JEE Main 2026 Chemistry Question Paper with Answer and Solution

(B) $(1)$. Wavelength of $A = 400 \ nm = 400 \times 10^{-9} \ m$.
$(2)$. Wavelength of $B$ $(\lambda) = \frac{c}{\nu} = \frac{3 \times 10^{8} \ m/s}{10^{16} \ s^{-1}} = 3 \times 10^{-8} \ m = 30 \ nm$.
$(3)$. Wavelength of $C$ $(\lambda) = \frac{1}{\bar{\nu}} = \frac{1}{10^{4} \ cm^{-1}} = 10^{-4} \ cm = 10^{-6} \ m = 1000 \ nm$.
Comparing wavelengths: $\lambda_{C} (1000 \ nm) > \lambda_{A} (400 \ nm) > \lambda_{B} (30 \ nm)$.
Since Energy $(E) = \frac{hc}{\lambda}$,$E \propto \frac{1}{\lambda}$.
Therefore,the order of energy is $E_{B} > E_{A} > E_{C}$.

(D) Manganate ion: $MnO_4^{2-}$; Permanganate ion: $MnO_4^-$.
$(A)$ Both $MnO_4^{2-}$ and $MnO_4^-$ have tetrahedral geometry due to $d^3s$ hybridization of $Mn$. This statement is correct.
$(B)$ In $MnO_4^{2-}$,$Mn$ is in $+6$ oxidation state. In $MnO_4^-$,$Mn$ is in $+7$ oxidation state. The statement says $+7$ and $+6$ respectively,which is incorrect.
$(C)$ Oxidation of $Mn^{2+}$ by peroxodisulphate $(S_2O_8^{2-})$ yields permanganate $(MnO_4^-)$,not manganate. This statement is incorrect.
$(D)$ $MnO_4^{2-}$ ($d^1$ configuration) is paramagnetic. $MnO_4^-$ ($d^0$ configuration) is diamagnetic. This statement is correct.
$(E)$ Acidified permanganate acts as an oxidizing agent,not a reducing agent. It oxidizes oxalate,nitrite,and iodide ions. This statement is incorrect.
Therefore,only statements $(A)$ and $(D)$ are correct.

(C) Let us analyze each reaction:
$(A)$ $CH_{4} + O_{2} \xrightarrow{Mo_{2}O_{3}} HCHO$ (Formaldehyde is produced,not an alcohol).
$(B)$ $2CH_{3}CH_{3} + 3O_{2} \xrightarrow[\Delta]{(CH_{3}COO)_{2}Mn} 2CH_{3}COOH$ (Ethanoic acid is produced).
$(C)$ $(CH_{3})_{3}CH \xrightarrow{KMnO_{4}} (CH_{3})_{3}COH$ (tert-Butyl alcohol is produced).
$(D)$ $2CH_{4} + O_{2} \xrightarrow[523 \ K, 100 \ atm.]{Cu} 2CH_{3}OH$ (Methanol is produced).
$(E)$ $CH_{3}CH=CHCH_{3} \xrightarrow{KMnO_{4}/H^{+}} 2CH_{3}COOH$ (Ethanoic acid is produced).
Thus,reactions $(C)$ and $(D)$ produce alcohol as the product.

(B) The relationship between equilibrium constant $K$ and temperature $T$ is given by the van't Hoff equation: $\ln K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}$.
Converting to base $10$ logarithm: $log_{10} K = -\frac{\Delta H^{\circ}}{2.303RT} + \frac{\Delta S^{\circ}}{2.303R}$.
Comparing this with the linear equation $y = mx + c$,where $y = log_{10} K$ and $x = \frac{1}{T}$:
The slope $m = -\frac{\Delta H^{\circ}}{2.303R}$.
The $y$-intercept $c = \frac{\Delta S^{\circ}}{2.303R}$.
Thus,the intercept and slope are $\frac{\Delta S^{\circ}}{2.303R}$ and $-\frac{\Delta H^{\circ}}{2.303R}$ respectively.

(D) For the reaction $A_{(g)} \rightleftharpoons B_{(g)}$,the equilibrium constant $K_c = \frac{[B]}{[A]} = \frac{0.375}{0.5} = 0.75$.
Initial moles of $A$ in $1 \text{ L}$ flask $= 0.5 \text{ mol}$.
Initial moles of $B$ in $1 \text{ L}$ flask $= 0.375 \text{ mol}$.
After adding $0.1 \text{ mol}$ of $A$,total moles of $A = 0.5 + 0.1 = 0.6 \text{ mol}$.
Let $x$ be the amount of $A$ reacted to reach new equilibrium.
New $[A] = 0.6 - x$ and New $[B] = 0.375 + x$.
$K_c = \frac{0.375 + x}{0.6 - x} = 0.75$.
$0.375 + x = 0.45 - 0.75x$.
$1.75x = 0.075 \Rightarrow x = 0.0428$.
New $[A] = 0.6 - 0.0428 = 0.557 \text{ M}$ and New $[B] = 0.375 + 0.0428 = 0.418 \text{ M}$.

(C) Metallic character increases down a group and decreases across a period from left to right. Among the given elements ($N$,$P$,$O$,$S$,$Cl$,$F$),$P$ (Phosphorus) is the most metallic as it is the furthest to the left and down in the periodic table compared to the others. Its valence shell configuration is $3s^2 3p^3$,so it has $5$ valence electrons.
$F$ (Fluorine) is the most electronegative and least metallic element in the list. Its valence shell configuration is $2s^2 2p^5$,so it has $7$ valence electrons.
Therefore,the number of valence electrons for the most metallic $(P)$ and least metallic $(F)$ elements are $5$ and $7$ respectively.

(D) The $-NO_{2}$ group is a strong electron-withdrawing group due to its $-I$ and $-M$ effects.
$1$. It reduces the electron density of the benzene ring,thereby deactivating it towards electrophilic aromatic substitution (Statement $B$ is correct).
$2$. It increases the susceptibility of the ring towards nucleophilic attack by stabilizing the intermediate carbanion (Meisenheimer complex) through its $-M$ effect,thereby activating the ring towards nucleophilic aromatic substitution (Statement $C$ is correct).
Therefore,statements $B$ and $C$ are correct.

(C) The dissociation equilibrium is $HA \rightleftharpoons H^{+} + A^{-}$.
For a weak acid,the concentration of $H^{+}$ is given by $[H^{+}] = \sqrt{K_{a} \times c}$.
Taking negative logarithm on both sides,we get $pH = \frac{1}{2} (pK_{a} - \log c)$.
Given $pK_{a} = 4$ and concentration $c = 10 \ mM = 10 \times 10^{-3} \ M = 10^{-2} \ M$.
Substituting the values: $pH = \frac{1}{2} [4 - \log(10^{-2})]$.
$pH = \frac{1}{2} [4 - (-2)] = \frac{1}{2} [6] = 3$.

(D) $1$. Calculate percentage of $C$:
Moles of $CO_2 = \frac{1.32 \ g}{44 \ g \ mol^{-1}} = 0.03 \ mol$.
Mass of $C = 0.03 \ mol \times 12 \ g \ mol^{-1} = 0.36 \ g$.
$\% C = \frac{0.36 \ g}{1.0 \ g} \times 100 = 36\%$.
$2$. Calculate percentage of $Br$:
Moles of $AgBr = \frac{0.75 \ g}{188 \ g \ mol^{-1}} \approx 0.00399 \ mol$.
Mass of $Br = 0.00399 \ mol \times 80 \ g \ mol^{-1} \approx 0.3192 \ g$.
$\% Br = \frac{0.3192 \ g}{0.53 \ g} \times 100 \approx 60.2\%$.
$3$. Calculate percentage of $H$:
$\% H = 100 - (\% C + \% Br) = 100 - (36 + 60.2) = 3.8\%$.
Rounding to the nearest integer,we get $4\%$.

(C) First,balance the redox reaction in acidic medium:
Oxidation half-reaction: $BH_{4}^{-} + 3H_{2}O \rightarrow H_{2}BO_{3}^{-} + 8H^{+} + 8e^{-}$
Reduction half-reaction: $ClO_{3}^{-} + 6H^{+} + 6e^{-} \rightarrow Cl^{-} + 3H_{2}O$
To balance the electrons,multiply the oxidation half-reaction by $3$ and the reduction half-reaction by $4$:
$3BH_{4}^{-} + 9H_{2}O \rightarrow 3H_{2}BO_{3}^{-} + 24H^{+} + 24e^{-}$
$4ClO_{3}^{-} + 24H^{+} + 24e^{-} \rightarrow 4Cl^{-} + 12H_{2}O$
Adding these,we get the balanced equation: $3BH_{4}^{-} + 4ClO_{3}^{-} \rightarrow 3H_{2}BO_{3}^{-} + 4Cl^{-} + 3H_{2}O$
In the Nernst equation,$n$ represents the number of moles of electrons transferred in the balanced redox reaction.
From the balanced equation,$n = 24$.

(C) In a period,moving from left to right,the atomic radius decreases due to an increase in effective nuclear charge.
In period $4$,the elements range from $K$ (Group $1$) to $Kr$ (Group $18$).
Excluding noble gases (which have larger van der Waals radii),the element with the largest atomic radius is $K$ (Potassium) and the element with the smallest atomic radius is $Br$ (Bromine).
Therefore,the correct pair is $K$ & $Br$.

(C) node is a region where the probability of finding an electron is zero. For a radial wave function,this corresponds to the point where the wave function $\psi(r)$ crosses the x-axis,i.e.,$\psi(r) = 0$.
In the radial wave function plot for a $2s$ orbital,the curve crosses the x-axis at a specific distance from the nucleus. Based on standard representations of this graph,point $C$ typically marks the intersection where $\psi(r) = 0$,representing the spherical node.

(A) For the Balmer series,the wave number $\bar{v}$ is given by the formula: $\bar{v} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} \right]$,where $n = 3, 4, 5, \dots$
For $n = 3$: $\bar{v} = R \left[ \frac{1}{4} - \frac{1}{9} \right] = R \left[ \frac{9-4}{36} \right] = \frac{5R}{36}$
For $n = 4$: $\bar{v} = R \left[ \frac{1}{4} - \frac{1}{16} \right] = R \left[ \frac{4-1}{16} \right] = \frac{3R}{16}$
For $n = 5$: $\bar{v} = R \left[ \frac{1}{4} - \frac{1}{25} \right] = R \left[ \frac{25-4}{100} \right] = \frac{21R}{100}$
Thus,the set of spectral lines is $\frac{5R}{36}, \frac{3R}{16}, \frac{21R}{100}$.

