JEE Main 2021 Chemistry Question Paper with Answer and Solution

(B) Isostructural species have the same geometry and hybridization.
$A.$ $SO_{4}^{2-}$ and $CrO_{4}^{2-}$ are both tetrahedral ($sp^3$ hybridization).
$B.$ $SiCl_{4}$ and $TiCl_{4}$ are both tetrahedral ($sp^3$ hybridization).
$C.$ $NH_{3}$ is trigonal pyramidal $(sp^3)$,while $NO_{3}^{-}$ is trigonal planar $(sp^2)$.
$D.$ $BCl_{3}$ is trigonal planar $(sp^2)$,while $BrCl_{3}$ is $T$-shaped $(sp^3d)$.
Therefore,the isostructural pairs are $A$ and $B$.

(C) In equation $(A)$: $HOCl + H_2O_2 \rightarrow H_3O^{+} + Cl^{-} + O_2$. The oxidation state of $Cl$ changes from $+1$ in $HOCl$ to $-1$ in $Cl^{-}$. Thus,$HOCl$ is reduced and $H_2O_2$ acts as a reducing agent.
In equation $(B)$: $I_2 + H_2O_2 + 2 OH^{-} \rightarrow 2 I^{-} + 2 H_2O + O_2$. The oxidation state of $I$ changes from $0$ in $I_2$ to $-1$ in $I^{-}$. Thus,$I_2$ is reduced and $H_2O_2$ acts as a reducing agent.
Therefore,in both equations $(A)$ and $(B)$,$H_2O_2$ acts as a reducing agent.

(C) The reaction of $HI$ with $3,3$-dimethyl-$1$-butene $(CH_3-C(CH_3)_2-CH=CH_2)$ proceeds via a carbocation intermediate.
First,protonation of the alkene forms a secondary carbocation: $CH_3-C(CH_3)_2-C^+H-CH_3$.
To increase stability,a $1,2$-methyl shift occurs from the adjacent quaternary carbon,resulting in a more stable tertiary carbocation: $CH_3-C^+(CH_3)-CH(CH_3)-CH_3$.
Finally,the nucleophilic attack of the iodide ion $(I^-)$ on this tertiary carbocation yields the major product,$2$-iodo-$2,3$-dimethylbutane $(CH_3-CI(CH_3)-CH(CH_3)-CH_3)$.

(C) The controlled oxidation of alkanes to aldehydes is a specific industrial process.
The reaction $CH_{3}CH_{2}CH_{3} \stackrel{Mo_{2}O_{3}}{\longrightarrow} CH_{3}CH_{2}CHO$ involves the catalytic oxidation of propane to propanal.
The reagent used for this transformation is Molybdenum oxide $(Mo_{2}O_{3})$.

(D) The gas $CH_4$ (methane) is released during the anaerobic degradation of vegetation. $CH_4$ is a potent greenhouse gas that contributes to global warming,and some studies suggest it may be associated with health risks,including potential carcinogenic effects in specific contexts.

(A) $1$. The reaction of $1-$methylcyclopentene with dilute alkaline $KMnO_4$ at $273 \ K$ (Baeyer's reagent) is a syn-hydroxylation reaction,which adds two hydroxyl groups across the double bond to form $1-$methylcyclopentane$-1,2-$diol $(A)$.
$2$. The product $A$ is a vicinal diol. The secondary alcohol group is more easily oxidized than the tertiary alcohol group. Treatment with $CrO_3$ (Jones reagent or similar chromium-based oxidant) selectively oxidizes the secondary alcohol to a ketone,resulting in $2-$hydroxy$-2-$methylcyclopentanone $(B)$.

(C) The reaction is $Cl_{2(g)} \rightleftharpoons 2Cl_{(g)}$.
Let the number of moles of $Cl_2$ molecules and $Cl$ atoms be $n$ each.
The total number of moles in the mixture is $n + n = 2n$.
The mole fraction of $Cl_2$ is $X_{Cl_2} = \frac{n}{2n} = 0.5$.
The mole fraction of $Cl$ is $X_{Cl} = \frac{n}{2n} = 0.5$.
Given the total pressure $P = 1 \ atm$,the partial pressures are:
$P_{Cl_2} = X_{Cl_2} \times P = 0.5 \times 1 = 0.5 \ atm$.
$P_{Cl} = X_{Cl} \times P = 0.5 \times 1 = 0.5 \ atm$.
The equilibrium constant $K_P$ is given by:
$K_P = \frac{(P_{Cl})^2}{P_{Cl_2}} = \frac{(0.5)^2}{0.5} = 0.5$.
We are given $K_P = x \times 10^{-1}$,so $0.5 = x \times 10^{-1}$.
Therefore,$x = 5$.

(D) Amphoteric compounds are those that can react with both acids and bases.
$BeO$ (Beryllium oxide) is amphoteric.
$Be(OH)_2$ (Beryllium hydroxide) is amphoteric.
$BaO$ (Barium oxide) is basic.
$Sr(OH)_2$ (Strontium hydroxide) is basic.
Therefore,there are $2$ amphoteric compounds in the given list.

(C) The reaction is a disproportionation reaction of $S_8$ in an alkaline medium.
Reduction half-reaction: $S_8 + 16e^{-} \rightarrow 8S^{2-}$
Oxidation half-reaction: $S_8 + 12H_2O \rightarrow 4S_2O_3^{2-} + 24H^{+} + 16e^{-}$
Adding both half-reactions: $2S_8 + 12H_2O \rightarrow 8S^{2-} + 4S_2O_3^{2-} + 24H^{+}$
To balance in basic medium,add $24OH^{-}$ to both sides:
$2S_8 + 12H_2O + 24OH^{-} \rightarrow 8S^{2-} + 4S_2O_3^{2-} + 24H_2O$
Simplifying the equation:
$2S_8 + 24OH^{-} \rightarrow 8S^{2-} + 4S_2O_3^{2-} + 12H_2O$
Dividing by $2$ to get the simplest integer coefficients:
$S_8 + 12OH^{-} \rightarrow 4S^{2-} + 2S_2O_3^{2-} + 6H_2O$
Comparing this with the given equation,the value of '$a$' is $12$.

(C) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta_{r}G^{\circ} = -RT \ln K_p$
Given that $T = 300 \ K$,$K_p = 100.0$,and $\ln 10 = 2.3$.
Substituting the values: $\Delta_{r}G^{\circ} = -R \times 300 \times \ln(100)$
Since $\ln(100) = \ln(10^2) = 2 \ln(10) = 2 \times 2.3 = 4.6$.
Therefore,$\Delta_{r}G^{\circ} = -R \times 300 \times 4.6 = -1380 R$.
Comparing this with $-xR$,we get $x = 1380$.

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.
For a proton $(P)$ and $Li^{3+}$ nucleus accelerated by the same potential $V$:
$\frac{\lambda_{Li}}{\lambda_{P}} = \sqrt{\frac{m_{P} \times q_{P}}{m_{Li} \times q_{Li}}} = \sqrt{\frac{m_{P} \times e}{8.3 m_{P} \times 3e}} = \sqrt{\frac{1}{8.3 \times 3}} = \sqrt{\frac{1}{24.9}} \approx \sqrt{0.04016} \approx 0.2004$.
Given $\frac{\lambda_{Li}}{\lambda_{P}} = x \times 10^{-1}$,we have $0.2004 = x \times 10^{-1}$,so $x \approx 2.004$.
Rounding to the nearest integer,$x = 2$.

(C) .

Chemical Species	Bond Order
$He_{2}^{+}$	$0.5$
$He_{2}^{-}$	$0.5$
$Be_{2}$	$0$
$O_{2}^{2-}$	$1$

According to $M.O.T.$,if the bond order of a chemical species is zero,then that chemical species does not exist.

(B) In the reaction $2I^{-} + H_{2}O_{2} + 2H^{+} \rightarrow I_{2} + 2H_{2}O$,the oxidation state of iodine increases from $-1$ to $0$,which indicates that $I^{-}$ is oxidized to $I_{2}$ by $H_{2}O_{2}$.
Thus,$H_{2}O_{2}$ acts as an oxidizing agent.
In the other reactions,$H_{2}O_{2}$ acts as a reducing agent.
Hence,the correct answer is $(B)$.

