JEE Main 2022 Chemistry Question Paper with Answer and Solution

(A) The characteristic flame colors and corresponding wavelengths for alkali metals are as follows:

Metal	Wavelength $(\lambda / nm)$
$Li$	$670.8$
$Na$	$589.2$
$Rb$	$780.0$
$Cs$	$455.5$

Matching the values:
$A (Li) - I (670.8)$
$B (Na) - II (589.2)$
$C (Rb) - III (780.0)$
$D (Cs) - IV (455.5)$
Therefore,the correct match is $A-I, B-II, C-III, D-IV$.

(A) $Cs$ (Caesium) is used in photoelectric cells due to its low ionization energy.
$Ga$ (Gallium) has a high boiling point and is used in high temperature thermometers.
$B$ (Boron) fibres are used in making bullet-proof vests.
$Si$ (Silicon) based silicones are used in water repellent sprays due to their non-polar alkyl groups.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(A) The Taj Mahal is made of marble,which is primarily calcium carbonate $(CaCO_3)$.
Acid rain contains sulfuric acid $(H_2SO_4)$,which reacts with the marble to form calcium sulfate $(CaSO_4)$,water,and carbon dioxide.
This reaction causes the surface of the marble to corrode and lose its luster.
The chemical reaction is: $CaCO_3 + H_2SO_4 \rightarrow CaSO_4 + H_2O + CO_2$.

(D) The molecular formula $C_4H_8$ corresponds to alkenes.
Oxidation of alkenes with acidic $KMnO_4$ leads to cleavage of the double bond.
$2-$methylpropene $(CH_3)_2C=CH_2$ on oxidation with acidic $KMnO_4$ gives acetone $(CH_3)_2C=O$ and $CO_2$ gas,which causes effervescence.
The reaction is: $(CH_3)_2C=CH_2 + [O] \xrightarrow{KMnO_4/H^+} (CH_3)_2C=O + CO_2 + H_2O$.
Thus,compound $A$ is $2-$methylpropene.

(B) Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M}$ and $d = \frac{m}{V}$,we get $P = \frac{dRT}{M}$.
Rearranging for molar mass: $M = \frac{dRT}{P}$.
Given values:
$d = 0.46 \, g \, L^{-1}$
$R = 0.082 \, L \, atm \, K^{-1} \, mol^{-1}$
$T = 257 + 273 = 530 \, K$
$P = \frac{100}{760} \, atm$
Substituting these values:
$M = \frac{0.46 \times 0.082 \times 530 \times 760}{100}$
$M = 151.93 \, g \, mol^{-1}$.
Rounding to the nearest integer,we get $152$.

(B) The molar mass of $NH_3$ is $14 + (3 \times 1) = 17 \ g \ mol^{-1}$.
Number of moles in $17.0 \ g$ of $NH_3$ is $n = \frac{17.0 \ g}{17 \ g \ mol^{-1}} = 1 \ mol$.
The enthalpy change for $1 \ mol$ is given as $23.4 \ kJ \ mol^{-1}$.
Number of moles in $85 \ g$ of $NH_3$ is $n = \frac{85 \ g}{17 \ g \ mol^{-1}} = 5 \ mol$.
The enthalpy change for $5 \ mol$ is $5 \times 23.4 \ kJ = 117 \ kJ$.

(D) The reaction for the neutralization of ammonia by sulfuric acid is: $2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$.
Number of milliequivalents $(meq)$ of $H_2SO_4$ used = $Molarity \times Volume \times n\text{-factor} = 1 \ M \times 12.5 \ mL \times 2 = 25 \ meq$.
Since $1 \ meq$ of $NH_3$ reacts with $1 \ meq$ of $H_2SO_4$,the $meq$ of $NH_3$ evolved = $25 \ meq$.
Mass of nitrogen = $\frac{meq \times 14}{1000} = \frac{25 \times 14}{1000} = 0.35 \ g$.
Percentage of nitrogen = $\frac{\text{Mass of nitrogen}}{\text{Mass of compound}} \times 100 = \frac{0.35}{0.55} \times 100 = 63.63\%$.
Rounding to the nearest integer,we get $64\%$.

(A) The hybridization and shape of the given molecules are as follows:
$A. XeO_3$: Steric number = $3$ (bond pairs) + $1$ (lone pair) = $4$. Hybridization is $sp^3$ and shape is pyramidal.
$B. XeF_2$: Steric number = $2$ (bond pairs) + $3$ (lone pairs) = $5$. Hybridization is $sp^3d$ and shape is linear.
$C. XeOF_4$: Steric number = $5$ (bond pairs) + $1$ (lone pair) = $6$. Hybridization is $sp^3d^2$ and shape is square pyramidal.
$D. XeF_6$: Steric number = $6$ (bond pairs) + $1$ (lone pair) = $7$. Hybridization is $sp^3d^3$ and shape is distorted octahedral.
Thus,the correct matching is $A-I, B-II, C-III, D-IV$.

(D) The ionization constant for reaction $(a)$ is $K_{a_1} = \frac{[H^{+}][HC_2O_4^-]}{[H_2C_2O_4]}$.
The ionization constant for reaction $(b)$ is $K_{a_2} = \frac{[H^{+}][C_2O_4^{2-}]}{[HC_2O_4^-]}$.
The reaction $(c)$ is the sum of reactions $(a)$ and $(b)$.
For reaction $(c)$,the equilibrium constant $K_{a_3} = \frac{[H^{+}]^2[C_2O_4^{2-}]}{[H_2C_2O_4]}$.
Multiplying $K_{a_1}$ and $K_{a_2}$ gives: $K_{a_1} \times K_{a_2} = \frac{[H^{+}][HC_2O_4^-]}{[H_2C_2O_4]} \times \frac{[H^{+}][C_2O_4^{2-}]}{[HC_2O_4^-]} = \frac{[H^{+}]^2[C_2O_4^{2-}]}{[H_2C_2O_4]} = K_{a_3}$.
Thus,$K_{a_3} = K_{a_1} \times K_{a_2}$.

(D) The electronic configurations are: $Be (2s^2)$,$B (2s^2 2p^1)$,$N (2s^2 2p^3)$,$O (2s^2 2p^4)$.
Due to the stable fully-filled $2s$ orbital,$Be$ has a higher ionization enthalpy than $B$.
Due to the stable half-filled $2p$ orbital,$N$ has a higher ionization enthalpy than $O$.
The overall order of first ionization enthalpy is $B < Be < O < N$.

(C) High purity $(>99.95\,\%)$ dihydrogen is obtained by the electrolysis of warm aqueous $Ba(OH)_2$ solution between nickel electrodes.

(C) In the $IIA$ group (alkaline earth metals),the density decreases from $Be$ to $Ca$ and then increases from $Ca$ to $Ba$.
The densities of these elements are: $Be (1.85 \ g/cm^3)$,$Mg (1.74 \ g/cm^3)$,$Ca (1.55 \ g/cm^3)$,and $Sr (2.63 \ g/cm^3)$.
Therefore,the correct order of density is $Sr > Be > Mg > Ca$.

(B) Neutral oxides are: $NO$,$N_2O$,and $CO$.
Acidic oxides are: $B_2O_3$,$N_2O_5$,$SO_3$,and $P_4O_{10}$.
Therefore,the total number of acidic oxides is $4$.

(C) . Sulphate - $III$. Laxative effect
$B$. Fluoride - $II$. Bending of bones
$C$. Nicotine - $I$. Pesticide
$D$. Sodium arsinite - $IV$. Herbicide
Therefore,the correct matching is $A-III, B-II, C-I, D-IV$.

