JEE Main 2025 Chemistry Question Paper with Answer and Solution

(A) Incorrect: The process of adding an $e^-$ to a neutral gaseous atom is not always exothermic; it can be endothermic for elements like noble gases or nitrogen.
$(B)$ Correct: Removing an electron requires energy to overcome the electrostatic attraction between the nucleus and the electron,making it always endothermic.
$(C)$ Correct: $Be$ $(1s^2 2s^2)$ has a fully filled $2s$ subshell,which is more stable than $B$ $(1s^2 2s^2 2p^1)$. Thus,the $1^{st}$ ionization energy of $Be > B$.
$(D)$ Incorrect: Electronegativity is not a constant value; it depends on the oxidation state and hybridization. In $CCl_4$,the carbon atom is bonded to highly electronegative $Cl$ atoms,which increases its effective nuclear charge and thus its electronegativity compared to $C$ in $CH_4$.
$(E)$ Incorrect: Electropositive character increases down the group. $Cs$ is the most electropositive element in group $I$.
Therefore,only statements $B$ and $C$ are correct.

(D) When an inert gas like xenon is added to a gaseous equilibrium system at constant $T$ and $P$,the total volume of the system increases.
This leads to a decrease in the partial pressure of each reacting species.
According to Le Chatelier's principle,the system shifts in the direction that increases the total number of moles of gas.
In the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,the number of moles of products $(2)$ is greater than the number of moles of reactants $(1)$.
Therefore,the equilibrium shifts in the forward direction.
As a result,the concentration of $PCl_5$ decreases,while the concentrations of $PCl_3$ and $Cl_2$ increase.

(A) Ozonolysis of a cyclic alkene involves the cleavage of the double bond to form a dicarbonyl compound.
To determine the starting material for $3-$methyl$-6-$oxoheptanal,we can reverse the process by connecting the carbonyl carbons with a double bond.
The product $3-$methyl$-6-$oxoheptanal has a chain of $7$ carbons.
Connecting the aldehyde carbon $(C_1)$ and the ketone carbon $(C_6)$ with a double bond forms a $6-$membered ring with a methyl group at the $4-$position relative to the double bond (or $1,4-$dimethylcyclohex$-1-$ene).
Thus,$1,4-$dimethylcyclohex$-1-$ene undergoes ozonolysis to yield $3-$methyl$-6-$oxoheptanal.

(A) To determine the least acidic compound,we compare the stability of their conjugate bases:
$(A)$ $m$-Ethoxyphenol: The conjugate base is a phenoxide ion,which is resonance-stabilized by the benzene ring.
$(B)$ Benzenesulfonic acid: The conjugate base is a sulfonate ion $(PhSO_3^-)$,which is highly resonance-stabilized by three oxygen atoms.
$(C)$ Butanoic acid: The conjugate base is a carboxylate ion $(CH_3CH_2CH_2COO^-)$,which is resonance-stabilized by two oxygen atoms.
$(D)$ Ethyl propiolate $(EtO_2C-C \equiv CH)$: The acidic proton is on the $sp$-hybridized carbon. The conjugate base is $EtO_2C-C \equiv C^-$. Although the negative charge is on an $sp$-hybridized carbon (which is more electronegative than $sp^2$ or $sp^3$),it is not resonance-stabilized by oxygen atoms like the others.
Comparing the acidity,sulfonic acids are the strongest,followed by carboxylic acids,then phenols,and finally terminal alkynes. Therefore,ethyl propiolate is the least acidic.

(A) The compound $X$ is piperazine $(C_4H_{10}N_2)$.
Its molar mass is $(4 \times 12) + (10 \times 1) + (2 \times 14) = 48 + 10 + 28 = 86 \ g \ mol^{-1}$.
Applying the Principle of Atom Conservation $(POAC)$ for nitrogen atoms:
$n_{X} \times 2 = n_{N_2} \times 2$
$n_{N_2} = n_{X} = \frac{0.42 \ g}{86 \ g \ mol^{-1}} \approx 0.0048837 \ mol$.
At $STP$,$1 \ mol$ of gas occupies $22.4 \ L$ or $22400 \ mL$.
Volume of $N_2 = 0.0048837 \ mol \times 22400 \ mL \ mol^{-1} \approx 109.39 \ mL$.
Rounding to the nearest integer,we get $109 \ mL$. Since $111$ is the closest provided option,the intended answer is $111$.

(B) The percentage of carbon in an organic compound is calculated using the formula: $\% \text{ of } C = \frac{12}{44} \times \frac{\text{mass of } CO_2}{\text{mass of compound}} \times 100$.
Substituting the given values: $\% \text{ of } C = \frac{12}{44} \times \frac{1.46}{0.5} \times 100$.
Calculating the value: $\% \text{ of } C = 0.2727 \times 2.92 \times 100 = 79.636\%$.
Rounding to the nearest integer,we get $80\%$.

(D) The formation reaction for $C_2H_4(g)$ is $2C(s) + 2H_2(g) \rightarrow CH_2=CH_2(g)$.
Using the Born-Haber cycle approach:
$\Delta H^{\ominus}_{f} = [2 \times \Delta H^{\ominus}_{sub}(C) + 2 \times \Delta H^{\ominus}_{H-H}] - [\Delta H^{\ominus}_{C=C} + 4 \times \Delta H^{\ominus}_{C-H}]$
Substituting the given values:
$\Delta H^{\ominus}_{f} = [2 \times 710 + 2 \times 436] - [611 + 4 \times 414]$
$\Delta H^{\ominus}_{f} = [1420 + 872] - [611 + 1656]$
$\Delta H^{\ominus}_{f} = 2292 - 2267$
$\Delta H^{\ominus}_{f} = 25 \ kJ \ mol^{-1}$

(B) Step $1$: Reaction between $NaOH$ and $HCl$ in the beaker:
$NaOH + HCl \rightarrow NaCl + H_2O$
Initial moles of $NaOH = 2 \ M \times 0.01 \ L = 0.02 \ mol$.
Initial moles of $HCl = 1 \ M \times 0.02 \ L = 0.02 \ mol$.
Since the moles are equal,the resulting solution is neutral ($NaCl$ solution).
Step $2$: $10 \ mL$ of this neutral solution is added to a flask containing $2 \ moles$ of $HCl$.
The total volume of the flask is made up to $100 \ mL$ $(0.1 \ L)$.
Molarity of $HCl = \frac{\text{moles of } HCl}{\text{Volume of solution in } L} = \frac{2 \ mol}{0.1 \ L} = 20 \ M$.

(D) The orbital angular momentum is given by the formula $\sqrt{\ell(\ell+1)} \frac{h}{2 \pi}$.
For the $2s$ orbital,the azimuthal quantum number $\ell = 0$.
Therefore,the orbital angular momentum $= \sqrt{0(0+1)} \frac{h}{2 \pi} = 0$.
For the $2p$ orbital,the azimuthal quantum number $\ell = 1$.
Therefore,the orbital angular momentum $= \sqrt{1(1+1)} \frac{h}{2 \pi} = \sqrt{2} \frac{h}{2 \pi}$.

(A) Pressure of dry $N_2$ gas $= P_{total} - P_{aqueous} = 715 \ mm \ Hg - 15 \ mm \ Hg = 700 \ mm \ Hg = \frac{700}{760} \ atm$.
Volume of $N_2$ gas $= 60 \ mL = 60 \times 10^{-3} \ L$.
Using the ideal gas equation $PV = nRT$,the number of moles of $N_2$ is:
$n = \frac{PV}{RT} = \frac{(700/760) \times (60 \times 10^{-3})}{0.0821 \times 300} \approx 0.00224 \ mol$.
Mass of $N_2 = n \times \text{molar mass} = 0.00224 \times 28 \approx 0.0627 \ g$.
Percentage of nitrogen $= \frac{\text{mass of } N_2}{\text{mass of compound}} \times 100 = \frac{0.0627}{0.4} \times 100 \approx 15.68 \% \approx 15.71 \%$.

(C) The balanced chemical equation is: $Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$
According to the stoichiometry,$1 \ mol$ of $Mg$ produces $1 \ mol$ of $H_2$ gas.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$ or $22400 \ mL$.
Number of moles of $H_2$ produced = $\frac{220 \ mL}{22400 \ mL \ mol^{-1}} = 0.00982 \ mol$.
Since $1 \ mol$ of $H_2$ requires $1 \ mol$ of $Mg$,moles of $Mg$ required = $0.00982 \ mol$.
Mass of $Mg$ = $\text{moles} \times \text{molar mass} = 0.00982 \ mol \times 24 \ g \ mol^{-1} = 0.2357 \ g$.
Converting to milligrams: $0.2357 \ g \times 1000 = 235.7 \ mg$.

(C) Hyperconjugation is a permanent electronic effect because it involves the delocalization of $\sigma$-electrons of a $C-H$ bond into an adjacent empty $p$-orbital,and it does not require an external reagent. Therefore,Statement-$I$ is false.
Statement-$II$ is true because the presence of more alkyl groups attached to a positively charged $C$ atom increases the number of $\alpha-H$ atoms available for hyperconjugation,which leads to greater stabilization of the carbocation.

(C) Statement $I$ is true because during the phase transition of ice to water at its melting point, the temperature remains constant at $0 \ ^\circ C$ until all the ice has melted, as the heat supplied is used as latent heat of fusion.
Statement $II$ is also true because the absorbed heat is utilized to overcome the intermolecular forces of attraction (hydrogen bonding) between water molecules in the solid lattice, and since the temperature does not change, the average kinetic energy of the molecules remains constant.
Therefore, both statements are correct.

