IIT JEE 1996 Chemistry Question Paper with Answer and Solution

(B) According to Molecular Orbital Theory, bond order is given by $\frac{1}{2}(N_b - N_a)$.
For $N_2$ ($14$ electrons), the bond order is $3$. When it forms $N_2^+$, an electron is removed from a bonding molecular orbital, so the bond order decreases to $2.5$. Since bond distance is inversely proportional to bond order, the $N-N$ bond distance increases.
For $O_2$ ($16$ electrons), the bond order is $2$. When it forms $O_2^+$, an electron is removed from an antibonding molecular orbital ($\pi^*$), so the bond order increases to $2.5$. Consequently, the $O-O$ bond distance decreases.

(C) To identify isostructural pairs,we determine the hybridization and geometry of each species:
$1$. $NF_3$: $N$ is $sp^3$ hybridized with one lone pair,resulting in a pyramidal geometry.
$2$. $H_3O^+$: $O$ is $sp^3$ hybridized with one lone pair,resulting in a pyramidal geometry.
$3$. $NO_3^-$: $N$ is $sp^2$ hybridized with no lone pairs,resulting in a trigonal planar geometry.
$4$. $BF_3$: $B$ is $sp^2$ hybridized with no lone pairs,resulting in a trigonal planar geometry.
$5$. $HN_3$: This is hydrazoic acid,which has a linear/bent structure depending on resonance,but it is not isostructural with the others.
Thus,the isostructural pairs are $[NF_3, H_3O^+]$ (both pyramidal) and $[NO_3^-, BF_3]$ (both trigonal planar).
Therefore,option $C$ is correct.

(B) According to Graham's law of effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.
Since $r = \frac{V}{t}$,for the same volume $V$,we have $\frac{V}{t} \propto \frac{1}{\sqrt{M}}$,which implies $t \propto \sqrt{M}$.
Therefore,the ratio of times is $\frac{t_1}{t_2} = \sqrt{\frac{M_1}{M_2}}$.
Given $t_{H_2} = 5 \ s$ and $M_{H_2} = 2 \ g/mol$:
For $He$ $(M = 4)$: $t = 5 \sqrt{\frac{4}{2}} = 5\sqrt{2} \ s \approx 7.07 \ s$.
For $O_2$ $(M = 32)$: $t = 5 \sqrt{\frac{32}{2}} = 5 \times 4 = 20 \ s$.
For $CO$ $(M = 28)$: $t = 5 \sqrt{\frac{28}{2}} = 5\sqrt{14} \ s \approx 18.7 \ s$.
For $CO_2$ $(M = 44)$: $t = 5 \sqrt{\frac{44}{2}} = 5\sqrt{22} \ s \approx 23.45 \ s$.
Thus,the correct match is $20 \ s$ for $O_2$.

(C) The root mean square velocity $(U_{rms})$ is given by the formula $U_{rms} = \sqrt{\frac{3RT}{M}}$.
For $H_2$ at $50 \ K$: $U_{H_2} = \sqrt{\frac{3R \times 50}{2}}$.
For $O_2$ at $800 \ K$: $U_{O_2} = \sqrt{\frac{3R \times 800}{32}}$.
The ratio is $\frac{U_{H_2}}{U_{O_2}} = \sqrt{\frac{50}{2} \times \frac{32}{800}} = \sqrt{25 \times 0.04} = \sqrt{1} = 1$.

(A) $CO$ is a neutral oxide.
Neutral oxides do not react with either acids or bases.
Other examples of neutral oxides include $N_2O$ and $NO$.
$SnO_2$ and $ZnO$ are amphoteric oxides,while $SiO_2$ is an acidic oxide.

(C) The reaction is $KF + HF \to KHF_2$.
$KHF_2$ is an ionic compound consisting of the potassium cation $(K^{+})$ and the hydrogen difluoride anion $([HF_2]^{-})$.
In the solid state and in solution,it dissociates as $KHF_2 \to K^{+} + [HF_2]^{-}$.
Therefore,the correct species present are $K^{+}$ and $[HF_2]^{-}$.

