JEE Main 2014 Chemistry Question Paper with Answer and Solution

(B) According to Fajan's rule,the ionic character of a bond depends on the polarizing power of the cation.
Greater size of the cation and smaller charge on the cation lead to less polarization,resulting in more ionic character.
Comparing the cations: $Rb^+$ is the largest with a $+1$ charge,making $RbCl$ the most ionic.
$Be^{2+}$ is the smallest with a $+2$ charge,leading to the highest polarization and the least ionic (most covalent) character.
Therefore,$RbCl$ has the greatest ionic character and $BeCl_{2}$ has the least ionic character.

(C) According to Fajan's rule,ionic character depends on the polarising power of the cation.
$Rb^+$ has the largest size among the given cations,resulting in the least polarising power and thus the greatest ionic character in $RbCl$.
$Be^{2+}$ has the smallest size and highest charge density,giving it the highest polarising power,which results in the greatest covalent character and the least ionic character in $BeCl_2$.
Therefore,the compounds with the greatest and least ionic character are $RbCl$ and $BeCl_2$ respectively.

(B) Given $X = \{ 4^n - 3n - 1 : n \in N \}$.
For $n=1$,$4^1 - 3(1) - 1 = 0$.
For $n=2$,$4^2 - 3(2) - 1 = 16 - 6 - 1 = 9$.
For $n=3$,$4^3 - 3(3) - 1 = 64 - 9 - 1 = 54$.
Using the binomial expansion,$4^n = (1+3)^n = 1 + n(3) + \frac{n(n-1)}{2}(3^2) + \dots + 3^n$.
Thus,$4^n - 3n - 1 = 9 \times [\frac{n(n-1)}{2} + \dots + 3^{n-2}]$.
This shows that every element in $X$ is a multiple of $9$ (including $0$).
$Y = \{ 9(n-1) : n \in N \} = \{ 0, 9, 18, 27, \dots \}$,which is the set of all non-negative multiples of $9$.
Since all elements of $X$ are multiples of $9$,$X \subseteq Y$.
Therefore,$X \cup Y = Y$.

(B) We need to evaluate the limit: $\mathop {\lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}}$
Using the identity $\cos^2 x = 1 - \sin^2 x$,we have:
$\mathop {\lim}\limits_{x \to 0} \frac{{\sin \left( {\pi (1 - \sin^2 x)} \right)}}{{{x^2}}}$
$= \mathop {\lim}\limits_{x \to 0} \frac{{\sin \left( {\pi - \pi \sin^2 x} \right)}}{{{x^2}}}$
Since $\sin(\pi - \theta) = \sin \theta$,the expression becomes:
$= \mathop {\lim}\limits_{x \to 0} \frac{{\sin \left( {\pi \sin^2 x} \right)}}{{{x^2}}}$
Multiply and divide by $\pi \sin^2 x$:
$= \mathop {\lim}\limits_{x \to 0} \left( \frac{{\sin \left( {\pi \sin^2 x} \right)}}{{\pi \sin^2 x}} \cdot \frac{{\pi \sin^2 x}}{{{x^2}}} \right)$
$= \left( \mathop {\lim}\limits_{x \to 0} \frac{{\sin \left( {\pi \sin^2 x} \right)}}{{\pi \sin^2 x}} \right) \cdot \pi \cdot \left( \mathop {\lim}\limits_{x \to 0} \frac{{\sin x}}{x} \right)^2$
Using the standard limit $\mathop {\lim}\limits_{\theta \to 0} \frac{{\sin \theta}}{\theta} = 1$:
$= 1 \cdot \pi \cdot (1)^2 = \pi$

(D) The first $50$ even natural numbers are $2, 4, 6, \dots, 100$.
The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
Here,$n = 50$. The sum of the first $n$ even numbers is $n(n+1) = 50 \times 51 = 2550$. Thus,the mean $\bar{x} = \frac{2550}{50} = 51$.
The sum of squares is $\sum x_i^2 = 2^2 + 4^2 + \dots + 100^2 = 2^2(1^2 + 2^2 + \dots + 50^2)$.
Using the formula $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,we get $\sum x_i^2 = 4 \times \frac{50(51)(101)}{6} = 4 \times \frac{257550}{6} = 4 \times 42925 = 171700$.
Now,$\frac{\sum x_i^2}{n} = \frac{171700}{50} = 3434$.
Variance $\sigma^2 = 3434 - (51)^2 = 3434 - 2601 = 833$.

(D) Given that $\alpha$ and $\beta$ are the roots of $px^2 + qx + r = 0$,we have $\alpha + \beta = -\frac{q}{p}$ and $\alpha\beta = \frac{r}{p}$.
Since $p, q, r$ are in $A.P.$,we have $2q = p + r$.
Given $\frac{1}{\alpha} + \frac{1}{\beta} = 4$,which implies $\frac{\alpha + \beta}{\alpha\beta} = 4$,so $\alpha + \beta = 4\alpha\beta$.
Substituting the values,$-\frac{q}{p} = 4(\frac{r}{p}) \Rightarrow q = -4r$.
Using $2q = p + r$,we get $2(-4r) = p + r$ $\Rightarrow -8r = p + r$ $\Rightarrow p = -9r$.
Now,$\alpha + \beta = -\frac{-4r}{-9r} = -\frac{4}{9}$ and $\alpha\beta = \frac{r}{-9r} = -\frac{1}{9}$.
Using the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$,we get $(\alpha - \beta)^2 = (-\frac{4}{9})^2 - 4(-\frac{1}{9}) = \frac{16}{81} + \frac{4}{9} = \frac{16 + 36}{81} = \frac{52}{81}$.
Therefore,$|\alpha - \beta| = \sqrt{\frac{52}{81}} = \frac{\sqrt{4 \times 13}}{9} = \frac{2\sqrt{13}}{9}$.

(C) We are given the equations for direction cosines $l, m, n$ as $l + m + n = 0$ and $l^2 = m^2 + n^2$.
From the first equation,$n = -(l + m)$.
Substituting this into the second equation: $l^2 = m^2 + (-(l + m))^2$.
$l^2 = m^2 + l^2 + m^2 + 2lm$.
$0 = 2m^2 + 2lm$.
$2m(m + l) = 0$.
This gives two cases: $m = 0$ or $m = -l$.
Case $1$: If $m = 0$,then $l + 0 + n = 0 \Rightarrow n = -l$. The direction ratios are $(l, 0, -l)$,which is proportional to $(1, 0, -1)$. Let $\vec{a} = \hat{i} - \hat{k}$.
Case $2$: If $m = -l$,then $l + (-l) + n = 0 \Rightarrow n = 0$. The direction ratios are $(l, -l, 0)$,which is proportional to $(1, -1, 0)$. Let $\vec{b} = \hat{i} - \hat{j}$.
The angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (1)(1) + (0)(-1) + (-1)(0) = 1$.
$|\vec{a}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}$.
$|\vec{b}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}$.
$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(B) We have $4^n - 3n - 1 = (1 + 3)^n - 3n - 1$.
Using the binomial expansion,$(1 + 3)^n = 1 + n(3) + \frac{n(n-1)}{2}(3^2) + \dots + 3^n$.
So,$4^n - 3n - 1 = 1 + 3n + 9 \binom{n}{2} + 27 \binom{n}{3} + \dots + 3^n - 3n - 1 = 9 \left[ \binom{n}{2} + 3 \binom{n}{3} + \dots + 3^{n-2} \right]$.
This shows that $4^n - 3n - 1$ is a multiple of $9$ for all $n \ge 2$.
For $n = 1$,$4^1 - 3(1) - 1 = 0$.
For $n = 2$,$4^2 - 3(2) - 1 = 9$.
Thus,$X = \{ 0, 9, 27, 54, \dots \}$.
For $Y = \{ 9(n-1) : n \in N \}$,we have $Y = \{ 0, 9, 18, 27, 36, 45, 54, \dots \}$.
Since every element of $X$ is a multiple of $9$,$X \subseteq Y$.
Therefore,$X \cup Y = Y$.

