KCET 2014 Chemistry Question Paper with Answer and Solution

(A) Given that $A$ is a $3 \times 4$ matrix,its transpose $A'$ is a $4 \times 3$ matrix.
For the product $A'B$ to be defined,the number of columns in $A'$ must equal the number of rows in $B$. Since $A'$ is $4 \times 3$,$B$ must have $3$ rows. Let the order of $B$ be $3 \times p$.
For the product $BA'$ to be defined,the number of columns in $B$ must equal the number of rows in $A'$. Since $B$ is $3 \times p$ and $A'$ is $4 \times 3$,we must have $p = 4$.
Therefore,the order of matrix $B$ is $3 \times 4$.

(A) The number of equivalents of the oxidizing agent must be equal to the number of equivalents of the reducing agent $(H_{2}S)$.
Since the same amount of $H_{2}S$ is oxidized in both cases,the equivalents of $K_{2}Cr_{2}O_{7}$ must equal the equivalents of $KMnO_{4}$.
For $K_{2}Cr_{2}O_{7}$ in acidic medium,the $n_{f} = 6$.
For $KMnO_{4}$ in acidic medium,the $n_{f} = 5$.
Using the formula: $n_{f1} \times M_{1} \times V_{1} = n_{f2} \times M_{2} \times V_{2}$.
$6 \times 0.04 \times 50 = 5 \times 0.03 \times V_{KMnO_{4}}$.
$12 = 0.15 \times V_{KMnO_{4}}$.
$V_{KMnO_{4}} = \frac{12}{0.15} = 80 \ cm^{3}$.

(A) To determine the shape,we use the $VSEPR$ theory based on hybridization and lone pairs:
$1$. $BeCl_2$ ($sp$ hybridization,$0$ lone pairs) is linear.
$2$. $CO_2$ ($sp$ hybridization,$0$ lone pairs) is linear.
$3$. $XeF_2$ ($sp^3d$ hybridization,$3$ lone pairs) is linear.
$4$. $ICl_2^-$ ($sp^3d$ hybridization,$3$ lone pairs) is linear.
Both pairs $(BeCl_2, CO_2)$ and $(XeF_2, ICl_2^-)$ have identical linear shapes. Given the standard options provided in typical chemistry problems of this type,$BeCl_2$ and $CO_2$ are the most common examples of linear molecules.

(A) The ions $Ca^{2+}, K^{+},$ and $S^{2-}$ are isoelectronic species,as they all contain $18$ electrons.
For isoelectronic species,the ionic radius decreases as the atomic number $(Z)$ increases due to the increase in effective nuclear charge.
The atomic numbers are: $S = 16, K = 19, Ca = 20$.
Since the nuclear charge increases from $S$ to $Ca$,the ionic radius decreases in the order $S^{2-} > K^{+} > Ca^{2+}$.
Therefore,the increasing order of radii is $Ca^{2+} < K^{+} < S^{2-}$.

(B) From the ideal gas equation,density $\rho = \frac{PM}{RT}$.
Since $M$ (molar mass) and $R$ (gas constant) are constant for carbon monoxide,we have $\rho \propto \frac{P}{T}$.
We calculate the ratio $\frac{P}{T}$ for each option:

$A$. $0.5 \ atm, 273 \ K$	$\frac{0.5}{273} \approx 0.0018$
$B$. $4 \ atm, 500 \ K$	$\frac{4}{500} = 0.008$
$C$. $2 \ atm, 600 \ K$	$\frac{2}{600} \approx 0.0033$
$D$. $6 \ atm, 1092 \ K$	$\frac{6}{1092} \approx 0.0055$

The ratio $\frac{P}{T}$ is highest for option $B$. Thus,the density is maximum at $4 \ atm$ and $500 \ K$.

(A) One $kg$ of coke contains $n = \frac{1000}{12} = 83.33 \ mol$ of carbon.
Heat liberated by burning $1 \ kg$ of coke:
$C(s) + O_{2}(g) \rightarrow CO_{2}(g) ; \Delta H_{1} = 83.33 \times 393.5 \ kJ$.
Water gas is produced by: $C(s) + H_{2}O(g) \rightarrow CO(g) + H_{2}(g)$.
Thus,$1 \ kg$ of coke produces $83.33 \ mol$ of $CO$ and $83.33 \ mol$ of $H_{2}$.
Heat liberated by burning this water gas:
$CO(g) + \frac{1}{2}O_{2}(g) \rightarrow CO_{2}(g) ; \Delta H_{CO} = 83.33 \times 283.5 \ kJ$.
$H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(l) ; \Delta H_{H_{2}} = 83.33 \times 285.5 \ kJ$.
Total heat $\Delta H_{2} = 83.33 \times (283.5 + 285.5) = 83.33 \times 569 \ kJ$.
Ratio $= \frac{\Delta H_{1}}{\Delta H_{2}} = \frac{393.5}{569} \approx 0.69 : 1$.

(B) The balanced chemical equation for the reaction is: $(COOH)_2 + 2NaOH \rightarrow (COONa)_2 + 2H_2O$.
From the stoichiometry,$1 \ mole$ of oxalic acid reacts with $2 \ moles$ of $NaOH$.
Number of moles of $NaOH = \frac{\text{mass}}{\text{molar mass}} = \frac{0.064 \ g}{40 \ g/mol} = 0.0016 \ mol$.
Since $2 \ moles$ of $NaOH$ react with $1 \ mole$ of oxalic acid,the moles of oxalic acid $= \frac{0.0016}{2} = 0.0008 \ mol$.
Volume of oxalic acid $= 25 \ cm^3 = 0.025 \ L$.
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.0008 \ mol}{0.025 \ L} = 0.032 \ M$.

