JEE Main 2025 Chemistry Question Paper with Answer and Solution

(C) According to Raoult's Law for non-volatile solutes,the relative lowering of vapor pressure is equal to the mole fraction of the solute: $\frac{\Delta P}{P^0} = X_{solute}$.
Given $\Delta P_1 = 10 \ mm \ Hg$ and $X_{solute, 1} = 0.2$.
So,$\frac{10}{P^0} = 0.2 \implies P^0 = \frac{10}{0.2} = 50 \ mm \ Hg$.
Now,for a new decrease $\Delta P_2 = 20 \ mm \ Hg$,the new mole fraction of solute is $X_{solute, 2} = \frac{\Delta P_2}{P^0} = \frac{20}{50} = 0.4$.
The mole fraction of the solvent is $X_{solvent} = 1 - X_{solute, 2} = 1 - 0.4 = 0.6$.

(B) According to Raoult's Law,the relative lowering of vapor pressure is equal to the mole fraction of the solute: $\frac{\Delta P}{P^\circ} = X_{\text{solute}}$.
Given,$\Delta P_1 = 10 \ mm$ when $X_{\text{solute}, 1} = 0.2$.
So,$\frac{10}{P^\circ} = 0.2$,which gives $P^\circ = \frac{10}{0.2} = 50 \ mm$.
Now,for $\Delta P_2 = 20 \ mm$,the new mole fraction of the solute is $X_{\text{solute}, 2} = \frac{\Delta P_2}{P^\circ} = \frac{20}{50} = 0.4$.
The mole fraction of the solvent is $X_{\text{solvent}} = 1 - X_{\text{solute}, 2} = 1 - 0.4 = 0.6$.

$ \frac{9 \times 10^9 \times 6.67 \times 10^{-19} \times 9.6 \times 10^{-10}}{6.67 \times 10^{-11} \times 19.2 \times 10^{-27} \times 9 \times 10^{-27}}$
$ \frac{1}{2} \times 10^{45}$
Charge is not integral multiple of electron.

(D) The given molecule is $pent-3-en-2-ol$,which is $CH_3-CH(OH)-CH=CH-CH_3$.
This molecule has one chiral center at $C-2$ (the carbon attached to the $-OH$ group).
It also has a carbon-carbon double bond at $C-3$ and $C-4$,which can exhibit geometrical isomerism ($cis$ and $trans$ forms).
Since there is one chiral center $(n=1)$,there are $2^1 = 2$ optical isomers ($R$ and $S$ configurations).
Since there is one double bond capable of geometrical isomerism,there are $2$ geometrical isomers ($cis$ and $trans$).
Total stereoisomers = (Number of optical isomers) $\times$ (Number of geometrical isomers) = $2 \times 2 = 4$.

(A) Electronegativity generally increases across a period from left to right and decreases down a group.
The electronegativity values (on the Pauling scale) for the elements in option $A$ are:
$Mg = 1.2$,$Al = 1.5$,$B = 2.0$,$N = 3.0$.
Therefore,the correct increasing order is $Mg < Al < B < N$.
The given order $Al < Mg < B < N$ is incorrect.

(C) . $Al^{3+} < Mg^{2+} < Na^{+} < F^{-}$ represents the order of ionic radii for isoelectronic species,where size increases as nuclear charge decreases. Thus,$A-IV$.
$B$. $B < C < O < N$ represents the order of first ionisation enthalpy,where $N$ has a higher value than $O$ due to stable half-filled $p$-orbitals. Thus,$B-I$.
$C$. $B < Al < Mg < K$ represents the order of metallic character,which increases down a group and decreases across a period. Thus,$C-II$.
$D$. $Si < P < S < Cl$ represents the order of electronegativity,which increases across a period. Thus,$D-III$.
Therefore,the correct match is $A-IV, B-I, C-II, D-III$.

(C) Vitamin-$C$ is known as Ascorbic acid.

(A) Since the vessel is thermally insulated,there is no heat exchange with the surroundings,so $q = 0$.
Mechanical stirring from outside implies that work is being done on the system,so $w > 0$.
According to the $I^{st}$ law of thermodynamics,$\Delta U = q + w$.
Substituting the values,$\Delta U = 0 + w$,which means $\Delta U = w$.
Since $w > 0$,it follows that $\Delta U > 0$.

(D) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = a_0 \cdot \frac{n^2}{Z}$.
For the first excited state of $He^+$,the principal quantum number $n = 2$ and the atomic number $Z = 2$.
Substituting these values: $r = a_0 \cdot \frac{2^2}{2} = a_0 \cdot \frac{4}{2} = 2 a_0$.

(A) Statement $I$: Propyne $(CH_3-C \equiv CH)$ has an acidic hydrogen atom attached to the $sp$ hybridized carbon. It reacts with sodium metal as follows:
$CH_3-C \equiv CH + Na \rightarrow CH_3-C \equiv C^- Na^+ + \frac{1}{2} H_2 \uparrow$
Thus,$1 \ mole$ of propyne liberates $0.5 \ mole$ of $H_2$ gas. Statement $I$ is correct.
Statement $II$: Molar mass of propyne $(C_3H_4)$ is $40 \ g/mol$.
Moles of propyne = $\frac{4 \ g}{40 \ g/mol} = 0.1 \ mole$.
The reaction with $NaNH_2$ is:
$CH_3-C \equiv CH + NaNH_2 \rightarrow CH_3-C \equiv C^- Na^+ + NH_3 \uparrow$
$0.1 \ mole$ of propyne will liberate $0.1 \ mole$ of $NH_3$ gas.
Volume of $NH_3$ at $STP$ = $0.1 \ mole \times 22400 \ mL/mol = 2240 \ mL$.
The statement says $224 \ mL$,which is incorrect. Therefore,Statement $II$ is incorrect.

(B) The chemical reaction is: $CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$
Initially,the pressure of $CO_2$ is $0.5 \ atm$.
Let $x \ atm$ be the pressure of $CO_2$ that reacts.
At equilibrium,the partial pressures are: $P_{CO_2} = (0.5 - x) \ atm$ and $P_{CO} = 2x \ atm$.
The total pressure at equilibrium is given as $0.8 \ atm$.
$P_{\text{total}} = P_{CO_2} + P_{CO} = (0.5 - x) + 2x = 0.5 + x = 0.8 \ atm$.
Solving for $x$,we get $x = 0.3 \ atm$.
Now,calculate $K_P$:
$K_P = \frac{(P_{CO})^2}{P_{CO_2}} = \frac{(2x)^2}{(0.5 - x)} = \frac{(2 \times 0.3)^2}{(0.5 - 0.3)} = \frac{(0.6)^2}{0.2} = \frac{0.36}{0.2} = 1.8 \ atm$.