(A) Optical isomerism requires the presence of at least one chiral carbon atom (a carbon atom bonded to four different groups).
Maleic acid $(HOOC-CH=CH-COOH)$ contains $sp^2$ hybridized carbons in a double bond and lacks a chiral carbon atom,so it does not exhibit optical isomerism.
Tartaric acid,lactic acid,and $\alpha$-amino acids all contain at least one chiral carbon atom (indicated by $*$ in the structure),allowing them to exhibit optical isomerism.

(A) The electronic configuration of Rubidium $(Z=37)$ is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1$.
The valence electron is present in the $5s$ orbital.
For the $5s$ orbital,the principal quantum number $n=5$.
Since it is an $s$-orbital,the azimuthal quantum number $l=0$.
For $l=0$,the magnetic quantum number $m=0$.
The spin quantum number $s$ can be $+1/2$ or $-1/2$. Thus,the set is $(5, 0, 0, +1/2)$.

(D) For $p$-dichlorobenzene $(i)$ and $p$-dicyanobenzene $(ii)$,the dipole moments of the individual bonds cancel each other out because they are equal in magnitude and opposite in direction. Hence,the resultant dipole moment of these molecules is zero.
For $p$-hydroquinone $(iii)$ and $p$-benzenedithiol $(iv)$,the $O-H$ and $S-H$ bond dipoles do not cancel each other out because they are not in opposite directions due to the existence of different conformations. Hence,the resultant dipole moment of these molecules is non-zero $(\mu \ne 0)$.

(B) The van der Waals equation for $1 \ mol$ of a gas is given by:
$(P + \frac{a}{V^2})(V - b) = RT$
At low pressure,the volume $V$ is very large,so $V >> b$. Therefore,the term $(V - b)$ can be approximated as $V$.
The equation simplifies to:
$(P + \frac{a}{V^2})V = RT$
Expanding the equation:
$PV + \frac{a}{V} = RT$
Rearranging to solve for $PV$:
$PV = RT - \frac{a}{V}$
Dividing both sides by $RT$ to express in terms of the compressibility factor $Z = \frac{PV}{RT}$:
$\frac{PV}{RT} = 1 - \frac{a}{VRT}$
Thus,$Z = 1 - \frac{a}{VRT}$.

(B) Let the mass of $O_2$ be $w$ and the mass of $N_2$ be $4w$.
The molar mass of $O_2$ is $32 \ g/mol$ and the molar mass of $N_2$ is $28 \ g/mol$.
The number of molecules is proportional to the number of moles $(n = \frac{\text{mass}}{\text{molar mass}})$.
Number of molecules of $O_2$ $(N_{O_2})$ = $\frac{w}{32} \times N_A$
Number of molecules of $N_2$ $(N_{N_2})$ = $\frac{4w}{28} \times N_A$
Ratio $\frac{N_{O_2}}{N_{N_2}} = \frac{w/32}{4w/28} = \frac{w}{32} \times \frac{28}{4w} = \frac{28}{128} = \frac{7}{32}$.

(B) The reaction for the formation of $2 \ mol$ of $NH_3$ is: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$.
The enthalpy change for this reaction is $\Delta H = 2 \times \Delta_f H^{\circ}(NH_3) = 2 \times (-46.0) = -92 \ kJ \ mol^{-1}$.
Using bond enthalpies,$\Delta H = \sum \text{Bond energies of reactants} - \sum \text{Bond energies of products}$.
Given bond energies (as enthalpy of formation from atoms is negative of bond dissociation energy): $BE(N \equiv N) = 712 \ kJ \ mol^{-1}$ and $BE(H-H) = 436 \ kJ \ mol^{-1}$.
Let $x$ be the bond enthalpy of $N-H$ bond. There are $6$ $N-H$ bonds in $2 \ mol$ of $NH_3$.
$-92 = [BE(N \equiv N) + 3 \times BE(H-H)] - 6x$.
$-92 = [712 + 3 \times 436] - 6x$.
$-92 = [712 + 1308] - 6x$.
$-92 = 2020 - 6x$.
$6x = 2020 + 92 = 2112$.
$x = 2112 / 6 = +352 \ kJ \ mol^{-1}$.

(A) The reaction is: $C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$
Given internal energy change $\Delta U = -1364.47 \ kJ \ mol^{-1}$ (heat released in bomb calorimeter).
Temperature $T = 25 + 273 = 298 \ K$.
Change in moles of gaseous species $\Delta n_g = (n_{products, gas} - n_{reactants, gas}) = 2 - 3 = -1$.
Gas constant $R = 8.314 \ J \ K^{-1} \ mol^{-1} = 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.
The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.
Substituting the values: $\Delta H = -1364.47 + (-1 \times 8.314 \times 10^{-3} \times 298)$.
$\Delta H = -1364.47 - 2.477572 \approx -1366.95 \ kJ \ mol^{-1}$.

(B) The relationship between $K_P$ and $K_C$ is given by $K_P = K_C(RT)^{\Delta n_g}$.
For the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$,the change in the number of gaseous moles is calculated as $\Delta n_g = n_{p(g)} - n_{r(g)}$.
Here,$n_{p(g)} = 1$ (for $SO_3$) and $n_{r(g)} = 1 + 0.5 = 1.5$ (for $SO_2$ and $O_2$).
Therefore,$\Delta n_g = 1 - 1.5 = -0.5$.
Comparing this with the given equation $K_P = K_C(RT)^x$,we get $x = -0.5$.

(D) reducing agent is a substance that undergoes oxidation by losing electrons.
In the reaction $H_2O_2 + 2OH^- \rightarrow O_2 + 2H_2O + 2e^-$,the oxidation state of oxygen in $H_2O_2$ increases from $-1$ to $0$ in $O_2$.
Since $H_2O_2$ loses electrons and is oxidized,it acts as a reducing agent.

(B) The formula for the percentage of nitrogen is: $\% \ N = \frac{1.4 \times \text{meq. of acid consumed}}{\text{mass of organic compound}}$
Step $1$: Calculate milliequivalents $(meq)$ of $H_2SO_4$ taken:
$meq \text{ of } H_2SO_4 = 60 \ mL \times \frac{1}{10} \ M \times 2 \text{ (basicity)} = 12 \ meq$
Step $2$: Calculate $meq$ of $NaOH$ used for back titration:
$meq \text{ of } NaOH = 20 \ mL \times \frac{1}{10} \ M \times 1 \text{ (acidity)} = 2 \ meq$
Step $3$: Calculate $meq$ of acid consumed by ammonia:
$meq \text{ of acid consumed} = 12 - 2 = 10 \ meq$
Step $4$: Calculate percentage of nitrogen:
$\% \ N = \frac{1.4 \times 10}{1.4} = 10 \%$

(B) $CsI_3$ is an ionic compound that dissociates in the solid state as $CsI_3 \rightarrow Cs^{+} + I_3^{-}$.
Here,$Cs^{+}$ is the cesium cation and $I_3^{-}$ is the triiodide anion.

(C) The reaction of $1, 1, 1-$trichloroethane $(CH_3-CCl_3)$ with silver powder $(Ag)$ is a dehalogenation reaction.
When $2$ moles of $1, 1, 1-$trichloroethane react with $6$ moles of silver powder,it undergoes coupling to form $2-$butyne.
The chemical equation is: $2 CH_3-CCl_3 + 6 Ag \rightarrow CH_3-C \equiv C-CH_3 + 6 AgCl$.

(D) Given the equation: $I = (e^{1000V/T} - 1) \text{ mA}$.
At $I = 5 \text{ mA}$,we have $5 = e^{1000V/T} - 1$,which implies $e^{1000V/T} = 6$.
Differentiating the current $I$ with respect to voltage $V$:
$\frac{dI}{dV} = e^{1000V/T} \cdot \frac{1000}{T}$.
Substituting the value of $e^{1000V/T} = 6$ and $T = 300 \text{ K}$:
$\frac{dI}{dV} = 6 \cdot \frac{1000}{300} = 6 \cdot \frac{10}{3} = 20 \text{ mA/V}$.
The error in current $\Delta I$ is given by $\Delta I = \left| \frac{dI}{dV} \right| \cdot \Delta V$.
Given $\Delta V = 0.01 \text{ V}$,we get:
$\Delta I = 20 \cdot 0.01 = 0.2 \text{ mA}$.

