AP EAMCET 2019 Chemistry Question Paper with Answer and Solution

(D) The reaction of an alkyne with sodium in liquid ammonia $(Na/Liq. NH_3)$ is known as Birch reduction,which results in the stereoselective anti-addition of hydrogen to the triple bond to form a $trans$-alkene. The double bond present in the cyclohexene ring remains unaffected under these conditions. Therefore,the product is a $trans$-alkene. Thus,option $(D)$ is correct.

(D) Key Idea: Functional groups that are electron-withdrawing due to resonance (often containing a double bond to a more electronegative atom) are meta-directing. Groups that donate electrons via resonance (due to the presence of lone pairs on the atom directly attached to the benzene ring) are ortho/para-directing.
$(A)$ $-COOH$: The carbon atom attached to the benzene ring is bonded to an oxygen atom via a double bond. It is electron-withdrawing and thus meta-directing.
$(B)$ $-NO_2$: The nitrogen atom attached to the benzene ring is bonded to an oxygen atom via a double bond. It is electron-withdrawing and thus meta-directing.
$(C)$ $-CHO$: The carbon atom attached to the benzene ring is bonded to an oxygen atom via a double bond. It is electron-withdrawing and thus meta-directing.
$(D)$ $-OCH_3$: The oxygen atom attached to the benzene ring has lone pairs of electrons,which it donates to the ring via resonance ($+M$ effect). Therefore,it is ortho/para-directing,not meta-directing.
Thus,option $(D)$ is the correct answer.

(B) Ortho and para directing groups are electron-donating groups that increase electron density at the ortho and para positions of the benzene ring.
Groups like $-NHCOCH_3$ $(II)$ and $-OCH_3$ $(III)$ have lone pairs on the atom directly attached to the benzene ring,exhibiting a $+M$ effect. Therefore,they are ortho/para directing.
In contrast,$-CHO$ $(I)$ and $-SO_3H$ $(IV)$ are electron-withdrawing groups and are meta-directing.
Thus,the correct option is $(b)$.

(C) The reactivity of benzene derivatives towards electrophilic substitution depends on the nature of the substituent attached to the ring.
Electron-donating groups (activating groups) increase the electron density in the ring,thereby increasing reactivity.
Electron-withdrawing groups (deactivating groups) decrease the electron density in the ring,thereby decreasing reactivity.
- The $-OH$ group exerts a strong $+M$ (mesomeric) effect,which strongly activates the ring.
- The $-CH_3$ group exerts $+I$ (inductive) and $+H$ (hyperconjugation) effects,which moderately activate the ring.
- Benzene $(A)$ has no substituent.
- The $-NO_2$ group exerts a strong $-M$ and $-I$ effect,which strongly deactivates the ring.
Thus,the order of reactivity is: Phenol $(C)$ > Toluene $(B)$ > Benzene $(A)$ > Nitrobenzene $(D)$.

(A) $(I)$ In a nitrating mixture $(HNO_3 + H_2SO_4)$,$HNO_3$ acts as a base because it accepts a proton from $H_2SO_4$ to form the nitronium ion $(NO_2^+)$. Thus,statement $I$ is incorrect.
$(II)$ The $\sigma$-complex (arenium ion) is the intermediate formed during the electrophilic aromatic substitution of benzene when an electrophile $(E^+)$ attacks the benzene ring. In this complex,one carbon atom becomes $sp^3$-hybridized. Thus,statement $II$ is correct.
$(III)$ During the Friedel-Crafts alkylation of benzene with $n$-propyl chloride,the primary carbocation $(CH_3CH_2CH_2^+)$ rearranges to a more stable secondary carbocation $(CH_3CH^+CH_3)$. Consequently,the major product formed is isopropyl benzene. Thus,statement $III$ is correct.
Therefore,statements $II$ and $III$ are correct,making option $(A)$ the correct answer.

(B) Benzene reacts with a mixture of conc. $HNO_3$ and conc. $H_2SO_4$ at $333 \ K$ $(X)$ to undergo nitration,forming nitrobenzene.
Benzene reacts with fuming sulphuric acid $(H_2SO_4 + SO_3)$ $(Y)$ to undergo sulphonation,forming benzene sulphonic acid.

(C) Bond length depends upon the electrostatic force of attraction between the nuclei and the shared pair of electrons.
Since isotopes have the same number of protons and electrons,the electronic environment remains identical.
Therefore,the bond length is the same for both $H_2$ and $D_2$.

(D) non-metal displacement reaction involves a metal displacing a non-metal (usually hydrogen or oxygen) from its compound.
In reaction $(A)$,$Ca$ displaces $H$ from $H_2O$. This is a non-metal displacement reaction.
In reaction $(B)$,$Ca$ displaces $V$ from $V_2O_5$. This is a metal displacement reaction.
In reaction $(C)$,$Fe$ displaces $H$ from $H_2O$. This is a non-metal displacement reaction.
In reaction $(D)$,$Al$ displaces $Cr$ from $Cr_2O_3$. This is a metal displacement reaction.
Therefore,$(A)$ and $(C)$ are non-metal displacement reactions.

(C) $(I)$ $B_2H_6$ (diborane) is an electron deficient hydride. This is due to the presence of $3C-2e^-$ bridge bonds.
$(II)$ $NH_3$ is an electron rich hydride due to the presence of a lone pair of electrons on the nitrogen atom.
$(III)$ $NaH$ is an ionic (saline) hydride; hydrides of alkali metals are ionic in nature. Thus,statement $III$ is incorrect.
$(IV)$ $YbH_{2.55}$ is an interstitial (non-stoichiometric) hydride. Here,hydrogen atoms occupy the interstitial spaces in the metal lattice. Thus,statement $IV$ is correct.
Therefore,statements $I$,$II$,and $IV$ are correct. The correct option is $(C)$.

(A) For vessel $A$:
$10 \ vol \ H_2O_2$ means $1 \ mL$ of $H_2O_2$ solution gives $10 \ mL \ O_2$ at $NTP$.
$2 \ H_2O_2 \longrightarrow 2 \ H_2O + O_2$.
$2 \ mol \ (68 \ g)$ of $H_2O_2$ gives $22400 \ mL \ O_2$ at $NTP$.
$O_2$ evolved from $500 \ mL$ of $10 \ vol \ H_2O_2 = 500 \times 10 = 5000 \ mL$.
Weight of $H_2O_2$ in $A = \frac{68 \times 5000}{22400} \approx 15.18 \ g$.
For vessel $B$:
$O_2$ evolved from $100 \ mL$ of $30 \ vol \ H_2O_2 = 100 \times 30 = 3000 \ mL$.
Weight of $H_2O_2$ in $B = \frac{68 \times 3000}{22400} \approx 9.11 \ g$.
For vessel $C$:
$Molarity = 2 \ M$,$Volume = 250 \ mL = 0.25 \ L$.
$Moles \ of \ H_2O_2 = Molarity \times Volume = 2 \times 0.25 = 0.5 \ mol$.
Weight of $H_2O_2$ in $C = 0.5 \ mol \times 34 \ g/mol = 17 \ g$.
Comparing the weights: $C (17 \ g) > A (15.18 \ g) > B (9.11 \ g)$.
Thus,the order is $C > A > B$.

(D) Bronsted acid is a proton $(H^{+})$ donor,and a Bronsted base is a proton $(H^{+})$ acceptor. Amphoteric species can act as both.
$1$. $HCl$: Only acts as an acid $(HCl \rightarrow H^{+} + Cl^{-})$.
$2$. $ClO_4^{-}$: Only acts as a base $(ClO_4^{-} + H^{+} \rightarrow HClO_4)$.
$3$. $OH^{-}$: Only acts as a base $(OH^{-} + H^{+} \rightarrow H_2O)$.
$4$. $H^{+}$: Only acts as an acid.
$5$. $H_2O$: Acts as an acid $(H_2O \rightarrow H^{+} + OH^{-})$ and as a base $(H_2O + H^{+} \rightarrow H_3O^{+})$.
$6$. $HSO_4^{-}$: Acts as an acid $(HSO_4^{-} \rightarrow H^{+} + SO_4^{2-})$ and as a base $(HSO_4^{-} + H^{+} \rightarrow H_2SO_4)$.
$7$. $SO_4^{2-}$: Only acts as a base $(SO_4^{2-} + H^{+} \rightarrow HSO_4^{-})$.
$8$. $H_2SO_4$: Only acts as an acid $(H_2SO_4 \rightarrow H^{+} + HSO_4^{-})$.
$9$. $Cl^{-}$: Only acts as a base $(Cl^{-} + H^{+} \rightarrow HCl)$.
Thus,only $H_2O$ and $HSO_4^{-}$ can act as both Bronsted acids and bases. The total number is $2$.

(D) The reaction between $NaOH$ and acetic acid $(CH_3COOH)$ is: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$.
Initial moles of $CH_3COOH = 0.06 \ M \times 0.050 \ L = 0.003 \ mol$.
Initial moles of $NaOH = 0.02 \ M \times 0.050 \ L = 0.001 \ mol$.
After the reaction,$0.001 \ mol$ of $CH_3COONa$ is formed and $0.002 \ mol$ of $CH_3COOH$ remains.
Since a buffer solution is formed,we use the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
$pH = 4.76 + \log \frac{0.001}{0.002} = 4.76 + \log (0.5)$.
$pH = 4.76 - 0.30 = 4.46$.