(B) $CaC_2$ is calcium carbide,which contains the acetylide ion $[C \equiv C]^{2-}$.
In this ion,the two carbon atoms are connected by a triple bond.
$A$ triple bond consists of one sigma $(\sigma)$ bond and two pi $(\pi)$ bonds.

(D) $PbI_4$ is the least stable halide among the given options due to the following reasons:
$(I)$ The size of the iodine atom is very large,which makes the $Pb-I$ bond weak.
$(II)$ Due to the inert pair effect,the $+2$ oxidation state of $Pb$ is significantly more stable than the $+4$ oxidation state. Consequently,$PbI_4$ tends to decompose into $PbI_2$ and $I_2$.

(A) The acidic strength of hypohalous acids $(HOX)$ depends on the electronegativity of the halogen atom $(X)$.
As the electronegativity of the halogen increases,it pulls the electron density from the $O-H$ bond more effectively,making the $O-H$ bond weaker and facilitating the release of $H^+$ ions.
The electronegativity order of halogens is $Cl > Br > I$.
Therefore,the order of acidic strength is $ClOH > BrOH > IOH$,which corresponds to $I > II > III$.

(C) $CsBr_3$ is an ionic compound that crystallizes in a structure containing $Cs^{+}$ cations and $Br_3^{-}$ anions. The $Br_3^{-}$ ion is a linear tri-bromide ion.

(B) The orbital angular momentum of an electron is given by the formula: $\text{Angular momentum} = \sqrt{l(l + 1)} \frac{h}{2\pi}$.
For an $s$ orbital,the azimuthal quantum number $l = 0$.
Substituting $l = 0$ into the formula: $\text{Angular momentum} = \sqrt{0(0 + 1)} \frac{h}{2\pi} = 0$.
Therefore,the orbital angular momentum of an electron in an $s$ orbital is zero.

(B) $CaC_2$ is an ionic compound consisting of $Ca^{2+}$ and $C_2^{2-}$ ions.
The structure of the acetylide ion $[C_2]^{2-}$ is $[:C \equiv C:]^{2-}$.
In a triple bond,there is one sigma $(\sigma)$ bond and two pi $(\pi)$ bonds.

(A) Since $f$ is an even function,$f(-x) = f(x)$ for all $x \in (-5, 5)$.
Given $f(x) = f\left( \frac{x + 1}{x + 2} \right)$.
Since $f(x) = f(-x)$,we have $f(x) = f\left( \frac{-x + 1}{-x + 2} \right)$.
Case $1$: $x = \frac{-x + 1}{-x + 2} \Rightarrow -x^2 + 2x = -x + 1 \Rightarrow x^2 - 3x + 1 = 0$.
Solving for $x$,we get $x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$.
Case $2$: $-x = \frac{x + 1}{x + 2} \Rightarrow -x^2 - 2x = x + 1 \Rightarrow x^2 + 3x + 1 = 0$.
Solving for $x$,we get $x = \frac{-3 \pm \sqrt{9 - 4}}{2} = \frac{-3 \pm \sqrt{5}}{2}$.
Thus,the four values are $\frac{-3 - \sqrt{5}}{2}, \frac{-3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2}$.

(C) The total binding energy of the initial reactants (two deuterons) is: $2 \times (2 \times 1.112 \text{ MeV}) = 4.448 \text{ MeV}$.
The total binding energy of the final product (one $\alpha$-particle) is: $4 \times 7.047 \text{ MeV} = 28.188 \text{ MeV}$.
The energy $Q$ released in the reaction is the difference between the total binding energy of the products and the total binding energy of the reactants:
$Q = [\text{Total B.E. of products}] - [\text{Total B.E. of reactants}]$
$Q = 28.188 \text{ MeV} - 4.448 \text{ MeV} = 23.74 \text{ MeV}$.
Rounding to the nearest provided option,the energy released is $23.8 \text{ MeV}$.