(D) $1$. Identify the principal functional group. The carboxylic acid $(-COOH)$ group has higher priority than the ester $(-COOCH_3)$ group. Thus,the parent chain is a hexanoic acid derivative.
$2$. Number the chain starting from the carbon of the $-COOH$ group as $C-1$.
$3$. The chain is $6$ carbons long: $CH_3-CH(COOCH_3)-CH_2-CH_2-CH(CH_3)-COOH$ is not the structure. Looking at the image,the structure is $CH_3-CH(COOH)-CH_2-CH_2-CH(COOCH_3)-CH_3$.
$4$. Numbering from the $-COOH$ end: $C-1$ is the carboxyl carbon,$C-2$ has a methyl group,$C-5$ has a methoxycarbonyl group $(-COOCH_3)$ and a methyl group.
$5$. The name is $5-$methoxycarbonyl$-2,5-$dimethylhexanoic acid.

(A) Analysis of the given statements:
$(A)$ Propene $(CH_3-CH=CH_2)$ does not show geometrical isomerism because one of the doubly bonded carbons is attached to two identical hydrogen atoms.
$(B)$ This is a correct statement. In a trans isomer,identical groups are on opposite sides of the double bond.
$(C)$ This is a correct statement. Cis$-$but$-2-$ene has a non-zero dipole moment due to the same side orientation of methyl groups,whereas trans$-$but$-2-$ene has a zero dipole moment due to symmetry.
$(D)$ $2-$methylbut$-2-$ene $(CH_3-C(CH_3)=CH-CH_3)$ does not show geometrical isomerism because one of the doubly bonded carbons is attached to two identical methyl groups.
$(E)$ This is an incorrect statement. Generally,the trans isomer has a higher melting point than the cis isomer due to better packing in the crystal lattice.
Thus,the incorrect statements are $(A)$,$(D)$,and $(E)$.

(A) The reaction between $CO_2$ and $Ca(OH)_2$ is: $Ca(OH)_2 + CO_2 \rightarrow CaCO_3 + H_2O$.
Let the moles of $CO_2$ be $n$.
According to the stoichiometry,$n$ moles of $CO_2$ react with $n$ moles of $Ca(OH)_2$.
Given that this amount of $CO_2$ reacted with exactly half the initial amount of $Ca(OH)_2$,the total initial moles of $Ca(OH)_2 = 2n$.
Excess $Ca(OH)_2 = 2n - n = n$.
This excess $Ca(OH)_2$ is neutralized by $HCl$: $Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O$.
Equivalents of $Ca(OH)_2 = $ Equivalents of $HCl$.
$n \times 2 = 0.1 \times 0.040 \times 1$.
$2n = 0.004 \implies n = 0.002 \ mol$.
At $STP$ $(1 \ atm, 273 \ K)$,$1 \ mol$ of gas occupies $22400 \ cm^3$.
Volume $x = 0.002 \times 22400 = 44.8 \ cm^3$.
Rounding to the nearest integer,$x = 45 \ cm^3$.

(B) The molar mass of $AgCl = 108 + 35.5 = 143.5 \ g \ mol^{-1}$.
The number of moles of $AgCl$ produced $= \frac{143.5 \times 10^{-3} \ g}{143.5 \ g \ mol^{-1}} = 10^{-3} \ mol$.
Since $1 \ mol$ of $AgCl$ contains $1 \ mol$ of $Cl$,the mass of chlorine $= 10^{-3} \ mol \times 35.5 \ g \ mol^{-1} = 35.5 \times 10^{-3} \ g = 35.5 \ mg$.
The percentage of chlorine $= \frac{\text{mass of } Cl}{\text{mass of organic compound}} \times 100 = \frac{35.5 \ mg}{180 \ mg} \times 100 = 19.72 \% \approx 20 \%$.

(C) To determine the linear geometry,we analyze the hybridization and lone pairs on the central atom:
$1$. $BeCl_2$: $sp$ hybridized,$0$ lone pairs,linear.
$2$. $CO_2$: $sp$ hybridized,$0$ lone pairs,linear.
$3$. $N_3^{-}$: $sp$ hybridized,$0$ lone pairs,linear.
$4$. $XeF_2$: $sp^3d$ hybridized,$3$ lone pairs on $Xe$,linear.
$5$. $NO_2^{+}$: $sp$ hybridized,$0$ lone pairs,linear.
$6$. $I_3^{-}$: $sp^3d$ hybridized,$3$ lone pairs on central $I$,linear.
Other species:
- $SO_2$: Bent ($sp^2$,$1$ lone pair).
- $NO_2$: Bent ($sp^2$,$1$ unpaired electron).
- $F_2O$: Bent ($sp^3$,$2$ lone pairs).
- $O_3$: Bent ($sp^2$,$1$ lone pair).
Total number of linear species is $6$.

(A) The dipole moments $(\mu)$ of the given compounds are as follows:
$NF_3$: $0.24 \ D$
$HBr$: $0.79 \ D$
$H_2S$: $0.95 \ D$
$CHCl_3$: $1.04 \ D$
Therefore,the increasing order of dipole moment is $NF_3 < HBr < H_2S < CHCl_3$.

(D) The reaction of $sec$-butylcyclohexane with bromine in the presence of sunlight $(hv)$ proceeds via a free radical mechanism.
Bromination is highly selective and prefers the formation of the most stable free radical intermediate.
In $sec$-butylcyclohexane,the hydrogen atom at the $sec$-butyl carbon (the carbon attached to the cyclohexane ring and the ethyl group) is a tertiary hydrogen.
Removal of this hydrogen creates a tertiary free radical,which is more stable than secondary or primary radicals.
Therefore,the bromine atom attacks this tertiary carbon to form the major product,which is $1$-bromo-$1$-cyclohexylbutane (also named $1$-bromo-$1$-cyclohexyl-$1$-methylpropane depending on nomenclature,but based on the structure provided in option $D$,it is the tertiary bromide).

(B) disproportionation reaction is one in which the same element in a given oxidation state is simultaneously oxidized and reduced.
For an element to undergo disproportionation,it must be in an intermediate oxidation state.
In $ClO_4^{-}$,the oxidation state of $Cl$ is calculated as: $x + (-2 \times 4) = -1$,which gives $x = +7$.
Since $+7$ is the maximum possible oxidation state for Chlorine,it cannot be further oxidized.
Therefore,$ClO_4^{-}$ cannot undergo disproportionation.

(B) $(A) \ dG = VdP - SdT$. At constant pressure,$dP = 0$,so $dG = -SdT$,which gives $\left(\frac{\partial G}{\partial T}\right)_{P} = -S$.
$(B) \ dH = nC_{P}dT$. Therefore,$\left(\frac{\partial H}{\partial T}\right)_{P} = C_{P}$.
$(C) \ dG = VdP - SdT$. At constant temperature,$dT = 0$,so $dG = VdP$,which gives $\left(\frac{\partial G}{\partial P}\right)_{T} = V$.
$(D) \ dU = nC_{V}dT$. Therefore,$\left(\frac{\partial U}{\partial T}\right)_{V} = C_{V}$.
Thus,the correct matching is $A-II, B-I, C-IV, D-III$.