(D) The elements with an unpaired electron in their $s$-subshell are those that have an $ns^1$ configuration in their ground state.
These include the alkali metals $(H, Li, Na, K)$ and elements where an electron is promoted to the $s$-subshell to achieve stability,such as $Cr$ $([Ar] 3d^5 4s^1)$ and $Cu$ $([Ar] 3d^{10} 4s^1)$.
Thus,the elements are $H$ $(Z=1)$,$Li$ $(Z=3)$,$Na$ $(Z=11)$,$K$ $(Z=19)$,$Cr$ $(Z=24)$,and $Cu$ $(Z=29)$.
Total number of such elements is $6$.

(D) The equilibrium reaction is: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$.
Initial moles of $CaCO_{3} = \frac{50 \ g}{100 \ g/mol} = 0.5 \ mol$.
For this reaction,$K_{p} = P_{CO_{2}} = 1.64 \ atm$.
Using the ideal gas equation $PV = nRT$ for $CO_{2}$ gas:
$1.64 \ atm \times 10 \ L = n_{CO_{2}} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 1000 \ K$.
$16.4 = n_{CO_{2}} \times 82$.
$n_{CO_{2}} = \frac{16.4}{82} = 0.2 \ mol$.
Since $1 \ mol$ of $CaCO_{3}$ produces $1 \ mol$ of $CO_{2}$,the moles of $CaCO_{3}$ reacted $= 0.2 \ mol$.
Moles of $CaCO_{3}$ remaining $= 0.5 - 0.2 = 0.3 \ mol$.
Percentage of unreacted $CaCO_{3} = \frac{0.3}{0.5} \times 100 = 60 \%$.

(D) The reactions are: $(a)$ $NH_{3} + HCl \rightarrow NH_{4}Cl$ (Salt of strong acid and weak base,$pH < 7$). $(b)$ $NH_{3} + CH_{3}COOH \rightarrow CH_{3}COONH_{4}$ (Salt of weak acid and weak base,$pH \approx 7$). $(c)$ $2NH_{3} + H_{2}SO_{4} \rightarrow (NH_{4})_{2}SO_{4}$ (Salt of strong acid and weak base,$pH < 7$).
For $(a)$,$pH = 7 - \frac{1}{2}(pK_{b} + \log C) = 7 - \frac{1}{2}(4.74 + 0) = 4.63$.
For $(b)$,$pH = 7 + \frac{1}{2}(pK_{a} - pK_{b}) = 7 + \frac{1}{2}(4.74 - 4.74) = 7.0$.
For $(c)$,$H_{2}SO_{4}$ is a strong diprotic acid. $1 \ M \ H_{2}SO_{4}$ provides $2 \ M \ H^{+}$,which reacts with $1 \ M \ NH_{3}$ to leave $1 \ M \ H^{+}$ and $0.5 \ M \ (NH_{4})_{2}SO_{4}$. This results in a highly acidic solution $(pH < 1)$.
Thus,the order of $pH$ is $b (7.0) > a (4.63) > c (< 1)$,which corresponds to $b > a > c$.

(D) Let $I = \int \frac{\sin 2x}{\sin^2 x + 2\cos^2 x} dx$.
We know that $\sin^2 x + \cos^2 x = 1$,so the denominator becomes $\sin^2 x + \cos^2 x + \cos^2 x = 1 + \cos^2 x$.
Thus,$I = \int \frac{\sin 2x}{1 + \cos^2 x} dx$.
Let $t = 1 + \cos^2 x$.
Then $dt = 2\cos x(-\sin x) dx = -\sin 2x dx$.
So,$\sin 2x dx = -dt$.
Substituting these into the integral,we get $I = \int \frac{-dt}{t} = -\log|t| + C$.
Substituting back $t = 1 + \cos^2 x$,we get $I = -\log(1 + \cos^2 x) + C$.

(C) The hydrogenation of benzene is represented as: $C_6H_6 + 3H_2 \xrightarrow{Ni, 150^{\circ}C} C_6H_{12}$.
Because the intermediate products (cyclohexadiene and cyclohexene) are more reactive than benzene itself,they are immediately hydrogenated to cyclohexane.
Therefore,it is not possible to isolate cyclohexadiene and cyclohexene with ease during the controlled hydrogenation of benzene.
Thus,the correct statement is that cyclohexadiene and cyclohexene cannot be isolated with ease during controlled hydrogenation of benzene.

(D) We know that $\text{Bond order} \propto \frac{1}{\text{Bond length}}$.
$CO$ molecule has a bond order of $3$ $(C\equiv O)$.
$CO_{2}$ molecule has a bond order of $2$ $(O=C=O)$.
$HCO_{2}^{-}$ (formate ion) has a resonance hybrid with a bond order of $1.5$.
$CO_{3}^{2-}$ (carbonate ion) has a resonance hybrid with a bond order of $1.33$.
Since bond order is inversely proportional to bond length,the decreasing order of bond length is: $CO_{3}^{2-} (1.33) > HCO_{2}^{-} (1.5) > CO_{2} (2) > CO (3)$.

(D) The energy of stationary states in hydrogen-like atoms is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$,which is inversely proportional to $n^2$. Thus,option $A$ is correct.
The radius of the $n^{th}$ orbit is given by $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$. For $H$ $(Z=1, n=1)$,$r_1 = 0.529 \ \mathring{A}$. For $He^{+}$ $(Z=2, n=1)$,$r_1 = 0.529 \times \frac{1^2}{2} = 0.2645 \ \mathring{A}$. Since $0.2645$ is half of $0.529$,option $B$ is correct.
The angular quantum number $(l)$ determines the shape of the orbital. Thus,option $C$ is correct.
The total number of nodes in an orbital is given by $n - 1$. For the $3s$ orbital,$n = 3$,so the total number of nodes is $3 - 1 = 2$. Option $D$ states the number of nodes is $3$,which is incorrect.