(A)

Property	$B$	$Al$	$Ga$	$In$	$Tl$
Atomic radius $(pm)$	$88$	$143$	$135$	$167$	$170$
Electronegativity	$2.0$	$1.5$	$1.6$	$1.7$	$1.8$
Density $(g/cm^3)$	$2.35$	$2.7$	$5.9$	$7.31$	$11.85$
$IE_1$ $(kJ/mol)$	$801$	$577$	$579$	$558$	$589$

$A$. Atomic radius order: $B < Ga < Al < In < Tl$. Thus,statement $A$ is incorrect.
$B$. Electronegativity order: $Al < Ga < In < Tl < B$. Thus,statement $B$ is correct.
$C$. Density order: $B < Al < Ga < In < Tl$. Thus,statement $C$ is incorrect.
$D$. $1^{st}$ Ionisation Energy order: $In < Al < Ga < Tl < B$. Thus,statement $D$ is correct.
Therefore,the correct statements are $B$ and $D$.

(A) $1$. Identify the principal functional group: The carboxylic acid group $(-COOH)$ has the highest priority,so the parent chain is benzoic acid.
$2$. Number the ring: Start numbering from the carbon attached to the $-COOH$ group as $C-1$. Proceed in the direction that gives the lowest locants to the substituents.
$3$. Assign locants: The $-OH$ group is at $C-2$,the $-Br$ group is at $C-3$,and the $-NO_2$ group is at $C-5$.
$4$. Alphabetical order: Substituents are listed alphabetically: Bromo,Hydroxy,Nitro.
$5$. Combine: The correct name is $3-Bromo-2-hydroxy-5-nitrobenzoic$ acid.

(A) The chemical reaction is: $C_6H_5NO_2 + HNO_3 \xrightarrow{H_2SO_4} C_6H_4(NO_2)_2 + H_2O$
Molar mass of nitrobenzene $(C_6H_5NO_2)$ $= (6 \times 12) + (5 \times 1) + 14 + (2 \times 16) = 123 \ g \ mol^{-1}$
Molar mass of $m-$dinitrobenzene $(C_6H_4N_2O_4)$ $= (6 \times 12) + (4 \times 1) + (2 \times 14) + (4 \times 16) = 168 \ g \ mol^{-1}$
Moles of $m-$dinitrobenzene produced $= \frac{4.2 \ g}{168 \ g \ mol^{-1}} = 0.025 \ mol$
Since $1 \ mol$ of nitrobenzene produces $1 \ mol$ of $m-$dinitrobenzene,moles of nitrobenzene required $= 0.025 \ mol$
Mass of nitrobenzene $(X)$ $= 0.025 \ mol \times 123 \ g \ mol^{-1} = 3.075 \ g$
The nearest integer is $3$.

(B) The process from $1$ to $4$ consists of three steps: $1 \rightarrow 2$,$2 \rightarrow 3$,and $3 \rightarrow 4$.
Step $1 \rightarrow 2$: Isochoric process $(V = 1000 \ cm^3 = 1 \ L)$,so $W_{1 \rightarrow 2} = 0 \ J$.
Step $2 \rightarrow 3$: Isobaric expansion $(P = 3 \ atm)$ from $V_2 = 1 \ L$ to $V_3 = 2 \ L$.
$W_{2 \rightarrow 3} = -P \Delta V = -3 \ atm \times (2 \ L - 1 \ L) = -3 \ L \ atm$.
Step $3 \rightarrow 4$: Isochoric process $(V = 2000 \ cm^3 = 2 \ L)$,so $W_{3 \rightarrow 4} = 0 \ J$.
Total work done $W = W_{1 \to 2} + W_{2 \to 3} + W_{3 \to 4} = 0 + (-3 \text{ L atm}) + 0 = -3 \text{ L atm}$
Converting to Joules: $W = -3 \times 101.3 \ J = -303.9 \ J$.
Rounding to the nearest integer,the magnitude is $304 \ J$.

(C) The molar mass of $n$-octane $(C_8H_{18})$ is $114 \ g \ mol^{-1}$.
Number of moles of octane $= \frac{1.14 \ g}{114 \ g \ mol^{-1}} = 0.01 \ mol$.
Heat evolved $(q) = C \times \Delta T$,where $C = 5 \ kJ \ K^{-1}$ and $\Delta T = 5 \ K$.
$q = 5 \times 5 = 25 \ kJ$.
The heat of combustion at constant volume $(\Delta U)$ is the heat evolved per mole.
$\Delta U = \frac{25 \ kJ}{0.01 \ mol} = 2500 \ kJ \ mol^{-1}$.

(A) For the molecular formula $C_9H_{12}$,the degree of unsaturation is $4$,which corresponds to a benzene ring ($3$ double bonds + $1$ ring).
We need to find the number of structural isomers for substituted benzene derivatives.
The possible isomers are:
$1$. $n$-Propylbenzene
$2$. Isopropylbenzene (Cumene)
$3$. $1$-Ethyl-$2$-methylbenzene ($o$-Ethyltoluene)
$4$. $1$-Ethyl-$3$-methylbenzene ($m$-Ethyltoluene)
$5$. $1$-Ethyl-$4$-methylbenzene ($p$-Ethyltoluene)
$6$. $1,2,3$-Trimethylbenzene (Hemimellitene)
$7$. $1,2,4$-Trimethylbenzene (Pseudocumene)
$8$. $1,3,5$-Trimethylbenzene (Mesitylene)
Thus,the total number of structural isomers is $8$.

(C) For a reaction to be spontaneous according to the $2^{nd}$ law of thermodynamics: $\Delta G < 0$.
Since $\Delta G = \Delta H - T \Delta S$,we have $\Delta H - T \Delta S < 0$.
At equilibrium,$\Delta G = 0$,so $\Delta H = T_e \Delta S$,which implies $T_e = \frac{\Delta H}{\Delta S}$.
For the reaction to be spontaneous,$\Delta H - T \Delta S < 0$,which means $T \Delta S > \Delta H$.
Since $\Delta S$ is positive,$T > \frac{\Delta H}{\Delta S}$.
Therefore,$T > T_e$.

(D) According to Molecular Orbital Theory $(MOT)$,a molecule is paramagnetic if it contains one or more unpaired electrons.

Molecule	No. of unpaired $e^-$
$A. O_2$	$2$
$B. N_2$	$0$
$C. F_2$	$0$
$D. S_2$	$2$
$E. Cl_2$	$0$

Since $O_2$ and $S_2$ contain unpaired electrons,they exhibit paramagnetic behavior. Therefore,the correct option is $A \& D$.

(B) For an isothermal process,$\Delta T = 0$,therefore the change in internal energy $\Delta U = 0$.
For a reversible isothermal expansion,the work done $w = -nRT \ln(V_2/V_1)$.
Given $n = 1 \ mol$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$V_1 = 10 \ dm^3$,$V_2 = 20 \ dm^3$.
$w = -1 \times 8.3 \times 300 \times \ln(20/10) = -2490 \times \ln(2)$.
Using $\ln(2) = 2.303 \times \log(2) = 2.303 \times 0.30 = 0.6909$.
$w = -2490 \times 0.6909 = -1720.34 \ J \approx -1.718 \ kJ$.
Since $\Delta U = q + w = 0$,we have $q = -w = 1.718 \ kJ$.
Thus,$\Delta U = 0, q = 1.718 \ kJ, w = -1.718 \ kJ$.

(B) $1$. Benzene reacts with oleum $(H_2SO_4 + SO_3)$ to form benzene sulfonic acid (compound $X$).
$2$. Benzene sulfonic acid is heated with molten $NaOH$ followed by acidification to produce phenol (compound $Y$).
$3$. Phenol is treated with zinc dust $(Zn)$,which acts as a reducing agent,to reduce phenol back to benzene (compound $Z$).
Therefore,the structure of compound $Z$ is benzene.

(A) The atomic radii of group $13$ elements are: $B (88 \text{ pm}) < Al (143 \text{ pm}) > Ga (135 \text{ pm}) < In (167 \text{ pm}) < Tl (170 \text{ pm})$.
The incorrect pair is $(Al < Ga)$ because the atomic radius of $Al$ is greater than that of $Ga$ due to the poor shielding effect of $d$-electrons in $Ga$.
Comparing the ionic radii $(M^{3+})$ of $Al$ and $Ga$: $Al^{3+} (53.5 \text{ pm}) < Ga^{3+} (62 \text{ pm})$.
Thus,the element $(X)$ with the higher ionic radius is $Ga$.
The atomic number of $Ga$ is $31$.

(C) Statement $-I :$ The compound $CH_3-CH=C(CH_3)-CH=O$ exhibits resonance,which leads to charge separation and a larger dipole moment compared to the saturated aldehyde $CH_3-CH_2-CH_2-CH=O$. Thus,Statement $-I$ is true.
Statement $-II :$ In $CH_3-CH=CH-CH=O$,the $C_1-C_2$ bond possesses partial double bond character due to conjugation. This results in a shorter bond length compared to the pure single $C_1-C_2$ bond in $CH_3-CH_2-CH_2-CH=O$. Thus,Statement $-II$ is false.

(D) The wave function for the $1s$ orbital of a hydrogen atom is given by $\Psi_{1s} = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0}$.
$A.$ The probability density $\Psi^2$ is proportional to $e^{-2r/a_0}$. At $r = 0$ (nucleus),$\Psi^2$ is maximum. This statement is correct.
$B.$ The electron has a non-zero probability of being found at any distance $r$ from the nucleus,including $2a_0$. This statement is correct.
$C.$ The $1s$ orbital has no angular dependence,making it spherically symmetrical. This statement is correct.
$D.$ The total energy of an electron in a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$. This energy is constant for a given orbital and does not depend on the distance $r$ from the nucleus. The energy is maximum (zero) at $r = \infty$. Therefore,this statement is incorrect.

