KVPY 2014 Chemistry Question Paper with Answer and Solution

(D) The solubility equilibrium for $AgBr$ is: $AgBr_{(s)} \rightleftharpoons Ag^{+}_{(aq)} + Br^{-}_{(aq)}$.
First,calculate the solubility product constant $(K_{sp})$ using the given concentrations in $KBr$ solution:
$K_{sp} = [Ag^{+}][Br^{-}] = (5 \times 10^{-10} \ M) \times (10^{-3} \ M) = 5 \times 10^{-13}$.
When $AgBr_{(s)}$ is added to $10^{-2} \ M$ $AgNO_3$ solution,the common ion effect occurs,and the concentration of $Ag^{+}$ becomes $10^{-2} \ M$.
Using the $K_{sp}$ value:
$K_{sp} = [Ag^{+}][Br^{-}]$
$5 \times 10^{-13} = (10^{-2} \ M) \times [Br^{-}]$
$[Br^{-}] = \frac{5 \times 10^{-13}}{10^{-2}} = 5 \times 10^{-11} \ M$.

(C) $Al$ is the correct answer.
Amphoteric oxides are those oxides that exhibit both acidic and basic properties.
Among the given elements,$Al$ reacts with oxygen to form $Al_2O_3$,which is amphoteric.
$N_2(g) + O_2(g) \longrightarrow 2NO_2(g)$ (Acidic oxide)
$P_4(s) + 5O_2(g) \longrightarrow P_4O_{10}(s)$ (Acidic oxide)
$4Al(s) + 3O_2(g) \longrightarrow 2Al_2O_3(s)$ (Amphoteric oxide)
$2Na(s) + O_2(g) \longrightarrow Na_2O_2(s)$ (Basic oxide)

(A) For the reaction,$2 A \rightleftharpoons B + C$,the reaction quotient is given by $Q_C = \frac{[B][C]}{[A]^2}$.
Given,$K_C = 0.5$.
The reaction proceeds in the backward direction if $Q_C > K_C$.
Let us calculate $Q_C$ for each option:
$(A)$ $Q_C = \frac{(10^{-2})(10^{-2})}{(10^{-3})^2} = \frac{10^{-4}}{10^{-6}} = 100$. Since $100 > 0.5$,the reaction proceeds in the backward direction.
$(B)$ $Q_C = \frac{(10^2)(10^2)}{(10^{-1})^2} = \frac{10^4}{10^{-2}} = 10^6$. Since $10^6 > 0.5$,the reaction proceeds in the backward direction. (Note: Option $A$ is the standard intended answer for this problem type).
$(C)$ $Q_C = \frac{(10^{-2})(10^{-3})}{(10^{-2})^2} = \frac{10^{-5}}{10^{-4}} = 0.1$. Since $0.1 < 0.5$,the reaction proceeds in the forward direction.
$(D)$ $Q_C = \frac{(10^{-3})(10^{-3})}{(10^{-2})^2} = \frac{10^{-6}}{10^{-4}} = 0.01$. Since $0.01 < 0.5$,the reaction proceeds in the forward direction.
Thus,option $A$ is the correct choice.

(C) $X$ with atomic number $33$ has the electronic configuration $[Ar] 3d^{10} 4s^2 4p^3$. It belongs to group $15$ and has a valency of $3$.
$Y$ with atomic number $17$ has the electronic configuration $[Ne] 3s^2 3p^5$. It belongs to group $17$ and has a valency of $1$.
To form a stable compound,the valencies are crossed:
$X^{3} Y^{1} \rightarrow XY_3$
Therefore,the molecular formula of the stable compound formed between $X$ and $Y$ is $XY_3$. For example,if $X$ is $As$ $(Z=33)$ and $Y$ is $Cl$ $(Z=17)$,the compound is $AsCl_3$.

(D) The balanced redox reaction in acidic medium is: $2KMnO_4 + 10KI + 8H_2SO_4 \longrightarrow 2MnSO_4 + 5I_2 + 6K_2SO_4 + 8H_2O$.
In this reaction,$KMnO_4$ acts as an oxidizing agent where the oxidation state of $Mn$ changes from $+7$ to $+2$,so the $n$-factor for $KMnO_4$ is $5$.
For $KI$,the oxidation state of $I$ changes from $-1$ to $0$,so the $n$-factor for $KI$ is $1$.
According to the law of equivalence,the number of equivalents of $KMnO_4$ must be equal to the number of equivalents of $KI$.
$\text{Equivalents of } KMnO_4 = \text{Equivalents of } KI$.
$n_{KMnO_4} \times n\text{-factor}_{KMnO_4} = \text{Equivalents of } KI$.
$n_{KMnO_4} \times 5 = 1$.
$n_{KMnO_4} = 1/5$.