(A) The dissociation of zirconium phosphate is represented as: $Zr_3(PO_4)_4(s) \rightleftharpoons 3Zr^{4+}(aq) + 4PO_4^{3-}(aq)$
Let the molar solubility be $s$. Then,the concentration of $Zr^{4+}$ is $3s$ and the concentration of $PO_4^{3-}$ is $4s$.
The solubility product expression is: $K_{sp} = [Zr^{4+}]^3 [PO_4^{3-}]^4$
Substituting the values: $K_{sp} = (3s)^3 (4s)^4$
$K_{sp} = (27s^3) \times (256s^4) = 6912s^7$
Solving for $s$: $s^7 = \frac{K_{sp}}{6912}$
Therefore,$s = (\frac{K_{sp}}{6912})^{\frac{1}{7}}$

(A) The stability of carbocations is determined by the electronic effects of the substituents attached to the benzene ring.
$1$. In $p$-methoxybenzyl carbocation,the $-OCH_3$ group at the para position exerts a strong $+M$ (mesomeric) effect,which significantly stabilizes the positive charge through resonance.
$2$. In $p$-methylbenzyl carbocation,the $-CH_3$ group provides stability through the $+H$ (hyperconjugation) effect.
$3$. In $m$-methoxybenzyl carbocation,the $-OCH_3$ group exerts a $-I$ (inductive) effect because it is at the meta position,which destabilizes the carbocation.
$4$. The benzyl carbocation has no additional substituents for stabilization.
Therefore,the order of stability is: $p$-methoxybenzyl carbocation > $p$-methylbenzyl carbocation > benzyl carbocation > $m$-methoxybenzyl carbocation.
The most stable carbocation is the $p$-methoxybenzyl carbocation.

(A) Statement-$I$ is false because elements on the extreme left of the periodic table are alkali and alkaline earth metals,which form basic oxides.
Statement-$II$ is true because non-metals on the extreme right of the periodic table form acidic oxides,which react with water to form acids (e.g.,$SO_3 + H_2O \rightarrow H_2SO_4$).
As we move from left to right in the periodic table,the non-metallic character increases,leading to an increase in the acidic strength of the oxides.

(D) Degenerate orbitals are orbitals that have the same energy level.
In a hydrogen-like atom,$2p_x$ and $2p_y$ orbitals possess the same energy,making them degenerate.
$A$ spectral line is observed only when an electron transitions between orbitals of different energy levels (i.e.,$\Delta E \neq 0$).
Since $2p_x$ and $2p_y$ have the same energy,the transition $2p_x \rightarrow 2p_y$ involves no change in energy $(\Delta E = 0)$,and thus no spectral line is observed.
Therefore,Statement-$I$ is false and Statement-$II$ is true.

(C) Statement $I$ is true because Lassaigne's test is used to detect nitrogen,sulphur,halogens,and phosphorus in organic compounds.
Statement $II$ is false because in Lassaigne's test,the organic compound is fused with metallic sodium,not magnesium,to convert the elements from covalent form into ionic form.

(C) secondary $(2^{\circ})$ hydrogen is a hydrogen atom attached to a secondary carbon atom (a carbon atom bonded to two other carbon atoms).
Let us analyze the structures:
$A$. $4-$Ethyl$-3,4-$dimethyloctane: This structure has five $2^{\circ}$ carbons,resulting in $10$ secondary hydrogens.
$B$. $2,2,4,4-$Tetramethylhexane: This structure has two $2^{\circ}$ carbons ($C3$ and $C5$),resulting in $4$ secondary hydrogens.
$C$. $2,2,3,3-$Tetramethylpentane: The structure is $CH_3-C(CH_3)_2-C(CH_3)_2-CH_2-CH_3$. The only $2^{\circ}$ carbon is at $C4$,which has two hydrogens attached to it. Thus,it has $2$ secondary hydrogens.
$D$. $2,2,4,5-$Tetramethylheptane: This structure has two $2^{\circ}$ carbons ($C3$ and $C6$),resulting in $4$ secondary hydrogens.
Therefore,the correct option is $C$.

(A) The molecular formula $C_6H_6$ corresponds to a degree of unsaturation of $4$ $(C_n H_{2n+2} - C_n H_m / 2 = (2(6)+2-6)/2 = 4)$.
This means the molecule has $4$ double bonds or equivalent unsaturation (rings/triple bonds).
The compound takes up $4$ moles of $H_2$ for complete hydrogenation,confirming it has $4$ $\pi$ bonds.
Each $\pi$ bond contains $2$ $\pi$ electrons,so $4$ $\pi$ bonds contain $4 \times 2 = 8$ $\pi$ electrons.
The compound is $1,3,5$-trimethylenecyclohexane or a similar isomer that is highly symmetric to yield only one monobromo derivative.
Thus,the number of $\pi$ electrons is $8$.

(D) The formation reaction for $C_{2}H_{6(g)}$ is: $2 C(\text{graphite}) + 3 H_{2(g)} \rightarrow C_{2}H_{6(g)}$,$\Delta H_{f}^{\circ} = ?$
Using Hess's Law:
$\Delta H_{f}^{\circ} = [2 \Delta H_{2}^{\circ} + 3 \Delta H_{3}^{\circ}] - \Delta H_{1}^{\circ}$
$\Delta H_{f}^{\circ} = 2(-393.5) + 3(-286) - (-1550)$
$\Delta H_{f}^{\circ} = -787 - 858 + 1550$
$\Delta H_{f}^{\circ} = -1645 + 1550 = -95 \ kJ \ mol^{-1}$
The magnitude of $\Delta H_{f}^{\circ}$ is $|-95| = 95 \ kJ \ mol^{-1}$.

(A) The formula for the final concentration $(M_F)$ of a mixture is $M_F = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2}$.
Given: $M_1 = 2 \ M$,$V_1 = 20 \ mL$,$M_2 = 0.5 \ M$,$V_2 = 400 \ mL$.
Substituting the values: $M_F = \frac{2 \times 20 + 0.5 \times 400}{20 + 400} = \frac{40 + 200}{420} = \frac{240}{420} \approx 0.5714 \ M$.
Converting to the required format: $0.5714 \ M = 57.14 \times 10^{-2} \ M$.
Rounding to the nearest integer,we get $57 \times 10^{-2} \ M$.

(A) Palladium $(Pd)$ belongs to the $5^{th}$ period.
Iridium $(Ir)$,Osmium $(Os)$,and Platinum $(Pt)$ all belong to the $6^{th}$ period.
Therefore,Palladium is the element that does not belong to the same period as the others.

(A) Given wavelength $\lambda = 900 \ nm = 900 \times 10^{-7} \ cm = 9 \times 10^{-5} \ cm$.
Rydberg constant $R_{H} = 10^5 \ cm^{-1}$.
Using Rydberg equation for $H$-atom $(Z=1)$: $\frac{1}{\lambda} = R_{H} \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
Substituting the values: $\frac{1}{9 \times 10^{-5}} = 10^5 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
$\frac{1}{9} = \frac{1}{n_1^2} - \frac{1}{n_2^2}$.
For this equality to hold,$n_1 = 3$ and $n_2 = \infty$.
This corresponds to the Paschen series transition from $\infty \rightarrow 3$.