(D) The reaction involves the electrophilic addition of $HBr$ to the two double bonds present in the molecule,which is allyl acrylate $(CH_2=CH-COO-CH_2-CH=CH_2)$.
$1$. The first $HBr$ molecule adds to the more reactive double bond. The double bond in the allyl group (alkene part) is more electron-rich than the double bond conjugated with the carbonyl group (acrylate part). Thus,$HBr$ adds to the allyl double bond following Markovnikov's rule to form $CH_2=CH-COO-CH_2-CH(Br)-CH_3$.
$2$. The second $HBr$ molecule adds to the remaining double bond in the acrylate part. This addition follows anti-Markovnikov's rule (if peroxide is present,though usually,in the absence of peroxide,it follows Markovnikov's rule,but here the electron-withdrawing carbonyl group directs the addition to the terminal carbon to form the more stable carbocation or via conjugate addition). The addition to the $CH_2=CH-$ group results in the formation of $Br-CH_2-CH_2-COO-CH_2-CH(Br)-CH_3$.
Therefore,the final product is $1-$bromopropan$-2-$yl $3-$bromopropanoate.

(A) Methyl orange is an acid-base indicator that changes color depending on the $pH$ of the solution.
In an acidic medium (at the end point of a titration involving an acid),methyl orange undergoes protonation.
The structure changes from the benzenoid form (yellow color in basic medium) to the quinonoid form (red/pink color in acidic medium).
As shown in the reaction,the addition of $H^+$ ions leads to the formation of the quinonoid structure,which is responsible for the color change observed at the end point.

(C) The balanced chemical equation for the reaction is:
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$
According to the stoichiometry of the reaction,$1 \ volume$ of $N_2$ produces $2 \ volumes$ of $NH_3$.
Therefore,to produce $20 \ L$ of $NH_3$,the volume of $N_2$ required is:
$V_{N_2 \text{ consumed}} = \frac{1}{2} \times 20 \ L = 10 \ L$
The initial volume of $N_2$ is $56.0 \ L$.
The volume of unused $N_2$ is:
$V_{N_2 \text{ unused}} = 56.0 \ L - 10 \ L = 46 \ L$
Thus,the correct option is $C$.

(C) $1$. Calculate the number of moles of propane $(C_{3}H_{8})$:
Molar mass of $C_{3}H_{8} = (3 \times 12) + (8 \times 1) = 44 \, g \, mol^{-1}$.
Moles $(n)$ = $\frac{11 \, g}{44 \, g \, mol^{-1}} = 0.25 \, mol$.
$2$. Use the ideal gas equation $PV = nRT$:
Given $P = 2 \, MPa = 2 \times 10^{6} \, Pa$,$V = 2 \, dm^{3} = 2 \times 10^{-3} \, m^{3}$,$n = 0.25 \, mol$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$.
$2 \times 10^{6} \times 2 \times 10^{-3} = 0.25 \times 8.3 \times T$.
$4000 = 2.075 \times T$.
$T = \frac{4000}{2.075} \approx 1927.71 \, K$.
$3$. Convert temperature to Celsius:
$T(^{\circ}C) = T(K) - 273.15 = 1927.71 - 273.15 = 1654.56 \, ^{\circ}C$.
Rounding to the nearest integer,we get $1655 \, ^{\circ}C$.

(C) The maximum number of emission lines produced when an electron drops from an excited state $n$ to the ground state is calculated using the formula:
$\text{Number of lines} = \frac{n(n-1)}{2}$
Given $n = 5$,substituting the value into the formula:
$\text{Number of lines} = \frac{5(5-1)}{2} = \frac{5 \times 4}{2} = \frac{20}{2} = 10 \text{ lines.}$

(B) The enthalpy of neutralization of a strong acid with a strong base is $-57.3 \ kJ \ mol^{-1}$.
For the weak acid $CH_{3}COOH$,the enthalpy of neutralization is the sum of the enthalpy of neutralization of $H^+$ and $OH^-$ ions and the enthalpy of ionization of the weak acid.
$\Delta H_{\text{neutralization}} = \Delta H_{\text{ionization}} + \Delta H_{\text{neutralization of strong acid/base}}$
$-55.3 \ kJ \ mol^{-1} = \Delta H_{\text{ionization}} + (-57.3 \ kJ \ mol^{-1})$
$\Delta H_{\text{ionization}} = -55.3 - (-57.3) = 2 \ kJ \ mol^{-1}$.

(B) To find the number of lone pairs on the central atom $(Xe)$ in each molecule:
$1$. In $XeO_3$: The $Xe$ atom has $8$ valence electrons. It forms $3$ double bonds with $3$ oxygen atoms (using $6$ electrons). Thus,$8 - 6 = 2$ electrons remain,which form $1$ lone pair.
$2$. In $XeOF_4$: The $Xe$ atom has $8$ valence electrons. It forms $1$ double bond with oxygen ($2$ electrons) and $4$ single bonds with fluorine atoms ($4$ electrons). Total electrons used = $6$. Thus,$8 - 6 = 2$ electrons remain,which form $1$ lone pair.
$3$. In $XeF_6$: The $Xe$ atom has $8$ valence electrons. It forms $6$ single bonds with fluorine atoms ($6$ electrons). Thus,$8 - 6 = 2$ electrons remain,which form $1$ lone pair.
Sum of lone pairs = $1 + 1 + 1 = 3$.

(D) The $R_{f}$ value is defined as the ratio of the distance travelled by the substance to the distance travelled by the solvent front.
$R_{f(A)} = \frac{2.08}{3.25}$
$R_{f(B)} = \frac{1.05}{3.25}$
The ratio of $R_{f}$ values of $A$ to $B$ is $\frac{R_{f(A)}}{R_{f(B)}} = \frac{2.08 / 3.25}{1.05 / 3.25} = \frac{2.08}{1.05} \approx 1.98 \approx 2$.

(D) The alkanes with molecular formula $C_{5}H_{12}$ are $n$-pentane,isopentane,and neopentane.
$1$. $n$-Pentane $(CH_{3}CH_{2}CH_{2}CH_{2}CH_{3})$ has $3$ non-equivalent sets of hydrogen atoms,giving $3$ monobromo derivatives.
$2$. Isopentane $((CH_{3})_{2}CHCH_{2}CH_{3})$ has $4$ non-equivalent sets of hydrogen atoms,giving $4$ monobromo derivatives.
$3$. Neopentane $((CH_{3})_{4}C)$ has $1$ set of equivalent hydrogen atoms,giving $1$ monobromo derivative.
Total number of monobromo derivatives = $3 + 4 + 1 = 8$.

(C) Let the number of moles of $SO_{2}Cl_{2}$ be $x$.
According to the reaction: $SO_{2}Cl_{2} + 2H_{2}O \rightarrow H_{2}SO_{4} + 2HCl$.
From the stoichiometry,$1 \, mole$ of $SO_{2}Cl_{2}$ produces $1 \, mole$ of $H_{2}SO_{4}$ and $2 \, moles$ of $HCl$.
Therefore,$x \, moles$ of $SO_{2}Cl_{2}$ produce $x \, moles$ of $H_{2}SO_{4}$ and $2x \, moles$ of $HCl$.
Total moles of $H^{+}$ ions produced = $(2 \times n(H_{2}SO_{4})) + n(HCl) = (2 \times x) + 2x = 4x$.
For complete neutralization,$n(H^{+}) = n(OH^{-})$.
Given $n(NaOH) = 16 \, moles$,so $n(OH^{-}) = 16$.
$4x = 16 \implies x = 4$.

(C) The allowed values for the azimuthal quantum number $l$ are $0, 1, 2, \dots, (n-1)$.
For $n=3$,the possible values of $l$ are $0, 1, 2$.
Therefore,$l=3$ is not possible for $n=3$ because $l$ must be less than $n$.