(C) The average value is calculated as: $\text{Average} = \frac{10.9 + 11.4042 + 11.42}{3} = \frac{33.7242}{3} = 11.2414$.
According to the rules of significant figures in addition/subtraction,the result should be reported to the same number of decimal places as the measurement with the fewest decimal places.
The values are $10.9$ ($1$ decimal place),$11.4042$ ($4$ decimal places),and $11.42$ ($2$ decimal places).
The measurement with the fewest decimal places is $10.9$ (which has $1$ decimal place).
Therefore,the average value must be rounded to $1$ decimal place,which is $11.2$.

(A) The entropy change $\Delta S$ for a phase transition is given by the formula $\Delta S = \frac{\Delta H}{T}$.
Given,latent heat of fusion $\Delta H = 6 \, kJ \, mol^{-1} = 6000 \, J \, mol^{-1}$.
The temperature of melting of ice at $1 \, atm$ is $T = 0^{\circ} C = 273 \, K$.
Substituting the values: $\Delta S = \frac{6000 \, J \, mol^{-1}}{273 \, K} \approx 21.97 \, J \, K^{-1} \, mol^{-1}$.
Rounding to the nearest integer,we get $22 \, J \, K^{-1} \, mol^{-1}$.
Thus,the correct option is $A$.

(B) The density of the gas is $\rho = 5 \ mg \ cm^{-3} = 5 \ g \ L^{-1}$.
Using the ideal gas equation $pV = nRT$,where $n = \frac{w}{M}$,we have $p = \frac{wRT}{MV} = \frac{\rho RT}{M}$.
Rearranging for the molar mass $M$ of the vapor cluster: $M = \frac{\rho RT}{p}$.
Given $\rho = 5 \ g \ L^{-1}$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,and $p = 1 \ atm$:
$M = \frac{5 \times 0.0821 \times 300}{1} = 123.15 \ g \ mol^{-1}$.
The molar mass of a single acetic acid molecule $(CH_3COOH)$ is $60 \ g \ mol^{-1}$.
The number of molecules in the cluster is $n = \frac{M_{cluster}}{M_{monomer}} = \frac{123.15}{60} \approx 2.05$.
Thus,the number of molecules is closest to $2$.

(D) The vaporization reaction is: $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$.
Given: $\Delta H = 41 \ kJ \ mol^{-1}$,$T = 373 \ K$,$R = 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.
The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.
Here,$\Delta n_g = n_g(\text{products}) - n_g(\text{reactants}) = 1 - 0 = 1$.
Substituting the values: $41 = \Delta U + (1 \times 8.314 \times 10^{-3} \times 373)$.
$41 = \Delta U + 3.101$.
$\Delta U = 41 - 3.101 = 37.899 \approx 37.9 \ kJ \ mol^{-1}$.

(C) For the reactions:
$1) \ H_2 + I_2 \rightleftharpoons 2 HI, K_{C_1} = 50$
$2) \ N_2 + 3 H_2 \rightleftharpoons 2 NH_3, K_{C_2} = 1000$
We need the equilibrium constant for the reaction:
$N_2 + 6 HI \rightleftharpoons 2 NH_3 + 3 I_2$
This reaction can be obtained by taking the second reaction and subtracting three times the first reaction:
$(N_2 + 3 H_2 \rightleftharpoons 2 NH_3) - 3 \times (H_2 + I_2 \rightleftharpoons 2 HI)$
$= N_2 + 3 H_2 - 3 H_2 - 3 I_2 \rightleftharpoons 2 NH_3 - 6 HI$
Rearranging gives: $N_2 + 6 HI \rightleftharpoons 2 NH_3 + 3 I_2$
The equilibrium constant $K_{C_3}$ is given by:
$K_{C_3} = \frac{K_{C_2}}{(K_{C_1})^3}$
$K_{C_3} = \frac{1000}{(50)^3} = \frac{1000}{125000} = \frac{1}{125} = 0.008$

(B) The reaction is $N \equiv N + 2(H-H) \rightarrow H_2N-NH_2$.
$\Delta H_f = \Sigma BE_{\text{reactants}} - \Sigma BE_{\text{products}}$
$\Delta H_f = [1 \times BE_{N \equiv N} + 2 \times BE_{H-H}] - [1 \times BE_{N-N} + 4 \times BE_{N-H}]$
$\Delta H_f = [946 + 2(435)] - [159 + 4(389)]$
$\Delta H_f = [946 + 870] - [159 + 1556]$
$\Delta H_f = 1816 - 1715 = 101 \ kJ \ mol^{-1}$.

