WBJEE 2018 Chemistry Question Paper with Answer and Solution

(B) Acid rain is primarily caused by the presence of oxides of sulphur and nitrogen in the atmosphere.
Among these,the oxides of sulphur ($SO_2$ and $SO_3$) are the major contributors,accounting for approximately $60-70\%$ of the acidity.
These oxides react with atmospheric water to form sulphuric acid $(H_2SO_4)$,which is the most abundant acid in acid rain.

(D) $NBS$ $(N-bromosuccinimide)$ performs allylic bromination via a free radical mechanism.
For $but-1-ene$ $(CH_3-CH_2-CH=CH_2)$,the allylic hydrogen is at the $C-3$ position.
Abstraction of this hydrogen generates an allylic radical: $CH_3-\dot{C}H-CH=CH_2 \leftrightarrow CH_3-CH=CH-\dot{C}H_2$.
Reaction of this resonance-stabilized radical with $Br \cdot$ leads to two constitutional isomers:
$1$. $3-bromobut-1-ene$ $(CH_3-CH(Br)-CH=CH_2)$,which contains a chiral center and thus exists as a pair of enantiomers ($d$ and $l$).
$2$. $1-bromobut-2-ene$ $(CH_3-CH=CH-CH_2Br)$,which exhibits geometrical isomerism ($cis$ and $trans$).
Therefore,the total number of possible organobromine compounds is $2$ (enantiomers) $2$ (geometrical isomers) = $4$.

(A) The given reaction sequence is:
$1$. $[P]$ is ethene $(CH_{2}=CH_{2})$. Reaction with $Br_{2}$ gives $1,2-dibromoethane$ $(CH_{2}Br-CH_{2}Br)$.
$2$. $1,2-dibromoethane$ reacts with $NaNH_{2}/NH_{3}$ (dehydrohalogenation) to form ethyne $(HC \equiv CH)$ as $[Q]$.
$3$. Ethyne $[Q]$ reacts with $20 \% H_{2}SO_{4}$ and $Hg^{2+}$ (Kucherov reaction) to form ethanal $(CH_{3}CHO)$ as $[R]$.
$4$. Ethanal $[R]$ undergoes Clemmensen reduction $(Zn-Hg/HCl)$ to form ethane $(CH_{3}CH_{3})$ as $[S]$.
Thus,$[P]$ = ethene,$[Q]$ = ethyne,$[R]$ = ethanal,$[S]$ = ethane.

(B, C) The correct statements for the peroxide ion $(O_{2}^{2-})$ are that it is diamagnetic and it has a bond order of $1$.
Electronic configuration of $O_{2}^{2-}$ is: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, (\pi 2p_{x}^{2} = \pi 2p_{y}^{2}), (\pi^{*} 2p_{x}^{2} = \pi^{*} 2p_{y}^{2})$.
Bond order $= \frac{1}{2} (\text{Number of bonding electrons} - \text{Number of anti-bonding electrons})$.
$BO = \frac{10 - 8}{2} = \frac{2}{2} = 1$.
Since the number of unpaired electrons is $0$,$O_{2}^{2-}$ is diamagnetic in nature.

(C) The strength of a hydrogen bond depends on the electronegativity of the atoms involved in the bond. The greater the electronegativity difference between the hydrogen atom and the atom it is bonded to,the stronger the hydrogen bond will be.
Fluorine $(F)$ is the most electronegative element in the periodic table.
Therefore,the $F-H...F$ bond is the strongest among the given options because fluorine has the highest electronegativity compared to oxygen $(O)$ and sulphur $(S)$.
The electronegativity order is $F > O > S$.