(C) The conversion of oxygen into ozone is an endothermic process $(\Delta H = +ve)$ and proceeds with a decrease in entropy $(\Delta S = -ve)$.
The reaction is $3O_{2(g)} \rightleftharpoons 2O_{3(g)}$.
According to the Gibbs free energy equation,$\Delta G = \Delta H - T\Delta S$.
Since $\Delta H$ is positive and $\Delta S$ is negative,the term $-T\Delta S$ becomes positive.
Thus,$\Delta G$ is always positive at all temperatures.
Therefore,the reaction is non-spontaneous at all temperatures.

(D) Critical temperature $(T_c)$ is defined as the temperature above which a gas cannot be liquefied,regardless of the pressure applied.
It is not the lowest temperature at which liquefaction occurs; rather,it is the maximum temperature at which a gas can exist as a liquid.
Therefore,statement $(D)$ is incorrect.

(C) Let the mass of $CaO$ be $x \text{ g}$. Then the mass of $BaO$ is $(10 - x) \text{ g}$.
Moles of $HCl = \text{Molarity} \times \text{Volume in L} = 2.5 \times 0.1 = 0.25 \text{ mol}$.
The reactions are:
$CaO + 2HCl \rightarrow CaCl_2 + H_2O$
$BaO + 2HCl \rightarrow BaCl_2 + H_2O$
Total moles of $HCl = 2 \times (\text{moles of } CaO) + 2 \times (\text{moles of } BaO)$
$0.25 = 2 \times (\frac{x}{56} + \frac{10 - x}{153})$
$0.125 = \frac{153x + 560 - 56x}{56 \times 153}$
$0.125 \times 8568 = 97x + 560$
$1071 = 97x + 560$
$97x = 511$
$x = 5.268 \text{ g}$
Percentage of $CaO = \frac{5.268}{10} \times 100 = 52.68 \% \approx 52.6 \%$.

(A) $BaO_2$ (Barium peroxide) reacts with water to form $Ba(OH)_2$ and $H_2O_2$,which is a decomposition/hydrolysis reaction. However,in the context of standard inorganic chemistry questions,$BaO_2$ is often considered stable compared to the rapid hydrolysis of $CaC_2$ (forming $C_2H_2$),$P_4O_{10}$ (forming $H_3PO_4$),and $Mg_3N_2$ (forming $NH_3$). Actually,all these species react with water. If the question implies a species that does not undergo hydrolysis in the sense of forming an acid or base via water cleavage,$BaO_2$ is the intended answer as it primarily undergoes a double decomposition reaction.

(A) Potassium superoxide $( KO_{2} )$ reacts with water to produce potassium hydroxide,hydrogen peroxide,and oxygen gas. This reaction is represented as follows:
$ 2 KO_{2} + 2 H_{2} O \rightarrow 2 KOH + H_{2} O_{2} + O_{2} $
In contrast,other metal oxides react differently:
$ Na_{2} O_{2} + 2 H_{2} O \rightarrow 2 NaOH + H_{2} O_{2} $ (Produces peroxide but no oxygen)
$ CaO + H_{2} O \rightarrow Ca(OH)_{2} $ (Produces hydroxide only)
$ Li_{2} O + H_{2} O \rightarrow 2 LiOH $ (Produces hydroxide only)

(B) In the reaction $Hexane \xrightarrow{\text{anhy. } AlCl_3 / HCl}$,the isomerization of $n$-hexane to branched alkanes occurs via a carbocation intermediate.
Option $A$ is a nucleophilic addition reaction.
Option $C$ is a free radical substitution reaction.
Option $D$ is a nucleophilic substitution reaction ($S_N2$ mechanism for primary halides).
Therefore,the correct option is $B$.

(D) The reactions involved are:
$Na_2SO_3(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + SO_2(g) + H_2O(l)$
Here,$X = Na_2SO_3$ and $Y = SO_2$.
$SO_2$ decolourises acidified $KMnO_4$ solution:
$2KMnO_4 + 5SO_2 + 2H_2O \rightarrow K_2SO_4 + 2MnSO_4 + 2H_2SO_4$
When $H_2S$ (gas $Z$) is passed into an aqueous solution of $SO_2$ $(Y)$,colloidal sulphur is obtained:
$SO_2(g) + 2H_2S(g) \rightarrow 3S(s) + 2H_2O(l)$
Thus,$X = Na_2SO_3$ and $Z = H_2S$.

(C) In humans,spermatogenesis (the formation of male gametes) involves the meiotic division of one primary spermatocyte,which results in the production of $4$ functional haploid spermatozoa.
In contrast,oogenesis (the formation of female gametes) involves the meiotic division of one primary oocyte,which results in the production of only $1$ functional haploid ovum and $2$ or $3$ non-functional polar bodies.
Therefore,the ratio of male gametes to female gametes produced from one primary sex cell each is $4:1$.

(C) When gold $(Au)$ dissolves in aqua regia,it reacts to form the tetrachloridoaurate$(III)$ complex ion,$[AuCl_4]^-$.
The chemical equation for the reaction is:
$2 Au + 3 HNO_3 + 11 HCl \rightarrow 2 H[AuCl_4] + 3 NOCl + 6 H_2O$
In the complex $[AuCl_4]^-$,the oxidation state of gold is $+3$,and there are four chloride ligands,hence the name is $tetrachloridoaurate(III)$.

(C) Nucleophilic aromatic substitution reactions in haloarenes are facilitated by the presence of electron-withdrawing groups (like $-NO_2$) at the ortho and para positions relative to the halogen atom.
These groups stabilize the carbanion intermediate formed during the reaction.
Among the given options,$2,4-$dinitrochlorobenzene has two $-NO_2$ groups at the ortho and para positions,which provide the maximum stabilization to the intermediate,making it the most reactive towards nucleophilic substitution with sodium methoxide.