(C) The binding energy per nucleon for deuteron $({}_1^2H)$ is $1.112 \, MeV$. Since it has $2$ nucleons,the total binding energy of one deuteron is $2 \times 1.112 \, MeV = 2.224 \, MeV$.
In the reaction ${}_1^2H + {}_1^2H \to {}_2^4He + Q$,the initial total binding energy is $2 \times (2.224 \, MeV) = 4.448 \, MeV$.
The $\alpha$-particle $({}_2^4He)$ has $4$ nucleons,and its binding energy per nucleon is $7.047 \, MeV$. Thus,the total binding energy of the final product is $4 \times 7.047 \, MeV = 28.188 \, MeV$.
The energy released $Q$ is the difference between the total binding energy of the products and the reactants: $Q = [B.E.]_{final} - [B.E.]_{initial}$.
$Q = 28.188 \, MeV - 4.448 \, MeV = 23.74 \, MeV$.
Rounding to the nearest provided option,$Q \approx 23.8 \, MeV$.

(D) The acidity of phenols is determined by the stability of the phenoxide ion formed after the loss of a proton.
Electron-withdrawing groups $(EWG)$ like $-NO_2$ increase acidity by stabilizing the negative charge through inductive $(-I)$ and resonance $(-M)$ effects.
Electron-donating groups $(EDG)$ like $-CH_3$ decrease acidity by destabilizing the negative charge through inductive $(+I)$ and hyperconjugation effects.
$(IV)$ $p$-Nitrophenol: The $-NO_2$ group at the para position exerts both $-I$ and $-M$ effects,making it the most acidic.
$(III)$ $m$-Nitrophenol: The $-NO_2$ group at the meta position exerts only the $-I$ effect,making it less acidic than the para isomer but more acidic than phenol.
$(I)$ Phenol: Acts as the reference compound.
$(II)$ $p$-Cresol: The $-CH_3$ group at the para position is an $EDG$,which decreases the acidity compared to phenol.
Therefore,the correct order of acidity is $IV > III > I > II$.

(C) The binding energy of a nucleus is given by the product of the number of nucleons and the binding energy per nucleon.
For the reactant side,we have two deuterons $(_{1}^{2}H)$:
Total binding energy of two deuterons $= 2 \times (2 \times 1.112 \, MeV) = 4 \times 1.112 \, MeV = 4.448 \, MeV$.
For the product side,we have one alpha particle $(_{2}^{4}He)$:
Total binding energy of one alpha particle $= 4 \times 7.047 \, MeV = 28.188 \, MeV$.
The energy released $Q$ in the fusion reaction is the difference between the total binding energy of the products and the total binding energy of the reactants:
$Q = (28.188 \, MeV) - (4.448 \, MeV) = 23.74 \, MeV$.
Rounding to the nearest provided option,we get $23.8 \, MeV$.

(B) The dipole moment $(mu)$ depends on the vector sum of individual bond dipoles.
$1$. For $p$-dichlorobenzene $(IV)$,the two $C-Cl$ bond dipoles are equal and opposite,so $\mu = 0 \ D$.
$2$. For toluene $(I)$,the methyl group is electron-donating,resulting in a small dipole moment $(\mu \approx 0.4 \ D)$.
$3$. For $m$-dichlorobenzene $(II)$,the angle between the two $C-Cl$ bonds is $120^{\circ}$,so $\mu = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2 \cos(120^{\circ})} = \mu_{C-Cl} \approx 1.48 \ D$.
$4$. For $o$-dichlorobenzene $(III)$,the angle between the two $C-Cl$ bonds is $60^{\circ}$,so $\mu = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2 \cos(60^{\circ})} = \sqrt{3} \mu_{C-Cl} \approx 2.54 \ D$.
Thus,the increasing order is $IV < I < II < III$.