(B) The target reaction is the oxidation of $2 \text{ mol}$ of $NH_{3}$ to $NO$:
$2 NH_{3} + \frac{5}{2} O_{2} \rightleftharpoons 2 NO + 3 H_{2} O$
Given equations:
$(i) N_{2} + 3 H_{2} \rightleftharpoons 2 NH_{3} ; K_{1}$
$(ii) N_{2} + O_{2} \rightleftharpoons 2 NO ; K_{2}$
$(iii) H_{2} + \frac{1}{2} O_{2} \rightleftharpoons H_{2} O ; K_{3}$
To obtain the target reaction:
$1$. Reverse equation $(i)$: $2 NH_{3} \rightleftharpoons N_{2} + 3 H_{2} ; K' = \frac{1}{K_{1}}$
$2$. Keep equation $(ii)$ as it is: $N_{2} + O_{2} \rightleftharpoons 2 NO ; K_{2}$
$3$. Multiply equation $(iii)$ by $3$: $3 H_{2} + \frac{3}{2} O_{2} \rightleftharpoons 3 H_{2} O ; K'' = K_{3}^{3}$
Adding these three equations:
$(2 NH_{3} + N_{2} + O_{2} + 3 H_{2} + \frac{3}{2} O_{2}) \rightleftharpoons (N_{2} + 3 H_{2} + 2 NO + 3 H_{2} O)$
Simplifying gives: $2 NH_{3} + \frac{5}{2} O_{2} \rightleftharpoons 2 NO + 3 H_{2} O$
The equilibrium constant $K$ is:
$K = K' \cdot K_{2} \cdot K'' = \frac{1}{K_{1}} \cdot K_{2} \cdot K_{3}^{3} = K_{2} \cdot \frac{K_{3}^{3}}{K_{1}}$

(B) $H_{2}SO_{4}$ (sulphuric acid) is present in the maximum amount in 'acid rain'.
Oxides of nitrogen and sulphur,released into the atmosphere from industries,automobiles,and thermal power plants,are the main sources of acid rain.
These oxides,upon oxidation followed by hydrolysis,produce sulphuric acid and nitric acid,which,along with $HCl$,are responsible for the acidity of rain.
The oxidation reaction is catalysed by particulate matter present in the polluted atmosphere.
$2SO_{2(g)} + O_{2(g)} + 2H_{2}O_{(l)} \longrightarrow 2H_{2}SO_{4(aq)}$
$4NO_{2(g)} + O_{2(g)} + 2H_{2}O_{(l)} \longrightarrow 4HNO_{3(aq)}$
Typically,$H_{2}SO_{4}$ contributes about $70\%$ of the total acidity in acid rain.

(B) The reaction described is hydroboration-oxidation,which follows anti-Markovnikov addition of water across the double bond.
To obtain $2-butanol$ $(CH_{3}CH(OH)CH_{2}CH_{3})$,the starting alkene must be $1-butene$ $(CH_{3}CH_{2}CH=CH_{2})$ or $2-butene$ $(CH_{3}CH=CHCH_{3})$.
For $1-butene$: The anti-Markovnikov addition of $H_{2}O$ yields $1-butanol$.
For $2-butene$ ($cis$ or $trans$): The anti-Markovnikov addition of $H_{2}O$ yields $2-butanol$ because both carbons of the double bond are equivalent.
Thus,only $2-butene$ (both $cis$ and $trans$ isomers) produces $2-butanol$ as the product.
Therefore,the number of such alkenes is $2$.

(A) The reaction of an alkene with $Br_2$ proceeds via the formation of a cyclic bromonium ion intermediate. The $Br^-$ ion then attacks the bromonium ion from the side opposite to the bridged bromine atom,resulting in anti-addition. For the given alkene,$CH_3-CH=CH-C_2H_5$ (pent$-2-$ene),the anti-addition of $Br_2$ leads to the formation of a pair of enantiomers as the major products. These are represented by the structures in options $A$ and $D$ (or equivalent representations of the enantiomeric pair).