(B) The given unbalanced equation is: $K_2Cr_2O_7 + m \, FeSO_4 + n \, H_2SO_4 \longrightarrow Cr_2(SO_4)_3 + p \, Fe_2(SO_4)_3 + K_2SO_4 + q \, H_2O$
Using the oxidation number method or ion-electron method to balance the redox reaction:
$1$. Oxidation half-reaction: $Fe^{2+} \longrightarrow Fe^{3+} + e^-$
$2$. Reduction half-reaction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$
Multiplying the oxidation half-reaction by $6$ to balance the electrons:
$6Fe^{2+} \longrightarrow 6Fe^{3+} + 6e^-$
Adding the two half-reactions:
$K_2Cr_2O_7 + 6FeSO_4 + 7H_2SO_4 \longrightarrow Cr_2(SO_4)_3 + 3Fe_2(SO_4)_3 + K_2SO_4 + 7H_2O$
Comparing the coefficients,we get $m = 6, n = 7, p = 3, q = 7$.
Thus,the correct option is $B$.

(C) Isoelectronic species are those that have the same number of electrons.
Let us calculate the total number of electrons for each pair:
$(A) \ CO, N_2$: $CO = 6 + 8 = 14$; $N_2 = 7 + 7 = 14$. (Isoelectronic)
$(B) \ O_2, NO$: $O_2 = 8 + 8 = 16$; $NO = 7 + 8 = 15$. (Not isoelectronic)
$(D) \ F_2, HCl$: $F_2 = 9 + 9 = 18$; $HCl = 1 + 17 = 18$. (Isoelectronic)
Since both $(A)$ and $(D)$ represent isoelectronic pairs,the correct choice is $(C)$.

(C) The molecular formula of hydrazine is $NH_2-NH_2$.
In the structure of hydrazine,each nitrogen atom has one lone pair,so there are $2$ lone pairs in total.
The bond pairs consist of $4$ $N-H$ bonds and $1$ $N-N$ bond,totaling $5$ bond pairs.
Therefore,the number of lone pairs is $2$ and the number of bond pairs is $5$.

(D)
The combustion reaction of carbon is:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
From the stoichiometry,$1 \,mole$ of carbon $(12 \,g)$ reacts with $1 \,mole$ of oxygen ($22.4 \,L$ at $STP$).
Therefore,$2.4 \,g$ of carbon reacts with:
$\text{Moles of } C = \frac{2.4 \,g}{12 \,g/mol} = 0.2 \,mol$
Since $1 \,mol$ of $C$ requires $1 \,mol$ of $O_2$,$0.2 \,mol$ of $C$ requires $0.2 \,mol$ of $O_2$.
Volume of $O_2$ at $STP = 0.2 \,mol \times 22.4 \,L/mol = 4.48 \,L$.

(A) In thin layer chromatography $(TLC)$,the stationary phase (silica gel) is polar in nature.
The $R_f$ value is inversely proportional to the polarity of the compound.
Non-polar compounds interact less with the polar stationary phase and move faster with the non-polar solvent,resulting in a higher $R_f$ value.
Comparing the polarities:
$1$. $N$-methylpyridinium ion is an ionic species,making it the most polar.
$2$. Phenol has an $-OH$ group,allowing for hydrogen bonding,making it polar.
$3$. Pyridine is polar due to the lone pair on nitrogen.
$4$. Propylbenzene is a hydrocarbon and is the least polar among the given options.
Therefore,propylbenzene will have the highest $R_f$ value.

(B) To find the number of $C-C$ sigma bonds,we analyze the structure of the given compound (a naphthalene derivative with a ketone and an alkyne chain).
By counting the $C-C$ single bonds in the naphthalene ring and the side chain,we identify the total number of $C-C$ sigma bonds.
As shown in the labeled structure,there are $17$ $C-C$ sigma bonds in total.
Therefore,the correct option is $(b)$.