(B) For a weak base $NH_3$,the ionisation constant $K_b = 2.5 \times 10^{-5}$.
For the conjugate acid $NH_4^{+}$,the ionisation constant $K_a$ is given by $K_a \times K_b = K_w = 10^{-14}$.
$K_a = \frac{10^{-14}}{2.5 \times 10^{-5}} = 4 \times 10^{-10}$.
For $0.01 \ M$ $NH_3$ solution,$[OH^{-}] = \sqrt{K_b \times C} = \sqrt{2.5 \times 10^{-5} \times 0.01} = \sqrt{2.5 \times 10^{-7}} = \sqrt{25 \times 10^{-8}} = 5 \times 10^{-4} \ M$.
$pOH = -\log[OH^{-}] = -\log(5 \times 10^{-4}) = 4 - \log 5 = 4 - 0.7 = 3.3$.
$pH = 14 - pOH = 14 - 3.3 = 10.7$.
Thus,the $pH$ is $10.7$ and the ionisation constant of the conjugate acid is $4 \times 10^{-10}$.

(D) The number of millimoles of acetic acid $(CH_3COOH)$ $= 10 \ mL \times 1.0 \ M = 10 \ mmol$.
The number of millimoles of sodium acetate $(CH_3COONa)$ $= 20 \ mL \times 0.5 \ M = 10 \ mmol$.
According to the Henderson-Hasselbalch equation:
$pH = pK_a + \log \left( \frac{[salt]}{[acid]} \right)$.
Since both components are in the same total volume $(100 \ mL)$,the ratio of their concentrations is equal to the ratio of their millimoles:
$pH = 4.76 + \log \left( \frac{10 \ mmol}{10 \ mmol} \right)$.
$pH = 4.76 + \log(1)$.
Since $\log(1) = 0$,we get $pH = 4.76$.
Therefore,option $(D)$ is correct.

(A) For a basic buffer,the Henderson-Hasselbalch equation is given by: $pOH = pK_b + \log \frac{[\text{salt}]}{[\text{base}]}$.
Given $pH = 10.5$,we calculate $pOH$ as: $pOH = 14 - 10.5 = 3.5$.
Substituting the values into the equation: $3.5 = pK_b + \log \frac{0.01}{0.1}$.
$3.5 = pK_b + \log(10^{-1})$.
$3.5 = pK_b - 1$,which gives $pK_b = 4.5$.
Using the relation $pK_a + pK_b = 14$,we find the $pK_a$ of the conjugate acid: $pK_a = 14 - 4.5 = 9.5$.

(B) For the salt $A_2B$,the dissociation equilibrium is $A_2B \rightleftharpoons 2A^{+} + B^{2-}$.
Let the solubility be $s \ mol \cdot L^{-1}$.
The concentrations of the ions are $[A^{+}] = 2s$ and $[B^{2-}] = s$.
The solubility product expression is $K_{sp} = [A^{+}]^2 [B^{2-}] = (2s)^2(s) = 4s^3$.
Given $K_{sp} = 3.2 \times 10^{-11}$.
Equating the two: $4s^3 = 3.2 \times 10^{-11}$.
$s^3 = \frac{3.2 \times 10^{-11}}{4} = 0.8 \times 10^{-11} = 8 \times 10^{-12}$.
Taking the cube root: $s = \sqrt[3]{8 \times 10^{-12}} = 2 \times 10^{-4} \ mol \cdot L^{-1}$.

(C) Let $m$ be the mass of the body,$\theta$ be the angle of inclination,and $\mu$ be the coefficient of friction.
The force required to move the body up the inclined plane is $F_{up} = mg \sin \theta + \mu mg \cos \theta$.
The force required to prevent the body from sliding down the inclined plane is $F_{down} = mg \sin \theta - \mu mg \cos \theta$.
According to the problem,$F_{up} = 2 F_{down}$.
Substituting the expressions,we get:
$mg \sin \theta + \mu mg \cos \theta = 2(mg \sin \theta - \mu mg \cos \theta)$
Dividing both sides by $mg$:
$\sin \theta + \mu \cos \theta = 2 \sin \theta - 2 \mu \cos \theta$
Rearranging the terms to solve for $\mu$:
$3 \mu \cos \theta = \sin \theta$
$\mu = \frac{1}{3} \tan \theta$
Given $\theta = 60^{\circ}$,we have:
$\mu = \frac{1}{3} \tan 60^{\circ} = \frac{1}{3} \times \sqrt{3} = \frac{1}{\sqrt{3}}$.

(C) For upward motion,the force required is $F_{up} = mg(\sin \theta + \mu \cos \theta)$.
For downward motion,the force required to prevent sliding is $F_{down} = mg(\sin \theta - \mu \cos \theta)$.
According to the problem,$F_{up} = 2 F_{down}$.
Substituting the expressions,we get:
$mg(\sin \theta + \mu \cos \theta) = 2 mg(\sin \theta - \mu \cos \theta)$
$\sin \theta + \mu \cos \theta = 2 \sin \theta - 2 \mu \cos \theta$
$3 \mu \cos \theta = \sin \theta$
$\mu = \frac{1}{3} \tan \theta$
Given $\theta = 60^{\circ}$,we have $\tan 60^{\circ} = \sqrt{3}$.
Therefore,$\mu = \frac{1}{3} \times \sqrt{3} = \frac{1}{\sqrt{3}}$.

(C) The magnetic force on a moving charge is given by $F = qvB \sin \theta$.
For a transverse magnetic field,$\theta = 90^{\circ}$,so $F = qvB$.
The velocity $v$ of an electron accelerated through a potential difference $V$ is given by $\frac{1}{2}m_e v^2 = eV$,which implies $v = \sqrt{\frac{2eV}{m_e}}$.
Substituting this into the force equation: $F = e \left( \sqrt{\frac{2eV}{m_e}} \right) B$.
This shows that $F \propto \sqrt{V}$.
When the potential is increased to $2V$,the new force $F'$ is:
$F' = e \left( \sqrt{\frac{2e(2V)}{m_e}} \right) B = \sqrt{2} \cdot e \left( \sqrt{\frac{2eV}{m_e}} \right) B = \sqrt{2}F$.

(B) From the given chemical reactions:
$1$. The reaction of diborane with excess ammonia at high temperature produces inorganic benzene $(B_3N_3H_6)$,so $X = B_3N_3H_6$.
$2$. The reaction of sodium hydride with boron trifluoride produces diborane $(B_2H_6)$,so $Y = B_2H_6$.
$3$. The hydrolysis of diborane produces orthoboric acid $(H_3BO_3)$,so $Z = H_3BO_3$.
Therefore,the correct sequence is $X = B_3N_3H_6, Y = B_2H_6, Z = H_3BO_3$.
Hence,option $(b)$ is correct.

(A) Diborane $(B_2H_6)$ can be prepared by the following reactions:
$I. 4BF_3 + 3LiAlH_4 \xrightarrow{\text{ether}} 2B_2H_6 + 3LiF + 3AlF_3$
$II. 2BF_3 + 6NaH \xrightarrow{450 \ K} B_2H_6 + 6NaF$
$IV. 2NaBH_4 + I_2 \longrightarrow B_2H_6 + 2NaI + H_2$
Reaction $III$ $(Na_2B_4O_7 + H_2O)$ is a hydrolysis reaction of borax which does not produce diborane.
Thus,reactions $I, II,$ and $IV$ are used to prepare diborane.

(A) Sodium hydride reacts with diborane in the presence of diethyl ether to produce sodium tetrahydroborate$(III)$.
The balanced chemical equation is:
$2NaH + B_2H_6 \xrightarrow{\text{Diethyl ether}} 2Na[BH_4]$
Therefore,$X$ is $Na[BH_4]$.

(A) The explanations of the given statements are as follows:
$(i)$ $H_3BO_3$ (orthoboric acid) is a monobasic Lewis acid. In aqueous solution,it accepts an $OH^-$ ion from water to form $[B(OH)_4]^-$ and releases $H^+$,acting as a Lewis acid.
$B(OH)_3 + H_2O \rightleftharpoons [B(OH)_4]^- + H^+$
$(ii)$ The correct formula for borax is $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$. This statement is correct.
$(iii)$ $NaBH_4$ (sodium borohydride) is a well-known reducing agent used in organic chemistry to reduce aldehydes and ketones to alcohols. This statement is correct.
Since all three statements are correct,the correct option is $A$.

(B) $(I)$ $Ga_2O_3$ is an amphoteric oxide. This is a correct statement as it reacts with both acids and bases:
Reaction with an acid: $Ga_2O_3 + 6 HCl \longrightarrow 2 GaCl_3 + 3 H_2O$
Reaction with a base: $Ga_2O_3 + 2 NaOH \longrightarrow 2 NaGaO_2 + H_2O$
$(II)$ The dimer of aluminium chloride $(Al_2Cl_6)$ has only two $Al-Cl-Al$ bridge bonds,not three. The structure consists of two $AlCl_4$ tetrahedra sharing a common edge.
$(III)$ Boron is a very hard refractory solid with a high melting temperature $(2349 \ K)$,which is a correct statement.
Therefore,statements $I$ and $III$ are correct. The correct option is $(b)$.