(D) $CH_{3}C \equiv CMgBr$ is a Grignard reagent derived from a terminal alkyne.
It can be prepared by the acid-base reaction between propyne $(CH_{3}C \equiv CH)$ and methylmagnesium bromide $(CH_{3}MgBr)$.
The terminal hydrogen of the alkyne is acidic and reacts with the basic Grignard reagent to form the alkynyl magnesium bromide and methane gas.
The chemical equation is:
$CH_{3}C \equiv CH + CH_{3}MgBr \longrightarrow CH_{3}C \equiv CMgBr + CH_{4}$

(B) The correct set of oxides arranged in the order of basic,amphoteric,and acidic is $BaO, Al_{2}O_{3}, SO_{2}$.
$BaO$ is a metal oxide of an alkaline earth metal,which is strongly basic.
$Al_{2}O_{3}$ is a well-known amphoteric oxide,reacting with both acids and bases.
$SO_{2}$ is a non-metal oxide,which is acidic in nature.

(D) The main reason that $SiCl_{4}$ is easily hydrolysed compared to $CCl_{4}$ is that silicon $(Si)$ has vacant $3d$-orbitals,which allow it to extend its coordination number beyond four.
Water molecules can coordinate with the vacant $d$-orbitals of the $Si$ atom in $SiCl_{4}$,facilitating the hydrolysis process.
In contrast,carbon $(C)$ lacks $d$-orbitals and cannot expand its coordination number beyond four,making $CCl_{4}$ resistant to hydrolysis by water.

(D) The thermal stability of alkaline earth metal carbonates increases down the group as the electropositive character of the metal increases.
The order of thermal stability is: $BeCO_3 < MgCO_3 < CaCO_3 < SrCO_3 < BaCO_3$.
$BeCO_3$ is the least thermally stable because the small size of the $Be^{2+}$ ion leads to high polarization of the carbonate ion,making it unstable.
It is so unstable that it must be stored in an atmosphere of $CO_2$ to prevent decomposition: $BeCO_3 \rightarrow BeO + CO_2$.

(D) Let the weight of both ethane $(C_2H_6)$ and hydrogen $(H_2)$ be $W \ g$.
Number of moles of $C_2H_6$ $(n_{C_2H_6})$ = $\frac{W}{30}$.
Number of moles of $H_2$ $(n_{H_2})$ = $\frac{W}{2}$.
According to Dalton's law of partial pressures, the fraction of total pressure exerted by a gas is equal to its mole fraction $(\chi)$.
$\chi_{H_2} = \frac{n_{H_2}}{n_{H_2} + n_{C_2H_6}} = \frac{\frac{W}{2}}{\frac{W}{2} + \frac{W}{30}}$.
$\chi_{H_2} = \frac{\frac{W}{2}}{\frac{15W + W}{30}} = \frac{W}{2} \times \frac{30}{16W} = \frac{15}{16}$.
Thus, the fraction of total pressure exerted by hydrogen is $\frac{15}{16}$ or $15: 16$.

(D) Mass of one electron = $9.108 \times 10^{-31} \ kg$.
Number of electrons in $1 \ mole = 6.023 \times 10^{23}$.
Mass of $1 \ mole$ of electrons = $(9.108 \times 10^{-31}) \times (6.023 \times 10^{23}) \ kg = 9.108 \times 6.023 \times 10^{-8} \ kg$.
Number of moles that weigh $1 \ kg = \frac{1}{9.108 \times 6.023 \times 10^{-8}} = \frac{1}{9.108 \times 6.023} \times 10^8 \ moles$.

(B) According to the Dulong-Petit law,the approximate atomic mass of the metal is $\approx \frac{6.4}{\text{Specific heat}} = \frac{6.4}{0.16} = 40 \ g \ mol^{-1}$.
In the metal chloride,the percentage of chlorine is $65\%$,so the percentage of metal $M$ is $35\%$.
The equivalent weight of the metal is calculated as $\frac{\text{Mass of metal}}{\text{Mass of chlorine}} \times 35.5 = \frac{35}{65} \times 35.5 \approx 19.11$.
The valency $n$ is given by $\frac{\text{Approximate atomic mass}}{\text{Equivalent weight}} = \frac{40}{19.11} \approx 2.09 \approx 2$.
Thus,the formula of the metal chloride is $MCl_2$.