(A) The molar mass of $CO_2$ is $44 \ g/mol$.
Initial moles of $CO_2$ = $\frac{x \times 10^{-3} \ g}{44 \ g/mol} = \frac{x \times 10^{-3}}{44} \ mol$.
Moles of $CO_2$ removed = $\frac{10^{21}}{6.02 \times 10^{23}} \approx 1.66 \times 10^{-3} \ mol$.
Moles left = $\text{Initial moles} - \text{Removed moles}$.
$2.8 \times 10^{-3} = \frac{x \times 10^{-3}}{44} - 1.66 \times 10^{-3}$.
$2.8 \times 10^{-3} + 1.66 \times 10^{-3} = \frac{x \times 10^{-3}}{44}$.
$4.46 \times 10^{-3} = \frac{x \times 10^{-3}}{44}$.
$x = 4.46 \times 44 = 196.24 \ mg$.

(B) The total entropy change $\Delta S_{\text{total}}$ is the sum of entropy changes for each step of the process:
$1$. Heating ice from $-5^{\circ} C$ $(268 \ K)$ to $0^{\circ} C$ $(273 \ K)$: $\Delta S_1 = \int_{268 \ K}^{273 \ K} \frac{C_{p,m}(\text{ice})}{T} \ dT$
$2$. Melting ice at $0^{\circ} C$ $(273 \ K)$: $\Delta S_2 = \frac{\Delta H_{m, \text{fusion}}}{273 \ K}$
$3$. Heating water from $0^{\circ} C$ $(273 \ K)$ to $100^{\circ} C$ $(373 \ K)$: $\Delta S_3 = \int_{273 \ K}^{373 \ K} \frac{C_{p,m}(\text{water})}{T} \ dT$
$4$. Vaporizing water at $100^{\circ} C$ $(373 \ K)$: $\Delta S_4 = \frac{\Delta H_{m, \text{vaporisation}}}{373 \ K}$
$5$. Heating steam from $100^{\circ} C$ $(373 \ K)$ to $110^{\circ} C$ $(383 \ K)$: $\Delta S_5 = \int_{373 \ K}^{383 \ K} \frac{C_{p,m}(\text{steam})}{T} \ dT$
Summing these gives the expression in option $B$.

(A) To determine the stability,we analyze the aromaticity of each species:
$1$. Species $p$ is the cyclopropenyl anion. It has $4n$ $\pi$ electrons ($4$ $\pi$ electrons),making it antiaromatic,which is highly unstable.
$2$. Species $q$ is the cyclopentadienyl anion. It has $4n+2$ $\pi$ electrons ($6$ $\pi$ electrons),making it aromatic,which is highly stable.
$3$. Species $r$ is cyclooctatetraene. It is non-aromatic because it adopts a non-planar tub-shaped conformation to avoid antiaromaticity,making it more stable than antiaromatic species but less stable than aromatic ones.
Therefore,the stability order is $q$ (aromatic) $> r$ (non-aromatic) $> p$ (antiaromatic).
The correct option is $A$.

(D) The chlorination of propane $(CH_3-CH_2-CH_3)$ gives two dichloro products: $1,2-dichloropropane$ $(CH_3-CHCl-CH_2Cl)$ and $1,3-dichloropropane$ $(ClCH_2-CH_2-CH_2Cl)$.
Among these,$1,2-dichloropropane$ is optically active because the carbon at position $2$ is chiral (bonded to $-H, -CH_3, -Cl, -CH_2Cl$). Thus,$x$ is $1,2-dichloropropane$ $(CH_3-CHCl-CH_2Cl)$.
When $x$ $(CH_3-CHCl-CH_2Cl)$ is further chlorinated,we look for structural isomers of trichloropropane formed by replacing one $H$ atom with a $Cl$ atom:
$1.$ Replacing $H$ at $C-1$: $CH_3-CHCl-CHCl_2$ $(1,1,2-trichloropropane)$
$2.$ Replacing $H$ at $C-2$: $CH_3-CCl_2-CH_2Cl$ $(1,2,2-trichloropropane)$
$3.$ Replacing $H$ at $C-3$: $ClCH_2-CHCl-CH_2Cl$ $(1,2,3-trichloropropane)$
Thus,there are $3$ structural isomers of trichloropropane obtained from $x$.

(D) Statement $I$ is true because in Lassaigne's test,covalent organic compounds are fused with sodium metal to form ionic salts like $NaCN$,$Na_2S$,and $NaSCN$.
Statement $II$ is false because an organic compound containing both $N$ and $S$ forms sodium thiocyanate $(NaSCN)$. When this extract is treated with $Fe^{3+}$ ions (usually from $FeCl_3$),it forms ferric thiocyanate,which gives a blood-red colour,not Prussian blue. Prussian blue is obtained only when $N$ is present without $S$.

(A) The condition for precipitation is $Q_{ip} > K_{sp}$.
For $A(OH)_{2}$:
$[A^{2+}][OH^{-}]^2 > 9 \times 10^{-10}$
Given $[A^{2+}] = 1 \ M$,we have $[OH^{-}]^2 > 9 \times 10^{-10}$,which implies $[OH^{-}] > 3 \times 10^{-5} \ M$.
For $B(OH)_{3}$:
$[B^{3+}][OH^{-}]^3 > 27 \times 10^{-18}$
Given $[B^{3+}] = 1 \ M$,we have $[OH^{-}]^3 > 27 \times 10^{-18}$,which implies $[OH^{-}] > 3 \times 10^{-6} \ M$.
Since the concentration of $OH^{-}$ required for the precipitation of $B(OH)_{3}$ $(3 \times 10^{-6} \ M)$ is lower than that required for $A(OH)_{2}$ $(3 \times 10^{-5} \ M)$,$B(OH)_{3}$ will precipitate first.

(A) $A \rightarrow IV$: $CCl_4$ and $CO_2$ follow the octet rule as all atoms have $8$ electrons in their valence shell.
$B \rightarrow II$: $BCl_3$ and $AlCl_3$ have incomplete octets with only $6$ electrons around the central atom.
$C \rightarrow I$: $NO$ and $NO_2$ are odd-electron molecules containing an unpaired electron.
$D \rightarrow III$: $H_2SO_4$ and $PCl_5$ have expanded octets where the central atom has more than $8$ electrons.
Therefore,the correct matching is $A-IV, B-II, C-I, D-III$.

(A) The dissociation reaction of ethylamine is: $C_2H_5NH_2 + H_2O \rightleftharpoons C_2H_5NH_3^+ + OH^-$
Given concentration $C = 1 \ mM = 10^{-3} \ M$.
Given $pH = 9$,so $pOH = 14 - 9 = 5$.
Thus,$[OH^-] = 10^{-pOH} = 10^{-5} \ M$.
Since $[OH^-] = [C_2H_5NH_3^+] = 10^{-5} \ M$ and $[C_2H_5NH_2] \approx C = 10^{-3} \ M$ (as degree of ionization is neglected),
$K_b = \frac{[C_2H_5NH_3^+][OH^-]}{[C_2H_5NH_2]} = \frac{10^{-5} \times 10^{-5}}{10^{-3}} = 10^{-7}$.
Comparing $10^{-7}$ with $10^{-x}$,we get $x = 7$.

