Can one predict the entropy change for aqueous phase reactions without calculation? I have seen several sources that claim aqueous solutions always have more entropy than pure liquids or solids, however, some aqueous ions have negative standard entropy.  This opens the possibility of solubility reactions with negative standard entropy change.  For example,
$$\begin{align}
\ce{NaF (s) &<=> Na+ (aq) + F- (aq)}&&\Delta S^{\circ} =-6.26 \ \mathrm{J/(mol\cdot K)}\\
\ce{AlCl3 (s) &<=> Al^3+ (aq) + 3 Cl- (aq)}&&\Delta S^{\circ} =-261.5 \ \mathrm{J/(mol\cdot K)}\\
\ce{AlF3 (s) &<=> Al^3+ (aq) + 3 F- (aq)}&&\Delta S^{\circ} =-429.58 \ \mathrm{J/(mol\cdot K)}
\end{align}$$

I have used thermodynamic data from this source.

So it seems to me that it is impossible to easily predict whether the entropy change for a reaction involving an aqueous phase is either positive or negative without calculation (unless a gas phase is involved, in which case the aqueous phase is moot)

Am I missing something, or are all the sources that claim an aqueous phase always has more entropy than a liquid phase oversimplifying the situation or just plain wrong?

I brought up the example of aluminum chloride to a chemistry instructor who then said that it is possible for molecular compounds to have negative entropy change when dissolved, but never for ionic compounds.  Yet both sodium fluoride and aluminum fluoride are ionic, and seem to have a negative entropy change when dissolved.

I understand why the standard entropies are negative, and about solvation shells. What I'm asking is whether the change in entropy can be predicted at a glance instead of looking up values and calculating as I have done.  Many textbooks claim all of these reactions should have a positive $\Delta S^{\circ}$ simply because "aqueous solutions have more entropy than solids or liquids." The instructor I talked to claimed that $\ce{AlCl3}$ will have negative $\Delta S^{\circ}$ because it is a molecular solid whereas $\ce{NaF}$ and $\ce{AlF3}$ should have positive $\Delta S^{\circ}$ because they are ionic solids. Are the textbooks and this instructor incorrect?

Some ions when dissolved in water may form a solvation shell that orders the solvent molecules in a way that the entropy decrease through this ordering outweighs the entropy increase through the ions leaving the crystal structure. Then you would have an overall decrease in entropy when dissolving the salt.

Small ions with high charge have a tendency for this.

The order of decreasing entropy from the state with highest to lowest entropy is- gas>aqueous>liquid>solid. Hence, just know that the entropy is positive when state changes from either liquid or solid to aqueous and it is negative when there is change of state from gas to aqueous.


Confusion with E/Z configuration Could someone please illustrate how to determine the E/Z configuration of this molecule?

What I understand so far is that if the two highest priority substituent on each side are on the same side, then it's Z and vice versa.

Now, how can I handle such a structure?
I've considered it and I thought it might be $\ce{CH-CH3}$ on one side (up), and $\ce{CH2CH2}$ on the other. However, I don't think it's right.

Bottom line is that I don't understand how to fractionate the ring into two substituents.

The CH3 and C2H5 groups are both smaller than the rest of the cycle. So these two groups can be forgotten and this molecule is Z.


Polarising Power and Covalent Character Higher the positive oxidation state of the central atom, more will be its polarising power which, in turn, increases the covalent character of bond formed between the central atom and the other atom.

This seems counter-intuitive to me. When the polarising power of the central atom increases, shouldn't the bond be more ionic instead of covalent? If not, can you tell me where I am wrong?

I believe it is about perspective. Note that "covalent molecule with ionic character" and "ionic compound with covalent character" are different.

If we are looking at a covalent molecule and assessing its ionic character, we cannot talk about polarising power, because in a covalent molecule, there would be no cation and anion to polarise in the first place. Thus, we would go by the rule: the bigger the electronegativity difference, the higher the ionic character and therefore, the more polarised the bond.

If we are looking at a ionic compound and assessing its covalent character, then we will go by the rule: the more polarising the central atom, the higher the covalent character.

Now, from the way you phrased your question, I believe that you are conflating the concepts "big electronegativity difference" and "polarising power". They are distinct concepts and should be thought of separately.

Although bond polarity is associated with electronegativity difference, "polarising power" is about the polarisation of the electron cloud of the anion by the cation and not the polarisation of the bond. This is also why "polarising power" is a concept associated with ionic compounds with covalent character.

Hope this clarifies your doubt! If you are still unclear and require some examples for clearer understanding, please leave a comment on my post.

Kindly note, we are talking about polarising power of 'cation'. This means there already exists an ionic bond and electron transfer has already taken place and the ions are formed. Now that cation again attracts the electron cloud of anion.
So more polarising power means that the among ions of ionic bond, the cation is attracting electron cloud of anion and hence inducing a 'covalent character'.


Why is the mass of 1 mole of an atom/molecule equal to its gram atomic/molecular mass?(proof) Why do we take the mass of 1 mole of an atom to be equal to its gram atomic mass and the mass of 1 molecule to be equal to its gram molecular mass? I want a mathematical rigorous proof. There are other questions on this site dealing with this question but none of them have the "proof".

$\def\p{\mathrm{p^+}}\def\n{\mathrm{n}}$
Mass of one proton = $m_\p$ and mass of one neutron = $m_\n$. (Masses are in unified atomic mass unit-$\mathrm{u}$)

C-12 isotope consists of 6 $\p$ and 6 $\n$. Therefore, mass of C-12 isotope is equal to 6 $m_\p$ + 6 $m_\n$. Since $m_\p$≈ $m_\n$, the mass of C-12 can be replaced by 12 $m_\p$. (electrons are not taken into account since they have negligible mass)

1 $\mathrm{u}$ is defined as 1/12 of the atomic mass of C-12 isotope.

⟹1 $\mathrm{u}$ = 1/12 × (6 $m_\p$ + 6 $m_\n$)

⟹ 1 $\mathrm{u}$ ≈  1/12 ×12 $m_\p$

⟹ 1 $\mathrm{u}$ ≈  1 $m_\p$   ...equation(1)

Now,

12 g of C-12 has $\pu{N_A}$ particles and constitutes 1 mole of it. (according to the definition of 1 mole)

⟹ 1 mole of (6 $\p$ + 6 $\n$) have a mass of 12 g.

⟹ 1 mole of (12 $\p$) have a mass of 12 g.

⟹ 1 mole of $\p$ have a mass of 1 g

⟹ 1 mole of $\p$ × 1 $m_\p$ = 1 g

⟹ 1 mole of $\p$ × 1 $\mathrm{u}$ = 1 g

⟹ 1 mole of $n$ $\p$ × 1 $\mathrm{u}$ = $n$ g

Therefore, mass of 1 mole of $n$ $\p$ is equal to $n$ g

Now, the final part,

Let an atom have $x$ $\p$  and $y$ $\n$ . The atom can be considered to be $(x+y)$ $\p$  for our purposes.

1 mole of $(x+y)$ $\p$ have a mass of $(x+y)$ g.

Thus the mass of 1 mole of atoms of an element is equal to its gram atomic mass.

Molecular mass of an element or a compound is equal to the sum of the atomic masses of each type of it's constituent atoms.
Thus the mass of 1 mole of molecules of an element or a compound is equal to the gram molecular mass.


Cahn-Ingold Prelog rules concerning tie breaking on second level comparing the cahn-ingold prelog priority of the above groups, I should first compare the atoms attached to the first carbon.
So we get:
(A): {O,C,H}
(B): {O,C,H}
(C): {O,C,H}
(D): {O,C,H}
therefore we have to consider the second level, and this is where I get confused:
Do I compare it this way:
(A):{C,H,H,H}
(B):{O,H,H,H}
(C):{F,C,H,H}
(D):{C,H,H,H}
I did this by rearranging all the "second level atoms" in descending order of atomic numbers.
since A and D are still in a tie, I rearrange all the "third level atoms" in descending order of atomic numbers:
(A):{H,H,H}
(D):{F,H,H}
so the priority is C>B>D>A
Is that correct?

look, all i can say is that $ D > C > A $

why ?

Lets compare A and C
$ OCH$3 and $ H $ are common to both. one has CH3 and the other has CH2F. of course, C > A.

compare A and D in the same way and find that D > A

compare D and C now.
there is clear proof that D > C because it has F in two places while C has H in those two exact positions.
thus it is seen that D > C > A

now see B and C
in B : after O we have CH3 and after C we have { H , H , F }
in C : after O we have H and after C we have { H , H , O }

looking at second layer after oxygens : B > C
looking at second layer after carbons : C > B.
thus even i am confused when it comes to B
but definitely D > C > A


What is the difference between an element and an atom? First, I would like to quote sentences from a book introducing elements and atoms:

An element is a fundamental (pure) form of matter that cannot be broken down to a simpler form.
Elements are made up of particles called atoms. An atom is the smallest unit of any element [...]

The first quote states that element cannot be broken down to a simpler form, but the second quote says elements are formed by atoms, which means element can be split into a simpler form (atoms)?
The difference between atom and element confuses me.

The wording in your quote is a bit unclear, I agree. A very nice definition is given by the Jefferson Lab page What is the simplest way of explaining what atoms, elements, compounds and mixtures are?

Atoms

  are the smallest bits of ordinary matter

Whereas an element is

  a substance that is made entirely from one type of atom

Atoms themselves can be 'broken down' into smaller sub-atomic particles (protons, neutrons and electrons) - the amounts of these (particularly the number of protons) define each element.

For example, in their un-ionised form, the element carbon is made entirely from carbon atoms, which have 6 protons and 6 electrons each, and cadmium atoms, which have 48 protons and 48 electrons, make up the element cadmium.

The amount of neutrons can vary with atoms of a particular element, forming isotopes, which (from the link):

  For example, carbon-12, carbon-13 and carbon-14 are three isotopes of the element carbon with mass numbers 12, 13 and 14 respectively. The atomic number of carbon is 6, which means that every carbon atom has 6 protons, so that the neutron numbers of these isotopes are 6, 7 and 8 respectively.

Different arrangements of atoms of single elements can form allotropes, such as for carbon, with diamond and graphite (among others - shown below). They both are composed of atoms of the element carbon, but in different configurations.

Below are diagrams of the allotropes of carbon featured in the article Work Function Engineering of Graphene (Garg et al. 2014)

An atom is a thing; an element is a type of thing.

An atom is a collection of protons, neutrons and electrons.  A single, isolated atom in its neutral state has some number of protons, the same number of electrons and some number neutrons (about as many as protons for the lighter elements, up to about 50% more, for heavier elements).  The number of neutrons or protons in an atom only changes as a result of radioactive processes or very high-energy interactions such as you get in particle accelerators.  And I mean really high energy: even if you think about blowing up sticks of dynamite, that isn't nearly enough energy to start messing with the protons and neutrons. Chemistry happens when atoms come together and share electrons or give electrons to each other.  Chemical reactions happen all the time and many of them don't need a whole lot of energy: moving electrons from atom to atom is often very easy.