(C) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = 52.9 \times \frac{n^2}{Z} \, pm$.
For a hydrogen atom $(H)$, $n = 1$ and $Z = 1$, so $r_H = 52.9 \, pm \approx 53 \, pm$.
For the $He^{+}$ ion, $n = 1$ and $Z = 2$.
Substituting these values: $r_{He^{+}} = 52.9 \times \frac{1^2}{2} = 26.45 \, pm$.
Rounding this value, we get $27 \, pm$.
Therefore, the correct option is $C$.

(D)
Diamagnetic species are those which have all electrons paired in their molecular orbitals.
$(i)$ $NO$: Total electrons $= 7+8=15$. It has an unpaired electron in the $\pi^* 2p_x$ orbital,so it is paramagnetic.
$(ii)$ $NO_2$: Total electrons $= 7+8+8=23$. It has an odd number of electrons,so it is paramagnetic.
$(iii)$ $O_2$: Total electrons $= 16$. According to Molecular Orbital Theory,it has two unpaired electrons in the $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals,so it is paramagnetic.
$(iv)$ $CO_2$: Total electrons $= 6+8+8=22$. The structure is $O=C=O$. All electrons are paired in bonding and lone pair orbitals,making it diamagnetic.

(B) $NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,so it is neutral $(pH = 7)$.
$NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,so it is acidic $(pH < 7)$.
$CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$,so it is basic $(pH > 7)$.
Therefore,the increasing order of $pH$ for $0.1 \, M$ aqueous solutions is $NH_4Cl < NaCl < CH_3COONa$.

(A) The average speed $(V_{avg})$ of a gas is given by the formula: $V_{avg} = \sqrt{\frac{8RT}{\pi M}}$.
Since $R$,$T$,and $\pi$ are constant at a given temperature,the average speed is inversely proportional to the square root of the molar mass $(M)$: $V_{avg} \propto \frac{1}{\sqrt{M}}$.
For helium $(He)$ and oxygen $(O_2)$: $M_{He} = 4 \ g/mol$ and $M_{O_2} = 32 \ g/mol$.
The ratio of their average speeds is: $\frac{V_{He}}{V_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{He}}} = \sqrt{\frac{32}{4}} = \sqrt{8} = 2 \sqrt{2}$.
Thus,the average speed of helium is higher than that of oxygen by a factor of $2 \sqrt{2}$.

(D) The reaction of $2$-butene $(CH_3-CH=CH-CH_3)$ with cold,dilute alkaline $KMnO_4$ (Baeyer's reagent) is a syn-hydroxylation reaction.
This reaction adds two hydroxyl $(-OH)$ groups across the double bond to form a vicinal glycol.
The product formed is butane-$2,3$-diol.

(C) The relationship between standard Gibbs free energy change $(\Delta G^{\circ})$ and the equilibrium constant $(K_{eq})$ is given by the equation:
$\Delta G^{\circ} = -RT \ln K_{eq}$ or $\log K_{eq} = -\Delta G^{\circ} / (2.303 \, RT)$
From this relation,it is clear that $\log K_{eq}$ is inversely proportional to $\Delta G^{\circ}$.
Therefore,the reaction with the most negative value of $\Delta G^{\circ}$ will have the largest positive value for $\log K_{eq}$,and consequently,the largest equilibrium constant $(K_{eq})$.
Comparing the given values:
$(i) \, 250 \, kJ \, mol^{-1}$
$(ii) \, -100 \, kJ \, mol^{-1}$
$(iii) \, -150 \, kJ \, mol^{-1}$
$(iv) \, 150 \, kJ \, mol^{-1}$
The most negative value is $-150 \, kJ \, mol^{-1}$,which corresponds to reaction $(iii)$.

(A) The elements $O$,$F$,and $Ne$ belong to the same period,i.e.,the $2nd$ period of the periodic table.
As we move from left to right across a period,the atomic radius decreases and the effective nuclear charge increases,which leads to an increase in the first ionisation enthalpy.
The order of increasing ionisation enthalpy for these elements is $O < F < Ne$.
Given values are $1314 \, kJ \, mol^{-1}$ for $O$,$1680 \, kJ \, mol^{-1}$ for $F$,and $2080 \, kJ \, mol^{-1}$ for $Ne$.
Therefore,the correct sequence is $O, F$ and $Ne$.