(A) $(I)$ In diamond,each carbon atom is $sp^3$ hybridised. This statement is correct.
$(II)$ Graphite consists of planar hexagonal layers of carbon atoms where each carbon is $sp^2$ hybridised. This statement is correct.
$(III)$ Silicones are organosilicon polymers with the general formula $(R_2SiO)_n$. Due to the presence of non-polar alkyl groups,they are hydrophobic (water-repelling) in nature. This statement is correct.
$(IV)$ The tendency for catenation in group $14$ elements decreases down the group due to the decrease in bond dissociation energy of the element-element bond. The correct order is $C >> Si > Ge \approx Sn$. Thus,statement $(IV)$ is incorrect.
Therefore,the correct statements are $(I)$,$(II)$,and $(III)$.

(C) Dehydration of formic acid $(HCOOH)$ with conc. $H_2SO_4$ is a laboratory method,not a commercial one.
$(B)$ Direct oxidation of carbon in limited supply of oxygen is difficult to control on a commercial scale.
$(C)$ Passing steam over hot coke at high temperatures produces water gas $(CO + H_2)$,which is a standard commercial process:
$C(s) + H_2O(g) \xrightarrow{1273 \ K} CO(g) + H_2(g)$
$(D)$ Heating limestone $(CaCO_3)$ produces $CO_2$,not $CO$.
Therefore,option $(C)$ is the correct answer.

(C) The mixture of $CO$ and $H_2$ is known as water gas or synthesis gas $(syn \ gas)$.
$C_{(s)} + H_2O_{(g)} \rightarrow CO_{(g)} + H_{2(g)}$ is the coal gasification reaction.
The production of dihydrogen can be increased by reacting carbon monoxide with steam in the presence of iron chromate as a catalyst at $673 \ K$.
$CO_{(g)} + H_2O_{(g)} \xrightarrow[\text{iron chromate}]{673 \ K} CO_{2(g)} + H_{2(g)}$
This specific reaction is known as the water-gas shift reaction.

(D) The given reaction represents the industrial preparation of methanol from a mixture of carbon monoxide and hydrogen,also known as synthesis gas or syngas.
In this process,the gaseous mixture is subjected to a pressure of $200-300 \ atm$ and passed over a heated catalyst mixture of $ZnO$ and $Cr_2 O_3$ maintained at $573-673 \ K$.
This reaction results in the formation of methanol vapors,which are subsequently condensed to a liquid state.
$CO + 2 H_2 \xrightarrow[\substack{200-300 \ atm, 573-673 \ K}]{ZnO-Cr_2 O_3(X)} CH_3 OH$

(B) In $C_{60}$, all carbon atoms are $sp^2$-hybridised.
It contains $12$ rings of five carbons each and $20$ rings of six carbons each.
It is an aromatic compound due to the delocalization of $\pi$-electrons in a spherical cage structure.
It is a pure form of carbon (allotrope).
The $C-C$ bond lengths are $143.5 \ pm$ and $138.3 \ pm$.
It is prepared by heating graphite in an electric arc under an inert atmosphere (e.g., $He$ gas), not in the presence of oxygen.
Thus, statements $(i)$, $(ii)$, $(iv)$, and $(v)$ are correct.

(C) Statement $(I)$ is correct: Germanium exists only in traces $(1.5 \ ppm)$ and is mainly recovered from flue dust arising from the roasting of zinc ores.
Statement $(II)$ is incorrect: The electronegativity of group $14$ elements decreases from carbon to silicon and remains approximately constant from silicon to $Sn$ $(Si \approx Ge \approx Sn)$.
Statement $(III)$ is correct: All the group $14$ elements (Carbon,Silicon,Germanium,Tin,Lead) are available in the solid state at room temperature.
Therefore,statements $(I)$ and $(III)$ are correct.

(D) Given,$y = \log_e \tan \left(\frac{\pi}{4} + \frac{x}{2}\right)$
Using the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have:
$ \quad=\log _e\left(\frac{\tan \left(\frac{\pi}{4}\right)+\tan \frac{x}{2}}{1-\tan \left(\frac{\pi}{4}\right) \tan \left(\frac{x}{2}\right)}\right)=\log _e\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)$
Taking the exponential on both sides:
$e^y = \frac{1 + \tan(x/2)}{1 - \tan(x/2)}$
Using the componendo and dividendo rule:
$\frac{e^y - 1}{e^y + 1} = \frac{(1 + \tan(x/2)) - (1 - \tan(x/2))}{(1 + \tan(x/2)) + (1 - \tan(x/2))} = \frac{2 \tan(x/2)}{2} = \tan \left(\frac{x}{2}\right)$
Since $\tanh \left(\frac{y}{2}\right) = \frac{e^y - 1}{e^y + 1}$,we get:
$\tanh \left(\frac{y}{2}\right) = \tan \left(\frac{x}{2}\right)$

(C) The given equation of the pair of lines is $x^2+2 \sqrt{2} x y+2 y^2+4 x+4 \sqrt{2} y+1=0$.
This can be written as $(x+\sqrt{2}y)^2 + 4(x+\sqrt{2}y) + 1 = 0$.
Let $X = x+\sqrt{2}y$. Then the equation becomes $X^2 + 4X + 1 = 0$.
Solving for $X$ using the quadratic formula $X = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $X = \frac{-4 \pm \sqrt{16-4}}{2} = -2 \pm \sqrt{3}$.
So,the two parallel lines are $x+\sqrt{2}y + 2 - \sqrt{3} = 0$ and $x+\sqrt{2}y + 2 + \sqrt{3} = 0$.
The distance $d$ between two parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is given by $d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.
Here,$A=1, B=\sqrt{2}, C_1=2-\sqrt{3}, C_2=2+\sqrt{3}$.
$d = \frac{|(2-\sqrt{3})-(2+\sqrt{3})|}{\sqrt{1^2+(\sqrt{2})^2}} = \frac{|-2\sqrt{3}|}{\sqrt{1+2}} = \frac{2\sqrt{3}}{\sqrt{3}} = 2$.

(C) The given pair of lines is $8 x^2-14 x y+3 y^2+10 x+10 y-25=0$.
Factoring the expression,we get $(4 x-y-5)(2 x-3 y+5)=0$.
The lines are $L_1: 4 x-y-5=0$ and $L_2: 2 x-3 y+5=0$.
The intersection point of $L_1$ and $L_2$ is $(2,3)$.
Let the pair of lines $S=0$ be $(4 x-y+c_1)(2 x-3 y+c_2)=0$.
Since these lines form a parallelogram with $L_1$ and $L_2$,the diagonals intersect at $(3,2)$.
The intersection of the new pair of lines is $(x_1, y_1)$. The midpoint of $(2,3)$ and $(x_1, y_1)$ is $(3,2)$.
Thus,$\frac{x_1+2}{2}=3 \Rightarrow x_1=4$ and $\frac{y_1+3}{2}=2 \Rightarrow y_1=1$.
Substituting $(4,1)$ into the lines $4 x-y+c_1=0$ and $2 x-3 y+c_2=0$,we get $c_1=-15$ and $c_2=-5$.
The equation $S=0$ is $(4 x-y-15)(2 x-3 y-5)=0$.
Expanding this,we get $8 x^2-14 x y+3 y^2-50 x+50 y+75=0$.

(D) Given the circles:
$S_1 \equiv x^2+y^2+6x-2y+k=0$
$S_2 \equiv x^2+y^2+2x-6y-15=0$
Since $S_1$ bisects the circumference of $S_2$,the common chord of $S_1$ and $S_2$ must be the diameter of $S_2$.
The equation of the common chord is given by $S_1 - S_2 = 0$:
$(x^2+y^2+6x-2y+k) - (x^2+y^2+2x-6y-15) = 0$
$4x + 4y + k + 15 = 0$
The center of circle $S_2$ is $(-g, -f) = (-1, 3)$.
Since the common chord is the diameter of $S_2$,it must pass through the center of $S_2$ $(-1, 3)$.
Substituting $(-1, 3)$ into the chord equation:
$4(-1) + 4(3) + k + 15 = 0$
$-4 + 12 + k + 15 = 0$
$8 + k + 15 = 0$
$k + 23 = 0$
$k = -23$

(D) Given,the equations of the circles are:
$S: x^2+y^2+6x-2y+k=0$
$S': x^2+y^2+2x-6y-15=0$
The circle $S$ bisects the circumference of circle $S'$,which means the common chord of the two circles passes through the center of circle $S'$.
The equation of the common chord is given by $S - S' = 0$:
$(x^2+y^2+6x-2y+k) - (x^2+y^2+2x-6y-15) = 0$
$4x + 4y + k + 15 = 0$ $(i)$
The center of circle $S'$ is $(-g, -f)$,where $2g=2$ and $2f=-6$,so the center is $(-1, 3)$.
Since the common chord passes through the center $(-1, 3)$,we substitute these coordinates into equation $(i)$:
$4(-1) + 4(3) + k + 15 = 0$
$-4 + 12 + k + 15 = 0$
$8 + k + 15 = 0$
$k + 23 = 0$
$k = -23$
Thus,option $D$ is correct.

(C) Given circles are:
$S_1: x^2+y^2+6x-2y+k=0$
$S_2: x^2+y^2+2x-6y-15=0$
If circle $S_1$ bisects the circumference of circle $S_2$,then the common chord of $S_1$ and $S_2$ must pass through the center of circle $S_2$.
The equation of the common chord is given by $S_1 - S_2 = 0$:
$(x^2+y^2+6x-2y+k) - (x^2+y^2+2x-6y-15) = 0$
$4x+4y+k+15 = 0$ ... $(i)$
The center of circle $S_2$ is $(-g, -f) = (-1, 3)$.
Since the common chord $(i)$ passes through $(-1, 3)$,we substitute these coordinates into the equation:
$4(-1) + 4(3) + k + 15 = 0$
$-4 + 12 + k + 15 = 0$
$8 + k + 15 = 0$
$k + 23 = 0$
$k = -23$
Thus,the correct option is $C$.

