WBJEE 2020 Chemistry Question Paper with Answer and Solution

(B) Let $S_i$ be the number of students who gave wrong answers to at least $i$ questions. We are given $S_i = 2^{n-i}$ for $i = 1, 2, \dots, n$.
The total number of wrong answers is the sum of the number of students who answered at least $i$ questions wrongly for all $i$ from $1$ to $n$.
Total wrong answers $= \sum_{i=1}^{n} S_i = \sum_{i=1}^{n} 2^{n-i}$.
This is a geometric series: $2^{n-1} + 2^{n-2} + \dots + 2^0$.
The sum of this geometric series is $\frac{2^n - 1}{2 - 1} = 2^n - 1$.
Given that the total number of wrong answers is $2047$,we have:
$2^n - 1 = 2047$
$2^n = 2048$
$2^n = 2^{11}$
Therefore,$n = 11$.

(A) $NaBH_4$ is a selective reducing agent that reduces aldehydes and ketones to alcohols but does not reduce carboxylic acids. In the given reaction,the ketone group is reduced to a hydroxyl group,while the carboxylic acid group remains unchanged. The reduction of the ketone in a substituted cyclohexane ring typically leads to the formation of the more stable isomer. The $trans$ isomer,where the bulky $-OH$ and $-CO_2H$ groups are in a more stable configuration,is the major product.

(B) The bond order is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For $He_{2}$ $(4 \ e^{-})$: Configuration is $\sigma(1s)^{2}, \sigma^{*}(1s)^{2}$. $B.O. = \frac{2-2}{2} = 0$.
For $He_{2}^{+}$ $(3 \ e^{-})$: Configuration is $\sigma(1s)^{2}, \sigma^{*}(1s)^{1}$. $B.O. = \frac{2-1}{2} = 0.5$.
For $He_{2}^{2+}$ $(2 \ e^{-})$: Configuration is $\sigma(1s)^{2}$. $B.O. = \frac{2-0}{2} = 1$.
Thus,the bond orders are $0, 0.5, 1$ respectively.

(C) The molecule is $O_2$.
$1$. $O_2$ has two unpaired electrons in its antibonding $\pi^*{2p}$ orbitals,which corresponds to a $2$-electron magnetic moment (paramagnetic).
$2$. The one-electron reduced species is the superoxide ion $(O_2^-)$ and the two-electron reduced species is the peroxide ion $(O_2^{2-})$. Both can act as oxidizing and reducing agents.
$3$. When $O_2$ is one-electron oxidized,it forms the dioxygenyl cation $(O_2^+)$. In $O_2$,the bond order is $2.0$ (electrons in antibonding orbitals). Removing one electron from the antibonding orbital increases the bond order to $2.5$,which causes the bond length to decrease.

(A) Given reactions:
$(i) 2 A \rightleftharpoons B + C, K_{1} = 1$
$(ii) 2 B \rightleftharpoons C + D, K_{2} = 16$
$(iii) 2 C + D \rightleftharpoons 2 P, K_{3} = 25$
We need the equilibrium constant for $P \rightleftharpoons A + \frac{1}{2} B$.
Reverse reaction $(iii)$ and divide by $2$: $P \rightleftharpoons C + \frac{1}{2} D, K' = \sqrt{\frac{1}{K_{3}}} = \frac{1}{5}$.
Reverse reaction $(ii)$ and divide by $2$: $\frac{1}{2} C + \frac{1}{2} D \rightleftharpoons B, K'' = \sqrt{\frac{1}{K_{2}}} = \frac{1}{4}$.
Reverse reaction $(i)$ and divide by $2$: $\frac{1}{2} B + \frac{1}{2} C \rightleftharpoons A, K''' = \sqrt{\frac{1}{K_{1}}} = 1$.
Adding these reactions: $P + \frac{1}{2} C + \frac{1}{2} D + \frac{1}{2} B + \frac{1}{2} C \rightleftharpoons C + \frac{1}{2} D + B + A$.
Simplifying: $P \rightleftharpoons A + \frac{1}{2} B$.
The equilibrium constant $K_{final} = K' \times K'' \times K''' = \frac{1}{5} \times \frac{1}{4} \times 1 = \frac{1}{20}$.