(D) The reaction involves the aromatization of an alkane using $Mo_{2}O_{3}$ as a catalyst at $773 \ K$ and $10-20 \ atm$ pressure.
The reactant is $2$-methylhexane,which undergoes cyclization and dehydrogenation to form toluene $(A)$.

(C) The mass of carbon in $2.64 \, g$ of $CO_2$ is: $m_C = \frac{12}{44} \times 2.64 = 0.72 \, g$.
The mass of hydrogen in $1.08 \, g$ of $H_2O$ is: $m_H = \frac{2}{18} \times 1.08 = 0.12 \, g$.
The mass of oxygen in the compound is: $m_O = \text{Total mass} - (m_C + m_H) = 1.80 - (0.72 + 0.12) = 1.80 - 0.84 = 0.96 \, g$.
The percentage of oxygen is: $\% \, O = \frac{0.96}{1.80} \times 100 = 53.33 \%$.

(A) In $B_{2}H_{6}$,the terminal $B-H$ bonds are $sp^{3}$ hybridized,while the bridging $B-H-B$ bonds involve $3c-2e$ (three-center two-electron) bonds.
According to Bent's rule,more $p$-character is directed towards the more electronegative atom or the bond with a smaller bond angle.
Since the bond angle $\theta_{2}$ (terminal $H-B-H$) is greater than $\theta_{1}$ (bridging $B-H-B$ angle),the terminal $B-H$ bonds have more $s$-character and less $p$-character compared to the bridging bonds.
Thus,option $A$ is correct.
$BH_{3}$ is an electron-deficient species and acts as a Lewis acid,not a base.

(C) The number of radial nodes for an orbital is given by the formula: $\text{Number of radial nodes} = n - \ell - 1$.
For the $3s$ orbital,the principal quantum number $n = 3$ and the azimuthal quantum number $\ell = 0$.
Substituting these values: $\text{Number of radial nodes} = 3 - 0 - 1 = 2$.
$A$ radial distribution function plot for an orbital with $2$ radial nodes must show $2$ points where the probability density touches the $r$-axis (excluding $r=0$ and $r=\infty$).
Looking at the provided plots:
- Plot $(A)$ has $0$ radial nodes (corresponds to $1s$).
- Plot $(B)$ has $1$ radial node (corresponds to $2s$).
- Plot $(C)$ has $0$ radial nodes.
- Plot $(D)$ has $2$ radial nodes.
Therefore,the correct plot for the $3s$ orbital is $(D)$.

(D) Statement $I$ is false. Reducing smog (classical smog) is a mixture of smoke,fog,and sulphur dioxide $(SO_2)$. It does not involve an allotrope of oxygen as an intermediate.
Statement $II$ is true. Photochemical smog is formed by the action of sunlight on nitrogen oxides $(NO_x)$ and hydrocarbons present in the troposphere. Sulphur dioxide is primarily associated with reducing smog,not photochemical smog.
Therefore,Statement $I$ is false and Statement $II$ is true.

(B) The dissolution reaction is $AgCN_{(s)} + H^{+}_{(aq)} \rightleftharpoons Ag^{+}_{(aq)} + HCN_{(aq)}$.
The equilibrium constant $K_{eq} = \frac{K_{sp}(AgCN)}{K_a(HCN)} = \frac{2.2 \times 10^{-16}}{6.2 \times 10^{-10}} \approx 3.55 \times 10^{-7}$.
Since $s = [Ag^{+}] = [HCN]$ and $[H^{+}] = 10^{-pH} = 10^{-3}$,we have $K_{eq} = \frac{s^2}{[H^{+}]}$.
$s^2 = K_{eq} \times [H^{+}] = 3.55 \times 10^{-7} \times 10^{-3} = 3.55 \times 10^{-10}$.
$s = \sqrt{3.55 \times 10^{-10}} \approx 1.88 \times 10^{-5} \approx 1.9 \times 10^{-5}$.

(B) The balanced redox reaction is:
$8CrO_{4}^{2-} + 3S_{2}O_{3}^{2-} + 10H_{2}O \rightarrow 8Cr(OH)_{4}^{-} + 6SO_{4}^{2-} + 2OH^{-}$
From the stoichiometry,the $n$-factor for $CrO_{4}^{2-}$ ($Cr^{+6}$ to $Cr^{+3}$) is $3$,and for $S_{2}O_{3}^{2-}$ ($S^{+2}$ to $S^{+6}$),the total change in oxidation state is $2 \times (6-2) = 8$.
Using the law of equivalence: $n_{1}M_{1}V_{1} = n_{2}M_{2}V_{2}$
$3 \times 0.154 \times V = 8 \times 0.25 \times 40$
$0.462 \times V = 80$
$V = \frac{80}{0.462} \approx 173.16 \ mL$
Rounding to the nearest integer,we get $173 \ mL$.

(D) Assuming the volume of the tyre remains constant, according to Gay-Lussac's Law, $P \propto T$ (where $T$ is in Kelvin).
Given: $P_1 = 35 \ psi$, $T_1 = 27 + 273 = 300 \ K$, $P_2 = 40 \ psi$.
Using the formula $\frac{P_2}{P_1} = \frac{T_2}{T_1}$:
$\frac{40}{35} = \frac{T_2}{300}$
$T_2 = \frac{40 \times 300}{35} = \frac{12000}{35} \approx 342.86 \ K$.
Converting back to Celsius: $T_2(^{\circ} C) = 342.86 - 273 = 69.86^{\circ} C$.
Rounding to the nearest integer, we get $70^{\circ} C$.

(B) The reaction is: $NH_{2}CN_{(s)} + \frac{3}{2} O_{2(g)} \rightarrow N_{2(g)} + CO_{2(g)} + H_{2}O_{(l)}$
The change in the number of moles of gaseous species is $\Delta n_{g} = (n_{products, g} - n_{reactants, g}) = (1 + 1) - \frac{3}{2} = 2 - 1.5 = 0.5 \ mol$.
Using the relation $\Delta H = \Delta U + \Delta n_{g} RT$:
$\Delta H = -742.24 \ kJ \ mol^{-1} + (0.5 \ mol) \times (8.314 \times 10^{-3} \ kJ \ mol^{-1} K^{-1}) \times (298 \ K)$
$\Delta H = -742.24 + 1.238786 \approx -741.001 \ kJ \ mol^{-1}$.
The magnitude of $\Delta H$ is $741 \ kJ \ mol^{-1}$.

(C) The $R_{f}$ value is calculated using the formula:
$R_{f} = \frac{\text{Distance travelled by compound from base line}}{\text{Distance travelled by solvent from base line}}$
From the chromatogram:
Distance travelled by compound $A$ from the base line $= 2 \, cm$
Distance travelled by solvent front from the base line $= 5 \, cm$
Therefore,$R_{f} = \frac{2}{5} = 0.4$
Expressing this as $\times 10^{-1}$:
$0.4 = 4 \times 10^{-1}$
Thus,the value is $4$.

(D) The reaction is $Na_{(g)} + Br_{(g)} \rightarrow NaBr_{(s)}$.
According to the Born-Haber cycle,the enthalpy change for this reaction is given by the sum of the ionization enthalpy of $Na_{(g)}$,the electron gain enthalpy of $Br_{(g)}$,and the lattice enthalpy of $NaBr_{(s)}$.
$\Delta H = IE_{1} + \Delta H_{eg} + LE$
$\Delta H = 495.8 \ kJ \ mol^{-1} + (-325.0 \ kJ \ mol^{-1}) + (-728.4 \ kJ \ mol^{-1})$
$\Delta H = 495.8 - 325.0 - 728.4 = -557.6 \ kJ \ mol^{-1}$
To express this in the form $x \times 10^{-1} \ kJ \ mol^{-1}$,we have:
$-557.6 \ kJ \ mol^{-1} = -5576 \times 10^{-1} \ kJ \ mol^{-1}$
Thus,the value is $5576$.