(B) The equation of the parabola is $y^2=4ax$,where $a=1$.
The equation of a normal to the parabola $y^2=4ax$ at point $t$ is $y = -tx + 2at + at^3$.
Since the normal passes through $(5,0)$,we have $0 = -5t + 2(1)t + (1)t^3$.
$t^3 - 3t = 0 \implies t(t^2 - 3) = 0$.
The values of $t$ are $t_1 = 0, t_2 = \sqrt{3}, t_3 = -\sqrt{3}$.
The feet of the normals are $(at^2, 2at)$,which are:
$P_1 = (1(0)^2, 2(1)(0)) = (0,0)$.
$P_2 = (1(\sqrt{3})^2, 2(1)(\sqrt{3})) = (3, 2\sqrt{3})$.
$P_3 = (1(-\sqrt{3})^2, 2(1)(-\sqrt{3})) = (3, -2\sqrt{3})$.
The centroid $C$ of the triangle formed by these points is $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
$C = \left(\frac{0+3+3}{3}, \frac{0+2\sqrt{3}-2\sqrt{3}}{3}\right) = \left(\frac{6}{3}, 0\right) = (2,0)$.

(B) The equation of a tangent to the parabola $y^2=4ax$ (where $a=1$) is $y=mx+\frac{a}{m}$,which simplifies to $y=mx+\frac{1}{m}$.
Since the tangent passes through the point $(1,4)$,we substitute these coordinates into the equation:
$4 = m(1) + \frac{1}{m}$
$4 = m + \frac{1}{m}$
$m^2 - 4m + 1 = 0$
Let the slopes of the two tangents be $m_1$ and $m_2$. From the quadratic equation,we have:
$m_1 + m_2 = 4$ and $m_1 m_2 = 1$.
The angle $\theta$ between the two tangents is given by:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$
Using the identity $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$:
$(m_1 - m_2)^2 = (4)^2 - 4(1) = 16 - 4 = 12$
$|m_1 - m_2| = \sqrt{12} = 2\sqrt{3}$
Therefore,$\tan \theta = \left| \frac{2\sqrt{3}}{1 + 1} \right| = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Since $\tan \theta = \sqrt{3}$,we have $\theta = \frac{\pi}{3}$.

(C) In neutral or faintly alkaline medium,$KMnO_4$ oxidises thiosulphate $(S_2O_3^{2-})$ to sulphate $(SO_4^{2-})$: $8MnO_4^- + 3S_2O_3^{2-} + H_2O \longrightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^-$. Thus,$x$ is neutral or faintly alkaline.
In acidic medium,$KMnO_4$ oxidises nitrite $(NO_2^-)$ to nitrate $(NO_3^-)$: $2MnO_4^- + 5NO_2^- + 6H^+ \longrightarrow 2Mn^{2+} + 5NO_3^- + 3H_2O$. Thus,$y$ is acidic.
Therefore,$x$ is neutral and $y$ is acidic. The correct option is $C$.

(A) $HCOOH \xrightarrow{\text{Conc. } H_2SO_4} \underset{(B)}{CO} + \underset{(A)}{H_2O}$
$3 \underset{(B)}{CO} + Fe_2O_3 \xrightarrow{\Delta} \underset{(D)}{2Fe} + \underset{(C)}{3CO_2}$
$CaCO_3 + 2HCl \text{ (dil.)} \longrightarrow \underset{(C)}{CO_2} + H_2O + CaCl_2$
From the reactions,$B$ is $CO$ and $C$ is $CO_2$.
The reduction of $Fe_2O_3$ by $CO$ yields $Fe$ $(D)$ and $CO_2$ $(C)$.
Therefore,$D$ is $Fe$.

(C) Aqueous sulphite reacts with dilute sulphuric acid to produce sulphur dioxide gas $(SO_{2(g)})$,which is $X_{(g)}$.
When $SO_{2(g)}$ is passed into an acidified $KMnO_4$ solution,it acts as a reducing agent and reduces $Mn^{7+}$ to $Mn^{2+}$.
The chemical equations are:
$(I)$ $SO_3^{2-} + 2H^+ \longrightarrow SO_2 + H_2O$
$(II)$ $5SO_2 + 2KMnO_4 + 2H_2O \longrightarrow K_2SO_4 + 2MnSO_4 + 2H_2SO_4$
In the product $MnSO_4$,the oxidation state of $Mn$ is $+2$.
Therefore,the correct option is $(c)$.

(B) The disproportionation reaction of $MnO_4^{2-}$ in acidic medium is given by the following half-reactions:
$2e^{-} + 4H^{+} + MnO_4^{2-} \longrightarrow MnO_2 + 2H_2O$
$(MnO_4^{2-} \longrightarrow MnO_4^{-} + e^{-}) \times 2$
Adding these,the overall balanced equation is:
$3MnO_4^{2-} + 4H^{+} \longrightarrow MnO_2 + 2MnO_4^{-} + 2H_2O$
From the stoichiometry,$3 \ mol$ of $MnO_4^{2-}$ produces $1 \ mol$ of $MnO_2$ and $2 \ mol$ of $MnO_4^{-}$.
Therefore,for $1 \ mol$ of $MnO_4^{2-}$,the products are $\frac{1}{3} \ mol$ of $MnO_2$ and $\frac{2}{3} \ mol$ of $MnO_4^{-}$.
Thus,option $(B)$ is correct.

(C) $(i)$ When chlorine oxidises sulphur dioxide in the presence of water,it gives $H_2SO_4$ as oxyacid $A$. The reaction is:
$Cl_2 + SO_2 + 2H_2O \longrightarrow H_2SO_4 + 2HCl$
In $H_2SO_4$,the oxidation state of $S$ is calculated as:
$2(+1) + x + 4(-2) = 0$ $\Rightarrow x - 6 = 0$ $\Rightarrow x = +6$.
$(ii)$ When chlorine oxidises iodine in the presence of water,it gives $HIO_3$ as oxyacid $B$. The reaction is:
$5Cl_2 + I_2 + 6H_2O \longrightarrow 2HIO_3 + 10HCl$
In $HIO_3$,the oxidation state of $I$ is calculated as:
$1 + x + 3(-2) = 0$ $\Rightarrow x - 5 = 0$ $\Rightarrow x = +5$.
Thus,the oxidation states of $S$ and $I$ are $+6$ and $+5$ respectively. Therefore,option $C$ is correct.

(B) To find the oxidation state of $S$ in the given compounds:
$1$. $H_2S_2O_8$ (Peroxodisulphuric acid): $2(+1) + 2(x) + 8(-2) = 0$ $\Rightarrow 2x = 14$ $\Rightarrow x = +6$.
$2$. $H_2SO_5$ (Peroxomonosulphuric acid): $2(+1) + x + 5(-2) = 0 \Rightarrow x = +8$ (Incorrect,structure shows $S$ is $+6$ due to peroxide linkage).
$3$. $H_2SO_3$ (Sulphurous acid): $2(+1) + x + 3(-2) = 0 \Rightarrow x = +4$.
$4$. $H_2SO_4$ (Sulphuric acid): $2(+1) + x + 4(-2) = 0 \Rightarrow x = +6$.
$5$. $H_2S_2O_7$ (Pyrosulphuric acid): $2(+1) + 2(x) + 7(-2) = 0$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$.
$6$. $SO_2Cl_2$ (Sulphuryl chloride): $x + 2(-2) + 2(-1) = 0 \Rightarrow x = +6$.
$7$. $SOCl_2$ (Thionyl chloride): $x + (-2) + 2(-1) = 0 \Rightarrow x = +4$.
Thus,the compounds with $S$ in $+6$ oxidation state are $H_2S_2O_8, H_2SO_5, H_2SO_4, H_2S_2O_7, SO_2Cl_2$. The total count is $5$.

(D) The balanced chemical equation for the reaction of $MnO_4^{-}$ with $I^{-}$ in acidic medium is:
$2 \ MnO_4^{-} + 10 \ I^{-} + 16 \ H^{+} \longrightarrow 2 \ Mn^{2+} + 5 \ I_2 + 8 \ H_2O$
The $n$-factor for $MnO_4^{-}$ (reduction from $+7$ to $+2$) is $5$.
The $n$-factor for $I^{-}$ (oxidation from $-1$ to $0$ in $I_2$) is $1$ per $I^{-}$ ion.
Using the law of equivalence:
$n_1 \times M_1 \times V_1 = n_2 \times M_2 \times V_2$
$5 \times 0.02 \times V = 1 \times 0.01 \times 60$
$0.1 \times V = 0.6$
$V = \frac{0.6}{0.1} = 6 \ mL$

(B) Moles of metal $(M) = \frac{3.2 \ g}{64 \ g/mol} = 0.05 \ mol$.
Weight of oxygen $= 3.6 \ g - 3.2 \ g = 0.4 \ g$.
Moles of oxygen $= \frac{0.4 \ g}{16 \ g/mol} = 0.025 \ mol$.
Ratio of moles of $M : O = 0.05 : 0.025 = 2 : 1$.
Therefore,the formula of the oxide is $M_2O$.
Hence,the correct option is $(B)$.