(D) The stability of carbocations is determined by electronic effects:
$1$. In $II$ $(^+CH_2-OCH_3)$,the oxygen atom has lone pairs that can donate electron density through resonance ($+M$ effect),completing the octet of the carbon atom,making it the most stable.
$2$. In $III$ $(^+CH_2-CH_3)$,the ethyl group provides stability through hyperconjugation and the inductive effect ($+I$ effect) of the methyl group.
$3$. In $I$ $(^+CH_2-COCH_3)$,the carbonyl group $(-C=O)$ is a strong electron-withdrawing group ($-I$ and $-M$ effect),which destabilizes the positive charge,making it the least stable.
Therefore,the order of stability is $I < III < II$.

(C) molecule is asymmetric if it lacks any element of symmetry,such as a plane of symmetry or a center of inversion.
$A$: $cis-1-bromo-3-chlorocyclobutane$ has a plane of symmetry passing through $C1$ and $C3$.
$B$: $1-bromo-2,2-dichlorocyclopropane$ has a plane of symmetry passing through the $C1-Br$ bond and the $C2$ atom.
$C$: $trans-1-bromo-3-chlorocyclobutane$ lacks both a plane of symmetry and a center of symmetry,making it asymmetric.
$D$: $1-bromo-2,2-dichlorocyclopropane$ (as shown in the image) also possesses a plane of symmetry.
Therefore,the correct option is $C$.

(C) The reaction of $3,3-dimethylbut-1-ene$ with $HBr$ proceeds via the formation of a carbocation intermediate.
Initially,the protonation of the double bond yields a secondary carbocation,which undergoes a $1,2-methyl$ shift to form a more stable tertiary carbocation.
$1$. The direct addition of $Br^-$ to the secondary carbocation gives $2-bromo-3,3-dimethylbutane$. This molecule has a chiral center,so it exists as a pair of enantiomers ($R$ and $S$).
$2$. The addition of $Br^-$ to the rearranged tertiary carbocation gives $2-bromo-2,3-dimethylbutane$,which is achiral.
Thus,the total number of alkyl bromides formed is $2$ (enantiomers) $+ 1$ (rearranged product) $= 3$.

(B) meso-compound is an achiral molecule that contains stereocenters but is superimposable on its mirror image due to an internal plane of symmetry.
$1$. $trans-hex-3-ene + Br_2$ undergoes anti-addition to form a racemic mixture of $(3R, 4R)$ and $(3S, 4S)-3,4-dibromohexane$.
$2$. $cis-hex-3-ene + H_2 \xrightarrow{Pd-C}$ undergoes syn-addition to form a meso-compound,$(3R, 4S)-hexane-3,4-diol$ (if it were dihydroxylation) or in this case,the product of hydrogenation is $hexane$,which is achiral but not a meso-compound in the stereochemical sense. However,looking at the options provided,the question asks for the reaction that yields a meso-compound.
$3$. $hex-3-yne + H_2 \xrightarrow{Lindlar's \ catalyst}$ gives $cis-hex-3-ene$,which is not a meso-compound.
$4$. $cyclohexene + Br_2 \xrightarrow{CCl_4}$ gives $trans-1,2-dibromocyclohexane$,which is a racemic mixture,not a meso-compound.
Re-evaluating the options based on standard textbook examples: The hydrogenation of $cis-3,4-dimethylhex-3-ene$ (if that were the substrate) would yield a meso-compound. Given the provided image for option $B$ shows $cis-3,4-dimethylhex-3-ene$,the product of syn-addition is indeed a meso-compound.

