# Solved Question Papers - IIT JEE/ChemSol1996
## Q1 – Objective Questions
### Ideal Gas Mixtures

For a student who missed the tricky part of this question, it is going to be very confusing, because he would be left wondering if there is anything that has to be taken upto liquid helium temperature to be liquefied. The point here is that ideal gases are assumed to have zero intermolecular forces. No matter what the conditions are, they just do not have any IMF whatsoever to be liquefied! Check back the assumptions for ideal gas.

### Balmer Series

The Balmer series corresponds to the energy level transition from some orbit to n=2. The shortest wavelength would be the one with the most energetic transition – naturally, n  ∞
$v=R_{H}\left({\frac {1}{2^{2}}}-{\frac {1}{\infty ^{2}}}\right)={\frac {R_{H}}{4}}$
If we were to find the wavelength, using 1240/E (simplified equation of E = hc/λ) would’ve been advisable. 27419.25 how do you this find cm$^{-1}$

#### A little warning

The point here is, people mess up in calculation. They might use the latter equation and then take its reciprocal – this at a time when they just require plugging in the value of Rhydberg's constant from the question paper!
This is the point of such easy questions, people who just browse the theory and equations before an examination under an illusion that the numerical problems would require only value substitution will slog here! The following applies to the students studying Science at SCHOOL in a laid back environment.
Personally, I have met many many people who take the Phy/Chem textbook and start reading it, skipping even the easiest in-text problems! Remember, we skip those questions because we do not want to think about them, because it is painful to think and hinder your pace when you want to READ pages in a hurry. These are those students who have a hilarious criterion of what they understand and what they don’t! If they have heard some terms from a chapter and have an idea of what the topics are, they feel they're ready for the chapter in the examination. If they have never bothered to read it, they say they are yet to study it! In the end, each and every examination they answer is disastrous.
Now, isn't it funny? And believe me; a lot of such students are people who have scored 90+ percents in their SSC, signifying that they are quite capable of studying HSC. You make the choice. Why ever do you want to read a Physics/Chemistry chapter when you do not even want to think about it? And why do you not want to think about it? Is it because you do not want to know that you did not understand what you read just five minutes ago?
My personal advice is very simple – give a text one reading at the most before you move onto the exercises. The most productive part of your studying would be searching the textbook for details when you need them in exercises!

### Rate Constant

Just recall the Arrhenius equation. K = A(e)^(-Ea/RT) When T tends towards infinity, the exponential term will tend towards 1, giving the value of the rate constant at infinite temperature = A.

### Ratio of RMS velocities

Simple application of RMS = ${\sqrt {\frac {3RT}{M}}}$
Taking the ratio, The required answer is ${\sqrt {\frac {M_{\left(oxygen\right)}T_{\left(hydrogen\right)}}{T_{\left(oxygen\right)}M_{\left(hydrogen\right)}}}}=1$

#### Note

By the way, most people, including relatively better placed students; make a little mistake of substituting molar masses in gm/mol in the equation of RMS speed. Whether or not you convert it into Kg/Mol, it will not matter here (because its a ratio that we calculated) but will introduce a BIG error (if at all you can call it an error and not a mistake!) in the calculation of the absolute value.

### Orbital Angular Momentum

This is a trivial piece of information. The orbital angular momentum is ${\frac {{\sqrt {\left(l\right)}}\left(l+1\right)h}{2\pi }}$.
(nh/2π) is the angular momentum.
At this point of time, you might want to revise all the quantum numbers and their significance, including names etc. Such important but trivial info is very valuable for some rank deciding marks at the JEE. For example, checkout the Matrix Match question of JEE 2008 on quantum numbers. Not many people get it right because of lack of revision.

#### Note to editors

Write a note distinguishing the two quantities and supply convincing information, maybe from wikipedia.
The editor can be anyone – you, some professor or even me when I am free.