(D) The given reactions are part of the Solvay process for the production of sodium carbonate.
$1$. $2NH_3 + H_2O(A) + CO_2 \rightarrow (NH_4)_2CO_3$
$2$. $(NH_4)_2CO_3 + H_2O + CO_2(B) \rightarrow 2NH_4HCO_3$
$3$. $NH_4HCO_3 + NaCl \rightarrow NaHCO_3(C) + NH_4Cl$
Comparing these with the given equations,we identify $A = H_2O$,$B = CO_2$,and $C = NaHCO_3$.

(D) The presence of ozone in the troposphere is harmful as it acts as a pollutant and generates photochemical smog.

(A) The electronic configurations correspond to the following elements:
$(a)$ $1s^{2} 2s^{2} \rightarrow Be$ (Beryllium)
$(b)$ $1s^{2} 2s^{2} 2p^{4} \rightarrow O$ (Oxygen)
$(c)$ $1s^{2} 2s^{2} 2p^{3} \rightarrow N$ (Nitrogen)
$(d)$ $1s^{2} 2s^{2} 2p^{1} \rightarrow B$ (Boron)
Ionization enthalpy ($\Delta_{i}H$) values are influenced by electronic stability:
$Be$ $(2s^{2})$ has a fully filled orbital,making its $IE$ higher than $B$ $(2s^{2} 2p^{1})$.
$N$ $(2p^{3})$ has a half-filled $p$-orbital,making its $IE$ higher than $O$ $(2p^{4})$.
The values are:
$N = 1402 \ kJ\ mol^{-1}$
$O = 1314 \ kJ\ mol^{-1}$
$Be = 899 \ kJ\ mol^{-1}$
$B = 801 \ kJ\ mol^{-1}$
Matching:
$(a) \rightarrow (ii)$
$(b) \rightarrow (iii)$
$(c) \rightarrow (iv)$
$(d) \rightarrow (i)$
Therefore,the correct option is $(a)$ $\rightarrow (ii), (b)$ $\rightarrow (iii), (c)$ $\rightarrow (iv), (d)$ $\rightarrow (i)$.

(A) Assertion $A$ is false. Hydrogen bonding is a special type of dipole-dipole interaction,but it is not the only non-covalent interaction that can be involved in molecular associations. Furthermore,the statement implies that only dipole-dipole interactions form hydrogen bonds,which is a restrictive definition.
Reason $R$ is true. Fluorine is indeed the most electronegative element. In the solid state,$HF$ molecules form a zig-zag chain structure where the hydrogen bonds are symmetrical,represented as $[F-H...F]^-$. The hydrogen atom is located symmetrically between two fluorine atoms.

(A) Statement $A$ is correct: Heavy water $(D_2O)$ is used in exchange reactions to study reaction mechanisms.
Statement $B$ is correct: Heavy water is prepared by the exhaustive electrolysis of ordinary water.
Statement $C$ is correct: The boiling point of $D_2O$ $(374.4 \ K)$ is higher than that of $H_2O$ $(373 \ K)$ due to stronger hydrogen bonding.
Statement $D$ is incorrect: The viscosity of $D_2O$ $(1.107 \ \text{centipoise})$ is greater than that of $H_2O$ $(0.89 \ \text{centipoise})$.
Therefore,statements $A, B,$ and $C$ are correct.

(D) The number of radial nodes is given by the formula $n-l-1$ and the number of angular nodes is given by $l$.
Given that radial nodes = $2$ and angular nodes = $2$,we have $l = 2$.
Substituting $l = 2$ into the radial node formula: $n-2-1 = 2$,which gives $n = 5$.
Since $l = 2$ corresponds to the $d$ orbital,the orbital is $5d$.

(D) Statement $I :$ The boiling point $(B.P.)$ of chloroform is $334 \ K$ and the $B.P.$ of aniline is $457 \ K$. Since there is a large difference in their boiling points,they can be separated by simple distillation. Thus,Statement $I$ is true.
Statement $II :$ In steam distillation,the mixture boils at a temperature lower than the boiling point of either component. Aniline is steam volatile and is separated from water by this method,where it boils at a temperature lower than its normal $B.P.$ of $457 \ K$. Thus,Statement $II$ is true.

(D) Step $1$: Dehydrohalogenation of $CH_{3}-CH=CHBr$ with $NaNH_{2}$ removes $HBr$ to form propyne $(CH_{3}-C \equiv CH)$.
Step $2$: Cyclic polymerization of $3$ moles of propyne in the presence of a red hot iron tube at $873 \text{ K}$ yields $1,3,5$-trimethylbenzene (mesitylene).
Therefore,the major product $A$ is $1,3,5$-trimethylbenzene.

(B) The $Carius$ method is primarily used for the estimation of halogens,sulphur,and phosphorus in organic compounds.
It is $NOT$ used for the estimation of nitrogen.
Nitrogen is typically estimated using the $Dumas$ method or the $Kjeldahl$ method.
Therefore,the statement that the $Carius$ method is used for the estimation of nitrogen is $FALSE$.

(D) $o$-Nitrophenol exhibits intramolecular hydrogen bonding,which reduces the intermolecular forces of attraction between its molecules. This makes it steam volatile.
Conversely,$p$-Nitrophenol exhibits intermolecular hydrogen bonding,leading to association of molecules and a higher melting point. $o$-Nitrophenol has a relatively low melting point compared to its isomers due to the lack of intermolecular hydrogen bonding.
Therefore,Statement $I$ is true and Statement $II$ is false.

(C) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G^{0}$ must be less than $0$.
$\Delta G^{0} = \Delta_{r} H^{0} - T \Delta_{r} S^{0} < 0$
Given $\Delta_{r} H^{0} = 80 \, kJ \, mol^{-1} = 80000 \, J \, mol^{-1}$ and $\Delta_{r} S^{0} = 2T \, J \, K^{-1} \, mol^{-1}$.
Substituting the values:
$80000 - T(2T) < 0$
$80000 - 2T^{2} < 0$
$2T^{2} > 80000$
$T^{2} > 40000$
$T > \sqrt{40000}$
$T > 200 \, K$
Therefore,the minimum temperature at which the reaction becomes spontaneous is $200 \, K$.

(B) The number given is $50000.020 \times 10^{-3}$.
According to the rules for significant figures:
$1$. All non-zero digits are significant.
$2$. Zeros between two non-zero digits are significant.
$3$. Trailing zeros in a number with a decimal point are significant.
In the number $50000.020$,the digits are $5, 0, 0, 0, 0, ., 0, 2, 0$.
All these $8$ digits are significant.
The exponential term $10^{-3}$ does not affect the number of significant figures.
Therefore,the total number of significant figures is $8$.

(C) The compressibility factor $Z$ is defined as $Z = \frac{PV_{m}}{RT}$.
Given the equation of state: $P(V_{m}-b) = RT$,we can write $PV_{m} - Pb = RT$.
Dividing by $RT$,we get $\frac{PV_{m}}{RT} - \frac{Pb}{RT} = 1$,which simplifies to $Z - \frac{Pb}{RT} = 1$,or $Z = 1 + \frac{Pb}{RT}$.
Now,we differentiate $Z$ with respect to $P$ at constant temperature $T$: $(\frac{\partial Z}{\partial P})_{T} = \frac{\partial}{\partial P}(1 + \frac{Pb}{RT}) = 0 + \frac{b}{RT} = \frac{1 \times b}{RT}$.
Comparing this with the given expression $\frac{xb}{RT}$,we find that $x = 1$.

(B) The reaction is $AB_{2(g)} \rightleftharpoons A_{(g)} + 2B_{(g)}$.
Initial moles: $AB_{2} = 1$,$A = 0$,$B = 0$.
Moles at equilibrium: $AB_{2} = 1-\alpha$,$A = \alpha$,$B = 2\alpha$.
Total moles at equilibrium $n_{T} = (1-\alpha) + \alpha + 2\alpha = 1 + 2\alpha$.
Using the ideal gas law $PV = nRT$ at equilibrium:
$1.9 \times 25 = n_{T} \times 0.0821 \times 300$.
$n_{T} = \frac{1.9 \times 25}{0.0821 \times 300} \approx 1.93$.
$1 + 2\alpha = 1.93 \implies 2\alpha = 0.93 \implies \alpha = 0.465$.
Equilibrium moles: $n_{AB_{2}} = 0.535$,$n_{A} = 0.465$,$n_{B} = 0.93$.
Partial pressures: $P_{i} = \frac{n_{i}}{n_{T}} \times P_{total}$.
$P_{AB_{2}} = \frac{0.535}{1.93} \times 1.9 = 0.526 \ atm$.
$P_{A} = \frac{0.465}{1.93} \times 1.9 = 0.458 \ atm$.
$P_{B} = \frac{0.93}{1.93} \times 1.9 = 0.916 \ atm$.
$K_{P} = \frac{P_{A} \times (P_{B})^{2}}{P_{AB_{2}}} = \frac{0.458 \times (0.916)^{2}}{0.526} \approx 0.73 \ atm^{2}$.
$K_{P} = 73 \times 10^{-2}$. Thus,$x = 73$ (closest integer option is $74$ due to rounding variations).