(A) Step $1$: The reduction of but-$2$-yne with $Na/NH_3(liq)$ in the presence of ethanol is a Birch reduction,which yields $trans$-but-$2$-ene as the intermediate $X$.
Step $2$: The reaction of $trans$-but-$2$-ene with dilute alkaline $KMnO_4$ (Baeyer's reagent) proceeds via $syn$-dihydroxylation.
Step $3$: $Syn$-dihydroxylation of $trans$-alkenes results in the formation of a racemic mixture of enantiomers.
Step $4$: The product is a racemic mixture of $(2R, 3R)$-butane-$2,3$-diol and $(2S, 3S)$-butane-$2,3$-diol,which corresponds to the structures shown in option $A$.

(C) During the electrolysis of water,$H_{2}O$ molecules are decomposed into $H_{2}$ and $O_{2}$ gas more readily than $D_{2}O$ molecules due to the kinetic isotope effect.
As a result,the concentration of heavy water $(D_{2}O)$ in the remaining liquid increases over time.
Since the density of $D_{2}O$ $(1.107 \ g/cm^3)$ is higher than that of $H_{2}O$ $(1.00 \ g/cm^3)$,the overall density of the remaining liquid water increases.

(D) For $SrCO_3$,the solubility product expression is $K_{sp} = [Sr^{2+}][CO_3^{2-}]$.
Given $[CO_3^{2-}] = 1.2 \times 10^{-3} \ M$ and $K_{sp}(SrCO_3) = 7.0 \times 10^{-10}$,we calculate the concentration of $Sr^{2+}$:
$[Sr^{2+}] = \frac{7.0 \times 10^{-10}}{1.2 \times 10^{-3}} = 5.83 \times 10^{-7} \ M$.
For $SrF_2$,the solubility product expression is $K_{sp} = [Sr^{2+}][F^{-}]^2$.
Substituting the values: $7.9 \times 10^{-10} = (5.83 \times 10^{-7}) \times [F^{-}]^2$.
$[F^{-}]^2 = \frac{7.9 \times 10^{-10}}{5.83 \times 10^{-7}} = 1.355 \times 10^{-3}$.
$[F^{-}] = \sqrt{1.355 \times 10^{-3}} \approx 3.68 \times 10^{-2} \ M \approx 3.7 \times 10^{-2} \ M$.

(A, C, D) $SiO_{2}$ is an acidic oxide and reacts with both strong bases and specific fluorinating agents.
$1$. $SiO_{2(s)} + 2F_{2(g)} \rightarrow SiF_{4(g)} + O_{2(g)}$
$2$. $SiO_{2(s)} + 6HF(aq.) \rightarrow H_{2}SiF_{6}(aq.) + 2H_{2}O(\ell)$
$3$. $SiO_{2(s)} + 2NaOH(aq.) \rightarrow Na_{2}SiO_{3}(aq.) + H_{2}O(\ell)$
Since $HF$,hot $NaOH$,and $F_{2}$ all react with $SiO_{2}$,the question implies identifying reagents that attack it. Given the standard multiple-choice format,this is a multiple-correct type question.

(B, C) . $SO_{2}$ is oxidized to $SO_{3}$ by $O_{3}$ in the atmosphere,so it acts as a sink. This statement is correct.
$B$. $FGD$ (Flue Gas Desulfurization) is a process used to remove $SO_{2}$ from flue gases,not $NO_{2}$. This statement is incorrect.
$C$. $NO$ is not easily removed by alkaline scrubbing because it is neutral. $NO_{2}$ can be removed by alkaline scrubbing. This statement is incorrect.
$D$. The reaction $CCl_{4} + HF \xrightarrow{SbF_{5}} CF_{4} + HCl$ is a standard fluorination reaction using $SbF_{5}$ as a catalyst. This statement is correct.
Therefore,statements $B$ and $C$ are incorrect.