(A) The balanced chemical equation for the reaction between $H_2O_2$ and acidified $KMnO_4$ is:
$2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \longrightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$
From the stoichiometry,$2 \text{ moles of } KMnO_4$ react with $5 \text{ moles of } H_2O_2$.
Moles of $KMnO_4 = \text{Molarity} \times \text{Volume (in L)} = 0.02 \times 0.5 = 0.01 \text{ mol}$.
Moles of $H_2O_2$ required $= \frac{5}{2} \times 0.01 = 0.025 \text{ mol}$.
Normality of $20 \text{ vol } H_2O_2 = \frac{20}{5.6} \approx 3.57 \text{ N}$.
Since the $n$-factor for $H_2O_2$ in this reaction is $2$,Molarity of $H_2O_2 = \frac{3.57}{2} = 1.785 \text{ M}$.
Volume of $H_2O_2 = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.025}{1.785} \approx 0.014 \text{ L} = 14 \text{ mL}$.
Alternatively,using the law of equivalence: $N_1V_1 = N_2V_2$.
Normality of $KMnO_4 = 0.02 \times 5 = 0.1 \text{ N}$.
$3.57 \times V_1 = 0.1 \times 500 \implies V_1 = \frac{50}{3.57} \approx 14 \text{ mL}$.
Thus,option $(A)$ is the correct answer.

(C) Given:
Volume of $HCl$ solution $(V_1) = 30 \ mL$
Volume of sodium carbonate solution $(Na_2CO_3)$ $(V_2) = 20 \ mL$
Concentration of $Na_2CO_3$ solution $(M_2) = 0.1 \ M$
Volume of $NaOH$ solution $(V_3) = 30.0 \ mL$
Concentration of $NaOH$ solution $(M_3) = 0.2 \ M$
For the reaction between $HCl$ and $Na_2CO_3$ $(Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2)$:
$n_{HCl} = 2 \times n_{Na_2CO_3}$
$M_1V_1 = 2 \times M_2V_2$
$M_1 = \frac{2 \times M_2 \times V_2}{V_1} = \frac{2 \times 0.1 \times 20}{30} = \frac{4}{30} \ M$
Also,for the reaction between $HCl$ and $NaOH$ $(NaOH + HCl \rightarrow NaCl + H_2O)$:
$M_1 V_f = M_3 V_3$ (where $V_f$ is the volume of $HCl$ solution required)
$V_f = \frac{M_3 \times V_3}{M_1} = \frac{0.2 \times 30}{4 / 30} = \frac{6}{4 / 30} = \frac{180}{4} = 90 \ mL$
Hence,option $C$ is the correct answer.

(D) The reaction between $Fe^{2+}$ and $Cr_2O_7^{2-}$ in acidic medium is: $Cr_2O_7^{2-} + 14H^{+} + 6Fe^{2+} \longrightarrow 2Cr^{3+} + 7H_2O + 6Fe^{3+}$.
Here,the $n$-factor for $K_2Cr_2O_7$ is $6$.
The reaction between $Fe^{2+}$ and $MnO_4^-$ in acidic medium is: $MnO_4^- + 8H^{+} + 5Fe^{2+} \longrightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}$.
Here,the $n$-factor for $KMnO_4$ is $5$.
Since the same amount of $Fe^{2+}$ reacts in both cases,the number of equivalents of $K_2Cr_2O_7$ must equal the number of equivalents of $KMnO_4$:
$M_1 \times V_1 \times n_1 = M_2 \times V_2 \times n_2$
Given: $M_1 = 0.01 \ M$,$V_1 = 20 \ mL$,$n_1 = 6$,$V_2 = 20 \ mL$,$n_2 = 5$.
$0.01 \times 20 \times 6 = M_2 \times 20 \times 5$
$M_2 = \frac{0.01 \times 6}{5} = 0.012 \ M$.
Thus,the molarity of $KMnO_4$ is $0.012 \ M$.

(A) In the Castner-Kellner cell,the electrolysis of brine ($NaCl$ solution) is performed.
At the anode (carbon),chloride ions are oxidized: $Cl^{-} \longrightarrow \frac{1}{2} Cl_2 + e^{-}$.
At the cathode (mercury),sodium ions are reduced to form a sodium amalgam: $Na^{+} + e^{-} + Hg \longrightarrow Na-Hg$.
The sodium amalgam then reacts with water to produce $NaOH$: $2Na-Hg + 2H_2O \longrightarrow 2NaOH + 2Hg + H_2 \uparrow$.
Thus,$A$ is $NaOH$ and $B$ is $NaCl$.
Therefore,option $(A)$ is correct.

(A) Gallium $(Ga)$ has a very low melting point of approximately $29.76 \ ^\circ C$ $(302.91 \ K)$.
Since the summer temperature in many regions often exceeds this value,$Ga$ exists in a liquid state during the summer season.

(D) German silver is an alloy composed of Copper $(Cu)$,Zinc $(Zn)$,and Nickel $(Ni)$.
Brass is an alloy composed primarily of Copper $(Cu)$ and Zinc $(Zn)$.
Comparing the two,Copper $(Cu)$ and Zinc $(Zn)$ are common to both alloys.
Iron $(Fe)$ is not a primary component of either,but Nickel $(Ni)$ is present in German silver and is not a standard component of brass.
Therefore,the metal that is not common to both is Nickel $(Ni)$.
Hence,option $(D)$ is the correct answer.

(C) Iodide ions $(I^-)$ are strong reducing agents. When $Cu^{2+}$ reacts with $I^-$,the $Cu^{2+}$ ions oxidize the iodide ions to molecular iodine $(I_2)$ and are themselves reduced to $Cu^+$ ions. Consequently,$CuI_2$ is unstable and does not exist in nature. The reaction is: $2Cu^{2+} + 4I^- \rightarrow 2CuI + I_2$.

(C) The electronic configurations are as follows:
$(I)$ Lanthanum ($La$,$Z=57$): $[Xe]_{54} \cdot 5d^1 \cdot 6s^2$ (No $f$-electrons)
$(II)$ Uranium ($U$,$Z=92$): $[Rn]_{86} \cdot 5f^3 \cdot 6d^1 \cdot 7s^2$ (Has $f$-electrons)
$(III)$ Lawrencium ($Lr$,$Z=103$): $[Rn]_{86} \cdot 5f^{14} \cdot 6d^1 \cdot 7s^2$ (Has $f$-electrons)
$(IV)$ Thorium ($Th$,$Z=90$): $[Rn]_{86} \cdot 6d^2 \cdot 7s^2$ (No $f$-electrons)
$(V)$ Actinium ($Ac$,$Z=89$): $[Rn]_{86} \cdot 6d^1 \cdot 7s^2$ (No $f$-electrons)
$(VI)$ Cerium ($Ce$,$Z=58$): $[Xe]_{54} \cdot 4f^1 \cdot 5d^1 \cdot 6s^2$ (Has $f$-electrons)
Based on the electronic configurations,the elements $La$,$Ac$,and $Th$ do not possess $f$-electrons.
Therefore,option $(c)$ is the correct answer.

(D) $(I)$ The ionic radius of lanthanoids decreases with an increase in atomic number due to lanthanoid contraction. The order is $Pr^{3+} (1.013 \ \mathring{A}) > Sm^{3+} (0.964 \ \mathring{A}) > Dy^{3+} (0.908 \ \mathring{A})$. Thus,statement $(I)$ is incorrect.
$(II)$ $Eu^{2+}$ $(Z = 63)$ has a stable $4f^7$ configuration but easily oxidizes to $Eu^{3+}$ $(4f^6)$,making it a strong reducing agent. Thus,statement $(II)$ is correct.
$(III)$ $Pu$ $(Z = 94)$ has the electronic configuration $[Rn] 5f^6 6d^0 7s^2$. It can exhibit oxidation states up to $+7$ by utilizing its $5f, 6d,$ and $7s$ electrons. Thus,statement $(III)$ is correct.
Therefore,the correct option is $(D)$.

(A) From the Nernst equation:
$E = E^{\circ} - \frac{0.059}{n} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$
Here,$n = 2$ and $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34 \ V - (-0.76 \ V) = 1.1 \ V$.
Substituting the values:
$E = 1.1 - \frac{0.059}{2} \log \left( \frac{C_2}{C_1} \right)$
To make $E$ maximum,the term $\frac{0.059}{2} \log \left( \frac{C_2}{C_1} \right)$ must be as small as possible (most negative).
This occurs when $\log \left( \frac{C_2}{C_1} \right)$ is the most negative value.
For option $A$: $\log \left( \frac{0.01}{0.1} \right) = \log(0.1) = -1$.
For option $B$: $\log \left( \frac{0.1}{0.01} \right) = \log(10) = 1$.
For option $C$: $\log \left( \frac{0.2}{0.1} \right) = \log(2) \approx 0.301$.
For option $D$: $\log \left( \frac{0.1}{0.2} \right) = \log(0.5) \approx -0.301$.
The value is minimum for option $A$,hence $E$ is maximum.