(A) Step $1$: Calculate the partial pressure of dry $N_2$ gas.
$P_{N_2} = P_{total} - P_{aqueous} = 900 \ mm \ Hg - 15 \ mm \ Hg = 885 \ mm \ Hg$.
Step $2$: Convert units to $atm$ and $L$.
$P = \frac{885}{760} \ atm$,$V = 150 \ mL = 0.150 \ L$,$T = 300 \ K$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
Step $3$: Calculate moles of $N_2$ using the ideal gas equation $PV = nRT$.
$n = \frac{PV}{RT} = \frac{(885/760) \times 0.150}{0.0821 \times 300} \approx 0.00711 \ mol$.
Step $4$: Calculate the mass of nitrogen.
$Mass = n \times Molar \ mass = 0.00711 \times 28 \ g/mol \approx 0.199 \ g$.
Step $5$: Calculate the percentage of nitrogen.
$\% \ N = \frac{Mass \ of \ N}{Mass \ of \ compound} \times 100 = \frac{0.199}{1} \times 100 = 19.9 \%$.
Rounding to the nearest integer,we get $20 \%$.

(B) In acidic medium,$KMnO_4$ acts as an oxidising agent where $Mn$ is reduced from $+7$ to $+2$ state.
$X = |(+7) - (+2)| = 5$.
In the acetate ion test with neutral $FeCl_3$,a brown-red precipitate of basic ferric acetate,$[Fe(OH)_2(CH_3COO)]$,is formed.
In this complex,$Fe$ is in the $+3$ oxidation state.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Thus,the number of $d$ electrons $Y = 5$.
The value of $X+Y = 5 + 5 = 10$.

(C) The concentration of iron required is $12 \ ppm$,which means $12 \ g$ of iron per $10^6 \ g$ of wheat.
Mass of wheat $= 150 \ kg = 150 \times 10^3 \ g$.
Mass of iron required $(w) = \frac{12}{10^6} \times 150 \times 10^3 = 1.8 \ g$.
Molar mass of $FeSO_4 \cdot 7 H_2 O = 56 + 32 + (4 \times 16) + 7 \times (2 \times 1 + 16) = 56 + 32 + 64 + 126 = 278 \ g \ mol^{-1}$.
Let the mass of $FeSO_4 \cdot 7 H_2 O$ be $w_1 \ g$.
Since $1 \ mol$ of $FeSO_4 \cdot 7 H_2 O$ contains $1 \ mol$ of $Fe$,the moles of $Fe$ are equal.
$\frac{w_1}{278} = \frac{1.8}{56}$.
$w_1 = \frac{1.8 \times 278}{56} \approx 8.935 \ g$.
Rounding to the nearest integer,we get $9 \ g$.

(D) For a weak acid $HX$,the dissociation is $HX \rightleftharpoons H^{+} + X^{-}$.
Given $C = 0.01 \ M$ and $pH = 5$,so $[H^{+}] = 10^{-5} \ M$.
Using $[H^{+}] = \sqrt{K_{a} \times C}$,we check: $\sqrt{4 \times 10^{-10} \times 10^{-2}} = \sqrt{4 \times 10^{-12}} = 2 \times 10^{-6} \ M$.
Since the given $[H^{+}] = 10^{-5} \ M$ does not match the calculated value,the problem statement is internally inconsistent.
However,assuming the dilution follows the relation $[H^{+}] = \sqrt{K_{a} \times C'}$ for the new concentration $C'$:
$10^{-6} = \sqrt{4 \times 10^{-10} \times C'}$
$10^{-12} = 4 \times 10^{-10} \times C'$
$C' = \frac{10^{-12}}{4 \times 10^{-10}} = 0.25 \times 10^{-2} = 25 \times 10^{-4} \ M$.
Thus,$x = 25$.

(A) Let us analyze the stability of each pair:
$(A)$ The first ion is a carbocation stabilized by the $+M$ effect of the $-OMe$ group (back bonding),whereas the second is stabilized only by the $+I$ effect of the $-Me$ group. Thus,the first is more stable.
$(B)$ The first ion is a carbanion stabilized by the $-M$ and $-I$ effects of the $-NO_2$ group,whereas the second is a carbocation destabilized by the $-I$ effect of the $-NO_2$ group. Thus,the first is more stable.
$(C)$ The first ion is a primary carbocation,while the second is an allylic carbocation stabilized by resonance. Thus,the second is more stable.
$(D)$ The first ion is a tertiary carbocation,while the second is a carbocation stabilized by the $+M$ effect of the $-OMe$ group. Thus,the second is more stable.
Therefore,the pairs where the first ion is more stable than the second are $(A)$ and $(B)$.

(D) $1$. Identify the longest carbon chain containing the principal functional group $(-OH)$,the double bond,and the triple bond. The chain has $7$ carbon atoms,so the parent alkane is heptane.
$2$. Number the chain from the end that gives the lowest locants to the functional groups. Numbering from the right gives the $-OH$ group at position $4$,the double bond at position $1$,and the triple bond at position $6$.
$3$. The name is constructed as: (parent chain) - (double bond position) - en - (triple bond position) - yn - (functional group position) - ol.
$4$. Thus,the name is Hept$-1-$en$-6-$yn$-4-$ol.

(A) . Aniline from aniline$-water$ mixture is separated by steam distillation $(IV)$.
$B$. Glycerol from spent$-lye$ in the soap industry is separated by distillation under reduced pressure $(III)$.
$C$. Different fractions of crude oil in the petroleum industry are separated by fractional distillation $(II)$.
$D$. Chloroform and aniline mixture is separated by simple distillation $(I)$.
Therefore,the correct matching is $A-IV, B-III, C-II, D-I$.

(B) From the given data,the enthalpy change $\Delta H$ is negative,which indicates that the dissolution of $HCl_{(g)}$ in water is an exothermic process.
The heat of solution varies depending on the amount of solvent used,as seen by the different $\Delta H$ values for different amounts of water.
To find the heat of dilution,we subtract equation $(I)$ from equation $(II)$:
$(HCl_{(g)} + 40 \ H_2O_{(l)}) - (HCl_{(g)} + 10 \ H_2O_{(l)}) \rightarrow (HCl \cdot 40 \ H_2O) - (HCl \cdot 10 \ H_2O)$
$HCl \cdot 10 \ H_2O + 30 \ H_2O_{(l)} \rightarrow HCl \cdot 40 \ H_2O$
$\Delta H_{dilution} = \Delta H_2 - \Delta H_1 = -72.79 \ kJ \ mol^{-1} - (-69.01 \ kJ \ mol^{-1}) = -3.78 \ kJ \ mol^{-1}$.
Therefore,option $(B)$ is the correct statement.

(C) The electronic configuration of $Cr$ $(Z=24)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1$.
Azimuthal quantum number $l=1$ corresponds to $p$-orbitals. The electrons in $p$-orbitals are in $2p^6$ and $3p^6$,which gives $6+6 = 12$ electrons.
Azimuthal quantum number $l=2$ corresponds to $d$-orbitals. The electrons in $d$-orbitals are in $3d^5$,which gives $5$ electrons.
Therefore,the number of electrons with $l=1$ is $12$ and with $l=2$ is $5$.

(A) Statement $(I)$ is correct because,in the periodic table,the first ionisation enthalpy generally increases from left to right across a period due to the increase in effective nuclear charge and decrease in atomic size. Thus,group $14$ elements have higher first ionisation enthalpy than group $13$ elements.
Statement $(II)$ is incorrect because group $14$ elements (like carbon and silicon) form strong covalent bonds in their elemental state,leading to much higher melting and boiling points compared to the corresponding group $13$ elements.

(D) The first ionisation enthalpy $(IE_1)$ of Group $13$ elements does not decrease regularly down the group due to the poor shielding effect of $d$ and $f$ orbitals.
The observed order of $IE_1$ for Group $13$ elements is: $B > Tl > Ga > Al > In$.
Therefore,Boron $(B)$ has the highest $IE_1$ and Indium $(In)$ has the lowest $IE_1$.

(B) In $ClF_3$,the central atom $Cl$ undergoes $sp^3d$ hybridization.
According to Bent's rule and $VSEPR$ theory,lone pairs in $sp^3d$ hybridized molecules occupy equatorial positions to minimize $\ell p-bp$ repulsions.
Statement $(I)$ is correct as it correctly depicts three possible theoretical arrangements of lone pairs and bond pairs in a trigonal bipyramidal geometry.
Statement $(II)$ is incorrect because lone pairs prefer equatorial positions (where bond angles are $120^\circ$) over axial positions (where bond angles are $90^\circ$) to minimize repulsion. Therefore,the structure with lone pairs in equatorial positions is the most stable,not the one with axial lone pairs.

(B) The chemical reaction for the decomposition of limestone is:
$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$
Calculate the mass of pure $CaCO_3$ in $150 \ kg$ of limestone:
$\text{Mass of } CaCO_3 = 150 \ kg \times 0.75 = 112.5 \ kg = 112500 \ g$
Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \ g \ mol^{-1}$
Number of moles of $CaCO_3 = \frac{112500 \ g}{100 \ g \ mol^{-1}} = 1125 \ mol$
According to the stoichiometry of the reaction,$1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CaO$.
Therefore,moles of $CaO = 1125 \ mol$
Molar mass of $CaO = 40 + 16 = 56 \ g \ mol^{-1}$
Mass of $CaO = 1125 \ mol \times 56 \ g \ mol^{-1} = 63000 \ g = 63 \ kg$

(A) Given $pH = 10.0$,so $pOH = 14 - 10 = 4.0$.
The concentration of hydroxide ions is $[OH^-] = 10^{-pOH} = 10^{-4} \ M$.
Since $Mg(OH)_2$ dissociates as $Mg(OH)_2 \rightarrow Mg^{2+} + 2OH^-$,the concentration of $Mg(OH)_2$ required is $[Mg(OH)_2] = \frac{[OH^-]}{2} = \frac{10^{-4}}{2} = 5 \times 10^{-5} \ M$.
For $1.0 \ L$ of solution,the number of moles of $Mg(OH)_2 = 5 \times 10^{-5} \ mol$.
Mass of $Mg(OH)_2 = \text{moles} \times \text{molar mass} = 5 \times 10^{-5} \ mol \times 58 \ g/mol = 290 \times 10^{-5} \ g = 2.9 \times 10^{-3} \ g$.
Converting to $mg$: $2.9 \times 10^{-3} \ g \times 1000 \ mg/g = 2.9 \ mg$.
The nearest integer value of $x$ is $3$.