## Question 3 – On Organic Chemistry
### (i) Mechanism of dehydration of alcohol

For those who have a general degree of awareness in Organic Chemistry, the answer is protonation of alcohol, followed by generation of a 2 degree carbocation. This intermediate immediately changes to the more stable three degree carbocation by a methyl shift. This carbocation is the one which eliminates the beta Proton to give the alkene.
The reaction given is the generation of alkene in the presence of acid from an alcohol. More precisely, it is the dehydration of that alcohol by the acid.
The mechanism is very simple,
* First the alcohol is protonated and H2O leaves.
* Then, since the carbocation so formed has no nucleophile to react with, it eliminates a β proton to give the alkene. Now, the product that we would expect is 2,2dimethylbut-1-ene.
The produced product is not that. Infact, the actual product that is obtained has a different carbon skeleton than the alcohol! This is because the formed carbocation is 2 degree. This particular carbon skeleton permits the formation of a more stable carbocation (3 degree) through a methyl shift. This is the major point.
* β deprotonation occurs on the rearranged carbocation and we get the alkene.

#### Checkpoints

We made a little point that deprotonation occurs because the carbocation has no nucleophile to react with? But, isn't the lone pair on oxygen in some other alcohol molecule a nucleophile? Will it not react with the carbocation? You give us the answer. [Hint: Check the mechanisms of dehydration of alcohol, there are two possible products under different reaction conditions.]
Also, some people are of the opinion that a two degree carbocation does not form at all! that is to say, it rearranges as soon as it forms to give a three degree carbocation. Now, the logic conveyed is a very obvious one and cannot be explained in more simple words. The point is, can you visualize whatever that was said? Think of a transition state mechanism as in your SN2 reactions.

### Order of acidity

The order of acidity is decided qualitatively by the stability of the conjugate base. But the sad thing is, many many students do not even know what a conjugate acid/base pair is. Being unable to solve Ionic equilibrium questions is one thing and not knowing the basics is another!
The conjugate bases of the substituted phenols will all have a negative charge on the whole molecule. We say it is on the whole molecule and not on the oxygen atom because the anion is a resonance hybrid of various canonical structures with the negative charge on different positions. The resonance hybrid with maximum dispersion of the negative charge will be the most stable.
Warning: - The following explanation provides the most basic reasoning that sprouts from the actual discussion on resonance hybrid. It will be beneficial for those who study their textbooks.
* Phenoxide - draw the resonance structures and see, the negative charge is dispersed on the two ortho positions, the para position and the oxygen atom.
* Phenoxide with methyl on para position – Draw the resonance structure with negative charge on para position. The negative charge is on the carbon bonded to the methyl group, which is an electron donating group. This canonical structure is destabilized compared to the unsubstituted structure. So, its contribution is less towards the resonance hybrid, meaning that the resonance hybrid has less charge dispersion. 2<1 (acidity)
* m-nitrophenoxide – See below
* p-nitrophenoxide – In the resonance structure with negative charge para to oxygen, the nitro group withdraws the electron density (how? That's another resonance structure! -m effect), thus stabilizing that particular resonance structure. Contrast this with (2) and you'll realize that (4)>(1)>(2)
* m-nitrophenoxide – There is no resonance structure with the negative charge on the meta position, so NO2 cannot withdraw electron density through -m effect. But then, -I effect due to its electronegativity still operates, and stabilizes it more than the phenoxide ion, although weakly.
So, the answer is (4)>(3)>(1)>(2)

### Cannizarro reaction

Factual – it is the hydride ion transfer. Note carefully that a hydride ion is transferred and not a proton!
NCERT behaved very strangely and eliminated the mechanisms of Cannizarro and Aldol Condensation reactions and much more from its course. Remember that Organic Chemistry is the only subject whose syllabus for AIEEE/JEE demands greater detail than the CBSE syllabi. In fact, studying organic chem on the surface has made it dull and rote-based for the normal student!
The mechanism of the cannizarro reaction is: -
* Nucleophillic addition of the hydroxide ion on the carbonyl carbon.
* The anion becomes an -ate anion by eliminating a hydride ion and the acidic H.
* The hydride ion acts as a nucleophile for addition over some other carbonyl carbon, creating an alcohol.

#### From Wikipedia
### Hoffmann isocyanide test

Well, that says it. This is a direct question! ...Direct question even by board exam standards! However, since they demand a structure, you need to provide the location of bonds et-al. This is the only hurdle, but is not a big challenge if you have drawn the resonance structures of the cyanide ion some times.

