IIT JEE 1987 Chemistry Question Paper with Answer and Solution

(B) If two compounds have the same empirical formula but different molecular formula,they must have different molecular weights.
For example,$CH_2O$ (empirical formula) and $C_6H_{12}O_6$ (molecular formula) have the same empirical formula but different molecular formulas,and consequently,they have different molecular weights.

(D) The dipole moment of a molecule is the vector sum of the individual bond dipoles.
In $NH_3$ and $H_2O$,the molecules have a net dipole moment due to their bent or pyramidal geometry and the presence of lone pairs.
In $cis-1,2-dichloroethene$,the two $C-Cl$ bonds are on the same side,leading to a non-zero resultant dipole moment.
In $trans-1,2-dichloroethene$,the two $C-Cl$ bonds are in opposite directions,and the two $C-H$ bonds are also in opposite directions. The bond dipoles cancel each other out,resulting in a net dipole moment of zero.

(A) Hydrogen bonding is maximum in ethanol $(C_2H_5OH)$.
Hydrogen bonding occurs when an $H$ atom is covalently bonded to a highly electronegative atom like $N, O,$ or $F$.
In ethanol,the $H$ atom is bonded to an oxygen atom,which is highly electronegative,allowing for strong intermolecular hydrogen bonding.
In triethylamine $((C_2H_5)_3N)$,although $N$ is electronegative,there is no $H$ atom directly attached to the nitrogen atom,so it cannot form hydrogen bonds with itself.
Diethyl ether and ethyl chloride do not have an $H$ atom attached to $N, O,$ or $F$,so they do not exhibit hydrogen bonding.

(A) The atomic radius of fluorine is determined by its covalent radius,which is approximately $0.72 \ \mathring{A}$.
Neon is a noble gas and does not form covalent bonds under normal conditions,so its atomic radius is measured as the Van der Waal's radius,which is approximately $1.60 \ \mathring{A}$.
Since the Van der Waal's radius is always larger than the covalent radius,the atomic radius of neon is significantly larger than that of fluorine.
Therefore,the values are $0.72 \ \mathring{A}$ and $1.60 \ \mathring{A}$ respectively.

(C) The electronegativity values (Pauling scale) for the given elements are approximately: $Si (1.90) < P (2.19) < C (2.55) < N (3.04)$.
In the periodic table,electronegativity increases from left to right across a period and decreases down a group.
Comparing the elements: $Si$ and $P$ are in the $3^{rd}$ period,while $C$ and $N$ are in the $2^{nd}$ period.
Since $Si$ and $P$ are below $C$ and $N$,they have lower electronegativity values.
Within the same period,electronegativity increases from left to right ($Si < P$ and $C < N$).
Therefore,the correct order is $Si < P < C < N$.

(A) The electronic configuration of Nitrogen $(N)$ is $1s^2 2s^2 2p^3$,which has a stable half-filled $p$-orbital.
The electronic configuration of Oxygen $(O)$ is $1s^2 2s^2 2p^4$.
Due to the extra stability of the half-filled $2p$ subshell in Nitrogen,it requires more energy to remove an electron compared to Oxygen.
Therefore,the first ionization potential of Nitrogen $(14.6 \ eV)$ is higher than that of Oxygen $(13.6 \ eV)$.

(B) The metallic lustre exhibited by sodium is explained by the oscillation of loose electrons.
When light falls on the surface of sodium,the free electrons present in the metallic lattice start oscillating at their mean positions due to the interaction with the incident light.
These electrons get excited to higher energy levels and,upon returning to their lower energy levels,they emit light in all directions.
This phenomenon of light reflection and emission gives sodium its characteristic metallic lustre.

(B) The empirical formula represents the simplest whole-number ratio of atoms in a compound.
If two compounds have the same empirical formula but different molecular formulae,their molecular weights must be different because the molecular formula is an integer multiple of the empirical formula $(Molecular \ Formula = n \times Empirical \ Formula)$.
Since the molecular weights are proportional to the molecular formulae,they will differ accordingly.

