are isoelectronic with it. (iii) $M g^{2+}$ has $10(12-2)$ electrons, therefore, the species $N^{3-}, O^{2-}, F^{-}, N e, N a^{+}, A l^{3+}$ etc. each one of which contains 10 electrons, are isoelectronic with it. (iv) $\quad R b^{+}$ has $36(37-1)$ electrons. Therefore, the species bromide ion, $B r^{-}(35+1),$ krypton, $K r(36+0)$ and strontium $S r^{2+}(38-2)$ each one of which has 36 electrons, are isoelectronic with it.