(C) $1$. Given $P = (5, 3)$. Since $PQ$ is parallel to the $x-$ axis,the $y-$ coordinate of $Q$ is the same as $P$,so $Q = (x_Q, 3)$.
$2$. $Q$ lies on the line $RQ: x - 2y = 2$. Substituting $y = 3$,we get $x - 2(3) = 2$,which implies $x - 6 = 2$,so $x = 8$. Thus,$Q = (8, 3)$.
$3$. $R$ lies on the $x-$ axis,so its $y-$ coordinate is $0$. Let $R = (x_R, 0)$. Since $R$ lies on $x - 2y = 2$,we have $x_R - 2(0) = 2$,so $x_R = 2$. Thus,$R = (2, 0)$.
$4$. The centroid $G$ of $\Delta PQR$ is given by $(\frac{x_P + x_Q + x_R}{3}, \frac{y_P + y_Q + y_R}{3}) = (\frac{5 + 8 + 2}{3}, \frac{3 + 3 + 0}{3}) = (5, 2)$.
$5$. We check which line equation is satisfied by $(5, 2)$:
- For $A: 5 - 2(2) + 1 = 5 - 4 + 1 = 2 \neq 0$.
- For $B: 2(5) + 2 - 9 = 10 + 2 - 9 = 3 \neq 0$.
- For $C: 2(5) - 5(2) = 10 - 10 = 0$. This is satisfied.
- For $D: 5(5) - 2(2) = 25 - 4 = 21 \neq 0$.
$6$. Therefore,the centroid lies on the line $2x - 5y = 0$.

(D) The first $50$ even natural numbers are $2, 4, 6, \ldots, 100$.
These can be written as $2(1), 2(2), 2(3), \ldots, 2(50)$.
The variance of a set of numbers $x_i$ multiplied by a constant $k$ is given by $\text{Var}(kx_i) = k^2 \text{Var}(x_i)$.
Here,$k = 2$ and the set is the first $50$ natural numbers $(1, 2, 3, \ldots, 50)$.
The variance of the first $n$ natural numbers is $\frac{n^2 - 1}{12}$.
For $n = 50$,$\text{Var} = \frac{50^2 - 1}{12} = \frac{2500 - 1}{12} = \frac{2499}{12}$.
Therefore,the variance of the first $50$ even natural numbers is $2^2 \times \frac{2499}{12} = 4 \times \frac{2499}{12} = \frac{2499}{3} = 833$.

(C) The dissociation of zirconium phosphate is given by:
$Zr_3(PO_4)_4 (s) \rightleftharpoons 3 Zr^{4+} (aq) + 4 PO_4^{3-} (aq)$
If $S$ is the molar solubility,then $[Zr^{4+}] = 3S$ and $[PO_4^{3-}] = 4S$.
The solubility product expression is:
$K_{sp} = [Zr^{4+}]^3 [PO_4^{3-}]^4$
$K_{sp} = (3S)^3 (4S)^4$
$K_{sp} = (27 S^3) (256 S^4)$
$K_{sp} = 6912 S^7$
Therefore,$S^7 = K_{sp} / 6912$
$S = (K_{sp} / 6912)^{1/7}$

(A) The slope of the tangent to the curve at any point is given by $\tan \theta = \frac{dy}{dx}$.
Given $y = \frac{x^3}{6}$,we have $\frac{dy}{dx} = \frac{3x^2}{6} = \frac{x^2}{2}$.
Thus,$\tan \theta = \frac{x^2}{2}$.
For the block to remain in equilibrium without slipping,the condition is $\mu \ge \tan \theta$.
The maximum height corresponds to the limiting case where $\mu = \tan \theta$.
Given $\mu = 0.5 = \frac{1}{2}$,we set $\frac{1}{2} = \frac{x^2}{2}$,which gives $x^2 = 1$,so $x = 1 \text{ m}$.
Substituting $x = 1$ into the equation for $y$,we get the maximum height:
$y = \frac{(1)^3}{6} = \frac{1}{6} \text{ m}$.

(A) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\sqrt{n(n+2)} = 1.73$,which corresponds to $n = 1$.
The electronic configuration of vanadium $(V)$ is $[Ar] 3d^3 4s^2$.
To have $n = 1$ unpaired electron,vanadium must be in the $+4$ oxidation state $(V^{4+})$,which has the configuration $[Ar] 3d^1$.
Therefore,the compound is $VCl_4$.

(A) The Van der Waals equation is given by $\left( P + \frac{n^2a}{V^2} \right)(V - nb) = nRT$.
At low pressure,the volume $V$ is very large,making the correction term $\frac{n^2a}{V^2}$ negligible compared to $P$.
At high temperature,the volume $V$ is also large,making the correction term $nb$ negligible compared to $V$.
Under these conditions,the equation simplifies to $PV = nRT$,which is the ideal gas equation.

(A) The dissociation reaction of $HI$ is:
$2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$
Let the initial moles of $HI$ be $1$.
Degree of dissociation $\alpha = 50\% = 0.5$.
At equilibrium:
Moles of $HI = 1 - \alpha = 1 - 0.5 = 0.5$
Moles of $H_2 = \frac{\alpha}{2} = \frac{0.5}{2} = 0.25$
Moles of $I_2 = \frac{\alpha}{2} = \frac{0.5}{2} = 0.25$
The equilibrium constant $K_c$ is given by:
$K_c = \frac{[H_2][I_2]}{[HI]^2}$
Since the number of moles on both sides is the same,volume $V$ cancels out.
$K_c = \frac{0.25 \times 0.25}{(0.5)^2} = \frac{0.0625}{0.25} = 0.25$

(B) Given:
Mass of solute $(w) = 120 \ g$
Mass of solvent $= 1000 \ g$
Molar mass of solute $(M_w) = 60 \ g/mol$
Density of solution $(d) = 1.12 \ g/mL$
Total mass of solution $= 1000 \ g + 120 \ g = 1120 \ g$
Volume of solution $(V) = \frac{\text{Mass of solution}}{d} = \frac{1120 \ g}{1.12 \ g/mL} = 1000 \ mL = 1 \ L$
Moles of solute $(n) = \frac{w}{M_w} = \frac{120 \ g}{60 \ g/mol} = 2 \ mol$
Molarity $(M) = \frac{n}{V(L)} = \frac{2 \ mol}{1 \ L} = 2 \ M$

(C) The energy of an electron in a hydrogen-like species is given by the formula $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For $Li^{2+}$ ion,the atomic number $Z = 3$.
The first excited state corresponds to $n = 2$.
Substituting these values into the formula:
$E_2 = -13.6 \times \frac{3^2}{2^2} \ eV$
$E_2 = -13.6 \times \frac{9}{4} \ eV$
$E_2 = -13.6 \times 2.25 \ eV = -30.6 \ eV$.

(D) The root mean square speed $(V_{rms})$ is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that $V_{rms(O_2)} = V_{rms(He)}$,we equate the expressions:
$\sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{He}}{M_{He}}}$
Squaring both sides and simplifying,we get $\frac{T_{O_2}}{M_{O_2}} = \frac{T_{He}}{M_{He}}$.
Here,$M_{O_2} = 32 \ g/mol$,$M_{He} = 4 \ g/mol$,and $T_{He} = 300 \ K$.
Substituting the values: $\frac{T_{O_2}}{32} = \frac{300}{4}$.
$T_{O_2} = \frac{300 \times 32}{4} = 300 \times 8 = 2400 \ K$.

