KVPY 2020 Chemistry Question Paper with Answer and Solution

(B)
The correct stability order of the given carbocations is $II > I > III$.
In structure $II$,the presence of the electron-releasing $-NMe_2$ group significantly stabilizes the carbocation through the $+M$ effect.
In structure $I$,the carbocation is stabilized by resonance with the benzene ring.
In structure $III$,due to $SIR$ (Steric Inhibition of Resonance),the $-C(Me)_2^+$ group is forced out of the plane of the benzene ring by the two ortho-methyl groups. This loss of conjugation significantly decreases its stability compared to $I$ and $II$.

(C) The $pH$ at the equivalence point for the titration of a weak acid with a weak base is given by the formula:
$pH = \frac{1}{2}(pK_{w} + pK_{a} - pK_{b})$
From this expression,it is clear that the $pH$ is independent of the concentration of the acid and the base (statement $ii$ is correct,$i$ is incorrect).
The expression also shows that the $pH$ depends on the $pK_{a}$ of the acid and the $pK_{b}$ of the base (statement $iii$ is correct,$iv$ is incorrect).
Therefore,the correct statements are $(ii)$ and $(iii)$.

(B) For a chemical reaction to be spontaneous at constant temperature and pressure,the change in Gibbs energy of the system must be negative,i.e.,$\Delta G_{\text{sys}} < 0$. This corresponds to statement $(i)$.
According to the second law of thermodynamics,for any spontaneous process,the total entropy change of the universe (system + surroundings) must be positive,i.e.,$\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0$. This corresponds to statement $(iv)$.
Statement $(ii)$ is incorrect because the total Gibbs energy change for the universe is not a standard criterion for spontaneity.
Statement $(iii)$ is not necessarily true because a reaction can be spontaneous even if the entropy change of the system is negative,provided the enthalpy change is sufficiently negative to make $\Delta G$ negative.
Therefore,statements $(i)$ and $(iv)$ are always true.

(C) The treatment of alkaline $KMnO_{4}$ solution with $KI$ solution oxidizes iodide $(I^{-})$ to iodate $(IO_{3}^{-})$.
The balanced chemical equation is as follows:
$2MnO_{4}^{-} + I^{-} + H_{2}O \longrightarrow 2MnO_{2} + IO_{3}^{-} + 2OH^{-}$

(D) The electronic configuration of the $C_{2}$ molecule is $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \pi 2p_{y}^{2} = \pi 2p_{x}^{2}$.
According to the molecular orbital theory for $C_{2}$,the next available molecular orbital after the $\pi 2p$ orbitals is the $\sigma 2p_{z}$ orbital.
Therefore,if an extra electron is added to the $C_{2}$ molecule,it will occupy the $\sigma 2p_{z}$ molecular orbital.
Thus,the correct option is $D$.

(B) According to Bohr's model,the velocity of an electron in a hydrogen-like species is given by the formula: $v_n = v_0 \times \frac{Z}{n}$,where $Z$ is the atomic number and $n$ is the orbit number.
For $He^{+}$,the atomic number $Z = 2$.
In the first orbit $(n=1)$,the velocity is $v = v_0 \times \frac{2}{1} = 2v_0$. Thus,$v_0 = \frac{v}{2}$.
In the second orbit $(n=2)$,the velocity $v^{\prime}$ is given by: $v^{\prime} = v_0 \times \frac{2}{2} = v_0$.
Substituting the value of $v_0$ from the first step: $v^{\prime} = \frac{v}{2} = 0.5v$.

(A) $1$. The molecular formula of $X$ is $C_{11}H_{14}$.
$2$. Ozonolysis of $X$ gives $P$ and $Q$. $P$ gives a positive iodoform test $(I_2/NaOH)$ and does not reduce Tollen's reagent,identifying $P$ as a methyl ketone (specifically acetophenone,$Ph-CO-CH_3$).
$3$. $Q$ gives a positive Fehling's test but no iodoform test,identifying $Q$ as an aliphatic aldehyde (specifically propanal,$CH_3CH_2CHO$).
$4$. Combining these fragments ($Ph-CO-CH_3$ and $CH_3CH_2CHO$) via ozonolysis reversal,the structure of $X$ is $Ph-C(CH_3)=CH-CH_2-CH_3$.
$5$. Hydrogenation of $X$ $(Ph-C(CH_3)=CH-CH_2-CH_3 + H_2 \rightarrow Ph-CH(CH_3)-CH_2-CH_2-CH_3)$ produces a chiral center at the carbon attached to the phenyl group,making the product optically active.