(D) The reaction of dichromate ion with a base is given by: $Cr_{2}O_{7}^{2-} + 2OH^{-} \longrightarrow 2CrO_{4}^{2-} + H_{2}O$.
In the product chromate ion $(CrO_{4}^{2-})$,let the oxidation state of $Cr$ be $x$.
$x + 4(-2) = -2$
$x - 8 = -2$
$x = +6$.
Thus,the oxidation number of $Cr$ in the product is $6$.

(D) According to Bohr's theory :
$A$. $KE = 13.6 \frac{Z^{2}}{n^{2}} \ eV/atom \Rightarrow KE \propto \frac{Z^{2}}{n^{2}}$ (Correct)
$B$. Speed of $e^{-} \propto \frac{Z}{n} \Rightarrow v \times n \propto Z$ (Incorrect)
$C$. Frequency of revolution of $e^{-} = \frac{v}{2 \pi r}$. Since $v \propto \frac{Z}{n}$ and $r \propto \frac{n^{2}}{Z}$,then Frequency $\propto \frac{Z/n}{n^{2}/Z} = \frac{Z^{2}}{n^{3}}$ (Incorrect)
$D$. $F = \frac{kZe^{2}}{r^{2}}$. Since $r \propto \frac{n^{2}}{Z}$,then $F \propto \frac{Z}{(n^{2}/Z)^{2}} = \frac{Z^{3}}{n^{4}}$ (Correct)
Thus,statements $A$ and $D$ are correct.

(D) Alkali metal salts impart characteristic colors to the flame due to the excitation of electrons to higher energy levels and their subsequent return to the ground state.
The characteristic wavelengths for these metals are:
$Li$: $670.8 \text{ nm}$ (Crimson Red)
$Na$: $589.2 \text{ nm}$ (Yellow)
$Rb$: $780.0 \text{ nm}$ (Red-violet)
$Cs$: $455.5 \text{ nm}$ (Blue)
Therefore,the correct matching is: $(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)$.

(C) The hydration of alkynes in the presence of $HgSO_4$ and $H_2SO_4$ follows Markovnikov's rule.
$1$. $HC \equiv CH + H_2O \xrightarrow{HgSO_4, H_2SO_4} CH_2=CH(OH) \rightleftharpoons CH_3-CHO$ (Ethanal).
$2$. Cyclohexylacetylene + $H_2O \xrightarrow{HgSO_4, H_2SO_4} \text{Cyclohexyl methyl ketone}$.
$3$. $CH_3-C \equiv CH + H_2O \xrightarrow{HgSO_4, H_2SO_4} CH_3-C(OH)=CH_2 \rightleftharpoons CH_3-CO-CH_3$ (Propanone).
$4$. $CH_3-C \equiv C-CH_3 + H_2O \xrightarrow{HgSO_4, H_2SO_4} CH_3-C(OH)=CH-CH_3 \rightleftharpoons CH_3-CO-CH_2-CH_3$ (Butan$-2-$one).
$CH_3-CH_2-CHO$ (Propanal) is an aldehyde that cannot be formed by the hydration of any alkyne because the hydration of terminal alkynes (other than ethyne) always yields a ketone due to the formation of a more stable secondary enol intermediate.

(A) To determine the aromaticity of a compound,we use $H$ückel's rule,which states that a planar,cyclic,fully conjugated system with $(4n+2) \pi$ electrons is aromatic,while one with $4n \pi$ electrons is antiaromatic. Nonaromatic compounds are those that do not satisfy these conditions (e.g.,they may not be planar or fully conjugated).
$1$. Cyclooctatetraene: It has $8 \pi$ electrons. It is not planar (it adopts a tub-shaped conformation) to avoid antiaromaticity,making it nonaromatic.
$2$. Furan: It is a cyclic,planar,fully conjugated system with $6 \pi$ electrons (including the lone pair on oxygen),making it aromatic.
$3$. Cyclobutadiene dication: It has $2 \pi$ electrons $(n=0)$,making it aromatic.
$4$. Anthracene: It is a polycyclic aromatic hydrocarbon with $14 \pi$ electrons,making it aromatic.
Therefore,cyclooctatetraene is nonaromatic.

(C) Statement $I$ is true because $BOD$ is a measure of the amount of dissolved oxygen required by bacteria to decompose organic matter in water,which directly impacts aquatic life.
Statement $II$ is false because clean water typically has a $BOD$ value of less than $5 \ ppm$,while highly polluted water has a $BOD$ value of $17 \ ppm$ or more.
There is no single 'optimum' value of $6.5 \ ppm$ defined for $BOD$.

(D) In $I_{3}^{-}$ ion,the central iodine atom has $7$ valence electrons. It forms $2$ bonds with other iodine atoms and has $3$ lone pairs of electrons.
Total electron pairs = $2$ (bond pairs) + $3$ (lone pairs) = $5$.
According to $VSEPR$ theory,the hybridization is $sp^{3}d$ and the electron geometry is trigonal bipyramidal.
The $3$ lone pairs occupy the equatorial positions to minimize repulsion.
Therefore,the $2$ iodine atoms occupy the axial positions,resulting in a linear shape with an $I-I-I$ bond angle of $180^{\circ}$.

(D) Assertion $A$ is true because hydrogen is indeed the most abundant element in the universe,but nitrogen $(N_2)$ is the most abundant gas in the Earth's troposphere.
Reason $R$ is true because hydrogen is the lightest element with atomic number $Z = 1$.
However,the fact that hydrogen is the lightest element is not the reason why it is not the most abundant gas in the troposphere. The low abundance of hydrogen in the atmosphere is due to its low density and high reactivity,which allows it to escape the Earth's gravitational pull.
Therefore,both $A$ and $R$ are true,but $R$ is not the correct explanation of $A$.

(A) The melting point $(M.P.)$ of ionic compounds is directly proportional to their lattice energy $(L.E.)$.
Lattice energy depends on the charge of the ions and the distance between them $(L.E. \propto \frac{q_1 q_2}{r_+ + r_-})$.
For the pair $LiF$ and $LiCl$,$LiF$ has a smaller anion $(F^-)$ compared to $Cl^-$,resulting in a shorter interionic distance and higher lattice energy,so $LiF > LiCl$.
For the pair $MgO$ and $NaCl$,$MgO$ consists of $Mg^{2+}$ and $O^{2-}$ ions (charges $\pm 2$),while $NaCl$ consists of $Na^+$ and $Cl^-$ ions (charges $\pm 1$). Higher charges in $MgO$ lead to significantly higher lattice energy,so $MgO > NaCl$.
Therefore,the correct order is $LiF > LiCl$ and $MgO > NaCl$.

(C) The combustion reaction for a hydrocarbon $C_{x}H_{y}$ is given by:
$C_{x}H_{y(g)} + (x + \frac{y}{4}) O_{2(g)} \rightarrow x CO_{2(g)} + \frac{y}{2} H_{2}O(\ell)$
According to Avogadro's Law,the volume ratio is equal to the stoichiometric coefficient ratio.
Given that $1$ volume of $C_{x}H_{y}$ produces $4$ volumes of $CO_{2}$,we have $x = 4$.
Given that $1$ volume of $C_{x}H_{y}$ requires $6$ volumes of $O_{2}$,we have $(x + \frac{y}{4}) = 6$.
Substituting $x = 4$ into the equation:
$4 + \frac{y}{4} = 6$
$\frac{y}{4} = 2$
$y = 8$.