(B) At anode: $Mg \longrightarrow Mg^{2+} + 2e^{-}$
At cathode: $Sn^{2+} + 2e^{-} \longrightarrow Sn$
The overall cell reaction is: $Mg + Sn^{2+} \longrightarrow Mg^{2+} + Sn$
$n = 2$
$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = -0.14 \ V - (-2.34 \ V) = 2.20 \ V$
Using the Nernst equation:
$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \frac{[Mg^{2+}]}{[Sn^{2+}]}$
$E_{\text{cell}} = 2.20 - \frac{0.0591}{2} \log \frac{0.01}{0.1}$
$E_{\text{cell}} = 2.20 - 0.02955 \times \log(10^{-1})$
$E_{\text{cell}} = 2.20 - 0.02955 \times (-1) = 2.20 + 0.02955 \approx 2.23 \ V$

(C) Given reactions are:
$(i) M^{2+}_{(aq)} + e^- \rightarrow M^{+}_{(aq)}, E_1^\circ = +0.15 \text{ V}$
$(ii) M^{+}_{(aq)} + e^- \rightarrow M_{(s)}, E_2^\circ = +0.50 \text{ V}$
The target reaction is:
$(iii) M^{2+}_{(aq)} + 2e^- \rightarrow M_{(s)}, E_3^\circ = ?$
Using the relation $\Delta G_3^\circ = \Delta G_1^\circ + \Delta G_2^\circ$:
$-n_3FE_3^\circ = -n_1FE_1^\circ - n_2FE_2^\circ$
$n_3E_3^\circ = n_1E_1^\circ + n_2E_2^\circ$
$2 \cdot E_3^\circ = (1 \cdot 0.15) + (1 \cdot 0.50)$
$2 \cdot E_3^\circ = 0.65$
$E_3^\circ = \frac{0.65}{2} = 0.325 \text{ V}$.

(D) For the reaction,$A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$
Using the relation $\Delta G^{\circ} = -RT \ln K_c = -nFE^{\circ}_{cell}$
Here,$n = 2$ (number of electrons transferred).
At $298 \ K$,the relation simplifies to $\log K_c = \frac{n E^{\circ}_{cell}}{0.0591}$.
Substituting the values: $\log K_c = \frac{2 \times 0.59}{0.059} = 20$.
Therefore,$K_c = 10^{20} = 1.0 \times 10^{20}$.
Hence,option $(D)$ is the correct answer.

(B) The $Leclanche$ cell consists of a zinc container which acts as the anode.
The cathode is a carbon (graphite) rod surrounded by a mixture of powdered manganese dioxide $(MnO_2)$ and carbon.
The space between the cathode and anode is filled by a moist paste of ammonium chloride $(NH_4Cl)$ and zinc chloride $(ZnCl_2)$.

(B) The reaction at the cathode is: $Cu^{2+} + 2e^{-} \rightarrow Cu_{(s)}$
For $63 \ g$ of $Cu$,the electricity required is $2 \ F$ $(2 \times 96500 \ C)$.
So,for $0.4725 \ g$ of $Cu$,the electricity required $(Q)$ is: $Q = \frac{2 \times 96500 \times 0.4725}{63} = 1447.5 \ C$.
Using Faraday's law: $Q = I \times t$,where $t = 15 \times 60 \ s = 900 \ s$.
$I = \frac{Q}{t} = \frac{1447.5}{900} = 1.608 \ A$.

(C) The general structure for these derivatives is $C=N-Z$.
For oximes,the derivative is formed by the reaction of a carbonyl compound with hydroxylamine $(NH_2OH)$. Thus,the group $Z$ attached to the nitrogen is $-OH$.
For semicarbazones,the derivative is formed by the reaction of a carbonyl compound with semicarbazide $(NH_2NHCONH_2)$. Thus,the group $Z$ attached to the nitrogen is $-NHCONH_2$.
Therefore,for oxime $(I)$,$Z = -OH$ and for semicarbazone $(II)$,$Z = -NHCONH_2$.

(B) The structures of the given compounds are as follows:
$I$. Cinnamaldehyde ($3$-phenylprop-$2$-enal): $C_6H_5-CH=CH-CHO$
$II$. Salicylaldehyde ($2$-hydroxybenzaldehyde): $A$ benzene ring with a $-CHO$ group at position $1$ and a $-OH$ group at position $2$.
$III$. Vanillin ($4$-hydroxy-$3$-methoxybenzaldehyde): $A$ benzene ring with a $-CHO$ group at position $1$,a $-OH$ group at position $4$,and a $-OCH_3$ group at position $3$.
Comparing these with the given options,option $B$ correctly represents the structures of cinnamaldehyde,salicylaldehyde,and vanillin in that order.

(B) $I$. Acidity: The presence of electron-withdrawing groups ($-I$ effect) increases the acidity of carboxylic acids,while electron-donating groups ($+I$ effect or hyperconjugation) decrease it. The order $NCCH_2COOH > FCH_2COOH > H_3CCH_2COOH$ is correct because $-CN$ has a stronger $-I$ effect than $-F$,and $-CH_3$ is electron-donating.
$II$. Reactivity (towards nucleophilic addition): Aldehydes are generally more reactive than ketones due to less steric hindrance and electronic factors. Among aldehydes,$CH_3CH_2CHO$ is more reactive than $PhCHO$ (due to resonance stabilization of the carbonyl group by the phenyl ring). Thus,the correct order is $CH_3CH_2CHO > PhCHO > PhCOCH_3$. The given order is incorrect.
$III$. Boiling points: Alcohols have the highest boiling points due to intermolecular hydrogen bonding. Among carbonyl compounds of similar molecular weight,aldehydes generally have higher boiling points than ketones due to greater dipole-dipole interactions and less steric hindrance. The order $H_3COCH_2CH_3$ (ketone) $< H_3CCH_2CHO$ (aldehyde) $< H_3CCH_2CH_2OH$ (alcohol) is correct.
Therefore,sets $I$ and $III$ are correct.

(C) Alkenes $(X)$ are electron-rich due to the presence of a $\pi$-electron cloud,making them susceptible to attack by electrophiles. Therefore,they undergo electrophilic addition reactions.
Carbonyl compounds $(Y)$ contain a polar $C=O$ bond where the carbon atom is electron-deficient (electrophilic). Consequently,they undergo nucleophilic addition reactions.

(B) Carbonyl compounds (aldehydes and ketones) form solid crystalline addition products with sodium bisulphite $(NaHSO_3)$.
These solid bisulphite addition products can be separated from the reaction mixture by filtration,leaving impurities behind.
The pure carbonyl compound is then regenerated from the solid addition product by treatment with dilute acid or base.
The reaction is as follows:
$CH_3CHO + NaHSO_3 \rightarrow CH_3CH(OH)SO_3Na$ (Solid addition product)
$CH_3CH(OH)SO_3Na + H_2O \rightarrow CH_3CHO + NaHSO_3$ (Pure aldehyde)

(D) The chemical formulas for the given ores are:
$(I)$ Magnetite: $Fe_3O_4$ (Oxide ore)
$(II)$ Kaolinite: $Al_2Si_2O_5(OH)_4$ (Silicate ore)
$(III)$ Siderite: $FeCO_3$ (Carbonate ore)
$(IV)$ Calamine: $ZnCO_3$ (Carbonate ore)
Since both Siderite and Calamine contain the carbonate group $(CO_3^{2-})$,they are classified as carbonate ores.
Therefore,the correct option is $(D)$.

(D) The liquation method is used for refining metals that have low melting points,such as $Pb$,$Sn$,$Bi$,and $Hg$,compared to their impurities.
High melting point metals are not refined by this method.
Therefore,the statement that high melting metals are refined by liquation is incorrect.
Thus,option $(D)$ is the correct answer.

(C) Vapour phase refining involves the conversion of a metal into a volatile compound,which is subsequently decomposed to obtain the pure metal.
$1$. Nickel is purified by the Mond process: Impure nickel is heated in a stream of carbon monoxide to form a volatile complex,$Ni(CO)_4$,which is then decomposed to obtain pure nickel.
$Ni + 4 CO$ $\xrightarrow{350 \ K} Ni(CO)_4$ $\xrightarrow{470 \ K} Ni \text{ (pure)} + 4 CO$
$2$. Zirconium is purified by the Van Arkel method: Zirconium is heated in iodine vapours to form a volatile complex,$ZrI_4$,at $870 \ K$,which is further decomposed to obtain pure metal.
$Zr + 2 I_2$ $\xrightarrow{870 \ K} ZrI_4$ $\xrightarrow{1800 \ K} Zr \text{ (pure)} + 2 I_2$
Thus,the pair of metals refined by this method is $Zr$ and $Ni$.

(C) The van Arkel method for the refining of $Ti$ metal involves the formation of a volatile iodide,$TiI_4$,which is subsequently decomposed to yield pure metal. Thus,$X_2$ is $I_2$.
$Y_2$ does not liberate $O_2$ from water and does not form $HY$ and $HOY$ with water. $I_2$ is the only halogen that does not react with water to produce $O_2$ or form $HX$ and $HOX$ species under standard conditions. Therefore,$Y_2$ is also $I_2$.
The reactions are:
$(i)$ $\text{Impure } Ti + 2I_2$ $\rightarrow TiI_4$ $\xrightarrow{\Delta} Ti + 2I_2$
(ii) $I_2 + H_2O \rightarrow \text{No reaction to form } O_2, HI, \text{ or } HOI$.
Thus,both $X_2$ and $Y_2$ are $I_2$. The correct option is $(C)$.