(C) For a reversible first-order reaction $A \underset{K_b}{\stackrel{K_f}{\rightleftharpoons}} B$,the rate law is given by: $-\frac{d[A]}{dt} = K_f[A] - K_b[B]$.
At equilibrium,$-\frac{d[A]}{dt} = 0$,so $K_f[A_{\text{eq}}] = K_b[B_{\text{eq}}]$. Since $[B_{\text{eq}}] = A_0 - A_{\text{eq}}$,we have $K_b = \frac{K_f A_{\text{eq}}}{A_0 - A_{\text{eq}}}$.
The integrated rate equation for this system is $(K_f + K_b)t = \ln \left( \frac{A_0 - A_{\text{eq}}}{A_t - A_{\text{eq}}} \right)$.
Given $A_t = \frac{A_0 + A_{\text{eq}}}{2}$,we substitute this into the expression:
$A_t - A_{\text{eq}} = \frac{A_0 + A_{\text{eq}}}{2} - A_{\text{eq}} = \frac{A_0 - A_{\text{eq}}}{2}$.
Substituting this into the integrated rate equation:
$(K_f + K_b)t = \ln \left( \frac{A_0 - A_{\text{eq}}}{(A_0 - A_{\text{eq}})/2} \right) = \ln(2)$.
Therefore,$t = \frac{\ln 2}{K_f + K_b}$.

(A) The standard Gibbs free energy change for the reaction is calculated as: $\Delta G^{\circ} = \Delta G_{f}^{\circ}(CaO) + \Delta G_{f}^{\circ}(CO_{2}) - \Delta G_{f}^{\circ}(CaCO_{3})$.
Substituting the given values: $\Delta G^{\circ} = (-603.501) + (-394.389) - (-1128.79) = +130.9 \, kJ \, mol^{-1} = 130900 \, J \, mol^{-1}$.
Using the relation $\Delta G^{\circ} = -RT \ln K_{p}$,we have $\ln K_{p} = -\frac{\Delta G^{\circ}}{RT}$.
$\ln K_{p} = -\frac{130900}{8.314 \times 298} \approx -52.834$.
Converting to base $10$ logarithm: $\log_{10} K_{p} = \frac{-52.834}{2.303} \approx -22.941$.
Thus,$K_{p} = 10^{-22.941} \approx 1.13 \times 10^{-23}$.
Since $K_{p} = P_{CO_{2}}$,the partial pressure of $CO_{2(g)}$ is $1.13 \times 10^{-23} \, atm$.

(B) When the partition is removed,each gas expands to occupy the total volume $V_{total} = V_{1} + V_{2}$.
For an ideal gas undergoing isothermal expansion,the change in entropy is given by $\Delta S = n R \ln \left( \frac{V_{final}}{V_{initial}} \right)$.
Since entropy is an extensive property,the total change in entropy is the sum of the entropy changes for each gas:
$\Delta S_{total} = \Delta S_{He} + \Delta S_{Ne}$
$\Delta S_{total} = n_{1} R \ln \left( \frac{V_{1}+V_{2}}{V_{1}} \right) + n_{2} R \ln \left( \frac{V_{1}+V_{2}}{V_{2}} \right)$
Thus,the correct option is $B$.

(B) The reaction of sodium borohydride with iodine produces diborane,which is a Lewis acid $(X = B_2H_6)$.
$2 NaBH_4 + I_2 \xrightarrow{\text{Diglyme}} B_2H_6(X) + H_2 + 2 NaI$
Heating diborane with ammonia in a $1:2$ molar ratio produces borazine,which is a cyclic compound $(Y = B_3N_3H_6)$ and hydrogen gas $(Z = H_2)$.
$3 B_2H_6 + 6 NH_3 \xrightarrow{\Delta} 2 B_3N_3H_6(Y) + 12 H_2(Z) \uparrow$

(A) compound exhibits optical activity if it is chiral,meaning it lacks any internal elements of symmetry (like a plane of symmetry or a center of inversion) and is non-superimposable on its mirror image.
$I$: This is a cyclobutane derivative with a center of inversion. It is achiral.
$II$: This is a $1,2$-diphenylcyclopropane derivative. The trans-isomer is chiral and optically active.
$III$: This is an allene derivative. Since the terminal carbons have different substituents ($H, H$ on one side and $Me, Cl$ on the other),it is chiral and optically active.
$IV$: This is a sugar derivative (like a pentose) with chiral centers. It is chiral and optically active.
$V$: This is $3$-methylhexane. The $C3$ atom is bonded to four different groups $(-H, -CH_3, -CH_2CH_3, -CH_2CH_2CH_3)$,making it a chiral center. It is optically active.
Based on the provided options and standard chemical analysis,compounds $II$,$III$,$IV$,and $V$ are optically active. Given the options,the most appropriate choice is $A$ (assuming $III$ is included in the set or the question implies specific structures).