So, the chemistry of an atom depends on the number of electrons and the number of electrons in an isolated atom depends directly on the number of protons. Electrons are so easy to add and remove from atoms (just rub a balloon against your hair: the static electricity is because you transferred electrons between your hair and the balloon) so we classify atoms according to the number of protons in them.  Neutrons aren't so relevant: I'll put a note about them at the end.

So, the element of an atom is determined by the number of protons.  All atoms of hydrogen have one proton, and all atoms with one proton are hydrogen. Two protons is helium, three is lithium, seventeen is chlorine, 79 is gold, etc.  A pure sample of an element contains only atoms of that type: for example, a pure sample of iron contains only atoms with 26 protons.  Water, on the other hand, is not an element: a molecule of water consists of two hydrogen atoms (one proton each) sharing electrons with an oxygen atom (eight protons).

Now, what does it mean to claim that an element "cannot be broken down into a simpler form," and why aren't atoms "a simpler form"?  Well, they're not a simpler form because an iron atom is iron: it's the same form, not simpler.  Think of it this way.  If I give you a lump of pure iron, all you can do is break it into smaller lumps of iron, or make it into a more complex substance, for example by letting it rust – rust is formed of iron and oxygen.  The smallest possible lump of iron you could make is a single iron atom, but that's still just an unbelievably tiny lump of iron. If you wanted to break down a lump of iron beyond single iron atoms, you'd need to use a nuclear reactor or a particle accelerator or something like that and then, finally, you'd be able to get something that wasn't iron, because you'd have changed the number of protons in the atoms.

Let's contrast that with water.  If I give you a bucket of pure water then, just like the lump of iron, you can divide it into smaller and smaller samples, ultimately ending up with a single water molecule.  But you can do something else: if you pass electricity through the water, it splits into pure hydrogen and pure oxygen.  These are "simpler" substances because each consists of atoms of just one element, whereas water has atoms of two elements.

What about neutrons? Well, in terms of chemistry, they don't do a whole lot and atoms that have the same number of protons but different numbers of neutrons are much more alike (they have essentially the same chemistry, for example) than atoms that have the same number of neutrons but different numbers of protons.  It makes much more sense to classify by the number of protons, since that determines the number of electrons and that determines the chemistry.

Suppose you tried instead to classify atoms according to the number of neutrons. Well, most argon atoms (18 protons) have 22 neutrons, but some chlorine atoms (17 protons) and a decent fraction of potassium atoms (19 protons) also have 22 neutrons. As you probably know, argon, chlorine and potassium are absolutely nothing like each other. On the other hand, potassium atoms with 22 neutrons behave almost identically to the most common kind of potassium atoms, which have 21 neutrons.


Why is SiO2 a solid while CO2 is a gas? I was under the impression that chemistry almost exclusively involves valence electrons because there isn't enough energy to strip off electrons located closer to the nucleus.

If that is true, and elements in the same period have similar properties because they have the same number of valence electrons, then why is $\ce{SiO2}$ a solid and $\ce{CO2}$ a gas? Surely a difference in mass by a factor of 2.5 can't be that big a deal.

Is it because of a van der Waals force difference, because silicon has extra electrons which result in compounds being formed from it being more symmetrical than those formed from carbon?

The reason why carbon dioxide is a gas and silicon dioxide is a solid is because their chemical structures are different.

Carbon dioxide is a linear structure with two double bonds between carbon and oxygen. It is a small molecule and non-polar with only weak bonds between the molecules. Hence it is a gas.

Silicon dioxide is not formed of small molecules. It consists of an infinite array of silicons where each silicon is bonded to four separate oxygens (and each oxygen is shared between two silicons). This creates a strong refractory solid (glass and sand are mostly silicon dioxide aka silica). So the same apparent overall formula doesn't describe the actual structure of the compounds at all. But the structures explain the difference in behaviour.

Of course this doesn't explain why silicon prefers to bond with four oxygens when carbon prefers just two. This is not completely simple and results from the relative bond strengths of carbon-oxygen bonds, carbon-oxygen double bonds and the equivalent bonds for silicon and oxygen. The simple version is that silicon oxygen bonds are strong relative to their double-bond equivalents whereas carbon-oxygen double bonds are strong relative to their single bond equivalents. Or, more precisely, if we could make a carbon-oxygen network solid with the equivalent structure to silica, it would tend to fall apart into carbon dioxide. If we could make silicon dioxide molecules, they would react with the release of energy to give silica.

Deeper explanations would need to look at why the relative strengths of double and single bonds turn out that way, but that would get into molecular quantum mechanics and would not be much more useful as an explanation.

The simplest explanation is the fact that the structures are different.

It is because of the structure of the $\ce{CO2}$. Two of carbon's valence electrons hybridize into two $sp$ hybrid orbitals. As a result, the  molecule is one dimensional with an angle of 180$^\circ$ between bonds and completely non-polar. The $\ce{Si}$, on the other hand, does not form such bonds and the angle is far from 180°, which in conjunction with oxygens high electronegativity makes it quite polar. Thus the interaction between neighboring $\ce{Si}$ and $\ce{O}$ atoms of different $\ce{SiO2}$ molecules is much higher and as a consequence you need much more energy to break the solid, giving it an increased melting point.
The mass (as discussed in various comments) does not play any role here since it is a matter of interaction or forces. The gravitational pull of single atoms or molecules is ridiculously small and never finds any considerations in such calculations (as it should!).


Why doesn't KI(aq) react with HCl(aq)? When concentrated sulfuric acid is added to anhydrous potassium chloride and the fumes produced are bubbled into aqueous potassium iodide solution, the observed solution would be colourless solution.

I think the first reaction is:
$$\ce{2KCl(aq) + H2SO4(aq) -> K2SO4 + 2HCl(g)}$$

I assumed the second reaction would be:
$$\ce{HCl + KI -> KCl + \frac{1}{2}I2 + \frac{1}{2}H2 }$$

However the second reaction is wrong according to the answer book as $\ce{HCl}$ won't react with $\ce{KI}$. From what I know, $\ce{Cl}$ is a stronger oxidising agent than $\ce{I}$, so shouldn't $\ce{I-}$ in $\ce{KI}$ be oxidized to $\ce{I2}$? I am an A-levels student so would appreciate a simpler answer.

In your first reaction, you added concentrated sulfuric acid to anhydrous potassium chloride. In your equation, you wrote $\ce{KCl (aq)}$, which is incorrect. Under dilute conditions, that reaction would not take place because all reagents and products would be in aqueous ionic state, meaning there is no reaction.
In your second reaction, the equation is completely wrong. It is true that elemental $\ce{Cl2}$ a stronger oxidizing agent than elemental $\ce{Br2}$ or elemental $\ce{I2}$. But in current situation is ionic $\ce{Cl-}$ versus ionic $\ce{I-}$ where both are aqueous. If all reagents are in ionic form and expected products are also ionic, there is no reaction. See your reaction below:
$$\ce{H+(aq) + Cl-(aq) +K+(aq) + I-(aq) <=> H+(aq) + I-(aq) +K+(aq) + Cl-(aq)}$$
There is no gas, liquid or solid formed, but all ions. Therefore, there is no reaction.

What happens in the second reaction solution will depend on the concentration of the reagents. When the hydrogen iodide concentration is high, it can form iodine. This is not a reaction of HI with an oxidant. What happens is

2HI ---> H2 + I2

If the HCl concentration in your experiment was very low in the second aqueous solution then I would not expect the reaction to occur. But if you look at an old bottle of concentrated HI, it will always be very dark due to the fact that it contains HI3. I know this as I have used HI solution to convert alcohols into alkyl iodides in lab.

Let us think for a moment about solvent extraction of metals from nitric acid, one of the problems is that the nitrate concentration is not equal to the anayltical nitric acid concentration (by titration) this is due to the fact that the Ka of nitric acid is not infinity. In the same way HI does not have an infinite Ka value.

As the anayltical concentration of HI in a solution increases, as the solvent properties will change and as Ka even in a very dilute solution is not infinity then the concentration of HI (intact HI) will increase. Also the concentration of the HI in the head space above the liquid will also increase.

The thermodynamics of formation of HI from I2 and H2 suggests that it should break down in to its elements. While Max Bondenstein did his work in the gas phase at a much higher temperture, it may be worth looking at.

Sulfuric acid will not form chlorine gas when it reacts with potassium chloride, it will form hydrogen chloride only. I know this as I have done this type of reaction myself.

When I was a teenaged boy I heated sodium chloride and sodium hydrogen sulfate in a test tube, it formed hydrogen chloride gas and not chlorine gas.


Does a freezing point of a substance always increase as its boiling point increases? I am studying for an introductory to physical sciences test and came upon a chart describing substances found in petroleum and the distillation process.

Each freezing point increases as each boiling point increases.  Each of these properties also increases with the rise in density.

Is this just true for substances in petroleum (methane, ethane, propane, butane, pentane...) or is this consistent throughout all substances?

Update:

(Pure substances)

Yes and no.

Yes in the sense that boiling point and freezing point usually correlate. If there is a line of homologous compounds, than this properties rise in general, as both are increased with increase in number of dispersion interactions involved. However, it is likely to find several compounds with odd behavior, especially around beginning of the line.

As a counterexample. Toluene has higher boiling point than benzene, but much lower freezing point. This is because toluene has less symmetrical form and consequently has harder time packing into strictly ordered lattice.

Another counterexample. Gallium melts around 30 celsium, but boils well above 2000 celsium. However, Tallium melts around 300 celsium, but boils around 1700.

However, for similar compounds the values usually correlate in the series. Say, for alkanes both values steadily grow with increase of number of carbons... give or take occasional peaks due to details in crystal packing.

However, if you consider compounds of different families, it is not predictable without looking into structures of the compounds.

This is true for all solutions and mixtures. Its what is called Colligative properties.
Freezing/melting point and Boiling of a homogeneously (evenly) mixed solution depends on the number of dissolved particles in the solution. How much these values change also depends specifically on the mixture.

If you think about it makes sense. Freezing point decreases because its harder for the similar species in the mixture to form nice regular crystal without other junk getting in the way. Similar boiling point increase is the result of solutes lowering the the vapor pressure of a solvent.

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch15/colligative.php


Is there a difference between negative/positive controls and control/experimental groups? First of all, just so I can fully understand, what is the correct definition of a control regarding science experiments?

Is there a difference between the definitions of a control/experiment group and negative/positive controls?

I know that the control group is the group that will not be tested on while the experimental group is the group that will be experimented on, but that seems very similar to negative and positive control.

Is it true that negative controls are the controls in the control group while positive controls are the controls in the experiment group?