(A) Let the mass of ethane $(C_2H_6)$ and hydrogen $(H_2)$ be $w \ g$ each.
Molar mass of $C_2H_6 = 30 \ g/mol$.
Molar mass of $H_2 = 2 \ g/mol$.
Number of moles of $C_2H_6$ $(n_{C_2H_6})$ = $\frac{w}{30}$.
Number of moles of $H_2$ $(n_{H_2})$ = $\frac{w}{2}$.
Total moles $(n_{total})$ = $n_{C_2H_6} + n_{H_2} = \frac{w}{30} + \frac{w}{2} = \frac{w + 15w}{30} = \frac{16w}{30}$.
Mole fraction of $H_2$ $(X_{H_2})$ = $\frac{n_{H_2}}{n_{total}} = \frac{w/2}{16w/30} = \frac{w}{2} \times \frac{30}{16w} = \frac{15}{16}$.
According to Dalton's law of partial pressure,the fraction of total pressure exerted by a gas is equal to its mole fraction.
Therefore,the fraction of total pressure exerted by hydrogen is $\frac{15}{16}$ or $15:16$.

(D) To find the number of atoms,we use the formula: $\text{Number of atoms} = \frac{\text{mass}}{\text{molar mass}} \times \text{atomicity} \times N_{A}$.
For $A$: $1 \ g$ of $Ag$ (atomic mass = $108 \ g/mol$): $\text{Atoms} = \frac{1}{108} \times 1 \times N_{A} \approx 0.0092 \ N_{A}$.
For $B$: $1 \ g$ of $Fe$ (atomic mass = $56 \ g/mol$): $\text{Atoms} = \frac{1}{56} \times 1 \times N_{A} \approx 0.0178 \ N_{A}$.
For $C$: $1 \ g$ of $Cl_{2}$ (molar mass = $71 \ g/mol$): $\text{Atoms} = \frac{1}{71} \times 2 \times N_{A} \approx 0.0281 \ N_{A}$.
For $D$: $1 \ g$ of $Mg$ (atomic mass = $24 \ g/mol$): $\text{Atoms} = \frac{1}{24} \times 1 \times N_{A} \approx 0.0416 \ N_{A}$.
Comparing the values,$1 \ g$ of $Mg$ has the largest number of atoms.

(B) The mass of one atom of ${}^{12}C$ is defined as exactly $12 \ \text{a.m.u.}$
Since $1 \ \text{a.m.u.} = 1.66056 \times 10^{-24} \ \text{g}$,
The mass of one atom of ${}^{12}C = 12 \times 1.66056 \times 10^{-24} \ \text{g}$
$= 1.99267 \times 10^{-23} \ \text{g} \approx 1.9923 \times 10^{-23} \ \text{g}$.

(B) The balanced chemical equation for the reaction is:
$Pb(NO_{3})_{2}(aq) + 2KI(aq) \rightarrow PbI_{2}(s) + 2KNO_{3}(aq)$
Calculate the initial millimoles of each reactant:
Millimoles of $Pb(NO_{3})_{2} = 5 \ mL \times 0.1 \ M = 0.5 \ mmol$
Millimoles of $KI = 10 \ mL \times 0.02 \ M = 0.2 \ mmol$
According to the stoichiometry,$1 \ mol$ of $Pb(NO_{3})_{2}$ reacts with $2 \ mol$ of $KI$. Therefore,$0.2 \ mmol$ of $KI$ will react with $0.1 \ mmol$ of $Pb(NO_{3})_{2}$.
Since $KI$ is the limiting reagent,the amount of $PbI_{2}$ formed depends on the amount of $KI$.
From the stoichiometry,$2 \ mmol$ of $KI$ produces $1 \ mmol$ of $PbI_{2}$.
So,$0.2 \ mmol$ of $KI$ will produce $0.1 \ mmol$ of $PbI_{2}$.
Convert millimoles to moles:
$0.1 \ mmol = 0.1 \times 10^{-3} \ mol = 10^{-4} \ mol$.

