IIT JEE 1991 Chemistry Question Paper with Answer and Solution

(C) Since the given sulphate is isomorphous with $ZnSO_4 \cdot 7H_2O$,its formula is $MSO_4 \cdot 7H_2O$.
Let $m$ be the atomic weight of $M$.
The molecular weight of $MSO_4 \cdot 7H_2O = m + 32 + (4 \times 16) + 7 \times (2 \times 1 + 16) = m + 32 + 64 + 126 = m + 222$.
The percentage of $M$ is given by $\frac{m}{m + 222} \times 100 = 9.87$.
$100m = 9.87(m + 222)$.
$100m = 9.87m + 2191.14$.
$90.13m = 2191.14$.
$m = \frac{2191.14}{90.13} \approx 24.3$.

(D) The molar mass of $CuSO_4 \cdot 5H_2O$ is calculated as: $63.5 + 32 + (4 \times 16) + 5 \times (2 + 16) = 63.5 + 32 + 64 + 90 = 249.5 \ g/mol$.
$6.022 \times 10^{23}$ molecules of $CuSO_4 \cdot 5H_2O$ weigh $249.5 \ g$.
Therefore,the weight of $1 \times 10^{22}$ molecules is:
$\text{Weight} = \frac{249.5 \times 1 \times 10^{22}}{6.022 \times 10^{23}} \ g$.
$\text{Weight} = \frac{249.5}{60.22} \ g \approx 4.143 \ g$.
Since $4.143 \ g$ is not exactly $4.159 \ g$,the correct option is $D$.

(C) In the molecule $HC\equiv C-CH=CH_2$,let us analyze the hybridization of the carbon atoms involved in the single bond between the second and third carbons.
$1$. The first carbon is $sp$ hybridized due to the triple bond.
$2$. The second carbon is also $sp$ hybridized due to the triple bond.
$3$. The third carbon is $sp^2$ hybridized due to the double bond.
$4$. The fourth carbon is $sp^2$ hybridized due to the double bond.
$5$. The single bond in question is between the second carbon $(sp)$ and the third carbon $(sp^2)$.
Therefore,the hybridization of the carbon atoms in the $C-C$ single bond is $sp-sp^2$.

(A) The structures are as follows:
$NO_2^+$: $[O=N=O]^+$
$NCO^-$: $[N \equiv C-O]^-$
$CS_2$: $S=C=S$
In $CS_2$,the central carbon atom is $sp$-hybridised,resulting in a linear geometry.
$NCO^-$ and $NO_2^+$ are isoelectronic with $CS_2$ (each having $16$ valence electrons),and they also exhibit a linear geometry.

(C) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n}$.
For $K_p = K_c$,the value of $\Delta n$ must be $0$.
$\Delta n$ is defined as the difference between the sum of the stoichiometric coefficients of gaseous products and gaseous reactants.
For option $(C)$: $H_{2_{(g)}} + Cl_{2_{(g)}} \rightleftharpoons 2HCl_{(g)}$,$\Delta n = 2 - (1 + 1) = 0$.
Therefore,for this reaction,$K_p = K_c$.

(C) For the reaction,$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} Cl_{2(g)}$,the total number of moles of gaseous products is $2$ and reactants is $1$.
According to Le Chatelier's principle,for a reaction where the number of moles of gaseous products is greater than the number of moles of gaseous reactants $(\Delta n_g > 0)$:
$1$. Introducing an inert gas at constant volume does not change the partial pressures of the reacting species,so the equilibrium remains unaffected.
$2$. Introducing chlorine gas $(Cl_2)$ increases the concentration of products,which shifts the equilibrium in the backward direction.
$3$. Introducing an inert gas at constant pressure increases the volume of the container. Since the number of moles of products is greater than reactants,the system shifts towards the side with more moles to counteract the increase in volume,thus favouring the forward reaction.
$4$. Decreasing the volume of the container increases the total pressure,which shifts the equilibrium towards the side with fewer moles (backward direction).

(A) The relationship between enthalpy change $(\Delta H)$ and internal energy change ($\Delta U$ or $\Delta E$) is given by: $\Delta H - \Delta U = \Delta n_g RT$.
For the reaction $2C_6H_{6(l)} + 15O_{2(g)} \to 12CO_{2(g)} + 6H_2O_{(l)}$,the change in the number of moles of gaseous species is: $\Delta n_g = \sum n_{g(products)} - \sum n_{g(reactants)} = 12 - 15 = -3$.
Given $T = 25\,^{\circ}C = 298\,K$ and $R = 8.314\,J\,K^{-1}\,mol^{-1}$.
Substituting the values: $\Delta H - \Delta U = -3 \times 8.314 \times 298 = -7432.7\,J$.
Converting to $kJ$: $-7432.7\,J \approx -7.43\,kJ$.