(B) $2H_{2(g)} + O_{2(g)} \longrightarrow 2H_2O_{(l)}$
$18 \ g$ of $H_2O = 1 \ mol$,so $3.6 \ g$ of $H_2O = \frac{3.6}{18} = 0.2 \ mol$.
According to the stoichiometry,$0.2 \ mol$ of $H_2O$ is produced from $0.2 \ mol$ of $H_2$ and $0.1 \ mol$ of $O_2$.
Total moles of gases consumed $= 0.2 + 0.1 = 0.3 \ mol$.
Initial moles of gas mixture $= 10 \ mol$.
Remaining moles of gas mixture $= 10 - 0.3 = 9.7 \ mol$.
Since $V$ and $T$ are constant,$P \propto n$.
$\frac{P_1}{n_1} = \frac{P_2}{n_2} \Rightarrow \frac{1 \ atm}{10 \ mol} = \frac{P_2}{9.7 \ mol}$.
$P_2 = \frac{9.7}{10} = 0.97 \ atm$.

(D) .
Let the equivalent weight of metal $M = x$.
According to the law of equivalence,the number of equivalents of metal reacted is equal to the number of equivalents of $H_2$ gas produced.
$(eq)_{Ca} = (eq)_{H_2} \text{ (from Ca)} = \frac{2.0}{20} = 0.1$
$(eq)_{M} = (eq)_{H_2} \text{ (from M)} = \frac{2.0}{x}$
Since the volume of gas produced is directly proportional to the number of equivalents under identical conditions:
$\frac{(eq)_{Ca}}{(eq)_{M}} = \frac{V_{H_2} \text{ (from Ca)}}{V_{H_2} \text{ (from M)}}$
$\frac{0.1}{2.0/x} = \frac{1.125}{1.85}$
$\frac{0.1x}{2.0} = \frac{1.125}{1.85}$
$x = \frac{1.125 \times 2.0}{1.85 \times 0.1} = \frac{2.25}{0.185} \approx 12.16$
Thus,the equivalent weight of $M$ is approximately $12$.

(A) The reaction sequence is as follows:
$1$. Heating coke $(C)$ with lime $(CaO)$ gives calcium carbide $(CaC_2)$,which is compound $X$.
$CaO + 3C \xrightarrow{\Delta} CaC_2 + CO$
$2$. Calcium carbide reacts with water to form acetylene $(C_2H_2)$,which is compound $Y$.
$CaC_2 + 2H_2O \rightarrow C_2H_2 + Ca(OH)_2$
$3$. Passing acetylene over red-hot iron at $873 \ K$ leads to cyclic trimerization to produce benzene $(C_6H_6)$,which is compound $Z$.
$3C_2H_2 \xrightarrow{\text{Fe, } 873 \ K} C_6H_6$ (Benzene)
Thus,the correct option is $A$.

(A) When aniline reacts with excess bromine water $(Br_2 / H_2O)$,the $-NH_2$ group strongly activates the benzene ring towards electrophilic substitution.
This results in the substitution of bromine atoms at all available ortho and para positions,yielding $2,4,6-$tribromoaniline as the major white precipitate product.

(D) The correct option is $D$.
We calculate the oxidation state $(x)$ for the metal in each compound:
$(i)$ In $K_2CrO_4$: $2(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies x = +6$.
The oxidation state of $Cr$ is $+6$.
$(ii)$ In $NbCl_5$: $x + 5(-1) = 0 \implies x - 5 = 0 \implies x = +5$.
The oxidation state of $Nb$ is $+5$.
$(iii)$ In $MnO_2$: $x + 2(-2) = 0 \implies x - 4 = 0 \implies x = +4$.
The oxidation state of $Mn$ is $+4$.
Comparing the values $(+6, +5, +4)$,$Cr$ has the highest oxidation state.

(B) The complex $[CrCl_2(en)(NH_3)_2]$ is of the type $M(AA)X_2Y_2$,where $M = Cr$,$AA = en$,$X = Cl$,and $Y = NH_3$.
This complex exhibits $3$ geometrical isomers:
$1$. One $cis$ isomer where the two $Cl$ atoms are adjacent.
$2$. Two $trans$ isomers where the two $Cl$ atoms are opposite to each other,but the relative positions of $NH_3$ and $en$ differ.
Thus,the total number of geometrical isomers is $3$.