(C) Step $1$: Calculate millimoles of reactants.
$n(NH_{4}OH) = 20\,mL \times 0.1\,M = 2\,mmol$.
$n(HCl) = 40\,mL \times 0.05\,M = 2\,mmol$.
Step $2$: Reaction stoichiometry.
$NH_{4}OH + HCl \rightarrow NH_{4}Cl + H_{2}O$.
Since both reactants are $2\,mmol$,they react completely to form $2\,mmol$ of $NH_{4}Cl$ (a salt of weak base and strong acid).
Step $3$: Calculate concentration of salt.
Total volume $= 20 + 40 = 60\,mL$.
$[NH_{4}Cl] = C = \frac{2\,mmol}{60\,mL} = \frac{1}{30}\,M$.
Step $4$: Calculate $pH$ of salt solution.
$pH = \frac{1}{2}[pK_{w} - pK_{b} - \log C]$.
$pK_{w} = 14$,$pK_{b} = -\log(10^{-5}) = 5$.
$pH = \frac{1}{2}[14 - 5 - \log(1/30)] = \frac{1}{2}[9 + \log 30] = \frac{1}{2}[9 + \log 3 + \log 10] = \frac{1}{2}[9 + 0.48 + 1] = \frac{10.48}{2} = 5.24$.
The nearest value is $5.2$.

(D) The atomic number of $Rn$ (Radon) is $86$.
Adding the electrons from the configuration: $86 + 14 (5f) + 1 (6d) + 2 (7s) = 103$.
The element with atomic number $103$ is $Lawrencium$.
According to $IUPAC$ nomenclature for elements with $Z > 100$:
$1 = un$,$0 = nil$,$3 = tri$.
Therefore,the name is $Unniltrium$.

(A) In an acidic medium,potassium permanganate $(KMnO_{4})$ acts as a strong oxidizing agent.
When it reacts with hydrogen peroxide $(H_{2}O_{2})$,the permanganate ion $(MnO_{4}^{-})$ is reduced to the manganese$(II)$ ion $(Mn^{2+})$.
The balanced chemical equation is:
$2MnO_{4}^{-} + 5H_{2}O_{2} + 6H^{+} \rightarrow 2Mn^{2+} + 5O_{2} + 8H_{2}O$
Therefore,the main product formed from the manganese species is $Mn^{2+}$.

(A) The density of alkali metals generally increases down the group as the atomic mass increases more significantly than the atomic volume. However,$K$ is an exception because of an unusual increase in its atomic size. The correct order of density is $Li < K < Na < Rb < Cs$.

(B) The given reactions are:
$2BF_{3} + 6NaH \xrightarrow{450 \ K} B_{2}H_{6} + 6NaF$
$B_{2}H_{6} + 2NMe_{3} \rightarrow 2[BH_{3} \leftarrow NMe_{3}]$
Here,$A$ is $B_{2}H_{6}$ (diborane) and $B$ is the adduct $[BH_{3} \leftarrow NMe_{3}]$.
In the adduct $[BH_{3} \leftarrow NMe_{3}]$,the boron atom is bonded to three hydrogen atoms and one nitrogen atom.
The boron atom is $sp^{3}$ hybridized and possesses a tetrahedral geometry.

(C) Photochemical smog is formed by the action of sunlight on nitrogen oxides $(NO_x)$ and volatile organic compounds $(VOCs)$ such as formaldehyde $(HCHO)$.
It typically contains ozone $(O_3)$,nitrogen dioxide $(NO_2)$,nitric oxide $(NO)$,formaldehyde $(HCHO)$,acrolein,and peroxyacetyl nitrate $(PAN)$.
Sulfur dioxide $(SO_2)$ is a primary component of classical smog (London smog),not photochemical smog.

(D) The compound $A$ is $2,4,4-trimethylpent-1-ene$.
Reaction with $O_3, H_2O / Zn$ (reductive ozonolysis) gives formaldehyde $(HCHO)$ as by-product $a$ and a ketone.
Reaction with $KMnO_4 / H^{+}$ (oxidative cleavage) gives formic acid $(HCOOH)$ as by-product $b$ and a ketone.
Formaldehyde $(a)$ on oxidation gives formic acid,which is found in ants.
Thus,$X$ is $O_3, H_2O / Zn$ and $Y$ is $KMnO_4 / H^{+}$.

(C) Statement $I$ is correct: Glycerol undergoes dehydration when heated with $KHSO_4$ to form acrolein $(CH_2=CH-CHO)$.
Statement $II$ is incorrect: Acrolein has a pungent,suffocating,and irritating odour,not a fruity odour. While it is used to detect the presence of glycerol,the description of its odour is wrong.

(C) To determine diamagnetism,we check the number of electrons in each species:
$N_{2}$ ($14$ $e^-$): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$ (All paired,Diamagnetic)
$N_{2}^{+}$ ($13$ $e^-$): Paramagnetic
$N_{2}^{-}$ ($15$ $e^-$): Paramagnetic
$N_{2}^{2-}$ ($16$ $e^-$): Paramagnetic
$O_{2}$ ($16$ $e^-$): Paramagnetic
$O_{2}^{+}$ ($15$ $e^-$): Paramagnetic
$O_{2}^{-}$ ($17$ $e^-$): Paramagnetic
$O_{2}^{2-}$ ($18$ $e^-$): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$ (All paired,Diamagnetic)
Thus,the species showing diamagnetism are $N_{2}$ and $O_{2}^{2-}$.
The total number of diamagnetic species is $2$.

(C) The formation reaction for propane is: $3C_{(gr)} + 4H_{2(g)} \rightarrow C_{3}H_{8(g)}$
The enthalpy of formation $\Delta H_f$ is calculated using the formula: $\Delta H_f = \sum \Delta H_{c}(\text{reactants}) - \sum \Delta H_{c}(\text{products})$
$\Delta H_f = [3 \times \Delta H_{c}(C) + 4 \times \Delta H_{c}(H_{2})] - \Delta H_{c}(C_{3}H_{8})$
$\Delta H_f = [3(-393.5) + 4(-285.8)] - (-2220.0)$
$\Delta H_f = [-1180.5 - 1143.2] + 2220.0$
$\Delta H_f = -2323.7 + 2220.0 = -103.7 \ kJ \ mol^{-1}$
The magnitude of the enthalpy of formation is $|-103.7| = 103.7 \ kJ \ mol^{-1}$.
Rounding to the nearest integer,we get $104 \ kJ \ mol^{-1}$.

(C) The total pressure of a moist gas is the sum of the partial pressure of the dry gas and the vapour pressure of water: $P_{total} = P_{gas} + P_{H_2O}$.
Given $P_{total} = 4 \ atm$ and $P_{H_2O} = 0.4 \ atm$,so $P_{gas} = 4 - 0.4 = 3.6 \ atm$.
According to Boyle's Law,at constant temperature,$P_1V_1 = P_2V_2$. When the volume is doubled $(V_2 = 2V_1)$,the new partial pressure of the dry gas becomes $P_{gas, new} = P_{gas} / 2 = 3.6 / 2 = 1.8 \ atm$.
The vapour pressure of water remains constant at a given temperature,so $P_{H_2O} = 0.4 \ atm$.
The new total pressure is $P_{total, new} = 1.8 + 0.4 = 2.2 \ atm$.
Since $2.2 \ atm = 22 \times 10^{-1} \ atm$,the value of $x$ is $22$.