Consider a chemical test, such as a test for the presence of lead in drinking water. You would use both experimental and control samples. Whenever you run the test on experimental samples, you also need to run it on both positive and negative control samples--positive samples are prepared deliberately with a certain known concentration of lead, and negative samples are known not to contain any lead (you have to be careful here!). If you get the incorrect results for the control samples, then you know that the results for the experimental samples may be off, so you have to repeat the test. You also might want to periodically have the analyst run a proficiency test with the standard positive and negative controls as well as samples containing various known amounts of lead.

Experimental and control groups might more typically refer to situations such as drug trials, where you are comparing a new drug to another drug that is already approved and used to treat an illness. The experimental group would be the group of subjects having the illness that are given the new drug and the control group would also have the illness but would be given the approved drug (preferably in a double-blind study design--where neither the subjects nor the people administering the drug know which treatment is which). Note that both the experimental group and control group are "experimented on" - otherwise they could tell whether they received the treatment or not, and the results could be erroneous due to the placebo effect.


Does oxygen gas form hydrogen bonds? I learned that hydrogen bonding is formed from a 'sandwich' of hydrogen and either oxygen, nitrogen, or fluorine where nitrogen, fluorine, or oxygen is the bread.

Will oxygen gas and water form hydrogen bonds? The oxygen-hydrogen sandwich could be made, but oxygen gas has no hydrogen attached. Is it necessary for both molecules to contain hydrogen for hydrogen bonding to occur?

To make it clear, the question is: do both molecules have to contain hydrogen in order for hydrogen bonding to occur?

To make it clear, the question is: do both molecules have to contain hydrogen in order for hydrogen bonding to occur?

No, as noted by Prakhar in his answer, tertiary amines are one class of compounds that can participate in hydrogen bonding (as an acceptor) when there is no hydrogen atom directly bound to the participating atom.

As an example, trimethylamine can hydrogen bond just fine with water:

More information can be found, e.g., here at ChemGuide (scroll down to "Solubility in water").

Yes both the molecules should have Hydrogen atoms for Hydrogen bonding to take place.

For instance there would be no hydrogen bonding between Water(H2O) and Oxygen(O2).
Since in oxygen both the atoms are equally electronegative therefore there is no partial charge developed.

Whereas in water a partial charge is developed due to higher electronegativity of Oxygen compared to Hydrogen. Thus Oxygen becomes partially negatively charged whereas the hydrogen becomes partially positively charged.

But water and oxygen wont form hydrogen bonds due to lack of partial charge separation in oxygen.

Here since both the oxygens are equally electronegative there is no charge separation.

NOTE:This is mostly true except for very rare cases such as hydrogen bonding between water and tertiary amines.


Inductive effect of phenyl ring I completely understand how phenyl group 'releases' or 'donates' electron through resonance, but how is that a phenyl group shows a weak -I effect, despite having electrons literally smeared on itself?

As the $\mathrm{s}$ characteristic of an atom increases, so too does its electronegativity. Proof of this can be seen when comparing the $\mathrm{p}K_\mathrm{a}$'s of 2-carbon hydrocarbons. The $\mathrm{p}K_\mathrm{a}$'s of $\ce{CH3 -CH3}$, $\ce{CH2 =CH2}$, and $\ce{CH ≡CH}$ are $51$, $44$, and $25$ respectively.

When considering phenyl group for what it is, a ring of six $\mathrm{sp^2}$-hybridized carbons, it is easy to see why it shows a strong electron withdrawing effect, and a weak electron donating effect. The electronegativity of the carbons make benzene a nucleophilic structure that does not easily give up its electron cloud.

in phenyl group the carbons are sp^2 hybridised which are having less s character and are thus electronegative in nature hence shows -I effect


Why is absolute zero unattainable? We were dealing with the Third Law of Thermodynamics in class, and my teacher mentioned  something that we found quite fascinating:

  It is physically impossible to attain a temperature of zero kelvin (absolute zero).

When we pressed him for the rationale behind that, he asked us to take a look at the graph for Charles' Law for gases:

His argument is, that when we extrapolate the graph to -273.15 degrees Celsius (i.e. zero kelvin), the volume drops down all the way to zero; and "since no piece of matter can occupy zero volume ('matter' being something that has mass and occupies space), from the graph for Charles' Law, it is very clear that it is not possible to attain the temperature of zero kelvin".

However, someone else gave me a different explanation: "To reduce the temperature of a body down to zero kelvin, would mean removing all the energy associated with the body. Now, since energy is always associated with mass, if all the energy is removed there won't be any mass left. Hence it isn't possible to attain absolute zero."

Who, if anybody, is correct?

Edit 1: A note-worthy point made by @Loong a while back:

  (From the engineer's perspective) To cool something to zero kelvin, first you'll need something that is cooler than zero kelvin.

Edit 2: I've got an issue with the 'no molecular motion' notion that I seem to find everywhere (including @Ivan's fantastic answer) but I can't seem to get cleared.

The notion:

  At absolute zero, all molecular motion stops. There's no longer any kinetic energy asscoiated with molecules/atoms.

The problem? I quote Feynman:

  As we decrease the temperature, the vibration decreases and decreases until, at absolute zero, there is a minimum amount of motion that atoms can have, but not zero.

He goes on to justify this by bringing in Heisenberg's Uncertainity Principle:

  Remember that when a crystal is cooled to absolute zero, the atoms do not stop moving, they still 'jiggle'. Why? If they stopped moving, we would know were they were and that they had they have zero motion, and that is against the Uncertainity Principle. We cannot know where they are and how fast they are moving, so they must be continually wiggling in there!

So, can anyone account for Feynman's claim as well? To the not-so-hardcore student of physics that I am (high-schooler here), his argument seems quite convincing.

So to make it clear; I'm asking for two things in this question:

1) Which argument is correct? My teacher's or the other guy's?

2) At absolute zero, do we have zero molecular motion as most sources state, or do atoms go on "wiggling" in there as Feynman claims?

There was a story in my days about a physical chemist who was asked to explain some effect, illustrated by a poster on the wall. He did that, after which someone noticed that the poster was hanging upside down, so the effect appeared reversed in sign. Undaunted, the guy immediately explained it the other way around, just as convincingly as he did the first time.

Cooking up explanations on the spot is a respectable sport, but your teacher went a bit too far. What's with that Charles' law? See, it is a gas law; it is about gases. And even then it is but an approximation. To make it exact, you have to make your gas ideal, which can't be done. As you lower the temperature, all gases become less and less ideal. And then they condense, and we're left to deal with liquids and solids, to which the said law never applied, not even as a very poor approximation. Appealing to this law when we are near the absolute zero is about as sensible as ruling out certain reaction mechanism on the grounds that it requires atoms to move faster than allowed by the road speed limit in the state of Hawaii.

The energy argument is even more ridiculous. We don't have to remove all energy, but only the kinetic energy. The $E=mc^2$ part remains there, so the mass is never going anywhere.

All that being said, there is no physical law forbidding the existence of matter at absolute zero. It's not like its existence will cause the world to go down with error 500. It's just that the closer you get to it, the more effort it takes, like with other ideal things (ideal vacuum, ideally pure compound, crystal without defects, etc). If anything, we're doing a pretty decent job at it. Using sophisticated techniques like laser cooling or magnetic evaporative cooling, we've long surpassed the nature's record in coldness.

Feynman is correct. Stop and think a moment. What does it mean to be "at absolute zero" if we accept the proposition that absolute zero is unobtainable?  Assume that absolute zero is unobtainable. Which of these statements is more false:1. At absolute zero, there is no motion. or 2. At absolute zero, there is no mass. Both are equally false; since the predicate is false, the statements are, logically, meaningless. (Which means both statements are equally true, too. But while #1 has often been used, I sure wouldn't use #2 unless I wanted to be pelted with rotten fruit. #1 is "close enough" to be a "useful" (if false) explanation while #2 has no physical justification.).  Classically, the Laws of Thermodynamics are empirical. They is what they is because they is. No "why" there. However, we can derive them from classical statistical mechanics (which was invented many years after the acceptance of the 2nd Law) or even better after the invention of quantum mechanics (1925 on) and its application to statistical mechanics. Temperature is a property of large populations of quantum particles, an atom does not have a (classically defined) temperature. So, in order to understand why this population can't have a temperature of 0 K, we need to understand how temperature is derived from the quantum mechanical properties of the particles that the system is composed of. That is a semester course in college, and not an Introductory one! If you understand Feynman's explanation, accept it (although it's not perfect - it was, after all, targeted at college freshmen with no previous exposure to quantum mechanics)


The definition of n-factor, what was it when it was introduced? I realize that n-factor is an outdated concept however I am interested in its formal definition when it was introduced.
We have many definitions now on how to calculate n-factor depending on the type of reaction and compounds involved.

For acids, n-factor is defined as the number of H+ ions replaced by 1 mole of acid in a reaction.
For bases, n-factor is defined as the number of OH– ions replaced by 1 mole of base in a reaction.
The n-factor for such salts is defined as the total moles of catioinic/anionic charge replaced in 1 mole of the salt
The n-factor of such salts is defined as the number of moles of electrons exchanged (lost or gained) by one mole of the salt.

However, none of these definitions explain what n-factor actually is. Is there a formal definition for n-factor?

Instead of asking what n-factor really was, a better question would have been what an equivalent was (But that would have been a duplicate)
According to the above question, the formal definition of an equivalent is as follows:

An equivalent (symbol: officially equiv; unofficially but often Eq) is the amount of a substance that reacts with (or is equivalent to) an arbitrary amount of another substance in a given chemical reaction.
In a more formal definition, the equivalent is the amount of a substance needed to do one of the following:

react with or supply one mole of hydrogen ions ($\ce{H+}$) in an acid-base reaction.
react with or supply one mole of electrons in a redox reaction.

Therefore given any chemical reaction, one equivalent of each reactant reacts to form one equivalent of each product, irrespective of the stoichiometric coefficients
$$\ce{aA + bB -> cC + dD}$$
In all such scenarios, the n-factor of the compound is defined as,
$$\textit{n}\text{-factor} = \frac{M_\mathrm{molar}}{M_\mathrm{equiv}}$$
However, why are there so many different definitions for calculating the same thing. This is because an equivalent doesn't have one fixed definition, rather depending upon the situation, the definition of an equivalent changes and hence the n-factor definition changes as well.
Therefore there is no formal definition per se for the n-factor, simply because it was never formally introduced. As M.Farooq pointed out in the comments,

I have only seen n-factor in South Asian textbooks. This is predominantly a South Asian college term for calculating chemical equivalents in current curricula. Here n stands for normality. You will not find this usage in British/US books. The n-factor is simply a coversion factor for molarity to normality concentration. Since the definition of an equivalent is dependent on the type of reaction we need to be aware of several scenarios which you have listed viz. acid-base, salts, redox reactions.