(D) The ideal gas equation is given by $PV = nRT = \frac{w}{M} RT$.
Rearranging for density $(d = \frac{w}{V})$,we get $d = \frac{PM}{RT}$.
Thus,the molar mass $M = \frac{dRT}{P}$.
Given: $d = 1.964 \ g \ L^{-1}$,$T = 273 \ K$,$P = 76 \ cm \ Hg = 1 \ atm$,and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $M = \frac{1.964 \times 0.0821 \times 273}{1} \approx 44 \ g \ mol^{-1}$.
The molar mass of $CO_2$ is $12 + 2 \times 16 = 44 \ g \ mol^{-1}$.
Therefore,the gas is $CO_2$.

(A) The velocity of an electron in the $n^{th}$ Bohr orbit of a hydrogen-like atom is given by the formula: $v_n = 2.188 \times 10^8 \times \frac{Z}{n} \ cm \ s^{-1}$.
For the first Bohr orbit of a hydrogen atom,$Z = 1$ and $n = 1$.
Substituting these values: $v_1 = 2.188 \times 10^8 \times \frac{1}{1} \ cm \ s^{-1} = 2.188 \times 10^8 \ cm \ s^{-1}$.

(A) The orbital angular momentum is given by the formula: $\text{Orbital angular momentum} = \sqrt{\ell(\ell+1)} \frac{h}{2 \pi}$.
For a $4f$ orbital,the azimuthal quantum number $\ell = 3$. Therefore,the orbital angular momentum is $\sqrt{3(3+1)} \frac{h}{2 \pi} = \sqrt{12} \frac{h}{2 \pi} = 2 \sqrt{3} \frac{h}{2 \pi}$.
For a $4s$ orbital,the azimuthal quantum number $\ell = 0$. Therefore,the orbital angular momentum is $\sqrt{0(0+1)} \frac{h}{2 \pi} = 0$.
The difference between the two is $2 \sqrt{3} \frac{h}{2 \pi} - 0 = 2 \sqrt{3} \frac{h}{2 \pi}$.

(D) The quantum numbers $n=3$ and $\ell=2$ correspond to the $3d$ subshell.
Since the last electron is filled in the $3d$ subshell,we look for the maximum number of electrons possible when the $3d$ subshell is completely filled.
The electronic configuration for a completely filled $3d$ subshell is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10}$.
Counting the total number of electrons: $2+2+6+2+6+2+10 = 30$.
This corresponds to the element Zinc $(Zn)$ with atomic number $30$.

(C) For an adiabatic process,the heat exchange $dq = 0$.
Since the gas expands against a vacuum,the external pressure $P_{ext} = 0$,so the work done $dw = -P_{ext} \times dV = 0$.
According to the first law of thermodynamics,$dU = dq + dw$.
Substituting the values,$dU = 0 + 0 = 0$.
For an ideal gas,internal energy $U$ is a function of temperature only $(U = f(T))$,so if $\Delta U = 0$,then $\Delta T = 0$.
Therefore,the correct statement is $\Delta U = 0$.

(C) The reaction is a Cannizzaro reaction. In the mechanism,the nucleophilic attack of $OH^-$ on the carbonyl carbon of $Ph-CDO$ forms a tetrahedral intermediate. In the rate-determining step,a hydride (or deuteride) ion is transferred from this intermediate to another molecule of $Ph-CDO$. Since the intermediate contains a $D$ atom attached to the carbon,a deuteride ion $(D^-)$ is transferred to the second $Ph-CDO$ molecule. This results in the formation of $Ph-CD_2-O^-$ and $Ph-COO^-$. Upon workup (or proton exchange with the solvent),the alkoxide $Ph-CD_2-O^-$ becomes $Ph-CD_2-OH$.

(D) The reaction of $1$-bromo-$4$-chlorobenzene with $Mg$ in diethyl ether forms a di-Grignard reagent,$ClMg-C_6H_4-MgBr$.
When this di-Grignard reagent reacts with $2$ equivalents of formaldehyde $(CH_2O)$,it undergoes nucleophilic addition at both the $ClMg-$ and $-MgBr$ sites.
Subsequent acidic hydrolysis $(H_3O^+)$ converts both $-CH_2OMgX$ groups into $-CH_2OH$ groups.
Thus,the final product is $1,4$-benzenedimethanol (also known as $p$-xylylene glycol).