(C) For a given value of $n$,the azimuthal quantum number $l$ can have values from $0$ to $n-1$.
For a given value of $l$,the magnetic quantum number $m$ can have values from $-l$ to $+l$ including $0$.
In option $C$,$n=2$ and $l=0$.
For $l=0$,the only possible value for $m$ is $0$.
Therefore,$m=-1$ is not possible for $l=0$.

(B) The atomic number of $Ni$ is $28$.
Its electronic configuration is: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8$.
In the $3d$ subshell,there are $8$ electrons.
According to Hund's rule,these $8$ electrons are filled in the $5$ $d$-orbitals as follows:
First,$5$ electrons are filled singly,and then the remaining $3$ electrons are paired.
This results in $3$ paired orbitals and $2$ singly occupied (unpaired) orbitals.
Therefore,the number of unpaired electrons in $Ni$ is $2$.

(A, C, D) Extensive properties are those whose values depend on the quantity or size of matter present in the system.
$H$ (Enthalpy),$E$ (Internal energy),and $V$ (Volume) are all extensive properties because they scale with the amount of substance.
$p$ (Pressure) is an intensive property as it is independent of the amount of matter.
Therefore,$H$,$E$,and $V$ are extensive variables.

(A) For a reversible adiabatic process,the relation between pressure $(P)$ and temperature $(T)$ is given by $P^{1-\gamma} T^\gamma = \text{constant}$.
This can be rearranged as $P \propto T^{\frac{\gamma}{\gamma-1}}$.
Given: $P \propto T^3$.
Therefore,$\frac{\gamma}{\gamma-1} = 3$.
$\gamma = 3\gamma - 3$.
$2\gamma = 3$.
$\gamma = \frac{3}{2}$.
Since $\gamma = \frac{C_p}{C_V}$,the ratio is $\frac{3}{2}$.

(A) The heat of neutralisation for the reaction between a strong acid and a strong base is defined as the heat released when $1 \ mole$ of $H^+$ ions reacts with $1 \ mole$ of $OH^-$ ions to form $1 \ mole$ of $H_2O$,which is $13.7 \ kcal$.
The balanced chemical equation is:
$HCl + NaOH \longrightarrow NaCl + H_2O; \Delta H = -13.7 \ kcal \text{ per mole of } H_2O \text{ formed}$.
Given amounts:
$HCl = 0.6 \ mole$
$NaOH = 0.25 \ mole$
Since $NaOH$ is the limiting reagent,the amount of $H_2O$ formed is determined by the amount of $NaOH$ present,which is $0.25 \ mole$.
Heat released $= \text{Heat of neutralisation} \times \text{moles of } H_2O \text{ formed}$
Heat released $= 13.7 \ kcal/mole \times 0.25 \ mole = 3.425 \ kcal$.

(C) Given $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log f(x) + c$.
Differentiating both sides with respect to $x$,we get:
$f(x) \sin x \cos x = \frac{1}{2(b^2 - a^2)} \cdot \frac{f'(x)}{f(x)}$.
Using $\sin x \cos x = \frac{\sin 2x}{2}$,we have:
$f(x) \cdot \frac{\sin 2x}{2} = \frac{1}{2(b^2 - a^2)} \cdot \frac{f'(x)}{f(x)}$.
Multiplying by $2$,we get:
$f(x) \sin 2x = \frac{1}{b^2 - a^2} \cdot \frac{f'(x)}{f(x)}$.
Rearranging the terms:
$(b^2 - a^2) \sin 2x \, dx = \frac{f'(x)}{(f(x))^2} \, dx$.
Integrating both sides:
$(b^2 - a^2) \int \sin 2x \, dx = \int (f(x))^{-2} f'(x) \, dx$.
$(b^2 - a^2) \left( -\frac{\cos 2x}{2} \right) = -\frac{1}{f(x)} + C_1$.
Assuming the constant of integration $C_1 = 0$ for the functional form:
$\frac{(b^2 - a^2) \cos 2x}{2} = \frac{1}{f(x)}$.
Therefore,$f(x) = \frac{2}{(b^2 - a^2) \cos 2x}$.

