WBJEE 2013 Chemistry Question Paper with Answer and Solution

(D) Initially,the charge on the capacitor $C_0$ is $Q_{initial} = C_0 V_0$.
When the switch $S$ is closed,the charge redistributes between the two capacitors until they reach a common potential $V$.
According to the law of conservation of charge,the total charge remains constant:
$Q_{initial} = Q_{final}$
$C_0 V_0 = C_0 V + C V$
Rearranging the equation to solve for $C$:
$C_0 V_0 - C_0 V = C V$
$C_0(V_0 - V) = C V$
$C = \frac{C_0(V_0 - V)}{V}$

(C) The benzoin condensation involves the reaction of two molecules of benzaldehyde in the presence of a cyanide ion $(CN^-)$ catalyst to form benzoin.
The mechanism begins with the nucleophilic attack of $CN^-$ on the carbonyl carbon of benzaldehyde.
This is followed by a proton transfer (tautomerization) to form a carbanion intermediate,which is stabilized by the electron-withdrawing $CN$ group and the phenyl ring.
The structure of this key intermediate is $Ph-C^-(OH)(CN)$.
This carbanion then attacks another molecule of benzaldehyde to continue the reaction,eventually leading to the formation of benzoin after the elimination of the $CN^-$ catalyst.

(A) Ozone $(O_{3})$ exhibits resonance,where the two $O-O$ bonds are equivalent due to the delocalization of $\pi$ electrons.
The actual structure is a resonance hybrid of two canonical forms.
The experimental value for the average $O-O$ bond length in ozone is $1.28 \mathring{A}$,which is intermediate between the single bond length $(1.48 \mathring{A})$ and the double bond length $(1.21 \mathring{A})$.

(D) The molecule $SOCl_{2}$ (thionyl chloride) has a trigonal pyramidal geometry due to the presence of a lone pair on the sulfur atom.
According to $VSEPR$ theory,the lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion.
In $SOCl_{2}$,the $Cl-S-Cl$ bond angle is approximately $96^{\circ}$ and the $Cl-S-O$ bond angle is approximately $106^{\circ}$.

(D) The bond angles for the given species are as follows:
$1.$ $NO_2^+$: It has $sp$ hybridization and a linear geometry,resulting in a bond angle of $180^\circ$.
$2.$ $NO_2$: It has $sp^2$ hybridization with one unpaired electron,resulting in a bond angle of approximately $134^\circ$.
$3.$ $NO_2^-$: It has $sp^2$ hybridization with one lone pair,resulting in a bond angle of approximately $115^\circ$.
Comparing these values,the increasing order of the bond angle is $NO_2^- < NO_2 < NO_2^+$.
Since this specific order is not listed in options $A$,$B$,or $C$,the correct answer is $D$.

(A) The $CO$ molecule has a total of $14$ electrons ($6$ from $C$ and $8$ from $O$).
According to Molecular Orbital Theory,for molecules with $14$ or fewer electrons,the energy order of molecular orbitals is $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z$.
Filling the $14$ electrons into these orbitals:
$1\sigma^2, 2\sigma^2, 3\sigma^2, 4\sigma^2, 1\pi^4, 5\sigma^2$.
In many textbooks,the notation is simplified as $1\sigma^2 2\sigma^2 1\pi^4 3\sigma^2$ where $1\sigma$ is $\sigma 1s$,$2\sigma$ is $\sigma^* 1s$,$3\sigma$ is $\sigma 2s$,$4\sigma$ is $\sigma^* 2s$,$1\pi$ is $\pi 2p$,and $5\sigma$ is $\sigma 2p_z$.
Thus,the correct configuration is $1\sigma^2 2\sigma^2 3\sigma^2 4\sigma^2 1\pi^4 5\sigma^2$,which corresponds to option $A$ when simplified.

(A) According to $Fajan's$ rule,the ionic character is inversely proportional to the polarising power of the cation.
Polarising power depends on the charge density of the cation.
As the size of the cation increases $(B^{3+} < Al^{3+} < Ga^{3+})$,the polarising power decreases.
Therefore,the ionic character increases as the size of the cation increases.
The correct order of ionic character is $BCl_{3} < AlCl_{3} < GaCl_{3}$.

(A) Baeyer's reagent is $1 \%$ cold dilute alkaline potassium permanganate $(KMnO_4)$.
It is used to identify unsaturation in organic compounds.
When added to unsaturated compounds,the purple colour of the reagent disappears.