(D) $(i)$ In the Mac-Arthur Forrest cyanide process,$Zn$ acts as a reducing agent to displace $Ag$ and $Au$ from their cyanide complexes: $2[M(CN)_2]^- + Zn \rightarrow [Zn(CN)_4]^{2-} + 2M$ (where $M = Ag$ or $Au$).
$(ii)$ Zinc is a low boiling point metal,so it is refined by distillation.
$(iii)$ Malachite is an ore of copper $(CuCO_3 \cdot Cu(OH)_2)$,not nickel.
Therefore,statements $(i)$ and $(ii)$ are correct.

(A) The reaction of ethylbenzene with $Br_2$ in the presence of $UV$ light is a free radical substitution reaction. The bromine radical abstracts a hydrogen atom from the benzylic carbon because the resulting benzylic radical is resonance-stabilized by the phenyl ring. The radical formed at the $\alpha$-carbon (benzylic position) is more stable than the radical at the $\beta$-carbon. Therefore,the major product is $1$-bromo-$1$-phenylethane $(C_6H_5CH(Br)CH_3)$.

(D) The reactivity of alkyl halides towards $S_{N}2$ reactions depends on the steric hindrance around the electrophilic carbon atom.
$S_{N}2$ reactivity order is: $1^{\circ} > 2^{\circ} > 3^{\circ}$ alkyl halide.
$(P) \ (CH_3)_2C(Br)CH_2CH_3$ is a $3^{\circ}$ alkyl halide (highly sterically hindered).
$(Q) \ BrCH_2(CH_2)_3CH_3$ is a $1^{\circ}$ alkyl halide (least sterically hindered).
$(R) \ CH_3CH(Br)(CH_2)_2CH_3$ is a $2^{\circ}$ alkyl halide.
Therefore,the correct order of reactivity is $Q > R > P$.

(C) The given reaction proceeds as follows:
$1$. The starting material is a chiral $2^{\circ}$ haloalkane,specifically $(S)-2-$bromobutane.
$2$. The reaction with $KCN$ proceeds via an $S_{N}2$ mechanism. The nucleophile $CN^-$ attacks from the side opposite to the leaving group $(Br^-)$,resulting in a Walden inversion of the configuration at the chiral center.
$3$. This produces $(R)-2-$methylbutanenitrile as product $X$.
$4$. Subsequent acid-catalyzed hydrolysis of the nitrile group $(-CN)$ converts it into a carboxylic acid group $(-COOH)$ without affecting the chiral center. Thus,the configuration remains inverted relative to the starting material,yielding $(R)-2-$methylbutanoic acid as product $Y$.
$5$. Comparing this with the given options,option $C$ correctly represents the structures of $X$ and $Y$.

(A) The $S_N1$ mechanism involves the formation of a planar carbocation intermediate.
In the case of $2-$bromooctane,the loss of the bromide ion leads to a planar carbocation intermediate.
Because the carbocation is planar,the nucleophile (e.g.,$OH^-$) can attack from either the front or the back side with equal probability.
This leads to the formation of a racemic mixture,which is represented as $(\pm)-$octan$-2-$ol.
Therefore,both the Assertion $(A)$ and the Reason $(R)$ are correct,and the Reason $(R)$ is the correct explanation for the Assertion $(A)$.

(A) $C_2H_4Br_2$ undergoes dehydrohalogenation with alc. $KOH$ followed by $NaNH_2$ to form ethyne $(X)$: $C_2H_4Br_2 \xrightarrow{\text{alc. } KOH, NaNH_2} HC \equiv CH (X)$.
Ethyne undergoes cyclic polymerization in the presence of red hot $Fe$ tube at $873 \ K$ to form benzene $(Y)$: $3C_2H_2 \xrightarrow{\text{Red hot Fe tube}} C_6H_6 (Y)$.
Benzene reacts with acetic anhydride in the presence of anhydrous $AlCl_3$ (Friedel-Crafts acylation) to form acetophenone $(Z)$: $C_6H_6 + (CH_3CO)_2O \xrightarrow{AlCl_3} C_6H_5COCH_3 (Z)$.

$A$. Correct: Due to resonance, the $C-Cl$ bond in chlorobenzene acquires partial double bond character, making it shorter $(169 \ pm)$ than the $C-Cl$ bond in chloromethane $(178 \ pm)$.
$B$. Correct: The partial double bond character in chlorobenzene makes the $C-Cl$ bond stronger and harder to break compared to the single $C-Cl$ bond in benzyl chloride.
$C$. Correct: Resonance in chlorobenzene involves the lone pair of $Cl$ delocalizing into the ring, creating partial double bond character.
$D$. Incorrect: The $Cl$ atom is $o/p$-directing, so chlorination of chlorobenzene yields $o$-dichlorobenzene and $p$-dichlorobenzene, not $m$-dichlorobenzene.
Therefore, statements $A, B,$ and $C$ are correct.

(B) The organic compound $X$ is benzyl chloride $(C_6H_5CH_2Cl)$.
When benzyl chloride reacts with $KCN$ in $C_2H_5OH$,it undergoes a nucleophilic substitution reaction to form benzyl cyanide $(C_6H_5CH_2CN)$ as the major product $Y$.
When benzyl cyanide $(Y)$ is reduced with $LiAlH_4$,it forms $2-$phenylethan$-1-$amine $(C_6H_5CH_2CH_2NH_2)$ as product $Z$.

(B) The given reaction involves the conversion of an aryl halide to an alkylbenzene by reacting it with an alkyl halide $(RX)$ and sodium metal in the presence of dry ether. This specific reaction is known as the Wurtz-Fittig reaction. The general equation is: $Ar-X + R-X + 2Na \xrightarrow{\text{dry ether}} Ar-R + 2NaX$. Therefore,the reagent $Z$ is $RX / Na / \text{dry } (C_2H_5)_2O$.

(B) $1$. For the reaction $X \xrightarrow{H_2/Ni} CH_3CH_2CH_2NH_2$,the reduction of a nitrile $(R-CN)$ with $H_2/Ni$ adds four hydrogen atoms to produce a primary amine $(R-CH_2NH_2)$. To obtain $CH_3CH_2CH_2NH_2$ (propan$-1-$amine),$X$ must be $CH_3CH_2CN$ (propanenitrile).
$2$. For the reaction $Y \xrightarrow{Br_2/NaOH} CH_3CH_2CH_2NH_2$,this is the Hofmann bromamide degradation reaction,which converts an amide $(R-CONH_2)$ into a primary amine $(R-NH_2)$ with one less carbon atom. To obtain $CH_3CH_2CH_2NH_2$ (propan$-1-$amine),$Y$ must be $CH_3CH_2CH_2CONH_2$ (butanamide).
$3$. Therefore,$X = CH_3CH_2CN$ and $Y = CH_3CH_2CH_2CONH_2$.

(C) The nitration mixture consists of concentrated $HNO_3$ and concentrated $H_2SO_4$. The reaction between them is as follows:
$HNO_3 + 2H_2SO_4 \rightarrow NO_2^{+} + H_3O^{+} + 2HSO_4^{-}$.
From the reaction,it is evident that $NO_2^{+}$,$H_3O^{+}$,and $HSO_4^{-}$ are present in the mixture.
$SO_4^{2-}$ ions are not formed in this reaction.

(B) The hydration of $2-$methyl$-2-$butene $(CH_3-C(CH_3)=CH-CH_3)$ follows Markovnikov's rule to form $2-$methyl$-2-$butanol $(X)$,which is a tertiary alcohol $(CH_3-C(OH)(CH_3)-CH_2-CH_3)$.
An isomer of $X$ is $3-$methyl$-2-$butanol $(CH_3-CH(OH)-CH(CH_3)_2)$,which is a secondary alcohol.
This isomer can be prepared by the reaction of acetaldehyde $(CH_3CHO)$ with isopropylmagnesium bromide $((CH_3)_2CHMgBr)$ followed by hydrolysis,as this is a Grignard reaction forming a secondary alcohol.
Thus,option $(B)$ is correct.

(A) Ethene reacts with oxygen in the presence of palladium$(II)$ chloride as a catalyst to form acetaldehyde.
This process is known as the Wacker process.
The chemical reaction is: $CH_2=CH_2 + \frac{1}{2}O_2 \xrightarrow{Pd(II), Cu(II)} CH_3CHO$.

(A) The conversion of an arenediazonium salt $(ArN_2^+Cl^-)$ to an arene $(ArH)$ is a reduction reaction. This can be achieved using ethanol $(CH_3CH_2OH)$ or hypophosphorous acid $(H_3PO_2)$ in the presence of water.
In the given reaction scheme,$X$ represents the reagent $CH_3CH_2OH$ and $Y$ represents the reagent $H_3PO_2/H_2O$,which are both standard reagents for the deamination of diazonium salts to arenes.
Therefore,the correct option is $A$.

(D) The reaction sequence is as follows:
$1$. The starting material is $4$-amino-$3$-bromoethylbenzene. Treatment with $NaNO_2/HCl$ at $273-278 \ K$ performs diazotization of the $-NH_2$ group to form the diazonium salt,$X$ ($4$-ethyl-$2$-bromobenzenediazonium chloride).
$2$. Reaction of $X$ with ethanol $(C_2H_5OH)$ reduces the diazonium group to a hydrogen atom,yielding $Y$ ($1$-bromo-$2$-ethylbenzene).
$3$. Finally,oxidation of the ethyl group with alkaline $KMnO_4$ converts the benzylic carbon into a carboxylic acid group,resulting in $Z$ ($2$-bromobenzoic acid).
Thus,the correct structure for $Z$ is $2$-bromobenzoic acid,which corresponds to option $(D)$.