(B) The molar mass of $BaSO_4 = 137 + 32 + (4 \times 16) = 233 \ g \ mol^{-1}.$
Millimoles of $BaSO_4 = \frac{466 \ mg}{233 \ g \ mol^{-1}} = 2 \ mmol.$
Since $1 \ mol$ of $BaSO_4$ contains $1 \ mol$ of $S$,the moles of $S = 2 \ mmol = 2 \times 10^{-3} \ mol.$
Mass of $S = 2 \times 10^{-3} \ mol \times 32 \ g \ mol^{-1} = 0.064 \ g = 64 \ mg.$
Percentage of $S = \frac{\text{Mass of } S}{\text{Mass of organic compound}} \times 100 = \frac{64 \ mg}{160 \ mg} \times 100 = 40 \%.$

(A) The reaction is $N_2O_4(g) \rightarrow 2NO_2(g)$.
Given: $\Delta H^{\circ} = 55.0 \ kJ \ mol^{-1} = 55000 \ J \ mol^{-1}$.
Given: $\Delta S^{\circ} = 175.0 \ J \ K^{-1} \ mol^{-1}$.
Temperature $T = 25 + 273 = 298 \ K$.
The Gibbs free energy change is given by the formula: $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$.
Substituting the values: $\Delta G^{\circ} = 55000 \ J \ mol^{-1} - (298 \ K \times 175.0 \ J \ K^{-1} \ mol^{-1})$.
$\Delta G^{\circ} = 55000 - 52150 = 2850 \ J \ mol^{-1}$.

(D) The spontaneity of a reaction is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T \Delta S$.
For a reaction to be spontaneous,$\Delta G < 0$.
$1$. If $\Delta H > 0$ and $\Delta S < 0$,then $\Delta G$ is always positive,so the reaction is non-spontaneous at all temperatures ($A$ is correct).
$2$. If $\Delta H > 0$ and $\Delta S > 0$,then $\Delta G < 0$ only at high temperatures ($B$ is incorrect).
$3$. If $\Delta H < 0$ and $\Delta S < 0$,then $\Delta G < 0$ only at low temperatures ($C$ is correct).
$4$. If $\Delta H < 0$ and $\Delta S > 0$,then $\Delta G$ is always negative,so the reaction is spontaneous at all temperatures ($D$ is correct).
Thus,conditions $A, C,$ and $D$ are correct. However,based on the provided options,the combination $A$ and $C$ is the most appropriate match.

(A) For the reaction: $X_2Y_{(g)} \rightleftharpoons X_{2(g)} + \frac{1}{2}Y_{2(g)}$
Initial moles: $1, 0, 0$
Moles at equilibrium: $(1-x), x, \frac{x}{2}$
Total moles at equilibrium: $1-x+x+\frac{x}{2} = 1+\frac{x}{2}$
Partial pressures: $P_{X_2Y} = \frac{1-x}{1+x/2} P$,$P_{X_2} = \frac{x}{1+x/2} P$,$P_{Y_2} = \frac{x/2}{1+x/2} P$
$Kp = \frac{P_{X_2} \cdot (P_{Y_2})^{1/2}}{P_{X_2Y}} = \frac{[x/(1+x/2)] P \cdot [x/2(1+x/2) P]^{1/2}}{[(1-x)/(1+x/2)] P}$
Since $x \ll 1$,we approximate $1-x \approx 1$ and $1+x/2 \approx 1$.
$Kp \approx \frac{x \cdot (x/2)^{1/2} \cdot P^{1/2}}{1} = \frac{x^{3/2} \cdot P^{1/2}}{\sqrt{2}}$
$Kp^2 = \frac{x^3 \cdot P}{2} \implies x^3 = \frac{2 Kp^2}{P}$
$x = \left( \frac{2 Kp^2}{P} \right)^{1/3}$

(C) Statement $I$ is true: For a given principal quantum number $n$,the total number of orbitals in a shell is calculated as $n^2$.
Statement $II$ is true: The magnetic quantum number $m_l$ determines the spatial orientation of orbitals and ranges from $-l$ to $+l$ including zero,which corresponds to the number of orbitals in a subshell.

(B) Ozonolysis involves the cleavage of double bonds to form carbonyl compounds. By analyzing the structure of each isomer in List-$I$ and performing the oxidative cleavage of the double bonds,we can determine the corresponding products in List-$II$.
$(A)$ The isomer has two double bonds,which upon ozonolysis yield a dialdehyde with a methyl substituent. This matches product $(III)$.
$(B)$ The isomer has two double bonds,which upon ozonolysis yield a keto-aldehyde. This matches product $(IV)$.
$(C)$ The isomer has two double bonds,which upon ozonolysis yield a dialdehyde. This matches product $(I)$.
$(D)$ The isomer has two double bonds,which upon ozonolysis yield a keto-aldehyde. This matches product $(II)$.
Thus,the correct matching is $A-III, B-IV, C-I, D-II$.

(D) The melting point of group $14$ elements decreases as we move down the group from $C$ to $Sn$,but $Pb$ has a slightly higher melting point than $Sn$ due to its metallic structure and bonding characteristics. The order of melting points is $C > Si > Ge > Pb > Sn$.
Atomic numbers and their corresponding elements:
$Z=6$ $(C)$: $3730 \ ^{\circ}C$
$Z=14$ $(Si)$: $1410 \ ^{\circ}C$
$Z=32$ $(Ge)$: $937 \ ^{\circ}C$
$Z=50$ $(Sn)$: $232 \ ^{\circ}C$
$Z=82$ $(Pb)$: $327 \ ^{\circ}C$
Comparing the given options $(Z=14, 6, 82, 50)$,the element with the lowest melting point is $Sn$ $(Z=50)$.

(A) The auto-ionization of water is an endothermic process: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.
As temperature increases,the equilibrium constant $K_W$ increases according to Le Chatelier's principle.
Since $K_W = [H^+][OH^-]$ and at neutrality $[H^+] = [OH^-] = \sqrt{K_W}$,the concentration of $H^+$ ions increases.
Therefore,$pH = -\log[H^+]$ decreases as the temperature rises.

(C) According to Moseley's law,the frequency $\nu$ of the characteristic $X$-rays emitted by an element is related to its atomic number $Z$ by the equation: $\sqrt{\nu} = a(Z - b)$,where $a$ and $b$ are constants.
This implies that a plot of $\sqrt{\nu}$ versus atomic number $(Z)$ is a straight line.
Statement $(I)$ is false because the plot of $\sqrt{\nu}$ versus atomic mass is not a straight line.
Statement $(II)$ is false because the plot of $\nu$ (not $\sqrt{\nu}$) versus atomic number is not a straight line; rather,$\sqrt{\nu}$ versus atomic number is a straight line.
Therefore,both statements are false.