(C) The formation reaction is: $\frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \rightarrow NH_3(g)$; $\Delta H_f = -46.0 \, kJ/mol$.
Bond dissociation energies are the negative of the enthalpy of formation from atoms:
$BE(N \equiv N) = +712 \, kJ/mol$
$BE(H-H) = +436 \, kJ/mol$.
Using the formula:
$\Delta H_f = [\frac{1}{2} BE(N \equiv N) + \frac{3}{2} BE(H-H)] - [3 \times BE(N-H)]$.
Substituting the values:
$-46.0 = [\frac{1}{2}(712) + \frac{3}{2}(436)] - 3 \times BE(N-H)$.
$-46.0 = [356 + 654] - 3 \times BE(N-H)$.
$-46.0 = 1010 - 3 \times BE(N-H)$.
$3 \times BE(N-H) = 1010 + 46 = 1056$.
$BE(N-H) = 1056 / 3 = 352 \, kJ/mol$.

(B) $1 \, \text{mole}$ of $H_2O$ contains $1 \, \text{mole}$ of oxygen atoms,which is $16 \, g$ of oxygen.
$\therefore$ $3.6 \, \text{moles}$ of $H_2O$ contains $3.6 \times 16 \, g$ of oxygen.
$= 57.6 \, g$.

(B) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n + 2)} \, BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 1.73 \, BM$,we have $1.73 = \sqrt{n(n + 2)}$.
Squaring both sides,$3 = n(n + 2)$,which gives $n^2 + 2n - 3 = 0$.
Solving for $n$,we get $(n + 3)(n - 1) = 0$,so $n = 1$.
Vanadium $(Z = 23)$ has the electronic configuration $[Ar] 3d^3 4s^2$.
For $n = 1$ (one unpaired electron),Vanadium must be in the $+4$ oxidation state $(V^{4+})$,which has the configuration $[Ar] 3d^1$.
Therefore,the formula of the chloride is $VCl_4$.

(A) The balanced chemical equation for the reaction between acidic $K_2Cr_2O_7$ and $H_2S$ is:
$K_2Cr_2O_7 + 4H_2SO_4 + 3H_2S \longrightarrow K_2SO_4 + Cr_2(SO_4)_3 + 7H_2O + 3S$
In this redox reaction,the dichromate ion is reduced to chromium$(III)$ sulfate $(Cr_2(SO_4)_3)$,and $H_2S$ is oxidized to elemental sulfur $(S)$.
$K_2SO_4$ is also formed as a byproduct.
$CrSO_4$ (chromium$(II)$ sulfate) is not formed in this reaction.

(B) The molecular orbital configuration for $O_2^-$ (total $17$ electrons) is: $KK(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_x)^2(\pi 2p_y)^2(\pi 2p_z)^2(\pi^* 2p_y)^2(\pi^* 2p_z)^1$.
Since there is one unpaired electron in the $\pi^* 2p_z$ orbital,$O_2^-$ is paramagnetic.
For $N_2$ ($14$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_y)^2(\pi 2p_z)^2(\sigma 2p_x)^2$ (all paired).
For $N_2^{2+}$ ($12$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_y)^2(\pi 2p_z)^2$ (all paired).
For $O_2^{2-}$ ($18$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_x)^2(\pi 2p_y)^2(\pi 2p_z)^2(\pi^* 2p_y)^2(\pi^* 2p_z)^2$ (all paired).

(B) Amphoteric substances are those that can react with both acids and bases.
$Al_2O_3 \cdot xH_2O$ is an amphoteric oxide.
It reacts with a base $(OH^{-})$ to form aluminate:
$Al_2O_3 \cdot xH_2O_{(s)} + 2OH^{-}{(aq)} \longrightarrow 2AlO_2^{-}{(aq)} + (x+1)H_2O_{(l)}$
It reacts with an acid $(H^{+})$ to form aluminum ions:
$Al_2O_3 \cdot xH_2O_{(s)} + 6H^{+}_{(aq)} \longrightarrow 2Al^{3+}_{(aq)} + (x+3)H_2O_{(l)}$
Therefore,Set $1$ (reaction with base) and Set $3$ (reaction with acid) best demonstrate its amphoteric nature.

(B) The $CaC_2$ compound consists of $Ca^{2+}$ and $C_2^{2-}$ ions.
In the $C_2^{2-}$ ion,the two carbon atoms are linked by a triple bond.
$A$ triple bond consists of one $\sigma$ bond and two $\pi$ bonds.
Therefore,the $C_2^{2-}$ ion contains one $\sigma$ bond and two $\pi$ bonds.

(C) The hydroboration-oxidation of propene follows anti-Markovnikov addition of water across the double bond.
$CH_3CH=CH_2 \xrightarrow{1. B_2H_6, 2. H_2O_2, NaOH} CH_3CH_2CH_2OH$
Thus,the product formed is propan$-1-$ol $(CH_3CH_2CH_2OH)$.

(D) To determine the stability of free radicals,we compare the factors affecting their stability such as resonance,hyperconjugation,and inductive effects.
In option $B$,the radical on the right is a tertiary alkyl radical,which is stabilized by hyperconjugation from $9$ $\alpha$-hydrogens,whereas the radical on the left is a secondary alkyl radical,which is stabilized by hyperconjugation from $4$ $\alpha$-hydrogens. Thus,the tertiary radical $(B)$ is more stable than the secondary radical $(A)$.
In option $C$,the radical on the right is a primary alkyl radical,while the radical on the left is a cyclopropyl radical. The cyclopropyl radical is less stable due to high angle strain.
In option $D$,$Ph_{3}\dot{C}$ is more stable than $(CH_{3})_{3}\dot{C}$ because resonance stabilisation in $Ph_{3}\dot{C}$ is much more effective than the hyperconjugation in $(CH_{3})_{3}\dot{C}$.
Therefore,in pair $D$,$A$ is more stable than $B$.

(B) The cyclic polymerization of acetylene $(CH \equiv CH)$ at high temperature (red hot iron tube) yields benzene $(C_6H_6)$,which is a poly-ene (a compound with multiple double bonds in a ring structure).
$3CH \equiv CH \xrightarrow{\Delta, \text{Fe tube}} C_6H_6$ (Benzene).
Similarly,propyne undergoes cyclic polymerization to form $1,3,5$-trimethylbenzene.

(C) According to Einstein's photoelectric equation,the kinetic energy of the ejected electron is given by:
$E = W + K.E.$
$\frac{hc}{\lambda} = \frac{hc}{\lambda_0} + \frac{1}{2}mv^2$
Rearranging for the kinetic energy term:
$\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$
$\frac{1}{2}mv^2 = hc \left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right) = hc \left(\frac{\lambda_0 - \lambda}{\lambda \lambda_0}\right)$
Solving for velocity $v$:
$v^2 = \frac{2hc}{m} \left(\frac{\lambda_0 - \lambda}{\lambda \lambda_0}\right)$
$v = \sqrt{\frac{2hc}{m} \left(\frac{\lambda_0 - \lambda}{\lambda \lambda_0}\right)}$

(D) Sodium acetate is a salt of a strong base $(NaOH)$ and a weak acid $(CH_3COOH)$.
The formula for the $pH$ of a salt of a strong base and a weak acid is given by:
$pH = 7 + \frac{1}{2}pK_a + \frac{1}{2} \log c$
Given:
$c = 0.1 \ M = 10^{-1} \ M$
$K_a = 1.0 \times 10^{-5}$
Step $1$: Calculate $pK_a$
$pK_a = - \log K_a = - \log (10^{-5}) = 5$
Step $2$: Calculate $\log c$
$\log c = \log (10^{-1}) = -1$
Step $3$: Calculate $pH$
$pH = 7 + \frac{1}{2}(5) + \frac{1}{2}(-1)$
$pH = 7 + 2.5 - 0.5$
$pH = 9.0$

(D) Solutions which resist the change in the value of $pH$ when a small amount of acid or base is added to them are known as buffer solutions.
Since $pH = -\log[H_3O^+]$,a constant $pH$ implies that the concentration of $H_3O^+$ remains effectively constant.