(A) The molar mass of acetylene $(C_2H_2)$ is $26 \, g \, mol^{-1}$.
Number of moles $(n)$ $= \frac{4.75 \, g}{26 \, g \, mol^{-1}} \approx 0.1827 \, mol$.
Temperature $(T)$ $= 50 + 273 = 323 \, K$.
Pressure $(P)$ $= \frac{740}{760} \, atm \approx 0.9737 \, atm$.
Using the ideal gas equation $PV = nRT$,we get $V = \frac{nRT}{P}$.
$V = \frac{0.1827 \times 0.0826 \times 323}{0.9737} \approx 5.0059 \, L$.
Rounding off to the nearest integer,we get $5 \, L$.

(D) The solubility product of $PbI_{2}$ is given by $K_{sp} = [Pb^{2+}][I^{-}]^{2} = 8.0 \times 10^{-9}$.
In $0.1 \ M$ $Pb(NO_{3})_{2}$ solution,$Pb(NO_{3})_{2}$ dissociates completely as $Pb(NO_{3})_{2} \rightarrow Pb^{2+} + 2NO_{3}^{-}$.
Thus,the concentration of $Pb^{2+}$ ions from $Pb(NO_{3})_{2}$ is $0.1 \ M$.
Let the solubility of $PbI_{2}$ be $s \ mol/L$. The dissociation of $PbI_{2}$ is $PbI_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2I^{-}(aq)$.
The total concentration of $Pb^{2+}$ is $(0.1 + s) \approx 0.1 \ M$ (since $s$ is very small) and $[I^{-}] = 2s$.
Substituting these into the $K_{sp}$ expression: $8.0 \times 10^{-9} = (0.1)(2s)^{2}$.
$8.0 \times 10^{-9} = 0.1 \times 4s^{2} = 0.4s^{2}$.
$s^{2} = \frac{8.0 \times 10^{-9}}{0.4} = 20 \times 10^{-9} = 2.0 \times 10^{-8}$.
$s = \sqrt{2.0 \times 10^{-8}} = \sqrt{2} \times 10^{-4} = 1.41 \times 10^{-4} = 141 \times 10^{-6} \ mol/L$.
Comparing this with $x \times 10^{-6} \ mol/L$,we get $x = 141$.

(C) $1$. The reaction of aniline with $NaNO_2/HCl$ at $0-5^{\circ}C$ is a diazotization reaction,which produces benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$.
$2$. The reaction of benzene diazonium chloride with $KCN$ (in the presence of $CuCN$) yields benzonitrile $(C_6H_5CN)$. Thus,$(A)$ is benzonitrile.
$3$. The reduction of benzonitrile with $SnCl_2/HCl$ followed by hydrolysis $(H_3O^+)$ is the Stephen reduction,which converts the nitrile group $(-CN)$ into an aldehyde group $(-CHO)$. Thus,$(B)$ is benzaldehyde.
$4$. Therefore,$(A)$ is benzonitrile and $(B)$ is benzaldehyde.

(A) The concentration of sulphide ores is often carried out by the froth flotation process.
In this process,depressants are used to selectively prevent one sulphide ore from forming froth.
For an ore containing $ZnS$ (Sphalerite) and $PbS$ (Galena),sodium cyanide $(NaCN)$,which is a group $1$ cyanide salt,is used as a depressant.
$NaCN$ selectively prevents $ZnS$ from coming to the froth by forming a complex,$Na_{2}[Zn(CN)_{4}]$,while $PbS$ forms froth.
Therefore,$ZnS$ (Sphalerite) is the ore associated with this process.

(B) In the Bayer's process,$Al_{2}O_{3}$ is leached with $NaOH$ to form sodium aluminate,$X = Na[Al(OH)_{4}]$.
Passing $CO_{2}$ gas $(Y)$ through the solution of $X$ results in the precipitation of hydrated alumina,$Z = Al_{2}O_{3} \cdot xH_{2}O$.
The chemical reactions are:
$Al_{2}O_{3}(s) + 2NaOH(aq) + 3H_{2}O(l) \rightarrow 2Na[Al(OH)_{4}](aq)$
$2Na[Al(OH)_{4}](aq) + CO_{2}(g) \rightarrow Al_{2}O_{3} \cdot xH_{2}O(s) + 2NaHCO_{3}(aq)$

(A) The reaction of a secondary alcohol with $HCl$ proceeds via an $S_N1$ mechanism involving the formation of a carbocation intermediate.
$1$. Protonation of the $-OH$ group by $HCl$ converts it into a good leaving group $(-OH_2^+)$.
$2$. Loss of water molecule generates a secondary carbocation at the carbon atom attached to the cyclohexane ring.
$3$. This secondary carbocation undergoes a $1,2$-hydride shift to form a more stable tertiary carbocation at the ring carbon.
$4$. Finally,the nucleophilic chloride ion $(Cl^-)$ attacks the tertiary carbocation to form the major product,$1$-ethyl-$1$-chloro-$2$-methylcyclohexane.

(C) In the presence of concentrated $H_2SO_4$,aniline is protonated to form the anilinium ion $(-NH_3^+)$.
The $-NH_3^+$ group is strongly electron-withdrawing due to its $-I$ effect and is meta-directing.
Therefore,during the nitration of aniline,a significant amount of the meta-nitro product is formed due to the presence of the anilinium ion.

(D) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = K(P)^{1/n}$
Taking the logarithm on both sides:
$\log \left(\frac{x}{m}\right) = \log K + \frac{1}{n} \log P$
This equation is in the form of a straight line $y = mx + c$,where:
$y = \log \left(\frac{x}{m}\right)$
$x = \log P$
$c = \log K$
$m = \frac{1}{n}$ (slope)
In the Freundlich adsorption isotherm,the value of $\frac{1}{n}$ ranges between $0$ and $1$ (i.e.,$0 \leq \frac{1}{n} \leq 1$).
Thus,the slope of the line is $\frac{1}{n}$ where $\frac{1}{n}$ lies between $0$ and $1$.

(B) Statement $I$ is false because cupric metaborate $(Cu(BO_2)_2)$ is blue in color,not colourless.
Statement $II$ is false because heating boric anhydride $(B_2O_3)$ and copper sulphate $(CuSO_4)$ in a non-luminous (oxidizing) flame produces cupric metaborate $(Cu(BO_2)_2)$,which is blue. Cuprous metaborate $(CuBO_2)$ is formed in a luminous (reducing) flame.

(B) The $\alpha$-helix structure of proteins is stabilized by intramolecular hydrogen bonding between the carbonyl oxygen $(C=O)$ of one amino acid residue and the amide hydrogen $(N-H)$ of the fourth amino acid residue along the polypeptide chain.

(C) Caprolactum is the monomeric unit of polymer Nylon$-6$.
$(b)$ $2-$Chloro$-1,3-$butadiene is the monomeric unit of polymer Neoprene.
$(c)$ Isoprene ($2-$methylbuta$-1,3-$diene) is the monomeric unit of Natural rubber.
$(d)$ Acrylonitrile $(CH_2=CH-CN)$ is one of the monomeric units of polymer Buna$-N$.
Therefore,the correct matching is: $(a)$ $\rightarrow (iii), (b)$ $\rightarrow (iv), (c)$ $\rightarrow (i), (d)$ $\rightarrow (ii)$.

(B) The major components in "Gun Metal" are:
$Cu : 87 \%$
$Sn : 10 \%$
$Zn : 3 \%$
Thus,the correct composition is $Cu, Sn$ and $Zn$.

(D) The electrode potential $E^{\Theta}$ for the reduction of $M^{2+}$ to $M$ in the $3d$-series is generally negative,except for copper.
Copper $(Cu)$ has a positive standard electrode potential $(E^{\Theta} = +0.34 \ V)$ because the high energy required to transform $Cu(s)$ to $Cu^{2+}(aq)$ (sum of enthalpy of atomization and ionization enthalpy) is not compensated by its hydration enthalpy.
Therefore,the correct option is $D$.