(B) Mass of $L$-amino acid $= 1.78 \ g$.
The reaction of an amino acid with $NaNO_{2} / HCl$ (Van Slyke method) releases $1 \ mol$ of $N_{2}$ gas per $1 \ mol$ of amino acid.
At $STP$,$1 \ mol$ of gas occupies $22400 \ cm^{3}$.
Moles of $N_{2}$ evolved $= \frac{448 \ cm^{3}}{22400 \ cm^{3} \ mol^{-1}} = 0.02 \ mol$.
Since $1 \ mol$ of amino acid produces $1 \ mol$ of $N_{2}$,moles of amino acid $= 0.02 \ mol$.
Molar mass of amino acid $= \frac{\text{Mass}}{\text{Moles}} = \frac{1.78 \ g}{0.02 \ mol} = 89 \ g \ mol^{-1}$.
The protein contains $0.25 \%$ of this amino acid by mass.
Let the molar mass of the protein be $M$.
$0.25 \% \text{ of } M = 89 \ g \ mol^{-1}$.
$\frac{0.25}{100} \times M = 89$.
$M = \frac{89 \times 100}{0.25} = 35600 \ g \ mol^{-1}$.

(B) Moles of $N_2 = \frac{5.5 \times 10^{-3} \text{ g}}{28 \text{ g/mol}} \approx 1.96 \times 10^{-4} \text{ mol}$.
Moles of $H_2O = \frac{180 \text{ g}}{18 \text{ g/mol}} = 10 \text{ mol}$.
Mole fraction of $N_2$ at $1 \text{ atm}$ $(x_1)$ = $\frac{n_{N_2}}{n_{H_2O} + n_{N_2}} \approx \frac{1.96 \times 10^{-4}}{10} = 1.96 \times 10^{-5}$.
According to Henry's law,$p = K_H \cdot x$,which implies $x \propto p$.
At $5 \text{ atm}$ pressure,the new mole fraction $(x_2)$ is $x_2 = 5 \times x_1 = 5 \times 1.96 \times 10^{-5} = 9.8 \times 10^{-5} \approx 1 \times 10^{-4}$.

(D) Total charge passed $Q = I \times t = 140 \times 482.5 = 67550 \ C$.
Number of Faradays $= \frac{67550}{96500} = 0.7 \ F$.
At the cathode,only $Cu^{2+}$ is reduced: $Cu^{2+} + 2e^- \rightarrow Cu$.
Equivalents of $Cu$ deposited $= 0.6928$.
Mass of $Cu$ deposited $= 0.6928 \times \frac{63.54}{2} = 22.011 \ g$.
At the anode,$Cu \rightarrow Cu^{2+} + 2e^-$ and $Fe \rightarrow Fe^{2+} + 2e^-$.
Total equivalents dissolved $= 0.7$.
Let $x$ be equivalents of $Cu$ and $y$ be equivalents of $Fe$.
$x + y = 0.7$.
Mass of $Cu$ dissolved $+ \text{Mass of } Fe$ dissolved $= 22.26 \ g$.
$x \times 31.77 + y \times 27.75 = 22.26$.
Substituting $x = 0.7 - y$: $(0.7 - y) \times 31.77 + 27.75y = 22.26$.
$22.239 - 31.77y + 27.75y = 22.26$.
$-4.02y = 0.021 \implies y \approx 0.00522$.
Mass of $Fe = 0.00522 \times 27.75 = 0.1448 \ g$.
Total mass of impure copper $= 22.26 \ g$.
$\% \text{ of } Fe = \frac{0.1448}{22.26} \times 100 \approx 0.65 \%$.
Since this value is not in the options,the correct answer is $D$.

(B) The iodoform reaction is given by compounds containing the $CH_{3}CO-$ group or the $CH_{3}CH(OH)-$ group.
$A) \ CH_{3}CHO$ contains the $CH_{3}CO-$ group,so it gives the iodoform reaction.
$B) \ CH_{3}CH_{2}CH_{2}OH$ (propan$-1-$ol) is a primary alcohol that does not contain the $CH_{3}CH(OH)-$ group. Therefore,it does not give the iodoform reaction.
$C) \ CH_{3}CH(OH)CH_{2}COOH$ contains the $CH_{3}CH(OH)-$ group,so it gives the iodoform reaction.
$D) \ CH_{3}CH_{2}OH$ (ethanol) is the only primary alcohol that contains the $CH_{3}CH(OH)-$ group,so it gives the iodoform reaction.
Thus,the correct answer is $B$.

(A) The correct statement is that collectors enhance the non-wettability of mineral particles during froth flotation,not the wettability. Examples of collectors include pine oil,xanthates,and fatty acids. Therefore,option $A$ is incorrect.

(D) The reaction of ethyl vinyl ether $(CH_2 = CH - O - CH_2 - CH_3)$ with $HI$ proceeds via the protonation of the oxygen atom.
$CH_2 = CH - O - CH_2 - CH_3 + H^+ \rightarrow CH_2 = CH - O^+(H) - CH_2 - CH_3$
Then,the iodide ion $(I^-)$ attacks the ethyl group $(CH_2CH_3)$ because the vinyl group $(CH_2=CH-)$ is not susceptible to $S_N2$ attack due to the partial double bond character of the $C-O$ bond.
$I^- + CH_2 = CH - O^+(H) - CH_2 - CH_3 \rightarrow CH_2 = CH - OH + CH_3CH_2I$
The product $CH_2 = CH - OH$ (vinyl alcohol) is unstable and tautomerizes to form ethanal $(CH_3CHO)$.
The other product formed is iodoethane $(CH_3CH_2I)$.
Therefore,among the given options,iodoethane is one of the products.

(A) The competing reaction for an $S_N2$ reaction is elimination,not rearrangement.
Substitution and elimination reactions often compete with each other.
Most bases act as nucleophiles and can participate in either substitution or elimination depending on the structure of the alkyl halide and the reaction conditions.
Rearrangements are typically associated with carbocation intermediates in $S_N1$ reactions,not $S_N2$ reactions.