(A) Moles of $H_2O$ produced $= \frac{0.9 \ g}{18 \ g \ mol^{-1}} = 0.05 \ mol$.
Moles of $H$ atoms $= 0.05 \ mol \times 2 = 0.1 \ mol$.
Mass of $H$ atoms in $0.01 \ mol$ of $(X) = 0.1 \ mol \times 1 \ g \ mol^{-1} = 0.1 \ g$.
Mass of $H$ atoms in $1 \ mol$ of $(X) = \frac{0.1 \ g}{0.01 \ mol} = 10 \ g \ mol^{-1}$.
Given that the compound contains $10\%$ hydrogen by mass,we have:
$10\% = \frac{\text{Mass of } H \text{ in } 1 \ mol \text{ of } (X)}{\text{Molar mass of } (X)} \times 100$.
$10 = \frac{10}{M} \times 100$.
$M = 100 \ g \ mol^{-1}$.

(C) The balanced chemical equation is: $4Al + 3O_2 \longrightarrow 2Al_2O_3$.
Moles of $Al = \frac{81.0 \ g}{27.0 \ g \ mol^{-1}} = 3.0 \ mol$.
Moles of $O_2 = \frac{128.0 \ g}{32.0 \ g \ mol^{-1}} = 4.0 \ mol$.
According to the stoichiometry,$4 \ mol$ of $Al$ requires $3 \ mol$ of $O_2$.
For $3.0 \ mol$ of $Al$,required $O_2 = \frac{3}{4} \times 3.0 = 2.25 \ mol$.
Since $4.0 \ mol$ of $O_2$ is available,$Al$ is the limiting reagent.
Moles of $Al_2O_3$ produced $= \frac{2}{4} \times 3.0 = 1.5 \ mol$.
Molar mass of $Al_2O_3 = (2 \times 27.0) + (3 \times 16.0) = 54.0 + 48.0 = 102.0 \ g \ mol^{-1}$.
Mass of $Al_2O_3 = 1.5 \ mol \times 102.0 \ g \ mol^{-1} = 153.0 \ g$.

(A) The reaction at the cathode for the reduction of aluminium is: $Al^{3+} + 3e^- \rightarrow Al(s)$.
According to Faraday's first law of electrolysis,the mass of the substance deposited is given by $m = \frac{M \times I \times t}{n \times F}$.
Here,$M = 27 \ g \ mol^{-1}$,$I = 2 \ A$,$t = 30 \times 60 \ s = 1800 \ s$,$n = 3$,and $F = 96500 \ C \ mol^{-1}$.
Substituting the values: $m = \frac{27 \times 2 \times 1800}{3 \times 96500} = \frac{97200}{289500} \approx 0.336 \ g$.

(C) Radioactive decay is a first-order nuclear process.
The decay constant $(\lambda)$ is a characteristic property of the radioactive nucleus and is independent of external physical conditions such as temperature,pressure,or chemical environment.
Therefore,the statement that the decay constant increases with an increase in temperature is false.

(A) The electronic configurations of the given ions are:
$Eu^{2+} (Z=63): [Xe] 4f^7 6s^0$
$Gd^{3+} (Z=64): [Xe] 4f^7 5d^0 6s^0$
$Eu^{3+} (Z=63): [Xe] 4f^6 6s^0$
$Tb^{3+} (Z=65): [Xe] 4f^8 6s^0$
$Sm^{2+} (Z=62): [Xe] 4f^6 6s^0$
Thus,only $Eu^{2+}$ and $Gd^{3+}$ have a $4f^7$ configuration.

(A) The compound $(CH_3)_3C-CH_2-Cl$ is neopentyl chloride.
Statement $I$ is incorrect because neopentyl chloride is a primary halide and forms a highly unstable primary carbocation upon the loss of the chloride ion,making $S_N1$ reactions extremely unfavorable.
Statement $II$ is correct because,although it is a primary halide,the bulky tert-butyl group adjacent to the reaction center causes significant steric hindrance,which makes the $S_N2$ transition state very difficult to achieve,resulting in a very slow reaction rate.

(C) The electrolysis of concentrated sulphuric acid $(H_2SO_4)$ is used to produce peroxydisulphuric acid $(H_2S_2O_8)$.
At the anode,the oxidation of hydrogen sulphate ions $(HSO_4^-)$ occurs:
$2HSO_4^{-} \rightarrow H_2S_2O_8 + 2e^{-}$

(C) Fehling's test is given by aliphatic aldehydes. Aromatic aldehydes do not give this test.
$(A)$ $C_6H_5CHO$ is an aromatic aldehyde,so it gives a negative test.
$(B)$ $C_6H_5COCH_3$ is a ketone,so it gives a negative test.
$(C)$ $HOCH_2-CO-(CHOH)_3-CH_2OH$ is a ketose sugar (fructose). In alkaline medium,it isomerizes to aldose,thus giving a positive Fehling's test.
$(D)$ $CH_3CHO$ is an aliphatic aldehyde,so it gives a positive test.
$(E)$ $C_6H_5CH_2CHO$ is an aliphatic aldehyde (the aldehyde group is not directly attached to the benzene ring),so it gives a positive test.
Therefore,compounds $(C)$,$(D)$,and $(E)$ give a positive Fehling's test.

(D) The $CFSE$ for an octahedral complex is calculated as: $CFSE = (-0.4 \times n_{t_{2g}} + 0.6 \times n_{e_g}) \Delta_0$.
For $K_3[Fe(SCN)_6]$,the central metal ion is $Fe^{3+}$,which has a $d^5$ configuration.
$SCN^-$ is a weak field ligand,so the electrons are arranged as $t_{2g}^3 e_g^2$.
$CFSE = (-0.4 \times 3 + 0.6 \times 2) \Delta_0 = (-1.2 + 1.2) \Delta_0 = 0 \Delta_0 = 0$.

(A) The elevation in boiling point is given by $\Delta T_{b} = i \cdot K_{b} \cdot m$,where $i$ is the van't Hoff factor and $m$ is the molality (proportional to concentration $C$).
Since $K_{b}$ is constant for the solvent,$\Delta T_{b} \propto i \cdot C$.

Solution	$i \cdot C$
$(i) \ 10^{-4} \ M \ NaCl$	$2 \times 10^{-4}$
$(ii) \ 10^{-4} \ M \ \text{Urea}$	$1 \times 10^{-4}$
$(iii) \ 10^{-3} \ M \ NaCl$	$2 \times 10^{-3}$
$(iv) \ 10^{-2} \ M \ NaCl$	$2 \times 10^{-2}$

Comparing the values of $i \cdot C$: $1 \times 10^{-4} < 2 \times 10^{-4} < 2 \times 10^{-3} < 2 \times 10^{-2}$.
Therefore,the order of increasing boiling points is $(ii) < (i) < (iii) < (iv)$.