(A) The reaction of phenols with phthalic anhydride in the presence of concentrated $H_2SO_4$ is a condensation reaction that forms phthalein dyes.
For a phenol to form a phthalein dye (which shows a pink colour in basic medium),it must have a free para-position relative to the hydroxyl $(-OH)$ group to allow for the electrophilic aromatic substitution $(EAS)$ by the phthalic anhydride intermediate.
In $2$-propylphenol,the para-position is free,allowing the reaction to proceed to form the corresponding phthalein derivative,which turns pink upon treatment with $NaOH$.

(C) Molar mass of $ClCH_2COOH = 94.5 \text{ g mol}^{-1}$.
Molality $(m) = \frac{9.45 \text{ g}}{94.5 \text{ g mol}^{-1} \times 0.5 \text{ kg}} = 0.2 \text{ m}$.
Using the formula $\Delta T_f = i \cdot K_f \cdot m$:
$0.5 = i \times 1.86 \times 0.2$.
$i = \frac{0.5}{0.372} \approx 1.344$.
For the dissociation $ClCH_2COOH \rightleftharpoons ClCH_2COO^- + H^+$,the van't Hoff factor is $i = 1 + \alpha$.
$\alpha = i - 1 = 0.344$.
The dissociation constant $K_a = \frac{C\alpha^2}{1-\alpha} = \frac{0.2 \times (0.344)^2}{1 - 0.344} = \frac{0.2 \times 0.118336}{0.656} \approx 0.0361 = 36.1 \times 10^{-3}$.
Thus,$x \approx 36$.

(B) The number of moles of compound $A$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{4.5 \ g}{90 \ g/mol} = 0.05 \ mol$.
The volume of the solution in liters is: $V = 250 \ mL = 0.250 \ L$.
The molarity $(M)$ is given by: $M = \frac{n}{V} = \frac{0.05 \ mol}{0.250 \ L} = 0.2 \ M$.
Expressing this in the form $x \times 10^{-1}$: $0.2 = 2 \times 10^{-1}$.
Therefore,the value of $x$ is $2$.

(B) The overall formation constant $K_{f}$ is the product of the stepwise stability constants:
$K_{f} = K_{1} \times K_{2} \times K_{3} \times K_{4}$
$K_{f} = 10^{4} \times (1.58 \times 10^{3}) \times (5 \times 10^{2}) \times 10^{2}$
$K_{f} = 7.9 \times 10^{11}$
The dissociation constant $K_{d}$ is the reciprocal of the formation constant $K_{f}$:
$K_{d} = \frac{1}{K_{f}} = \frac{1}{7.9 \times 10^{11}}$
$K_{d} \approx 1.26 \times 10^{-12}$
Given $K_{d} = x \times 10^{-12}$,we have $x = 1.26$.
Rounding to the nearest integer,$x = 1$.

(A) In a body-centered cubic $(BCC)$ unit cell,an atom is present at the center of the cube.
This central atom is surrounded by $8$ corner atoms,each of which is at an equal distance from the center.
Therefore,the coordination number of an atom in a $BCC$ structure is $8$.

(C) The reaction is a first order process.
The integrated rate equation for a first order reaction is given by:
$Kt = \ln \frac{[A]_{0}}{[A]_{t}}$
Given that the reaction proceeds $40\%$ to completion,the remaining concentration $[A]_{t}$ is $60\%$ of the initial concentration $[A]_{0}$.
So,$[A]_{t} = 0.60 [A]_{0}$ or $\frac{[A]_{0}}{[A]_{t}} = \frac{100}{60} = \frac{5}{3}$.
Substituting the values:
$3.3 \times 10^{-4} \ s^{-1} \times t = \ln \left(\frac{100}{60}\right)$
$3.3 \times 10^{-4} \times t = \ln(1.6667)$
$3.3 \times 10^{-4} \times t = 0.5108$
$t = \frac{0.5108}{3.3 \times 10^{-4}} \ s$
$t = 1547.95 \ s$
To convert the time into minutes:
$t = \frac{1547.95}{60} \ min$
$t = 25.799 \ min$
Rounding off to the nearest integer,we get $26 \ min$.

(C) The $+3$ oxidation state of lanthanoids is the most stable. Therefore,lanthanoids in the $+4$ oxidation state have a strong tendency to gain an electron and convert into the $+3$ state,acting as strong oxidizing agents.
For example,$Ce^{+4}$ in $CeO_{2}$ is used to oxidize alcohols,aldehydes,and ketones.
Lanthanoids in the $+2$ oxidation state have a strong tendency to lose an electron and convert into the $+3$ oxidation state,thus acting as strong reducing agents.
Therefore,$EuSO_{4}$ (containing $Eu^{2+}$) acts as a strong reducing agent.
Both statements are true.

(B) In basic or neutral medium,$N-N$ coupling is favourable,while in slightly acidic medium,$C-N$ coupling is favourable to form $p-$aminoazobenzene.
Reaction $A$: Nitrobenzene is reduced to aniline by $Sn/HCl$,then diazotized by $HNO_2$ to form benzene diazonium chloride,which couples with aniline in acidic medium to give $p-$aminoazobenzene.
Reaction $B$: $NaBH_4$ is a selective reducing agent and does not reduce nitrobenzene to aniline. Therefore,no diazonium salt is formed,and $p-$aminoazobenzene is not produced.
Reaction $C$: Aniline is directly diazotized by $HNO_2$ to form benzene diazonium chloride,which then couples with aniline in acidic medium to give $p-$aminoazobenzene.
Thus,only reaction $B$ will not give $p-$aminoazobenzene.

(D) $CH_3-CH_2-OH$ on oxidation with $CrO_3 - H_2SO_4$ (Jones reagent) gives acetic acid $(CH_3-COOH)$ because it is a strong oxidizing agent.
The other reactions produce acetaldehyde $(CH_3-CHO)$:
$(A)$ Dehydrogenation of ethanol: $CH_3-CH_2-OH \xrightarrow[573 \ K]{Cu} CH_3-CHO + H_2$
$(B)$ Reduction of nitrile: $CH_3-CN \xrightarrow[(ii) H_2O]{(i) DIBAL-H} CH_3-CHO$
$(C)$ Wacker process: $CH_2=CH_2 + O_2 \xrightarrow[H_2O]{Pd(II)/Cu(II)} CH_3-CHO$

(A) The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$. Therefore,$Cr^{+}$ is $[Ar] 3d^5$.
The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$. Therefore,$Mn^{2+}$ is $[Ar] 3d^5$.
Both ions have the same outermost electronic configuration of $3d^5$.

(A) Lactose is a disaccharide composed of $\beta-D-galactose$ and $\beta-D-glucose$.
In lactose,the glycosidic linkage is formed between the $C-1$ of the galactose unit and the $C-4$ of the glucose unit.
This is specifically a $\beta-1,4-glycosidic$ linkage.

(C) Sodium bicarbonate $(NaHCO_3)$ is a weak base. It reacts with acids that are stronger than carbonic acid ($H_2CO_3$,$pK_a \approx 6.35$) to liberate $CO_2$ gas.
$1$. $B$ (Benzoic acid,$pK_a \approx 4.2$) is a stronger acid than $H_2CO_3$,so it reacts with $NaHCO_3$ to liberate $CO_2$.
$2$. $C$ (Picric acid,$pK_a \approx 0.38$) is a very strong acid due to the electron-withdrawing effect of three $-NO_2$ groups. It is much stronger than $H_2CO_3$ and thus liberates $CO_2$ with $NaHCO_3$.
$3$. $A$ ($2$,$4$,$6$-triamino phenol) is a very weak acid because the $-NH_2$ groups are electron-donating by resonance,which destabilizes the phenoxide ion. It is weaker than $H_2CO_3$ and does not liberate $CO_2$ with $NaHCO_3$.
Therefore,both $B$ and $C$ will liberate $CO_2$.

(C) For $[Mn(CN)_6]^{4-}$:
$Mn$ is in $+2$ oxidation state ($3d^5$ configuration).
$CN^-$ is a strong field ligand $(SFL)$,so it causes pairing of electrons.
Thus,the configuration becomes $t_{2g}^5 e_g^0$,leaving one unpaired electron.
The hybridization is $d^2sp^3$ and it is paramagnetic.
For $[Fe(CN)_6]^{3-}$:
$Fe$ is in $+3$ oxidation state ($3d^5$ configuration).
$CN^-$ is a strong field ligand $(SFL)$,causing pairing of electrons.
Thus,the configuration becomes $t_{2g}^5 e_g^0$,leaving one unpaired electron.
The hybridization is $d^2sp^3$ and it is paramagnetic.
Therefore,both complexes have $d^2sp^3$ hybridization and are paramagnetic.