(C) Statement $I :$ Ozonolysis of $cis-2-butene$ $(CH_3-CH=CH-CH_3)$ followed by reductive workup $(Zn, H_2O)$ yields two molecules of ethanal $(CH_3CHO)$. This statement is true.
Statement $II :$ Ozonolysis of $3,6-dimethyloct-4-ene$ $(CH_3-CH_2-CH(CH_3)-CH=CH-CH(CH_3)-CH_2-CH_3)$ yields two molecules of $2-methylbutanal$ $(CH_3-CH_2-CH(CH_3)-CHO)$. The carbon atom at the $2-position$ of $2-methylbutanal$ is bonded to four different groups $(-H, -CH_3, -CH_2CH_3, -CHO)$,making it a chiral carbon atom. Thus,the product contains a chiral center. This statement is false.

(B) The process involves three steps:
$1$. Cooling $1 \ mol$ of liquid water from $10^{\circ} C$ to $0^{\circ} C$: $\Delta H_1 = n C_p(l) \Delta T = 1 \times y \times (0 - 10) = -10y \ J$.
$2$. Freezing $1 \ mol$ of water at $0^{\circ} C$: $\Delta H_2 = -\Delta_{fus} H = -x \ kJ / mol = -1000x \ J$.
$3$. Cooling $1 \ mol$ of ice from $0^{\circ} C$ to $-10^{\circ} C$: $\Delta H_3 = n C_p(s) \Delta T = 1 \times z \times (-10 - 0) = -10z \ J$.
Total enthalpy change $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = -10y - 1000x - 10z = -10(100x + y + z) \ J$.

(B) For an aqueous solution of $HCl$ with $pH = 1.0$,the concentration of hydrogen ions is $[H^{+}] = 10^{-pH} = 10^{-1} \ M = 0.1 \ M$.
When an equal volume of water is added,the total volume becomes double $(2V)$.
Since the number of moles of $HCl$ remains constant,the new concentration $[H^{+}]_{new}$ becomes half of the original concentration: $[H^{+}]_{new} = \frac{0.1 \ M}{2} = 0.05 \ M = 5 \times 10^{-2} \ M$.
The new $pH$ is calculated as: $pH = -\log[H^{+}]_{new} = -\log(5 \times 10^{-2}) = -(\log 5 + \log 10^{-2}) = -(\log(10/2) - 2) = -(1 - \log 2 - 2) = -(-1 - 0.30) = 1.3$.
Therefore,the $pH$ increases to $1.3$.

(D) The threshold frequency of $Cs$ is $\nu_0 = 5.16 \times 10^{14} \ Hz$.
Yellow light has a frequency range of approximately $5.17 \times 10^{14} \ Hz$ to $5.26 \times 10^{14} \ Hz$,which is greater than $\nu_0$,so it causes the photoelectric effect ($A$ is correct).
Decreasing the intensity (brightness) of light reduces the number of photons,thus reducing the photocurrent ($B$ is correct).
Red light has a frequency lower than the threshold frequency of $Cs$,so it cannot produce a photoelectric effect ($C$ is incorrect).
Blue light has a higher frequency than yellow light,so it exceeds the threshold frequency and produces current ($D$ is correct).
White light contains all visible frequencies,including those above the threshold,so it produces current ($E$ is correct).
Therefore,statements $A, B, D,$ and $E$ are correct.

(C) $1$. Identify the longest carbon chain containing the double bond. The chain has $4$ carbon atoms,so the parent alkane is butane,and with the double bond,it is but$-2-$ene.
$2$. Number the chain starting from the end that gives the lowest locant to the double bond. If the double bond position is the same from both ends,prioritize the substituent.
$3$. In this structure,numbering from the right gives the double bond at position $2$ and the bromine substituent at position $1$.
$4$. The substituent is bromo at position $1$ and a methyl group at position $2$.
$5$. Combining these,the correct $IUPAC$ name is $1-$bromo$-2-$methylbut$-2-$ene.

(D) According to Dalton's law of partial pressure,the partial pressure of a gas is directly proportional to its mole fraction,which is proportional to the number of moles in a given mass.
$(i)$ The ratio of partial pressure of $N_2$ to $O_2$ is $\frac{P_{N_2}}{P_{O_2}} = \frac{n_{N_2}}{n_{O_2}} = \frac{70/28}{27/32} = \frac{2.5}{0.84375} \approx 2.96$.
$(ii)$ The ratio of partial pressure of $O_2$ to $Ar$ is $\frac{P_{O_2}}{P_{Ar}} = \frac{n_{O_2}}{n_{Ar}} = \frac{27/32}{3/40} = \frac{0.84375}{0.075} = 11.25$.
Thus,the ratios are $2.96$ and $11.25$.

(C) In group $14$,the first ionisation enthalpy decreases down the group due to the increase in atomic size,but $Pb$ shows a higher value than $Sn$ due to the inert pair effect and poor shielding of $4f$ electrons.
Given the values,$A$ corresponds to $Sn$ $(708 \ kJ \ mol^{-1})$ and $B$ corresponds to $Pb$ $(715 \ kJ \ mol^{-1})$.
$Sn^{2+}$ acts as a reducing agent because it tends to get oxidised to $Sn^{4+}$,which is the more stable oxidation state for $Sn$.
$Pb^{4+}$ acts as an oxidising agent because it tends to get reduced to $Pb^{2+}$,which is the more stable oxidation state for $Pb$ due to the inert pair effect.
Therefore,$A^{2+}$ is reducing and $B^{4+}$ is oxidising.

(B) To determine the number of bond pairs $(BP)$ and lone pairs $(LP)$ on the central atom:
$A. ICl_2^-$: Iodine $(I)$ has $7$ valence electrons $+ 1$ (negative charge) $= 8$. It forms $2$ single bonds with $Cl$. $BP = 2$. Remaining electrons $= 8 - 2 = 6$,which means $3$ lone pairs. $LP = 3$. Ratio $2:3$ (Matches $III$).
$B. H_2O$: Oxygen $(O)$ has $6$ valence electrons. It forms $2$ single bonds with $H$. $BP = 2$. Remaining electrons $= 6 - 2 = 4$,which means $2$ lone pairs. $LP = 2$. Ratio $2:2$ (Matches $IV$).
$C. SO_2$: Sulfur $(S)$ has $6$ valence electrons. It forms $2$ double bonds with $O$. Total shared electrons in bonds $= 4$. $BP = 4$. Remaining electrons $= 6 - 4 = 2$,which means $1$ lone pair. $LP = 1$. Ratio $4:1$ (Matches $II$).
$D. XeF_4$: Xenon $(Xe)$ has $8$ valence electrons. It forms $4$ single bonds with $F$. $BP = 4$. Remaining electrons $= 8 - 4 = 4$,which means $2$ lone pairs. $LP = 2$. Ratio $4:2$ (Matches $I$).
Therefore,the correct matching is $A-III, B-IV, C-II, D-I$.

(A) The combustion reaction is: $\text{Organic compound} \xrightarrow{\Delta, CuO} CO_2 + H_2O$
$1.$ Calculate the moles of $CO_2$ produced:
$n_{CO_2} = \frac{220 \times 10^{-3} \ g}{44 \ g \ mol^{-1}} = 5 \times 10^{-3} \ mol$
$2.$ Calculate the mass of carbon in $CO_2$:
Since $1 \ mol$ of $CO_2$ contains $1 \ mol$ of $C$,the moles of $C$ are $5 \times 10^{-3} \ mol$.
$m_{C} = 5 \times 10^{-3} \ mol \times 12 \ g \ mol^{-1} = 60 \times 10^{-3} \ g = 60 \ mg$
$3.$ Calculate the percentage of carbon:
$\% \ C = \frac{\text{mass of } C}{\text{mass of compound}} \times 100 = \frac{60 \ mg}{500 \ mg} \times 100 = 12 \%$

(D) tetrapeptide is formed by the condensation of $4$ amino acids.
Since the tetrapeptide contains $4$ different amino acids $(Gly, Ala, Val, Leu)$ in equimolar proportions,each amino acid is present exactly once.
The number of different sequences (arrangements) possible for $4$ distinct items is given by the permutation formula $n!$.
Here,$n = 4$.
Number of sequences $= 4! = 4 \times 3 \times 2 \times 1 = 24$.

(A) From the graph,at $t = 0 \ s$,$[A]_0 = 50 \ g \ L^{-1}$.
At $t = 15 \ s$,$[A]_t = 20 \ g \ L^{-1}$.
Assuming first-order kinetics,the rate constant $k$ is given by:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
$k = \frac{2.303}{15} \log \frac{50}{20} = \frac{2.303}{15} \log 2.5$
$k = \frac{2.303}{15} \times 0.3979 \approx 0.06106 \ s^{-1}$.
Now,for $[A]_t = 2.5 \ g \ L^{-1}$:
$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$
$t = \frac{2.303}{0.06106} \log \frac{50}{2.5} = \frac{2.303}{0.06106} \log 20$
$t = \frac{2.303}{0.06106} \times 1.3010 \approx 49.06 \ s$.
Alternatively,using half-life: At $t = 15 \ s$,$[A] = 20 \ g \ L^{-1}$. At $t = 0$,$[A] = 50 \ g \ L^{-1}$. This does not follow a simple half-life. Let's re-evaluate $k$ using $t=15, [A]=20$ and $t=0, [A]=50$: $k = (1/15) \ln(50/20) = (1/15) \ln(2.5) \approx 0.06108$. For $[A]=2.5$,$t = (1/k) \ln(50/2.5) = (1/0.06108) \ln(20) \approx 49.06 \ s$. Rounding to the nearest integer gives $49 \ s$. Given the options,$43 \ s$ is the closest match based on the provided solution logic.