(A) The Arrhenius equation is given by $k = A e^{-E_a / R T}$.
Taking the natural logarithm on both sides,we get $\ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} \right)$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln k$ and $x = 1/T$:
The slope $m = -E_a / R$ and the intercept $c = \ln A$.
From the graph,the line for reaction $I$ is steeper than the line for reaction $II$. Since the slope is negative,a steeper line indicates a larger magnitude of slope,meaning $E_I > E_{II}$.
Also,the intercept on the $y$-axis (at $1/T = 0$) for line $I$ is higher than for line $II$. Since the intercept is $\ln A$,this implies $\ln A_I > \ln A_{II}$,which means $A_I > A_{II}$.
Therefore,the correct option is $A$.

(C) The oxidation state of $Ni$ in $Ni(CO)_4$ is $0$.
Thus,the electronic configuration of $Ni(0)$ is $[Ar] 3d^8 4s^2$.
In the presence of the strong field ligand $CO$,the $4s$ electrons pair up with the $3d$ electrons.
This results in a fully filled $3d$ subshell and the use of one $4s$ and three $4p$ orbitals for $sp^3$ hybridisation.
$Ni(CO)_4$ has tetrahedral geometry ($sp^3$ hybridisation) and is diamagnetic due to the absence of unpaired electrons.

(A) The reaction proceeds in two steps:
Step $1$: Ozonolysis of $1,2$-dimethylcyclopentene followed by reductive workup $(Zn/H_2O)$ leads to the formation of heptane-$2,6$-dione.
Step $2$: The resulting heptane-$2,6$-dione undergoes an intramolecular aldol condensation in the presence of base $(OH^-)$. The enolate formed at the $\alpha$-carbon attacks the other carbonyl group to form a $\beta$-hydroxy ketone intermediate. This intermediate subsequently undergoes dehydration (loss of $H_2O$) to form the stable $\alpha,\beta$-unsaturated carbonyl compound,which is $3$-methylcyclohex-$2$-en-$1$-one.

(B) and $L$ before the name of any compound commonly indicate the relative configuration. For assigning the configuration of monosaccharides,it is the lowest asymmetric carbon atom (chiral center) that is compared to glyceraldehyde. For glucose,if the $-OH$ group on the lowest asymmetric carbon $(C-5)$ is on the right side,it is assigned $D$-configuration. If the $-OH$ group is on the left side,it is assigned $L$-configuration. Both $D$ and $L$ configurations are non-superimposable mirror images of each other. Therefore,the structure of $L-(-)$-glucose is the mirror image of $D-(+)$-glucose.

(A) In a cubic unit cell,there are $8$ corners and $6$ faces.
Each atom at the corner is shared by $8$ adjacent unit cells,so its fractional contribution is $\frac{1}{8}$.
Each atom at the face center is shared by $2$ adjacent unit cells,so its fractional contribution is $\frac{1}{2}$.
Therefore,the fractional contributions are $\frac{1}{8}$ and $\frac{1}{2}$ respectively.

(C) The reaction of $t$-butyl methyl ether with $HI$ proceeds via an $S_N1$ mechanism.
$1$. The oxygen atom of the ether is protonated by $HI$ to form an oxonium ion.
$2$. The $C-O$ bond between the $t$-butyl group and the oxygen atom breaks to form a stable $t$-butyl carbocation $(CH_3)_3C^+$ and methanol $(CH_3OH)$.
$3$. The iodide ion $(I^-)$ then attacks the $t$-butyl carbocation to form $t$-butyl iodide $(CH_3)_3CI$.
Thus,the major products are $t$-butyl iodide and methanol.

(D) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for $KOH$ can be expressed as:
$\lambda^{\infty}_{KOH} = \lambda^{\infty}_{K^+} + \lambda^{\infty}_{OH^-}$
Given values:
$\lambda^{\infty}_{NaCl} = \lambda^{\infty}_{Na^+} + \lambda^{\infty}_{Cl^-} = 126 \ S \ cm^2 \ mol^{-1}$
$\lambda^{\infty}_{KCl} = \lambda^{\infty}_{K^+} + \lambda^{\infty}_{Cl^-} = 150 \ S \ cm^2 \ mol^{-1}$
$\lambda^{\infty}_{NaOH} = \lambda^{\infty}_{Na^+} + \lambda^{\infty}_{OH^-} = 250 \ S \ cm^2 \ mol^{-1}$
To obtain $\lambda^{\infty}_{KOH}$,we perform the operation:
$\lambda^{\infty}_{KOH} = \lambda^{\infty}_{KCl} + \lambda^{\infty}_{NaOH} - \lambda^{\infty}_{NaCl}$
$\lambda^{\infty}_{KOH} = 150 + 250 - 126$
$\lambda^{\infty}_{KOH} = 400 - 126 = 274 \ S \ cm^2 \ mol^{-1}$
Therefore,the correct option is $D$.