(C) $NaBH_4$ is a selective reducing agent that reduces ketones and aldehydes to alcohols but does not reduce esters under mild conditions.
Ethyl $3$-oxobutanoate contains both a ketone group and an ester group.
Therefore,$NaBH_4$ will selectively reduce the ketone group at the $C-3$ position to a hydroxyl group,while the ester group at the $C-1$ position remains unaffected.
The reaction is:
$CH_3COCH_2COOC_2H_5 \xrightarrow{NaBH_4, CH_3OH} CH_3CH(OH)CH_2COOC_2H_5$
Thus,the product is ethyl $3$-hydroxybutanoate.

(A) The reaction is a $Perkin$ condensation. It involves the reaction of an aromatic aldehyde with an acid anhydride in the presence of the corresponding carboxylate salt.
In this case,$p$-nitrobenzaldehyde reacts with propanoic anhydride $(CH_3CH_2CO)_2O$ in the presence of sodium propionate $(CH_3CH_2COONa)$.
The $\alpha$-hydrogen of the propanoic anhydride is abstracted by the propionate ion to form a nucleophilic enolate.
This enolate attacks the carbonyl carbon of $p$-nitrobenzaldehyde.
Subsequent steps involve the formation of a $\beta$-hydroxy acid derivative,followed by dehydration to form an $\alpha,\beta$-unsaturated acid.
The product formed is $2$-methyl-$3$-($4$-nitrophenyl)prop$-2-$enoic acid. The major product is the $(E)$-isomer due to steric considerations,which corresponds to option $A$.

(B) The compounds are:
$I$: $p$-Nitrophenol
$II$: $p$-Cresol ($4$-Methylphenol)
$III$: $m$-Nitrophenol
$IV$: $p$-Methoxybenzoic acid
Step $1$: Compare the acidity of phenols and carboxylic acids. Carboxylic acids are significantly more acidic than phenols. Thus,$IV$ is the most acidic.
Step $2$: Compare the acidity of the phenols $(I, II, III)$.
- $II$ ($p$-Cresol) has a $-CH_3$ group,which is an electron-donating group ($+I$ and hyperconjugation),decreasing acidity. Thus,$II$ is the least acidic.
- $I$ ($p$-Nitrophenol) has a $-NO_2$ group at the para position,which exerts a strong electron-withdrawing effect ($-I$ and $-M$).
- $III$ ($m$-Nitrophenol) has a $-NO_2$ group at the meta position,which exerts only an electron-withdrawing inductive effect $(-I)$.
- The $-M$ effect of the $-NO_2$ group in $p$-Nitrophenol $(I)$ makes it more acidic than $m$-Nitrophenol $(III)$.
Step $3$: Combining these,the order of acidity is $II < III < I < IV$.

(A) Alkaline hydrolysis of esters is a nucleophilic acyl substitution reaction. The rate of this reaction depends on the electrophilicity of the carbonyl carbon.
Electron-withdrawing groups $(EWG)$ increase the electrophilicity of the carbonyl carbon,thereby increasing the rate of hydrolysis.
Electron-donating groups $(EDG)$ decrease the electrophilicity of the carbonyl carbon,thereby decreasing the rate of hydrolysis.
Analysis of substituents:
$(I)$ $-NO_2$ group: Strong $-R$ and $-I$ effect (Strong $EWG$).
$(II)$ $-OCH_3$ group: Strong $+R$ and weak $-I$ effect (Strong $EDG$).
$(III)$ $-CH_3$ group: $+I$ effect and hyperconjugation (Weak $EDG$).
Order of electrophilicity: $-NO_2 > -CH_3 > -OCH_3$.
Therefore,the order of rates of alkaline hydrolysis is $I > III > II$.