(D) The reaction $Ph-CH=CH_2 + HBr \xrightarrow{(PhCOO)_2} Ph-CH_2-CH_2-Br$ is an Anti-Markovnikov addition.
$A$. Correct: The reaction proceeds via a free radical mechanism where the more stable benzylic radical intermediate is formed.
$B$. Incorrect: The role of peroxide is to generate the benzoyloxy radical,which then abstracts $H$ from $HBr$ to generate the bromine radical $(\dot{Br})$,not the hydrogen radical $(\dot{H})$.
$C$. Correct: As shown in the mechanism,the benzoyloxy radical decomposes to form $Ph\dot{}$ radical,which abstracts $H$ from $HBr$ to form benzene $(Ph-H)$ as a byproduct.
$D$. Incorrect: $1-\text{Bromo}-2-\text{phenylethane}$ is the major product in this Anti-Markovnikov addition.
$E$. Correct: In the absence of peroxide,the reaction follows the electrophilic addition mechanism,which proceeds via a carbocation intermediate (Markovnikov addition).
Therefore,statements $A, C,$ and $E$ are correct.

(A) Statement $I$: $BF_4^-$ ($sp^3$,tetrahedral),$SiF_4$ ($sp^3$,tetrahedral),and $XeF_4$ ($sp^3d^2$,square planar) have identical bond lengths due to symmetry. $SF_4$ ($sp^3d$,see-saw) has axial and equatorial bonds of different lengths. Thus,only $1$ species $(SF_4)$ has unequal bond lengths. Statement $I$ is false.
Statement $II$: Bond orders are: $O_2^+$ $(2.5)$,$O_2^-$ $(1.5)$,$O_2^{2-}$ $(1.0)$,$F_2$ $(1.0)$. $O_2^+$ has the highest bond order,not $O_2^-$. Statement $II$ is false.

Species	Bond Order
$O_2^+$	$2.5$
$O_2^-$	$1.5$
$O_2^{2-}$	$1.0$
$F_2$	$1.0$

(B) $1$. Compound $'X'$ $(C_3H_6Cl_2)$ reacts with excess $NaNH_2$ to form propyne $(CH_3-C \equiv CH)$ as compound $'Y'$.
$2$. Hydration of propyne $(Y)$ with $dil. H_2SO_4/Hg^{2+}$ gives acetone $(CH_3-CO-CH_3)$ as compound $'Z'$. Acetone contains a methyl ketone group and gives a yellow precipitate of iodoform $(CHI_3)$ with $NaOI$. Thus,Statement $I$ is true.
$3$. Cyclic trimerization of propyne $(Y)$ using a red hot iron tube gives $1,3,5$-trimethylbenzene (mesitylene) as compound $'Q'$ $(C_9H_{12})$.
$4$. In $1,3,5$-trimethylbenzene,there are $3$ aromatic $H$ atoms (on the ring) and $9$ aliphatic $H$ atoms (in the three methyl groups). The ratio of aromatic to aliphatic $H$ atoms is $3:9 = 1:3$. Thus,Statement $II$ is true.

(C) For a buffer solution of a weak base and its salt,the Henderson-Hasselbalch equation is: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Given $pH = 9$,so $pOH = 14 - 9 = 5$.
Given $pK_b = 5.699$,so $5 = 5.699 + \log \frac{[Salt]}{[Base]}$.
$\log \frac{[Salt]}{[Base]} = 5 - 5.699 = -0.699$.
Since $\log 5 = 0.699$,then $\log \frac{[Salt]}{[Base]} = -\log 5 = \log \frac{1}{5}$.
Therefore,$\frac{[Salt]}{[Base]} = \frac{1}{5}$.
Reaction: $B + HCl \rightarrow BH^+ + Cl^-$.
Initial moles: $n_B = 0.02y$,$n_{HCl} = 0.02x$.
After reaction: $n_{BH^+} = 0.02x$,$n_{B, remaining} = 0.02y - 0.02x$.
Ratio: $\frac{0.02x}{0.02y - 0.02x} = \frac{1}{5}$ $\Rightarrow \frac{x}{y-x} = \frac{1}{5}$ $\Rightarrow 5x = y - x$ $\Rightarrow y = 6x$.
Total volume: $x + y = 100 \ mL$.
Substituting $y$: $x + 6x = 100$ $\Rightarrow 7x = 100$ $\Rightarrow x = 14.28 \approx 14.3 \ mL$.
$y = 100 - 14.3 = 85.7 \ mL$.

(D) The reaction of benzene with $Br_2$ in the presence of $FeBr_3$ gives bromobenzene $(A)$.
Nitration of bromobenzene with conc. $HNO_3$ and conc. $H_2SO_4$ yields a mixture of ortho-bromonitrobenzene $(B)$ and para-bromonitrobenzene $(C)$.
Since ortho- and para-isomers have different boiling points due to differences in their polarity and intermolecular forces,they are separated by fractional distillation.

(C) The stability of a carbanion is directly proportional to the percentage of $s$-character in the hybrid orbital of the carbon atom bearing the negative charge.
$1$. In $CH\equiv C^{-}$,the carbon is $sp$ hybridized ($50\% \ s$-character).
$2$. In $CH_2=CH^{-}$,the carbon is $sp^2$ hybridized ($33.3\% \ s$-character).
$3$. In $CH_3-CH_2^{-}$,the carbon is $sp^3$ hybridized ($25\% \ s$-character).
Higher $s$-character means the electrons are held closer to the nucleus,increasing stability.
Therefore,the correct order of stability is: $CH\equiv C^{-} > CH_2=CH^{-} > CH_3-CH_2^{-}$.

(A) For an isothermal process involving an ideal gas, the internal energy change $\Delta U$ and enthalpy change $\Delta H$ are zero because they depend only on temperature, which remains constant.
For an isothermal reversible expansion, the work done $w$ is given by $w = -nRT \ln(V_2/V_1) = -P_1V_1 \ln(P_1/P_2)$.
Given $P_1 = 0.5 \ MPa = 0.5 \times 10^6 \ Pa$, $V_1 = 20.0 \ dm^3 = 20.0 \times 10^{-3} \ m^3$, and $P_2 = 0.2 \ MPa$.
$w = -(0.5 \times 10^6 \ Pa) \times (20.0 \times 10^{-3} \ m^3) \times \ln(0.5/0.2)$.
$w = -10000 \ J \times \ln(2.5) = -10000 \times 2.303 \times \log(2.5)$.
Since $\log(2.5) = \log(5/2) = \log(5) - \log(2) = 0.6989 - 0.3010 = 0.3979$.
$w = -10000 \times 2.303 \times 0.3979 \approx -9166 \ J \approx -9.1 \ kJ$.
Since $\Delta U = q + w = 0$, then $q = -w = 9.1 \ kJ$.

(C) The balanced chemical equation for the oxidation of $I^{-}$ to $I_{2}$ by acidified $K_{2}Cr_{2}O_{7}$ is: $Cr_{2}O_{7}^{2-} + 14H^{+} + 6I^{-} \rightarrow 2Cr^{3+} + 3I_{2} + 7H_{2}O$. Here,$6$ electrons are involved,so $X = 6$.
The balanced chemical equation for the oxidation of $S^{2-}$ to $S$ by acidified $K_{2}Cr_{2}O_{7}$ is: $Cr_{2}O_{7}^{2-} + 3S^{2-} + 14H^{+} \rightarrow 3S + 2Cr^{3+} + 7H_{2}O$. Here,$6$ electrons are involved,so $Y = 6$.
Therefore,the sum $X + Y = 6 + 6 = 12$.

(B) Step $1$: Calculate the pressure of dry $N_2$ gas. $P_{N_2} = P_{total} - P_{aqueous} = 715 \ mm - 15 \ mm = 700 \ mm = \frac{700}{760} \ atm$.
Step $2$: Use the ideal gas equation $PV = nRT$ to find the moles of $N_2$. $n_{N_2} = \frac{PV}{RT} = \frac{(700/760 \ atm) \times (0.070 \ L)}{0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} \times 300 \ K} \approx 0.00263 \ mol$.
Step $3$: Calculate the mass of $N_2$. $W_{N_2} = n_{N_2} \times 28 \ g/mol = 0.00263 \times 28 \approx 0.07364 \ g$.
Step $4$: Calculate the percentage of nitrogen. $\% \ N = \frac{W_{N_2}}{W_{compound}} \times 100 = \frac{0.07364 \ g}{0.50 \ g} \times 100 \approx 14.73 \%$.
Rounding to the nearest integer,the answer is $15 \% $.

(B) The energy of a spectral line is given by $\Delta E = 13.6 \times Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \text{ eV}$.
For the Lyman series,the lowest energy line $(L_1)$ corresponds to the transition from $n_2 = 2$ to $n_1 = 1$: $\Delta E(L_1) = 13.6 \times 1^2 (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 \times \frac{3}{4} \text{ eV}$.
For the Balmer series,the lowest energy line $(B_1)$ corresponds to the transition from $n_2 = 3$ to $n_1 = 2$: $\Delta E(B_1) = 13.6 \times 1^2 (\frac{1}{2^2} - \frac{1}{3^2}) = 13.6 \times (\frac{1}{4} - \frac{1}{9}) = 13.6 \times \frac{5}{36} \text{ eV}$.
Given $\Delta E(L_1) = x \times \Delta E(B_1)$,we have $x = \frac{\Delta E(L_1)}{\Delta E(B_1)} = \frac{3/4}{5/36} = \frac{3}{4} \times \frac{36}{5} = \frac{27}{5} = 5.4$.
To express $x$ as $x \times 10^{-1}$,we have $5.4 = 54 \times 10^{-1}$.
Therefore,the value is $54$.

(D) For the precipitation of $YS_{(s)}$,the condition is $[Y^{2+}][S^{2-}] \geq K_{sp}(YS)$.
Given $[Y^{2+}] = 0.01 \ M$ and $K_{sp}(YS) = 4 \times 10^{-16}$,we have $[S^{2-}] \geq \frac{4 \times 10^{-16}}{0.01} = 4 \times 10^{-14} \ M$.
For the dissociation of $H_2S$,the equilibrium is $H_2S_{(aq)} \rightleftharpoons 2H^{+}_{(aq)} + S^{2-}_{(aq)}$.
The equilibrium constant expression is $\frac{[S^{2-}][H^{+}]^2}{[H_2S]} = K_{a1} \times K_{a2} = 1.0 \times 10^{-21}$.
Substituting the values: $\frac{(4 \times 10^{-14})[H^{+}]^2}{0.1} = 1.0 \times 10^{-21}$.
$[H^{+}]^2 = \frac{1.0 \times 10^{-22}}{4 \times 10^{-14}} = 0.25 \times 10^{-8} = 25 \times 10^{-10}$.
$[H^{+}] = 5 \times 10^{-5} \ M$.
$pH = -\log[H^{+}] = -\log(5 \times 10^{-5}) = 5 - \log 5 = 5 - 0.70 = 4.3$.
The nearest integer value for the pH is $4$.