(B) The formula for the percentage of nitrogen in Kjeldahl's method is: $\% \text{N} = \frac{1.4 \times \text{Molarity of acid} \times \text{Basicity of acid} \times \text{Volume of acid in mL}}{\text{Mass of organic compound in g}}$
Here,$\text{Molarity} = 2 \ M$,$\text{Basicity of } H_2SO_4 = 2$,$\text{Volume} = 2.5 \ mL$,and $\text{Mass} = 0.25 \ g$.
Substituting the values:
$\% \text{N} = \frac{1.4 \times 2 \times 2 \times 2.5}{0.25}$
$\% \text{N} = \frac{1.4 \times 10}{0.25} = \frac{14}{0.25} = 56$
Therefore,the percentage of nitrogen is $56\%$.

(A) The degree of unsaturation $(DU)$ for $C_{4}H_{5}N$ is calculated as:
$DU = C + 1 - \frac{H - N}{2} = 4 + 1 - \frac{5 - 1}{2} = 5 - 2 = 3$.
$A$ possible acyclic structure with $DU = 3$ is $CH_{2}=C=CH-CH=NH$.
In this structure,all carbon atoms are either $sp$ or $sp^{2}$ hybridised.
Specifically,the carbons are $sp^{2}$,$sp$,$sp^{2}$,and $sp^{2}$ hybridised respectively.
Therefore,the number of $sp^{3}$ hybridised carbons is $0$.

(C) Mass of $Fe$ in $3.3 \ g$ of hemoglobin $= \frac{0.34}{100} \times 3.3 \ g = 0.01122 \ g$.
Number of moles of $Fe = \frac{\text{mass}}{\text{atomic mass}} = \frac{0.01122 \ g}{56 \ g/mol} \approx 2.0036 \times 10^{-4} \ mol$.
Number of $Fe$ atoms $= \text{moles} \times N_A = 2.0036 \times 10^{-4} \times 6.022 \times 10^{23} \approx 1.206 \times 10^{20}$ atoms.

(A) According to Fajan's rule,the covalent character of an ionic bond increases as the size of the anion increases.
The order of the size of the anions is $F^- < Cl^- < Br^- < I^-$.
Therefore,the increasing order of covalent character is $CaF_{2} < CaCl_{2} < CaBr_{2} < CaI_{2}$.

(C) For a basic buffer,the $pH$ is related to $pOH$ by the equation: $pH + pOH = 14$.
Given $pH = 8.26$,so $pOH = 14 - 8.26 = 5.74$.
The Henderson-Hasselbalch equation for a basic buffer is: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Substituting the given values: $5.74 = 4.74 + \log \frac{[NH_4^+]}{0.2}$.
$1 = \log \frac{[NH_4^+]}{0.2}$,which implies $\frac{[NH_4^+]}{0.2} = 10^1 = 10$.
Therefore,$[NH_4^+] = 10 \times 0.2 = 2 \, M$.
Since the volume is $1 \, L$,the number of moles of $NH_4Cl$ required is $2 \, mol$.
Mass of $NH_4Cl = \text{moles} \times \text{molar mass} = 2 \, mol \times 53.5 \, g \, mol^{-1} = 107 \, g$.

(D) In an acidic medium,hydrogen peroxide $(H_{2}O_{2})$ acts as a reducing agent towards potassium permanganate $(KMnO_{4})$.
The balanced chemical equation for the reaction is:
$6H^{+} + 2MnO_{4}^{-} + 5H_{2}O_{2} \longrightarrow 2Mn^{2+} + 8H_{2}O + 5O_{2}$
Thus,the products formed are $Mn^{2+}$,$H_{2}O$,and $O_{2}$.

(C) Assertion $A$ is true: $LiF$ is sparingly soluble in water due to its exceptionally high lattice energy compared to its hydration energy.
Reason $R$ is false: Although the $Li^{+}$ ion has the smallest ionic radius among its group members,it actually possesses the highest hydration enthalpy,not the least,because hydration enthalpy is inversely proportional to the ionic size $(\Delta H_{hyd} \propto 1/r)$.

(A) Boric acid ($H_3BO_3$ or $B(OH)_3$) is a weak monobasic Lewis acid.
It does not act as a proton donor (Brønsted acid) by releasing $H^{+}$ ions directly.
Instead,it acts as a Lewis acid by accepting a lone pair of electrons from the $OH^{-}$ ion of a water molecule.
The reaction is: $B(OH)_3 + 2H_2O \rightleftharpoons [B(OH)_4]^- + H_3O^+$.
This process releases $H^{+}$ ions (as $H_3O^+$) into the solution,which is why it exhibits acidic properties.
Therefore,both Assertion $A$ and Reason $R$ are correct,and $R$ is the correct explanation of $A$.

(A) The correct matching is as follows:
$A$. Microorganisms are primarily found in $II$. Domestic sewage.
$B$. Plant nutrients (like nitrates and phosphates) are commonly found in $III$. Chemical fertilizers.
$C$. Toxic heavy metals are often released by $IV$. Chemical factories.
$D$. Sediment is caused by $I$. Strip mining activities.
Therefore,the correct sequence is $A-II, B-III, C-IV, D-I$.

(B) According to the $IUPAC$ priority rules for functional groups,the order of decreasing priority is: $ -COOH > -SO_3H > -COOR > -COCl > -CONH_2 > -CN > -CHO > C=O > -OH > -NH_2 > -OR > -R $.
Comparing the given options with the standard $IUPAC$ priority table:
Option $B$ is $-SO_3H > -COCl > -CONH_2 > -CN$,which follows the correct decreasing order of priority among the groups listed.

(B) Benzenoid compounds are organic compounds that contain at least one benzene ring.
$A$ (Phenanthrene),$C$ (Naphthalene),and $D$ (Aniline) all contain benzene rings in their structures.
$B$ (Cyclooctatetraene) is an eight-membered cyclic polyene $(C_8H_8)$ that does not contain a benzene ring.
Therefore,cyclooctatetraene is not a benzenoid compound.

(D) Phenolphthalein $(HPh)$ is a weak organic acid.
In an acidic medium,the equilibrium $HPh \rightleftharpoons H^{+} + Ph^{-}$ shifts to the left due to the common ion effect of $H^{+}$ ions,keeping the concentration of the unionized form $(HPh)$ high,which is colourless.
In a basic medium,the $OH^{-}$ ions react with $H^{+}$ to form water,shifting the equilibrium to the right. This increases the concentration of the ionized form $(Ph^{-})$,which is pink in colour.
Therefore,Assertion $A$ is true,but Reason $R$ is false because phenolphthalein does dissociate in a basic medium.

(A) Given: $P = 6 \ bar = 6 \times 10^{5} \ Pa$,$V = 0.0125 \ m^{3}$,$T = 27 + 273 = 300 \ K$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$.
Using the ideal gas equation $PV = n_{mix}RT$:
$n_{mix} = \frac{PV}{RT} = \frac{6 \times 10^{5} \times 0.0125}{8.3 \times 300} = \frac{7500}{2490} \approx 3.012 \ mol \approx 3 \ mol$.
Let the moles of $He$ be $x$ and moles of $H_{2}$ be $(3-x)$.
The total mass is $10 \ g$,so: $4x + 2(3-x) = 10$.
$4x + 6 - 2x = 10 \implies 2x = 4 \implies x = 2 \ mol$.
Mass of $He = n \times M = 2 \ mol \times 4 \ g \ mol^{-1} = 8 \ g$.

(A) For the ion ${}_{22}^{48}X^{3-}$:
Atomic number $(Z)$ = $22$.
Mass number $(A)$ = $48$.
Number of neutrons = $A - Z = 48 - 22 = 26$.
Number of electrons = $Z + 3 = 22 + 3 = 25$.
We are given that the nucleus contains '$a$'$\%$ more neutrons than the number of electrons.
Percentage difference = $\frac{\text{Number of neutrons} - \text{Number of electrons}}{\text{Number of electrons}} \times 100$.
Percentage difference = $\frac{26 - 25}{25} \times 100 = \frac{1}{25} \times 100 = 4$.
Thus,the value of '$a$' is $4$.