Yes,the definitions are correct.
n-factor is required when finding the equivalent weight of the substance.
In case of acids
equivalent wt.=molecular weight of acid/basicity(no. of replaceable hydrogen atoms)
here,the basicity is the n-factor
important observation:replaceable atoms of hydrogen are those which are attached to oxygen atoms.
And in the same way,
in case of bases
equivalent wt.=molecular weight of base/acidity(no. of replaceable hydroxyl ions
in case of bases
equivalent wt.=molecular weight of salt/no. of cations *valency of the cation
And so on,
The denominator in each case is the n factor .The numerator (molecular wt.)being the same factor in the equivalent weights of different types of substances,the denominator plays an important role of changing the equivalent wt. according to the substance.
Equivalent weight of a substance is that amount of the substance that reacts with or can displace 1.008g or 1.008 parts  of hydrogen or 35.5 g or 35.5 parts  of chlorine or 8 g or 8 parts of oxygen.
important observation:the equivalent wt. of hydrogen is 1.008g or parts ,of oxygen =8g or parts,of chlorine =35.5g or parts
And equivalent wt. in turn is required to find the normality of a substance.
normality=Amount of solute in 1000ml or 1L of solvent/equivalent wt. of solute


Can the ions of dissolved salts be physically separated with a strong enough external electric field? Say you had a arbitrary amount of sodium chloride dissolved in water. Could an external electric field (e.g. the electrically charged plates in Millikan's oil drop experiment) physically pull apart the dissolved ions in the water if it was strong enough? What forces would it need to overcome?

If it is possible, how powerful would the electric field have to be? What would be the outcome if the $\ce{Na+}$ and $\ce{Cl-}$ ions were then physically separated into two different containers?

Could an external electric field (e.g. the electrically charged plates in Millikan's oil drop experiment) physically pull apart the dissolved ions in the water if it was strong enough?
What forces would it need to overcome?

Yes, this is the principle behind electrophoresis and electric double layer supercapacitors where charge is separated by an electrostatic field shown below.

Diagram of Double Layer Capacitor charging and Discharging

What would be the outcome if the $\ce{Na+}$ and $\ce{Cl-}$ ions were then physically separated into two different containers?

Assuming you could so this without any interaction with the container, you would have 2 Leyden Jars that were opposite in charged and storing a lot of energy.


How to test for methanol? I wanted to make some ethanol from baker's yeast and I used sucrose. I've made it and put it in a glass jar. I left it on a warm place (near the radiator) until the yeast died and after they died I distilled the yellowish liquid using simple distillation. When I smelled the product in the receiving flask I was confused. It resembled ethanol a little. The smell was like sweet and medium-intense ethanol. I think that it might be methanol. If there are any reagents or tests other than smelling that I can use at home, please do tell.

Sorry @ JavaScriptCoder, but I've to agree with aventurin. I'd say idoform test should have been done in a certified laboratory because of safety concern. However, I think fractional distillation might be helpful for this identification (or simply verification). OP can do this at home since he/she has done simple distillation at home. Ethanol/water binary system give an azeotrope, which boils at $\pu{78.1 ^\circ C}$ and gives $95\%$ ethanol. Methanol, on the other hand, make no azeotrope with water, and boils at $\pu{64.7 ^\circ C}$. The temperature range of $\pu{13.4 ^\circ C}$ need only about ≤20 theoretical plates to separate. If you get first distillate below, say, $\pu{67 ^\circ C}$, then you got methanol or something other than ethanol. So, don't drink it.

For more information, read the Wikipedia page having Azeotrope tables.

I have decided to format my comment into an answer.

To distinguish the ethanol from methanol, have three test tubes. In one of them, put $\pu{2 mL}$ known methanol, in one of them $\pu{2 mL}$ known ethanol, and one of them $\pu{2 mL}$ your mystery solution. Then, apply about $\pu{5 mL}$ of $\pu{0.5 M}$ of $\ce{I2}$, and about $\pu{10 mL}$ $\pu{1 M}$ $\ce{NaOH}$ to each test tube in that order. If your chemical is ethanol, the reaction will produce $\ce{CHI3}$, which is harmful (make sure to dispose of it properly). However your alcohol won't kill you unless you drink to excess. Otherwise, your chemical is methanol, which is poisonous (make sure to dispose of that properly also). Definitely don't drink that.

Source is from the Royal Society of Chemistry. It also linked this source.


How to measure a distance along the surface of a protein in a PDB I would like to measure the distance between two active sites in a protein complex (pdb), but I do not find a straightforward way to do it.

To be more precise I need the distance along the surface of the protein and not the linear distance that would go through the protein, and which the distance wizard will give.

I have access to Pymol and would prefer a solution using that software although I would consider other as well. Thanks.

I would use linear distances between a suitable amount of fixed points as an approximation. You cannot measure exact distances along a rough surface, as the distance will vary depending on minimum measuring size. This is known as the coastline length paradox in math. https://en.wikipedia.org/wiki/Coastline_paradox

In theory the length would approach infinity as the measuring size approaches 0.

In practice you can get around the issue if you use a step size compatible with the smallest stable feature.

A suitable measuring size would probaly be in the order of ~10^-11 meters. (1/10 the size of an carbon atom).


Reactivity of Alkanes, Alkenes, and Alkynes In my textbook, it says that alkanes are generally the least reactive of the three, alkenes are slightly more reactive, and alkynes are even more reactive. However, alkanes have single bonds, alkenes have double bonds, and alkynes have triple bonds. How can the molecule with triple bonds be the most reactive if triple bonds require the most energy to break?

Maybe it will help if you look at reactivity in terms of ability to attract electrons instead of bond enthalpies. Alkanes are $sp^3$ hybridized, and hence have $25$% $s$ orbital character and $75$% p character. $s$ orbitals are closer to the nucleus and thus have a 'contracting' effect on the hybrid orbital. Greater the $s$ character, 'smaller' the hybrid orbital. This means that electrons are more closely packed. Alkanes have only 25% $s$ character, hence the hybrid orbitals are comparatively larger, and the effective nuclear charge on outermost electrons is less. Alkenes ($sp^2$) and alkynes ($sp$) have 33.3% and 50% $s$ character respectively. Thus their hybrid orbitals are SMALLER; i.e. effective nuclear charge is more. This means that it is easier to accept electrons, as now the effective strength of nucleus is more. Hence, alkynes can easily accept electrons, followed by alkenes and then alkanes. Thus, alkynes are most reactive, followed by alkenes and alkanes. Even though increase in $s$ character increases bond enthalpy, but it also increases the electron withdrawing capacity; and it is the latter which predominates.

(In fact, in general, greater the $s$ character, more the electronegativity and more the acidic nature).

The three bonds consist of two pi and one sigma bond.. Pi are very easy to break and sigma is difficult.. So alkynes can be easily broken as they have a more number of pi bonds while alkanes have only one sigma bind which is difficult to break.


De-excitation of a molecule In our chemistry lecture today on UV/vis-spectroscopy, we discussed the origin of colour in conjugated molecules due to electronic transitions from lower-energy molecular orbitals to higher-energy molecular orbitals. One of the examples we talked about was the origin of the orange colour of carrot. The lecturer mentioned that carrots appear orange as it contains conjugated molecules that absorbs blue light, resulting in the complementary orange colour being emitted. This blue light is used to excite electrons from π MOs to π* MOs in the molecule. One of my classmates asked what happens to the excited electron when it falls back to the ground state, wouldn't it cause blue-light photons to be emitted?

To this, the lecturer replied that perhaps, there is de-excitation of the electrons through other pathways, such as through molecular vibrations (i.e. kinetic energy of the molecules). However, he also was not so sure about the matter. Thus, I would like to seek clarification on how these excited electrons de-excite after initial excitation.

After radiation is absorbed and the electron is at an excited state in the molecule there are several pathways for de-excitation to occur (see fig.). The pathway of choice depends on its rate, ie how fast it can happen. It turns out that the fastest de-excitation pathways are radiationless (wavy arrows in fig) such as internal conversion (IC) that happens through vibrational relaxation of the molecule and external conversion (EC) where the molecule returns to its ground state through collision with solvent molecules. This is why matter usually doenst emit light after absorbing it. There are instances however where this is actually possible because it happens that the radiationless pathways are not the most efficient way for the molecule to return to its ground state and that is when fluorescence (FL) and phosphorescence (Ph) occur.

What your lecturer (as many lecturers, unfortunately) did not tell you is that the de-excitation pathways depend on the substance and the wavelengths you consider.

You can't know beforehand if a substance will be fluorescent, phosphorescent, or de-excitates with or without radiations. That is why only some substances can be used to make lasers (they need to support stimulated emission of radiations)


Hybridisation of terminal nitrogen in diazomethane I have a few questions about the terminal nitrogen (highlighted in red) in diazomethane, $\ce{CH2N2}$.

Is that nitrogen $\mathrm{sp}$ or $\mathrm{sp^2}$ hybridised?
What type of orbitals do the lone pairs on that nitrogen reside in?
Does the negative charge on nitrogen undergo resonance?

As usual with these small molecules, the bonding situation is a lot more complicated than expected. In this very molecule this is to the extend that not only the Lewis formalism, but also the conventional resonance formalism breaks down.
However, Jan's description is sound, and provides a good starting point for further analysis.
Unfortunately formulating a clear cut valence bond description is difficult, because it involves strongly coupled electrons. This is in part due to the high symmetry of the molecule, $C_\mathrm{2v}$, and the even higher local symmetry of the $\ce{CNN}$ moiety, $C_\infty$. Because of the nature of molecular orbital theory, i.e. being better at describing delocalisation, it is actually easier to understand.

Here are the molecular orbitals computed at the DF-M06L/def2-SVP level of theory:

And here is the same thing in numbers

 Atomic contributions to Alpha molecular orbitals:
 Alpha occ 1 OE=-14.323 is N2-s=1.00
 Alpha occ 2 OE=-14.252 is N1-s=1.00
 Alpha occ 3 OE=-10.099 is C3-s=1.00
 Alpha occ 4 OE=-1.055 is N2-s=0.51 N1-s=0.29 N1-p=0.10
 Alpha occ 5 OE=-0.802 is N2-p=0.31 C3-s=0.31 N1-s=0.14 N2-s=0.12 C3-p=0.07
 Alpha occ 6 OE=-0.518 is C3-p=0.26 C3-s=0.24 H5-s=0.16 H4-s=0.16 N1-s=0.07 N2-s=0.06
 Alpha occ 7 OE=-0.487 is N2-p=0.45 C3-p=0.24 N1-p=0.14 H5-s=0.08 H4-s=0.08
 Alpha occ 8 OE=-0.450 is N2-p=0.61 N1-p=0.29 C3-p=0.09
 Alpha occ 9 OE=-0.407 is N1-s=0.44 N1-p=0.41 N2-p=0.08
 Alpha occ 10 OE=-0.366 is N1-p=0.28 C3-p=0.24 N2-p=0.16 H5-s=0.15 H4-s=0.15
 Alpha occ 11 OE=-0.206 is C3-p=0.57 N1-p=0.38
 Alpha vir 12 OE=-0.082 is N1-p=0.55 N2-p=0.32 H5-s=0.06 H4-s=0.06
 Alpha vir 13 OE=-0.005 is C3-p=0.33 N2-p=0.33 N1-p=0.32

From the visual inspection we can see that the terminal nitrogen has only one orbital that could be classified as a lone pair, i.e mo 9; the hybridisation according to this calculation is approximately sp. There are three bonding π orbitals, i.e. mo 7, 8, 10. And there is also one anti-bonding π orbital, i.e. mo 11. The sigma bond is formed by two bonding (mo 4, 5) and one anti-bonding orbital (mo 6).
This is no coincidence, but pretty much always the case for terminal atoms.