(D) Statement $I$ is true: Metallic character increases down a group and decreases across a period from left to right. $K$ ($Group$ $1$) is the most metallic,followed by $Mg$ ($Group$ $2$),$Al$ ($Group$ $13$),and $B$ ($Group$ $13$,metalloid). Thus,$K > Mg > Al > B$ is correct.
Statement $II$ is false: While atomic radius is greater than cationic radius,it is smaller than anionic radius. Therefore,it is not always greater than the ionic radius for any element.

(C) The stability of a carbanion is increased by electron-withdrawing groups $(EWG)$ and decreased by electron-donating groups $(EDG)$.
$IV$ $(p-CHO)$: $-CHO$ is a strong $EWG$ ($-M$ effect),providing maximum stability.
$I$ $(p-Br)$: $-Br$ is an $EWG$ ($-I$ effect),providing stability.
$II$ $(C_6H_5-CH_2^-)$: Reference structure.
$V$ $(p-CH_3)$: $-CH_3$ is an $EDG$ ($+I$ and hyperconjugation),decreasing stability.
$III$ $(p-CH_3O)$: $-OCH_3$ is a strong $EDG$ ($+M$ effect),providing minimum stability.
Thus,the decreasing order of stability is $IV > I > II > V > III$.

(D) For a reversible isothermal expansion,the work done is given by $w = -nRT \ln \frac{V_f}{V_i}$. Thus,$(A) \rightarrow (II)$.
$(B)$ For free expansion,the external pressure $p_{ex} = 0$,so $w = -p_{ex} \Delta V = 0$. Thus,$(B) \rightarrow (I)$.
$(C)$ For irreversible expansion,the work done is $w = -p_{ex}(V_f - V_i)$. Thus,$(C) \rightarrow (III)$.
$(D)$ For irreversible compression,the work done is $w = -p_{ex}(V_f - V_i) = -p_{ex}(-(V_i - V_f)) = -p_{ex}(V_i - V_f)$ is not standard,but based on the provided options,the work done is $w = -p_{ex}(V_f - V_i)$. However,looking at the options,$(D)$ corresponds to $(IV)$ as $w = -p_{ex}(V_i - V_f)$ is the expression for compression where $V_i > V_f$. Thus,$(D) \rightarrow (IV)$.

(D) $CuCl_{2}$ is a metal chloride that is soluble in water at room temperature.
$AgCl$ is known to be sparingly soluble in water at room temperature.
$PbCl_{2}$ is a metal chloride that is insoluble in cold water but soluble in hot water.
Therefore,the correct sequence is $x = CuCl_{2}$,$y = AgCl$,and $z = PbCl_{2}$.

(A) $1$. $IF_3$: Iodine has $7$ valence electrons. It forms $3$ bonds with $F$ atoms and has $2$ lone pairs. Steric number = $3 + 2 = 5$ ($sp^3d$ hybridization). Due to $2$ lone pairs,it is $T$-shaped. (Correct)
$2$. $IF_5$: Iodine has $7$ valence electrons. It forms $5$ bonds with $F$ atoms and has $1$ lone pair. Steric number = $5 + 1 = 6$ ($sp^3d^2$ hybridization). Due to $1$ lone pair,it is square pyramidal. (Correct)
$3$. $IF_7$: Iodine has $7$ valence electrons. It forms $7$ bonds with $F$ atoms and has $0$ lone pairs. Steric number = $7 + 0 = 7$ ($sp^3d^3$ hybridization). It is pentagonal bipyramidal. (Correct)
$4$. $ClO_4^-$: Chlorine has $7$ valence electrons. It forms $4$ bonds with $O$ atoms and has $0$ lone pairs. Steric number = $4 + 0 = 4$ ($sp^3$ hybridization). It is tetrahedral,not square planar. (Incorrect)
Thus,$A$,$B$,and $C$ are correct.

(B) Dipole moment: $NF_3 (0.24 \ D) < NH_3 (1.47 \ D)$. Statement $A$ is incorrect.
$(B)$ $BeH_2$ is $sp$ hybridized and linear,so its dipole moment is $0$. Statement $B$ is correct.
$(C)$ Bond order of $O_2^{2-}$ is $\frac{10-8}{2} = 1$. Bond order of $F_2$ is $\frac{8-6}{2} = 1$. Statement $C$ is correct.
$(D)$ In $O_3$,the central oxygen atom has a formal charge of $+1$. Statement $D$ is incorrect.
$(E)$ In $NO_2$,nitrogen has an odd number of electrons ($7$ valence electrons),so it does not follow the octet rule. Statement $E$ is incorrect.
Therefore,only $B$ and $C$ are correct.

(A) The stability of alkenes is determined by the number of hyperconjugative structures (alpha-hydrogens) and steric factors.
$I$: Tetrasubstituted alkene,$12 \ \alpha-H$,most stable.
$III$: Trisubstituted alkene,$9 \ \alpha-H$.
$II$: Disubstituted trans-alkene,$6 \ \alpha-H$,more stable than cis due to less steric hindrance.
$IV$: Disubstituted cis-alkene,$6 \ \alpha-H$,less stable than trans.
Thus,the decreasing order of stability is $I > III > II > IV$.

(A) Given:
$A \rightarrow D$; $\Delta_{t} H = (+ve) = E_{D} - E_{A} \therefore E_{D} > E_{A}$.
Mechanism:

$A \rightarrow B$; $\Delta_{r} H = (+ve) \Rightarrow E_{B} > E_{A}$

$B \rightarrow C$; $\Delta_{r} H = (-ve) \Rightarrow E_{C} < E_{B}$

$C \rightarrow D$; $\Delta_{r} H = (-ve) \Rightarrow E_{D} < E_{C}$

Since the overall reaction is endothermic,the final energy level of $D$ must be higher than the initial energy level of $A$. However,the intermediate steps involve exothermic processes. Based on the energy relations $E_{B} > E_{A}$,$E_{C} < E_{B}$,and $E_{D} < E_{C}$,the potential energy profile must show $B$ as a local maximum relative to $A$ and $C$,and $C$ as a local minimum relative to $B$ and $D$. The correct graph shows the energy increasing from $A$ to $B$,decreasing from $B$ to $C$,and decreasing further from $C$ to $D$,while maintaining $E_{D} > E_{A}$ overall.

(A) For a diprotic acid $H_2X$,the ionization steps are:
$H_2X \rightleftharpoons H^{+} + HX^{-}$,$K_{a1} = 2.5 \times 10^{-8}$
$HX^{-} \rightleftharpoons H^{+} + X^{2-}$,$K_{a2} = 1.0 \times 10^{-13}$
Since $K_{a1} \gg K_{a2}$,the concentration of $H^{+}$ is primarily determined by the first dissociation.
For $0.1 \ M$ $H_2X$: $[H^{+}] \approx \sqrt{K_{a1} \times C} = \sqrt{2.5 \times 10^{-8} \times 0.1} = \sqrt{2.5 \times 10^{-9}} = 5 \times 10^{-5} \ M$.
The second dissociation constant is $K_{a2} = \frac{[H^{+}][X^{2-}]}{[HX^{-}]}$.
Since $[H^{+}] \approx [HX^{-}]$,we have $[X^{2-}] = K_{a2} = 1.0 \times 10^{-13} \ M$.
Expressing this in the form $Y \times 10^{-15} \ M$:
$1.0 \times 10^{-13} = 100 \times 10^{-15} \ M$.
Thus,$Y = 100$.

(A) Statement $I$: The bond dissociation enthalpy of $F_{2}$ is lower than $Cl_{2}$ and $Br_{2}$ due to high inter-electronic repulsion between the lone pairs of small $F$ atoms. Thus,the correct order is $Cl_{2} > Br_{2} > F_{2} > I_{2}$. Statement $I$ is true.
Statement $II$: According to Fajan's rule,higher oxidation state of the metal cation leads to greater covalent character. Therefore,$SnCl_{4} > SnCl_{2}$ and $PbCl_{4} > PbCl_{2}$ are correct. However,for uranium halides,$UF_{6}$ has a higher oxidation state $(+6)$ than $UF_{4}$ $(+4)$,so $UF_{6} > UF_{4}$ is the correct order. The statement claims $UF_{4} > UF_{6}$,which is false. Thus,Statement $II$ is false.

(A) Statement-$I$: Compound $(X)$ contains a carboxylic acid group $(-COOH)$,which is acidic enough to react with $NaHCO_3$ to evolve $CO_2$ gas,thus it dissolves. It also has two chiral centers (marked with $*$),so Statement-$I$ is true.
Statement-$II$: Compound $(Y)$ is $CH_3-CH=C=O$. The structure is $CH_3-CH=C=O$. The first carbon $(-CH_3)$ is $sp^3$,the second carbon $(-CH=)$ is $sp^2$,the third carbon $(=C=)$ is $sp$,and the fourth carbon $(=O)$ is not a carbon atom. Thus,it has one $sp^3$,one $sp^2$,and one $sp$ hybridized carbon. Therefore,Statement-$II$ is false.

(A) The bond lengths of the given bonds are approximately as follows:
$C-H(A) \approx 107 \ pm$
$C\equiv N(D) \approx 116 \ pm$
$C=O(C) \approx 121 \ pm$
$C-O(B) \approx 143 \ pm$
Comparing these values, the increasing order of bond length is $A < D < C < B$.

(B) Statement-$I$: The atomic radius of $Mg$ $(160 \ pm)$ is greater than $Al$ $(143 \ pm)$. The ionic radius of $Mg^{2+}$ $(72 \ pm)$ is greater than $Al^{3+}$ $(54 \ pm)$. Thus, the correct order is $Mg > Al > Mg^{2+} > Al^{3+}$. Therefore, Statement-$I$ is false.
Statement-$II$: The magnitude of electron gain enthalpy generally increases across a period and decreases down a group. Due to small size and inter-electronic repulsion, oxygen $(O)$ has a lower electron gain enthalpy than sulfur $(S)$. The correct order of magnitude of electron gain enthalpy is $Cl > Br > S > O$. Therefore, Statement-$II$ is true.