(B) The given sequence of chemical reactions is:
$C_{4}H_{10}O$ $\xrightarrow{K_{2}Cr_{2}O_{7} / H_{2}SO_{4}} C_{4}H_{8}O$ $\xrightarrow{I_{2} / NaOH, \text{Warm}} CHI_{3}$
$(N)$
Step $1$: The compound $N$ is oxidized by $K_{2}Cr_{2}O_{7} / H_{2}SO_{4}$ to a ketone $(C_{4}H_{8}O)$. This indicates that $N$ is a secondary alcohol.
Step $2$: The ketone formed undergoes the iodoform reaction with $I_{2} / NaOH$ to produce iodoform $(CHI_{3})$. This confirms that the ketone must have a methyl group attached to the carbonyl carbon (i.e.,it is a methyl ketone).
Butan-$2$-ol $(CH_{3}CH(OH)CH_{2}CH_{3})$ is a secondary alcohol. Upon oxidation,it yields butan-$2$-one $(CH_{3}COCH_{2}CH_{3})$,which is a methyl ketone and gives a positive iodoform test.
Therefore,$N$ is butan-$2$-ol.

(B) Methoxybenzene (anisole) on treatment with $HI$ produces phenol and methyl iodide.
This reaction occurs because the $C_{aryl}-O$ bond is stronger than the $C_{alkyl}-O$ bond.
The carbon of the phenyl group is $sp^{2}$-hybridised,which gives the $C_{aryl}-O$ bond a partial double bond character,making it resistant to cleavage.
Consequently,the nucleophilic iodide ion attacks the less sterically hindered methyl group,resulting in the formation of phenol and methyl iodide $(CH_3I)$.

(A) The correct order of reactivity for the nucleophilic addition reaction of the given carbonyl compounds with ethylmagnesium iodide is $I > III > II > IV$.
This order is determined by two primary factors:
$(i)$ Inductive effect: Alkyl groups are electron-releasing. As the number of alkyl groups attached to the carbonyl carbon increases,the electron density on the carbonyl carbon increases,which reduces its electrophilicity and thus decreases reactivity towards nucleophilic attack.
$(ii)$ Steric effect: As the number and size of alkyl groups attached to the carbonyl carbon increase,the steric hindrance around the carbonyl carbon increases,making it more difficult for the nucleophile to attack.
Therefore,formaldehyde $(I)$ is the most reactive,followed by acetaldehyde $(III)$,acetone $(II)$,and finally di-tert-butyl ketone $(IV)$.

(D) When aniline is treated with conc. $H_{2}SO_{4}$,it first forms anilinium hydrogen sulphate.
Upon heating this salt at $200^{\circ}C$,it undergoes a rearrangement reaction (sulphonation) to form $p$-aminobenzenesulphonic acid,which is commonly known as sulphanilic acid.
The reaction proceeds through the formation of a zwitter ion.

(A AND C) The reaction of cyclobutyl amine with $HNO_{2}$ produces a diazonium salt,which is unstable and loses $N_{2}$ to form a cyclobutyl carbocation. This carbocation can undergo ring contraction to form a more stable cyclopropylmethyl carbocation. Both carbocations can then react with water to form their respective alcohols. Therefore,the products are cyclobutanol and cyclopropylmethanol. The reaction sequence is as follows:
$1. \text{Cyclobutylamine} + HNO_{2} \rightarrow \text{Cyclobutyl diazonium ion} + H_{2}O$
$2. \text{Cyclobutyl diazonium ion} \rightarrow \text{Cyclobutyl carbocation} + N_{2}$
$3. \text{Cyclobutyl carbocation} \xrightarrow{\text{Ring contraction}} \text{Cyclopropylmethyl carbocation}$
$4. \text{Cyclobutyl carbocation} + H_{2}O \rightarrow \text{Cyclobutanol} + H^{+}$
$5. \text{Cyclopropylmethyl carbocation} + H_{2}O \rightarrow \text{Cyclopropylmethanol} + H^{+}$

(A) The reactivity of carboxylic acid derivatives towards nucleophilic acyl substitution (such as hydrolysis) depends on the leaving group ability of the substituent attached to the acyl group.
The weaker the base,the better the leaving group.
The order of basicity of the leaving groups is $Cl^{-} < CH_3COO^{-} < C_2H_5O^{-} < NH_2^-$.
Therefore,the order of leaving group ability is $Cl^{-} > CH_3COO^{-} > C_2H_5O^{-} > NH_2^-$.
Thus,the order of ease of hydrolysis is: $CH_3COCl$ $(I) > (CH_3CO)_2O$ $(II) > CH_3COOC_2H_5$ $(III) > CH_3CONH_2$ $(IV)$.