Due to symmetry constraints, the hybridisation pattern at terminal atoms is usually:

for the orbital forming the single bond $\mathrm{sp}^{\geq1}$
two orthogonal (π) lone pairs $2\times\mathrm{p}$
one (σ) lone pair $\mathrm{sp}^{\leq1}$

Localisation of the orbitals is not easy for this compound. I performed a natural bond orbital analysis, which transforms the (delocalised) canonical orbitals into hybridorbital like ones, but it results in a low occupied carbon lone pair orbital and a partially occupied anti-bonding nitrogen-nitrogen π orbital.

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   1. - 3. core orbitals (omitted)
   4. (1.97583) LP ( 1) N  1            s( 67.37%)p 0.48( 32.60%)d 0.00(  0.03%)
   5. (1.33940) LP ( 1) C  3            s(  0.00%)p 1.00( 99.92%)d 0.00(  0.08%)
   6. (1.99655) BD ( 1) N  1- N  2
               ( 43.82%)   0.6620* N  1 s( 33.02%)p 2.02( 66.72%)d 0.01(  0.26%)
               ( 56.18%)   0.7495* N  2 s( 46.00%)p 1.17( 53.92%)d 0.00(  0.08%)
   7. (1.99587) BD ( 2) N  1- N  2
               ( 51.93%)   0.7206* N  1 s(  0.00%)p 1.00( 99.63%)d 0.00(  0.37%)
               ( 48.07%)   0.6933* N  2 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
   8. (1.98025) BD ( 3) N  1- N  2
               ( 39.11%)   0.6254* N  1 s(  0.00%)p 1.00( 99.54%)d 0.00(  0.46%)
               ( 60.89%)   0.7803* N  2 s(  0.00%)p 1.00( 99.89%)d 0.00(  0.11%)
   9. (1.99580) BD ( 1) N  2- C  3
               ( 65.65%)   0.8103* N  2 s( 53.76%)p 0.86( 46.18%)d 0.00(  0.05%)
               ( 34.35%)   0.5861* C  3 s( 30.11%)p 2.32( 69.75%)d 0.00(  0.14%)
  10. (1.95419) BD ( 1) C  3- H  4
               ( 60.92%)   0.7805* C  3 s( 34.95%)p 1.86( 65.03%)d 0.00(  0.02%)
               ( 39.08%)   0.6251* H  4 s( 99.91%)p 0.00(  0.09%)
  11. (1.95419) BD ( 1) C  3- H  5
               ( 60.92%)   0.7805* C  3 s( 34.95%)p 1.86( 65.03%)d 0.00(  0.02%)
               ( 39.08%)   0.6251* H  5 s( 99.91%)p 0.00(  0.09%)
 ---------------- non-Lewis ----------------------------------------------------
  12. (0.01076) BD*( 1) N  1- N  2
               ( 56.18%)   0.7495* N  1 s( 33.02%)p 2.02( 66.72%)d 0.01(  0.26%)
               ( 43.82%)  -0.6620* N  2 s( 46.00%)p 1.17( 53.92%)d 0.00(  0.08%)
  13. (0.64416) BD*( 2) N  1- N  2
               ( 48.07%)   0.6933* N  1 s(  0.00%)p 1.00( 99.63%)d 0.00(  0.37%)
               ( 51.93%)  -0.7206* N  2 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
  14. (0.07754) BD*( 3) N  1- N  2
               ( 60.89%)   0.7803* N  1 s(  0.00%)p 1.00( 99.54%)d 0.00(  0.46%)
               ( 39.11%)  -0.6254* N  2 s(  0.00%)p 1.00( 99.89%)d 0.00(  0.11%)

This is usually an indication of a strongly delocalised system such as a three-centre bond. Investigating the molecule with natural resonance theory gives the rather odd contribution of long range N1 - C3 bond.

 TOPO matrix for the leading resonance structure:

     Atom  1   2   3   4   5
     ---- --- --- --- --- ---
   1.  N   2   2   0   0   0
   2.  N   2   0   2   0   0
   3.  C   0   2   0   1   1
   4.  H   0   0   1   0   0
   5.  H   0   0   1   0   0

         Resonance
    RS   Weight(%)                  Added(Removed)
 ---------------------------------------------------------------------------
    1*     33.02
    2*(2)  32.31    N  1- N  2, ( N  2- C  3), ( N  1),  C  3
    3*(2)  28.81    N  1- C  3, ( N  2- C  3), ( N  1),  N  2
    minor contributions omitted
 ---------------------------------------------------------------------------
          100.00   * Total *                [* = reference structure]

According to NBO, the terminal nitrogen has an $\mathrm{sp}^{0.5}$ σ-type lone pair orbital, two $\mathrm{p}$ π-type bonding orbitals, and a $\mathrm{sp}^2$ σ-type bonding orbital.

Hybridisation is a mathematical tool to analyse a bonding situation. There is no correct way of applying it. It is also very important to understand an atom is never hybridised. (Even though organic chemists tend to use this jargon - even in textbooks.) The orbitals at an atom can be hybridised.

In da wikipedia  you can see that the C=N=N fragment is linear AND another resonance structure involves triple N-N bond. So, the answers

it is sp in one resonance structure and probably sp2  in another
resonance structure.
depends on the resonance structure
considered
yep.

Diazomethane typically reacts with its negatively charged carbon, so this is of little consequence though.


Is it possible to have same mass and volume for a solvent and its solution Is there any solute-solvent interaction possible, in such a way that a solvent with mass X grams and volume Y litres, and a solute with mass A grams and volume B litres, to form a solution that has exactly the same mass and volume as that of the solvent (X grams and Y litres)?

As Klaus suggests above, you can't add mass to another mass without it increasing.  That would defy the first law of thermodynamics that energy cannot be destroyed since mass and energy are directly linked.

However, much of this depends on what degree of precision you're measuring to.  If you're working with a cubic metre of water and only measuring to the first two significant figures, mass would be 1000kg and volume 1000l, and you add 40kg of sodium chloride to make a 4% saline solution, you would still only have 1000kg and 1000l because you're not counting the third significant figure.

It's all about context ;-)


Ionic or covalent bond Which bond is stronger, ionic or covalent? I have a lot of confusion about this.

Covalent and ionic bonds are best thought of as opposite ends of a spectrum, where electrons are shared evenly (as in the case of a symmetric molecule like N2) or unevenly (like HF).

That being said, bonds with more ionic character are generally considered stronger due to the significant electostatic contribution to bonding (the cationic and anionic components experience attraction).

This principle is the core concept of Pauling Electronegativities, where the degree to which electrons in a bond are "attracted" to one atom over the other (that is, the degree to which a bond is ionic) is ultimately derived from experimental bond dissociation energies (i.e. bond strengths).

Ionic bonds are stronger since they're between 2 charged particles and its because of the electrostatic force of attraction between the oppositively charged ions. They have higher melting and boiling points and require more energy than the covalent bonds break. Covalent bonds are as a result of Adams sharing electrons.  They can be easily broken into its primary structure as the atoms are close by to share the electron.


Is carbonyl-enol tautomerization only intermolecular or can it be intramolecular? During workup of a conjugate addition reaction, it is likely that protonation occurs at the oxygen of the enolate formed. See discussion here: Do enolates get protonated at the carbon or oxygen atom? In that discussion, it is essentially said "The oxygen is probably protonated, but it really doesn't matter." However, I think this could be important in select instances, specifically in the protonation of enolates with diastereotopic faces (which isn't all that rare). But this mostly depends on if tautomerization can occur intramolecularly.

An example of what I'm getting at:

This is an example I just made up. This reaction likely would not actually happen like this. In actually, I could see some γ-deprotonation happening.

In this reaction, one chiral center is formed, and an enolate is formed, and the most likely atom to be protonated upon addition of an acid is the oxygen:

That enol could go on to form one of two products. The intramolecular reaction, I think, would likely give much more of the thermodynamic product (shown below) than an intermolecular reaction, though I'm not certain on that. The reactions here are simplified; I realize both would most likely make some mixture of the two products:

My thoughts are that the four-membered intermediate required for the intramolecular reaction will just be too strained for this to occur. I realize that in basic or acidic solution, catalysis will occur, as the Wikipedia page mentions. I guess the best way to phrase my question is this: In the hypothetical situation that a single ester enol molecule (like that shown as the product of O-protonation above) is dissolved in pure hexane, would that molecule be stuck in in the enol form or would it be able to rearrange? I realize this is more feasible with an dienol, but I'm specifically trying to figure out if the enol can do this.

For the sake of this question, ignore the basicity of the nitrogen. Maybe I should have used the diphenylmethane anion or something. Doesn't matter, it wouldn't reach the proton, either.

tl;dr: The intramolecular variant is an impossible rearrangement. Only intermolecular proton transfer can be observed.

This is actually a very nice example to discuss sigmatropic rearrangements, because the desired [1,3]-rearrangement would be one. In general, these can proceed in four possible ways, depending on the configuration of the migrating group and whether it remains on the same side of the π system or not. We would have to draw orbital schemes to determine whether or not the orbital of the migrating group is able to interact both with the point of origin and the point of arrival favourably.

With regards to the configuration, the two possibilities are:

migration under retention
migration under inversion

A migration under retention is always possible because the corresponding orbital’s lobe is already pointing towards the π system it is migrating along. A migration under inversion is only possible if the (atomic) orbital whose bond is broken and later reformed has lobes on two different sides; typically for migrating atoms, this means a $\mathrm p$ orbital.

In the case of a proton shift, we must consider a hydrogen atom. Its only occupied orbital through which it bonds to the carbon chain is $\mathrm{1s}$. This orbital does not have two lobes, therefore a rearrangement mechanism under inversion is impossible; we need only consider the retention mechanisms.

With respect to the π system, the two possibilities are:

suprafacial (remaining on the same side)
antarafacial (switching sides)

Again, a suprafacial migration is practically always possible. For an antarafacial migration to be considered, the chain must be flexible enough so that both sides can interact with the migrating group; ideally at the same time.