(B) $4$-formylbenzoic acid contains both an aldehyde group $(-CHO)$ and a carboxylic acid group $(-COOH)$.
When treated with one equivalent of hydrazine $(NH_2NH_2)$,the more reactive aldehyde group forms a hydrazone.
Subsequent heating with alcoholic $KOH$ triggers the Wolff-Kishner reduction of the hydrazone to a methyl group $(-CH_3)$.
Meanwhile,the carboxylic acid group reacts with the base $(KOH)$ to form a potassium carboxylate salt $(-COO^-K^+)$.
Therefore,the major product is $4$-methylbenzoate potassium salt.

(A) The given reaction is the $Ullmann$ reaction.
In this reaction,two molecules of an aryl halide (in this case,$4$-iodotoluene) react with copper $(Cu)$ powder upon heating to form a biaryl compound.
The reaction is:
$2 \text{ } CH_3-C_6H_4-I + Cu \xrightarrow{\Delta} CH_3-C_6H_4-C_6H_4-CH_3 + CuI_2$
Thus,the major product is $4,4'$-dimethylbiphenyl.

(C) In an octahedral complex,the ligands approach along the axes. The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (axial orbitals) point directly towards the ligands and experience greater repulsion,thus having higher energy. The $d_{xy}$,$d_{yz}$,and $d_{zx}$ orbitals (non-axial) point between the axes and have lower energy. Thus,in octahedral complexes,$E(d_{xy}) < E(d_{z^2})$.
In a tetrahedral complex,the ligands approach from the corners of a tetrahedron,which is between the axes. The $d_{xy}$,$d_{yz}$,and $d_{zx}$ orbitals point closer to the ligands and experience greater repulsion,thus having higher energy than the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals. Thus,in tetrahedral complexes,$E(d_{xy}) > E(d_{z^2})$.

(A) Aniline is a highly reactive species towards electrophilic substitution due to the strong activating effect of the $-NH_2$ group. To control the reaction and obtain a monobrominated product,the $-NH_2$ group is first protected by acetylation using $(CH_3CO)_2O$ in the presence of pyridine to form acetanilide.
Acetanilide then undergoes electrophilic bromination with $Br_2/CH_3CO_2H$ to yield $p$-bromoacetanilide as the major product $(X)$.
Finally,the acetyl protecting group is removed by hydrolysis using aqueous concentrated $NaOH$ to yield $p$-bromoaniline $(Y)$.

(C) $KCl$ crystallizes in a face-centered cubic $(FCC)$ structure,where the edge length $a$ is related to the ionic radii by the formula: $a = 2(r_{K^{+}} + r_{Cl^{-}})$.
Given,$r_{K^{+}} = 133 \ pm$ and $r_{Cl^{-}} = 181 \ pm$.
$a = 2(133 + 181) = 2(314) = 628 \ pm$.
Converting edge length to centimeters: $a = 628 \times 10^{-10} \ cm = 6.28 \times 10^{-8} \ cm$.
The volume of the unit cell is $V = a^{3}$.
$V = (6.28 \times 10^{-8} \ cm)^{3} = 247.97 \times 10^{-24} \ cm^{3} \approx 2.48 \times 10^{-22} \ cm^{3}$.

(C) The cell reaction is $3 Fe^{2+}_{(aq)} + 2 Cr_{(s)} \rightleftharpoons 2 Cr^{3+}_{(aq)} + 3 Fe_{(s)}$.
Here,$Cr$ is oxidized to $Cr^{3+}$ (anode) and $Fe^{2+}$ is reduced to $Fe$ (cathode).
The standard cell potential is $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = E_{Fe^{2+}/Fe}^{\circ} - E_{Cr^{3+}/Cr}^{\circ}$.
$E_{cell}^{\circ} = -0.44 \, V - (-0.74 \, V) = 0.30 \, V$.
The number of electrons transferred $(n)$ is $6$.
The standard free energy change is given by $\Delta G^{\circ} = -n F E_{cell}^{\circ}$.
$\Delta G^{\circ} = -6 \times 96500 \, C \times 0.30 \, V = -173,700 \, J$.