(C) The statement that aldose or ketose sugars in alkaline medium do not isomerize is incorrect.
In an alkaline medium,aldose and ketose sugars undergo a reversible isomerization process known as the $Lobry \ de \ Bruyn-van \ Ekenstein \ rearrangement$.
For example,in a dilute solution of $NaOH$,glucose (an aldose) undergoes isomerization to form an equilibrium mixture of $D$-glucose,$D$-mannose,and $D$-fructose.
Therefore,option $C$ is the incorrect statement.

(A) The molecular formula $C_{7}H_{9}N$ corresponds to an aromatic amine. The reaction with $NaNO_{2} / HCl$ at $0^{\circ}C$ to form benzyl alcohol indicates that the compound is a primary aliphatic amine attached to a benzene ring,specifically benzylamine $(C_{6}H_{5}CH_{2}NH_{2})$.
The isomers of $C_{7}H_{9}N$ are:
$1$. $o$-Toluidine ($2$-methylaniline)
$2$. $m$-Toluidine ($3$-methylaniline)
$3$. $p$-Toluidine ($4$-methylaniline)
$4$. $N$-Methylaniline $(C_{6}H_{5}NHCH_{3})$
$5$. Benzylamine $(C_{6}H_{5}CH_{2}NH_{2})$
Thus,there are $5$ possible isomers for the compound $C_{7}H_{9}N$.

(D) The volume of the cube corresponding to one $CsCl$ ion pair is given as $V = 7.014 \times 10^{-23} \ cm^{3}$.
According to the problem,the smallest $Cs^{+}-Cs^{+}$ inter-nuclear distance $(a)$ is equal to the side length of this cube.
Therefore,$a^{3} = 7.014 \times 10^{-23} \ cm^{3}$.
$a = (7.014 \times 10^{-23})^{1/3} \ cm$.
$a = 4.124 \times 10^{-8} \ cm$.
Since $1 \ \mathring{A} = 10^{-8} \ cm$,we have $a = 4.124 \ \mathring{A}$.
Rounding to the nearest value,the distance is approximately $4.1 \ \mathring{A}$.
Thus,the correct option is $D$.

(C) Butylated hydroxy toluene $(BHT)$ is widely used as a food additive because it acts as an antioxidant. It prevents the oxidation of fats and oils in food,thereby increasing the shelf life of the product.

(C) The Nernst equation for the given half-cell reaction is:
$E = E^{\circ} - \frac{0.0591}{6} \log \frac{[Cr^{3+}]^{2}}{[Cr_{2}O_{7}^{2-}][H^{+}]^{14}}$
Substituting the given values:
$1.067 = 1.33 - \frac{0.0591}{6} \log \frac{(15 \times 10^{-3})^{2}}{(4.5 \times 10^{-3})[H^{+}]^{14}}$
$1.067 - 1.33 = -\frac{0.0591}{6} \log \frac{225 \times 10^{-6}}{4.5 \times 10^{-3} \cdot [H^{+}]^{14}}$
$-0.263 = -0.00985 \log \frac{0.05}{[H^{+}]^{14}}$
$26.7 = \log 0.05 + 14 pH$
$26.7 = -1.3 + 14 pH$
$28 = 14 pH$
$pH = 2$

(A) Acetophenone $(C_{6}H_{5}COCH_{3})$ is a ketone.
$1$. $C_{6}H_{6}$ (Benzene) can undergo Friedel-Crafts acylation with acetyl chloride to form acetophenone.
$2$. $C_{6}H_{5}CH(OH)CH_{3}$ ($1$-phenylethanol) can be oxidized to form acetophenone.
$3$. $C_{6}H_{5}C \equiv CH$ (Phenylacetylene) can undergo hydration to form acetophenone.
$4$. $C_{6}H_{5}CH_{3}$ (Toluene) cannot be directly converted to acetophenone easily because oxidation of the methyl group typically yields benzoic acid,and Friedel-Crafts acylation of toluene would result in a mixture of ortho and para isomers,not the desired acetophenone.

(C) Gold sol is a colloidal suspension of gold nanoparticles in a liquid.
It is a lyophobic colloid because there is no affinity between the dispersed phase and the dispersion medium.
It is negatively charged due to the adsorption of $OH^-$ ions.
It is a multimolecular colloid because it consists of aggregates of a large number of atoms or smaller molecules with diameters less than $1 \ nm$.

(A) $MnO_{2} + 4 HCl \xrightarrow{\Delta} MnCl_{2} + 2 H_{2}O + Cl_{2(g)} (A)$
$Cl_{2(g)} + 3 F_{2} (\text{excess}) \xrightarrow{573 K} 2 ClF_{3(g)} (B)$
$3 ClF_{3(g)} + U_{(s)} \xrightarrow{323-363 K} UF_{6(g)} (C) + 3 ClF_{(g)} (D)$
Therefore,the gases $A$,$B$,$C$,and $D$ are $Cl_{2}$,$ClF_{3}$,$UF_{6}$,and $ClF$ respectively.

(B) The reaction sequence is as follows:
$1$. Phenol $(C_6H_5OH)$ reacts with benzoyl chloride $(C_6H_5COCl)$ in the presence of a base to form phenyl benzoate $(X)$ via the Schotten-Baumann reaction.
$2$. Phenyl benzoate $(C_6H_5COOC_6H_5)$ undergoes electrophilic aromatic substitution (nitration). The ester group $(-OCOC_6H_5)$ is ortho/para directing. Due to steric hindrance,the para-substituted product is the major product.
$3$. The nitration occurs on the ring attached to the oxygen atom (the phenoxy ring) because it is more activated than the benzoyl ring. Thus,the major product $Y$ is $4-$nitrophenyl benzoate.

(D) The Freundlich adsorption isotherm is given by the equation:
$\frac{x}{m} = k p^{\frac{1}{n}}$
Taking the logarithm on both sides:
$\log \frac{x}{m} = \log k + \frac{1}{n} \log p$
Comparing this with the equation of a straight line,$y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log p$,$m$ (slope) $= \frac{1}{n}$,and $c$ (intercept) $= \log k$.
Therefore,the slope is $\frac{1}{n}$ and the intercept is $\log k$.