(C) $1$. Identify the principal functional group. The nitrile group $(-CN)$ has higher priority than the ketone group $(=O)$. Thus,the parent chain is a nitrile.
$2$. Number the chain starting from the carbon of the $-CN$ group as $C-1$.
$3$. The chain is $CH_3-C(CH_3)_2-CH_2-C(=O)-CH_3$. Numbering gives: $C-1$ (nitrile carbon),$C-2$ (quaternary carbon with two methyl groups),$C-3$ (methylene group),$C-4$ (ketone carbonyl),$C-5$ (terminal methyl).
$4$. The substituents are two methyl groups at $C-2$ and an oxo group at $C-4$.
$5$. Combining these,the $IUPAC$ name is $2,2-$dimethyl$-4-$oxopentanenitrile.

(C) An optically active molecule must lack a plane of symmetry or a center of inversion.
In option $(A)$,the molecule has a plane of symmetry,making it a meso compound.
In option $(B)$,the molecule has a plane of symmetry,making it a meso compound.
In option $(C)$,the molecule has different functional groups at the ends ($COOMe$ and $COOH$),which prevents the existence of a plane of symmetry,thus making it optically active.
In option $(D)$,the molecule has a plane of symmetry,making it a meso compound.
Therefore,only molecule $(C)$ is optically active.

(C) Molecules having an integer value for the degree of unsaturation $(DU)$ can exist in a free state as stable compounds.
$DU = \frac{\sum n(v-2)}{2} + 1$,where $n$ is the number of atoms and $v$ is the valency of the atom.
For $C_7H_9O$: $DU = \frac{7(4-2) + 9(1-2) + 1(2-2)}{2} + 1 = \frac{14-9+0}{2} + 1 = 3.5$.
For $C_8H_{12}O$: $DU = \frac{8(4-2) + 12(1-2) + 1(2-2)}{2} + 1 = \frac{16-12+0}{2} + 1 = 3.0$.
For $C_6H_{12}O$: $DU = \frac{6(4-2) + 12(1-2) + 1(2-2)}{2} + 1 = \frac{12-12+0}{2} + 1 = 1.0$.
For $C_{10}H_{17}O$: $DU = \frac{10(4-2) + 17(1-2) + 1(2-2)}{2} + 1 = \frac{20-17+0}{2} + 1 = 2.5$.
Since $C_8H_{12}O$ and $C_6H_{12}O$ have integer values,$C_6H_{12}O$ is a stable compound.

(B) Tautomerism requires the presence of at least one $\alpha$-hydrogen atom adjacent to a carbonyl group $(C=O)$.
$A$. $(Me_3CCO)_3CH$ has an $\alpha$-hydrogen atom on the central carbon,which is flanked by three carbonyl groups,making it highly acidic and capable of tautomerism.
$B$. Cyclohex$-2-$ene$-1,4-$dione has $\alpha$-hydrogen atoms at the $C-5$ and $C-6$ positions,allowing it to exhibit keto-enol tautomerism.
$C$. Cyclohexa$-2,5-$diene$-1,4-$dione (p-benzoquinone) has no $\alpha$-hydrogen atoms,so it cannot exhibit tautomerism.
$D$. Cyclohexane$-1,2-$dione has $\alpha$-hydrogen atoms at the $C-3$ and $C-6$ positions,allowing it to exhibit keto-enol tautomerism.
Given the options provided in the image,compounds $(a)$,$(b)$,and $(d)$ all possess $\alpha$-hydrogens and can exhibit tautomerism. However,in many standard contexts for this specific question,$(b)$ is often highlighted for its specific structural features.

(A) Lassaigne's test is used to detect elements like $N, S,$ and halogens in organic compounds.
$1$. $A$ blue colouration or precipitate (Prussian blue,$Fe_4[Fe(CN)_6]_3$) is formed when the compound contains both $C$ and $N$,which react with molten $Na$ to form $NaCN$.
$2$. $A$ white precipitate $(AgCl)$ is formed when the compound contains chlorine,which reacts with molten $Na$ to form $NaCl$,which then reacts with $AgNO_3$.
$3$. $N$-methylaniline $(C_6H_5NHCH_3)$ contains $C$ and $N$,thus it gives a blue colouration.
$4$. $2$-chlorobenzoic acid contains $Cl$,thus it gives a white precipitate with $AgNO_3$.
Therefore,the correct pair is $N$-methylaniline and $2$-chlorobenzoic acid.

(C) In $2-$methylpropane,there are $9$ primary hydrogens and $1$ tertiary hydrogen.
The relative reactivity of tertiary hydrogen towards chlorination is about $5.0$ to $5.5$ times that of primary hydrogen.
However,the number of primary hydrogens $(9)$ is much greater than the number of tertiary hydrogens $(1)$.
Therefore,the statistical factor $(9 \times 1 = 9)$ outweighs the reactivity factor $(1 \times 5.5 = 5.5)$.
As a result,$1-$chloro$-2-$methylpropane is formed as the major product.