(B) To name the compound $CH_2=CH-CH(CH_3)_2$ according to $IUPAC$ rules:
$1$. Identify the longest carbon chain containing the double bond. The longest chain has $4$ carbon atoms.
$2$. Number the chain starting from the end closer to the double bond. The double bond starts at carbon $1$.
$3$. The substituent (methyl group) is at position $3$.
$4$. Combining these,the name is $3-$methylbut$-1-$ene or $3-$methyl$-1-$butene.

(C) In the compound $N \equiv C - CH = CH_2$,the carbon atom $(1)$ is bonded to nitrogen by a triple bond,so it is $sp$ hybridised.
The carbon atom $(2)$ is bonded to carbon $(1)$ by a single bond and to the other carbon by a double bond,so it is $sp^2$ hybridised.
Therefore,the hybridisation of carbon $(1)$ and carbon $(2)$ is $sp$ and $sp^2$ respectively.

(B) $1.$ $CH_3-CH_2-CH_2-CH_2-CH_2-CH_3$ $(n-hexane)$
$2.$ $CH_3-CH(CH_3)-CH_2-CH_2-CH_3$ $(2-methylpentane)$
$3.$ $CH_3-CH_2-CH(CH_3)-CH_2-CH_3$ $(3-methylpentane)$
$4.$ $CH_3-CH(CH_3)-CH(CH_3)-CH_3$ $(2,3-dimethylbutane)$
$5.$ $CH_3-C(CH_3)_2-CH_2-CH_3$ $(2,2-dimethylbutane)$
There are a total of $5$ structural isomers for $C_6H_{14}$.

(A) The rotation about a $C-C$ bond is determined by the steric hindrance of the substituents attached to the carbon atoms.
$A$) $CH_3-CH_3$ (Ethane) has only small hydrogen atoms,allowing free rotation.
$B$) $CH_2=CH_2$ (Ethylene) has a double bond,which restricts rotation.
$C$) $CH \equiv CH$ (Acetylene) has a triple bond,which restricts rotation.
$D$) $C_2Cl_6$ (Hexachloroethane) has bulky chlorine atoms that cause significant steric hindrance,making rotation difficult.
Therefore,ethane has the least hindered rotation.

(B) The student can select at most $n$ books from $(2n + 1)$ books.
Let $T$ be the total number of ways to select at least one book.
$T = ^{2n+1}C_1 + ^{2n+1}C_2 + \dots + ^{2n+1}C_n = 63$.
We know that the sum of binomial coefficients is given by:
$^{2n+1}C_0 + ^{2n+1}C_1 + \dots + ^{2n+1}C_n + ^{2n+1}C_{n+1} + \dots + ^{2n+1}C_{2n+1} = 2^{2n+1}$.
Using the property $^{n}C_r = ^{n}C_{n-r}$,we have $^{2n+1}C_0 = ^{2n+1}C_{2n+1} = 1$ and $^{2n+1}C_1 = ^{2n+1}C_{2n}$,etc.
Thus,$^{2n+1}C_0 + 2(^{2n+1}C_1 + ^{2n+1}C_2 + \dots + ^{2n+1}C_n) = 2^{2n+1}$.
$1 + 2(T) = 2^{2n+1}$.
$1 + 2(63) = 2^{2n+1}$.
$1 + 126 = 2^{2n+1} \Rightarrow 127 = 2^{2n+1}$.
Wait,re-evaluating the problem statement: If the student selects at most $n$ books,the number of ways is $\sum_{k=1}^{n} {^{2n+1}C_k} = 63$.
Since $\sum_{k=0}^{2n+1} {^{2n+1}C_k} = 2^{2n+1}$,and $\sum_{k=0}^{n} {^{2n+1}C_k} = \sum_{k=n+1}^{2n+1} {^{2n+1}C_k} = \frac{1}{2} \times 2^{2n+1} = 2^{2n}$.
So,$\sum_{k=1}^{n} {^{2n+1}C_k} = 2^{2n} - ^{2n+1}C_0 = 2^{2n} - 1$.
Given $2^{2n} - 1 = 63$ $\Rightarrow 2^{2n} = 64$ $\Rightarrow 2^{2n} = 2^6$.
$2n = 6 \Rightarrow n = 3$.