(D) In a vacancy defect,some of the lattice sites are vacant,which leads to a decrease in the number of atoms per unit volume,thereby decreasing the density of the substance. Therefore,the statement that vacancy defect increases the density is incorrect.

(C) The potential difference between the fixed layer and the diffused layer of charges on the surface of a colloidal particle is known as the Zeta potential or electrokinetic potential.

(B) According to the Ellingham diagram,the line for the formation of $MgO$ $(2Mg O_{2} \rightarrow 2MgO)$ intersects the line for the formation of $Al_{2}O_{3}$ $(4/3 Al O_{2} \rightarrow 2/3 Al_{2}O_{3})$ at $1350^{\circ} C$.
Below $1350^{\circ} C$,the $MgO$ line lies below the $Al_{2}O_{3}$ line,meaning $Mg$ is a better reducing agent and can reduce $Al_{2}O_{3}$.
Above $1350^{\circ} C$,the $Al_{2}O_{3}$ line lies below the $MgO$ line,meaning $Al$ is a better reducing agent and can reduce $MgO$. Thus,Assertion $A$ is correct.
Reason $R$ states that the melting and boiling points of magnesium are lower than those of aluminium. While this is a factual statement,it is not the reason for the intersection of the Ellingham lines,which depends on the standard Gibbs free energy of formation $(\Delta G^{\circ})$ of the oxides. Thus,$R$ is correct but is not the correct explanation of $A$.

(B) Pure nitrogen gas is prepared by the thermal decomposition of sodium or barium azide. The reaction for barium azide is:
$Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2(g)$

(D) The acidic character of an oxide depends on the oxidation state of the central atom. As the oxidation state increases,the electronegativity of the central atom increases,which increases the acidic character of the oxide. Therefore,$E_2O_5$ is more acidic than $E_2O_3$.
Statement $-I$ is false.
Down the group,the metallic character increases and the non-metallic character decreases. Since acidic character is associated with non-metallic oxides,the acidic character of $E_2O_3$ decreases down the group.
Statement $-II$ is true.

(C) The stability of the divalent state in lanthanoids is related to the electronic configuration and the reduction potential $E^0_{M^{3+}/M^{2+}}$.
$Eu^{2+}$ has a stable half-filled $f$-orbital configuration $([Xe] 4f^7)$.
$Yb^{2+}$ has a stable fully-filled $f$-orbital configuration $([Xe] 4f^{14})$.
However,comparing the standard reduction potentials: $E^0_{Eu^{3+}/Eu^{2+}} = -0.35 \ V$ and $E^0_{Yb^{3+}/Yb^{2+}} = -1.05 \ V$.
$A$ less negative reduction potential indicates that $Eu^{3+}$ is more easily reduced to $Eu^{2+}$ than $Yb^{3+}$ is to $Yb^{2+}$,making $Eu^{2+}$ the most stable among the given options.

(C) $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand $(SFL)$,causing pairing,resulting in $dsp^2$ hybridization (square planar) and diamagnetic behavior.
$[Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing $4s$ electrons to pair into $3d$,resulting in $3d^{10}$ configuration,$sp^3$ hybridization (tetrahedral) and diamagnetic behavior.
$[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand $(WFL)$,resulting in $sp^3$ hybridization (tetrahedral) and paramagnetic behavior due to $2$ unpaired electrons.
Thus,Statement $I$ is true and Statement $II$ is false.

(D) pesticide is a substance used for destroying insects,fungi,or other organisms harmful to cultivated plants or to animals.
$DDT$ (dichlorodiphenyltrichloroethane) is a well-known organochlorine insecticide.
Organophosphates are a class of chemicals widely used as insecticides.
Dieldrin is an organochlorine insecticide.
Sodium arsenite $(NaAsO_2)$ is primarily used as a wood preservative,a herbicide,or in the manufacturing of glass and pigments,but it is not classified as a pesticide in the context of common agricultural insecticides like the others listed.

(A) The given reaction is a Cannizzaro reaction involving benzaldehyde $(PhCHO)$ in the presence of a base $(NaOD)$ and a deuterated solvent $(D_2O)$.
In the mechanism of the Cannizzaro reaction, the hydride ion $(H^-)$ is transferred from the tetrahedral intermediate of one aldehyde molecule to the carbonyl carbon of another aldehyde molecule.
Since the reaction is carried out in $D_2O$, the oxygen atom of the hydroxyl group in the resulting alcohol will exchange its hydrogen with deuterium, leading to the formation of an $-OD$ group.
However, the hydrogen atom attached to the alpha-carbon of the alcohol $(PhCH_2OD)$ comes from the aldehyde carbon $(PhCHO)$, which does not exchange with the solvent during the hydride transfer step.
Therefore, the product formed is benzyl alcohol with a deuterated hydroxyl group, which is $Ph-CH_2-OD$.

(C) $S_N1$ reactions proceed via the formation of a carbocation intermediate. The stability of the carbocation determines the reactivity.
In $1$-chlorobicyclo$[2.2.1]$heptane,the bridgehead carbon atom is involved in the formation of a carbocation. According to Bredt's rule,a double bond or a carbocation cannot exist at the bridgehead position of a small bicyclic system because it would introduce excessive strain and prevent the carbocation from achieving the required planar $sp^2$ hybridization.
Therefore,the formation of the carbocation at the bridgehead position is extremely difficult,making this compound inactive towards $S_N1$ reactions.

(C) The reaction sequence is as follows:
$1$. Aniline reacts with $Br_2$ water to form $2,4,6-$tribromoaniline as a white precipitate.
$2$. $2,4,6-$Tribromoaniline reacts with $NaNO_2/HCl$ at $0-5^{\circ}C$ to form $2,4,6-$tribromobenzenediazonium chloride.
$3$. $2,4,6-$Tribromobenzenediazonium chloride reacts with $H_3PO_2$ (hypophosphorous acid) and water to undergo reduction,replacing the diazonium group with a hydrogen atom,resulting in $1,3,5-$tribromobenzene.

(B) Primary aliphatic amines react with nitrous acid $(HNO_2)$ at low temperatures $(273-278 \, K)$ to form unstable aliphatic diazonium salts.
Upon raising the temperature to room temperature $(298 \, K)$,these diazonium salts decompose rapidly by losing nitrogen gas to form carbocations,which then react with water to yield alcohols.
The reaction sequence is: $R-NH_2 + HNO_2$ $\xrightarrow{273 \, K} [R-N_2^+Cl^-]$ $\xrightarrow{298 \, K} R-OH + N_2 + HCl$.

(B) copolymer is a polymer formed from two or more different types of monomer units.
Buna-$S$ is a copolymer of $1,3$-butadiene and styrene.
$PHBV$ (poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate) is a copolymer of $3$-hydroxybutanoic acid and $3$-hydroxypentanoic acid.
Butadiene-styrene is another name for Buna-$S$,which is a copolymer.
Neoprene is a homopolymer of chloroprene ($2$-chloro-$1,3$-butadiene).
Therefore,neoprene is not a copolymer.

(B) The $\alpha$-helix is one of the most common secondary structures of proteins.
It is stabilized by intramolecular hydrogen bonding between the carbonyl oxygen $(C=O)$ of one amino acid residue and the amide hydrogen $(N-H)$ of the fourth amino acid residue along the polypeptide chain.
This regular pattern of $H$-bonding gives the $\alpha$-helix its characteristic stability.