(B) The rate law is given by $r = k[A]^{\alpha}[B]^{\beta}$.
When the concentration of $B$ is doubled,the rate does not change,which implies that the reaction is of zero order with respect to $B$. Therefore,$\beta = 0$.
When the concentrations of both $A$ and $B$ are doubled,the rate increases by a factor of $4$:
$\frac{r_2}{r_1} = \frac{k[2A]^{\alpha}[2B]^{\beta}}{k[A]^{\alpha}[B]^{\beta}} = 4$
$2^{\alpha} \cdot 2^{\beta} = 4$
Since $\beta = 0$,we have $2^{\alpha} = 4$,which gives $\alpha = 2$.
The overall order of the reaction is $\alpha + \beta = 2 + 0 = 2$.
The unit of the rate constant for a second-order reaction is $L \ mol^{-1} \ s^{-1}$.

(D) The reaction $2 F_{2} + 2 H_{2}O \rightarrow 4 HF + O_{2}$ occurs because $F_{2}$ is a strong oxidizing agent.
$F$ is the most electronegative element in the periodic table,so the statement that $F$ is less electronegative than $O$ is false.
$F_{2}$ is a stronger oxidizing agent than $O_{2}$ because of its high electronegativity and the low bond dissociation energy of the $F-F$ bond compared to the $O=O$ bond.
The $H-F$ bond is stronger than the $H-O$ bond due to the high electronegativity of $F$.

(D) $1$. Identify the ligands: There are four $NH_3$ (ammine) ligands and two $Cl^-$ (chloro) ligands attached to the central $Cr$ atom.
$2$. Determine the geometry: The two $Cl^-$ ligands are adjacent to each other (at $90^\circ$ angle),which indicates a $cis$ configuration.
$3$. Apply $IUPAC$ nomenclature rules:
- List ligands alphabetically: $ammine$ comes before $chloro$.
- Use prefixes for the number of ligands: $tetra$ for four $NH_3$ and $di$ for two $Cl$.
- The name is $cis-tetraamminodichlorochromium(III)$ chloride.
$4$. Oxidation state of $Cr$: $x + 4(0) + 2(-1) = +1$ (since $Cl^-$ is outside the coordination sphere),so $x = +3$.

(C) The given reaction is a classic example of the Thermite process,which is used for the reduction of metal oxides using aluminum as a reducing agent.
In the Thermite process,aluminum reduces metal oxides like $Fe_{2}O_{3}$ to produce the metal and aluminum oxide $(Al_{2}O_{3})$.
The reaction is: $Fe_{2}O_{3(s)} + 2 Al_{(s)} \xrightarrow{\text{Heat}} Al_{2}O_{3(s)} + 2 Fe_{(s)}$.
Comparing this with the given equation $M_{2}O_{3} + 2 Al \rightarrow Al_{2}O_{3} + 2 M$,we can identify that $M$ is Iron $(Fe)$.

(B) The reaction involves the substitution of a chlorine atom in $CH_3-O-CH_2-Cl$ with a hydroxyl group.
In this substrate,the carbon atom attached to chlorine is adjacent to an oxygen atom with lone pairs.
If the $C-Cl$ bond breaks to form a carbocation,the intermediate $CH_3-O^+-CH_2$ is formed.
This carbocation is highly stabilized by resonance due to the donation of lone pair electrons from the oxygen atom $(CH_3-O^+=CH_2)$.
Due to this significant resonance stabilization,the reaction proceeds via the $S_{N}1$ mechanism.

(C) The sodium salt is sodium thiosulphate $(Na_2S_2O_3 \cdot 5H_2O)$,which is a colourless efflorescent salt.
When dilute acid is added to it,it undergoes disproportionation to form sulphur (white precipitate) and sulphur dioxide gas $(SO_2)$:
$Na_2S_2O_{3(s)} + 2H^+_{(aq)} \rightarrow S_{(s)} + SO_{2(g)} + H_2O_{(\ell)} + 2Na^+_{(aq)}$
$SO_2$ gas turns acidified potassium dichromate solution green due to the reduction of $Cr(VI)$ to $Cr(III)$:
$K_2Cr_2O_{7(\text{aq})} + 3SO_{2(\text{g})} + H_2SO_{4(\text{aq})} \rightarrow K_2SO_{4(\text{aq})} + Cr_2(SO_4)_{3(\text{aq})} + H_2O_{(\ell)}$
Thus,the correct option is $C$.