(B) $2, 2-$dimethylbutane is prepared using the Corey-House synthesis,which involves the reaction of a lithium dialkylcuprate with an alkyl halide.
To prepare $2, 2-$dimethylbutane $(CH_3-C(CH_3)_2-CH_2-CH_3)$,we need a tert-butyl group and an ethyl group.
The Corey-House reaction follows an $S_N2$ mechanism,which works best with primary alkyl halides.
Therefore,the reaction between $(Me_3C)_2CuLi$ and $MeCH_2Br$ is the most efficient route.
The reaction is: $(Me_3C)_2CuLi + MeCH_2Br \rightarrow Me_3C-CH_2CH_3 + Me_3CCu + LiBr$.

(B) $1$. The molecular formula $C_8H_{16}$ corresponds to the general formula $C_nH_{2n}$,which indicates an alkene.
$2$. Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
$3$. The problem states that one of the products is acetone,which is $CH_3COCH_3$. This implies the original alkene must contain a $(CH_3)_2C=$ group.
$4$. The compound must also be optically active,meaning it must contain a chiral center (a carbon atom bonded to four different groups).
$5$. Analyzing the options,the structure in option $B$ is $3,4$-dimethylhex-$2$-ene,but specifically,the structure shown is $2,3$-dimethylhex-$2$-ene. Let's re-examine the structure: the compound $2,3$-dimethylhex-$2$-ene has a chiral center at the $C_3$ position,which is bonded to a methyl group,an ethyl group,a hydrogen atom,and the $C=C$ group. Thus,it is optically active.
$6$. Ozonolysis of $2,3$-dimethylhex-$2$-ene yields acetone and $2$-methylbutanal. Therefore,the correct structure is the one shown in option $B$.

(D)

Medium	Colour of Litmus
Acidic	Red
Neutral	Violet
Basic	Blue

Litmus is a natural indicator extracted from lichens. In an acidic solution,it turns red. In a neutral solution,it retains its original purple/violet colour. In a basic solution,it turns blue.

(B) For a very dilute solution of a strong base,the contribution of $OH^{-}$ ions from the dissociation of water cannot be neglected.
$KOH \longrightarrow K^{+} + OH^{-}$
$[OH^{-}]_{KOH} = 10^{-8} \ M$
$[OH^{-}]_{water} = 10^{-7} \ M$
Total $[OH^{-}] = 10^{-8} + 10^{-7} = 10^{-7}(0.1 + 1) = 1.1 \times 10^{-7} \ M$
$pOH = -\log(1.1 \times 10^{-7}) = 7 - \log(1.1) \approx 7 - 0.0414 = 6.9586$
Since $pH + pOH = 14$ at $25^{\circ} C$,
$pH = 14 - 6.9586 = 7.0414 \approx 7.04$.
Given the options,the closest value is $7.02$.

(B) The initial moles of acetic acid and sodium acetate in $1 \ \text{L}$ of solution are $0.01 \ \text{mole}$ each.
When $1 \times 10^{-3} \ \text{mole}$ of $HCl$ is added,it reacts with the salt (sodium acetate) to form acetic acid:
$CH_{3}COO^{-} + H^{+} \longrightarrow CH_{3}COOH$
New moles of salt = $0.01 - 0.001 = 0.009 \ \text{mole}$.
New moles of acid = $0.01 + 0.001 = 0.011 \ \text{mole}$.
Using the Henderson-Hasselbalch equation:
$pH = pK_{a} + \log \frac{[\text{salt}]}{[\text{acid}]} = 4.75 + \log \frac{0.009}{0.011} = 4.75 + \log(0.818) \approx 4.75 - 0.087 = 4.663$
Thus,the final $pH$ is approximately $4.66$.

(B) For a salt of $MX_{2}$ type,the dissociation is given by: $MX_{2} \rightleftharpoons M^{2+} + 2X^-$.
If $s$ is the solubility in $mol / L$,then $[M^{2+}] = s$ and $[X^-] = 2s$.
The solubility product $K_{sp}$ is given by: $K_{sp} = [M^{2+}][X^-]^2 = (s)(2s)^2 = 4s^3$.
Given $K_{sp} = 3.2 \times 10^{-8}$.
So,$4s^3 = 3.2 \times 10^{-8}$.
$s^3 = \frac{3.2 \times 10^{-8}}{4} = 0.8 \times 10^{-8} = 8 \times 10^{-9}$.
Taking the cube root,$s = \sqrt[3]{8 \times 10^{-9}} = 2 \times 10^{-3} \ mol / L$.