(A) The conversion of $3$-methylaniline to $3$-nitrotoluene involves the following steps:
$1.$ Diazotization: $3-CH_3-C_6H_4-NH_2 + NaNO_2 + HCl \xrightarrow{273-278 \ K} 3-CH_3-C_6H_4-N_2^+Cl^-$ (Reagent $X = NaNO_2, HCl$).
$2.$ Reaction with $HBF_4$: $3-CH_3-C_6H_4-N_2^+Cl^- + HBF_4 \rightarrow 3-CH_3-C_6H_4-N_2^+BF_4^-$ (Reagent $Y = HBF_4$).
$3.$ Replacement by nitro group: $3-CH_3-C_6H_4-N_2^+BF_4^- + NaNO_2 \xrightarrow{Cu, \Delta} 3-CH_3-C_6H_4-NO_2$ (Reagent $Z = NaNO_2, Cu, \Delta$).

(D) The reaction sequence is as follows:
$1$. Cumene is oxidized and then hydrolyzed to form phenol $(X)$.
$2$. Phenol reacts with $NaOH$ followed by $CO_2$ and acidification (Kolbe-Schmitt reaction) to form salicylic acid $(Y)$,which is $2$-hydroxybenzoic acid.
$3$. Salicylic acid reacts with acetic anhydride $(CH_3CO)_2O$ to form acetylsalicylic acid,commonly known as aspirin $(Z)$.

(C) The correct matches are:
$A$. $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$ uses $Fe$ as a catalyst (Haber process) $(IV)$.
$B$. $2H_{2(g)} + O_{2(g)} \longrightarrow 2H_2O_{(l)}$ uses $Pt$ as a catalyst $(II)$.
$C$. $CO_{(g)} + 3H_{2(g)} \longrightarrow CH_{4(g)} + H_2O_{(g)}$ uses $Ni$ as a catalyst $(I)$.
$D$. $CO_{(g)} + 2H_{2(g)} \longrightarrow CH_3OH_{(g)}$ uses $ZnO-Cr_2O_3$ as a catalyst $(III)$.
Thus,the correct sequence is $A-IV, B-II, C-I, D-III$.

(A) Among the given statements,$(B)$ and $(C)$ are incorrect.
$(B)$ Silicon $(Si)$ and germanium $(Ge)$ are group $14$ elements and are intrinsic semiconductors,not impurities.
$(C)$ Phosphorus $(P)$ and arsenic $(As)$ are group $15$ elements. They have $5$ valence electrons. When doped into group $14$ elements,$4$ electrons form covalent bonds,and the $5$th electron remains free. Thus,they are electron-rich impurities,not electron-deficient.

(A) $P_4O_{10}$ is an acidic dehydrating agent. It cannot be used to dry ammonia $(NH_3)$ because ammonia is basic in nature and reacts with $P_4O_{10}$.
The chemical reaction is: $12NH_3 + P_4O_{10} + 6H_2O \rightarrow 4(NH_4)_3PO_4$.
Since $P_4O_{10}$ reacts with $NH_3$,it cannot be used as a drying agent for it.
Therefore,both $(A)$ and $(R)$ are correct and $(R)$ is the correct explanation of $(A)$.

(B) When white phosphorus $(P_4)$ is heated with concentrated $NaOH$ in an inert atmosphere of $CO_2$,it undergoes disproportionation to form phosphine $(PH_3)$ and sodium hypophosphite $(NaH_2PO_2)$.
$P_4 + 3NaOH + 3H_2O \rightarrow PH_3(A) + 3NaH_2PO_2(B)$
When phosphine $(A)$ is bubbled into aqueous $CuSO_4$ solution,it forms copper phosphide $(Cu_3P_2)$ and sulfuric acid $(H_2SO_4)$.
$3CuSO_4 + 2PH_3 \rightarrow Cu_3P_2 + 3H_2SO_4(C)$
Therefore,$B$ is $NaH_2PO_2$ and $C$ is $H_2SO_4$.

(D) Pyrophosphoric acid $(H_4P_2O_7)$ has the structure $(HO)_2P(O)-O-P(O)(OH)_2$. It contains $4$ $P-OH$ bonds.
Hypophosphoric acid $(H_4P_2O_6)$ has the structure $(HO)_2P(O)-P(O)(OH)_2$. It contains $4$ $P-OH$ bonds.
Therefore,the number of $P-OH$ bonds in both acids is $4$ and $4$ respectively.

(C) White phosphorus $(P_4)$ undergoes disproportionation reaction when heated with concentrated $NaOH$ solution to form phosphine $(PH_3)$ and sodium hypophosphite $(NaH_2PO_2)$.
The chemical equation is: $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$.
In the salt $NaH_2PO_2$ (sodium hypophosphite),let the oxidation state of $P$ be $x$.
The sum of oxidation states is: $(+1) + 2(+1) + x + 2(-2) = 0$.
$1 + 2 + x - 4 = 0$.
$x - 1 = 0$.
$x = +1$.

(A) The reaction of phosphorus with sulfuryl chloride $(SO_2Cl_2)$ is: $P_4 + 10 SO_2Cl_2 \longrightarrow 4 PCl_5 + 10 SO_2$ $(X = SO_2)$.
At $723 \ K$,the Deacon process for the manufacture of chlorine is: $4 HCl + O_2 \xrightarrow[CuCl_2]{723 \ K} 2 Cl_2 + 2 H_2O$ $(Y = Cl_2)$.
$Cl_2$ $(Y)$ reacts with $SO_2$ $(X)$ in the presence of water to form hydrochloric acid $(A)$ and sulfuric acid $(B)$: $SO_2 + 2 H_2O + Cl_2 \longrightarrow 2 HCl + H_2SO_4$.

(B) Phosphine $(PH_3)$ reacts with $CuSO_4$ solution to form a black precipitate of cupric phosphide $(Cu_3P_2)$,not $CuHPO_4$.
$3 CuSO_4 + 2 PH_3 \longrightarrow Cu_3P_2 + 3 H_2SO_4$
$PH_3$ is a very weak base.
It is prepared by the reaction of calcium phosphide with $HCl$:
$Ca_3P_2 + 6 HCl \longrightarrow 2 PH_3 + 3 CaCl_2$
$PH_3$ is used in smoke screens (Holm's signals).
Therefore,statement $(B)$ is incorrect.

(A) White phosphorus $(P_4)$ reacts with sulphuryl chloride $(SO_2Cl_2)$ to form phosphorus pentachloride $(PCl_5)$ and sulphur dioxide ($SO_2$,which is $X$).
$P_4 + 10 SO_2Cl_2 \longrightarrow 4 PCl_5 + 10 SO_2$ $(X)$
Chlorine $(Cl_2)$ reacts with sulphur dioxide ($SO_2$,$X$) in the presence of wood charcoal to form sulphuryl chloride ($SO_2Cl_2$,which is $Y$).
$SO_2$ $(X)$ $+ Cl_2 \xrightarrow{\text{Wood charcoal}} SO_2Cl_2$ $(Y)$
Therefore,$X$ is $SO_2$ and $Y$ is $SO_2Cl_2$.

(C) Reducing power decreases from $SO_2$ to $TeO_2$ on moving down the group.
$(B)$ The order of boiling points of hydrides of $16^{\text{th}}$ group elements is $H_2S < H_2Se < H_2Te < H_2O$. Due to $H$-bonding,$H_2O$ has a higher boiling point.
$(C)$ Both rhombic and monoclinic sulphur contain $S_8$ molecules. Therefore,statement $(C)$ is incorrect.
$(D)$ The bond angle of ozone is $117^{\circ}$.
Thus,option $(C)$ is the incorrect statement.

(D) Statement $(i)$ is correct: Sulphuric acid is manufactured by the contact process.
Statement $(ii)$ is correct: $SO_3$ dissolves in $H_2SO_4$ to form oleum or pyrosulphuric acid $(H_2S_2O_7)$.
Statement $(iii)$ is correct: $H_2SO_4$ is widely used in the manufacture of fertilisers like ammonium sulphate and super phosphate.
Statement $(iv)$ is incorrect: In the reaction $S + 2H_2SO_4 (\text{conc.}) \longrightarrow 3SO_2 + 2H_2O$,sulphur $(S)$ is oxidised from $0$ to $+4$ oxidation state in $SO_2$,while $H_2SO_4$ acts as an oxidising agent and is reduced to $SO_2$. Thus,$S$ is oxidised,not $H_2SO_4$.

(D) Liquid helium is used as a cryogenic agent for carrying out various experiments at low temperatures.
Therefore,statement $(d)$ is incorrect.
Helium is lighter than air but heavier than hydrogen,and it is non-flammable,making it safer for balloons.
It is used in gas-cooled nuclear reactors and for cooling superconducting magnets.

(B) The reaction $ArN_2^+ Cl^- + Cu + HCl \longrightarrow ArCl + N_2 + CuCl$ is known as the Gattermann reaction.
In this reaction,the diazonium salt reacts with copper metal in the presence of $HCl$ to produce chlorobenzene.
This reaction is a modification of the Sandmeyer reaction,where copper powder is used instead of copper$(I)$ chloride $(CuCl)$ or copper$(I)$ bromide $(CuBr)$.