(D) The correct answer is $(D)$.
Methylcyclohexane is a molecule that contains $1^{\circ}$,$2^{\circ}$,and $3^{\circ}$ carbon atoms.
In methylcyclohexane:
- The methyl group carbon is a $1^{\circ}$ carbon atom (attached to one other carbon).
- The five carbon atoms in the cyclohexane ring (excluding the one attached to the methyl group) are $2^{\circ}$ carbon atoms (each attached to two other carbons).
- The carbon atom in the cyclohexane ring that is attached to the methyl group is a $3^{\circ}$ carbon atom (attached to three other carbons: two in the ring and one methyl group).
Thus,it contains $1^{\circ}$,$2^{\circ}$,and $3^{\circ}$ carbon atoms.

(B)
Aniline can be purified by steam distillation. This method is used to separate compounds that are steam volatile and insoluble in water. In this process,steam from a generator is passed through a flask containing the liquid to be distilled. The resulting mixture of steam and the volatile organic compound is condensed and collected in a receiver.

(B) The acidity of a hydrocarbon is determined by the stability of its conjugate base. The more stable the conjugate base,the more acidic the hydrocarbon.
Option $(B)$ represents cyclopentadiene. Upon losing a proton $(H^+)$,it forms the cyclopentadienyl anion,which contains $6 \pi$ electrons ($4n+2$ where $n=1$). This makes the conjugate base aromatic and thus highly stable.
Other options do not form such stable aromatic conjugate bases upon deprotonation. Therefore,cyclopentadiene is the most acidic among the given compounds.

(B) At $STP$,$1 \, L$ of water gas contains $1:1$ ratio of $CO$ and $H_2$ gases.
Thus,$V_{CO} = 0.5 \, L$ and $V_{H_2} = 0.5 \, L$.
Volume of $O_2$ in $9 \, L$ of air $= 9 \times 0.20 = 1.8 \, L$.
Upon ignition,$CO$ reacts with $O_2$ according to the equation: $2CO_{(g)} + O_{2(g)} \longrightarrow 2CO_{2(g)}$.
Since $CO$ is the limiting reagent ($0.5 \, L$ of $CO$ requires $0.25 \, L$ of $O_2$),the volume of $CO_2$ produced is equal to the volume of $CO$ consumed,which is $0.5 \, L$.
Number of moles of $CO_2 = \frac{\text{Volume at } STP}{22.4 \, L/mol} = \frac{0.5}{22.4} \approx 0.022 \, mol$.

(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
For $\lambda_1 = 400 \ nm = 400 \times 10^{-9} \ m$:
$E_1 = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} \ J \approx 4.97 \times 10^{-19} \ J$.
Converting to $eV$: $E_1 = \frac{4.97 \times 10^{-19}}{1.602 \times 10^{-19}} \ eV \approx 3.1 \ eV$.
For $\lambda_2 = 800 \ nm = 800 \times 10^{-9} \ m$:
$E_2 = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{800 \times 10^{-9}} \ J \approx 2.48 \times 10^{-19} \ J$.
Converting to $eV$: $E_2 = \frac{2.48 \times 10^{-19}}{1.602 \times 10^{-19}} \ eV \approx 1.55 \ eV$.
Since the work function $\Phi = 2 \ eV$,photoelectrons are ejected only if the incident photon energy $E \ge \Phi$.
Here,$E_1 (3.1 \ eV) > 2 \ eV$ and $E_2 (1.55 \ eV) < 2 \ eV$.
Therefore,only $400 \ nm$ light leads to the ejection of photoelectrons.

(D) is the correct statement.
Explanation of why others are incorrect:
$(a)$ The equilibrium constant is temperature-dependent and changes with temperature for a given reaction.
$(b)$ The equilibrium constant provides information about the extent of the reaction,not the rate at which equilibrium is reached.
$(c)$ At equilibrium,the forward and backward reactions do not stop; they continue at equal rates,making the process dynamic,not static.