(A) The thermochemical equation for the formation of methane is: $C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)}$
The enthalpy of formation is given by: $\Delta_{f}H^{\Theta} = \sum \text{Bond Enthalpies of Reactants} - \sum \text{Bond Enthalpies of Products}$
$-X = [\Delta H_{sub}(C) + 2 \times B.E.(H-H)] - [4 \times B.E.(C-H)]$
Substituting the given values: $-X = Y + 2Z - 4 \times B.E.(C-H)$
Rearranging for $B.E.(C-H)$: $4 \times B.E.(C-H) = X + Y + 2Z$
$B.E.(C-H) = \frac{X+Y+2Z}{4}$

(A) The molar mass of magnesium pyrophosphate $(Mg_2P_2O_7)$ is: $2 \times 24 + 2 \times 31 + 7 \times 16 = 48 + 62 + 112 = 222 \ g \ mol^{-1}$.
The moles of $Mg_2P_2O_7$ formed = $\frac{1.79 \ g}{222 \ g \ mol^{-1}} \approx 0.008063 \ mol$.
Since each mole of $Mg_2P_2O_7$ contains $2 \ mol$ of phosphorus atoms,the moles of $P$ = $2 \times 0.008063 = 0.016126 \ mol$.
Mass of phosphorus = $0.016126 \ mol \times 31 \ g \ mol^{-1} \approx 0.4999 \ g$.
Percentage of phosphorus = $\frac{\text{Mass of } P}{\text{Mass of compound } X} \times 100 = \frac{0.4999}{1.00} \times 100 = 49.99\%$.
Rounding to the nearest integer,we get $50\%$.

(A) The balanced chemical equation for the reaction is: $MnO_{2(s)} + 4HCl_{(aq)} \rightarrow MnCl_{2(aq)} + Cl_{2(g)} + 2H_{2}O_{(l)}$
Calculate the molar mass of $MnO_{2}$: $55 + (2 \times 16) = 87 \ g \ mol^{-1}$.
Calculate the number of moles of $MnO_{2}$ used: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{8.7 \ g}{87 \ g \ mol^{-1}} = 0.1 \ mol$.
According to the stoichiometry of the reaction,$1 \ mol$ of $MnO_{2}$ produces $1 \ mol$ of $Cl_{2}$.
Therefore,$0.1 \ mol$ of $MnO_{2}$ will produce $0.1 \ mol$ of $Cl_{2}$.
Calculate the molar mass of $Cl_{2}$: $2 \times 35.5 = 71 \ g \ mol^{-1}$.
Calculate the weight of $Cl_{2}$ liberated: $0.1 \ mol \times 71 \ g \ mol^{-1} = 7.1 \ g$.

(A) . $2-$Methylpropene $(C_4H_8)$ and but$-1-$ene $(C_4H_8)$ differ in the carbon skeleton,so they are chain isomers $(III)$.
$B$. $Cis-$but$-2-$ene and $trans-$but$-2-$ene are geometric isomers,which are a type of stereoisomers $(I)$.
$C$. $2-$Butanol (alcohol) and diethyl ether (ether) have different functional groups,so they are functional group isomers $(IV)$.
$D$. But$-1-$ene and but$-2-$ene differ in the position of the double bond,so they are position isomers $(II)$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(D) The energy of a spectral line is given by $\Delta E = 13.6 \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \text{ eV}$. For hydrogen $(Z=1)$,$\Delta E \propto (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.

$A$. Paschen ($1^{st}$ line)	$n_1=3, n_2=4 \rightarrow \Delta E \propto (\frac{1}{9} - \frac{1}{16}) = \frac{7}{144} \approx 0.0486$
$B$. Balmer ($2^{nd}$ line)	$n_1=2, n_2=4 \rightarrow \Delta E \propto (\frac{1}{4} - \frac{1}{16}) = \frac{3}{16} = 0.1875$
$C$. Paschen ($3^{rd}$ line)	$n_1=3, n_2=6 \rightarrow \Delta E \propto (\frac{1}{9} - \frac{1}{36}) = \frac{3}{36} = 0.0833$
$D$. Bracket ($4^{th}$ line)	$n_1=4, n_2=8 \rightarrow \Delta E \propto (\frac{1}{16} - \frac{1}{64}) = \frac{3}{64} = 0.0468$

Comparing the values: $0.0468 (D) < 0.0486 (A) < 0.0833 (C) < 0.1875 (B)$.
Therefore,the correct ascending order is $D < A < C < B$.

(B) According to the Born-Haber cycle for the formation of $LiF(s)$:
$\Delta H_f^{\circ} = \Delta H_{sub}(Li) + \frac{1}{2} \Delta H_{diss}(F_2) + IE(Li) + EGE(F) + U$
Where $U$ is the lattice enthalpy.
Substituting the given values:
$-594 = 155 + \frac{150}{2} + 520 + (-313) + U$
$-594 = 155 + 75 + 520 - 313 + U$
$-594 = 437 + U$
$U = -594 - 437 = -1031 \ kJ \ mol^{-1}$
The magnitude of the lattice enthalpy is $|-1031| = 1031 \ kJ \ mol^{-1}$.

(C) The reaction sequence is as follows:
$1$. The compound $[P]$ is $1,1-dibromo-1-(4-methylphenyl)methane$ (also known as $4-methylbenzal$ bromide,but with the formula $C_9H_{10}Br_2$,it is $1-(1,1-dibromoethyl)-4-methylbenzene$ or similar gem-dibromide structure).
$2$. Treatment of $[P]$ with excess $NaNH_2$ followed by $HCl$ leads to dehydrohalogenation,forming the terminal alkyne $[Q]$,which is $1-ethynyl-4-methylbenzene$ $(p-methylphenylacetylene)$.
$3$. Hydration of the alkyne $[Q]$ with $HgSO_4$ and $dil. H_2SO_4$ (Kucherov reaction) yields the ketone $[R]$,which is $4-methylacetophenone$.
$4$. $4-methylacetophenone$ contains a $CH_3CO-$ group,thus it gives a positive Iodoform test. It is a ketone,so it gives a negative Tollen's test.
$5$. Therefore,$[P]$ is $1,1-dibromoethyl-4-methylbenzene$ (represented by option $C$).

(D) The energy of a transition in a hydrogen-like atom is given by $\Delta E = 13.6 \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \ eV$.
For the first Balmer line,the transition is from $n_2 = 3$ to $n_1 = 2$:
$x = 13.6 \times (1)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{5}{36} \right) \quad \dots(i)$
For the second Balmer line,the transition is from $n_2 = 4$ to $n_1 = 2$:
$\Delta E = 13.6 \times (1)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \left( \frac{3}{16} \right) \quad \dots(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{\Delta E}{x} = \frac{13.6 \times (3/16)}{13.6 \times (5/36)} = \frac{3}{16} \times \frac{36}{5} = \frac{3 \times 9}{4 \times 5} = \frac{27}{20} = 1.35$
Therefore,$\Delta E = 1.35x$.

(B) For a basic buffer,$pOH = pK_{b} + \log \frac{[Salt]}{[Base]}$.
Given $pH = 9.25$,so $pOH = 14 - 9.25 = 4.75$.
Substituting values: $4.75 = 4.75 + \log \frac{[Salt]}{[Base]}$,which implies $\log \frac{[Salt]}{[Base]} = 0$,so $[Salt] = [Base]$.
This means the number of millimoles of $NH_{4}Cl$ (salt) formed must equal the remaining millimoles of $NH_{4}OH$ (base).
Checking Option $(B)$:
$NH_{4}OH + HCl \rightarrow NH_{4}Cl + H_{2}O$
Initial millimoles of $NH_{4}OH = 0.2 \ M \times 500 \ mL = 100 \ mmol$.
Initial millimoles of $HCl = 0.1 \ M \times 500 \ mL = 50 \ mmol$.
After reaction,$50 \ mmol$ of $NH_{4}Cl$ is formed and $50 \ mmol$ of $NH_{4}OH$ remains.
Since $[Salt] = [Base]$,the $pH$ will be $9.25$.

(B) Statement-$I$: The first ionization enthalpy $(IE_1)$ generally increases across a period due to increasing $Z_{eff}$. However,$N$ $(2p^3)$ has a stable half-filled configuration,making its $IE_1$ higher than $O$ $(2p^4)$. The correct order is $C < O < N < F$. Thus,Statement-$I$ is true.
Statement-$II$: The magnitude of electron gain enthalpy $(|\Delta_{eg}H|)$ generally decreases down a group. However,oxygen $(O)$ has a very small size,leading to strong inter-electronic repulsions,which makes its electron gain enthalpy value lower than the rest of the group members. The correct order is $S > Se > Te > Po > O$. Thus,Statement-$II$ is true.

(C) The dipole moments of the given molecules are as follows:
$H_2S = 0.95 \ D$
$H_2O = 1.85 \ D$
$NF_3 = 0.23 \ D$
$NH_3 = 1.47 \ D$
$CHCl_3 = 1.04 \ D$
Comparing these values,$NF_3$ has the lowest dipole moment.
Therefore,the molecule $(X)$ is $NF_3$.
In $NF_3$,the central atom is Nitrogen $(N)$.
Nitrogen has $5$ valence electrons. It forms $3$ single bonds with $3$ Fluorine atoms,leaving $1$ lone pair on the Nitrogen atom.
Thus,the number of lone pairs on the central atom is $1$.

(A) $1$. Identify the functional group: The compound is an ester,so the suffix is -oate. The alkyl group attached to the oxygen atom is a propyl group.
$2$. Number the main chain: The carbon atom of the ester group is $C-1$. The longest chain containing the ester group has $7$ carbons,making it a heptanoate derivative.
$3$. Identify substituents: There is a bromo group at position $2$ and a methyl group at position $5$.
$4$. Combine: The name is propyl $2$-bromo-$5$-methylheptanoate.

(C) The number of equivalents of $H_2SO_4$ used = $Molarity \times Volume \times n-factor = 1 \times 0.015 \times 2 = 0.03 \ eq$.
Since $1 \ eq$ of $H_2SO_4$ reacts with $1 \ eq$ of $NH_3$,the moles of $NH_3$ evolved = $0.03 \ mol$.
Since each mole of $NH_3$ contains $1 \ mole$ of nitrogen atoms,the moles of nitrogen = $0.03 \ mol$.
Mass of nitrogen = $0.03 \times 14 \ g = 0.42 \ g$.
Percentage of nitrogen = $\frac{\text{Mass of nitrogen}}{\text{Mass of compound}} \times 100 = \frac{0.42}{1} \times 100 = 42\%$.

(A) In reaction $(1)$,the starting material is $3$-methylcyclohexene. Under radical bromination conditions $(Br_2, h\nu)$,the reaction proceeds via the formation of a free radical. The most stable radical is the tertiary $(3^\circ)$ radical formed at the $C-3$ position. Therefore,the major product should be $3$-bromo-$3$-methylcyclohexene,not $3$-(bromomethyl)cyclohexene. Thus,reaction $(1)$ is incorrectly represented.

Correct order of
($i$) Boiling point: $HF > HI > HBr > HCl$
($ii$) Melting point: $HI > HF > HBr > HCl$

(C) $X$: For square planar complex $[M(abcd)]$,the number of geometrical isomers is $3$ (three pairs of trans ligands).
$Y$: For octahedral complex $[M(aa)_2b_2]$,the $cis$-isomer is optically active (chiral) and the $trans$-isomer is optically inactive (achiral). Thus,$Y = 1$.
$Z$: For octahedral complex $[Ma_3b_3]$,the geometrical isomers are $fac$ (facial) and $mer$ (meridional). Thus,$Z = 2$.
$X+Y+Z = 3+1+2 = 6$.