Since the thought experiments are rather hard to perform, here are some depictions to make it easier.

Figure 1: Suprafacial [1,5] (left) and antarafacial [1,7] (right) hydrogen migrations. The depicted π orbital corresponds to the HOMO of a full π system including the $\ce{C-H}$ bond.

Generally, the full π system and the neighbouring $\ce{C-H}$ bond that is to be broken must be considered as I have done in figure 1. This π system can be drawn up with the standard rules for linear π systems. In the [1,5]-rearrangement, the system is penta-2,4-dien-1-yl while in the [1,7] case it is hepta-2,4,6-trien-1-yl. The $\ce{C-H}$ bond is considered part of the π system with the proton being the migrating group. Therefore, the HOMO (the orbital of importance) will be the one in question. (If the $\ce{C-H}$ bond is considered part of hydrogen, we must consider the π system’s LUMO, meaning that the orbital will remain the same. We must always have an occupied and an unoccupied orbital for consideration.)

The image shows us that the [1,5]-migration — over two double bonds — can only occur in a suprafacial manner with hydrogen. If we had a group capable of inversion, an antarafacial manner would be possible given the correct geometry. In the [1,7] case — over three double bonds —, hydrogen can only migrate in an antarafacial manner under retention; the only possible suprafacial manner would be under inversion which hydrogen cannot do.

This has been turned into a general rule of thumb: If your migration is [1,$4n+1$], it may proceed suprafacial under retention but if it is [1,$4n+3$] it must proceed in an antarafacial manner. In the [1,3]-case, only antarafacial migration would be possible, but there is no way that the π system can twist so it would allow for antarafacial migration. The π orbital in question is shown in figure 2 below.

Figure 2: The π orbital relevant for a [1,3]-migration.

Therefore, an intramolecular proton transfer is impossible and an external source would have to protonate.

One thing to note: the molecule itself is able to deprotonate/reprotonate another molecule of its type intermolecularly. So the compound would not be stuck in the enol form but a second molecule would provide the much desired proton to allow the keto form to be regenerated, even in pure hexane (if it is even soluble in that).

This is a very hypothetical situation, but let's say that you obtain that enol, and you dissolve it in an aprotic solvent, and that the enol is higher energy than the ketone, I would think a second molecule would act as the base and send your product to the lower energy state. There is no need for this to be intra-molecular, as the Pka of one enol is exactly the same as the Pka of the other enol, they will freely exchange protons up to the point where you will get protonation at the carbon.


How does the proticity of an acid affect the enthalpy change of its reaction with a base? For an experiment I've carried out a series of reactions between monoprotic bases (ammonia and potassium hydroxide) and several acids. Two of these acids were monoprotic (hydrochlori acid and ethanoic acid), another was  diprotic (sulfuric acid) and the other was triprotic (phosphoric acid). My results showed that the triprotic acid released more energy per mole than both the diprotic and monoprotic acids, and that the diprotic acid released more energy per mole than the monoprotic acids. This gave rise to my question: how does the proticity of an acid affect the enthalpy change of its reaction with a base?

To be clear I compared my results between weak acids, strong acids so the strength of the acid has already been accounted for.

You have to remember that enthalpy change is defined at kilojoules PER MOLE. The concentration won't affect enthalpy change because the same amount of heat is given off per mole of zinc , however the reaction will go faster with a higher concentration.


Why are tertiary carbocations the most reactive, if they're already stable? This may seem silly, but doesn't it seem weird for a compound that's stable (in this context, the tertiary carbocation) to be the most reactive?

I mean, wouldn't it be the least, given that it's already stable and wouldn't want to leave that stable configuration. On the same lines, the methyl carbocation sounds like the most reactive due to its high instability, right? wouldn't it want to react in order to form a stable compound?

This is an excellent question. Please correct me if I'm wrong, but I think this is what you are grasping at:

First, it is true that tertiary carbocations are generally more stable than primary carbocations (and secondary carbocations) due to having more inductively donating alkyl groups. The hyperconjugative effect can also be invoked to explain the relative stabilities of primary, secondary, and tertiary carbocations.

Second, transition states involving tertiary carbocations as opposed to primary carbocations are more favorable precisely because tertiary carbocations are more stable than primary carbocations.

Note that this does not mean that tertiary carbocations are more reactive. We don't generally say that something which exists in a transition state is reactive/unreactive. One reason is that the transition state complex cannot be directly captured or observed. Another reason is that it is the transition state ... of course it's going to be reactive - of course it'll change very rapidly!

We may however say that reactants and products are either reactive or unreactive. So if something goes through a transition state involving a tertiary carbocation, it might well be more reactive than something which goes through a transition state involving a primary carbocation.

Of course, before labeling anything as reactive or unreactive, you'd do well to note exactly what you mean by reactivity. To put it lightly, there are myriad reactions. An alkyne such as acetylene is reactive with regard to combustion; acetylene torches are commonplace. However, acetylene isn't going to be reactive with regard to, say, backside attack (for a variety of reasons).

Ultimately I think your issue lies with semantics rather than the chemistry ... still, a good question!

Good question.... Tertiary carbonation is more reactive than secondary carbonation which is in turn more reactive than primary carbocation due to stability... Primary carbocation is reactive and unstable so convert again back to a neutral reactant molecule instead of converting into products but tertiary carbocation is stable and there are more chances to form product by reacting with a nucleophile..


Crystal field splitting pattern with strong field and weak field ligands What will be crystal field splitting pattern for octahedral complexes if some of the ligands are strong and some are weak field ligands?

Well, the split would depend on:
The identity of the metal ion as well as the identity of the ligands themselves. You'd have to consult a spectrochemical series to determine the magnitude of the split caused by the ligands.

Alternatively, you could beam white light through the complex and using the light that appears, determine the wavelength of the light absorbed (the complementary colour) to promote electron(s) from the lower split d orbital to a higher one... This would give you more information as to the magnitude of the split (sometimes, the experimental value will provide an answer that theoretically cannot be determined).


What is the condensation point of water? The boiling point of water is $\pu{100^\circ C}$. However, we find that even at room temperature, water can evaporate into gas.
Therefore, water vapor can exist at temperatures of, say, $\pu{50^\circ C}$.

At what temperature, can you say for certain that all water vapor will turn into liquid?
In other words, what is the condensation point of water?

You're right - water vapor can exist at temperatures such as $\pu{50^\circ C}$. This is a phenomenon known as evaporative cooling, where molecules of water with higher kinetic energies tend to "release" themselves from the system, and as a result, less and less water molecules are held within that system.

Temperature is a sort of proxy for kinetic energy, and vice versa. Reference, also, the Maxwell-Boltzmann distribution, which reveals that at $\pu{100^\circ C}$, for example, not all molecules in the system possess the kinetic energy that they would at $\pu{100^\circ C}$, but rather the average exists at that temperature.

To answer your question, never in a real scenario.

I wasn't sure about this myself! But after thinking about it logically, I realized that when a gas condenses into a liquid at around 50 Degrees, I knew that when a liquid condensed, it was one step below, so when it condensed it would become a solid. When it freezes (condenses) into a solid, the temperature is around freezing (0 Degrees Celcius) and (32 Degrees Farenheit)


Thermodynamics, pressure volume work In thermodynamics work done by the system is negative but in case of expansion of gas it is positive because force and displacement are in same direction, why?

You are correct. The system does positive work. Textbooks introduce the negative to show that the system loses energy as it does expansion work. As the gas expands, it does positive work but it loses this energy in the process. Work, like heat, is basically a transfer of energy.

I think it is due to pressure which is negative in the pressure volume work equation actually here two types of pressure act at a time one is external other is internal so net effect of external is less then internal which is taken negative.


Can Fe(CO)5 adopt both square pyramidal and trigonal bipyramidal geometries? I know the hybridization of $\ce{Fe(CO)5}$ is $\mathrm{dsp^3}$. According to my book, coordination compounds with coordination number 5 can interchange between square pyramidal and trigonal bipyramidal geometries. Is it true for this compound? Doesn't seem so according to the picture:

Was my understanding wrong somewhere?

Your book was correct that a five coordinate metal complex is able to adopt both square pyramidal and trigonal bipyramidal geometries, and in both cases the sp3d hybridisation scheme applies (if you believe in hybridisation...).

Which geometry is adopted depends upon a combination of steric and electronic factors, and isn't necessarily trivial to predict, though MO theory and crystal field theory can help understand which might be more favourable.

The structure of iron pentacarbonyl is widely agreed to be trigonal bipyramidal, with three CO ligands occupying equatorial positions, and two CO ligands occupying axial positions.

  Trigonal bipyramidal structure of iron pentacarbonyl

Whilst both geometries are possible (and indeed observed, IF5 is square pyramidal), I suspect that your confusion arises from the fact that iron pentacarbonyl is able to undergo a phenomenon known as Berry psuedorotation.

  Mechanism of Berry psuedorotation

Berry psuedoroation is the process in which two of the equatorial ligands are switched out for the two axial ligands via a square planar intermediate. This process is sufficiently rapid that if observed by NMR, only one environment for the carbon of the CO ligand is observed (i.e. the psuedorotation is faster than the NMR timescale).

Here the hybridisation is dsp3 in $\ce{Fe(CO)5}$. It is basically an inner orbital complex. So, it is a spin-free orbital. The hybridisation plays a major role in determining the shape in these cases. So, when the hybridisation is dsp3 in most of the cases you get the shape as trigonal bipyramidal, whereas in sp3d we get square pyramidal.


Relativistic effects and lanthanide contraction So I understand than the lanthanide contraction is due to poor shielding of the 4f electrons which decreases the radius.
However, if Im not mistaken the relativistic effects lead to a contraction of the core orbitals (s,p) while the f and d orbitals expand. So what I do not understand how do they match, one of them decreases the radius and the other one should expand it as the f orbitals expand right?
They usually say relativistic effects are one of the reasons for the lanthanide contraction
Thanks

The lanthanide contraction is caused by two effects

The 4f orbitals are very diffuse and therefore result in poor screening of the electrons further out, those in the n=5 and n=6 orbitals
relativistic effects

This Wikipedia article presents a nice discussion of the lanthanide contraction.  The article estimates that the first effect, the screening effect, is the major factor, with relativistic effects playing only a minor role (~10%) in the observed contraction.

  what I do not understand how do they match, one of them decreases the
  radius and the other one should expand it as the f orbitals expand
  right?

That's correct, the outer s- and p-orbitals contract due to the lanthanide effect, while d- and f-orbitals expand (see this SE Chem post for a more detailed description).  In the lanthanide series, the 6s electrons are further from the nucleus than the 4f electrons, therefore they (the 6s electrons) determine the radius.  A contraction of the 6s orbital radius results in a contraction of the observed lanthanide electron radius, even though the 4f electrons have moved slightly further away from the nucleus.