(C) Colligative properties are those properties of a solution that depend only on the number of solute particles present in the solution,regardless of their nature.
There are four main colligative properties:
$1$. Relative lowering of vapour pressure.
$2$. Elevation in boiling point.
$3$. Depression in freezing point.
$4$. Osmotic pressure.
Among the given options,vapour pressure itself is a physical property,but the 'relative lowering of vapour pressure' is the colligative property. Therefore,'vapour pressure' is not a colligative property.

(B) In the brown ring complex $[Fe(H_2O)_5NO]SO_4$,the oxidation state of iron is calculated by considering the ligands.
Water $(H_2O)$ is a neutral ligand with a charge of $0$.
The nitric oxide $(NO)$ ligand in this specific complex acts as $NO^+$.
The sulfate ion $(SO_4^{2-})$ has a charge of $-2$.
Let the oxidation state of $Fe$ be $x$.
$x + 5(0) + 1(+1) + (-2) = 0$
$x + 1 - 2 = 0$
$x - 1 = 0$
$x = +1$.
Therefore,the oxidation state of iron is $+1$.

(A) $NF_3$ is the least basic among the nitrogen trihalides.
This is due to the high electronegativity of the three $F$ atoms,which strongly pull the electron density away from the nitrogen atom.
As a result,the lone pair of electrons present on the nitrogen atom is not easily available for donation to a Lewis acid.

(D) $N_2O$ (nitrous oxide) is a colourless gas.
$NO$ (nitric oxide) is a colourless gas.
$N_2O_5$ (dinitrogen pentoxide) is a colourless solid.
$NO_2$ (nitrogen dioxide) is a brown coloured gas.
Therefore,the correct option is $D$.

(D) When concentrated $H_2SO_4$ reacts with dry $KNO_3$,it produces $KHSO_4$ and $HNO_3$.
The $HNO_3$ produced decomposes upon heating to release brown fumes of nitrogen dioxide $(NO_2)$.
The reaction is: $2KNO_3 + H_2SO_4 \to 2KHSO_4 + 2NO_2 \uparrow + \frac{1}{2}O_2 \uparrow$ (Brown gas).

(B) The reaction of $n$-propyl bromide $(CH_3-CH_2-CH_2-Br)$ with ethanolic potassium hydroxide $(KOH)$ is a dehydrohalogenation reaction.
This is an elimination reaction ($E2$ mechanism) where the base removes a proton from the $\beta$-carbon and the bromide ion is eliminated.
$CH_3-CH_2-CH_2-Br + KOH \xrightarrow{C_2H_5OH} CH_3-CH=CH_2 + KBr + H_2O$.
The product formed is $Propene$.

(C) Colligative properties are those properties of solutions that depend only on the number of solute particles and not on their nature.
$A$,$B$,and $D$ are colligative properties.
$C$ Vapour pressure is not a colligative property; however,the relative lowering of vapour pressure is a colligative property.

(A) During the electrolysis of an aqueous solution of sodium sulphate $(Na_2SO_4)$,the ions present are $Na^+$,$SO_4^{2-}$,$H^+$,and $OH^-$.
At the cathode,$H^+$ ions are reduced in preference to $Na^+$ ions because the reduction potential of $H^+$ is higher than that of $Na^+$.
The reaction at the cathode is: $2H_2O(l) + 2e^- \to H_2(g) + 2OH^-(aq)$.
At the anode,$OH^-$ ions (or $H_2O$) are oxidized in preference to $SO_4^{2-}$ ions.
The reaction at the anode is: $2H_2O(l) \to O_2(g) + 4H^+(aq) + 4e^-$.
Therefore,the products at the cathode and anode are $H_2$ and $O_2$ respectively.

(D) The correct option is $(D)$.
During the discharge of a lead storage battery,the chemical reaction is:
$Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$
As shown in the equation,$H_2SO_4$ is consumed during the discharge process.

(C) Secondary alcohols on oxidation yield ketones.
$CH_3-CH(OH)-CH_2-CH_3$ ($2-$butanol) is a secondary alcohol.
Upon oxidation with an oxidizing agent like $KMnO_4$ or $K_2Cr_2O_7$,it forms $CH_3-CO-CH_2-CH_3$ (ethyl methyl ketone or butan$-2-$one).
The reaction is: $CH_3-CH(OH)-CH_2-CH_3 \xrightarrow{[O]} CH_3-CO-CH_2-CH_3 + H_2O$.