(C) In basic medium,the reaction between $KMnO_4$ and $KI$ is: $2MnO_4^{-} + I^{-} + H_2O \rightarrow 2MnO_2 + I_2 + 2OH^{-}$.
Here,$n$-factor for $KMnO_4$ is $3$ (change in oxidation state of $Mn$ from $+7$ to $+4$).
Equivalents of $KMnO_4 = 0.2 \times 0.5 \times 3 = 0.3 \ \text{eq}$.
Since $I_2$ is produced from $KMnO_4$,the equivalents of $I_2$ produced are equal to the equivalents of $KMnO_4$ reacted,which is $0.3 \ \text{eq}$.
The titration reaction is $I_2 + 2S_2O_3^{2-} \rightarrow 2I^{-} + S_4O_6^{2-}$.
Equivalents of $I_2 = \text{Equivalents of } Na_2S_2O_3$.
$0.3 = 0.1 \times V \times 1$.
$V = 3 \ L$.

(C) $1$. For $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. It forms $dsp^2$ hybridization,making it diamagnetic.
$2$. For $Ni(CO)_4$: $Ni$ is $3d^8 4s^2$ $(3d^{10})$. $CO$ is a strong field ligand. It forms $sp^3$ hybridization,making it diamagnetic.
$3$. For $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,so electrons do not pair. It forms $sp^3$ hybridization with two unpaired electrons,making it paramagnetic.
Therefore,$Ni(CO)_4$ and $[Ni(CN)_4]^{2-}$ are diamagnetic,while $[NiCl_4]^{2-}$ is paramagnetic.

(C) Statement $I$: Ions with $d^0$ or $d^{10}$ configuration are colourless. $Sc^{3+}$ $(d^0)$,$Ti^{4+}$ $(d^0)$,and $Zn^{2+}$ $(d^{10})$ are colourless.
- $[Sc^{3+}, Ti^{3+}]$: $Sc^{3+}$ is colourless,$Ti^{3+}$ is coloured.
- $[Mn^{2+}, Cr^{2+}]$: Both are coloured ($Mn^{2+}$ is $d^5$,$Cr^{2+}$ is $d^4$).
- $[Cu^{2+}, Zn^{2+}]$: $Cu^{2+}$ is coloured $(d^9)$,$Zn^{2+}$ is colourless.
- $[Ni^{2+}, Ti^{4+}]$: $Ni^{2+}$ is coloured $(d^8)$,$Ti^{4+}$ is colourless.
Only $1$ pair $([Mn^{2+}, Cr^{2+}])$ has both ions coloured. Thus,Statement $I$ is false.
Statement $II$: $Th^{4+}$ is the most stable oxidation state of Thorium and cannot be oxidized further,so it cannot act as a reducing agent. $Eu^{2+}$ is a strong reducing agent as it tends to change to $Eu^{3+}$. Thus,Statement $II$ is false.

(C) By analyzing the structure of the pentapeptide,we can identify the individual amino acids by breaking the peptide bonds through hydrolysis. The amino acids present are:
$1$. $N$-terminal: $\text{Threonine} (\text{Thr})$
$2$. Second: $\text{Serine} (\text{Ser})$
$3$. Third: $\text{Aspartic acid} (\text{Asp})$
$4$. Fourth: $\text{Glycine} (\text{Gly})$
$5$. $C$-terminal: $\text{Alanine} (\text{Ala})$
Thus,the sequence is $\text{Thr} - \text{Ser} - \text{Asp} - \text{Gly} - \text{Ala}$.
Among these,$\text{Threonine}$ is an essential amino acid. Therefore,$(Y) = \text{Threonine}$ and the sequence is $\text{Thr} - \text{Ser} - \text{Asp} - \text{Gly} - \text{Ala}$.

(D) $1$. In reaction $(A)$,nitration of benzanilide occurs primarily at the ortho and para positions of the aniline ring due to the activating effect of the $-NHCOPh$ group. The product shown in the image is incorrect.
$2$. In reaction $(B)$,aryl halides do not undergo nucleophilic substitution with potassium phthalimide under standard conditions,so no reaction occurs.
$3$. In reaction $(C)$,the carbylamine reaction produces an isocyanide,which upon reduction with $Sn/HCl$ yields a secondary amine ($N$-methylbenzylamine),not the primary amine shown.
$4$. In reaction $(D)$,the Hoffmann bromamide degradation of propanamide $(CH_3CH_2CONH_2)$ correctly yields ethylamine $(CH_3CH_2NH_2)$ along with the specified byproducts. This is a standard textbook reaction.

(B) The Griess-Ilosvay test is a standard analytical method used for the detection of nitrite ions $(NO_2^-)$.
In this test,sulphanilic acid reacts with nitrous acid $(HNO_2)$,which is generated from the nitrite ion in an acidic medium,to form a diazonium salt.
This diazonium salt then undergoes a coupling reaction with $\alpha$-naphthylamine to produce a red-colored azo-dye.
Both statements accurately describe the chemical principles of this test.
Therefore,both Statement $I$ and Statement $II$ are true.

(A) Stability: The bond dissociation enthalpy decreases down the group as the size of the central atom increases,so stability decreases: $NH_{3} > PH_{3} > AsH_{3} > SbH_{3} > BiH_{3}$. Thus,statement $A$ is correct.
Basicity: The electron density on the central atom decreases as the size increases,so basicity decreases: $NH_{3} > PH_{3} > AsH_{3} > SbH_{3} > BiH_{3}$. Thus,statement $B$ is correct.
Reducing character: As stability decreases,the tendency to release $H_{2}$ increases,so reducing character increases: $NH_{3} < PH_{3} < AsH_{3} < SbH_{3} < BiH_{3}$. Thus,statement $C$ is correct.
Boiling point: $NH_{3}$ has a higher boiling point than $PH_{3}$ due to hydrogen bonding. The trend is $PH_{3} < AsH_{3} < NH_{3} < SbH_{3} < BiH_{3}$. Thus,statement $D$ is incorrect.

(A) The structures are: $P$ ($1$-phenylprop$-2-$en$-1-$ol),$Q$ ($3$-phenylpropanal),$R$ ($1$-phenylpropan$-2-$one),$S$ ($1$-phenylpropan$-1-$one).
$A$. $2,4-DNP$ reacts with aldehydes and ketones. $Q$ (aldehyde),$R$ (ketone),and $S$ (ketone) give the test. Statement $A$ is correct.
$B$. Bayer's test is for unsaturation. Only $P$ has a double bond. Statement $B$ is incorrect.
$C$. Aromatic compounds give a sooty flame. All four compounds contain a phenyl ring,so all give a sooty flame. Statement $C$ is technically correct,but we evaluate based on the options provided.
$D$. $I_2/NaOH$ (Iodoform test) requires a $CH_3CO-$ group or $CH_3CH(OH)-$ group. $R$ has $CH_3CO-$ and $S$ does not. Statement $D$ is incorrect.
$E$. Tollen's reagent is for aldehydes. Only $Q$ is an aldehyde. Statement $E$ is correct.
Combining the correct statements $A, C, E$,the correct option is $A$.

(A) For a first-order reaction,the time $t$ taken for the concentration to reach $A_t$ from $A_0$ is given by $t = \frac{1}{k} \ln \frac{A_0}{A_t}$.
For $t_{1/8}$,the remaining concentration is $\frac{A_0}{8}$,so $t_{1/8} = \frac{1}{k} \ln \frac{A_0}{A_0/8} = \frac{1}{k} \ln 8$.
For $t_{1/10}$,the remaining concentration is $\frac{A_0}{10}$,so $t_{1/10} = \frac{1}{k} \ln \frac{A_0}{A_0/10} = \frac{1}{k} \ln 10$.
Taking the ratio: $\frac{t_{1/8}}{t_{1/10}} = \frac{\ln 8}{\ln 10} = \frac{\log 8}{\log 10} = \log 8$.
Since $\log 8 = \log 2^3 = 3 \log 2 = 3 \times 0.3 = 0.9$.
Therefore,$\frac{t_{1/8}}{t_{1/10}} \times 10 = 0.9 \times 10 = 9$.

(C) The reaction is the Kolbe-Schmitt reaction. The molecular weight of compound $(y)$ is $2 \times \text{vapour density} = 2 \times 47 = 94$.
Given the percentages: $C = 76.6\%$,$H = 6.38\%$,$O = 100 - (76.6 + 6.38) = 17.02\%$.
Moles of $C = 76.6/12 = 6.38$,$H = 6.38/1 = 6.38$,$O = 17.02/16 = 1.06$.
Ratio $C:H:O = 6:6:1$. The empirical formula is $C_6H_6O$,which corresponds to phenol $(C_6H_5OH)$.
Thus,compound $(x)$ is phenol and compound $(y)$ is salicylic acid ($2$-hydroxybenzoic acid).
Phenol $(x)$ is acidic and dissolves in $NaOH$. Salicylic acid $(y)$ is more acidic than phenol due to the electron-withdrawing $-COOH$ group and intramolecular hydrogen bonding,so it also dissolves in $NaOH$.
Salicylic acid $(y)$ reacts with $NaHCO_3$ to evolve $CO_2$ gas.
Both are aromatic compounds and burn with a sooty flame.
Statement $(c)$ is incorrect because salicylic acid $(y)$ is more acidic than phenol $(x)$.

(D) According to Raoult's law for an ideal solution,$P_S = X_A P_A^{\circ} + X_B P_B^{\circ}$.
For the first mixture: $X_A = 2/5$,$X_B = 3/5$,and $P_S = 320 \ mm \ Hg$.
$320 = P_A^{\circ} (2/5) + P_B^{\circ} (3/5) \implies 2 P_A^{\circ} + 3 P_B^{\circ} = 1600$ ...$(I)$
After adding $1$ mole of $A$ and $1$ mole of $B$,the new moles are $3$ moles of $A$ and $4$ moles of $B$. Total moles = $7$.
$X_A' = 3/7$,$X_B' = 4/7$,and $P_S' = 328.6 \ mm \ Hg$.
$328.6 = P_A^{\circ} (3/7) + P_B^{\circ} (4/7) \implies 3 P_A^{\circ} + 4 P_B^{\circ} = 2300.2$ ...$(II)$
Solving the system of equations:
Multiply $(I)$ by $3$: $6 P_A^{\circ} + 9 P_B^{\circ} = 4800$ ...$(III)$
Multiply $(II)$ by $2$: $6 P_A^{\circ} + 8 P_B^{\circ} = 4600.4$ ...$(IV)$
Subtracting $(IV)$ from $(III)$: $P_B^{\circ} = 199.6 \ mm \ Hg \approx 200 \ mm \ Hg$.
Substituting $P_B^{\circ} = 200$ into $(I)$: $2 P_A^{\circ} + 3(200) = 1600 \implies 2 P_A^{\circ} = 1000 \implies P_A^{\circ} = 500 \ mm \ Hg$.
Thus,the vapour pressures are $500 \ mm \ Hg$ and $200 \ mm \ Hg$.