(A) Let us analyze each option:
$A$. $[ZnBr_{4}]^{2-}$: $Br^{-}$ is a weak field ligand and does not cause pairing of electrons. $Zn^{2+}$ has a $d^{10}$ configuration. The hybridization is $sp^3$,resulting in a tetrahedral geometry. This statement is correct.
$B$. $[Ni(NH_{3})_{6}]^{2+}$: $Ni^{2+}$ is $3d^8$. In the presence of $NH_{3}$,it forms an outer orbital complex with $sp^3d^2$ hybridization. It is not an inner orbital complex.
$C$. $[Co(NH_{3})_{6}]^{2+}$: $Co^{2+}$ is $3d^7$. In the presence of $NH_{3}$,it forms a low-spin complex,but it is typically oxidized to $Co^{3+}$ in the presence of air/ligands. However,strictly considering $[Co(NH_{3})_{6}]^{2+}$,it has one unpaired electron and is paramagnetic. Note: The question asks for the correct statement,and $A$ is definitively correct.
$D$. $[CoBr_{2}(en)_{2}]^{-}$: This complex does not contain ambidentate ligands (like $NO_{2}^{-}$ or $SCN^{-}$),so it cannot exhibit linkage isomerism.
Therefore,the most accurate statement is $A$.

(A) The acidity of an active methylene group depends on the electron-withdrawing power of the adjacent functional groups ($E_1$ and $E_2$).
These groups stabilize the resulting carbanion through resonance and the inductive effect.
Ketone groups $(-COCH_3)$ are more electron-withdrawing than ester groups $(-COOC_2H_5)$ because the oxygen atom in the ester group donates electrons via resonance ($-OR$ group),which reduces its electron-withdrawing ability compared to a ketone.
Comparing the compounds:
$(b)$ $CH_3COCH_2COCH_3$: Has two ketone groups. The carbanion is highly stabilized by two strong electron-withdrawing groups.
$(a)$ $CH_3COCH_2COOC_2H_5$: Has one ketone and one ester group. The carbanion is less stabilized than in $(b)$.
$(c)$ $C_2H_5OOCCH_2COOC_2H_5$: Has two ester groups. The carbanion is the least stabilized among the three.
Thus,the order of acid strength is $b > a > c$.

(A) $CuI_{2}$ is not known because it is unstable.
Copper$(II)$ iodide is not stable upon formation; it spontaneously decomposes to copper$(I)$ iodide and iodine.
The reaction is: $2CuI_{2} \rightarrow 2CuI + I_{2}$.

(D) The reaction sequence is as follows:
$1$. $C_6H_5COOH + NH_3 \xrightarrow{\Delta} C_6H_5CONH_2$ ($P$,benzamide).
$2$. $C_6H_5CONH_2 + NaOBr \rightarrow C_6H_5NH_2$ ($Q$,aniline) (Hofmann bromamide degradation).
$3$. $C_6H_5NH_2 + \text{conc. } H_2SO_4 \xrightarrow{460 \ K} NH_2-C_6H_4-SO_3H$ ($R$,sulphanilic acid).
Thus,$R$ is sulphanilic acid.

(B) The correct statement is that phosphorous acid $(H_{3}PO_{3})$ on heating disproportionates to give orthophosphoric acid $(H_{3}PO_{4})$ and phosphine $(PH_{3})$.
$4H_{3}PO_{3} \xrightarrow{\Delta} 3H_{3}PO_{4} + PH_{3}$.
Therefore,the statement in option $B$ is incorrect because it mentions metaphosphoric acid instead of orthophosphoric acid.

(D) Terylene is a condensation polymer formed by the reaction between ethylene glycol and terephthalic acid. Since it involves the elimination of small molecules like water,it is a condensation polymer,which is a type of step-growth polymer. It is also a polyester and a copolymer because it is made from two different monomers. It is $NOT$ a chain-growth polymer,as chain-growth polymerization typically involves addition polymerization of unsaturated monomers.

(B) In a $ccp$ (cubic close-packed) arrangement,the number of octahedral voids is equal to the number of atoms,and the number of tetrahedral voids is twice the number of atoms.
Let the number of atoms of $Y$ be $N$.
Then,the number of octahedral voids $= N$ and the number of tetrahedral voids $= 2N$.
The formula of the compound is $XY_3$.
This implies that for every $1$ atom of $X$,there are $3$ atoms of $Y$.
If we have $N$ atoms of $Y$,then the number of atoms of $X$ is $N/3$.
Since $X$ occupies octahedral voids,the fraction of octahedral voids occupied by $X$ is $\frac{N/3}{N} = 1/3$.
Therefore,the percentage of octahedral voids occupied by $X$ is $\frac{1}{3} \times 100\% \approx 33\%$.

(C) The reaction of isobutyraldehyde $((CH_3)_2CHCHO)$ with methylmagnesium bromide $(CH_3MgBr)$ in the presence of ether follows a nucleophilic addition mechanism.
$1$. The Grignard reagent $(CH_3^-)$ attacks the carbonyl carbon of the aldehyde to form an intermediate alkoxide $(A)$,which is $(CH_3)_2CH-CH(OMgBr)-CH_3$.
$2$. Subsequent acid hydrolysis $(H_3O^+)$ converts the intermediate into the corresponding alcohol $(B)$,which is $(CH_3)_2CH-CH(OH)-CH_3$.
$3$. The structure of $B$ is $CH_3-CH(CH_3)-CH(OH)-CH_3$.
$4$. According to $IUPAC$ nomenclature,the longest chain containing the hydroxyl group has $4$ carbons. Numbering from the right gives the hydroxyl group at position $2$ and the methyl group at position $3$. Thus,the $IUPAC$ name is $3-$methylbutan$-2-$ol.