(D) Diborane $(B_2H_6)$ has a bridged structure where two $BH_2$ units are linked by two bridging hydrogen atoms.
Each bridge consists of a $3$-center-$2$-electron $(3c-2e)$ bond.
Since there are two such bridges,the total number of electrons involved in the bridging bonds is $2 \times 2 = 4$ electrons.
These are often referred to as banana bonds.

(A) The chemical formula of the borate anion in borax is $[B_4O_5(OH)_4]^{2-}$.
By examining the structure of the $[B_4O_5(OH)_4]^{2-}$ ion:
$1$. There are $5$ $B-O-B$ linkages forming the cyclic structure.
$2$. There are $4$ $B-OH$ bonds attached to the boron atoms.
Therefore,the number of $B-O-B$ links is $5$ and the number of $B-OH$ bonds is $4$.

(A) Silicones are synthetic polymers containing $R_2SiO$ repeating units.
Silicon oil is prepared by the hydrolysis and subsequent polymerisation of a mixture of $trimethylchlorosilane$ $( (CH_3)_3SiCl )$ and $dimethyldichlorosilane$ $( (CH_3)_2SiCl_2 )$.
The $trimethylchlorosilane$ acts as a chain terminator,controlling the chain length of the resulting silicone polymer,which determines its viscosity (forming oil).
These are widely used as lubricants and antifoaming agents.

(C) The $10$ millionth part is equivalent to $10^{-7}$.
$\therefore$ The volume of the part is $1.33 \ cm^{3} \times 10^{-7} = 1.33 \times 10^{-7} \ mL$.
For pure water at $25^{\circ} C$,$[H^{+}] = 10^{-7} \ mol/L$.
Since $1 \ L = 1000 \ mL$,$1 \ mL$ of water contains $10^{-7} / 1000 = 10^{-10} \ mol$ of $H^{+}$ ions.
Therefore,the number of moles of $H^{+}$ in $1.33 \times 10^{-7} \ mL$ is $1.33 \times 10^{-7} \times 10^{-10} = 1.33 \times 10^{-17} \ mol$.
Number of $H^{+}$ ions $= \text{moles} \times N_{A} = 1.33 \times 10^{-17} \times 6.022 \times 10^{23}$.
$= 8.009 \times 10^{6} \approx 8.01 \times 10^{6}$ or $8.01$ million.

(C) The van der Waals' equation for $n$ moles of a gas is $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
At very low pressure,the volume $V$ is very large,making the term $\frac{a}{V^2}$ negligible.
Also,the volume $V$ is much larger than the excluded volume $b$,so $(V - b) \approx V$.
Thus,the equation reduces to $PV = RT$,which is the ideal gas equation.
Therefore,a van der Waals' gas behaves ideally at very low pressure.

(B) The given lines are $x+y=4$ and $x-y=2$.
Solving these equations,we add them to get $2x=6$,so $x=3$. Substituting $x=3$ into $x+y=4$,we get $y=1$.
The point of intersection is $(3, 1)$.
The slope of the line is $m = \tan(\tan^{-1}(3/4)) = 3/4$.
The equation of the line passing through $(3, 1)$ with slope $3/4$ is $(y-1) = \frac{3}{4}(x-3)$,which simplifies to $4y-4 = 3x-9$,or $3x-4y=5$.
Thus,$y = \frac{3x-5}{4}$.
Substituting this into the parabola equation $y^{2}=4(x-3)$,we get $\left(\frac{3x-5}{4}\right)^{2} = 4(x-3)$.
$\frac{9x^{2}-30x+25}{16} = 4x-12$.
$9x^{2}-30x+25 = 64x-192$.
$9x^{2}-94x+217 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$,we have $x = \frac{94 \pm \sqrt{8836 - 7812}}{18} = \frac{94 \pm \sqrt{1024}}{18} = \frac{94 \pm 32}{18}$.
So,$x_{1} = \frac{126}{18} = 7$ and $x_{2} = \frac{62}{18} = \frac{31}{9}$.
Therefore,$|x_{1}-x_{2}| = |7 - \frac{31}{9}| = |\frac{63-31}{9}| = \frac{32}{9}$.

(D) According to the Pauli Exclusion Principle,no two electrons in an atom can have the same set of all four quantum numbers $(n, l, m, s)$.
Therefore,all four quantum numbers are necessary to uniquely identify and describe the state of an electron in an atom.
$n$ (Principal quantum number) defines the shell and energy.
$l$ (Azimuthal quantum number) defines the subshell and orbital shape.
$m$ (Magnetic quantum number) defines the spatial orientation.
$s$ (Spin quantum number) defines the electron spin direction.

(B) The work done $(W)$ in a pressure-volume system is defined as $W = -P \Delta V$.
At constant volume,the change in volume $(\Delta V)$ is $0$.
Therefore,$W = -P \times 0 = 0$.