(C) Step $1$: Calculate $O_2$ evolved during $Cu$ deposition.
$Eq$ of $Cu = Eq$ of $O_2$
$\frac{300 \times 10^{-3}}{63.54 / 2} = n_{O_2(1)} \times 4$
$n_{O_2(1)} = \frac{0.3}{63.54 \times 2} = 2.36 \times 10^{-3} \ mol$.
Step $2$: Calculate $O_2$ evolved during the additional $28 \ minutes$.
$Q = I \times t = 0.6 \ A \times (28 \times 60) \ s = 1008 \ C$.
$n_{e^-} = \frac{1008}{96500} = 0.01044 \ mol$.
Since $4 \ mol \ e^-$ produce $1 \ mol \ O_2$,$n_{O_2(2)} = \frac{0.01044}{4} = 2.611 \times 10^{-3} \ mol$.
Step $3$: Total $O_2$ volume at $STP$.
$n_{total} = (2.36 + 2.611) \times 10^{-3} = 4.971 \times 10^{-3} \ mol$.
$V = n \times 22400 \ mL = 4.971 \times 10^{-3} \times 22400 \approx 111.35 \ mL$.
Rounding to the nearest integer,we get $111 \ mL$.

(C) In reaction $A$,phenol reacts with sodium $(Na)$ to release hydrogen gas $(x = H_2)$.
Molar mass of $H_2 = 2 \ g/mol$.
In reaction $B$,benzoic acid reacts with sodium bicarbonate $(NaHCO_3)$ to release carbon dioxide gas $(y = CO_2)$.
Molar mass of $CO_2 = 44 \ g/mol$.
Sum of molar masses = $2 + 44 = 46$.

(D) Statement $I$: In $CH_2=CH-Cl$ (vinyl chloride),the lone pair on chlorine atom participates in resonance with the double bond,giving the $C-Cl$ bond a partial double bond character. This makes it shorter and stronger than the $C-Cl$ bond in $CH_3-CH_2-Cl$ (ethyl chloride),where only a single bond exists. Thus,Statement $I$ is true.
Statement $II$: The given molecule is a chiral alkyl halide. Hydrolysis of such a molecule typically proceeds via an $S_N1$ mechanism,which involves the formation of a planar carbocation intermediate. Nucleophilic attack on this carbocation occurs from both sides with equal probability,leading to the formation of a racemic mixture. $A$ racemic mixture is optically inactive and cannot rotate plane polarized light $(PPL)$. Thus,Statement $II$ is false.

(B) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.
For the catalysed reaction,$k_c = A e^{-E_{ac} / RT}$ and for the uncatalysed reaction,$k_u = A e^{-E_{au} / RT}$.
Given that the frequency factor $A$ is the same,the ratio is $\frac{k_c}{k_u} = e^{(E_{au} - E_{ac}) / RT} = e^{\Delta E_a / RT}$.
Taking the logarithm on both sides: $\log_{10} \left( \frac{k_c}{k_u} \right) = \frac{\Delta E_a}{2.303 RT}$.
Given $\Delta E_a = 10 \ kJ \ mol^{-1} = 10000 \ J \ mol^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,and $T = 27 + 273 = 300 \ K$.
$\log_{10} \left( \frac{k_c}{k_u} \right) = \frac{10000}{2.303 \times 8.314 \times 300} = \frac{10000}{5744.14} \approx 1.741$.

(D) During acetylation,each $-OH$ group is replaced by an $-OCOCH_3$ group.
This results in the replacement of one hydrogen atom (mass $1 \ u$) with an acetyl group ($-COCH_3$,mass $43 \ u$).
Therefore,the net increase in molar mass for each $-OH$ group is $43 - 1 = 42 \ g \ mol^{-1}$.
Let $n$ be the number of $-OH$ groups.
The increase in molar mass is $290 - 122 = 168 \ g \ mol^{-1}$.
Thus,$n \times 42 = 168$.
$n = \frac{168}{42} = 4$.
So,there are $4$ hydroxyl groups present in compound $(X)$.

(C) Tollen's reagent is an oxidizing agent that reacts with aldehydes to form a silver mirror.
Compounds that exist in equilibrium with an open-chain aldehyde form will give a positive Tollen's test.
Hemiacetals are in equilibrium with their open-chain aldehyde form in the presence of a base (present in Tollen's reagent).
Compound $C$ is a cyclic hemiacetal,which can open to form an aldehyde,thus giving a positive Tollen's test.
Compounds $A$ and $B$ are acetals (ethers),which do not open to form aldehydes under these conditions.
Compound $D$ is a cyclic alcohol,which does not contain the hemiacetal functional group required to form an aldehyde.
Therefore,the correct compound is $C$.

(A) Given:
$P^0 = 640 \ mm \ Hg$,$P_s = 600 \ mm \ Hg$.
Relative lowering of vapor pressure: $\frac{P^0 - P_s}{P^0} = i \cdot X_2$.
$\frac{640 - 600}{640} = i \cdot X_2 \implies i \cdot X_2 = \frac{40}{640} = \frac{1}{16}$ ... $(i)$.
Elevation in boiling point: $\Delta T_b = T_b - T_b^0 = 375 - 373 = 2 \ K$.
Formula: $\Delta T_b = i \cdot K_b \cdot m$.
$2 = i \cdot 0.52 \cdot \frac{W \times 1000}{M \times 100} \implies 2 = i \cdot 5.2 \cdot \frac{W}{M} \implies i = \frac{2 \cdot M}{5.2 \cdot W} = \frac{M}{2.6 \cdot W}$ ... $(ii)$.
Substituting $(ii)$ in $(i)$:
$\left(\frac{M}{2.6 \cdot W}\right) \cdot X_2 = \frac{1}{16}$.
$X_2 = \frac{2.6 \cdot W}{16 \cdot M} = \frac{1.3}{8} \cdot \frac{W}{M}$.

(C) . Vinyl chloride is $CH_2=CHCl$ $(III)$.
$B$. Benzyl chloride is $C_6H_5CH_2Cl$ $(IV)$.
$C$. Alkyl chloride is $CH_3-CH(Cl)CH_3$ $(II)$.
$D$. Allyl chloride is $CH_2=CH-CH_2Cl$ $(I)$.
Therefore,the correct matching is $A-III, B-IV, C-II, D-I$.

(A) Statement-$I$:
$1$. $[CoF_6]^{3-}$: $Co^{3+}$ $(d^6)$,$F^-$ is a weak field ligand,configuration is $t_{2g}^4 e_g^2$,$4$ unpaired electrons (Paramagnetic).
$2$. $[TiF_6]^{3-}$: $Ti^{3+}$ $(d^1)$,$1$ unpaired electron (Paramagnetic).
$3$. $V_2O_5$: $V^{5+}$ $(d^0)$,$0$ unpaired electrons (Diamagnetic).
$4$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ $(d^5)$,$CN^-$ is a strong field ligand,configuration is $t_{2g}^5 e_g^0$,$1$ unpaired electron (Paramagnetic).
Total paramagnetic species = $3$. Statement-$I$ is true.
Statement-$II$:
$1$. $K_4[Fe(CN)_6]$: $Fe^{2+}$ $(d^6)$,strong field,$0$ unpaired electrons.
$2$. $K_3[Fe(CN)_6]$: $Fe^{3+}$ $(d^5)$,strong field,$1$ unpaired electron.
$3$. $[Fe(H_2O)_6]SO_4 \cdot H_2O$: $Fe^{2+}$ $(d^6)$,weak field,$4$ unpaired electrons.
$4$. $[Fe(H_2O)_6]Cl_3$: $Fe^{3+}$ $(d^5)$,weak field,$5$ unpaired electrons.
Order: $0 < 1 < 4 < 5$. Statement-$II$ is true.

(A) The $4 \ L$ solution contains $0.4 \ mol$ of $[Co(NH_{3})_{5}SO_{4}]Br$ and $0.4 \ mol$ of $[Co(NH_{3})_{5}Br]SO_{4}$.
In $2 \ L$ of the mixture,there are $0.2 \ mol$ of each complex.
When $2 \ L$ of the mixture reacts with excess $AgNO_{3}$,only $[Co(NH_{3})_{5}SO_{4}]Br$ reacts to form $AgBr$ precipitate $(Y)$. Thus,$0.2 \ mol$ of $AgBr$ is formed.
When $2 \ L$ of the mixture reacts with excess $BaCl_{2}$,only $[Co(NH_{3})_{5}Br]SO_{4}$ reacts to form $BaSO_{4}$ precipitate $(Z)$. Thus,$0.2 \ mol$ of $BaSO_{4}$ is formed.

(A) The stability of benzene diazonium salts is influenced by the nature of the substituent attached to the benzene ring.
Electron-donating groups $(EDG)$ increase the stability of the diazonium cation by dispersing the positive charge,while electron-withdrawing groups $(EWG)$ decrease the stability by intensifying the positive charge.
In the given compounds:
$A$: $-OCH_3$ is a strong $EDG$ ($+M$ effect).
$C$: Benzene diazonium chloride (no substituent).
$D$: $-CN$ is an $EWG$ ($-M$ effect).
$B$: $-NO_2$ is a very strong $EWG$ ($-M$ effect).
Therefore,the stability order is $A > C > D > B$.

(B) Moles of solute '$A$' = $\frac{0.3 \ g}{60 \ g \ mol^{-1}} = 0.005 \ mol$.
Moles of solute '$B$' = $\frac{0.9 \ g}{180 \ g \ mol^{-1}} = 0.005 \ mol$.
Total moles of solute = $0.005 + 0.005 = 0.01 \ mol$.
Volume of solution = $100 \ mL = 0.1 \ L$.
Molarity $(C)$ = $\frac{\text{Total moles}}{\text{Volume in } L} = \frac{0.01 \ mol}{0.1 \ L} = 0.1 \ M$.
Temperature $(T)$ = $27 + 273 = 300 \ K$.
Using the formula for osmotic pressure: $\pi = CRT$.
$\pi = 0.1 \ mol \ L^{-1} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 300 \ K$.
$\pi = 2.46 \ atm$.