#### From wikipedia
## Question 5 – On Ions
### Cell potential

Now THIS, is a very very interesting problem! Simple to the end but very intriguing for the untrained mind!
There are quite a few problems relating the solubility of a salt with some cell potential. Out of experience I can say that those problems do not have a good reputation! So, what happens when you see a similar one in an examination like the JEE? Loss of confidence, what else! Keep cool! Tough questions at the JEE are all about common sense. Let's see why.
We have been given the standard cell potential of the Cu2+/Cu redox couple and we are to find its cell potential at pH 14. We keep one fact in mind that cell potentials depend upon the concentration of reactants as given in the Nernst equation. But then, what has a 1 Molar concentration of hydroxide ions got to do with the concentration of cupric ions? EVERYTHING! Cupric hydroxide is a precipitate and the Q value of its precipitation cannot exceed its k$_{SP}$. In other words, cupric and hydroxide ions are in equilibrium with the precipitate of cupric hydroxide.
At equilibrium, from the k$_{SP}$ of cupric hydroxide, we get the concentration of cupric ions at pH14 to be 10$^{(-19)}$ M. Plug this value in the Nernst equation and get your answer. Some calculation is required, so you are not home yet, but still, you can celebrate the gain of 3 marks which not everybody would have been able to fetch in examination conditions!
The only point to be kept in mind is that the number of electrons involved in this redox half cell is 2, so n=2 in the Nernst equation. The rest is just calculation.
The answer is -0.22 V

#### A little doubt

Now this is a doubt on my part. Any student or teacher who has the answer is very much welcome to answer it here!
The question has asked us to find the cell potential at pH 14. Now we used the pH 14 data to come to the conclusion that the concentration of cupric ions is 10$^{(-19)}$. If we give it a thought, the concentration can be anything less than this particular value then why did we carry on solving the problem with that particular value? Is it some assumption regarding cell potential that requires saturated solution? Or is it something much more trivial?

### On Salt hydrolysis

This one is a direct question again. Make sure that you get it right anytime you do it!
Let h be the degree of hydrolysis and C be the concentration. Set up a reaction table,
* The equilibrium constant would then be ${\frac {Ch^{2}}{1-h}}$
This is equal to the Kb of the cyanide ions, that is 10$^{(-4.7)}$. Remember that the sodium ions are spectator ions in the hydrolysis. Oh sorry, do I have to mention that cyanide ions will be in equilibrium with HCN, creating hydroxide ions? (I also apologize for any arrogant text! That just happens to be my native writing style!)
Look at the equation carefully. The denominator cannot be greater than one, so if the fraction as of the order 1/100000, it is only because degree of hydrolysis of cyanide ions itself is very small. That is to say, h<<1, or 1-h is approx 1.
We carry on with Ch$^2$ = K$_b$
$h={\sqrt {\frac {K_{b}}{C}}}ch={\sqrt {C}}K_{b}$
The expression Ch is nothing but the concentration of hydroxide ions in water.
Just take the logarithms of the equation. We get $-pOH={\frac {\log C}{2}}{\frac {\log K_{b}}{2}}$
Giving due respect to the signs of pKb and logKb, we get pOH = 2.5, pH = 11.5

#### Let us laugh a little

The general trend in such questions is to specify the pKa of the conjugate acid, hydrocyanic acid in this case. Most people are then trained to carrying on the question through K$_a$ values – using K$_b$ = K$_w$/K$_a$.
So when I came across an anonymous (and published) solution of this question by a JEE tutor, I was not surprised that he had elongated this simple problem by four more steps in giving the relations between pKa, pKb and pKw.
$h={\sqrt {\frac {K_{w}}{CK_{a}}}}$
Then after the logarithm steps, he replaced pKa with 14 – pKb (and using about three lines to explain this). It would take very subtle observation to realize he was spinning round in circles.

#### Disclaimer

I mention it clearly that the above section was not written with immodest intentions or under a superiority complex. It is just a modest way to make people realize that even JEE specific classes do not guarantee quality education. Very few are competent enough and even fewer stick to the basics while nurturing open thought. BCPL, Kota is one of the best in this regard. Please keep in mind that this is just a personal opinion!

## Question 6 - Chemical Bonding
### Alkali Earth Metal Carbonates

Stability of these carbonates increases as the size of the AE Metal increases. So, IV < II < III < I. Many sources serves this statement as a fact, however I could not explain this on the basis of Fajan’s rule or otherwise. You are welcome to edit it here on wikibooks if you can provide a legitimate explanation.