(D) Given: Half-life of $C^{14}$,$t_{1/2} = 5760 \ years$.
Initial amount,$N_0 = 200 \ mg$.
Final amount,$N_t = 25 \ mg$.
Radioactive decay follows first-order kinetics.
We can use the relation: $N_t = N_0 \times (1/2)^n$,where $n$ is the number of half-lives.
$25 = 200 \times (1/2)^n$
$1/8 = (1/2)^n$
$(1/2)^3 = (1/2)^n$
Therefore,$n = 3$.
Total time $t = n \times t_{1/2} = 3 \times 5760 \ years = 17280 \ years$.
Alternatively,as shown in the diagram:
$200 \ mg$ $\xrightarrow{t_{1/2}} 100 \ mg$ $\xrightarrow{t_{1/2}} 50 \ mg$ $\xrightarrow{t_{1/2}} 25 \ mg$.
Total time = $3 \times 5760 \ years = 17280 \ years$.

(C) The process described is electron capture,where the nucleus $^{64}_{29}Cu$ captures an orbital electron $(-1e^0)$.
In this process,the atomic number decreases by $1$ while the mass number remains constant.
The nuclear equation is: $^{64}_{29}Cu + _{-1}e^0 \longrightarrow ^{64}_{28}Ni$.
Thus,the product is $^{64}_{28}Ni$.

(D) $AgCl$ dissolves in excess $NH_4OH$ to form a soluble complex,diamminesilver$(I)$ chloride.
The chemical reaction is:
$AgCl(s) + 2NH_4OH(aq) \rightarrow [Ag(NH_3)_2]Cl(aq) + 2H_2O(l)$
The complex dissociates in solution as:
$[Ag(NH_3)_2]Cl \rightarrow [Ag(NH_3)_2]^+ + Cl^-$
Thus,the cation present in the resulting solution is $[Ag(NH_3)_2]^+$.

(D) The ferric ion $(Fe^{3+})$ reacts with potassium ferrocyanide $(K_4[Fe(CN)_6])$ to form a Prussian blue precipitate of ferric ferrocyanide,which is $Fe_4[Fe(CN)_6]_3$.
The chemical reaction is: $4Fe^{3+} + 3[Fe(CN)_6]^{4-} \rightarrow Fe_4[Fe(CN)_6]_3$.

(C) The reduction reaction at the cathode is given by:
$Al^{3+} + 3e^{-} \longrightarrow Al$
From the stoichiometry of the reaction,$1 \ mole$ of $Al^{3+}$ ions requires $3 \ moles$ of electrons to be reduced to $1 \ mole$ of $Al$ metal.
Since $1 \ mole$ of electrons carries a charge of $1 \ Faraday$ $(F)$,the total electricity required is $3 \ F$.

(B) The given cell reaction is $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$.
Oxidation half-reaction: $Zn \rightarrow Zn^{2+} + 2e^{-}$,$E^{\circ}_{ox} = +0.76 \ V$.
Reduction half-reaction: $Fe^{2+} + 2e^{-} \rightarrow Fe$,$E^{\circ}_{red} = -0.41 \ V$ (since $E^{\circ}_{ox}$ for $Fe$ is $+0.41 \ V$,the reduction potential $E^{\circ}_{red} = -E^{\circ}_{ox}$).
The standard e.m.f. of the cell is:
$E^{\circ}_{cell} = E^{\circ}_{ox} + E^{\circ}_{red}$
$E^{\circ}_{cell} = 0.76 \ V + (-0.41 \ V) = +0.35 \ V$.