(B) The Ellingham diagram is a plot of the Gibbs free energy change $(\Delta G)$ versus temperature $(T)$ for the formation of metal oxides from their respective elements and oxygen. It helps in predicting the feasibility of the reduction of metal oxides.

(B) The Freundlich adsorption isotherm is given by the equation $\frac{x}{m} = K P^{\frac{1}{n}}$.
At moderate pressure,the value of the exponent is $\frac{1}{n}$,where $n > 1$.
Comparing this with the given expression $\frac{x}{m} \propto P^x$,we find that $x = \frac{1}{n}$.

(D) The first step is the electrophilic addition of $HCl$ to the double bond of the cyclohexene ring. According to Markovnikov's rule,the $H^+$ adds to the carbon with more hydrogens,and $Cl^-$ adds to the other carbon,forming the chloro-substituted product $[A]$.
In the second step,the reaction with $NaI$ in dry acetone is a Finkelstein reaction,which is an $S_N2$ substitution where the chlorine atom is replaced by an iodine atom to form the iodo-substituted product $[B]$.

(A) The provided image shows the polymerization of $1,3-butadiene$ and styrene to form $Buna-S$ in the presence of nascent oxygen as an initiator.
$Buna-S$ is a synthetic elastomer (a type of rubber),not a thermosetting polymer.
$Neoprene$ is a homopolymer of $chloroprene$,not a copolymer.
$Buna-N$ is a synthetic copolymer,not a natural polymer.
Therefore,the statement that the synthesis of $Buna-S$ requires nascent oxygen is correct based on the provided reaction scheme.

(D) The reaction proceeds in three steps:
$1$. Acidic hydrolysis of propanenitrile $(CH_{3}CH_{2}CN)$ yields propanoic acid $(CH_{3}CH_{2}COOH)$.
$2$. Treatment of propanoic acid with $SOCl_{2}$ yields propanoyl chloride $(CH_{3}CH_{2}COCl)$.
$3$. Rosenmund reduction of propanoyl chloride using $Pd/BaSO_{4}$ and $H_{2}$ yields propanaldehyde $(CH_{3}CH_{2}CHO)$.

(D) $BF_3$,$SiCl_4$,and $PCl_5$ undergo hydrolysis because they have vacant orbitals or are highly reactive towards water.
$SF_6$ is inert towards hydrolysis due to steric hindrance and the fact that the sulfur atom is coordinatively saturated by six fluorine atoms,preventing the attack of water molecules.
Thus,the number of halides inert to hydrolysis is $1$.

(C) The electrolyte $A_{2}B_{3}$ dissociates as: $A_{2}B_{3} \rightarrow 2A^{3+} + 3B^{2-}$.
The number of ions produced per formula unit is $n = 2 + 3 = 5$.
The degree of dissociation $\alpha = 0.60$.
The van't Hoff factor $i = 1 + \alpha(n - 1) = 1 + 0.60(5 - 1) = 1 + 0.60(4) = 1 + 2.4 = 3.4$.
The elevation in boiling point is given by $\Delta T_{b} = i K_{b} m$.
Substituting the values: $\Delta T_{b} = 3.4 \times 0.52 \times 1 = 1.768 \ K$.
The boiling point of the solution $T_{b} = T_{b}^{\circ} + \Delta T_{b} = 373.15 + 1.768 = 374.918 \ K$.
Rounding off to the nearest integer,we get $375 \ K$.

(B) The Arrhenius equation is given by $\log k = \log A - \frac{E_a}{2.303 RT}$.
Comparing this with the equation of a straight line $y = mx + c$,the slope is $m = -\frac{E_a}{2.303 R}$.
From the given graph,the slope is $-10,000 \ K$.
Therefore,$\frac{E_a}{2.303 R} = 10,000 \ K$.
Using the Arrhenius equation in the form $\log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$:
Given $k_1 = 10^{-5} \ s^{-1}$ at $T_1 = 500 \ K$ and $k_2 = 10^{-4} \ s^{-1}$ at $T_2 = ?$.
Substituting the values: $\log \left( \frac{10^{-4}}{10^{-5}} \right) = 10,000 \left( \frac{1}{500} - \frac{1}{T_2} \right)$.
$\log(10) = 1 = 10,000 \left( \frac{1}{500} - \frac{1}{T_2} \right)$.
$\frac{1}{10,000} = \frac{1}{500} - \frac{1}{T_2}$.
$\frac{1}{T_2} = \frac{1}{500} - \frac{1}{10,000} = \frac{20 - 1}{10,000} = \frac{19}{10,000}$.
$T_2 = \frac{10,000}{19} \approx 526.31 \ K$.
Rounding to the nearest integer,$T_2 = 526 \ K$.

(B) Let $x$ be the milliequivalents of $NaOH$ and $y$ be the milliequivalents of $Na_{2}CO_{3}$.
At the first end point (phenolphthalein indicator),$NaOH$ is neutralized and $Na_{2}CO_{3}$ is converted to $NaHCO_{3}$:
$x + y = \frac{1}{10} \times 17.5 = 1.75 \text{ meq} \dots (1)$
At the second end point (methyl orange indicator),$NaHCO_{3}$ is neutralized:
$y = \frac{1}{10} \times 1.5 = 0.15 \text{ meq} \dots (2)$
Mass of $Na_{2}CO_{3} = \text{milliequivalents} \times \text{equivalent weight} \times 10^{-3}$
$= 0.15 \times 53 \times 10^{-3} = 0.00795 \ g$
Weight percentage of $Na_{2}CO_{3} = \frac{0.00795}{0.4} \times 100 = 1.9875 \%$
Rounding off to the nearest integer,we get $2 \%$.

(D) The reaction proceeds in two steps:
$1$. $3 CH \equiv CH \xrightarrow{\text{Red hot Fe tube, } 873 \text{ K}} C_6H_6$ (Benzene).
$2$. $C_6H_6 + CO + HCl \xrightarrow{AlCl_3} C_6H_5CHO$ (Benzaldehyde) via Gattermann-Koch reaction.
In benzaldehyde $(C_6H_5CHO)$,there are $6$ carbon atoms in the benzene ring,all of which are $sp^{2}$ hybridized.
The carbonyl carbon atom $(C=O)$ is also $sp^{2}$ hybridized.
Therefore,the total number of $sp^{2}$ hybridized carbon atoms is $6 + 1 = 7$.

(A) Neoprene is a polymer of chloroprene ($2-$chloro$-1,3-$butadiene). Its structure is formed by the polymerization of the monomer $CH_2=C(Cl)-CH=CH_2$. The resulting polymer is $n[CH_2=C(Cl)-CH=CH_2] \rightarrow [-CH_2-C(Cl)=CH-CH_2-]_n$.

(B) The chemical compositions of the given ores are as follows:
$a$. Kernite: $Na_2B_4O_7 \cdot 4H_2O$ (contains Boron).
$b$. Cassiterite: $SnO_2$ (contains Tin).
$c$. Calamine: $ZnCO_3$ (contains Zinc).
$d$. Cryolite: $Na_3AlF_6$ (contains Fluorine).
Matching these with the elements in List-$II$:
$a \rightarrow ii$
$b \rightarrow i$
$c \rightarrow iv$
$d \rightarrow iii$
Therefore,the correct matching is $a$ $\rightarrow ii, b$ $\rightarrow i, c$ $\rightarrow iv, d$ $\rightarrow iii$.

(B) $1$. Reaction with $Br_2$ in $CS_2$ at $273 \ K$: This is an electrophilic aromatic substitution reaction. Due to the low polarity of the solvent $CS_2$,the reaction is less polar,leading to the formation of the para-isomer as the major product. Thus,$A$ is $p$-bromophenol.
$2$. Reaction with $CHCl_3$ and $NaOH$ followed by $H_3O^+$: This is the Reimer-Tiemann reaction. It introduces a formyl group $(-CHO)$ at the ortho position of the phenol. Thus,$B$ is $o$-hydroxybenzaldehyde (salicylaldehyde).
Therefore,the major products are $p$-bromophenol and $o$-hydroxybenzaldehyde.