### Structure of molecules

* NF$_3$ – Same as NH$_3$. It is a tetrahedron with one vertex occupied by a lone pair so that the overall molecule appears pyramidal.
* Nitrate ion – The central atom is nitrogen in sp$^2$ hybrid state. One oxygen is bonded through one σ and one π bond; the second one with a co-ordinate bond and the last one has a negative charge on it and is bonded by one σ bond.
* BF$_3$ – Of course sp$^2$ hybrid with trigonal planar structure!
* Hydronium ion – Three bond pairs and a lone pair, makes it the same as Ammonia. That is, sp$^3$ hybrid with pyramidal shape. Actually, water has two lone pairs. But when a hydronium ion is formed from it, one lone pair of oxygen is shared with the 1S orbital of a proton making it the third bond pair.
* Hydrazoic acid – Might give you the illusion of ammonia like structure – but however can one think of a sp$^3$ hybrid hydrogen? Consider the structure below. One might want to refer the structure of the tri-iodide ion at the moment as well!
The answer is C.

#### Note

Note carefully that the molecules all look similar to the untrained student! And it is not an exaggeration to say that even sincere 12th students will not be able to tell the difference between these similar molecules! Hybridization’s theory is one thing and practicing it over a lot of molecules is another! I mention this here solely to emphasize that practice is the key! Most people make the mistake of reading the same text again and again. Why don’t read it once and then solve all the problems at the end of the chapter? Alternately, why not grab some question papers and search for questions from the particular topic? I devote a lot of space in this book just to emphasize that practice allows a student to take decisions at a much better speed! DECISION MAKING allows a student to solve questions at a better speed and also to avoid conflicting thoughts while attempting questions!

### Calcium Carbide

On hydrolysis, CaC$_2$ gives ethyne. This is enough to tell that the anionic part of calcium carbide is the same as ethyne minus two protons. The answer is then B

### Cesium Tribromide

Alkali metals having a trivalent cation is an unheard statement for me. Tribromide ion is a possibility because we know I$_3$$^-$ ion does exist. We just discussed its structure above. B

# = Q7 – Organic Chemistry
## Q8 – Inorganic Chemistry
### Dipole moments

Benzene is a non polar molecule, because the vector sum of all of the bond dipoles in the molecule is zero. When one of the hydrogen is replaced by a methyl group, the resultant dipole moment is equivalent to a single C-H dipole moment in a methane molecule. Use your existing knowledge of vectors to come to this conclusion.
In p-dichlorobenzene all bond moments add up to zero. In o-dichlorobenzene, only two bond moments of antiparallel C-H bonds cancels. The resultant is a little less than vector sum of two C-Cl bond dipoles at 60$^o$. [Why? Don’t forget the C-H bonds antiparallel to the C-Cl bonds!] Similarly, in m-dichlorobenzene the resultant dipole moment is a little less than two C-Cl bond dipoles placed at 120$^o$.
So, Toluene>Benzene. In other words, toluene>p-dichlorobenzene. o-dichlorobenzene>m-dichlorobenzene. Amongst the meta isomer and toluene, the meta isomer has a higher dipole moment. [Why?]
B

#### Notes

Regarding ortho/para

### Hypo
## Q9 – Organic Chemistry
### 3-ethyl-2-pentene

HBr in the presence of peroxide. Br$_2$/H$_2$O OMDM

### Catalytic hydrogenation
## Q10 – Objectives again
### Electrolysis

Obtaining sodium out of the electrolysis of any aqueous sodium salt is out of the question? Why would the Hall Heroultz process be important if it was so easy to obtain sodium? S$_2$O$_8$$^{2-}$ is the reduction product of sulphate ions – so we cant obtain it at the anode. A is the answer then!
Note: - We require a better answer over here! Editors, please!

### Effusion

Stick to the basics! We’ll simply use the fact that the ratio of rates of effusion of two gases is proportional to the square root of the inverse ratio of molar masses!
$r_{\left(H_{2}\right)}={\frac {VolumeX}{Timetaken}}={\frac {X}{5sec}}$
We’ll write relative rates of effusion as ratio of X/(Time taken by gas) and equate it with that given by graham’s law.
So for example, checking it with Oxygen, ${\ce {{\frac {\frac {X}{T_{H}_{2}}}{\frac {X}{T_{O}_{2}}}}={\sqrt {\frac {M_{O}_{2}}{M_{H}_{2}}}}}}$
This gives us T$_{O2}$ = 20 seconds, which is correct. Answer is B. This was just an example, the QP expects you to verify all options before arriving to the correct one.