(A) According to Boyle's law,for a fixed amount of gas at constant temperature,$P_1V_1 = P_2V_2$.
Given: $V_1 = 750.0 \ mL$,$P_1 = 840.0 \ mm \ Hg$,$P_2 = 360.0 \ mm \ Hg$.
Substituting the values: $840.0 \ mm \ Hg \times 750.0 \ mL = 360.0 \ mm \ Hg \times V_2$.
$V_2 = \frac{840.0 \times 750.0}{360.0} \ mL = 1750 \ mL$.
Converting to liters: $1750 \ mL = 1.750 \ L$.

(B) Given,$C_p = 10 \, cal \, K^{-1} \, mol^{-1}$.
$T_1 = 1000 \, K$,$T_2 = 100 \, K$.
Mass $m = 32 \, g$.
For $CD_2O$,molar mass $= 12 + 2(2) + 16 = 32 \, g \, mol^{-1}$.
Number of moles $n = \frac{32 \, g}{32 \, g \, mol^{-1}} = 1 \, mol$.
At constant pressure,the change in entropy is given by $\Delta S = n \times C_p \times \ln \frac{T_2}{T_1}$.
Substituting the values: $\Delta S = 1 \times 10 \times 2.303 \times \log \frac{100}{1000}$.
$\Delta S = 23.03 \times \log(10^{-1})$.
$\Delta S = 23.03 \times (-1) = -23.03 \, cal \, deg^{-1}$.

(A) The energy change for the transition from $n_1 = 1$ to $n_2 = 2$ is calculated as:
$\Delta E = - 2.0 \times 10^{-18} \times \left( \frac{1}{2^2} - \frac{1}{1^2} \right)$
$\Delta E = - 2.0 \times 10^{-18} \times \left( \frac{1}{4} - 1 \right) = - 2.0 \times 10^{-18} \times \left( - \frac{3}{4} \right) = 1.5 \times 10^{-18} \, J$
Using the relation $\Delta E = \frac{hc}{\lambda}$,the wavelength $\lambda$ is:
$\lambda = \frac{hc}{\Delta E} = \frac{6.625 \times 10^{-34} \times 3 \times 10^8}{1.5 \times 10^{-18}}$
$\lambda = 1.325 \times 10^{-7} \, m$

(C) In the periodic table,as we move down a group,the number of shells increases,which leads to an increase in the atomic radius.
$C$ $(Z=6)$ and $Ge$ $(Z=32)$ belong to Group $14$. Moving from $C$ to $Ge$ represents moving down the group,hence the atomic radius increases.
For option $A$,ionization enthalpy decreases down the group.
For option $B$,electron gain enthalpy (magnitude) generally decreases down the group.
For option $D$,noble character is a property of the group and does not increase significantly in the manner described.

(C) The bond order is directly proportional to the bond dissociation energy.
Calculating the bond order using Molecular Orbital Theory:
For $N_2$ ($14$ electrons): Bond order = $(10-4)/2 = 3$.
For $O_2$ ($16$ electrons): Bond order = $(10-6)/2 = 2$.
For $O_2^-$ ($17$ electrons): Bond order = $(10-7)/2 = 1.5$.
Since the bond order follows the order $N_2 (3) > O_2 (2) > O_2^- (1.5)$,the bond dissociation energy follows the same order: $N_2 > O_2 > O_2^-$.

(D) $Na_2O_2$ is the peroxide of sodium,not the superoxide.
The formula of sodium superoxide is $NaO_2$.
Therefore,the statement that $Na_2O_2$ is the superoxide of sodium is incorrect.

(C) In a $BCC$ unit cell of $CsCl$,the $Cs^{+}$ ion is at the body center and $Cl^{-}$ ions are at the corners.
The body diagonal of the cube is equal to $\sqrt{3} a$.
Along the body diagonal,the ions are in contact such that the distance is $r_{Cl^{-}} + 2r_{Cs^{+}} + r_{Cl^{-}} = \sqrt{3} a$.
This simplifies to $2(r_{Cs^{+}} + r_{Cl^{-}}) = \sqrt{3} a$.
Therefore,the correct expression is $r_{Cs^{+}} + r_{Cl^{-}} = \frac{\sqrt{3}}{2} a$.

(A) The osmotic pressure $\pi$ is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
For $C_2H_5OH$ (non-electrolyte): $i = 1$,$\pi = 1 \times 0.500 \times RT = 0.500 RT$.
For $Mg_3(PO_4)_2$ (dissociates into $3Mg^{2+} + 2PO_4^{3-}$,so $i = 5$): $\pi = 5 \times 0.100 \times RT = 0.500 RT$.
For $KBr$ (dissociates into $K^+ + Br^-$,so $i = 2$): $\pi = 2 \times 0.250 \times RT = 0.500 RT$.
For $Na_3PO_4$ (dissociates into $3Na^+ + PO_4^{3-}$,so $i = 4$): $\pi = 4 \times 0.125 \times RT = 0.500 RT$.
Since the osmotic pressure of all the given solutions is equal,they are isotonic.

(A) For the $0.2 \, M$ solution:
$R_1 = 50 \, \Omega$,$\kappa_1 = 1.4 \, S \, m^{-1}$.
Cell constant $G^* = R_1 \times \kappa_1 = 50 \times 1.4 = 70 \, m^{-1}$.
For the $0.5 \, M$ solution:
$R_2 = 280 \, \Omega$.
Specific conductance $\kappa_2 = \frac{G^*}{R_2} = \frac{70}{280} = 0.25 \, S \, m^{-1}$.
Molar conductivity $\Lambda_m = \frac{\kappa_2}{C} = \frac{0.25 \, S \, m^{-1}}{0.5 \, mol \, L^{-1}}$.
Convert concentration to $mol \, m^{-3}$: $0.5 \, mol \, L^{-1} = 0.5 \times 1000 \, mol \, m^{-3} = 500 \, mol \, m^{-3}$.
$\Lambda_m = \frac{0.25}{500} = 0.0005 = 5 \times 10^{-4} \, S \, m^2 \, mol^{-1}$.

(A) The given half-cell reactions are:
$(i) \ Mn^{2+} + 2e^{-} \rightarrow Mn; E^{\circ}_{red} = -1.18 \ V$
$(ii) \ Mn^{3+} + e^{-} \rightarrow Mn^{2+}; E^{\circ}_{red} = +1.51 \ V$
The target reaction is $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$.
This reaction is the sum of the reduction of $Mn^{2+}$ to $Mn$ and the oxidation of $Mn^{2+}$ to $Mn^{3+}$.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
Here,$Mn^{2+} \rightarrow Mn$ acts as the cathode $(E^{\circ} = -1.18 \ V)$ and $Mn^{2+} \rightarrow Mn^{3+} + e^{-}$ acts as the anode ($E^{\circ}_{ox} = -1.51 \ V$,so $E^{\circ}_{red} = +1.51 \ V$).
$E^{\circ}_{cell} = (-1.18 \ V) - (+1.51 \ V) = -2.69 \ V$.
Since $E^{\circ}_{cell}$ is negative,$\Delta G^{\circ} = -nFE^{\circ}_{cell}$ will be positive,meaning the reaction is non-spontaneous and will not occur.

(C) According to the Debye-$H$ückel-Onsager equation,the variation of molar (or equivalent) conductivity with concentration for strong electrolytes is given by the relation:
$\lambda_C = \lambda_{\infty} - B \sqrt{C}$
Here,$\lambda_C$ is the equivalent conductance at concentration $C$,$\lambda_{\infty}$ is the equivalent conductance at infinite dilution,and $B$ is a constant that depends on the nature of the solvent and temperature.