(A, C, D) White phosphorus $(P_{4})$ consists of a discrete tetrahedral molecule.
It contains $6$ $P-P$ single bonds and $4$ lone pairs of electrons (one on each phosphorus atom).
The $P-P-P$ bond angle in the tetrahedral structure is $60^{\circ}$.
Therefore,options $A$,$C$,and $D$ are correct characteristics of white phosphorus.

(D) $Cl_{2}O_{7}$ is the anhydride of $HClO_{4}$.
An acid anhydride is a compound that reacts with water to form an acid.
The reaction is:
$Cl_{2}O_{7} + H_{2}O \longrightarrow 2HClO_{4}$ (Perchloric acid)
$Cl_{2}O_{7}$ is the oxide of chlorine in its highest oxidation state $(+7)$,which corresponds to perchloric acid $(HClO_{4})$.

(C) Fluorine is a powerful oxidizing agent. It reacts vigorously with water at room temperature to produce oxygen gas and hydrofluoric acid $(HF)$. In an aqueous medium,$HF$ dissociates into $H^{+}$ and $F^{-}$ ions. The chemical equation for the reaction is:
$2F_2 + 2H_2O \rightarrow 4H^{+} + 4F^{-} + O_2$

(C) $Dacron$ is a condensation polymer. It is also known as $Terylene$.
It is a polymer obtained by the condensation reaction between $Ethylene \ glycol$ and $Terephthalic \ acid$ at $420-460 \ K$ in the presence of a zinc acetate-antimony trioxide catalyst.
The reaction is as follows:
$nHOCH_2CH_2OH + nHOOC-C_6H_4-COOH \xrightarrow{420-460 \ K} [-O-CH_2CH_2-O-CO-C_6H_4-CO-]_n + 2nH_2O$

(A) According to the question,$[X] +$ dil. $H_{2}SO_{4} \longrightarrow [Y]$ (a colourless,suffocating gas).
$[Y] + K_{2}Cr_{2}O_{7} + H_{2}SO_{4} \longrightarrow$ Green colouration of the solution.
Sulphites $(SO_{3}^{2-})$ react with dil. $H_{2}SO_{4}$ to release sulphur dioxide $(SO_{2})$ gas,which is colourless and has a suffocating odour.
The reaction is: $SO_{3}^{2-} + 2H^{+} \longrightarrow SO_{2} + H_{2}O$.
$SO_{2}$ acts as a reducing agent and reduces acidified potassium dichromate $(K_{2}Cr_{2}O_{7})$ from $Cr(VI)$ (orange) to $Cr(III)$ (green) as $Cr_{2}(SO_{4})_{3}$.
The reaction is: $K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + 3SO_{2} \longrightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + H_{2}O$.

(B) The highest oxidation state is achieved by $(n-1) d^5 n s^2$,which corresponds to a maximum oxidation state of $+7$. This is because the $(n-1) d$-electrons can participate in bonding along with $n s$-electrons,as their energy levels are comparable. The oxidation states for the given configurations are summarized below:
| Electronic configuration | Oxidation state |
| :--- | :--- |
| $(n-1) d^8 n s^2$ | $+2, +3, +4$ |
| $(n-1) d^3 n s^2$ | $+2, +3, +4, +5$ |
| $(n-1) d^5 n s^1$ | $+2, +3, +4, +5, +6$ |
| $(n-1) d^5 n s^2$ | $+2, +3, +4, +5, +6, +7$ |

(A) The number of $X$ atoms at the corners of the cube is $8$. Since each corner atom contributes $\frac{1}{8}$ to the unit cell,the total number of $X$ atoms per unit cell is $8 \times \frac{1}{8} = 1$.
The number of $Y$ atoms at the body centre is $1$. Since the body-centered atom is entirely within the unit cell,the total number of $Y$ atoms per unit cell is $1$.
Therefore,the ratio of $X:Y$ is $1:1$,and the formula of the compound is $XY$.