(A) $MnO_2 + 4HCl \longrightarrow MnCl_2 + Cl_2 + 2H_2O$
$Cl_2 + 2KI \longrightarrow 2KCl + I_2$
$I_2 + 2Na_2S_2O_3 \longrightarrow 2NaI + Na_2S_4O_6$
Equivalents of $MnO_2 = \text{Equivalents of } Cl_2 = \text{Equivalents of } I_2 = \text{Equivalents of } Na_2S_2O_3$
Equivalents of $Na_2S_2O_3 = \text{Molarity} \times \text{Volume (L)} \times n\text{-factor} = 0.1 \times 0.060 \times 1 = 6 \times 10^{-3} \, eq$
Since the $n$-factor of $MnO_2$ in the reaction $MnO_2 \longrightarrow Mn^{2+}$ is $2$,moles of $MnO_2 = \frac{6 \times 10^{-3}}{2} = 3 \times 10^{-3} \, mol$
Molar mass of $MnO_2 = 55 + 2 \times 16 = 87 \, g/mol$
Mass of $MnO_2 = 3 \times 10^{-3} \times 87 = 0.261 \, g$
$\% \, MnO_2 = \frac{0.261}{2.0} \times 100 = 13.05 \, \%$
Nearest integer is $13$.

(C) Given: $P_A^{\circ} = 50 \ Torr$,$P_B^{\circ} = 100 \ Torr$,$X_A = 0.3$,$X_B = 1 - 0.3 = 0.7$.
According to Raoult's Law,the partial pressures are:
$P_A = P_A^{\circ} X_A = 50 \times 0.3 = 15 \ Torr$.
$P_B = P_B^{\circ} X_B = 100 \times 0.7 = 70 \ Torr$.
Total pressure $P_{total} = P_A + P_B = 15 + 70 = 85 \ Torr$.
The mole fraction of $B$ in the vapour phase $(y_B)$ is given by:
$y_B = \frac{P_B}{P_{total}} = \frac{70}{85} = \frac{14}{17}$.
Comparing this with $\frac{x}{17}$,we get $x = 14$.

(B) The reduction half-reaction for the dichromate ion is:
$Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \longrightarrow 2Cr^{3+} + 7H_{2}O$
From the balanced equation,$1 \ mol$ of $Cr_{2}O_{7}^{2-}$ requires $6 \ mol$ of electrons for reduction.
Since $1 \ mol$ of electrons carries $1 \ Faraday$ of charge,the quantity of electricity required is $6 \ Faraday$.

(B) For a first order reaction,the time $t$ required for completion is given by $t = \frac{1}{k} \ln \left(\frac{[A]_0}{[A]_t}\right)$.
For $67 \, \%$ completion,$[A]_t = [A]_0 - 0.67[A]_0 = 0.33[A]_0 \approx \frac{1}{3}[A]_0$.
Thus,$t_{67 \, \%} = \frac{1}{k} \ln \left(\frac{1}{1/3}\right) = \frac{\ln 3}{k}$.
The half-life is $t_{1/2} = \frac{\ln 2}{k}$.
Therefore,$\frac{t_{67 \, \%}}{t_{1/2}} = \frac{\ln 3}{\ln 2} \approx \frac{1.0986}{0.6931} \approx 1.585$.
Given $t_{67 \, \%} = (x \times 10^{-1}) \times t_{1/2}$,we have $x \times 10^{-1} = 1.585$,which implies $x = 15.85$.
Rounding to the nearest integer,$x = 16$.

(D) Synergic bonding is a characteristic feature of metal carbonyl complexes where the metal-carbon bond has both $\sigma$-donor and $\pi$-acceptor character.
All three complexes,$[Cr(CO)_6]$,$[Mn(CO)_5]$,and $[Mn_2(CO)_{10}]$,contain metal-carbonyl bonds.
Therefore,all $3$ complexes exhibit synergic bonding.
The correct answer is $3$.

(A) The coagulating value is defined as the minimum concentration of an electrolyte in $m\,mol \, L^{-1}$ required to cause coagulation of a sol in $2 \, hours$.
$1$. Calculate the molar mass of $NaCl$: $M = 23.0 + 35.5 = 58.5 \, g \, mol^{-1}$.
$2$. Calculate the molarity of the given $42.12 \% (w/v)$ solution:
$42.12 \% (w/v)$ means $42.12 \, g$ of $NaCl$ in $100 \, mL$ of solution.
Concentration in $g \, L^{-1} = 42.12 \times 10 = 421.2 \, g \, L^{-1}$.
Concentration in $mol \, L^{-1} = \frac{421.2}{58.5} = 7.2 \, mol \, L^{-1} = 7200 \, m\,mol \, L^{-1}$.
$3$. The question asks for the coagulating value,which is the concentration required for coagulation. Given the data provided is insufficient to relate the time $(10 \, hours)$ to the standard definition $(2 \, hours)$ without a specific rate law or empirical relationship,the provided options suggest a calculation based on the concentration itself. However,based on standard textbook problems of this type,the answer is derived as $7200 \, m\,mol \, L^{-1}$. Since this is not an option,and the provided solution was 'Data insufficient',we maintain this assessment.

(A) The correct matches for the ores are as follows:
$A$. Siderite is $FeCO_3$ $(I)$.
$B$. Malachite is $CuCO_3 \cdot Cu(OH)_2$ $(II)$.
$C$. Sphalerite is $ZnS$ $(III)$.
$D$. Calamine is $ZnCO_3$ $(IV)$.
Therefore,the correct sequence is $A-I, B-II, C-III, D-IV$.

(C) The structure of $CuSO_{4} \cdot 5H_{2}O$ is $[Cu(H_{2}O)_{4}]SO_{4} \cdot H_{2}O$.
In this complex,the $Cu(II)$ ion is coordinated by four water molecules through oxygen atoms,forming $Cu-O$ coordinate bonds. Thus,Statement $I$ is correct.
Regarding Statement $II$,the ligands coordinating with the $Cu(II)$ ion are only water molecules,which are $O$-based ligands. The sulfate ion $(SO_{4}^{2-})$ is not directly coordinated to the $Cu(II)$ ion; it is linked via hydrogen bonding. Therefore,Statement $II$ is incorrect.

(C) $1$. Empirical formula calculation: Moles of $C = 52.2/12 = 4.35$,$H = 4.9/1 = 4.9$,$Br = 42.9/80 = 0.536$. Ratio: $C:H:Br = 8:9:1$. Empirical formula is $C_8H_9Br$. Molar mass is $184 \ g/mol$,which matches $C_8H_9Br$.
$2$. Oxidation with $KMnO_4$: Isomer $(A)$ gives benzoic acid,implying it has an alkyl group attached to the benzene ring at the alpha position (e.g.,$C_6H_5-CH(Br)-CH_3$). Isomer $(B)$ gives $p$-bromobenzoic acid,implying it has an ethyl group at the para position relative to the bromine atom (e.g.,$p-Br-C_6H_4-CH_2CH_3$).
$3$. Isomer $(A)$ is optically active: $C_6H_5-CH(Br)-CH_3$ has a chiral center.
$4$. Isomer $(A)$ gives a precipitate with $AgNO_3$: The benzylic bromide is reactive.
$5$. Therefore,$(A)$ is $1$-bromo-ethylbenzene $(H_3C-CHBr-C_6H_5)$ and $(B)$ is $1$-bromo$-4-$ethylbenzene ($p$-ethylbromobenzene).

(D) Aniline $(C_6H_5NH_2)$ is a Lewis base due to the lone pair of electrons on the nitrogen atom.
Friedel-Crafts catalysts like anhydrous $AlCl_3$ are Lewis acids.
When aniline reacts with $AlCl_3$,the lone pair on the nitrogen atom coordinates with the $AlCl_3$ to form a salt,resulting in a positively charged nitrogen atom $(C_6H_5NH_2^+-AlCl_3^-)$.
Because of this positive charge,the nitrogen atom becomes strongly electron-withdrawing,which deactivates the benzene ring towards electrophilic substitution.
Therefore,aniline does not undergo Friedel-Crafts alkylation or acylation.