(D) Let the rate of reaction be $\frac{d[C]}{dt} = k[A]^{x}[B]^{y}.$
From the given data:
$1.2 \times 10^{-3} = k[0.1]^{x}[0.1]^{y} \quad (i)$
$1.2 \times 10^{-3} = k[0.1]^{x}[0.2]^{y} \quad (ii)$
$2.4 \times 10^{-3} = k[0.2]^{x}[0.1]^{y} \quad (iii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{1.2 \times 10^{-3}}{1.2 \times 10^{-3}} = \frac{k[0.1]^{x}[0.1]^{y}}{k[0.1]^{x}[0.2]^{y}}$
$1 = (0.5)^{y} \Rightarrow y = 0.$
Dividing equation $(i)$ by $(iii)$:
$\frac{1.2 \times 10^{-3}}{2.4 \times 10^{-3}} = \frac{k[0.1]^{x}[0.1]^{y}}{k[0.2]^{x}[0.1]^{y}}$
$0.5 = (0.5)^{x} \Rightarrow x = 1.$
Thus,the rate law is $\frac{d[C]}{dt} = k[A]^{1}[B]^{0} = k[A].$

(B) Metals that are highly reactive,such as those in the $s$-block (e.g.,$Ca$,$Na$,$K$),have a very negative standard reduction potential.
When an aqueous solution of their salts is electrolyzed,water is reduced at the cathode instead of the metal ion.
The reaction at the cathode is $H_2O(l) + e^- \rightarrow \frac{1}{2} H_2(g) + OH^-(aq)$.
Therefore,these metals cannot be obtained by the electrolysis of their aqueous salt solutions; they are typically extracted by the electrolysis of their molten salts.

(C) The decreasing order of acid strength of chlorine oxoacids is: $HClO_4 > HClO_3 > HClO_2 > HOCl$.
Reasoning: Consider the stability of the conjugate bases ($ClO_4^-$,$ClO_3^-$,$ClO_2^-$,$ClO^-$).
The negative charge is more delocalized on $ClO_4^-$ due to resonance involving more oxygen atoms,making $ClO_4^-$ the most stable conjugate base and thus $HClO_4$ the strongest acid.
As the number of oxygen atoms around the $Cl$-atom increases,the oxidation state of the $Cl$-atom increases,which increases the electron-withdrawing effect,thereby facilitating the release of the $H^+$ ion.

(A) Nitric oxide $(NO)$ is paramagnetic in the gaseous state because of the presence of one unpaired electron in its antibonding molecular orbital.
The molecular orbital configuration of $NO$ ($15$ electrons) is: $\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2p_z}^{2} \pi_{2p_x}^{2} \pi_{2p_y}^{2} \pi_{2p_x}^{*1}$.
Since it has an unpaired electron,it is paramagnetic,not diamagnetic.
Therefore,the statement that it is diamagnetic is incorrect.

(B) The crystal field splitting energy $(\Delta_o)$ is inversely proportional to the wavelength $(\lambda)$ of the light absorbed: $\Delta_o = \frac{hc}{\lambda}$.
Stronger ligands cause greater splitting and thus absorb light of shorter wavelengths.
The order of wavelengths for the given colors is: $\text{Red} (L_1) > \text{Yellow} (L_3) > \text{Green} (L_2) > \text{Blue} (L_4)$.
Since ligand strength $\propto \frac{1}{\lambda}$,the increasing order of ligand strength is: $L_1 < L_3 < L_2 < L_4$.

(D) The reduction of iron oxides to metallic iron is a key step in metallurgy.
In option $D$,the sequence is:
$1. \ 3Fe + 2O_2 \xrightarrow{\Delta} Fe_3O_4$
$2. \ Fe_3O_4 + CO \xrightarrow{600^{\circ}C} 3FeO + CO_2$
$3. \ FeO + CO \xrightarrow{700^{\circ}C} Fe + CO_2$
This sequence correctly represents the extraction of iron from its oxide ore using carbon monoxide as a reducing agent at specific temperatures.

(B) The rate of $S_{N}2$ reactions is inversely proportional to the steric hindrance around the electrophilic carbon atom.
As the number of alkyl groups attached to the carbon bearing the halogen increases,the steric crowding increases,which hinders the approach of the nucleophile.
The order of reactivity for alkyl halides in $S_{N}2$ reactions is: $Methyl \ halide > Primary \ (1^{\circ}) \ halide > Secondary \ (2^{\circ}) \ halide > Tertiary \ (3^{\circ}) \ halide$.
Comparing the given compounds:
$CH_3Cl$ (Methyl halide) > $CH_3CH_2Cl$ (Primary) > $(CH_3)_2CHCl$ (Secondary) > $(CH_3)_3CCl$ (Tertiary).
Therefore,the correct order is $CH_3Cl > CH_3CH_2Cl > (CH_3)_2CHCl > (CH_3)_3CCl$.

(A) The reaction of sodium phenoxide with $CO_2$ at $125\,^{\circ}C$ and $5 \ atm$ pressure is the Kolbe-Schmidt reaction,which yields salicylic acid ($2$-hydroxybenzoic acid) as the intermediate $B$.
In the second step,salicylic acid undergoes acetylation with acetic anhydride $(Ac_2O)$ to form acetylsalicylic acid,commonly known as Aspirin,which is the major product $C$.
The structure of $C$ is $2$-acetoxybenzoic acid.

(D) $R-CH_2-OH \xrightarrow{PCC} R-CHO$
$PCC$ (Pyridinium chlorochromate) is a mild oxidizing agent that selectively oxidizes primary alcohols to aldehydes without further oxidation to carboxylic acids.
$KMnO_4$,$K_2Cr_2O_7$,and $CrO_3$ are strong oxidizing agents that typically oxidize primary alcohols directly to carboxylic acids.

(C) The reaction sequence is as follows:
$1.$ Reduction: $CH_3-COOH \xrightarrow{LiAlH_4} CH_3-CH_2-OH$ (Ethanol,$A$)
$2.$ Chlorination: $CH_3-CH_2-OH \xrightarrow{PCl_5} CH_3-CH_2-Cl$ (Ethyl chloride,$B$)
$3.$ Dehydrohalogenation: $CH_3-CH_2-Cl \xrightarrow{Alc. KOH} CH_2=CH_2$ (Ethylene,$C$)
Therefore,the product $C$ is Ethylene.

(D) This reaction is known as the $Carbylamine$ test,which is used to identify primary amines.
When an aliphatic primary amine $(R-NH_2)$ is heated with chloroform $(CHCl_3)$ and ethanolic potassium hydroxide $(KOH)$,it forms an alkyl isocyanide $(R-NC)$,which has a foul smell.
The chemical equation is: $R-NH_2 + CHCl_3 + 3KOH (alc.) \rightarrow R-NC + 3KCl + 3H_2O$.
This reaction involves the formation of a dichlorocarbene intermediate $(:CCl_2)$.

(A) The basic strength of amines in aqueous solution depends on the combined effect of inductive effect,solvation effect,and steric hindrance. For methyl-substituted amines,the order of basic strength is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > C_6H_5NH_2$.
Since $pK_b = -\log K_b$,a higher $K_b$ value corresponds to a smaller $pK_b$ value.
Therefore,$(CH_3)_2NH$ (dimethylamine) has the highest basic strength and consequently the smallest $pK_b$ value.
$C_6H_5NH_2$ (aniline) is the least basic due to the resonance of the lone pair with the benzene ring.

(A) Condensation polymers are formed by the repeated condensation reaction between two different bi-functional or tri-functional monomeric units,usually involving the loss of small molecules like $H_2O$,$HCl$,etc.
$Dacron$ (also known as $Terylene$) is a condensation polymer formed by the polymerization of ethylene glycol $(HO-CH_2-CH_2-OH)$ and terephthalic acid $(HOOC-C_6H_4-COOH)$.
$Neoprene$,$Teflon$,and $Acrylonitrile$ (which forms $PAN$) are addition polymers.
Therefore,the correct option is $A$.

(A) Quinoline is an alkaloid and is not present in $DNA$.
$DNA$ contains four nitrogenous bases: adenine,guanine,cytosine,and thymine.