(D) For an ideal gas,internal energy $(U)$ and enthalpy $(H)$ are functions of temperature only.
In an isothermal process,$\Delta T = 0$,which implies $\Delta U = 0$ and $\Delta H = 0$.
From the first law of thermodynamics,$\Delta U = Q + W$. Since $\Delta U = 0$,$Q = -W$.
For expansion,work is done,so $W \neq 0$,which means $Q \neq 0$.
Therefore,the correct combination is $\Delta U=0, Q \neq 0, W \neq 0$ and $\Delta H=0$.

(C) Entropy is a state function that measures the degree of randomness or disorder in a system.
For a reversible process,the change in entropy $(dS)$ is defined as the ratio of the heat exchanged reversibly $(\delta q_{rev})$ to the absolute temperature $(T)$ at which the exchange occurs.
Thus,the mathematical expression is $dS = \frac{\delta q_{rev}}{T}$.

(C) When two different ideal gases are mixed under isothermal reversible conditions,the process is spontaneous and leads to an increase in the randomness of the system.
The entropy of mixing,$\Delta S_{mix}$,is given by the formula $\Delta S_{mix} = -n R \sum x_i \ln x_i$,where $x_i$ is the mole fraction of each gas.
Since $x_i < 1$,$\ln x_i$ is negative,making $\Delta S_{mix}$ always positive.
Therefore,the entropy of the system increases.

(B) From the Gibbs-Helmholtz equation,$\Delta G = \Delta H - T \Delta S$.
For a process to be spontaneous at constant temperature and pressure,the change in Gibbs free energy of the system must be negative,i.e.,$\Delta G < 0$.
This implies a decrease or lowering of the Gibbs free energy of the system.

(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -RT \ln K_{eq}$.
Given: $T = 25^{\circ} C = 298 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,and $K_{eq} = 0.15$.
Substituting the values: $\Delta G^{\circ} = -8.314 \times 298 \times \ln(0.15)$.
Using $\ln(0.15) \approx -1.897$:
$\Delta G^{\circ} = -8.314 \times 298 \times (-1.897) \approx 4703 \ J \approx 4.7 \ kJ$.
Alternatively,using $\log_{10}$:
$\Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times \log_{10}(0.15) \approx -5705.8 \times (-0.8239) \approx 4701 \ J \approx 4.7 \ kJ$.

(C) The reaction of $2,6-$dideutero-fluorobenzene with $NaNH_2$ in liquid $NH_3$ proceeds via the benzyne mechanism.
$1$. The amide ion $(NH_2^-)$ abstracts a proton from the ortho position relative to the fluorine atom.
$2$. This leads to the elimination of $F^-$ and the formation of a benzyne intermediate,specifically $3-$deuterobenzyne.
$3$. The nucleophilic attack of $NH_2^-$ on the benzyne intermediate can occur at two non-equivalent positions due to the presence of the deuterium atom.
$4$. This results in a mixture of $2-$deuteroaniline and $3-$deuteroaniline as the final products.

(D) Direct nitration of aniline with concentrated $HNO_3$ and $H_2SO_4$ (nitrating mixture) leads to the formation of a significant amount of $m$-nitroaniline due to the protonation of the $-NH_2$ group to form the $-NH_3^+$ ion,which is meta-directing.
In acidic medium,aniline exists as anilinium ion $(C_6H_5NH_3^+)$,which is strongly deactivating and meta-directing.
Therefore,the major product obtained under these conditions is $m$-nitroaniline.

(D) Ribose is an aldopentose that reacts with phenylhydrazine $(Ph-NH-NH_2)$ to form a characteristic osazone derivative.
In $2$-deoxyribose,the $-OH$ group at the $C-2$ position is replaced by a hydrogen atom.
The formation of an osazone requires the presence of an $\alpha$-hydroxy carbonyl group (or a group that can be oxidized to it).
Since $2$-deoxyribose lacks the $-OH$ group at the $C-2$ position,it cannot form the stable osazone structure under standard conditions,allowing it to be differentiated from ribose.

(C) dipeptide is formed by the condensation of two amino acids,where the carboxyl group $(-COOH)$ of one amino acid reacts with the amino group $(-NH_2)$ of another to form a peptide bond $(-CO-NH-)$.
For the dipeptide gly-ala (Glycine-Alanine):
$1$. Glycine $(NH_2-CH_2-COOH)$ is the $N$-terminal amino acid.
$2$. Alanine $(NH_2-CH(CH_3)-COOH)$ is the $C$-terminal amino acid.
$3$. The peptide bond is formed between the $-COOH$ of Glycine and the $-NH_2$ of Alanine.
$4$. The resulting structure is $NH_2-CH_2-CO-NH-CH(CH_3)-COOH$.