(D) The reaction sequence is as follows:
$1$. $Nitrobenzene$ reacts with $Br_2$ in $AcOH$ to undergo electrophilic aromatic substitution,forming $m-bromonitrobenzene$ (since $-NO_2$ is a meta-directing group).
$2$. Reduction of $m-bromonitrobenzene$ with $Sn/HCl$ yields $m-bromoaniline$.
$3$. Treatment of $m-bromoaniline$ with $NaNO_2/HCl$ at $273-278 \ K$ results in diazotization,forming $m-bromobenzenediazonium \ chloride$.
$4$. Finally,reaction with ethanol $(C_2H_5OH)$ reduces the diazonium salt to $bromobenzene$ (specifically $m-bromobenzene$ in this context,which is simply $bromobenzene$ as the bromine position is fixed) and ethanol is oxidized to acetaldehyde $(CH_3CHO)$.
Thus,the final products are $bromobenzene$ and $CH_3CHO$.

(B) For $[NiCl_4]^{2-}$: The oxidation state of $Ni$ is $+2$. The electronic configuration is $[Ar] 3d^8$. Due to the weak field ligand $Cl^-$,it undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry with $2$ unpaired electrons,making it paramagnetic.
For $[Ni(CO)_4]$: The oxidation state of $Ni$ is $0$. The electronic configuration is $[Ar] 3d^8 4s^2$. In the presence of the strong field ligand $CO$,the electrons pair up to form $[Ar] 3d^{10} 4s^0$. It undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry with $0$ unpaired electrons,making it diamagnetic.

(D) For a first-order reaction,the rate constant $k$ is given by the Arrhenius equation: $k = A e^{-E_a/RT}$.
Substituting the given values: $k = 10^{20} \times e^{-\frac{191.48 \times 10^3}{8.314 \times 1000}}$.
$k = 10^{20} \times e^{-23.031} \approx 10^{20} \times e^{-\ln(10^{10})} = 10^{20} \times 10^{-10} = 10^{10} \ s^{-1}$.
The half-life $t_{1/2}$ is given by: $t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693}{10^{10}} = 6.93 \times 10^{-11} \ s$.
Converting to picoseconds $(1 \ ps = 10^{-12} \ s)$: $t_{1/2} = 69.3 \ ps$.
The nearest integer is $69$.

(A) The reaction sequence is as follows:
$1$. Nitrobenzene is reduced by $Sn + HCl$ to form aniline $(C_6H_5NH_2)$.
$2$. Aniline reacts with $NaNO_2 + HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.
$3$. Benzenediazonium chloride reacts with $Cu_2Cl_2$ (Sandmeyer reaction) to form chlorobenzene $(C_6H_5Cl)$.
$4$. Chlorobenzene reacts with $Na$ in dry ether (Wurtz-Fittig reaction) to form biphenyl $(C_6H_5-C_6H_5)$.
The final product $(A)$ is biphenyl $(C_{12}H_{10})$.
Molar mass of biphenyl = $(12 \times 12) + (10 \times 1) = 144 + 10 = 154 \ g \ mol^{-1}$.

(A) For a monosaccharide to be correlated to $D$-glyceraldehyde,the $-OH$ group on the lowest chiral carbon (the chiral carbon farthest from the carbonyl group) must be on the right-hand side in the Fischer projection.
Looking at the structures provided in the solution image:
In structure $(A)$,the $-OH$ group on the lowest chiral carbon is on the right.
In structure $(B)$,the $-OH$ group on the lowest chiral carbon is on the right.
In structure $(C)$,the $-OH$ group on the lowest chiral carbon is on the left ($L$-configuration).
In structure $(D)$,the $-OH$ group on the lowest chiral carbon is on the right.
Thus,structures $(A)$,$(B)$,and $(D)$ have the $D$-configuration.
The total number of such structures is $3$.

(D) The group $15$ elements are $N, P, As, Sb, Bi$. The $E-E$ bond strength decreases down the group due to the increase in atomic size,making the $N-N$ bond the strongest and the $Bi-Bi$ bond the weakest. However,among non-metallic elements,the $P-P$ bond is significantly stronger than the $N-N$ bond because the small size of the nitrogen atom leads to strong interelectronic repulsion between the lone pairs of the bonded nitrogen atoms. Thus,$N-N$ is the weakest single covalent bond among the non-metallic group $15$ elements.
Nitrogen has a maximum covalency of $4$ because it lacks $d$-orbitals in its valence shell,limiting its ability to expand its octet.

(C) From the given energy profile diagram:
$1$. The activation energy of the forward reaction $(E_{a,f})$ is the energy difference between the activated complex and the reactant,which is $E_1 + E_2$.
$2$. The energy of the product is higher than the energy of the reactant $(E_{product} > E_{reactant})$.
$3$. Since stability is inversely proportional to energy,the reactant is more stable than the product,or the product is less stable than the reactant.

(A) The correct matches are:
$A$. $R-CONH_2$ (Amide) is reduced to amine $(R-CH_2NH_2)$ using $LiAlH_4, H_2O$ $(III)$.
$B$. $C_6H_5NO_2$ (Nitrobenzene) is reduced to aniline $(C_6H_5NH_2)$ using $Sn, HCl$ $(IV)$.
$C$. $R-C \equiv N$ (Nitrile) is reduced to amine $(R-CH_2NH_2)$ using $H_2 / Ni$ $(II)$.
$D$. $N$-alkylphthalimide is converted to primary amine $(R-NH_2)$ using $NaOH$ (aqueous) $(I)$ in the Gabriel phthalimide synthesis.
Thus,the correct matching is $A-III, B-IV, C-II, D-I$.

(A) The reaction sequence $RBr$ $\xrightarrow{(i) Mg, \text{dry ether}} RMgBr$ $\xrightarrow{(ii) H_2O} RH$ represents the formation of an alkane from an alkyl bromide via a Grignard reagent.
To obtain $2-$methylbutane $(C_5H_{12})$ from $RBr$,the alkyl group $R$ must be a $2-$methylbutyl group.
The possible structural isomers of $2-$methylbutyl bromide are:
$1.$ $1-$bromo$-2-$methylbutane
$2.$ $2-$bromo$-2-$methylbutane
$3.$ $2-$bromo$-3-$methylbutane
$4.$ $1-$bromo$-3-$methylbutane
All these isomers,upon reaction with $Mg$ in dry ether followed by hydrolysis with $H_2O$,will yield $2-$methylbutane.
Therefore,the total number of such structural isomers is $4$.

(B) The crystal field stabilization energy $(CFSE)$ depends on two main factors: the oxidation state of the central metal ion and the strength of the ligand.
$1$. Oxidation state: Higher oxidation state leads to higher $CFSE$ (e.g.,$Co^{3+} > Co^{2+}$).
$2$. Ligand strength: Stronger ligands lead to higher $CFSE$ (e.g.,$en > NH_3$).
Comparing the complexes:
- $[Co(NH_3)_4]^{2+}$ has the lowest $CFSE$ due to lower coordination number and lower oxidation state.
- $[Co(NH_3)_6]^{2+}$ has a higher $CFSE$ than $[Co(NH_3)_4]^{2+}$ due to the higher coordination number.
- $[Co(NH_3)_6]^{3+}$ has a higher $CFSE$ than $[Co(NH_3)_6]^{2+}$ due to the higher oxidation state of $Co^{3+}$.
- $[Co(en)_3]^{3+}$ has the highest $CFSE$ because $en$ is a stronger ligand than $NH_3$.
Thus,the correct order is: $[Co(NH_3)_4]^{2+} < [Co(NH_3)_6]^{2+} < [Co(NH_3)_6]^{3+} < [Co(en)_3]^{3+}$.