(D) Option $A$: The reaction $Li_2O + 2KCl \rightarrow 2LiCl + K_2O$ is thermodynamically unfavorable because $Li_2O$ is more stable than $K_2O$.
Option $B$: The reaction $[CoCl(NH_3)_5]^{2+} + 5H^+ \rightarrow Co^{2+} + 5NH_4^+ + Cl^-$ is balanced in terms of atoms and charge ($+2 + 5 = +7$ on left,$+2 + 5 - 1 = +6$ on right; wait,let's re-check: $Co^{2+} + 5NH_4^+ + Cl^-$ gives $+2 + 5 - 1 = +6$. The charge is not balanced).
Option $C$: The reaction $[Mg(H_2O)_6]^{2+} + (EDTA)^{4-} \rightarrow [Mg(EDTA)]^{2-} + 6H_2O$ is balanced.
Option $D$: The reaction $CuSO_4 + 4KCN \rightarrow K_2[Cu(CN)_4] + K_2SO_4$ is balanced: $Cu$ $(1)$,$S$ $(1)$,$O$ $(4)$,$K$ $(4)$,$C$ $(4)$,$N$ $(4)$ on both sides.

(A) In a face centered cubic $(FCC)$ lattice:
Number of atoms $A$ at corners $= 8 \times \frac{1}{8} = 1$.
Number of face centered positions in an $FCC$ unit cell is $6$.
Since one atom $B$ is missing from one of the face centered points,the number of atoms $B$ present $= 5 \times \frac{1}{2} = \frac{5}{2}$.
The ratio of atoms $A:B = 1 : \frac{5}{2} = 2 : 5$.
Therefore,the formula of the ionic compound is $A_2B_5$.

(C) The deposition of metals during electrolysis occurs in the order of their reduction potentials,starting from the highest reduction potential to the lowest reduction potential.
Given standard reduction potentials: $E_A^\circ = -1.2 \ V$,$E_B^\circ = 0.6 \ V$,$E_C^\circ = 0.85 \ V$,and $E_D^\circ = -0.76 \ V$.
Arranging these in decreasing order: $0.85 \ V (C) > 0.6 \ V (B) > -0.76 \ V (D) > -1.2 \ V (A)$.
Thus,the sequence of deposition is $C, B, D, A$.

(C) Given half-life $t_{1/2} = 15 \ minutes$.
Total time $T = 1 \ hr = 60 \ minutes$.
The number of half-lives $n$ is calculated as $n = \frac{T}{t_{1/2}} = \frac{60}{15} = 4$.
The amount of substance remaining is given by the formula $\frac{N}{N_0} = (\frac{1}{2})^n$.
Substituting the value of $n$,we get $\frac{N}{N_0} = (\frac{1}{2})^4 = \frac{1}{16}$.
Therefore,the amount of substance left after $1 \ hour$ is $\frac{1}{16}$ of the original amount.

(C) The total charge $Q$ passed through the cell is given by $Q = i \times t$.
Given $i = 10.0 \, A$ and $t = 2.00 \, h = 2.00 \times 3600 \, s = 7200 \, s$.
$Q = 10.0 \, A \times 7200 \, s = 72,000 \, C$.
According to Faraday's law,the number of moles of electrons required to deposit $n$ moles of metal $X$ with oxidation state $x$ is $n \times x = \frac{Q}{F}$.
Here,$n = 0.250 \, mol$ and $F = 96,500 \, C \, mol^{-1}$.
$0.250 \times x = \frac{72,000}{96,500}$.
$0.250 \times x \approx 0.746$.
$x = \frac{0.746}{0.250} \approx 2.98 \approx 3$.
Therefore,the oxidation state of metal $X$ is $3+$.

(C) The reaction of $CaF_2$ and $SiO_2$ with concentrated $H_2SO_4$ produces silicon tetrafluoride gas $(SiF_4)$:
$2CaF_2 + SiO_2 + 2H_2SO_4 \longrightarrow SiF_4 + 2CaSO_4 + 2H_2O$
When $SiF_4$ is hydrolyzed,it forms hydrofluosilicic acid and a white gelatinous precipitate of silicic acid $(H_2SiO_3)$:
$3SiF_4 + 3H_2O \longrightarrow 2H_2SiF_6 + H_2SiO_3$ (white gelatinous precipitate of silicic acid).

(C) The atomic number of $Co$ is $27$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
Since the complex is octahedral and diamagnetic,the ligands must be strong field ligands that cause pairing of electrons in the $3d$ orbitals.
This results in two empty $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals,which hybridize to form six $d^2sp^3$ hybrid orbitals.
Thus,the hybridisation involved is $d^2sp^3$.

(C) The iron obtained from the blast furnace is known as $Pig \ Iron$. It contains about $4 \%$ carbon and many impurities like $S, P, Si, Mn$ in smaller amounts.

(A) For $[Fe(CN)_6]^{3-}$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$. Since $CN^-$ is a strong field ligand,it causes pairing of electrons,resulting in one unpaired electron $(n=1)$. Thus,it is paramagnetic.
For $[FeF_6]^{3-}$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$. Since $F^-$ is a weak field ligand,it does not cause pairing,resulting in five unpaired electrons $(n=5)$. Thus,it is also paramagnetic.
Therefore,both complexes are paramagnetic.

(D) The Cannizzaro reaction is a disproportionation reaction of aldehydes lacking $\alpha$-hydrogen atoms,where one molecule of the aldehyde is reduced to an alcohol and another is oxidized to a salt of a carboxylic acid. It does not involve the formation of a new carbon-carbon bond.
$2C_6H_5CHO + KOH \longrightarrow C_6H_5CH_2OH + C_6H_5COOK$
In contrast,the Reimer-Tiemann reaction,Friedel-Crafts acylation,and Wurtz reaction all involve the formation of new carbon-carbon bonds.

(A) The bond length depends on the size of the halogen atom.
As the size of the halogen atom increases down the group $(F < Cl < Br < I)$,the bond length of the $C$-halogen bond also increases.
Therefore,the order of increasing bond length is $CH_3F < CH_3Cl < CH_3Br < CH_3I$.

(B) is the correct option.
Allyl phenyl ether is prepared by the $Williamson$ ether synthesis.
In this reaction,sodium phenoxide $(C_6H_5-ONa)$ acts as a nucleophile and attacks the electrophilic carbon of allyl bromide $(CH_2=CH-CH_2-Br)$ via an $S_N2$ mechanism.
Aryl halides like bromobenzene $(C_6H_5-Br)$ do not undergo nucleophilic substitution under ordinary conditions due to the partial double bond character of the $C-X$ bond.
Reaction: $C_6H_5-ONa + Br-CH_2-CH=CH_2 \rightarrow C_6H_5-O-CH_2-CH=CH_2 + NaBr$

(C) The reaction uses $DMF$ (Dimethylformamide),which is a polar aprotic solvent,favoring the $S_N2$ mechanism.
$S_N2$ reactions proceed with complete inversion of configuration (Walden inversion).
Among the given options,$C_6H_5CH(CH_3)Br$ is a secondary alkyl bromide that is sterically accessible for $S_N2$ attack.
$C_6H_5CH_2Br$ is primary and also undergoes $S_N2$,but $C_6H_5CH(CH_3)Br$ is the classic example of an optically active secondary halide showing inversion.
$C_6H_5CH(C_6H_5)Br$ and $C_6H_5C(CH_3)(C_6H_5)Br$ are sterically hindered and likely to proceed via $S_N1$ or show mixed results.

(B) Buna-$S$ is a copolymer formed by the polymerization of $1,3$-butadiene $(CH_2=CH-CH=CH_2)$ and styrene $(C_6H_5-CH=CH_2)$.
Its structure consists of repeating units derived from these two monomers,represented as $-(CH_2-CH=CH-CH_2-CH(C_6H_5)-CH_2)_n-$.