(D) The expressions for most probable velocity $(V_{mp})$,average velocity $(\bar{V})$,and root mean square velocity $(V_{rms})$ are as follows:
$V_{mp} = \sqrt{\frac{2RT}{M}}$
$\bar{V} = \sqrt{\frac{8RT}{\pi M}} \approx \sqrt{\frac{2.55RT}{M}}$
$V_{rms} = \sqrt{\frac{3RT}{M}}$
Comparing these values,we get the order: $V_{mp} < \bar{V} < V_{rms}$.
In the Maxwell-Boltzmann distribution curve,the peak corresponds to $V_{mp}$,followed by $\bar{V}$,and then $V_{rms}$ on the right side of the peak.
Thus,the correct plot is the one where the order from left to right is $V_{mp}$,$\bar{V}$,$V_{rms}$.

(C) The electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 3d^{10} 4s^1$.
There is only one unpaired electron,which is located in the $4s$ orbital.
For the $4s$ orbital,the principal quantum number $n=4$.
The azimuthal quantum number $l=0$ for an $s$-orbital.
The magnetic quantum number $m=0$.
The spin quantum number $s$ can be $+\frac{1}{2}$ or $-\frac{1}{2}$.
Thus,the correct set of quantum numbers is $n=4, l=0, m=0, s=+\frac{1}{2}$.

(D) The correct answer is $(d)$.
$AsCl_3$ has a trigonal pyramidal geometry with a lone pair on the $As$ atom,which results in a non-zero net dipole moment,making it a polar molecule.
In contrast,$AlCl_3$ (trigonal planar),$CCl_4$ (tetrahedral),and $SeCl_6$ (octahedral) are highly symmetrical molecules where the individual bond dipoles cancel each other out,resulting in a net dipole moment of zero $(\mu = 0)$.

(A) According to Fajan's rule,the covalent character of an ionic compound increases with a decrease in the size of the cation.
As we move down the group $2$ $(Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+})$,the size of the cation increases.
Consequently,the polarising power of the cation decreases,leading to a decrease in the covalent character.
Therefore,the correct order of covalent character is $BaCl_2 < SrCl_2 < CaCl_2 < MgCl_2$.

(C) .
Statement $c$ is correct,whereas statements $a$,$b$,and $d$ are incorrect.
$2.005$ has four significant figures because zeros between two non-zero digits are significant.
Corrected statements:
$100$ has one significant figure.
$1.00 \times 10^2$ has three significant figures.
$0.0025$ has two significant figures.

(A) The thermodynamic cycle in the $p-V$ plane consists of two isothermal and two adiabatic processes.
$1. A \rightarrow B$: Isothermal expansion $(T = \text{constant})$,entropy increases ($S$ increases).
$2. B \rightarrow C$: Adiabatic expansion $(S = \text{constant})$,temperature decreases ($T$ decreases).
$3. C \rightarrow D$: Isothermal compression $(T = \text{constant})$,entropy decreases ($S$ decreases).
$4. D \rightarrow A$: Adiabatic compression $(S = \text{constant})$,temperature increases ($T$ increases).
In the $T-S$ plane,isothermal processes appear as horizontal lines $(T = \text{constant})$ and adiabatic processes appear as vertical lines $(S = \text{constant})$. Based on the directions of the processes,the correct representation is a rectangle in the $T-S$ plane with $A \rightarrow B$ at a higher temperature than $C \rightarrow D$.

(C) The ionisation potential $(IP)$ generally decreases as we move down a group in the periodic table due to the increase in atomic size and shielding effect.
$Na$ and $K$ both belong to Group $1$ (alkali metals).
Since $K$ is below $Na$ in the periodic table,the $IP$ of $K$ must be lower than the $IP$ of $Na$.
The $IP$ of $Na$ is $5.14 \ eV$.
Among the given options,only $4.3 \ eV$ is less than $5.14 \ eV$.
Therefore,the correct option is $C$.

(D) The hydrocarbon $(X)$ is but$-1-$yne $(CH_3-CH_2-C \equiv CH)$.
$1$. It decolourises bromine water due to the presence of a triple bond.
$2$. It forms a white precipitate with ethanolic $AgNO_3$ (Tollens' reagent) because it is a terminal alkyne,which has an acidic hydrogen.
$3$. Treatment of but$-1-$yne with $HgCl_2$ in aqueous $H_2SO_4$ (hydration) produces butan$-2-$one $(CH_3-CH_2-CO-CH_3)$.
$4$. Butan$-2-$one contains a methyl ketone group $(CH_3-CO-)$,which gives a positive iodoform test (yellow precipitate of $CHI_3$) when treated with $I_2$ and $NaOH$.