(B) The reaction of $BaO_2$ with dilute $H_2SO_4$ is given by:
$BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$
In the products,the most electronegative element is oxygen.
In $H_2O_2$,oxygen is in the peroxide state,so its oxidation state is $-1$.
In $BaSO_4$,oxygen is in the oxide state,so its oxidation state is $-2$.
Therefore,the oxidation states are $-1$ and $-2$.

(C) An oxidising agent is a substance that can be reduced (its oxidation state decreases),and a reducing agent is a substance that can be oxidised (its oxidation state increases).
In $Al_2O_3$,the oxidation state of $Al$ is $+3$,which is its maximum possible oxidation state (group $13$).
Therefore,$Al$ cannot be further oxidised,meaning $Al_2O_3$ cannot act as a reducing agent.
In the other compounds,the central atom is in an intermediate oxidation state,allowing them to act as both oxidising and reducing agents.

(B) Strength of the solution is given by the formula: $\text{Strength} = \text{Normality} \times \text{Equivalent mass}$.
The equivalent mass of $H_2O_2$ is $\frac{\text{Molar mass}}{n\text{-factor}} = \frac{34}{2} = 17 \ g \ eq^{-1}$.
Given,$\text{Normality} = 1.5 \ N$ and $\text{Volume} = 1 \ L$.
$\text{Strength} = 1.5 \times 17 = 25.5 \ g \ L^{-1}$.
Since the volume is $1 \ L$,the amount of $H_2O_2$ present is $25.5 \ g$.

(A) The oxidation state of $S$ in $SO_2$ is $+4$.
The range of oxidation states for $S$ is from $-2$ to $+6$.
Since the oxidation state of $S$ in $SO_2$ $(+4)$ lies between its minimum $(-2)$ and maximum $(+6)$ oxidation states,it can either be oxidized to $+6$ or reduced to $-2$.
Therefore,$SO_2$ can act as both an oxidizing and a reducing agent.

(D) The cyclobutadienyl anion $({C_4}{H_4})^{2-}$ consists of a four-membered ring with two double bonds and two negative charges on the carbons.
Each double bond contributes $2$ $\pi$ electrons,totaling $4$ $\pi$ electrons.
The two negative charges each represent a lone pair of electrons in a $p$-orbital,which also participate in the $\pi$ system.
Therefore,total $\pi$ electrons $= 4 \text{ (from double bonds)} + 2 \times 2 \text{ (from negative charges)} = 8$ $\pi$ electrons.
Thus,the correct option is $D$.

(C) In $CH_3CN$ (acetonitrile),the nitrogen atom has a lone pair of electrons,which allows it to act as a nucleophile.
Additionally,the carbon atom of the cyano group $(-CN)$ is bonded to a highly electronegative nitrogen atom,making the carbon atom electron-deficient and susceptible to nucleophilic attack,thus allowing it to act as an electrophile.
Therefore,$CH_3CN$ exhibits both nucleophilic and electrophilic character.

(B) The bond length depends on the hybridization of the carbon atoms involved.
As the $s$-character increases,the bond length decreases.
$C_2H_2$ ($sp$ hybridization,$1.20 \ \mathring{A}$) < $C_2H_4$ ($sp^2$ hybridization,$1.34 \ \mathring{A}$) < $C_6H_6$ (delocalized $sp^2$ hybridization,$1.39 \ \mathring{A}$) < $C_2H_6$ ($sp^3$ hybridization,$1.54 \ \mathring{A}$).
Thus,the correct order is $C_2H_2 < C_2H_4 < C_6H_6 < C_2H_6$.

(C) To balance the equation $NO_3^- + 4H^+ + x e^- \to 2H_2O + NO$,we must balance both the atoms and the charge.
In $NO_3^-$,the oxidation state of $N$ is $+5$.
In $NO$,the oxidation state of $N$ is $+2$.
The change in oxidation state is $5 - 2 = 3$.
Therefore,$3$ electrons are required to reduce $N^{+5}$ to $N^{+2}$.
The balanced equation is $NO_3^- + 4H^+ + 3e^- \to 2H_2O + NO$.

(D) The structures of the given oxoanions are as follows:
$1$. $S_2O_3^{2-}$ (Thiosulphate): Contains an $S-S$ bond.
$2$. $S_2O_4^{2-}$ (Dithionite): Contains an $S-S$ bond.
$3$. $S_2O_5^{2-}$ (Pyrosulphite): Contains an $S-S$ bond.
$4$. $S_2O_7^{2-}$ (Pyrosulphate or Disulphate): The structure is $O_3S-O-SO_3^{2-}$,which contains an $S-O-S$ linkage,not an $S-S$ bond.
Therefore,the correct option is $D$.