(C) Acetone $(CH_3COCH_3)$ reacts with iodine $(I_2)$ in the presence of a base like potassium hydroxide $(KOH)$ to undergo the iodoform reaction.
This reaction is a characteristic test for methyl ketones.
The chemical equation is:
$CH_3COCH_3 + 3I_2 + 4KOH \longrightarrow CHI_3 + CH_3COOK + 3KI + 3H_2O$
The major product formed is iodoform $(CHI_3)$.

(D) The appearance of colour in solid alkali metal halides is generally due to $F$-centres (Farbe centres).
These are anionic vacancies occupied by unpaired electrons.
When white light falls on these crystals,the electrons absorb energy from the visible region to get excited,which results in the crystal exhibiting a complementary colour.
For example,$NaCl$ with $F$-centres appears yellow,$KCl$ appears magenta,and $KBr$ appears blue.

(D) For the reaction $2SO_2 + O_2 \rightleftharpoons 2SO_3$,the rate of reaction is expressed as:
Rate $= -\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt}$
Given that $\frac{d[O_2]}{dt} = -2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
Equating the terms for $SO_2$ and $O_2$:
$-\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{d[O_2]}{dt}$
$\frac{d[SO_2]}{dt} = 2 \times \frac{d[O_2]}{dt}$
$\frac{d[SO_2]}{dt} = 2 \times (-2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1})$
$\frac{d[SO_2]}{dt} = -5.00 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

(B) The rate of reaction is defined as the rate of disappearance of reactant divided by its stoichiometric coefficient,which is equal to the rate of appearance of product divided by its stoichiometric coefficient.
For the reaction $2N_2O_5 \to 4NO_2 + O_2$,the rate expression is:
$-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt}$
Given that $-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$ and $\frac{d[NO_2]}{dt} = k'[N_2O_5]$,substitute these into the rate expression:
$\frac{1}{2} (k[N_2O_5]) = \frac{1}{4} (k'[N_2O_5])$
$\frac{k}{2} = \frac{k'}{4}$
$k' = 2k$ or $2k = k'$.

(B) The reducing power of a species is inversely proportional to its standard reduction potential $(E^o)$.
The given standard reduction potentials are:
$E^o (Al^{3+}/Al) = -1.66 \ V$
$E^o (Fe^{3+}/Fe^{2+}) = +0.77 \ V$
$E^o (Br_2/Br^-) = +1.09 \ V$
$A$ more negative $E^o$ value indicates a stronger reducing agent.
Comparing the values: $-1.66 \ V < +0.77 \ V < +1.09 \ V$.
Therefore,the order of reducing power is $Br^- < Fe^{2+} < Al$.

(B) The molecular geometry of $BrF_5$ is square pyramidal,not trigonal bipyramidal.
$Br$ in $BrF_5$ has $7$ valence electrons,forms $5$ bonds with $F$ atoms,and has $1$ lone pair,resulting in $sp^3d^2$ hybridization and square pyramidal geometry.

(B) For $[Mn(CN)_5]^{2-}$,let the oxidation state of $Mn$ be $x$.
$x + 5(-1) = -2$
$x - 5 = -2$
$x = +3$.
Thus,the correct $IUPAC$ name is Pentacyanomanganate$(III)$ ion.
Therefore,the combination in option $B$ is incorrect.

(C) In the formation of the coordination complex $[Co(NH_3)_6]^{3+}$,the central metal ion $Co^{3+}$ acts as the Lewis acid because it accepts lone pairs of electrons from the ligands.
$NH_3$ acts as the Lewis base because it donates lone pairs of electrons to the central metal ion.
Therefore,the reaction is: $\mathop {Co^{3+}}\limits_{\text{Lewis acid}} + \mathop {6NH_3}\limits_{\text{Lewis base}} \to \mathop {[Co(NH_3)_6]^{3+}}\limits_{\text{adduct}}$

(D) The given reaction is known as the Gattermann reaction (specifically,the Gattermann formylation of phenol).
In this reaction,an aromatic compound like phenol is treated with a mixture of $HCN$ and dry $HCl$ gas in the presence of a Lewis acid catalyst such as anhydrous $ZnCl_2$ or $AlCl_3$ to introduce a formyl $(-CHO)$ group into the aromatic ring.

(B) The reduction of $benzene-diazonium$ chloride $(C_6H_5N_2Cl)$ with $Zn/HCl$ is a strong reduction process.
It leads to the formation of $phenylhydrazine$ $(C_6H_5NHNH_2)$ as the final product.
Note: While mild reducing agents like $SnCl_2/HCl$ or $Na_2SO_3$ can produce $phenylhydrazine$,strong reduction with $Zn/HCl$ also yields $phenylhydrazine$ as the primary product.

(C) In the presence of concentrated $H_2SO_4$,two moles of chlorobenzene react with one mole of chloral $(CCl_3CHO)$ through a condensation reaction.
The oxygen atom from the aldehyde group of chloral and the hydrogen atoms from the para-positions of the two chlorobenzene rings are removed as a water molecule $(H_2O)$.
This results in the formation of $p,p'$-dichlorodiphenyltrichloroethane,commonly known as $DDT$.
The reaction is:
$2 C_6H_5Cl + CCl_3CHO \xrightarrow{H_2SO_4} (Cl-C_6H_4)_2CH-CCl_3 + H_2O$

(C) The Tishchenko reaction is a modification of the Cannizzaro reaction.
In the Tishchenko reaction,two molecules of an aldehyde (which lacks an $\alpha$-hydrogen atom) undergo disproportionation in the presence of an alkoxide catalyst (such as aluminium alkoxide or sodium alkoxide) to form an ester.
In the Cannizzaro reaction,the base used is typically a strong aqueous base like $NaOH$ or $KOH$,and the products are a carboxylic acid salt and an alcohol.
Both reactions involve the disproportionation of aldehydes lacking $\alpha$-hydrogens,making the Tishchenko reaction a variation of the Cannizzaro process.

(C) Diphenhydramine is a well-known antihistamine drug used to treat symptoms of allergy,such as itching,runny nose,and sneezing.

(C) The $C-O$ bond length in alcohol is $142 \ pm$ and in phenol it is $136 \ pm$. The $C-O$ bond length in phenol is shorter than that in methanol due to the conjugation of unshared pair of electrons on oxygen with the ring, which imparts double bond character to the $C-O$ bond. Thus, statement $C$ is incorrect.

(B) According to the Arrhenius equation:
$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 \ R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$
Given: $k_1 = 1.3 \times 10^{-4} \ M^{-1} \ s^{-1}$,$T_1 = 100 + 273 = 373 \ K$
$k_2 = 1.3 \times 10^{-3} \ M^{-1} \ s^{-1}$,$T_2 = 150 + 273 = 423 \ K$
$\log \left( \frac{1.3 \times 10^{-3}}{1.3 \times 10^{-4}} \right) = \frac{E_a}{2.303 \times 8.314} \left( \frac{423 - 373}{373 \times 423} \right)$
$\log(10) = \frac{E_a}{19.147} \left( \frac{50}{157779} \right)$
$1 = \frac{E_a \times 50}{3020755.7}$
$E_a = \frac{3020755.7}{50} \approx 60415 \ J \ mol^{-1} \approx 60.4 \ kJ \ mol^{-1}$
Thus,the closest value is $60 \ kJ \ mol^{-1}$.

(C) The reduction reaction at the cathode is: $Cu^{2+} (aq) + 2e^- \to Cu (s)$.
To deposit $1 \ mole$ of $Cu$ $(63.5 \ g)$,$2 \ moles$ of electrons are required.
Number of moles of $Cu$ to be deposited $= \frac{\text{Given mass}}{\text{Atomic mass}} = \frac{6.35 \ g}{63.5 \ g/mol} = 0.1 \ mol$.
Since $1 \ mole$ of $Cu$ requires $2 \ moles$ of electrons,$0.1 \ mole$ of $Cu$ requires $0.1 \times 2 = 0.2 \ moles$ of electrons.
Total number of electrons $= 0.2 \times N_A = \frac{2}{10} \times N_A = \frac{N_A}{5}$.