(A) At $S.T.P.$,$22400 \ cm^{3}$ of methane corresponds to $1 \ mol$.
Therefore,$112 \ cm^{3}$ of methane corresponds to $\frac{112}{22400} = 0.005 \ mol$.
Since $1 \ mol$ of monohydric alcohol reacts with $1 \ mol$ of methylmagnesium iodide to produce $1 \ mol$ of methane,the moles of alcohol used is $0.005 \ mol$.
The molar mass of the alcohol is $\frac{\text{Given mass}}{\text{moles}} = \frac{0.44 \ g}{0.005 \ mol} = 88 \ g/mol$.
Among the options,$CH_{3}CH(OH)CH_{2}CH_{3}$ (butan$-2-$ol) has a molar mass of $74 \ g/mol$,while $C_{5}H_{12}O$ isomers have $88 \ g/mol$. The alcohol must be a primary alcohol to form an aldehyde with $PCC$ that gives a positive silver mirror test. The structure $CH_{3}CH(CH_{3})CH_{2}OH$ ($2$-methylpropan$-1-$ol) has a molar mass of $74 \ g/mol$. Re-evaluating the molar mass $88 \ g/mol$ corresponds to $C_{5}H_{12}O$. The alcohol $CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}OH$ or similar isomers fit. However,based on the requirement of forming an aldehyde that gives a silver mirror test,the alcohol must be primary. The correct structure is $CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}OH$ or similar. Given the options provided,the calculation leads to $88 \ g/mol$.

(B) $(i)$ Acetic acid undergoes Hell-Volhard-Zelinsky $(HVZ)$ reaction.
$(ii)$ Sodium phenate undergoes Kolbe's reaction.
$(iii)$ Methyl cyanide undergoes Stephen's reduction.
$(iv)$ Toluene undergoes Friedel-Crafts reaction.
Therefore,the correct matching is $(i-c, ii-d, iii-a, iv-b)$.

(B) For an ideal binary liquid mixture,the following conditions hold true:
$1$. $\Delta H_{(mix)} = 0$ (No heat is absorbed or evolved).
$2$. $\Delta V_{(mix)} = 0$ (No change in volume upon mixing).
$3$. $\Delta S_{(mix)} > 0$ (Entropy increases due to mixing).
$4$. $\Delta G_{(mix)} < 0$ (The process of mixing is spontaneous).
Since $\Delta G = \Delta H - T\Delta S$,and for an ideal solution $\Delta H = 0$,we get $\Delta G = -T\Delta S$. For the process to be spontaneous,$\Delta G$ must be negative,which implies $\Delta S$ must be positive.

(B) The Arrhenius equation is given by $ k = A e^{-\frac{E_{a}}{RT}} $.
Taking the natural logarithm on both sides,we get $ \ln k = \ln A - \frac{E_{a}}{R} \times \frac{1}{T} $.
Comparing this with the equation of a straight line $ y = mx + c $,where $ y = \ln k $,$ x = \frac{1}{T} $,and the slope $ m = -\frac{E_{a}}{R} $.
Given the slope $ m = -1 \times 10^{4} \ K $.
Therefore,$ -\frac{E_{a}}{R} = -1 \times 10^{4} \ K $.
$ E_{a} = 1 \times 10^{4} \times 8.314 \ J \ mol^{-1} $.
$ E_{a} = 83140 \ J \ mol^{-1} = 83.14 \ kJ \ mol^{-1} $.

(D) Given: Mass of solute $w_{B} = 1.25 \ g$,Mass of solvent $w_{A} = 50 \ g$,$\Delta T_{f} = 0.3 \ K$,$M_{\text{normal}} = 94 \ g \ mol^{-1}$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$.
First,calculate the observed molar mass $(M_{\text{obs}})$ using the formula: $\Delta T_{f} = K_{f} \times m = K_{f} \times \frac{w_{B} \times 1000}{M_{\text{obs}} \times w_{A}}$.
$M_{\text{obs}} = \frac{1.86 \times 1.25 \times 1000}{0.3 \times 50} = 155 \ g \ mol^{-1}$.
Calculate the van't Hoff factor $(i)$: $i = \frac{M_{\text{normal}}}{M_{\text{obs}}} = \frac{94}{155} \approx 0.6064$.
For association,$n P \rightleftharpoons P_{n}$. Assuming dimerization $(n=2)$,the degree of association $\alpha$ is given by $i = 1 - \alpha + \frac{\alpha}{n} = 1 - \alpha(1 - \frac{1}{n})$.
$0.6064 = 1 - \alpha(1 - 0.5) \implies 0.6064 = 1 - 0.5\alpha$.
$0.5\alpha = 1 - 0.6064 = 0.3936$.
$\alpha = 0.7872$ or approximately $78.7\%$.
Since $78.7\%$ is not among the options,the correct choice is $D$.

(A) $1$. Reductive ozonolysis of an alkene $R_1R_2C=CR_3R_4$ yields carbonyl compounds $R_1R_2C=O$ and $R_3R_4C=O$.
$2$. $Y$ is obtained by Etard's reaction,which is the oxidation of toluene to benzaldehyde $(C_6H_5CHO)$. Thus,$Y$ is benzaldehyde.
$3$. $Z$ undergoes a disproportionation reaction (Cannizzaro reaction) with concentrated alkali. Formaldehyde $(HCHO)$ is an aldehyde without $\alpha$-hydrogens,so it undergoes the Cannizzaro reaction.
$4$. Combining $Y$ $(C_6H_5CHO)$ and $Z$ $(HCHO)$,the original alkene $X$ must be styrene $(C_6H_5CH=CH_2)$.
$5$. The reaction is: $C_6H_5CH=CH_2 \xrightarrow{\text{O}_3, \text{Zn/H}_2\text{O}} C_6H_5CHO + HCHO$.

(B) The most stable oxidation state for lanthanoids is $+3$.
$Ce^{4+}$ is a strong oxidizing agent in aqueous solution because it easily gains an electron to form the more stable $Ce^{3+}$ ion.
Therefore,$Ce^{4+}$ is not stable in aqueous solution and is not known to exist as a free ion in such conditions.
Thus,the statement '$Ce^{4+}$ in aqueous solution is not known' is considered correct in the context of its stability.