Only orbitals with l=0 (s orbitals) have radial components that are non-zero at the nucleus:

s-shells, including and inparticular the 6s shell, are much more sensitive to the relativistic effects (compared to l>0 shells) due to s-orbitals allowing electronc density AT the nucleus.  If s-shell electron density is non-zero at the nucleus, it's reasonable to suspect that s-shell electron density is contracted more by relativistic effects.


A simple equilibrium problem Take the following simple equilibrium:
$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$

Ethanoic acid will act as a typical Brønsted-Lowry acid and donate a proton to water, forming a conjugate acid-base pair. In this case, H3O+ is the conjugate acid, and CH3COO- is the conjugate base.

Logic tells us that if we have a solution of ethanoic acid and water, increasing the amount of acid will decrease the solution's pH. This makes sense, since there'd literally be a higher acid concentration.

However, the interpretation I get due to the equilibrium is the opposite. I see it as increasing the concentration of ethanoic acid will mean there's a larger concentration of conjugate base, hence the solution will become more basic.

Is this flawed logic or simply a misunderstanding of Le Chatelier's principle?

You are neglecting the hydronium ions produced during dissociation of acetic acid.

We can consider the following sequence of events when we add more acetic acid to solution. In reality they occur simultaneously, but it's easier to think about it sequentially.

We have more than the equilibrium concentration of acetic acid, so some of it dissociates to produce $\ce{H+}$ and acetate ion.
Now we have more than the equilibrium concentration of acetate ion, so some of it reacts with $\ce{H+}$ to form acetic acid again.

Your argument is that step 2 makes the solution more basic, but you are neglecting the effect of step 1. Every extra acetate ion produced in step 1 also involves the production of a $\ce{H+}$. Furthermore, step 2 has to occur to a lesser degree than step 1, because step 2 consumes the acteate ions produced in step 1. The acidity of the solution must therefore go up as a whole.*

*Of course it's also possible that steps 1 and 2 both occur at nearly the same rates, and then the pH of the solution doesn't really change much as a result. This is what happens in a buffer.

I will do my best here, If more acid is added, the pH should remain the same due to the continuous reactions with water creating base so that equilibrium is once again reached. This should remain the scenario until such time as more acid is added than the amount of water. When this stage is reached the additional acid will no longer be able to react with water and so will remain is its acidic state thus decreasing the pH.


stereochem of epoxide opening Here I have an epoxide opening reaction, but I was wondering that when a nucleophile attacks, How would the stereochemistry work? I have 2 products I predicted and I'm not sure which one would be correct. It's either the first one predicted in the reaction of the second picture I have attached. Or are they actually the same thing?

In these three-membered ring reactions, it's an SN2 substitution reaction that happens. Hence, there is an inversion of the configuration associated.
Edit: Perhaps the mechanism is more complicated than SN2, contrary to Klein's book which states that it's SN2, this online source says that the actual mechanism is between sn1 and SN2 (leaning to SN2 side) to explain the preference of nucleophile to attack tertiary carbon though more sterically hindered.

Reference: See page-431 of Organic Chemistry David-R-Klein


How does black CuO impart a green color to glazes and glass? I read from a source that cupric oxide (CuO) imparts green to blue colour to glazes and glass. But CuO is black in colour. How is this possible?

This is due to the oxidation of cupric oxide.

The main componds believed to cause this colour comprise a mixture of 3 compounds: $\ce{Cu4SO4(OH)6}$ (green); $\ce{Cu2CO3(OH)2}$ (green); and
$\ce{Cu3(CO3)2(OH)2}$(blue).

The following reactions are believed to take place:

Copper (I) oxidised to the black copper (II) sulfide ($\ce{CuS}$)  in the presence of sulfur impurities. Under accelerated conditions, the described process occur at a faster rate.

$\ce{CuO}$ and $\ce{CuS}$  reacts with carbon dioxide ($\ce{CO2}$) and hydroxide ions ($\ce{OH-}$) in water (in presence of air) to form $\ce{Cu2CO3(OH)2}$.

The extent of humidity and the level of sulfur have a significant impact on how fast the compounds develop, (under controlled conditions these reactants are varied to produce a favourable hue) as well as the relative
ratio of the three components.

$$\ce{2CuO + CO2 + H2O → Cu2CO3(OH)2~~~~~~~(1)}$$

$$\ce{3CuO + 2CO2 + H2O → Cu3(CO3)2(OH)2~~~~~~~~(2)}$$

$$\ce{4CuO + SO3 +3H2O → Cu4SO4(OH)6~~~~~~~~~(3)}$$

References

http://www.wskc.org/documents/281621/282063/ENGAGE_E3S_Chemistry_Statue+of+Liberty.pdf/e4f24c7e-3666-425e-9c41-7dbdd0065eb4

The green-blue color is common for copper compounds, because copper(II) ions in oxide neighborhood usually adsorb light in the red region thanks to relatively low energy of the $d-d$ transitions, that happens to be low because the d-orbitals of copper are small thank to the $d$-shrinking effect (large effective charge affective the $d$-orbitals). The best model for copper glasses/glases would be copper silicates
Here are links to description of two copper silicates
https://en.wikipedia.org/wiki/Egyptian_blue
https://en.wikipedia.org/wiki/Bayldonite
Please, note: these two are mere examples, the exact silicate phases formed in glasses and glazes may have very complex composition and structure or be actually amorphous. The point is that the color pallette is typical for isolated $Cu^{2+}$ ions in oxide neigbrohood, which is the neighborhood of copper in glasses in glazes

So, the actual question is where the black color of copper oxide comes from. Black color is a total adsorption in the visible spectrum, that comes from a family of electron transitions. Indeed, the paper on the band structure of copper oxide says that there are two rather wide bands with band gap ~1.4-1.7 eV (the energy of photons of  deep-red light is ~1.7) Why it has this particular energy is a much harder question with no simple answer.

Bands are the closest equivalent to orbitals known in quantum theory of solids. Bands are formed from individual overlapping atomic orbitals, but since they are formed from infinite number of atomic orbitals (in theory, but in crystals the size of the crystal makes the theory reasonably close to experiment) they form a continuous spectrum. Individual 'orbitals' in a band are known as states. Bands normally have finite width, thus it is possible to have band gaps - energy ranges with no corresponding electron state for this particular crystal. If the highest band is not completely filled with electrons, the crystal(usually) has metallic conductivity as the electrons may freely move from state to state.  If the highest band is completely filled with electrons and there is a gap between it and the next band, the crystal usually is a semiconducor (or an insulator if the gap is large enough). If the gap has wideness near visible spectrum, it may color the crystal. If the gap is larger than energy of violet photons (~3 eV), the crystal is transparent (or white)


Nucleophilicity in Polar Aprotic Solvent Why is it that fluoride is more nucleophilic than iodide in an aprotic solvent? The Clayden textbook suggests that lower-row elements of the periodic table are more nucleophilic since their valence electrons are in higher energy levels and are therefore more reactive. However according to the slides from my lecture class, the order of nucleophilicity for halides in an aprotic solvent goes from F- > Cl- > Br- > I- which is opposite from what is stated in the Clayden textbook. Can someone please explain this?

Thank you

It's because of the anion stability. Fluoride is less stable than iodide, therefore more likely to donate electrons.
In protic solvents, the fluoride was stabilised by the partially positive hydrogen end. Therefore, it is less nucleophilic then iodide.


Are p‐xylylene and fulvene aromatic? Are the following structures aromatic?

I have not seen this type of question anywhere so I do not know the answer.
This is just a problem that has been roaming around in my head that I may have seen somewhere.
My approach: since aromaticity leads to stability the carbon-carbon bond in both structure should be polar leading to aromaticity like this:

But this is just a thought and I do not know if it is right.
Is any of the two structures aromatic?

To decide if a molecule is aromatic or not you must know some rules-

The Molecule Must Be Cyclic and the system should be conjugated.
The molecule must follow Huckle's rule i.e. it should have (4n+2) Pi Electrons participating in resonance.

3.The molecule must be planar.

For details see-https://chem.libretexts.org/Textbook_Maps/Organic_Chemistry/Map%3A_Organic_Chemistry_(McMurry)/Chapter_15%3A_Benzene_and_Aromaticity/15.03_Aromaticity_and_the_Huckel_4n___2_Rule

Now if you try to draw the resonance structure for your 1st molecule then you will see that 4 electrons are participating in resonance at a time. So the system is not fully conjugated. It will be a cross conjugated system.So both the structures you have mentioned are non aromatic.

Stability is a result of aromaticity. You can not comment that if a compound is stable then it is due to aromaticity.

Here is an image showing resonance in para xylylene.

It can be seen from the resonating structures that some of the resonance structures are not aromatic making the molecule non-aromatic in total.

Any recommendations on making the image better are welcomed.


Why is the enthalpy of combustion higher for n-butane as compared to 2-methylpropane? N-butane has an enthalpy of formation of $\pu{-125.6 kJ/mol}$, whereas it is $\pu{-134.2kJ/mol}$ for 2-methylpropane. When comparing the heat set free at the combustion process, why is it higher for n-butane? Isn't the enthalpy of formation the negative of enthalpy of combustion and therefore 2-methylpropane should be higher?

Take 4 moles of carbon and 5 moles of H2. Synthesize a mole of n-butane and measure the heat evolved (125.6 kJ). Then repeat the process, but make 2-methyl propane (you'll measure 134.2 kJ). Then, with sufficient oxygen,  prepare 4 moles of CO2 and 5 moles of H2O from each of the previous two procedures and measure the heat of combustion (more for n-butane).

The total heat evolved to form the final oxidation products from carbon and H2 will be the same, but since the paths are different, the heats evolved in the two steps of each procedure will be different. n-Butane evolves less heat upon formation, therefore can (must) evolve more heat in the next step when it is oxidized to final products.

Isn't the enthalpy of formation the negative of enthalpy of combustion

No! That certainly isn't true.

Consider this equation: $\ce{CH4 + O2 -> CO2 + H2O}$. If you reverse it, you'll get: $\ce{CO2 + H2O -> CH4 + O2}$ which is clearly not an enthalpy of formation because:

the left hand side of the reaction does not have elements instead compounds
the right hand side of the reaction should have had a single product, but instead has two

I cannot comment on the second part of your question because I could not find standard data for heats of combustion of both the compounds.


Friedel–Crafts reaction of phenol Does phenol undergo Friedel–Crafts reactions or does it react with Lewis acids like aniline does?

Like aniline, phenol too reacts to a very less extent during Friedel-Crafts reaction.

The reason being that the oxygen atom of phenol has lone pair of electrons which coordinate with Lewis acid.

In fact most substituents with lone pair would give poor yield.