(A) The speed of a transverse wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
According to Hooke's law,the tension $T$ in the string is proportional to the extension $x$,i.e.,$T = kx$.
Therefore,the speed $v$ is proportional to $\sqrt{x}$ (since $\mu$ remains constant),so $v \propto \sqrt{x}$.
Given the initial extension is $x$ and speed is $v$,and the new extension is $1.5x$ with new speed $v'$,we have:
$\frac{v'}{v} = \sqrt{\frac{1.5x}{x}} = \sqrt{1.5} \approx 1.22$.
Thus,the new speed is $v' = 1.22v$.

(D) The acidity of phenols depends on the stability of the phenoxide ion formed after the loss of a proton.
Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the negative charge,while electron-donating groups $(EDG)$ decrease acidity by destabilizing it.
$(I)$ Phenol: Reference compound.
$(II)$ $p$-Cresol: The $-CH_3$ group is an electron-donating group ($+I$ and hyperconjugation),which destabilizes the phenoxide ion,making it the least acidic.
$(III)$ $m$-Nitrophenol: The $-NO_2$ group is a strong $EWG$ ($-I$ and $-M$ effects). At the meta position,only the $-I$ effect operates.
$(IV)$ $p$-Nitrophenol: The $-NO_2$ group at the para position exerts both $-I$ and $-M$ effects,providing maximum stabilization to the phenoxide ion.
Comparing the effects: The $-M$ effect of $-NO_2$ at the para position is stronger than the $-I$ effect at the meta position.
Therefore,the order of acidity is: $(IV) > (III) > (I) > (II)$.

(D) $Mg^{2+}$ $(Z=12)$: $1s^2 2s^2 2p^6$ (No unpaired electrons).
$Ti^{3+}$ $(Z=22)$: $[Ar] 3d^1$ ($1$ unpaired electron).
$V^{3+}$ $(Z=23)$: $[Ar] 3d^2$ ($2$ unpaired electrons).
$Fe^{2+}$ $(Z=26)$: $[Ar] 3d^6$ ($4$ unpaired electrons).
Therefore,$Fe^{2+}$ has the maximum number of unpaired electrons.

(B) The dipole moment $(\mu)$ depends on the vector sum of individual bond dipoles.
$(IV)$ $p$-dichlorobenzene: The two $C-Cl$ bond dipoles are equal and opposite, so $\mu = 0$.
$(I)$ Toluene: The methyl group is electron-donating, creating a small dipole moment.
$(II)$ $m$-dichlorobenzene: The angle between the two $C-Cl$ bonds is $120^{\circ}$, resulting in a net dipole moment.
$(III)$ $o$-dichlorobenzene: The angle between the two $C-Cl$ bonds is $60^{\circ}$, resulting in the largest net dipole moment due to the smaller angle.
Thus, the correct order is: $IV < I < II < III$.

(B) Sodium thiosulphate $(Na_2S_2O_3)$ is prepared by boiling a solution of sodium sulphite $(Na_2SO_3)$ with powdered sulphur $(S)$ in an alkaline medium.
The chemical reaction is: $Na_2SO_3 + S \xrightarrow{\Delta} Na_2S_2O_3$.

(A) The solubility of metal sulfides in $HNO_3$ depends on the solubility product $(K_{sp})$ of the sulfide and the oxidizing power of the acid.
$HgS$ has an extremely low $K_{sp}$ value $(10^{-52})$,which makes it insoluble in hot dilute $HNO_3$.
$PbS$,$CuS$,and $CdS$ have higher $K_{sp}$ values compared to $HgS$ and dissolve in hot dilute $HNO_3$ to form their respective nitrates and sulfur or $H_2S$ (which is then oxidized).
Therefore,$HgS$ does not dissolve.

(C) When the mixture is heated with excess $Na_2O_2$ (a strong oxidizing agent),the following reactions occur:
$1$. $Fe^{2+}$ is oxidized to $Fe^{3+}$,which then precipitates as brown $Fe(OH)_3$.
$2$. $Al^{3+}$ forms soluble $[Al(OH)_4]^-$ (aluminate) in the filtrate.
$3$. $Cr^{3+}$ (from chrome alum) is oxidized to $CrO_4^{2-}$ (chromate),which is yellow and soluble in the filtrate.
Therefore,the filtrate contains $CrO_4^{2-}$ (yellow) and $[Al(OH)_4]^-$,while the residue contains $Fe(OH)_3$ (brown).