(A) Dacron (also known as Terylene) is a condensation polymer formed by the polymerization of ethylene glycol $(HO-CH_2-CH_2-OH)$ and terephthalic acid $(HOOC-C_6H_4-COOH)$.
Since the linkage formed between the monomers is an ester linkage $(-COO-)$,it is classified as a polyester polymer.
Thus,Assertion $A$ is correct because Dacron is indeed a polyester.
Reason $R$ is also correct as it correctly identifies the monomers involved in the synthesis of Dacron.
Furthermore,Reason $R$ provides the correct explanation for Assertion $A$ because the presence of ester linkages resulting from the condensation of these specific monomers defines it as a polyester.

(C) The primary structure of a protein refers to the specific sequence of amino acids linked by peptide bonds. This sequence is determined by genetic information and is not disrupted by physical changes like heating or chemical changes,which only affect the higher-level folding (secondary,tertiary,and quaternary structures).

(A) The mixture of chloroxylenol and terpineol is commonly known as $Dettol$,which acts as an antiseptic.
Antiseptics are chemical substances applied to living tissues to prevent the growth of microorganisms.

(B) When $BaCl_{2}$ is added to a salt containing $SO_{3}^{2-}$,a white precipitate of $BaSO_{3}$ is formed: $Ba^{2+} + SO_{3}^{2-} \rightarrow BaSO_{3} \downarrow$.
Upon adding dilute $HCl$,the precipitate dissolves and releases $SO_{2}$ gas,which has a characteristic pungent smell of burning sulphur: $BaSO_{3} + 2HCl \rightarrow BaCl_{2} + H_{2}O + SO_{2} \uparrow$.

(D) For glucose,the van't Hoff factor $i = 1$.
The elevation in boiling point is given by $\Delta T_{b} = K_{b} \cdot m_{1} = 4 \, K$,where $m_{1} = 1.5 \, m$.
So,$K_{b} \cdot 1.5 = 4 \implies K_{b} = \frac{4}{1.5}$.
The depression in freezing point is given by $\Delta T_{f} = K_{f} \cdot m_{2} = 4 \, K$,where $m_{2} = 4.5 \, m$.
So,$K_{f} \cdot 4.5 = 4 \implies K_{f} = \frac{4}{4.5}$.
The ratio $\frac{K_{b}}{K_{f}} = \frac{4 / 1.5}{4 / 4.5} = \frac{4.5}{1.5} = 3$.

(B) The cell reaction is: $H_{2(g)} + Cu^{2+}_{(aq)} \rightarrow 2 H^{+}_{(aq)} + Cu_{(s)}$
Using the Nernst equation: $E_{cell} = E^{\ominus}_{cell} - \frac{0.06}{n} \log \frac{[H^{+}]^2}{[Cu^{2+}]}$
Here,$n = 2$,$E_{cell} = 0.31 \, V$,$E^{\ominus}_{cell} = E^{\ominus}_{Cu^{2+}/Cu} - E^{\ominus}_{H^{+}/H_2} = 0.34 - 0 = 0.34 \, V$.
Given $pH = 3$,so $[H^{+}] = 10^{-3} \, M$.
Substituting the values: $0.31 = 0.34 - \frac{0.06}{2} \log \frac{(10^{-3})^2}{[Cu^{2+}]}$
$-0.03 = -0.03 \log \frac{10^{-6}}{[Cu^{2+}]}$
$1 = \log \frac{10^{-6}}{[Cu^{2+}]}$
$10^1 = \frac{10^{-6}}{[Cu^{2+}]}$
$[Cu^{2+}] = 10^{-7} \, M$.
Comparing with $10^{-x} \, M$,we get $x = 7$.

(A) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.
Comparing this with the given equation $k = (6.5 \times 10^{12} \, s^{-1}) e^{-26000 \, K / T}$,we get $\frac{E_a}{R} = 26000 \, K$.
Substituting $R = 8.314 \, J \, K^{-1} \, mol^{-1}$,we have $E_a = 26000 \times 8.314 \, J \, mol^{-1}$.
$E_a = 216164 \, J \, mol^{-1} = 216.164 \, kJ \, mol^{-1}$.
Rounding to the nearest integer,the activation energy is $216 \, kJ \, mol^{-1}$.

(B) In $[MnBr_{6}]^{4-}$,the oxidation state of $Mn$ is $+2$.
Electronic configuration of $Mn^{2+}$ is $[Ar] 3d^{5}$.
Since $Br^{-}$ is a weak field ligand,the electrons remain unpaired in the $d$-orbitals.
Number of unpaired electrons $(n)$ = $5$.
The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ B.M.$
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.91 \ B.M.$
Rounding off to the closest integer,we get $6 \ B.M.$

(C) The reaction of coordination compounds with $AgNO_{3}$ depends on the number of chloride ions present outside the coordination sphere (ionizable chlorides).
The reaction is given by: $[Co(NH_{3})_{x}Cl_{3-n}]Cl_{n} + nAgNO_{3} \rightarrow nAgCl \downarrow + [Co(NH_{3})_{x}Cl_{3-n}](NO_{3})_{n}$.
Given that $2$ equivalents of $AgCl$ precipitate,it implies that there are $2$ ionizable chloride ions $(n = 2)$.
Thus,the complex is $[Co(NH_{3})_{5}Cl]Cl_{2}$.
Comparing this with the formula $CoCl_{3} \cdot xNH_{3}$,we find that $x = 5$.

(A) The isomers of $C_{4}H_{10}O$ that are alcohols are:
$1$. $CH_{3}CH_{2}CH_{2}CH_{2}OH$ (butan-$1$-ol): Achiral.
$2$. $CH_{3}CH(OH)CH_{2}CH_{3}$ (butan-$2$-ol): Chiral,as the $C_{2}$ carbon is bonded to four different groups $(-H, -OH, -CH_{3}, -CH_{2}CH_{3})$.
$3$. $(CH_{3})_{2}CHCH_{2}OH$ ($2$-methylpropan-$1$-ol): Achiral.
$4$. $(CH_{3})_{3}COH$ ($2$-methylpropan-$2$-ol): Achiral.
Thus,there is only $1$ chiral alcohol with the molecular formula $C_{4}H_{10}O$.

(A) The given reaction is an aldol condensation of cyclohexanone in the presence of a base $(OH^-)$ followed by heating,which leads to the formation of an $\alpha,\beta$-unsaturated ketone.
The product $'P'$ is $2$-cyclohexylidenecyclohexanone.
In this molecule,there is one $C=O$ double bond and one $C=C$ double bond.
Each double bond contains $2$ $\pi$ electrons.
Therefore,the total number of $\pi$ electrons in the product $'P'$ is $2 + 2 = 4$.

(D) Colloidal particles carry an electric charge,which is the primary factor for their stability.
Since all particles in a given colloidal solution carry the same type of charge,they repel each other,preventing them from coming close enough to aggregate or precipitate.
Therefore,the presence of equal and similar charges on colloidal particles provides stability to the colloidal solution.

(A) The atomic number of $Pt$ is $78$.
Following the Aufbau principle,the expected configuration is $[Xe]\, 4f^{14}\, 5d^{8}\, 6s^{2}$.
However,$Pt$ exhibits an exceptional electronic configuration due to the stability of the $d^{9}s^{1}$ arrangement.
Therefore,the correct electronic configuration is $[Xe]\, 4f^{14}\, 5d^{9}\, 6s^{1}$.