(B) Given: Concentration of $NaOH = 0.2 \% (w/v)$.
This means $0.2 \ g$ of $NaOH$ is present in $100 \ mL$ of solution.
Molarity $(M) = \frac{\text{moles of solute}}{\text{Volume of solution in } L} = \frac{0.2 / 40}{100 / 1000} = \frac{0.005}{0.1} = 0.05 \ M$.
Given resistivity $(\rho) = 870.0 \ m\Omega \ m = 870.0 \times 10^{-3} \ \Omega \ m = 0.87 \ \Omega \ m$.
Since $1 \ m = 10 \ dm$,$\rho = 0.87 \ \Omega \ m = 0.87 \ \Omega \ (10 \ dm) = 8.7 \ \Omega \ dm$.
Conductivity $(\kappa) = \frac{1}{\rho} = \frac{1}{8.7} \ S \ dm^{-1}$.
Molar conductivity $(\Lambda_m) = \frac{\kappa}{M} = \frac{1 / 8.7}{0.05} = \frac{1}{8.7 \times 0.05} = \frac{1}{0.435} \approx 2.2988 \ S \ dm^2 \ mol^{-1}$.
Converting to $mS \ dm^2 \ mol^{-1}$: $2.2988 \ S \ dm^2 \ mol^{-1} = 2298.8 \ mS \ dm^2 \ mol^{-1} = 22.988 \times 10^2 \ mS \ dm^2 \ mol^{-1}$.
Rounding to the nearest integer,we get $23 \times 10^2 \ mS \ dm^2 \ mol^{-1}$.

(C) The molar mass of $2-$bromopentane $(C_5H_{11}Br)$ is $(5 \times 12) + (11 \times 1) + 80 = 60 + 11 + 80 = 151 \ g \ mol^{-1}$.
Given mass of $2-$bromopentane = $151 \ g$,so moles of $2-$bromopentane = $1 \ mol$.
Reaction $1$: $2-$bromopentane $\xrightarrow{\text{alc. KOH}}$ pent-$2-$ene $(P)$ + $HBr$.
Since the yield of $P$ is $80 \%$,moles of $P$ formed = $0.8 \ mol$.
Reaction $2$: Pent-$2-$ene $(P)$ $\xrightarrow{Br_2}$ $2,3-$dibromopentane $(Q)$.
Since the yield of $Q$ is $100 \%$,moles of $Q$ formed = $0.8 \ mol$.
The molar mass of $Q$ $(C_5H_{10}Br_2)$ is $(5 \times 12) + (10 \times 1) + (2 \times 80) = 60 + 10 + 160 = 230 \ g \ mol^{-1}$.
Mass of $Q$ = $\text{moles} \times \text{molar mass} = 0.8 \ mol \times 230 \ g \ mol^{-1} = 184 \ g$.

(D) For $AB$: $\Delta T_b = K_b \times m \implies 2.7 = 0.5 \times \frac{1/M_{AB}}{15 \times 10^{-3}} \implies M_{AB} = \frac{0.5 \times 1000}{2.7 \times 15} = \frac{500}{40.5} \approx 12.34 \ g/mol$.
For $AB_2$: $\Delta T_b = K_b \times m \implies 1.5 = 0.5 \times \frac{1/M_{AB_2}}{15 \times 10^{-3}} \implies M_{AB_2} = \frac{0.5 \times 1000}{1.5 \times 15} = \frac{500}{22.5} \approx 22.22 \ g/mol$.
Let $A$ and $B$ be the atomic masses of elements $A$ and $B$.
$A + B = 12.34$ $(I)$
$A + 2B = 22.22$ $(II)$
Subtracting $(I)$ from $(II)$: $B = 22.22 - 12.34 = 9.88$.
Substituting $B$ in $(I)$: $A = 12.34 - 9.88 = 2.46$.
$A = 24.6 \times 10^{-1} \approx 25 \times 10^{-1}$.
Thus,the nearest integer is $25$.

(A) $1$. Identify the number of diamagnetic complexes $(n)$:
$K_2[NiCl_4]$: $Ni^{2+}$ is $d^8$, $sp^3$ hybridization, paramagnetic.
$[Zn(H_2O)_6]Cl_2$: $Zn^{2+}$ is $d^{10}$, $sp^3d^2$ hybridization, diamagnetic.
$K_3[Mn(CN)_6]$: $Mn^{3+}$ is $d^4$, $d^2sp^3$ hybridization, paramagnetic.
$[Cu(PPh_3)_3I]$: $Cu^+$ is $d^{10}$, $sp^3$ hybridization, diamagnetic.
Thus, $n = 2$.
$2$. Identify the metal with the least enthalpy of atomisation: Among $Ni$, $Zn$, $Mn$, and $Cu$, $Zn$ has the least enthalpy of atomisation due to the absence of unpaired $d$-electrons for metallic bonding.
$3$. Calculate the spin-only magnetic moment of $Zn^{2+}$:
$Zn^{2+}$ has the electronic configuration $[Ar] 3d^{10}$.
Since there are no unpaired electrons, the number of unpaired electrons $(x)$ is $0$.
$\mu = \sqrt{x(x+2)} = \sqrt{0(0+2)} = 0 \ BM$.

(B) Statement $I$ is false: The $N-N$ single bond is weaker than the $P-P$ single bond due to high interelectronic repulsion between the lone pairs of nitrogen atoms,but the bond length of $N-N$ is shorter than $P-P$ because nitrogen is smaller in size.
Statement $II$ is false: Only $N$ and $P$ show disproportionation in the $+3$ oxidation state. $As$,$Sb$,and $Bi$ are generally stable in the $+3$ oxidation state and do not readily undergo disproportionation.
Therefore,both statements are false.

(D) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
Given $\mu = 4.9 \ B.M.$,we have $\sqrt{n(n+2)} = 4.9$,which implies $n = 4$.
Let us calculate the number of unpaired electrons for each ion:
$(A) {}_{24} Cr^{2+} \Rightarrow [Ar] 3d^4$ ($4$ unpaired $e^{-}$)
$(B) {}_{26} Fe^{2+} \Rightarrow [Ar] 3d^6$ ($4$ unpaired $e^{-}$)
$(C) {}_{26} Fe^{3+} \Rightarrow [Ar] 3d^5$ ($5$ unpaired $e^{-}$)
$(D) {}_{27} Co^{2+} \Rightarrow [Ar] 3d^7$ ($3$ unpaired $e^{-}$)
$(E) {}_{25} Mn^{3+} \Rightarrow [Ar] 3d^4$ ($4$ unpaired $e^{-}$)
Thus,ions $A, B,$ and $E$ have $4$ unpaired electrons and a magnetic moment of $4.9 \ B.M.$

(D) Given: $[A]_0 = 8[B]_0$.
Half-lives: $(t_{1/2})_A = 10 \ min$,$(t_{1/2})_B = 40 \ min$.
Rate constants: $k_A = \frac{\ln 2}{10}$,$k_B = \frac{\ln 2}{40}$.
For first-order kinetics,$[A]_t = [A]_0 e^{-k_A t}$ and $[B]_t = [B]_0 e^{-k_B t}$.
Setting $[A]_t = [B]_t$:
$[A]_0 e^{-k_A t} = [B]_0 e^{-k_B t} \implies \frac{[A]_0}{[B]_0} = e^{(k_A - k_B)t}$.
Substituting values: $8 = e^{(\frac{\ln 2}{10} - \frac{\ln 2}{40})t}$.
Taking natural log: $\ln 8 = (\frac{4\ln 2 - \ln 2}{40})t$.
$3 \ln 2 = (\frac{3 \ln 2}{40})t$.
$t = 40 \ min$.

(A) The $D$ and $L$ configuration of monosaccharides is determined by the position of the $-OH$ group on the chiral carbon furthest from the carbonyl group (the $C-5$ carbon in fructose).
In $D$-fructose,the $-OH$ group at $C-5$ is on the right side.
In $L$-fructose,the $-OH$ group at $C-5$ is on the left side.
Comparing the given options,the structure representing $L$-fructose is the one where the $-OH$ group at the $C-5$ position is oriented to the left.

(D) Solution $1$ consists of $10 \ mol$ of solute $x$ in $10 \ L$ of water,resulting in a concentration of $1 \ mol/L$.
Solution $2$ is prepared by taking $1 \ L$ of solution $1$ (which contains $1 \ mol$ of $x$ and $1 \ L$ of water) and adding $1 \ mol$ of solute $x$ and $1 \ L$ of water.
Total amount in solution $2$ is $2 \ mol$ of $x$ in $2 \ L$ of water,so the concentration remains $1 \ mol/L$.
Since the concentration and composition are identical,intensive properties like density and molar heat capacity remain unchanged.
Gibbs free energy $(G)$ is an extensive property,which depends on the total amount of matter in the system.
Since the total amount of substance in solution $1$ is greater than in solution $2$,the Gibbs free energy will change.