(A) The percentage of sulfur in the organic compound is calculated as follows:
$\text{Percentage of } S = \frac{32 \times \text{mass of } BaSO_4}{233 \times \text{mass of organic compound}} \times 100$
$= \frac{32 \times 0.233}{233 \times 0.102} \times 100 = 31.37 \ \%$
Now,we calculate the percentage of sulfur in each option:
$A$: $1,4-\text{dithiane } (C_4H_8S_2)$,Molar mass $= 120 \ g/mol$. $\% S = \frac{64}{120} \times 100 = 53.33 \ \%$
$B$: $1,4-\text{dimethoxybut-2-ene}$,contains no sulfur.
$C$: $\text{tert-butylthiol } (C_4H_{10}S)$,Molar mass $= 90 \ g/mol$. $\% S = \frac{32}{90} \times 100 = 35.55 \ \%$
$D$: $\text{tetrahydrothiophene } (C_4H_8S)$,Molar mass $= 88 \ g/mol$. $\% S = \frac{32}{88} \times 100 = 36.36 \ \%$
Re-evaluating the calculation: The mass of $BaSO_4$ is $0.233 \ g$,which corresponds to $\frac{0.233}{233} = 0.001 \ mol$ of $S$. The mass of $S$ is $0.001 \times 32 = 0.032 \ g$. The percentage of $S$ is $\frac{0.032}{0.102} \times 100 \approx 31.37 \ \%$. Among the choices,$1,4-\text{dithiane}$ has two sulfur atoms. If the compound is $1,4-\text{dithiane}$,the theoretical percentage is $53.33 \ \%$. Given the options provided,$1,4-\text{dithiane}$ is the only sulfur-containing cyclic compound that fits the structural context of the options.

(C) The relationship between volume strength and normality for $H_2O_2$ is given by the formula: $\text{Volume strength} = 5.6 \times \text{Normality}$.
Given,$\text{Normality} = 1.79 \, N$.
Therefore,$\text{Volume strength} = 5.6 \times 1.79 = 10.024 \, \text{volume}$.
Rounding to the nearest whole number,the strength is $10 \, \text{volume}$.
Thus,the correct option is $C$.

(D)
All given statements are correct.
$(i)$ At temperatures above the critical temperature $(T_c)$,such as $T_1$,the gas cannot be liquefied by applying pressure alone.
$(ii)$ At $T_2$,point $B$ represents the onset of condensation where the gas begins to liquefy.
$(iii)$ $T_c$ (critical temperature) is defined as the highest temperature at which a gas can be liquefied by pressure.
$(iv)$ At point $A$,the substance is entirely in the liquid state; further compression leads to a sharp increase in pressure as liquids are nearly incompressible.

(A) . Polylactic acid is a biodegradable polymer.
It is a thermoplastic polyester.
It is obtained by the condensation polymerization of lactic acid.
The reaction is as follows:
$2CH_3CH(OH)COOH$ $\xrightarrow{-H_2O} \text{Lactide}$ $\xrightarrow{n} (-O-CH(CH_3)-CO-)_n$ (Polylactic acid).

(A) The reaction described is a crossed Cannizzaro reaction. Compounds that lack $\alpha$-hydrogen atoms undergo this reaction in the presence of concentrated $KOH$. Formaldehyde $(HCHO)$ acts as the reducing agent,being more reactive towards nucleophilic attack,and thus it gets oxidized to formate $(HCOO^-)$,while the other aldehyde is reduced to its corresponding alcohol. The compounds $II$ and $V$ are aldehydes that do not possess $\alpha$-hydrogens,allowing them to be reduced to alcohols by $HCHO$ in the presence of conc. $aq.$ $KOH$.