(C) The preparation of $Me_{3}CCN$ (pivalonitrile) from $Me_{3}CBr$ via $S_{N}2$ reaction with $NaCN$ is not feasible because $Me_{3}CBr$ is a tertiary alkyl halide,which undergoes elimination rather than substitution.
The reaction of $Me_{3}CMgBr$ (a Grignard reagent) with $ClCN$ (cyanogen chloride) is the most effective method for introducing a cyano group to a tertiary carbon atom.
The reaction is as follows:
$Me_{3}CMgBr + ClCN \rightarrow Me_{3}CCN + Mg(Cl)Br$

(D) For the reaction $2 NO_{2(g)} + F_{2(g)} \longrightarrow 2 NO_2F_{(g)}$,the rate of reaction is expressed in terms of the rate of change of partial pressure as:
Rate $= -\frac{1}{2} \left[ \frac{dp(NO_2)}{dt} \right] = -\left[ \frac{dp(F_2)}{dt} \right] = +\frac{1}{2} \left[ \frac{dp(NO_2F)}{dt} \right]$.
Comparing this with the given options,option $D$ is correct.

(B) The reaction is $CH_{3}COOC_{2}H_{5} + H_{2}O \xrightarrow{H^+} CH_{3}COOH + C_{2}H_{5}OH$.
In a standard pseudo-first order reaction,water is in excess,making the order with respect to ester $1$.
However,if the reaction is carried out with a large excess of ester,the concentration of the ester remains effectively constant throughout the reaction.
Therefore,the rate of reaction becomes independent of the concentration of the ester,resulting in a zero-order reaction with respect to the ester.

(D) The number of half-lives $n$ is calculated as: $n = \frac{t}{t_{1/2}} = \frac{22920}{5730} = 4$.
The amount of ${}^{14}C$ remaining after $n$ half-lives is given by: $N = N_0 (\frac{1}{2})^n = N_0 (\frac{1}{2})^4 = \frac{N_0}{16}$.
The fraction of ${}^{14}C$ that has decayed is: $\text{Decayed fraction} = \frac{N_0 - N}{N_0} = 1 - \frac{N}{N_0} = 1 - \frac{1}{16} = \frac{15}{16}$.

(C) From the Arrhenius equation:
$\ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} \left[ \frac{1}{T_{1}} - \frac{1}{T_{2}} \right]$
Given $T_{1} = 27 + 273 = 300 \ K$,$T_{2} = 327 + 273 = 600 \ K$,and $E_{a} = 600 R$.
Substituting the values:
$\ln \frac{k_{2}}{k_{1}} = \frac{600 R}{R} \left[ \frac{1}{300} - \frac{1}{600} \right]$
$\ln \frac{k_{2}}{k_{1}} = 600 \left[ \frac{2 - 1}{600} \right]$
$\ln \frac{k_{2}}{k_{1}} = 600 \times \frac{1}{600} = 1$
Therefore,$\frac{k_{2}}{k_{1}} = e^{1} = e$.

(A, B, C) The reaction of $Ni^{2+}$ with dimethylglyoxime $(DMG)$ in an ammoniacal (basic) medium produces a cherry-red precipitate of nickel$(II)$ dimethylglyoximate,$[Ni(DMG)_2]$.
In this complex,two dimethylglyoximate units are bound to one $Ni^{2+}$ ion.
The two dimethylglyoximate units are held together by strong intramolecular hydrogen bonding.
Each dimethylglyoximate unit forms a five-membered chelate ring with the $Ni^{2+}$ ion through the nitrogen atoms of the oxime groups.
Therefore,statements $A$,$B$,and $C$ are correct,while $D$ is incorrect because the chelate ring formed is five-membered,not six-membered.

(A) When excess potassium iodide $(KI)$ is added to mercuric chloride $(HgCl_{2})$,the reaction is as follows:
$HgCl_{2} + 4KI \longrightarrow K_{2}[HgI_{4}] + 2KCl$
In the complex $[HgI_{4}]^{2-}$,the central metal ion is $Hg^{2+}$.
The electronic configuration of $Hg$ $(Z=80)$ is $[Xe] 4f^{14} 5d^{10} 6s^{2}$.
For $Hg^{2+}$,the configuration is $[Xe] 4f^{14} 5d^{10} 6s^{0}$.
The four $I^-$ ligands donate electron pairs to the $6s$ and $6p$ orbitals of $Hg^{2+}$,resulting in $sp^{3}$ hybridization.
This leads to a tetrahedral geometry for the complex $K_{2}[HgI_{4}]$.