(D) The cyanide process (leaching) is primarily used for the extraction of $Gold$ $(Au)$ and $Silver$ $(Ag)$ from their ores.
In the froth flotation process for $Zinc$ sulfide $(ZnS)$,$NaCN$ or $KCN$ is used as a depressant to prevent $ZnS$ from forming froth.
However,$Copper$ $(Cu)$ extraction from its ores (like $CuFeS_2$) typically involves roasting and smelting,not a cyanide-based leaching process.

(B) The reaction of white phosphorus $(P_{4})$ with alkali $(NaOH)$ is given by:
$P_{4} + 3 NaOH + 3 H_{2}O \rightarrow PH_{3} + 3 NaH_{2}PO_{2}$
The product $NaH_{2}PO_{2}$ is the salt of hypophosphorous acid,also known as phosphinic acid $(H_{3}PO_{2})$.
In the structure of phosphinic acid $(H_{3}PO_{2})$,there are two $P-H$ bonds,one $P=O$ bond,and one $P-OH$ bond.
Therefore,the correct answer is phosphinic acid.

(B) The reaction of $2$-bromophenol with $2$ equivalents of an organolithium reagent (like $n-BuLi$ or $t-BuLi$) involves two steps:
$1$. Acid-base reaction: The phenolic $-OH$ proton is acidic and is deprotonated by the first equivalent of the organolithium reagent to form a lithium phenoxide salt.
$2$. Halogen-metal exchange: The second equivalent of the organolithium reagent reacts with the aryl bromide to form an aryllithium species.
Thus,the intermediate $A$ is the dilithium salt of $2$-bromophenol,which has the structure of a benzene ring with an $-OLi$ group at the ortho position relative to a $-Li$ group. This species then reacts with $CO_2$ followed by acidic workup $(H_3O^+)$ to yield salicylic acid.

(A) $R-OH + R'-COOH \longrightarrow R'-COOR + H_2O$
This reaction is a nucleophilic acyl substitution reaction where the alcohol acts as a nucleophile.
Electron withdrawing groups $(EWG)$ on the carboxylic acid increase the electrophilicity of the carbonyl carbon by pulling electron density away,which facilitates the attack of the nucleophile,thereby increasing the rate of esterification.

(A) $1$. In the first reaction,aniline reacts with $Br_2$ in water (excess). Due to the strongly activating nature of the $-NH_2$ group,electrophilic substitution occurs at all ortho and para positions,yielding $2,4,6$-tribromoaniline as the major product $A$.
$2$. In the second reaction,aniline is first acetylated using acetic anhydride to form acetanilide. This protects the $-NH_2$ group and reduces its activating effect. Subsequent bromination with $Br_2$ in $CH_3COOH$ occurs primarily at the para position due to steric hindrance at the ortho positions. Finally,hydrolysis with $HCl$ removes the acetyl group to yield $p$-bromoaniline as the major product $B$.

(C) Polymers that can be stretched and return to their original shape upon the release of force are known as elastomers.
$Buna-N$ is a synthetic rubber,which is an elastomer.
Therefore,it can be stretched and retains its original status on releasing the force.

(B) $DNA$ (Deoxyribonucleic acid) contains the sugar $\beta-D-2-\text{deoxyribose}$.
$RNA$ (Ribonucleic acid) contains the sugar $\beta-D-\text{ribose}$.

(C) The chemical structures of the given compounds are as follows:
$1$. Cimetidine $(C_{10}H_{16}N_6S)$ contains a sulphur atom.
$2$. Ranitidine $(C_{13}H_{22}N_4O_3S)$ contains a sulphur atom.
$3$. Histamine $(C_5H_9N_3)$ is a nitrogenous compound and does not contain a sulphur atom.
$4$. Saccharin $(C_7H_5NO_3S)$ contains a sulphur atom.
Therefore,Histamine is the correct answer.

(C) Phenols are weakly acidic,so Statement $I$ is correct.
Phenols are more acidic than alcohols and water because the phenoxide ion is resonance-stabilized.
While phenols are soluble in $NaOH$ due to the formation of sodium phenoxide,the claim that they are weaker acids than alcohols and water is false.
Therefore,Statement $II$ is incorrect.

(B) $1$. Calculate the mass of acetic acid: $Mass = Density \times Volume = 1.02 \, g \, mL^{-1} \times 1.2 \, mL = 1.224 \, g$.
$2$. Calculate the moles of acetic acid: $Moles = \frac{1.224 \, g}{60 \, g \, mol^{-1}} = 0.0204 \, mol$.
$3$. Calculate molality $(m)$: Since the solution is dilute,we assume the density of the solution is approximately $1 \, g \, mL^{-1}$,so $2 \, L$ of solution is $2 \, kg$ of water. $m = \frac{0.0204 \, mol}{2 \, kg} = 0.0102 \, mol \, kg^{-1}$.
$4$. Use the depression in freezing point formula: $\Delta T_{f} = i \times K_{f} \times m$.
$5$. For acetic acid dissociation $(CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H^{+})$,$i = 1 + \alpha$.
$6$. Substitute values: $0.0198 = (1 + \alpha) \times 1.85 \times 0.0102$.
$7$. Solve for $\alpha$: $1 + \alpha = \frac{0.0198}{1.85 \times 0.0102} \approx 1.04928$.
$8$. $\alpha = 0.04928$,which is $4.928 \% \approx 5 \%$. The nearest integer is $5$.

(A) The reactions at the electrodes are:
At anode: $2H_2O \rightarrow O_2(g) + 4H^+ + 4e^-$
At cathode: $4H^+ + 4e^- \rightarrow 2H_2(g)$
Total charge passed $Q = i \times t = 0.10 \ A \times 2 \times 3600 \ s = 720 \ C$.
Number of moles of electrons $n_e = \frac{Q}{F} = \frac{720}{96500} \approx 0.00746 \ mol$.
From the stoichiometry,$4 \ mol$ of electrons produce $1 \ mol$ of $O_2$ and $2 \ mol$ of $H_2$ (total $3 \ mol$ of gas).
Total moles of gas produced $n_{total} = \frac{n_e}{4} + \frac{n_e}{2} = \frac{3}{4} n_e = 0.75 \times 0.00746 = 0.005595 \ mol$.
Volume at $STP = n_{total} \times 22.7 \ L \ mol^{-1} = 0.005595 \times 22.7 \approx 0.127 \ L = 127 \ cm^3$.

(A) Using the Arrhenius equation: $\ln \left(\frac{k_{310}}{k_{300}}\right) = \frac{E_a}{R} \left(\frac{1}{T_{300}} - \frac{1}{T_{310}}\right)$
$\ln \left(\frac{k_{310}}{k_{300}}\right) = \frac{532611}{8.3} \times \left(\frac{310 - 300}{310 \times 300}\right)$
$\ln \left(\frac{k_{310}}{k_{300}}\right) = \frac{532611}{8.3} \times \frac{10}{93000} = 64170 \times \frac{10}{93000} \approx 6.9$
Since $\ln 10 = 2.3$,then $6.9 = 3 \times 2.3 = 3 \times \ln 10 = \ln 10^3$.
Therefore,$\frac{k_{310}}{k_{300}} = 10^3$,which means $k_{300} = 10^{-3} \, k_{310}$.
Comparing this with $k_{300} = x \times 10^{-3} \, k_{310}$,we get $x = 1$.

(A) The reaction $2Na_2CrO_4 + 2H^{+} \rightarrow Na_2Cr_2O_7 + 2Na^{+} + H_2O$ shows that the product $B$ is sodium dichromate $(Na_2Cr_2O_7)$.
The structure of the dichromate ion $(Cr_2O_7^{2-})$ consists of two $CrO_4$ tetrahedra sharing a common oxygen atom at one corner.
In this structure,there are $6$ terminal oxygen atoms (three bonded to each chromium atom) and $1$ bridging oxygen atom.
Therefore,the number of terminal oxygen atoms is $6$.