(A) The melting point of pure $CaCl_{2}$ is very high $(1045 \ K)$.
Adding $CaF_{2}$ to the electrolyte mixture lowers the melting point of the electrolyte.
This allows the electrolysis process to be carried out at a lower temperature,which saves energy and prevents the evaporation of the electrolyte.

(B) In a face-centred cubic $(FCC)$ lattice,the atoms touch each other along the face diagonal.
The length of the face diagonal is $a \sqrt{2}$.
Since the face diagonal consists of two radii of the corner atoms and one full diameter of the face-centred atom,the distance between the centres of two nearest atoms (the closest distance) is half of the face diagonal.
Therefore,the closest distance $d = \frac{a \sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.

(C) The depression in freezing point is given by $\Delta T_{f} = K_{f} \times m$.
Here,$10 \%$ mass of ethylene glycol ($C_2H_6O_2$,molar mass $= 62 \ g \ mol^{-1}$) means $10 \ g$ of solute in $90 \ g$ of water.
Molality $(m)$ $= \frac{\text{mass of solute}}{\text{molar mass of solute} \times \text{mass of solvent in kg}} = \frac{10}{62} \times \frac{1000}{90} \approx 1.79 \ mol \ kg^{-1}$.
$\Delta T_{f} = 1.86 \times 1.79 \approx 3.33^{\circ} C$.
Since the freezing point of pure water is $0^{\circ} C$,the temperature at which ice begins to separate is $0 - 3.33 = -3.33^{\circ} C$,which is approximately $-3.3^{\circ} C$.

(B) Given: Mole fraction of ethanol $(x_{EtOH})$ = $0.08$.
Since the sum of mole fractions is $1$,the mole fraction of water $(x_{H_2O})$ = $1 - 0.08 = 0.92$.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
The formula relating molality to mole fraction is:
$m = \frac{x_{solute}}{x_{solvent}} \times \frac{1000}{MW_{solvent}}$
Substituting the values:
$m = \frac{0.08}{0.92} \times \frac{1000}{18.02} \approx 4.83 \ mol \ kg^{-1}$.

(A) The electronic configuration of uranium $({ }_{92} U)$ is $[Rn] \ 5f^{3} \ 6d^{1} \ 7s^{2}$.
In the $5f$ subshell,there are $3$ unpaired electrons,and in the $6d$ subshell,there is $1$ unpaired electron.
The $7s$ subshell is fully filled.
Therefore,the total number of unpaired electrons is $3 + 1 = 4$.

(A) According to the $Hardy-Schulze$ law,the coagulating power of an ion depends on the magnitude of the charge carried by the ion.
$Fe(OH)_{3}$ sol is a positively charged sol.
Therefore,the ion with the highest negative charge will be most effective for its flocculation.
Comparing the charges of the given ions:
$PO_{4}^{3-}$ has a charge of $-3$.
$SO_{4}^{2-}$ has a charge of $-2$.
$SO_{3}^{2-}$ has a charge of $-2$.
$NO_{3}^{-}$ has a charge of $-1$.
Since $PO_{4}^{3-}$ carries the highest negative charge,it is the most effective ion for the flocculation of $Fe(OH)_{3}$ sol.

(A, B, D) For a process to be spontaneous,the Gibbs free energy change,$\Delta G$,must be negative.
Polymerization involves the association of monomer units into a long chain,which leads to a decrease in the randomness of the system,meaning $\Delta S$ is negative.
According to the Gibbs-Helmholtz equation,$\Delta G = \Delta H - T\Delta S$.
Since $\Delta G < 0$ and $\Delta S < 0$,for the process to be spontaneous,$\Delta H$ must be negative and its magnitude must be greater than $T\Delta S$.