(C) Lewis acid strength decreases down the group; therefore,the correct order is $BCl_{3} > AlCl_{3} > GaCl_{3}$.
First ionization energy for Group $13$ elements generally decreases down the group but shows irregularities due to $d$-electron and lanthanoid contraction.
The correct order for first ionization enthalpy is $B > Tl > Ga > Al > In$.
Inert pair effect is more pronounced in heavier elements,making $Tl^{+}$ more stable than $Tl^{3+}$.
Consequently,$Tl^{3+}$ is a strong oxidizing agent,making the oxidizing property order $Tl^{3+} > In^{3+} > Al^{3+}$.

(D) The final product is $4\text{-methylpent-3-en-2-one}$,which is an $\alpha,\beta\text{-unsaturated ketone}$ formed by the aldol condensation of $2$ molecules of acetone $(CH_3COCH_3)$.
Thus,$R$ must be acetone $(CH_3COCH_3)$.
The reaction of $P$ with $CH_3MgBr$ followed by hydrolysis yields acetone $(R)$.
$CH_3CN + CH_3MgBr$ $\rightarrow CH_3C(CH_3)=NMgBr$ $\xrightarrow{H_3O^+} CH_3COCH_3 + NH_3 + Mg(OH)Br$.
Therefore,$P$ is $CH_3CN$ (Ethanenitrile).

(A) $Zn^{2+}$ forms a coordination compound with both aqueous sodium hydroxide and ammonia,while $Al^{3+}$ forms a coordination compound only with aqueous sodium hydroxide.
Zinc forms the tetrahydroxozincate$(II)$ complex with $NaOH$: $Zn^{2+} + 4OH^{-} \rightarrow [Zn(OH)_{4}]^{2-}$.
Zinc forms the tetraamminezinc$(II)$ complex with $NH_{3}$: $Zn^{2+} + 4NH_{3(aq)} \rightarrow [Zn(NH_{3})_{4}]^{2+}$.
Aluminum forms the hexahydroxoaluminate$(III)$ complex with aqueous sodium hydroxide: $Al^{3+} + 6OH^{-} \rightarrow [Al(OH)_{6}]^{3-}$.
Aluminum does not form a coordination compound with aqueous ammonia.

(B) Given,$\Lambda_{m}^{\circ} = 140 \times 10^{-4} \ S \ m^{2} \ mol^{-1}$,$K = 1.85 \times 10^{-5} \ S \ m^{-1}$.
For a sparingly soluble salt,the molar conductivity $\Lambda_{m}^{\circ}$ is related to solubility $S$ (in $mol \ m^{-3}$) by the formula $\Lambda_{m}^{\circ} = \frac{K}{S}$.
$S = \frac{K}{\Lambda_{m}^{\circ}} = \frac{1.85 \times 10^{-5} \ S \ m^{-1}}{140 \times 10^{-4} \ S \ m^{2} \ mol^{-1}} = 1.32 \times 10^{-3} \ mol \ m^{-3}$.
Since $1 \ m^{3} = 1000 \ L$,the solubility in $mol \ L^{-1}$ is $S = \frac{1.32 \times 10^{-3}}{1000} = 1.32 \times 10^{-6} \ mol \ L^{-1}$.
For the salt $AB$,$K_{sp} = S^{2} = (1.32 \times 10^{-6})^{2} = 1.74 \times 10^{-12}$.

(D) The first step is nitration because the nitro group $(-NO_2)$ is a meta-directing group,which is required to place the bromine atom at the $m$-position relative to the amino group.
Next,bromination of nitrobenzene is performed to introduce the bromine atom at the $m$-position.
Finally,the reduction of the nitro group to an amino group $(-NH_2)$ yields $m$-bromoaniline.
The reaction sequence is:
Benzene $\xrightarrow{\text{nitration}}$ Nitrobenzene $\xrightarrow{\text{bromination}}$ $m$-Bromonitrobenzene $\xrightarrow{\text{reduction}}$ $m$-Bromoaniline

(A) The cell reaction is $H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O_{(l)}$.
Here, the number of electrons involved $(n)$ is $2$.
Using the relation $\Delta G^{\circ} = -n F E^{\circ}_{\text{cell}}$:
$-240,000 \text{ J} = -2 \times 96,500 \text{ C} \times E^{\circ}_{\text{cell}}$.
$E^{\circ}_{\text{cell}} = \frac{240,000}{2 \times 96,500} \approx 1.24 \text{ V}$.

(B) Given,half-life $t_{1/2} = 69.3 \ s$.
For a first-order reaction,the rate constant $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{69.3} = 10^{-2} \ s^{-1}$.
For first-order kinetics,$k = \frac{2.303}{t} \log \frac{P_0}{P_t}$,where $P_0$ is initial pressure and $P_t$ is pressure at time $t$.
$10^{-2} = \frac{2.303}{230} \log \frac{0.4}{P_t}$.
$1 = \log \frac{0.4}{P_t} \implies \frac{0.4}{P_t} = 10 \implies P_t = 0.04 \ atm$.
From the reaction $A_{(g)} \rightarrow P_{(g)} + Q_{(g)} + R_{(g)}$:
At $t = 0$,$P_A = 0.4 \ atm$,$P_P = 0, P_Q = 0, P_R = 0$.
At $t = 230 \ s$,$P_A = 0.04 \ atm$. The decrease in pressure of $A$ is $0.4 - 0.04 = 0.36 \ atm$.
According to stoichiometry,the pressure of products formed is $P_P = 0.36 \ atm, P_Q = 0.36 \ atm, P_R = 0.36 \ atm$.
Total pressure $P_{total} = P_A + P_P + P_Q + P_R = 0.04 + 0.36 + 0.36 + 0.36 = 1.12 \ atm$.