(B) In benzene,benzoic acid molecules undergo intermolecular hydrogen bonding to form dimers.
This association leads to a decrease in the number of particles in the solution.
Consequently,the observed molecular weight is higher than the theoretical value,which corresponds to the dimerization of benzoic acid.

(B) The atomic number $(Z)$ of $_{13}^{29}Al$ is $13$ and the number of neutrons $(n)$ is $29 - 13 = 16$.
The $\frac{n}{p}$ ratio is $\frac{16}{13} \approx 1.23$.
For light elements,the stable $\frac{n}{p}$ ratio is approximately $1$.
Since the $\frac{n}{p}$ ratio of $_{13}^{29}Al$ is higher than the stable ratio,it lies above the belt of stability.
To achieve stability,it converts a neutron into a proton by emitting a $\beta$-particle $(_{-1}^{0}e)$: $_{13}^{29}Al \rightarrow _{14}^{29}Si + _{-1}^{0}e$.

(B) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.
As $T \to \infty$,the term $\frac{E_a}{RT} \to 0$.
Therefore,the rate constant $k$ approaches the pre-exponential factor $A$ (Arrhenius parameter).
Given that the Arrhenius parameter $A = 6.0 \times 10^{14}\,s^{-1}$,the value of the rate constant as $T \to \infty$ is $6.0 \times 10^{14}\,s^{-1}$.

(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 reaction of copper sulphate with potassium cyanide proceeds in steps:
$1$. $CuSO_4 + 2KCN \to Cu(CN)_2 + K_2SO_4$
$2$. The unstable $Cu(CN)_2$ decomposes to give $Cu_2(CN)_2$ and cyanogen gas: $2Cu(CN)_2 \to Cu_2(CN)_2 + (CN)_2$
$3$. $Cu_2(CN)_2$ then reacts with excess $KCN$ to form the stable complex potassium tetracyanocuprate$(I)$: $Cu_2(CN)_2 + 6KCN \to 2K_3[Cu(CN)_4]$
Thus,the final product is $K_3[Cu(CN)_4]$.

(D) When $CuSO_4$ reacts with an excess of $KCN$ solution,it first forms a precipitate of $Cu(CN)_2$,which is unstable and decomposes to form $CuCN$ and $(CN)_2$.
$2CuSO_4 + 4KCN \to 2CuCN + (CN)_2 + 2K_2SO_4$
$CuCN$ then reacts with excess $KCN$ to form the stable complex $K_3[Cu(CN)_4]$.
$CuCN + 3KCN \to K_3[Cu(CN)_4]$
The overall reaction is:
$2CuSO_4 + 10KCN \to 2K_3[Cu(CN)_4] + (CN)_2 + 2K_2SO_4$
Thus,the final product is $K_3[Cu(CN)_4]$.

(D) The acidity of phenols depends on the nature of the substituents attached to the benzene ring.
$1$. Electron-withdrawing groups $(EWG)$ like $-NO_2$ increase acidity by stabilizing the phenoxide ion through inductive and resonance effects.
$2$. Electron-releasing groups $(ERG)$ like $-CH_3$ decrease acidity by destabilizing the phenoxide ion.
$3$. In $p$-nitrophenol $(IV)$,the $-NO_2$ group at the para position exerts both $-I$ and $-M$ effects,making it the most acidic.
$4$. In $m$-nitrophenol $(III)$,the $-NO_2$ group exerts only the $-I$ effect,making it less acidic than $p$-nitrophenol but more acidic than phenol.
$5$. Phenol $(I)$ is the reference compound.
$6$. In $p$-cresol $(II)$,the $-CH_3$ group exerts $+I$ and hyperconjugation effects,which decrease the acidity,making it the least acidic.
Therefore,the correct order of acidity is $IV > III > I > II$.