(D) Cetyltrimethyl ammonium bromide is commonly used for sanitizing surfaces.
It is a popular cationic detergent.
It is effective against vegetative bacteria,viruses,and some fungi.
The chemical structure is as follows:
$CH_3(CH_2)_{14}CH_2N^+(CH_3)_3Br^-$

(C) Aryl halides are generally less reactive towards nucleophilic substitution reactions.
However,the presence of an electron-withdrawing group $(-NO_2)$ at the ortho- or para-position significantly increases the reactivity of haloarenes towards nucleophilic substitution due to the stabilization of the carbanion intermediate.
In $I$ ($o$-nitrofluorobenzene),the $-NO_2$ group is at the ortho-position.
In $III$ ($p$-nitrofluorobenzene),the $-NO_2$ group is at the para-position.
In $II$ ($m$-nitrofluorobenzene),the $-NO_2$ group is at the meta-position,where it does not stabilize the intermediate carbanion through resonance as effectively as it does at the ortho- and para-positions.
Thus,the reactivity order is $I > III > II$.

(B) The correct option is $B$. The conversion of $2-$phenylpropanamide into $1-$phenylethylamine involves the loss of a carbonyl carbon atom,which is characteristic of the Hofmann bromamide degradation reaction.
The reaction is as follows:
$CH_{3}CH(Ph)CONH_{2} + Br_{2} + 4NaOH \rightarrow CH_{3}CH(Ph)NH_{2} + Na_{2}CO_{3} + 2NaBr + 2H_{2}O$
In this reaction,the amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide. The alkyl or aryl group migrates from the carbonyl carbon to the nitrogen atom,resulting in the formation of a primary amine with one carbon atom less than the starting amide.

(A) The given compound $(X)$ is acetonitrile $(CH_{3}CN)$.
Acidic hydrolysis of acetonitrile $(CH_{3}CN)$ yields acetic acid $(CH_{3}COOH)$.
Reduction of acetonitrile $(CH_{3}CN)$ using reducing agents like $LiAlH_{4}$ or $H_{2}/Ni$ yields ethanamine $(CH_{3}CH_{2}NH_{2})$.
The complete reaction scheme is:
$CH_{3}COOH \xleftarrow{\text{Acid hydrolysis}} CH_{3}CN (X) \xrightarrow{\text{Reduction}} CH_{3}CH_{2}NH_{2}$

(D) . $A$ nucleus $X$ captures a $\beta$-particle and then emits a neutron and $\gamma$-ray to form $Y$. The nuclear reaction is as follows:
${}_{Z}X^{A} + {}_{-1}e^{0}$ $\rightarrow {}_{Z-1}X^{A}$ $\xrightarrow{-{}_{0}n^{1}} {}_{Z-1}Y^{A-1} + \gamma$
Here,$A$ is the mass number of $X$ and $Z$ is the atomic number of $X$.
Number of neutrons in $X = A - Z$.
Number of neutrons in $Y = (A - 1) - (Z - 1) = A - Z$.
Since both $X$ and $Y$ have the same number of neutrons,they are isotones.

(C) The boiling point elevation is given by the formula $\Delta T_{b} = i \times K_{b} \times m$.
For $CuSO_{4} \cdot 5H_{2}O$,the van't Hoff factor $i = 2$ because it dissociates as $CuSO_{4} \rightarrow Cu^{2+} + SO_{4}^{2-}$.
Given $K_{b} = 0.512 \, K \, kg \, mol^{-1}$ and molality $m = 0.1 \, mol \, kg^{-1}$.
$\Delta T_{b} = 2 \times 0.512 \times 0.1 = 0.1024 \, K$.
The boiling point of pure water is $100^{\circ}C$.
Therefore,the boiling point of the solution is $T_{b} = 100 + 0.1024 = 100.1024^{\circ}C$,which is closest to $100.10^{\circ}C$.

(A) mixture of toluene and benzene forms an ideal solution. $p_{B}^{\circ}$ and $p_{T}^{\circ}$ are the vapour pressures of pure benzene and toluene,respectively.
According to Raoult's law,$p_{\text{total}} = \chi_{B} p_{B}^{\circ} + \chi_{T} p_{T}^{\circ}$.
We know that $\chi_{B} + \chi_{T} = 1$,so $\chi_{T} = 1 - \chi_{B}$.
Substituting this into the equation: $p_{\text{total}} = \chi_{B} p_{B}^{\circ} + (1 - \chi_{B}) p_{T}^{\circ}$.
Rearranging the terms: $p_{\text{total}} = p_{T}^{\circ} + (p_{B}^{\circ} - p_{T}^{\circ}) \chi_{B}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = p_{\text{total}}$ and $x = \chi_{B}$,the slope $m$ is equal to $(p_{B}^{\circ} - p_{T}^{\circ})$.