(C) In $K_{3}[Co(CO_{3})_{3}]$,$Co$ is in $+3$ oxidation state ($d^{6}$ configuration). $CO_{3}^{2-}$ is a weak field ligand,so it forms an outer orbital complex: $sp^{3}d^{2}$ hybridisation,octahedral geometry,and $4$ unpaired electrons. Magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{4(6)} = \sqrt{24} \approx 4.9 \ BM$. Statement-$I$ is true.
For $[Ni(CN)_{4}]^{2-}$,$Ni^{2+}$ is $d^{8}$. $CN^{-}$ is a strong field ligand,so $dsp^{2}$ hybridisation,square planar,$0$ unpaired electrons,$\mu = 0 \ BM$.
For $[MnBr_{4}]^{2-}$,$Mn^{2+}$ is $d^{5}$. $Br^{-}$ is a weak field ligand,so $sp^{3}$ hybridisation,tetrahedral,$5$ unpaired electrons,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.9 \ BM$.
For $[CoF_{6}]^{3-}$,$Co^{3+}$ is $d^{6}$. $F^{-}$ is a weak field ligand,so $sp^{3}d^{2}$ hybridisation,octahedral,$4$ unpaired electrons,$\mu = 4.9 \ BM$.
All values in Statement-$II$ are correct. Thus,both statements are true.

(A) The electrode reaction for the metal/metal insoluble salt electrode is: $MX_{(s)} + e^- \rightarrow M_{(s)} + X^-_{(aq)}$.
This can be expressed as the sum of two half-reactions:
$M^+_{(aq)} + e^- \rightarrow M_{(s)}$ $(E^{\circ} = 0.79 \ V)$
$MX_{(s)} \rightleftharpoons M^+_{(aq)} + X^-_{(aq)}$ $(K_{sp} = 10^{-10})$
Using the relation $E^{\circ}_{cell} = E^{\circ}_{M^+/M} + \frac{0.059}{n} \log K_{sp}$:
$E^{\circ}_{X^-/MX_{(s)}/M} = 0.79 + 0.059 \log(10^{-10})$
$E^{\circ} = 0.79 + 0.059 \times (-10)$
$E^{\circ} = 0.79 - 0.59 = 0.20 \ V$
Converting to $mV$: $0.20 \ V = 200 \ mV$.

(A) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $\mu > 3.0 \ BM$,we need $\sqrt{n(n+2)} > 3.0$,which implies $n(n+2) > 9$. This holds for $n \geq 3$.
Electronic configurations:
$V^{3+} (3d^2): n = 2, \mu = \sqrt{2(4)} = 2.83 \ BM$
$Ti^{2+} (3d^2): n = 2, \mu = \sqrt{2(4)} = 2.83 \ BM$
$Ni^{2+} (3d^8): n = 2, \mu = \sqrt{2(4)} = 2.83 \ BM$
$Fe^{2+} (3d^6): n = 4, \mu = \sqrt{4(6)} = 4.90 \ BM$
$Co^{2+} (3d^7): n = 3, \mu = \sqrt{3(5)} = 3.87 \ BM$
Thus,$Fe^{2+}$ and $Co^{2+}$ have $\mu > 3.0 \ BM$.
In high spin octahedral complexes:
$Fe^{2+} (3d^6)$ has $4$ unpaired electrons.
$Co^{2+} (3d^7)$ has $3$ unpaired electrons.
Sum of unpaired electrons = $4 + 3 = 7$.

(A) The elevation in boiling point is given by $\Delta T_{b} = K_{b} \times m$,where $m$ is the molality.
$0.5 = 5.0 \times m \implies m = 0.1 \ mol \ kg^{-1}$.
Molality $m = \frac{n_{X}}{W_{Y} (\text{in } kg)} = \frac{n_{X}}{0.150 \ kg} = 0.1 \implies n_{X} = 0.015 \ mol$.
The relative lowering in vapour pressure is given by $\frac{P^{o} - P_{s}}{P^{o}} = X_{X} = \frac{n_{X}}{n_{X} + n_{Y}}$.
$n_{Y} = \frac{150 \ g}{300 \ g \ mol^{-1}} = 0.5 \ mol$.
Since the solution is dilute,$n_{X} + n_{Y} \approx n_{Y} = 0.5 \ mol$.
Relative lowering $= \frac{0.015}{0.5} = 0.03 = 3 \times 10^{-2}$.
Thus,the value is $3$.

(A) For an isotonic solution,the osmotic pressure $(\pi)$ of the $NaCl$ solution must be equal to the osmotic pressure of the living cell.
$\pi = iCRT$
Given $\pi = 12 \ atm$,$T = 300 \ K$,$R = 0.08 \ L \ atm \ K^{-1} \ mol^{-1}$,and $i = 2$ (for $NaCl \rightarrow Na^{+} + Cl^{-}$).
$12 = 2 \times C \times 0.08 \times 300$
$12 = 48 \times C$
$C = \frac{12}{48} = 0.25 \ mol \ L^{-1}$
The molar mass of $NaCl = 23 + 35.5 = 58.5 \ g \ mol^{-1}$.
Strength of $NaCl$ solution = $C \times \text{Molar mass}$
$= 0.25 \times 58.5 = 14.625 \ g \ L^{-1}$.
Rounding to the nearest integer,we get $15 \ g \ L^{-1}$.

(B) The reaction between common salt $(NaCl)$,potassium dichromate $(K_{2}Cr_{2}O_{7})$,and concentrated sulfuric acid $(H_{2}SO_{4})$ is as follows:
$4 \ NaCl + K_{2}Cr_{2}O_{7} + 6 \ H_{2}SO_{4}$ $\longrightarrow 2 \ KHSO_{4} + 2 \ CrO_{2}Cl_{2} + 4 \ NaHSO_{4} + 3 \ H_{2}O$
The gas evolved is chromyl chloride $(CrO_{2}Cl_{2})$.
In $CrO_{2}Cl_{2}$,let the oxidation state of $Cr$ be $x$.
$x + 2(-2) + 2(-1) = 0$
$x - 4 - 2 = 0$
$x = +6$
Thus,the formula is $CrO_{2}Cl_{2}$ and the oxidation state is $+6$.

(B) The standard cell potential,denoted as $E_{cell}^0$,is a constant value for a given electrochemical cell at a specific temperature. It depends only on the standard reduction potentials of the half-cells involved and is independent of the concentrations of the reactants or the time the cell has been operating. Therefore,a plot of $E_{cell}^0$ versus time will be a horizontal line,indicating that the value remains constant over time.

(B) . $Mn$ in $Mn_2O_7$ has an oxidation state of $+7$,which is its highest oxidation state. This statement is correct.
$B$. Oxygen is a small,highly electronegative atom that can form multiple bonds ($p\pi-d\pi$ bonding) with transition metals,stabilizing them in higher oxidation states. This statement is correct.
$C$. $Mn_2O_7$ is a covalent oxide,not an ionic one. This statement is incorrect.
$D$. The structure of $Mn_2O_7$ consists of two $MnO_3$ tetrahedra sharing a common oxygen atom (bridged oxygen). This statement is correct.
Therefore,statements $A, B$ and $D$ are correct.

(A) $sec$-butyl chloride ($2$-chlorobutane) is the only compound among the given options that contains a chiral carbon atom and thus exhibits optical isomerism.
The molecular formula of $sec$-butyl chloride is $C_4H_9Cl$.
Molar mass of $C_4H_9Cl = (4 \times 12) + (9 \times 1) + 35.5 = 48 + 9 + 35.5 = 92.5 \ g/mol$.
Percentage of carbon $= \frac{\text{Mass of carbon}}{\text{Molar mass}} \times 100 = \frac{48}{92.5} \times 100 \approx 51.89 \%$.
Rounding to the nearest integer,we get $52 \%$.

(A) For $[Mn(H_2O)_6]^{2+}$,the configuration is $d^5$ with a weak field ligand $(H_2O)$,resulting in $t_{2g}^3 e_g^2$. The $CFSE$ is $(3 \times -0.4 \Delta_o) + (2 \times 0.6 \Delta_o) = 0 \Delta_o$.
For $[Cr(H_2O)_6]^{2+}$,the configuration is $d^4$ with a weak field ligand $(H_2O)$,resulting in $t_{2g}^3 e_g^1$. The $CFSE$ is $(3 \times -0.4 \Delta_o) + (1 \times 0.6 \Delta_o) = -0.6 \Delta_o$. Since $|-0.6| > 0$,Statement $I$ is true.
For potassium ferricyanide,$K_3[Fe(CN)_6]$,the iron is in $+3$ oxidation state $(d^5)$. $CN^-$ is a strong field ligand,leading to $t_{2g}^5 e_g^0$ with $1$ unpaired electron. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ B.M.$
For sodium ferrocyanide,$Na_4[Fe(CN)_6]$,the iron is in $+2$ oxidation state $(d^6)$. $CN^-$ is a strong field ligand,leading to $t_{2g}^6 e_g^0$ with $0$ unpaired electrons. $\mu = 0 \ B.M.$
Since $\sqrt{3} > 0$,Statement $II$ is true.

(D) The rate of an $S_{N}2$ reaction depends on the nucleophilicity of the nucleophile. Nucleophilicity is generally inversely proportional to the stability of the anion (conjugate base).
The stability order of the given anions is: $ClO_{4}^{-} > CH_{3}COO^{-} > PhO^{-} > {}^{-}OH$.
Since nucleophilicity is the reverse of the stability of the anion,the order of nucleophilicity (and thus the rate of reaction) is: ${}^{-}OH > PhO^{-} > CH_{3}COO^{-} > ClO_{4}^{-}$.

(C) . Enzymes lower the activation energy of a reaction,making the hydrolysis of sucrose faster compared to acid catalysis. This statement is correct.
$B$. Denaturation disrupts the hydrogen bonds and other interactions that maintain the secondary and tertiary structures of a protein,but the primary structure (amino acid sequence) remains intact. This statement is correct.
$C$. Nucleotides are joined by phosphodiester linkages between the $C_{3}$ and $C_{5}$ carbons of the pentose sugar,not glycosidic linkages. This statement is incorrect.
$D$. Quaternary structure refers to the arrangement of multiple polypeptide subunits in a protein complex,not just the folding of a single chain. This statement is incorrect.
Therefore,only statements $A$ and $B$ are correct.

(B) The reaction of benzyl halides with $KCN$ typically proceeds via an $S_{N}1$ mechanism in polar protic solvents or $S_{N}2$ depending on conditions. However,for benzyl halides,the rate is governed by the stability of the intermediate carbocation in $S_{N}1$ or steric/electronic factors in $S_{N}2$. In both cases,electron-donating groups $(EDG)$ increase reactivity,while electron-withdrawing groups $(EWG)$ decrease it.
$a$ has $-OH$ (strong $EDG$ via resonance),$b$ has $-NH_2$ (strong $EDG$ via resonance),$c$ has $-NO_2$ (strong $EWG$ at para position),and $d$ has $-NO_2$ (strong $EWG$ at meta position).
Comparing $a$ and $b$: $-NH_2$ is a stronger $EDG$ than $-OH$ due to lower electronegativity of $N$ compared to $O$,making $b$ more reactive than $a$.
Comparing $c$ and $d$: $-NO_2$ at the para position $(c)$ exerts a stronger destabilizing effect on the carbocation than at the meta position $(d)$ due to direct resonance destabilization.
Thus,the order is $b > a > d > c$.