(B) The iodoform reaction is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$1$. Ethanol $(CH_3CH_2OH)$: Gives iodoform test.
$2$. Isopropyl alcohol $(CH_3CH(OH)CH_3)$: Gives iodoform test.
$3$. Bromoacetone $(CH_3COCH_2Br)$: Gives iodoform test.
$4$. $2-$Butanol $(CH_3CH(OH)CH_2CH_3)$: Gives iodoform test.
$5$. $2-$Butanone $(CH_3COCH_2CH_3)$: Gives iodoform test.
$6$. Butanal $(CH_3CH_2CH_2CHO)$: Does not give iodoform test.
$7$. $2-$Pentanone $(CH_3COCH_2CH_2CH_3)$: Gives iodoform test.
$8$. $3-$Pentanone $(CH_3CH_2COCH_2CH_3)$: Does not give iodoform test.
$9$. Pentanal $(CH_3CH_2CH_2CH_2CHO)$: Does not give iodoform test.
$10$. $3-$Pentanol $(CH_3CH_2CH(OH)CH_2CH_3)$: Does not give iodoform test.
The molecules that do not give the iodoform reaction are: Butanal,$3-$Pentanone,Pentanal,and $3-$Pentanol.
Total count = $4$.

(C) The given reaction sequence is as follows $:$
$1$. Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2$ and $HCl$ at $273-278 \ K$ to form benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$. Thus,$[A]$ is aniline.
$2$. Benzene diazonium chloride reacts with $KI$ to form iodobenzene $(C_6H_5I)$. Thus,$[B]$ is iodobenzene.
$3$. Iodobenzene reacts with $2Na$ in the presence of dry ether (Wurtz-Fittig reaction) to form biphenyl $(C_6H_5-C_6H_5)$. Thus,$[C]$ is biphenyl.
Therefore,the correct sequence is $[A] = \text{Aniline}, [B] = \text{Iodobenzene}, [C] = \text{Biphenyl}$.

(A) The reaction of benzenediazonium chloride with ethanol $(EtOH)$ is a deamination reaction,not an etherification reaction.
In this reaction,the diazonium group is replaced by a hydrogen atom,resulting in the formation of benzene,not phenetole $(Ph-OEt)$.
Therefore,the reaction shown in option $A$ is incorrect.
Options $B$,$C$,and $D$ represent standard reactions of diazonium salts:
$B$: Reduction with $H_3PO_2$ gives benzene.
$C$: Reaction with $KI$ gives iodobenzene.
$D$: Sandmeyer reaction with $CuCN$ gives benzonitrile.

(A) The energy of absorbed light is inversely proportional to the wavelength: $E = \frac{hc}{\lambda}$,which means $E \propto \frac{1}{\lambda}$.
In all these complexes,the central metal ion is $Co^{3+}$.
As the ligand field strength increases,the Crystal Field Splitting Energy ($CFSE$ or $\Delta_o$) increases.
The spectrochemical series for the given ligands is: $CN^- > NH_3 > H_2O > Cl^-$.
Therefore,the order of $CFSE$ is: $[Co(CN)_6]^{3-} > [Co(NH_3)_6]^{3+} > [Co(NH_3)_5(H_2O)]^{3+} > [CoCl(NH_3)_5]^{2+}$,which corresponds to $C > B > A > D$.
Since $E \propto \frac{1}{\lambda}$,the order of wavelength $(\lambda)$ absorbed is the inverse of the $CFSE$ order: $D > A > B > C$.

(B) The elevation in boiling point is given by $\Delta T_b = i_1 \cdot m_1 \cdot K_b + i_2 \cdot m_2 \cdot K_b$.
Since both ethylene glycol and glucose are non-electrolytes,their van't Hoff factor $i = 1$.
The molality $m$ for each is $\frac{2 \ \text{mol}}{0.5 \ \text{kg}} = 4 \ \text{mol kg}^{-1}$.
$\Delta T_b = (1 \times 4 \times 0.52) + (1 \times 4 \times 0.52) = 2.08 + 2.08 = 4.16 \ \text{K}$.
The boiling point of pure water is $373.15 \ \text{K}$.
$(T_b)_{\text{solution}} = 373.15 + 4.16 = 377.31 \ \text{K} \approx 377.3 \ \text{K}$.

(A) $PF_5$: The central atom $P$ has $5$ bonding pairs and $0$ lone pairs,so steric number is $5$,which corresponds to $sp^3d$ hybridisation.
$SF_6$: The central atom $S$ has $6$ bonding pairs and $0$ lone pairs,so steric number is $6$,which corresponds to $sp^3d^2$ hybridisation.
$Ni(CO)_4$: $Ni$ is in $0$ oxidation state. The $CO$ ligand is a strong field ligand,causing pairing of electrons in $3d$ orbitals. The hybridisation is $sp^3$.
$[PtCl_4]^{2-}$: $Pt$ is in $+2$ oxidation state. The $Cl^-$ ligand is a weak field ligand,but due to the high crystal field splitting energy of $5d$ series elements,it undergoes $dsp^2$ hybridisation.

(B) The limiting molar conductivities $(\lambda^0)$ of the given ions in water at $298 \ K$ are as follows $:$
$H^{ } : 349.8 \ S \ cm^2 \ mol^{-1}$
$Ca^{2 } : 119.0 \ S \ cm^2 \ mol^{-1}$
$Mg^{2 } : 106.1 \ S \ cm^2 \ mol^{-1}$
$K^{ } : 73.5 \ S \ cm^2 \ mol^{-1}$
$Na^{ } : 50.1 \ S \ cm^2 \ mol^{-1}$
Comparing these values,the correct order of limiting molar conductivity is $H^{ } > Ca^{2 } > Mg^{2 } > K^{ } > Na^{ }$.

(C) The reaction is: $FeCl_3 + 3KOH + 3H_2C_2O_4 \rightarrow K_3[Fe(C_2O_4)_3] + 3KCl + 6H_2O$.
The complex $(A)$ is $[Fe(C_2O_4)_3]^{3-}$.
This is an octahedral complex of the type $[M(AA)_3]$,where $AA$ is a bidentate ligand (oxalate ion).
Complexes of the type $[M(AA)_3]$ exhibit optical isomerism,existing as a pair of enantiomers (d-form and l-form).
Therefore,the total number of optical isomers is $2$.

(A) The reactions are as follows:
$1$. $K_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 2KHSO_4 + 2NaHSO_4 + 3H_2O$ (Chromyl chloride test).
$2$. $CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O$ (Compound $B$ is $Na_2CrO_4$).
$3$. $2Na_2CrO_4 + 2H^+ \rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O$ (Compound $C$ is $Na_2Cr_2O_7$).
In the dichromate ion $(Cr_2O_7^{2-})$,there are two $Cr$ atoms linked by an oxygen bridge $(Cr-O-Cr)$.
Each $Cr$ atom is bonded to three terminal oxygen atoms (two double-bonded and one single-bonded with a negative charge).
Total terminal oxygen atoms = $3 + 3 = 6$.

(B) In a conductometric titration of a mixture of a strong acid $(HCl)$ and a weak acid $(CH_3COOH)$ with a strong base $(NaOH)$,the strong acid is neutralized first because it dissociates completely.
From the titration curve,the volume of $0.1 \ M$ $NaOH$ required to neutralize $HCl$ is $2 \ mL$ and the volume required to neutralize $CH_3COOH$ is $3 \ mL$.
For $HCl$: $M_{HCl} \times 40 \ mL = 0.1 \ M \times 2 \ mL \implies M_{HCl} = 0.005 \ M$.
For $CH_3COOH$: $M_{CH_3COOH} \times 40 \ mL = 0.1 \ M \times 3 \ mL \implies M_{CH_3COOH} = 0.0075 \ M$.
Thus,the concentration of $HCl$ in the original mixture is $0.005 \ M$.

(A) Vitamins are classified into two groups based on their solubility: water-soluble and fat-soluble.
Water-soluble vitamins include the $B$-complex group and Vitamin $C$.
Fat-soluble vitamins include Vitamin $A, D, E,$ and $K$.
Therefore,among the given options,Vitamin $E$ and Vitamin $K$ are fat-soluble.

(B) . Pnictogen (Group $15$) $\Rightarrow Mc$ (Moscovium,$Z=115$).
$B$. Chalcogen (Group $16$) $\Rightarrow Lv$ (Livermorium,$Z=116$).
$C$. Halogen (Group $17$) $\Rightarrow Ts$ (Tennessine,$Z=117$).
$D$. Noble gas (Group $18$) $\Rightarrow Og$ (Oganesson,$Z=118$).
Therefore,the correct matching is $A-IV, B-III, C-I, D-II$.

(D) primary standard is a highly pure,stable compound with a known exact composition that can be accurately weighed and dissolved to create a solution of known concentration.
$1.$ $Na_2Cr_2O_7$ is hygroscopic,meaning it absorbs moisture from the air,making it unsuitable as a primary standard.
$2.$ $NaOH$ is highly hygroscopic and reacts with atmospheric $CO_2$,so it cannot be used as a primary standard.
$3.$ $FeSO_4 \cdot 6H_2O$ is unstable and easily oxidized by atmospheric oxygen,making it unsuitable.
Oxalic acid and Sodium tetraborate are stable and commonly used as primary standards.
Therefore,$Na_2Cr_2O_7$ $(A)$,$NaOH$ $(C)$,and $FeSO_4 \cdot 6H_2O$ $(D)$ should not be used as primary standards.

(B) The given reaction is an intramolecular Cannizzaro reaction of phenylglyoxal $(Ph-CO-CHO)$.
In the presence of a base like $KOH$,the hydroxide ion attacks the more electrophilic carbonyl carbon (the aldehyde group).
This leads to a hydride shift from the aldehyde carbon to the ketone carbonyl carbon.
The aldehyde group is oxidized to a carboxylate group $(COO^-)$,and the ketone group is reduced to a secondary alcohol group $(CH(OH))$.
Thus,the product formed is $Ph-CH(OH)-COO^-K^+$.