(A) Species having unpaired electrons are paramagnetic.
$Ni$ has atomic number $28$,electronic configuration $[Ar] 3d^{8} 4s^{2}$.
$1$. $[Ni(H_{2}O)_{6}]^{2+}$: $Ni$ is in $+2$ oxidation state $(3d^{8})$. $H_{2}O$ is a weak field ligand,so no pairing occurs. It has $2$ unpaired electrons (Paramagnetic).
$2$. $[Ni(PPh_{3})_{2}Cl_{2}]$: $Ni$ is in $+2$ oxidation state $(3d^{8})$. It is a tetrahedral complex. Due to the presence of $Cl^{-}$ ligands,it remains high spin with $2$ unpaired electrons (Paramagnetic).
$3$. $[Ni(CO)_{4}]$: $Ni$ is in $0$ oxidation state $(3d^{8} 4s^{2})$. $CO$ is a strong field ligand,causing pairing of all electrons. It has $0$ unpaired electrons (Diamagnetic).
$4$. $[Ni(CN)_{4}]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^{8})$. $CN^{-}$ is a strong field ligand,causing pairing of electrons. It has $0$ unpaired electrons (Diamagnetic).
Thus,$[Ni(H_{2}O)_{6}]^{2+}$ and $[Ni(PPh_{3})_{2}Cl_{2}]$ are paramagnetic.

(B) The orange solid is ammonium dichromate,$(NH_{4})_{2}Cr_{2}O_{7}$.
Upon heating,it undergoes thermal decomposition to produce nitrogen gas (colourless),chromium$(III)$ oxide (green solid),and water vapor:
$(NH_{4})_{2}Cr_{2}O_{7} \xrightarrow{\Delta} N_{2} \uparrow + Cr_{2}O_{3} + 4H_{2}O$
The green solid,$Cr_{2}O_{3}$,can be reduced to metallic chromium by aluminium powder in a thermite reaction:
$Cr_{2}O_{3} + 2Al \longrightarrow 2Cr + Al_{2}O_{3}$

(B) The equivalent conductance $(\Lambda_{eq})$ at infinite dilution depends on the ionic mobility of the ions in the solvent.
In water,$H^{+}$ and $HO^{-}$ ions exhibit exceptionally high ionic mobilities due to the Grotthuss mechanism (proton hopping).
Among the given ions,$H^{+}$ has the highest mobility,followed by $HO^{-}$.
$K^{+}$ is a simple hydrated cation with moderate mobility,while $CH_{3}COO^{-}$ is a large polyatomic anion with relatively lower mobility.
Therefore,the correct order of equivalent conductances at infinite dilution is $H^{+} > HO^{-} > K^{+} > CH_{3}COO^{-}$.

(C) Given,specific conductance $k = 1.25 \times 10^{-3} \ S \ cm^{-1}$.
Resistance $R = 800 \ \Omega$.
The relationship between specific conductance $(k)$,resistance $(R)$,and cell constant $(G^*)$ is given by the formula: $k = (1/R) \times G^*$.
Rearranging the formula to solve for the cell constant: $G^* = k \times R$.
Substituting the given values: $G^* = (1.25 \times 10^{-3} \ S \ cm^{-1}) \times (800 \ \Omega)$.
$G^* = 1.00 \ cm^{-1}$.

(A) The molar conductivity at infinite dilution for the salt $[(COO^{-})_{2} Na^{+} K^{+}]$ is the sum of the molar conductivities of its constituent ions: $\lambda_{m}^{\infty} = \lambda_{m}^{\infty} (oxalate^{2-}) + \lambda_{m}^{\infty} (Na^{+}) + \lambda_{m}^{\infty} (K^{+})$.
Substituting the given values: $\lambda_{m}^{\infty} = 148.2 + 73.5 + 50.1 = 271.8 \ S \ cm^{2} \ mol^{-1}$.
The $n$-factor for the salt $[(COO^{-})_{2} Na^{+} K^{+}]$ is $2$ because the oxalate ion has a charge of $-2$.
Therefore,the equivalent conductivity at infinite dilution is $\lambda_{eq}^{\infty} = \frac{\lambda_{m}^{\infty}}{n-factor} = \frac{271.8}{2} = 135.9 \ S \ cm^{2} \ eq^{-1}$.

(D) The $SbCl_5$ acts as a Lewis acid and abstracts the chloride ion $(Cl^-)$ from the substrate,$(+)-2-$chloro$-2-$phenylethane.
This results in the formation of a planar,$sp^2$ hybridized carbocation intermediate.
Because the carbocation is planar,the chloride ion can attack from either the top or the bottom face with equal probability.
This leads to the formation of both enantiomers,resulting in a racemic mixture.