(B) The reaction is $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$.
Initially $(t=0)$: $P_A = 1 \ bar$,$P_B = 0$,$P_C = 0$. Total pressure $P_0 = 1 \ bar$.
At time $t = 100 \ min$: $P_A = 1 - x$,$P_B = x$,$P_C = x$.
Total pressure $P_t = (1 - x) + x + x = 1 + x$.
Given $P_t = 1.5 \ bar$,so $1 + x = 1.5 \implies x = 0.5 \ bar$.
Partial pressure of $A$ at $t = 100 \ min$ is $P_A = 1 - 0.5 = 0.5 \ bar$.
For a first-order reaction,$k = \frac{2.303}{t} \log \frac{P_0}{P_A}$.
$k = \frac{2.303}{100} \log \frac{1}{0.5} = \frac{2.303}{100} \log 2$.
Using $\log 2 = 0.3$,$k = \frac{2.303 \times 0.3}{100} = \frac{0.6909}{100} \approx 6.9 \times 10^{-3} \ min^{-1}$.

(D) The correct matches are as follows:
$A$. $H_2, Pd-BaSO_4$ is used in the Rosenmund reduction to convert acid chlorides to aldehydes $(A-II)$.
$B$. $SnCl_2, HCl$ is used in the Stephen reaction to reduce nitriles to imines,which are then hydrolyzed to aldehydes $(B-IV)$.
$C$. $CrO_2Cl_2, CS_2$ is used in the Etard reaction to oxidize toluene to benzaldehyde $(C-I)$.
$D$. $CO, HCl, \text{Anhyd. } AlCl_3$ is used in the Gatterman-Koch reaction to form benzaldehyde from benzene $(D-III)$.
Therefore,the correct sequence is $A-II, B-IV, C-I, D-III$.

(C) Step $1$: $Nitrobenzene$ reacts with $Br_2/FeBr_3$ to give $m-bromonitrobenzene$.
Step $2$: Reduction with $Sn/HCl$ converts the $-NO_2$ group to $-NH_2$,yielding $m-bromoaniline$.
Step $3$: $pH$ neutralisation is performed.
Step $4$: $m-bromoaniline$ reacts with $Br_2/H_2O$ to give $2,4,6-tribromo-3-bromoaniline$ (which is $2,3,4,6-tetrabromoaniline$).
Step $5$: Diazotization with $NaNO_2/HBr$ at $0-5^{\circ}C$ converts $-NH_2$ to $-N_2^+Br^-$.
Step $6$: Sandmeyer reaction with $CuBr/NaBr$ replaces the diazonium group with a $Br$ atom.
The final product is $1,2,3,4,5-pentabromobenzene$ (or $1,2,3,4,6-pentabromobenzene$ depending on numbering,but it contains $5$ bromine atoms).
Thus,the number of $Br$ atoms in the final product $(P)$ is $5$.

(B) The Arrhenius equation is given by $\log_{10} k = \log_{10} A - \frac{Ea_1}{2.303 RT}$.
Comparing this with the given equation $\log_{10} k = 14.34 - \frac{1.5 \times 10^4}{T}$,we get $\frac{Ea_1}{2.303 R} = 1.5 \times 10^4$.
$Ea_1 = 1.5 \times 10^4 \times 2.303 \times 8.314 \ J \ mol^{-1} = 287207 \ J \ mol^{-1} = 287.207 \ kJ \ mol^{-1}$.
Given that $Ea_2 = \frac{1}{5} Ea_1$,we have $Ea_2 = \frac{287.207}{5} = 57.44 \ kJ \ mol^{-1}$.
The nearest integer value is $57$.

(B) The reaction is: $C_6H_5NH_2 + C_6H_5COCl \rightarrow C_6H_5NHCOC_6H_5 + HCl$
Molar mass of aniline $(C_6H_5NH_2)$ = $6 \times 12 + 7 \times 1 + 14 = 93 \ g \ mol^{-1}$.
Molar mass of benzanilide $(C_6H_5NHCOC_6H_5)$ = $13 \times 12 + 11 \times 1 + 14 + 16 = 197 \ g \ mol^{-1}$.
Moles of aniline = $\frac{5.8 \ g}{93 \ g \ mol^{-1}} \approx 0.06237 \ mol$.
Theoretical moles of benzanilide = $0.06237 \ mol$.
Actual moles of benzanilide = $0.06237 \times 0.82 \approx 0.05114 \ mol$.
Mass of benzanilide = $0.05114 \ mol \times 197 \ g \ mol^{-1} \approx 10.07 \ g$.
Rounding to the nearest integer,the mass is $10 \ g$.

(C) Mixed oxides are those which are formed by the combination of two simple oxides of the same metal in different oxidation states.
Analyzing the given oxides:
$1. Mn_{3}O_{4} = MnO \cdot Mn_{2}O_{3}$ (Mixed oxide)
$2. Fe_{3}O_{4} = FeO \cdot Fe_{2}O_{3}$ (Mixed oxide)
$3. Co_{3}O_{4} = CoO \cdot Co_{2}O_{3}$ (Mixed oxide)
The other oxides like $Ti_{2}O_{3}, V_{2}O_{4}, Cr_{2}O_{3},$ and $Fe_{2}O_{3}$ are simple oxides.
Therefore,there are $3$ mixed oxides in the given list.

(B) The cell reaction is $M^{2+} + H_2O \rightarrow M + \frac{1}{2}O_2 + 2H^+$.
For the reaction to be spontaneous,$E_{cell} > 0$.
At the limiting condition,$E_{cell} = 0$,so $E_{cathode} = E_{anode}$.
$E_{cathode} = E^{\circ}_{M^{2+}/M} - \frac{0.059}{2} \log \frac{1}{[M^{2+}]} = 0.994 - 0 = 0.994 \ V$.
$E_{anode} = E^{\circ}_{O_2/H_2O} - \frac{0.059}{4} \log \frac{1}{[H^+]^4 P_{O_2}^{1/2}} = 1.23 + \frac{0.059}{4} \log ([H^+]^4 \times 1) = 1.23 + 0.059 \log [H^+] = 1.23 - 0.059 \times pH$.
Equating $E_{cathode} = E_{anode}$:
$0.994 = 1.23 - 0.059 \times pH$.
$0.059 \times pH = 1.23 - 0.994 = 0.236$.
$pH = \frac{0.236}{0.059} = 4$.
Thus,the nearest integer is $4$.

(B) The target molecule is $3, 3-$dimethyl$-2-$butanol,which has the structure $(CH_3)_3C-CH(OH)-CH_3$.
$A$. Reaction of $3, 3-$dimethylbutanal with $MeMgBr$ followed by $H_3O^+$ gives $3, 3-$dimethyl$-2-$butanol. This is a valid preparation.
$B$. Acid-catalyzed hydration of $3, 3-$dimethyl$-1-$butene involves a carbocation rearrangement (ethyl shift) to form a more stable carbocation,leading to $2, 3-$dimethyl$-2-$butanol,not the target alcohol. Thus,$B$ cannot prepare it.
$C$. Ozonolysis of $2, 3, 4, 4-$tetramethyl$-2-$pentene followed by reduction with $NaBH_4$ yields $3, 3-$dimethyl$-2-$butanol and acetone. This is a valid preparation.
$D$. Reduction of $3, 3-$dimethyl$-2-$butanone with $LiAlH_4$ gives $3, 3-$dimethyl$-2-$butanol. This is a valid preparation.
$E$. Hydration of $3, 3-$dimethyl$-1-$butyne with $Hg^{2+}/H^+$ gives $3, 3-$dimethyl$-2-$butanone,not the target alcohol. Thus,$E$ cannot prepare it.
Therefore,$B$ and $E$ cannot be used to prepare the target alcohol.

(B) Compound $A$ $(C_{8}H_{8}O_{2})$ undergoes cross-Aldol condensation with acetophenone to form a single product. This implies $A$ is an aldehyde without $\alpha$-hydrogens,such as $4-$methoxybenzaldehyde $(p-CH_3OC_6H_4CHO)$.
When $4-$methoxybenzaldehyde reacts with conc. $NaOH$,it undergoes the Cannizzaro reaction because it lacks $\alpha$-hydrogens.
The reaction produces $4-$methoxybenzyl alcohol $(p-CH_3OC_6H_4CH_2OH)$ and $4-$methoxybenzoate ion $(p-CH_3OC_6H_4COO^-)$.
Therefore,compound $A$ is $4-$methoxybenzaldehyde.

(A) The reducing power of a species is inversely proportional to its standard reduction potential $(E^{\circ})$.
Lower reduction potential indicates a stronger tendency to undergo oxidation,thus acting as a better reducing agent.
The reduction potentials are:
$Al^{3+}/Al = -1.66 \ V$
$Cr^{3+}/Cr = -0.74 \ V$
$Fe^{3+}/Fe^{2+} = +0.77 \ V$
$Co^{3+}/Co^{2+} = +1.81 \ V$
Arranging these in increasing order of reduction potential:
$Al < Cr < Fe^{2+} < Co^{2+}$
Therefore,the order of reducing power (decreasing) is:
$Al > Cr > Fe^{2+} > Co^{2+}$

(B) $[Ni(PPh_3)_2Cl_2]$ is a $d^8$ complex. Since it is paramagnetic,it must have a tetrahedral geometry.
$(A)$ Tetrahedral complexes of the type $[MA_2B_2]$ do not show geometrical isomerism. Thus,statement $A$ is incorrect.
$(B)$ The complex is blue in colour,not white. Thus,statement $B$ is incorrect.
$(C)$ For $Ni^{2+}$ $(d^8)$ in a tetrahedral field,the configuration is $e^4 t_2^4$,which has $2$ unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$. Thus,statement $C$ is correct.
$(D)$ The $CFSE$ for a tetrahedral complex is calculated as $CFSE = (-0.6 \times n_e + 0.4 \times n_{t_2}) \Delta_t$. For $d^8$,$CFSE = (-0.6 \times 4 + 0.4 \times 4) \Delta_t = -0.8 \Delta_t$. The statement mentions $-0.8 \Delta_0$,which is incorrect.
$(E)$ $Ni(CO)_4$ is tetrahedral,and $[Ni(PPh_3)_2Cl_2]$ is also tetrahedral. Thus,statement $E$ is correct.
The incorrect statements are $A, B,$ and $D$.