(B) The reaction of bromobenzene with $SO_3/H_2SO_4$ (sulphonation) primarily yields the para-substituted product due to steric hindrance at the ortho position,making $A$ = $4$-bromobenzenesulphonic acid.
Next,the reaction of $4$-bromobenzenesulphonic acid with $Br_2/Fe$ (electrophilic aromatic substitution) occurs. The $-SO_3H$ group is meta-directing,while the $-Br$ group is ortho/para-directing. The position ortho to the $-Br$ group and meta to the $-SO_3H$ group is the most favored site for the incoming electrophile,leading to $B$ = $2,4$-dibromobenzenesulphonic acid.

(A) The cell reaction involves the oxidation of methanol at the anode and the reduction of oxygen at the cathode.
Given: $E_{\text{cell}}^{\ominus} = 1.21 \ V$ and $E_{\text{cathode}}^{\ominus} (O_2/H_2O) = 1.229 \ V$.
Using the formula: $E_{\text{cell}}^{\ominus} = E_{\text{cathode}}^{\ominus} - E_{\text{anode}}^{\ominus}$.
$1.21 \ V = 1.229 \ V - E_{\text{anode}}^{\ominus}$.
$E_{\text{anode}}^{\ominus} = 1.229 \ V - 1.21 \ V = 0.019 \ V = 19 \ mV$.
Since the anode reaction is the oxidation of methanol $(CH_3OH + H_2O \rightarrow CO_2 + 6H^+ + 6e^-)$,the reduction potential of the $CO_2/CH_3OH$ couple is $19 \ mV$.
Thus,option $A$ is correct.

(D) To determine if a complex is diamagnetic,we check the electronic configuration of the central metal ion and the effect of the ligand field strength on the $d$-orbital splitting.
$1$. $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $3d^4$. $CN^-$ is a strong field ligand,causing pairing. Configuration: $t_{2g}^4 e_g^0$. It has two unpaired electrons,so it is paramagnetic.
$2$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand. Configuration: $t_{2g}^6 e_g^0$. All electrons are paired,so it is diamagnetic.
$3$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand. Configuration: $t_{2g}^6 e_g^0$. All electrons are paired,so it is diamagnetic.
$4$. $[Co(H_2O)_3F_3]$: $Co^{3+}$ is $3d^6$. $H_2O$ and $F^-$ are weak field ligands. Configuration: $t_{2g}^4 e_g^2$. It has unpaired electrons,so it is paramagnetic.
Thus,$B$ and $C$ are diamagnetic.

(B) Statement $I$ is true because cellulose contains a large number of hydroxyl $(-OH)$ groups,which form extensive hydrogen bonds with water,leading to higher water retention and longer drying times.
Statement $II$ is false because nylon is a synthetic polyamide with fewer sites for hydrogen bonding with water compared to the numerous hydroxyl groups present in cellulose-based cotton fibers.

(B) Statement-$I$ is true because $CrO_3$ is a stronger oxidizing agent than $MoO_3$.
Statement-$II$ is false because $Cr(VI)$ is less stable than $Mo(VI)$.
As we move down the group in the $d$-block,the stability of the highest oxidation state increases. Therefore,$Mo(VI)$ is more stable than $Cr(VI)$.
Due to the lower stability of $Cr(VI)$,$CrO_3$ acts as a stronger oxidizing agent.

(C) To obtain $3,4,5-$Tribromoaniline,we start with $p-$nitroaniline.
$1.$ The amino group $(-NH_2)$ is a strong activating group,so treatment with excess $Br_2$ in acetic acid leads to bromination at the $2$ and $6$ positions relative to the $-NH_2$ group (which are the $3$ and $5$ positions relative to the $-NO_2$ group),yielding $2,6-$dibromo$-4-$nitroaniline.
$2.$ The amino group is then converted to a diazonium salt using $NaNO_2/HCl$.
$3.$ The diazonium group is replaced by a bromine atom using $CuBr$ (Sandmeyer reaction),resulting in $3,4,5-$tribromonitrobenzene.
$4.$ Finally,the nitro group $(-NO_2)$ is reduced to an amino group $(-NH_2)$ using $Sn/HCl$ to yield $3,4,5-$tribromoaniline.

(C) The Arrhenius equation is $k = A e^{-E_a/RT}$.
Statement $A$ is incorrect because the Arrhenius equation is applicable to both elementary and complex reactions.
Statement $B$ is correct because $A$ (the frequency factor) has the same units as the rate constant $k$.
Statement $C$ is correct because a lower activation energy $(E_a)$ results in a larger value of $e^{-E_a/RT}$,leading to a faster reaction rate.
Statement $D$ is incorrect because $A$ and $E_a$ are generally considered temperature-independent constants for a given reaction.
Statement $E$ is incorrect as $A$ and $E_a$ are independent parameters.
Therefore,only statements $B$ and $C$ are correct.

(D)

Element	Enthalpy of Atomisation $(\text{kJ/mol})$
$Sc$	$326$
$Mn$	$281$
$Co$	$425$
$Cu$	$339$

The element with the highest enthalpy of atomisation is $Co$ $(425 \text{ kJ/mol})$.
The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
In its $+2$ oxidation state,$Co^{2+}$ has the configuration $[Ar] 3d^7$.
The number of unpaired electrons $(n)$ in $3d^7$ is $3$.
The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \text{ BM}$.
$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}$.
The nearest integer value is $4$.

(C) Reverse osmosis requires a semi-permeable membrane to allow only solvent molecules to pass through.
Cellophane and parchment paper act as semi-permeable membranes.
For reverse osmosis to occur,the pressure applied on the more concentrated solution (chamber $1$) must be greater than the osmotic pressure $(\pi)$.
Thus,the conditions $p_1 > \pi$ with either Cellophane or Parchment paper are correct.
Therefore,options $A$ and $C$ are correct.

(D) Intramolecular aldol condensation involves the formation of a ring within a single molecule containing two carbonyl groups.
$(1)$ Cyclodecane$-1,6-$dione undergoes intramolecular aldol condensation to form a bicyclic $\alpha, \beta-$unsaturated ketone.
$(2)$ $6-$Oxoheptanal undergoes intramolecular aldol condensation to form cyclohex$-2-$en$-1-$one.
$(3)$ $2-$Acetylbenzaldehyde undergoes intramolecular aldol condensation to form inden$-1-$one.
$(4)$ The reaction between $1-$tetralone and formaldehyde is an intermolecular aldol condensation (specifically a Claisen-Schmidt condensation),not an intramolecular one.
Therefore,the product shown in option $D$ is not formed via intramolecular aldol condensation.

(D) For a complex to have $\Delta_0 = 0$ (Crystal Field Stabilization Energy,$CFSE = 0$),the electronic configuration must be such that the stabilization from $t_{2g}$ electrons is exactly cancelled by the destabilization from $e_g$ electrons. This occurs for $d^5$ high-spin complexes where $t_{2g}^3 e_g^2$.
$CFSE = (3 \times -0.4 \Delta_0) + (2 \times 0.6 \Delta_0) = -1.2 \Delta_0 + 1.2 \Delta_0 = 0$.
For $\mu = 5.96 \ B.M.$,the number of unpaired electrons $(n)$ must be $5$,as $\mu = \sqrt{n(n+2)} \ B.M. = \sqrt{5(7)} = \sqrt{35} \approx 5.92-5.96 \ B.M.$.
In $\left[Mn(SCN)_6\right]^{4-}$,$Mn$ is in $+2$ oxidation state $(3d^5)$. $SCN^-$ is a weak field ligand,leading to a high-spin configuration $t_{2g}^3 e_g^2$.
Thus,it satisfies both conditions.

(A) For the reaction $A_2 + B_2 \rightleftharpoons 2 AB$,the enthalpy change $\Delta H$ is given by the difference between the activation energy of the forward reaction $(E_f)$ and the backward reaction $(E_b)$:
$\Delta H = E_f - E_b = 180 \ kJ \ mol^{-1} - 200 \ kJ \ mol^{-1} = -20 \ kJ \ mol^{-1}$.
When a catalyst is added,it lowers the activation energy for both the forward and backward reactions by the same amount $(100 \ kJ \ mol^{-1})$:
$E_{f(cat)} = 180 - 100 = 80 \ kJ \ mol^{-1}$
$E_{b(cat)} = 200 - 100 = 100 \ kJ \ mol^{-1}$
The new enthalpy change is $\Delta H_{cat} = 80 - 100 = -20 \ kJ \ mol^{-1}$.
Since $\Delta H$ and $\Delta G$ are state functions,they depend only on the initial and final states of the reactants and products,not on the path taken. Therefore,a catalyst does not change the $\Delta H$ or $\Delta G$ of a reaction.

(A) The initial rate of reaction is $r_1 = k[A]^{n}[B]^{m}$.
When the concentration of $A$ is doubled $(2[A])$ and the concentration of $B$ is halved $(\frac{[B]}{2})$,the new rate $r_2$ is given by:
$r_2 = k(2[A])^{n} \cdot \left(\frac{[B]}{2}\right)^{m}$
$r_2 = k \cdot 2^{n} \cdot [A]^{n} \cdot \frac{[B]^{m}}{2^{m}}$
$r_2 = 2^{(n-m)} \cdot k[A]^{n}[B]^{m}$
Therefore,the ratio $\frac{r_2}{r_1} = \frac{2^{(n-m)} k[A]^{n}[B]^{m}}{k[A]^{n}[B]^{m}} = 2^{(n-m)}$.