(D) The correct option is $D$. When a copper rod is dipped into an aqueous solution of $AgNO_{3}$,a displacement reaction occurs because copper is more reactive than silver.
$Cu_{(s)} + 2AgNO_{3(aq)} \longrightarrow Cu(NO_{3})_{2(aq)} + 2Ag_{(s)}$
In this reaction,$Cu$ is oxidized to $Cu^{2+}$ ions,which impart a blue color to the solution.
The half-cell reactions are:
$Cu_{(s)} \longrightarrow Cu^{2+}_{(aq)} + 2e^{-}; E^{\circ} = -0.34 \, V$
$Ag^{+}_{(aq)} + e^{-} \longrightarrow Ag_{(s)}; E^{\circ} = 0.80 \, V$
The overall cell potential is positive,indicating the reaction is spontaneous.

(C) The correct option is $(C)$. Among the given options,$PdCl_{4}^{2-}$ exhibits square planar geometry.

Complex	Hybridisation
$CdCl_{4}^{2-}$	$sp^{3}$
$Zn(CN)_{4}^{2-}$	$sp^{3}$
$PdCl_{4}^{2-}$	$dsp^{2}$
$Cu(CN)_{4}^{3-}$	$sp^{3}$

The $Pd^{2+}$ ion has a $4d^{8}$ configuration. In the presence of $Cl^{-}$ ligands,it undergoes $dsp^{2}$ hybridisation,which results in a square planar geometry.

(A) In metal carbonyl complexes,the $M-C$ $\sigma$-bond is formed by the donation of a lone pair of electrons from the carbonyl carbon into a vacant orbital of the metal.
The $M-C$ $\pi$-bond (back-bonding) is formed by the donation of a pair of electrons from a filled $d$-orbital of the metal (such as $d_{xy}$,$d_{yz}$,or $d_{zx}$) into the vacant antibonding $\pi^{*}$-orbital of carbon monoxide.
Therefore,the correct pair of orbitals involved in $\pi$-bonding is a metal $d$-orbital (like $d_{xy}$) and the carbonyl $\pi^{*}$-orbital.

(B) The complex $[Ni(dimethylglyoximate)_{2}]$ involves the $Ni^{2+}$ ion.
Dimethylglyoximate $(DMG^-)$ is a strong field chelating ligand.
In this square planar complex,the $Ni^{2+}$ ion ($d^8$ configuration) undergoes $dsp^2$ hybridization.
Due to the strong field nature of the ligand,all electrons are paired in the $d$-orbitals.
Since the number of unpaired electrons $(n)$ is $0$,the magnetic moment $(\mu)$ is calculated as $\mu = \sqrt{n(n+2)} = \sqrt{0(0+2)} = 0.00 \ \mu_{B}$.

(B) In a hexagonal closed pack $(HCP)$ structure,the number of atoms per unit cell is $6$.
Since the number of octahedral voids is equal to the number of atoms in the close-packed structure,the number of octahedral voids is $6$.
Given that $M$ occupies $2/3$ of the octahedral voids,the number of $M$ atoms per unit cell is $\frac{2}{3} \times 6 = 4$.
The ratio of $M$ atoms to $N$ atoms is $4:6$,which simplifies to $2:3$.
Therefore,the formula of the compound is $M_2 N_3$.

(C) The correct option is $C$. The transformation involves the conversion of $4$-oxo-$4$-phenylbutanoic acid to $\alpha$-tetralone. The steps are as follows:
$i$. $Zn(Hg)/HCl$ (Clemmensen reduction): Reduces the ketone group to a methylene group $(-CH_2-)$.
$ii$. $SOCl_2$: Converts the carboxylic acid group $(-COOH)$ to an acid chloride $(-COCl)$.
$iii$. Anhydrous $AlCl_3$ (Friedel-Crafts acylation): Facilitates intramolecular cyclization to form the cyclic ketone ($\alpha$-tetralone).

(D) The reaction shown is the industrial preparation of phenol from cumene-like hydroperoxidation. The starting material $X$ is cyclohexylbenzene. Upon oxidation with $O_2$ in the presence of a catalyst and heat,it forms a hydroperoxide intermediate. Subsequent treatment with aqueous acid leads to the cleavage of the hydroperoxide into phenol and cyclohexanone $(Y)$.
Thus,$X$ is cyclohexylbenzene and $Y$ is cyclohexanone.