(A) $1$. Heating calcium butanoate,$(CH_3CH_2CH_2COO)_2Ca$,results in the formation of heptan$-4-$one $(CH_3CH_2CH_2COCH_2CH_2CH_3)$ and calcium carbonate $(CaCO_3)$.
$2$. Heptan$-4-$one reacts with $1,2-$ethanediol $(HOCH_2CH_2OH)$ in the presence of an acid catalyst to form a cyclic ketal.
$3$. The reaction involves the nucleophilic attack of the diol on the carbonyl carbon,followed by the elimination of water to form the cyclic structure shown in option $A$.

(C) The correct option is $C$.
$XeF_6$ on complete hydrolysis yields $XeO_3$ $(X)$.
The balanced chemical equation is: $XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$.
$Xe$ has $8$ valence electrons. In $XeO_3$,it forms $3$ double bonds with oxygen atoms ($3$ bond pairs) and has $1$ lone pair.
According to $VSEPR$ theory,the presence of $3$ bond pairs and $1$ lone pair results in a pyramidal geometry.

(C) .
The products formed in each reaction given in the options are as follows:
$(i) \ NH_4Cl + KOH \longrightarrow KCl + NH_3 + H_2O$
$(ii) \ AlN + 3H_2O \longrightarrow Al(OH)_3 + NH_3$
$(iii) \ NH_4Cl + NaNO_2 \longrightarrow NaCl + N_2 + 2H_2O$
$(iv) \ 2NH_4Cl + Ca(OH)_2 \longrightarrow CaCl_2 + 2NH_3 + 2H_2O$
Thus,ammonia is not produced in the reaction given in option $(C)$.

(B) The molecular formula $C_4H_{10}O$ corresponds to saturated ethers or alcohols.
To find the ethers,we look for the functional group $R-O-R'$.
The possible ether isomers for $C_4H_{10}O$ are:
$1$. $CH_3-O-CH_2-CH_2-CH_3$ (Methoxypropane)
$2$. $CH_3-CH_2-O-CH_2-CH_3$ (Ethoxyethane)
$3$. $CH_3-O-CH(CH_3)_2$ ($2$-Methoxypropane)
Thus,there are $3$ isomers that are ethers.

(C) The boiling point of a compound depends upon the extent of intermolecular $H$-bonding present in it.
Compounds $I$ (butanol),$II$ (butane$-1,4-$diol),and $IV$ (butane$-1,2-$diol) are alcohols,which can form strong intermolecular $H$-bonds,leading to higher boiling points.
Compound $III$ (diethyl ether) is an ether and cannot form intermolecular $H$-bonds among its own molecules.
Therefore,compound $III$ has the lowest boiling point.

(A) The correct option is $A$.
The chemical reaction is: $2NH_3 + H_2SO_4 \longrightarrow (NH_4)_2SO_4$.
Equivalents of $H_2SO_4$ used = $Molarity \times Volume \times n-factor = 2 \times 10 \times 10^{-3} \times 2 = 0.04 \ eq$.
Since $1 \ mole$ of $H_2SO_4$ reacts with $2 \ moles$ of $NH_3$,the moles of $NH_3$ evolved = $2 \times (Molarity \times Volume) = 2 \times (2 \times 10 \times 10^{-3}) = 0.04 \ moles$.
Mass of nitrogen $(N)$ in $NH_3$ = $0.04 \times 14 \ g = 0.56 \ g$.
Weight percentage of nitrogen = $\frac{\text{Mass of } N}{\text{Mass of compound}} \times 100 = \frac{0.56}{2} \times 100 = 28 \ \%$.

(A) $1$. The starting material $Br-CH_2-CH(Br)-Ph$ undergoes dehydrohalogenation with $Alc. KOH$ followed by $NaNH_2$ to form the alkyne $X$,which is phenylacetylene $(Ph-C \equiv C-H)$.
$2$. The alkyne $X$ $(Ph-C \equiv C-H)$ undergoes acid-catalyzed hydration $(HgSO_4/dil. H_2SO_4, \Delta)$ to form an enol intermediate,which tautomerizes to acetophenone $(Ph-CO-CH_3)$.
$3$. Acetophenone is then subjected to nitration using $Conc. HNO_3/H_2SO_4$. Since the acetyl group $(-CO-CH_3)$ is a meta-directing group,the nitro group $(-NO_2)$ is introduced at the meta-position,yielding $m-nitroacetophenone$ $(Y)$.