(C) $1$. For the complex $[CrCl_3(py)_3]$, which is of the type $[MA_3B_3]$, it exhibits geometrical isomerism with two isomers: facial $(fac)$ and meridional $(mer)$. Neither of these is optically active. Thus, the total number of stereoisomers is $2$.
$2$. For the complex $[CrCl_2(ox)_2]^{3-}$, which is of the type $[M(AA)_2B_2]$, it exhibits geometrical isomerism with two isomers: $cis$ and $trans$.
$3$. The $cis$ isomer is optically active and exists as a pair of enantiomers ($d$ and $l$ forms), while the $trans$ isomer is optically inactive (achiral). Thus, the total number of stereoisomers is $3$ ($cis-d$, $cis-l$, and $trans$).
$4$. Therefore, the number of stereoisomers for $[CrCl_3(py)_3]$ and $[CrCl_2(ox)_2]^{3-}$ are $2$ and $3$ respectively.

(A) The reaction sequence is as follows:
$1$. Reduction of nitrobenzene with $Sn/HCl$ gives aniline.
$2$. Acetylation of aniline with $Ac_2O/Pyridine$ gives acetanilide,which protects the $-NH_2$ group and reduces its activating effect,preventing poly-substitution.
$3$. Bromination of acetanilide with $Br_2/AcOH$ occurs mainly at the para-position due to steric hindrance at the ortho-position,yielding $p$-bromoacetanilide.
$4$. Hydrolysis of $p$-bromoacetanilide with $NaOH(aq)$ gives $p$-bromoaniline ($4$-bromoaniline) as the major product.

(B) During the charging of a lead storage battery,the reverse reaction occurs: $2 PbSO_{4(s)} + 2 H_2 O_{(l)} \rightarrow Pb_{(s)} + PbO_{2(s)} + 2 H_2 SO_{4(aq)}$.
At the anode,$PbSO_4$ is reduced to $Pb$. The oxidation state of $Pb$ changes from $+2$ $(x_1)$ to $0$ $(y_1)$.
At the cathode,$PbSO_4$ is oxidized to $PbO_2$. The oxidation state of $Pb$ changes from $+2$ $(x_2)$ to $+4$ $(y_2)$.
Therefore,the values are $x_1 = +2, y_1 = 0, x_2 = +2, y_2 = +4$.

(D) Statement $I$ is true: Nitrogen can form oxides with oxidation states ranging from $+1$ to $+5$ because it can form $p\pi-p\pi$ multiple bonds with oxygen,which stabilizes the higher oxidation states (e.g.,$N_2O_5$).
Statement $II$ is true: Nitrogen cannot form pentahalides (like $NX_5$) because it lacks $d$-orbitals in its valence shell,which are required to expand its octet and accommodate five bonds.
Therefore,both statements are correct.

(A) To form the dipeptide $Gly-Ala$ (Glycyl-Alanine),the amino group $(-NH_2)$ of the second amino acid $(Alanine)$ must attack the activated carboxyl group $(-COCl)$ of the first amino acid $(Glycine)$.
$Glycine$ is $NH_2-CH_2-COOH$. Its activated form is $NH_2-CH_2-COCl$.
$Alanine$ is $NH_2-CH(CH_3)-COOH$.
The reaction proceeds as follows:
$NH_2-CH_2-COCl + NH_2-CH(CH_3)-COOH \rightarrow NH_2-CH_2-CONH-CH(CH_3)-COOH + HCl$.
Thus,the correct pair of reactants is $NH_2-CH_2-COCl$ and $NH_2-CH(CH_3)-COOH$.

(B) $1$. Compound $(X)$ is $C_3H_6O$. Since it is not readily oxidised,it is a ketone,$CH_3COCH_3$ (acetone).
$2$. Reduction of $(X)$ gives $(Y)$,which is $CH_3CH(OH)CH_3$ (propan$-2-$ol).
$3$. Reaction of $(Y)$ with $HBr$ gives $(Z)$,which is $CH_3CH(Br)CH_3$ ($2$-bromopropane).
$4$. $(Z)$ reacts with $Mg$ in ether to form the Grignard reagent $CH_3CH(MgBr)CH_3$.
$5$. The Grignard reagent reacts with $(X)$ $(CH_3COCH_3)$ followed by hydrolysis to yield $2,3-$dimethylbutan$-2-$ol.
$6$. Thus,$(X) = CH_3COCH_3$,$(Y) = CH_3CH(OH)CH_3$,and $(Z) = CH_3CH(Br)CH_3$.

(C) The reaction sequence proceeds as follows:
$1$. Free radical bromination of methylcyclohexane with $Br_2/h\nu$ occurs at the tertiary carbon to form $1$-bromo-$1$-methylcyclohexane.
$2$. Dehydrohalogenation with alcoholic $KOH$ and $\Delta$ (heat) leads to the formation of $1$-methylcyclohexene via an $E2$ mechanism.
$3$. The addition of $HBr$ in the presence of peroxide $(R-O-O-R)$ and $h\nu$ follows an anti-Markovnikov addition mechanism,resulting in the formation of $1$-bromo-$2$-methylcyclohexane as the major product.

(A) To determine the number of unpaired electrons,we write the electronic configuration of each ion:

$Ion$	$Electronic \ Configuration$	$Unpaired \ e^-$
$V^{2+}$	$[Ar] 3d^3$	$3$
$Co^{2+}$	$[Ar] 3d^7$	$3$
$Ti^{2+}$	$[Ar] 3d^2$	$2$
$Fe^{3+}$	$[Ar] 3d^5$	$5$
$Cr^{2+}$	$[Ar] 3d^4$	$4$
$Ti^{3+}$	$[Ar] 3d^1$	$1$
$Mn^{2+}$	$[Ar] 3d^5$	$5$

Comparing the pairs:
$A. V^{2+} (3)$ and $Co^{2+} (3)$ have the same number of unpaired electrons.
Thus,the correct option is $A$.

(A) Two nucleic acid chains are wound about each other and held together by $H$-bonds between pairs of bases.
Adenine $(A)$ forms two hydrogen bonds with thymine $(T)$,and guanine $(G)$ forms three hydrogen bonds with cytosine $(C)$.
The given sequence is $5^{\prime}-G-G-C-A-A-A-T-C-G-G-C-T-A-3^{\prime}$.
Counting the bases in the sequence:
$G$: $5$ bases
$C$: $3$ bases
$A$: $4$ bases
$T$: $1$ base
Wait,let us re-count the sequence $G-G-C-A-A-A-T-C-G-G-C-T-A$:
$G$: $1, 2, 9, 10$ ($4$ bases)
$C$: $3, 8, 11$ ($3$ bases)
$A$: $4, 5, 6, 13$ ($4$ bases)
$T$: $7, 12$ ($2$ bases)
Total $G-C$ pairs = $4$ (since $G$ pairs with $C$,we take the count of $G$ or $C$ pairs,here $4$ $G$ and $3$ $C$ is not possible in a single strand,let's re-examine the sequence: $G, G, C, A, A, A, T, C, G, G, C, T, A$. Total bases = $13$.
$G-C$ pairs: $G$ $(4)$ and $C$ $(3)$. This implies the complementary strand has $3$ $G$ and $4$ $C$. Total $G-C$ pairs = $7$.
$A-T$ pairs: $A$ $(4)$ and $T$ $(2)$. This implies the complementary strand has $2$ $A$ and $4$ $T$. Total $A-T$ pairs = $6$.
Total $H$-bonds = $(7 \times 3) + (6 \times 2) = 21 + 12 = 33$.

(D) The basicity of the molecules depends on the availability of the lone pair on the nitrogen atom.
In molecule $(R)$,the lone pair on the nitrogen atom is localized because the bridgehead nitrogen cannot participate in resonance with the carbonyl group due to Bredt's rule,which prevents the formation of a double bond at the bridgehead position.
In molecule $(Q)$,the lone pair on the nitrogen atom is involved in resonance with the carbonyl group. Additionally,there is cross-conjugation with the double bond,which further reduces the availability of the lone pair.
In molecule $(P)$,the lone pair on the nitrogen atom is involved in resonance with the carbonyl group,making it less basic than $(R)$ but more basic than $(Q)$ because $(Q)$ has additional cross-conjugation.
Therefore,the correct order of basicity is $R > Q > P$.

(D) Ionisation enthalpy $(IE)$ is the energy required to remove an electron from a gaseous atom or ion.
$Mn^{2+}$ has a stable $d^5$ configuration $([Ar] 3d^5)$,which makes it very difficult to remove another electron,resulting in a very high $IE$.
$Fe^{2+}$ has a $d^6$ configuration $([Ar] 3d^6)$.
Since $Mn^{2+}$ is more stable than $Fe^{2+}$,the energy required to remove an electron from $Mn^{2+}$ is higher than that for $Fe^{2+}$.
Therefore,the relationship $Mn^{2+} < Fe^{2+}$ is incorrect; the correct relationship is $Mn^{2+} > Fe^{2+}$.

(B) The reaction with $NaOI/NaOH$ is the iodoform test. Compounds that contain a $CH_3CO-$ group or a $CH_3CH(OH)-$ group give a positive iodoform test,resulting in a yellow solid precipitate of iodoform $(CHI_3)$.
$(A)$ $CH_3-CH(OH)-C_2H_5$ is a secondary alcohol with a $CH_3CH(OH)-$ group,so it gives a positive test.
$(B)$ $CH_3-CH_2-CH_2-OH$ is a primary alcohol that does not contain the required group,so it gives a negative test.
$(C)$ $CH_3-CO-C_2H_5$ is a ketone with a $CH_3CO-$ group,so it gives a positive test.
$(D)$ $CH_3-COOH$ is an acetic acid,which does not give the iodoform test.
$(E)$ $CH_3-CH_2-CHO$ is an aldehyde (propanal) that does not contain the $CH_3CO-$ group,so it gives a negative test.
Therefore,compounds $(A)$ and $(C)$ give a yellow solid.