(C) The acidic strength of substituted phenols depends on the electron-withdrawing effect ($-I$ and $-M$ effects) of the substituent group attached to the benzene ring.
Stronger electron-withdrawing groups stabilize the phenoxide ion more effectively,thereby increasing the acidic strength.
The electron-withdrawing tendency of the given substituents follows the order: $-F < -Cl < -NO_2$.
Therefore,the correct order of acidic strength is $p$-fluorophenol $ < $ $p$-chlorophenol $ < $ $p$-nitrophenol.

(A) The reaction between sodium nitroprusside $(Na_2[Fe(CN)_5NO])$ and sulphide ion $(S^{2-})$ is a standard test for the detection of sulphur.
The reaction is as follows:
$Na_2S + Na_2[Fe(CN)_5NO] \rightarrow Na_4[Fe(CN)_5NOS]$
This complex dissociates in solution as:
$Na_4[Fe(CN)_5NOS] \rightleftharpoons 4Na^+ + [Fe(CN)_5NOS]^{4-}$
The resulting complex $[Fe(CN)_5NOS]^{4-}$ is a tetranionic complex of iron $(II)$ where the $NO^+$ ligand is replaced by $NOS^{3-}$ (or effectively coordinates as $NOS^{3-}$ to the $Fe^{2+}$ center),resulting in a deep purple colouration.

(A) The chemical formula for the ore chromite is $FeCr_{2}O_{4}$.
It is an oxide mineral belonging to the spinel group.

(A, B, C, D) The Lintz and Donawitz $(L.D.)$ process is a highly efficient method for steel manufacturing.
Key advantages include:
$1$. The process is very quick due to the high-pressure oxygen blast.
$2$. It produces better quality steel by effectively removing impurities like $C$,$Si$,and $Mn$ as slag.
$3$. Scrap iron can be used as a raw material along with molten iron.
$4$. Operating costs are relatively low compared to older methods.
Since all the given options are valid advantages of the $L.D.$ process,the question implies selecting all that apply.

(B, C) $n-$butanol is a straight-chain alcohol,while $t-$butanol is a branched-chain alcohol.
Branching decreases the surface area of the molecule,which reduces the strength of intermolecular van der Waals forces,leading to a lower boiling point for $t-$butanol compared to $n-$butanol.
Additionally,branching increases the solubility of alcohols in water because it reduces the hydrophobic character of the alkyl group.
Therefore,$t-$butanol is more soluble in water and has a lower boiling point than $n-$butanol.
Thus,both statements $B$ and $C$ are correct.

(A) The industrial manufacture of nitric acid is carried out by the $Ostwald$ process. The steps are as follows:
$1$. Catalytic oxidation of ammonia: $4 NH_{3} + 5 O_{2} \xrightarrow[800-900^{\circ} C]{Pt \text{ gauge}} 4 NO + 6 H_{2} O$
$2$. Oxidation of nitric oxide to nitrogen dioxide: $2 NO + O_{2} \longrightarrow 2 NO_{2}$
$3$. Absorption of nitrogen dioxide in water: $4 NO_{2} + 2 H_{2} O + O_{2} \longrightarrow 4 HNO_{3}$
Thus,the intermediate compounds formed are nitric oxide $(NO)$ and nitrogen dioxide $(NO_{2})$.

(D) $H_{2}SO_{4}$ saturated with $SO_{3}$ is called $Oleum$ or $Sulphan$.
The chemical reaction is: $H_{2}SO_{4} + SO_{3} \longrightarrow H_{2}S_{2}O_{7}$.

(D) The reaction of chlorine gas with red hot calcium oxide is given by the following chemical equation:
$2 CaO + 2 Cl_2 \xrightarrow{\text{Red hot}} 2 CaCl_2 + O_2 \uparrow$
Thus,the products formed are calcium chloride and oxygen.

(A) The thermal decomposition of chloric acid $(HClO_{3})$ is given by the following reaction:
$3 HClO_{3} \stackrel{\Delta}{\longrightarrow} HClO_{4} + Cl_{2} + 2 O_{2} + H_{2}O$
Thus,the products formed are perchloric acid $(HClO_{4})$,chlorine $(Cl_{2})$,oxygen $(O_{2})$,and water $(H_{2}O)$.

(B) $\Delta T_{f} = i \times K_{f} \times m$
$\therefore i = \frac{\Delta T_{f}}{K_{f} \times m} = \frac{0.19}{1.86 \times 0.1} = 1.0215 \approx 1.02$
Again from,$\alpha = \frac{i-1}{n-1} = \frac{1.02-1}{2-1} = 0.02$
For $CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H^{+}$
$K_{a} = C \alpha^{2}$
$= 0.1 \times (0.02)^{2} = 0.1 \times 4 \times 10^{-4} = 4 \times 10^{-5}$