(C) When a sulfite or bisulfite salt is treated with warm dil. $H_{2}SO_{4}$,sulfur dioxide gas $(X = SO_{2})$ is evolved.
$SO_{2}$ acts as a reducing agent and reacts with acidified potassium dichromate $(K_{2}Cr_{2}O_{7})$ paper.
The reaction is: $K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + 3SO_{2} \longrightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + H_{2}O$.
The green compound $Y$ formed is chromium$(III)$ sulfate,$Cr_{2}(SO_{4})_{3}$.

(D) Vitamin $K$ is essential for the synthesis of clotting factors in the liver,which promotes blood coagulation. Conversely,substances that interfere with $Vitamin \ K$ or act as anticoagulants are used to delay blood clotting. Among the given options,$Vitamin \ K$ is the primary vitamin involved in the blood clotting process. However,if the question implies an anticoagulant effect,it is important to note that $Vitamin \ K$ promotes clotting,while its deficiency or specific inhibitors delay it.

(D) The compound $A$ $(C_{4}H_{8}Cl_{2})$ undergoes hydrolysis at $373 \text{ K}$ to form a geminal diol,which is unstable and loses a water molecule to form a carbonyl compound $B$ $(C_{4}H_{8}O)$.
Since $B$ reacts with hydroxylamine,it must be an aldehyde or a ketone.
Since $B$ does not give Tollen's test,it cannot be an aldehyde; therefore,$B$ must be a ketone.
$2,2-$Dichlorobutane on hydrolysis gives $2,2-$butanediol,which loses water to form $2-$butanone $(CH_{3}COCH_{2}CH_{3})$.
$2-$Butanone is a ketone,so it reacts with hydroxylamine and does not give Tollen's test.
Thus,$A$ is $2,2-$dichlorobutane and $B$ is $2-$butanone.

(A) $PbO_2$ is an amphoteric oxide and a strong oxidizing agent.
It acts as the cathode material in lead storage batteries.

(D) Most lanthanoids form $M_{2}O_{3}$ type oxides.
However,$Ce$,$Pr$,$Nd$,$Tb$,and $Dy$ can form $MO_{2}$ type oxides.
$Yb$ does not form $MO_{2}$ type oxide; it typically forms $Yb_{2}O_{3}$.

(C) The reaction of an alkyl-substituted benzene with $Br_2$ in the presence of $UV$ light (or heat) proceeds via a free radical mechanism.
This reaction is a free radical substitution that occurs at the benzylic position.
The benzylic hydrogen is abstracted to form a stable benzylic radical,which then reacts with $Br_2$ to form the monobrominated product.
In the given molecule,$3-\text{ethylbenzonitrile}$,the benzylic carbon is the $CH_2$ group attached to the benzene ring.
Therefore,the bromine atom will substitute one of the hydrogen atoms on this benzylic carbon,resulting in $3-(1-\text{bromoethyl})benzonitrile$ as the major product.

(D) The reaction of an amine with benzenesulphonyl chloride (Hinsberg reagent) produces a product that is insoluble in alkaline solution if the amine is a secondary $(2^{\circ})$ amine. This is because the resulting $N,N$-disubstituted sulphonamide does not contain any acidic hydrogen atom attached to the nitrogen atom.
The amine is prepared by the ammonolysis of ethyl chloride $(CH_3CH_2Cl)$,which implies the amine must contain an ethyl group $(CH_3CH_2-)$.
Among the options,$N$-propylaniline is a secondary amine,but it is not formed by the ammonolysis of ethyl chloride. The structure $CH_3CH_2CH_2NHCH_2CH_3$ ($N$-ethylpropylamine) is a secondary amine that can be formed by the reaction of propylamine with ethyl chloride. Therefore,the correct structure is $CH_3CH_2CH_2NHCH_2CH_3$.

(C) For an exothermic reaction,the relationship between activation energy of forward reaction $(E_{a,f})$,activation energy of reverse reaction $(E_{a,r})$,and enthalpy change ($\Delta H$ or $\Delta E$) is given by:
$\Delta E = E_{a,f} - E_{a,r}$
Given:
$E_{a,f} = 30 \ kJ \ mol^{-1}$
$\Delta E = -20 \ kJ \ mol^{-1}$
Substituting the values:
$-20 = 30 - E_{a,r}$
$E_{a,r} = 30 + 20 = 50 \ kJ \ mol^{-1}$
Therefore,the activation energy for the reverse reaction is $50 \ kJ \ mol^{-1}$.

(B) The given half-reaction is: $MnO_4^- + 8H^{+} + 5e^- \rightarrow Mn^{2+} + 4H_2O$
In this reaction,the reduction of $1 \ mol$ of $MnO_4^-$ involves the gain of $5 \ mol$ of electrons.
According to Faraday's laws,the charge required for $1 \ mol$ of electrons is $1 \ F$.
Therefore,the electricity required for $1 \ mol$ of $MnO_4^-$ is $5 \ F$.
For $5 \ mol$ of $MnO_4^-$,the total electricity required is $5 \ mol \times 5 \ F/mol = 25 \ F$.

(B) Step $1$: Calculate moles of reactants.
$n_{SO_2} = \frac{224 \, mL}{22400 \, mL/mol} = 0.01 \, mol$.
$n_{NaOH} = 0.1 \, M \times 0.1 \, L = 0.01 \, mol$.
Step $2$: Reaction stoichiometry.
$SO_2 + NaOH \rightarrow NaHSO_3$.
Since $n_{SO_2} = n_{NaOH} = 0.01 \, mol$, both are consumed to produce $0.01 \, mol$ of $NaHSO_3$.
Step $3$: Calculate lowering of vapour pressure.
$\Delta P = P^0_{H_2O} \times \chi_{solute} = P^0_{H_2O} \times \frac{n_{solute}}{n_{solute} + n_{solvent}}$.
$n_{solvent} = \frac{36 \, g}{18 \, g/mol} = 2 \, mol$.
$\Delta P = 24 \times \frac{0.01}{0.01 + 2} = 24 \times \frac{0.01}{2.01} \approx 24 \times 0.004975 = 0.1194 \, mm \, Hg$.
$\Delta P = 11.94 \times 10^{-2} \, mm \, Hg$.
Rounding to the nearest integer, $x = 12$.

(A) First,calculate the number of moles of $O_2$ adsorbed:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{3.12 \, g}{32 \, g/mol} = 0.0975 \, mol$.
Next,calculate the volume of $O_2$ using the ideal gas equation $PV = nRT$:
$V = \frac{nRT}{P} = \frac{0.0975 \, mol \times 0.0821 \, L \, atm \, K^{-1} \, mol^{-1} \times 300 \, K}{1 \, atm} = 2.40 \, L$.
Since $1.2 \, g$ of platinum adsorbs $2.40 \, L$ of $O_2$,the volume adsorbed per gram is:
$\frac{2.40 \, L}{1.2 \, g} = 2 \, L/g$.

(D) The structure of $[Mn_{2}(CO)_{10}]$ consists of two $Mn(CO)_{5}$ units joined by a $Mn-Mn$ metal-metal bond.
Each $Mn$ atom is bonded to five terminal $CO$ ligands.
There are no $CO$ ligands bridging the two $Mn$ atoms.
Therefore,the number of bridging $CO$ ligands is $0$.

(B) The conversion of nitrobenzene to $m$-dibromobenzene involves the following steps:
$1$. Bromination of nitrobenzene using $Br_2/Fe$ gives $m$-bromonitrobenzene.
$2$. Reduction of $m$-bromonitrobenzene using $Sn/HCl$ gives $m$-bromoaniline.
$3$. Diazotization of $m$-bromoaniline using $NaNO_2/HCl$ at $0-5^{\circ}C$ gives $m$-bromobenzenediazonium chloride.
$4$. Sandmeyer reaction of $m$-bromobenzenediazonium chloride with $CuBr/HBr$ replaces the diazonium group with a bromine atom to yield $m$-dibromobenzene.