(A) The packing fraction $(PF)$ is defined as the ratio of the area of the circles to the area of the rectangle.
$PF = \frac{\pi a^{2} + \pi b^{2}}{(2a + 2b)(2a)} = \frac{\pi(a^{2} + b^{2})}{4a(a + b)}$
Dividing by $a^{2}$,we get $PF = \frac{\pi(1 + r^{2})}{4(1 + r)}$,where $r = \frac{b}{a}$.
To minimize the packing fraction,we differentiate $PF$ with respect to $r$ and set it to zero:
$\frac{d(PF)}{dr} = \frac{\pi}{4} \left[ \frac{(1+r)(2r) - (1+r^{2})(1)}{(1+r)^{2}} \right] = 0$
$2r + 2r^{2} - 1 - r^{2} = 0$
$r^{2} + 2r - 1 = 0$
Using the quadratic formula $r = \frac{-B \pm \sqrt{B^{2} - 4AC}}{2A}$:
$r = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$
Since $r$ must be positive,$r = \sqrt{2} - 1 \approx 1.414 - 1 = 0.414$.
Thus,the value is closest to $0.41$.

(D) For the complex $[Co(NH_3)_3Cl_3]$,it exists in two geometric isomeric forms: $fac$ (facial) and $mer$ (meridional). Both are optically inactive,so there are $2$ stereoisomers.
For the complex $[Ni(en)_2Cl_2]$,it exists in $trans$ and $cis$ forms. The $trans$ form is achiral,while the $cis$ form exists as a pair of enantiomers. Thus,there are $3$ stereoisomers in total ($1$ trans + $2$ cis enantiomers).

(C) $X$ is $CrO_{2}Cl_{2}$ and $Y$ is $Na_{2}CrO_{4}$.
When a mixture of $NaCl$,$K_{2}Cr_{2}O_{7}$ and conc. $H_{2}SO_{4}$ is heated in a dry test tube,a red vapour $(X)$,$CrO_{2}Cl_{2}$ (chromyl chloride),is evolved.
This vapour $CrO_{2}Cl_{2}$ reacts with an aqueous solution of $NaOH$ to form $Na_{2}CrO_{4}$ (sodium chromate),which is yellow in color.
The chemical reactions are as follows:
$4Cl^{-} + Cr_{2}O_{7}^{2-} + 6H^{+} \longrightarrow 2CrO_{2}Cl_{2} (X) + 3H_{2}O$
$CrO_{2}Cl_{2} + 4OH^{-} \longrightarrow CrO_{4}^{2-} + 2Cl^{-} + 2H_{2}O$
The resulting $CrO_{4}^{2-}$ ions in the presence of $Na^{+}$ ions form $Na_{2}CrO_{4} (Y)$.

(C) The correct acidity order is $III > IV > I > II$.
In substituted phenols,electron-withdrawing groups $(EWG)$ like the nitro $(-NO_2)$ group increase acidity by stabilizing the phenoxide ion through delocalization of the negative charge.
$p$-Nitrophenol $(III)$ is more acidic than $m$-nitrophenol $(IV)$ because the $-NO_2$ group at the para position exerts both $-I$ and $-R$ effects,whereas at the meta position,it only exerts the $-I$ effect.
Phenol $(I)$ is less acidic than nitrophenols but more acidic than $p$-methoxyphenol $(II)$.
$p$-Methoxyphenol $(II)$ contains a methoxy $(-OCH_3)$ group,which is an electron-donating group $(EDG)$ due to the $+R$ effect,destabilizing the phenoxide ion and decreasing the acidity.

(A) Given,specific heat of substance $= 0.86 \,J \,g^{-1} \,K^{-1}$.
$1 \,molal$ solution means $1 \,mole$ of solute in $1000 \,g$ of solvent (water).
Mass of solute $= 1 \,mole \times 58 \,g \,mol^{-1} = 58 \,g$.
Total mass of solution $= 1000 \,g + 58 \,g = 1058 \,g$.
In $1058 \,g$ solution,mass of solute $= 58 \,g$ and mass of water $= 1000 \,g$.
For $10 \,g$ of solution:
Mass of solute $= (58 / 1058) \times 10 \approx 0.548 \,g$.
Mass of water $= (1000 / 1058) \times 10 \approx 9.452 \,g$.
Energy required $q = (m_{solute} \times c_{solute} \times \Delta T) + (m_{water} \times c_{water} \times \Delta T)$.
$q = (0.548 \times 0.86 \times 10) + (9.452 \times 4.2 \times 10)$.
$q = 4.7128 + 396.984 = 401.6968 \,J \approx 401.7 \,J$.