The two pathways involved in the reaction with phenol reduce the overall yeild:

Phenols are examples of bidentate nucleophiles, meaning that they can react at two positions:

on the aromatic ring giving an aryl ketone via C-acylation, a Friedel-Crafts reaction

or,

on the phenolic oxygen giving an ester via O-acylation, an esterification

phenol DOES NOT undergo Friedel Craft Alkylation or Acylation. This is because oxygen's lone pair in phenol makes co ordinate bond with AlCl3 (A Lewis acid) hence blocking it .
So alternative way is Fries Rearrangement.


Why is the reaction between potassium permanganate and hydrogen peroxide spontaneous? When hydrogen peroxide is mixed with potassium permanganate, oxygen gas and water vapour are formed, according to the reaction (source):

$$\ce{2MnO4- + 3H2O2 -> 2MnO2 + 2H2O + 3O2 + 2OH-}$$

This reaction is spontaneous, and exothermic. It is an example of a redox reaction, with the following half reactions occurring (data from Vanýsek):

$$
\begin{align}
\ce{MnO4- + 2 H2O + 3 e- &-> MnO2 + 4 OH-} &\quad E^\circ_\mathrm{red} &= 0.595~\mathrm{V} \\
\ce{H2O2 &-> O2 + 2 H+ + 2 e-} &\quad E^\circ_\mathrm{ox} &= -0.695~\mathrm{V}
\end{align}
$$

$E^\circ_\mathrm{cell}$ is equal to the sum of the oxidation potential and the reduction potential of the two half reactions; in this case, it would be $-0.1~\mathrm{V}$. A redox reaction is spontaneous if $E^\circ_\mathrm{cell}$ is positive — how can it be, then, that hydrogen peroxide spontaneously reacts with permanganate ions?

Using thermodynamical data (from NIST), I have calculated that the $\Delta G^\circ_\mathrm{m}$ of the reaction is $-463.576~\mathrm{kJ}$. The reaction should indeed be spontaneous. How can it be, then, that the results of the thermodynamical approach and the electrochemical one differ drastically?

For both half-reactions, the actual potentials depend on $\mathrm{pH}$; however, the values given in the question apply to different $\mathrm{pH}$.

The given reaction of hydrogen peroxide and the corresponding potential apply to $\mathrm{pH=0}$ or $\left[\ce{H+}\right]=1$:

$$\ce{O2 + 2 H+ + 2e- <=> H2O2}\qquad E^\circ=0.695\ \mathrm V$$

Whereas the given reaction of permanganate and the corresponding potential apply to $\mathrm{pH=14}$ or $\left[\ce{OH-}\right]=1$:

$$\ce{MnO4- + 2 H2O + 3e- <=> MnO2 + 4 OH-}\qquad E^\circ=0.595\ \mathrm V$$

The corresponding reaction and potential for $\mathrm{pH=0}$ or $\left[\ce{H+}\right]=1$ are:

$$\ce{MnO4- + 4 H+ + 3e- <=> MnO2 + 2H2O}\qquad E^\circ=1.695\ \mathrm V$$

The potential of a cell is the reduction potential minus the oxidation potential. In your case, it would be $E^\circ_\mathrm{cell} = 1.19 - (-2.08) = 3.27~\mathrm{V}$ for the whole reaction or $1.64~\mathrm{V/mol}$ of permanganate. You must remember to take into account the stoichiometric coefficients of the half reactions to obtain the overall potential


What is the substance turning salty water green when electrolysed? I am a student and I take O.L. Chemistry, in my knowledge I know if I electrolysed concentrated aqueous NaCl solution I should get Hydrogen evolved at the cathode and chlorine at the anode.

I did an experiment, I add a lot of table slat to tap water to get the electrolyte ready (aq. NaCl) then I used graphite electrodes (just got it from a mechanical pencil) and passed electricity through the salty water. I used of course direct current using a 6 Volts battery I got.

Well my suggestions were half correct, Hydrogen really evolved and bubbled violently at the cathode, but at anode I didn't see any yellowish green gas indicating presence of chlorine, yet only colorless bubbles continue to evolve from the anode but was less violently than the cathode. I thought this is Oxygen evolving so I increased the concentration of solution by adding table salt but also only colorless bubbles evolves at the anode. Well I didn't really calculated how much precisely I should add salt but I added insanely a lot of table salt.

Only to discover after around 5 minutes of electrolyzing the salty water, it became darkish green in color, I didn't really knew what is this and this is my question, I used graphite electrodes so those doesn't react like if I used iron, the solution was just nice tap water and not really bad or muddy. My only suggestion is that this is actually chlorine dissolved in water ... So is this correct ?? I knew that chlorine is soulable in water but I didn't thought it will turn the whole water greenish in color....

Can anyone help me in that ? And please if you're going to answer me, answer me as if I am somebody who knows nothing in chimestry .....

Thanks in advance.

My speculation is based on the words specifying the use of 'tap water' during the electrolysis of aqueous sodium chloride.
In particular, the interaction of formed chlorine and existing ferrous bicarbonate (found in tap water) could form small amounts of aqueous FeCl2. [EDIT TO QUOTE SOURCE]

Bacterial Iron in Well Water – Iron bacteria is usually identified by slime in places such as toilet reservoirs or by the presence of a slimy mass fouling softeners or filters.
Ferric Iron in Well Water – Also known as red water iron, ferric iron in well water is essentially clear water iron that has been exposed to oxygen – usually from the air, thereby oxidizing.  Carbon dioxide leaves the water and the oxygen combines with the iron to form ferric ions (Fe+++).  This gives the water a red rust coloring.
Ferrous Iron in Well Water – Often called clear-water iron because it is clear when poured, this substance is found in water that contains no oxygen. Typically, it comes from deeper wells and groundwater sources. Carbon dioxide acts on iron in the ground to form soluble ferrous bicarbonate. In water this forms ferrous ions (Fe++)

Per Wikipedia on Ferrous chloride, to quote:

The compound is highly soluble in water, giving pale green solutions.

Repeat the electrolysis experiment with distilled water and see if you get the same coloration issue.
[EDIT][EDIT] Interestingly, low solubility Fe(OH)2 is clear and only acquires a greenish tint in the presence of oxygen (per Wikipedia, to quote: "Iron(II) hydroxide is a white solid, but even traces of oxygen impart a greenish tinge.). Further, well-water is associated with high levels of bicarbonate and low O2 levels (see this significant work). To claim that Fe(HCO3)2 is at all green is without basis, as I also could not find any source anywhere asserting a coloration, other than that it is, in effect, 'clear' as was claimed above per the comment: 'Often called clear-water iron because it is clear when poured'. Theoretical formation of the usual hydration sphere for Fe(OH)2 interacting with bicarbonate in low solubility conditions with high HCO3-, low O2, significant CO2 presence (and its associated partial pressure) suggest reasons why it may not be a valid argument. If I am wrong, please provide a single source citing color, else it should be likely assumed that my supposition is correct and Fe(HCO3)2 is clear.  By the way, Atomistry on Ferrous bicarbonate, to quote:

Ferrous hydrogen carbonate or ferrous bicarbonate, FeH2(CO3)2, exists in solutions obtained by dissolving ferrous oxide or carbonate in water charged with excess of carbon dioxide. On exposure to air carbon dioxide is evolved, and a precipitate is obtained consisting essentially of ferric hydroxide.

which one could interpret as that clear-water iron, upon exiting the pipe, liberates CO2 and forms Fe(OH)2, which, with further oxygen exposure, may display a green tint (or, per the introduction of electrolysis formed chlorine).


Why is the cyclopropenium ion aromatic? I don't understand how the 'cyclopropenium' ion is aromatic. According to my understanding, the carbons within the central 3 carbon ring must each be sp2 hybridised with a single electron in a p orbital - since each carbon is bonded to 3 other atoms and has 4 valence electrons. How does that lead to aromaticity, since this doesn't fullfil Huckel's rule?

It does meet Huckel's rule.  When you have $m$ carbon atoms in a conjugated ring with a positive charge of $q$, there are $m-q$ pi electrons.  Here $m=3, q=1$ therefore the number of pi electrons is $m-q=2=4×0+2$.

A Huckel model calculation indicates that cyclopropenyl cation has roughly as much stabilization as benzene, but it's distributed over three instead of six carbon atoms.  Thus aromaticity is especially powerful in a cyclopropenyl cation ring.   This explains how a positive (formal) charge can be stabilized on as few as three carbon atoms and the ring holds together despite steric strain.

It's because the of the small size of the molecule. The hydrogen on the cation does not interfere with pi electrons orbiting around the cyclic molecule.


Why does having equivalent resonance structures give more stability? Although the phenolate ion has more resonance structures (4) compared to acetate ion (2), acetate is more stable because it has two equivalent resonance structures of same energy.

Why does having equivalent resonance structures give more stability? Shouldn't a compound with more resonance structures be more stable?

Preamble
One important thing to know is that what we call "resonance structure" is a byproduct of our chemical notation which can't describe the structure of some compounds effectively using only one chemical structure. Personally I find the old term mesomeric structure more appropriate (meso- Greek mésos in the middle;  -merism from Gk. merismos "dividing, partition") because resonance is already used for a phenomena, a real phenomena, in physics that has nothing to do with resonance in chemistry. Furthermore the compound is not continuously changing it's structure between the possible resonance structures, like the term resonance seems to suggest, but is a "normal steady" compound that simply has an electron distribution that cannot be described with a single chemical structure.
Why does having equivalent resonating structures give more stability?
The resonance energy is due to electron delocalization, so more the electrons are delocalized in the molecule higher is the resonance energy ( this means a decrease in the overall molecule energy). This is clear in the acetate ion - we can draw a resonance hybrid  like this:

Here the electrons are equally distributed (delocalized) between the two oxygen atoms.
However for some compounds the real structure is not always a structure right in the middle between the different resonance structures; the first assumption is true only if the two structures are two equivalent structures. So it can occur that the electrons are not well delocalized between all the atoms hence the resonance energy is lower and so the overall stability.
In the case of phenolate ion I think the first structure in the figure [2] gives a greater contribution to the final structure of the phenolate ion so the electrons are not well delocalized, but if I'm not wrong are mainly in the oxygen atom. Unfortunately  I can't find the energy of the different resonance structures.

Finally, however, I think that you should study case by case if the number of the resonance structures or their contributions are determinant to the overall resonance energy taking into account valency and electronegativity of the atoms involved.

You can estimate stability by the number of resonance structures you can draw. For the acetate ion, there are 2 resonance structures, while for the protonated form (acetic acid) only one structure is possible, which favors deprotonation. For the phenolate ion, you can formulate 4 structures, with the negative charge delocalized into the benzene ring. For phenol, you can draw 4 structures as well, although 3 of them have a positive formal charge at the oxygen. This is why phenol is weakly acidic, and quite acidic for an alcohol.

Edit: The content of this post can also help understanding the different acidities of phenol and acetic acid: Why is ethanoic acid more acidic than phenol?