(D) . In $S_2O_7^{2-}$ (pyrosulfate ion),the two sulfur atoms are linked via an oxygen atom ($S-O-S$ linkage). There is no direct $S-S$ bond.
The structures are as follows:
$1$. $S_2O_4^{2-}$: $O_2S-SO_2^{2-}$ (contains $S-S$ bond)
$2$. $S_2O_5^{2-}$: $O_2S-SO_3^{2-}$ (contains $S-S$ bond)
$3$. $S_2O_3^{2-}$: $S-SO_3^{2-}$ (contains $S-S$ bond)
$4$. $S_2O_7^{2-}$: $O_3S-O-SO_3^{2-}$ (contains $S-O-S$ bond,no $S-S$ bond)

(D) The thermal decomposition reactions for the given compounds are as follows:
$1$. $NH_4NO_2 \xrightarrow{\Delta} N_2 + 2H_2O$
$2$. $2NaN_3 \xrightarrow{\Delta} 2Na + 3N_2$
$3$. $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + Cr_2O_3 + 4H_2O$
Since all three compounds liberate $N_2$ gas upon thermal decomposition,the correct option is $D$.

(B) In the qualitative analysis of basic radicals,$NH_4Cl$ and $NH_4OH$ (aqueous $NH_3$) are used as group reagents for Group-$III$ cations.
$NH_4Cl$ suppresses the dissociation of $NH_4OH$ due to the common ion effect,thereby reducing the concentration of $OH^-$ ions.
This low concentration of $OH^-$ is sufficient to exceed the solubility product $(K_{sp})$ of $Al(OH)_3$,$Fe(OH)_3$,and $Cr(OH)_3$,causing them to precipitate.
Therefore,$Al^{3+}$ is precipitated by these reagents.

(C) $1$. In $Ni(CO)_4$,$Ni$ is in $0$ oxidation state. The configuration is $[Ar] 3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing of $4s$ electrons into $3d$ orbitals,resulting in $sp^3$ hybridization with no unpaired electrons. Thus,it is diamagnetic.
$2$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state. The configuration is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons in $3d$ orbitals,resulting in $dsp^2$ hybridization with no unpaired electrons. Thus,it is diamagnetic.
$3$. In $[NiCl_4]^{2-}$,$Ni$ is in $+2$ oxidation state. The configuration is $[Ar] 3d^8$. $Cl^-$ is a weak field ligand,so it does not cause pairing of electrons. This results in $sp^3$ hybridization with two unpaired electrons in $3d$ orbitals. Thus,it is paramagnetic.
Therefore,$Ni(CO)_4$ and $[Ni(CN)_4]^{2-}$ are diamagnetic,while $[NiCl_4]^{2-}$ is paramagnetic.

(A) The reaction is: $C_2H_5Br + AgNO_2 \rightarrow C_2H_5NO_2 + AgBr$.
$AgNO_2$ is a covalent compound where the $Ag-O$ bond is present.
However,due to the nature of the nucleophilic substitution,the attack occurs through the nitrogen atom because the $N$ atom is more nucleophilic than the $O$ atom in this specific covalent environment.
Thus,the major product formed is nitroethane $(C_2H_5NO_2)$.

(A) The reaction of $1-$chlorobutane with alcoholic $KOH$ is a dehydrohalogenation reaction (an elimination reaction).
In this reaction,a hydrogen atom is removed from the $\beta$-carbon and the chlorine atom is removed from the $\alpha$-carbon,resulting in the formation of an alkene.
$CH_3CH_2CH_2CH_2-Cl + KOH(\text{alc.}) \to CH_3CH_2CH=CH_2 + KCl + H_2O$.
The product formed is $1-$butene.

(D) $m$-chlorobenzaldehyde does not contain an $\alpha$-hydrogen atom. Therefore,it undergoes the Cannizzaro reaction in the presence of concentrated $KOH$.
In the Cannizzaro reaction,one molecule of the aldehyde is oxidized to the corresponding carboxylic acid salt,and another molecule is reduced to the corresponding alcohol.
Thus,$m$-chlorobenzaldehyde undergoes disproportionation to form potassium $m$-chlorobenzoate and $m$-chlorobenzyl alcohol.

(C) Aldol condensation requires the presence of at least one $\alpha$-hydrogen atom in the aldehyde or ketone molecule.
$1$. Benzaldehyde $(C_6H_5CHO)$ has no $\alpha$-hydrogen.
$2$. $2, 2-$dimethylpropionaldehyde $((CH_3)_3CCHO)$ has no $\alpha$-hydrogen.
$3$. Acetaldehyde $(CH_3CHO)$ has three $\alpha$-hydrogen atoms.
$4$. Formaldehyde $(HCHO)$ has no $\alpha$-hydrogen.
Therefore,only acetaldehyde can undergo base-catalyzed aldol condensation.