(D) Given: Molarity $(M)$ = $3 \ M$,Density $(d)$ = $1.25 \ g/mL$,Molar mass of $NaCl$ $(M_w)$ = $58.5 \ g/mol$.
The formula for molality $(m)$ is: $m = \frac{M \times 1000}{1000 \times d - M \times M_w}$.
Substituting the values: $m = \frac{3 \times 1000}{1000 \times 1.25 - 3 \times 58.5}$.
$m = \frac{3000}{1250 - 175.5}$.
$m = \frac{3000}{1074.5} \approx 2.79 \ m$.

(C) homoleptic complex contains only one type of ligand.
$(A) \ [Fe(CN)_5NO]^{2-}$: Heteroleptic (contains $CN^-$ and $NO$ ligands).
$(B) \ [CoF_6]^{3-}$: Homoleptic,$Co^{3+}$ $(3d^6)$,$F^-$ is a weak field ligand,so it is high spin.
$(C) \ [Fe(CN)_6]^{4-}$: Homoleptic,$Fe^{2+}$ $(3d^6)$,$CN^-$ is a strong field ligand,so it is low spin $(t_{2g}^6 e_g^0)$.
$(D) \ [Co(NH_3)_6]^{3+}$: Homoleptic,$Co^{3+}$ $(3d^6)$,$NH_3$ is a strong field ligand,so it is low spin $(t_{2g}^6 e_g^0)$.
$(E) \ [Cr(H_2O)_6]^{2+}$: Homoleptic,$Cr^{2+}$ $(3d^4)$,$H_2O$ is a weak field ligand,so it is high spin $(t_{2g}^3 e_g^1)$.
Thus,$(C)$ and $(D)$ are homoleptic and low spin complexes.

(D) $1$. Toluene reacts with $CrO_2Cl_2$ in $CS_2$ followed by hydrolysis $(H_3O^+)$ to form benzaldehyde $(C_6H_5CHO)$ via the Etard reaction.
$2$. Benzaldehyde then reacts with sodium bisulphite $(NaHSO_3)$ to form a crystalline addition product,which is the residue $(A)$. The structure of $(A)$ is $C_6H_5CH(OH)SO_3Na$.
$3$. When this residue $(A)$ is treated with dilute $HCl$,it undergoes hydrolysis to regenerate benzaldehyde,which is compound $(B)$.
$4$. Therefore,residue $(A)$ is $C_6H_5CH(OH)SO_3Na$ and compound $(B)$ is $C_6H_5CHO$.

(D) Statement $I$: Corrosion is an electrochemical process where the metal surface acts as an anode (undergoing oxidation) and the impurities or cathodic regions act as cathodes. This statement is correct.
Statement $II$: Corrosion,particularly the rusting of iron,is significantly accelerated in an acidic medium because the presence of $H^+$ ions facilitates the reduction reaction. In an alkaline medium,the rate of corrosion is generally lower. Therefore,Statement $II$ is false.
Conclusion: Statement $I$ is true,but Statement $II$ is false.

(B) The electronic configuration of Niobium $(Nb, Z=41)$ is $[Kr] \ 4d^4 \ 5s^1$. Thus,the number of electrons in the $4d$ orbital is $x = 4$.
The electronic configuration of Ruthenium $(Ru, Z=44)$ is $[Kr] \ 4d^7 \ 5s^1$. Thus,the number of electrons in the $4d$ orbital is $y = 7$.
Therefore,the value of $x+y = 4+7 = 11$.

(C) The chemical formula of the complex formed between $Ni^{2+}$ and dimethylglyoxime $(dmg)$ is $[Ni(dmg)_2]$.
Each dimethylglyoxime ligand has the formula $C_4H_7N_2O_2^-$.
In the complex $[Ni(dmg)_2]$,there are two such ligands.
Total number of $H$ atoms = $2 \times 7 = 14$.
These $14$ hydrogen atoms include $12$ hydrogen atoms from the four methyl groups $(-CH_3)$ and $2$ hydrogen atoms involved in intramolecular hydrogen bonding.

(D) $SO_2$ contains sulfur in the $+4$ oxidation state. \\ Since sulfur can exist in oxidation states ranging from $-2$ to $+6$,$SO_2$ can act as an oxidizing agent (by being reduced to $S$ or $H_2S$) as well as a reducing agent (by being oxidized to $SO_3$ or $SO_4^{2-}$). \\ Therefore,the statement that $SO_2$ cannot act as a reducing agent is incorrect.

(C) Given: $\Delta T_f = 0.558^{\circ} C$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,$m = 0.1 \ m$.
Using the formula for depression in freezing point: $\Delta T_f = i \times K_f \times m$.
Substituting the values: $0.558 = i \times 1.86 \times 0.1$.
Solving for $i$: $i = 0.558 / 0.186 = 3$.
Since the van't Hoff factor $i = 3$,the complex dissociates into $3$ ions.
For the complex $[Cr(NH_3)_5Cl]Cl_2$,the dissociation is $[Cr(NH_3)_5Cl]Cl_2 \rightarrow [Cr(NH_3)_5Cl]^{2+} + 2Cl^{-}$,which yields $1 + 2 = 3$ ions.
Thus,the complex is $[Cr(NH_3)_5Cl]Cl_2$.

(A) The reaction $FeO_4^{2-} \rightarrow Fe^{2+}$ involves a change in oxidation state from $+6$ to $+2$,so the number of electrons involved is $n_4 = 4$.
This process can be broken down into two steps:
$1) FeO_4^{2-} + 3e^- \rightarrow Fe^{3+} \quad (E_1^{\Theta} = 2.0 \ V, n_1 = 3)$
$2) Fe^{3+} + 1e^- \rightarrow Fe^{2+} \quad (E_2^{\Theta} = 0.8 \ V, n_2 = 1)$
The total Gibbs free energy change is given by $\Delta G_4^{\Theta} = \Delta G_1^{\Theta} + \Delta G_2^{\Theta}$.
Using $\Delta G^{\Theta} = -nFE^{\Theta}$,we get:
$-n_4 F E_4^{\Theta} = -n_1 F E_1^{\Theta} - n_2 F E_2^{\Theta}$
$n_4 E_4^{\Theta} = n_1 E_1^{\Theta} + n_2 E_2^{\Theta}$
$4 \times E_4^{\Theta} = (3 \times 2.0) + (1 \times 0.8)$
$4 E_4^{\Theta} = 6.0 + 0.8 = 6.8$
$E_4^{\Theta} = \frac{6.8}{4} = 1.7 \ V$.

(C) The correct matches are as follows:
$A.$ Swarts reaction: Used for the preparation of alkyl fluorides,e.g.,$CH_3CH_2F$ (Ethyl fluoride).
$B.$ Sandmeyer's reaction: Used to convert diazonium salts to aryl halides or cyanides,e.g.,$C_6H_5CN$ (Cyanobenzene).
$C.$ Wurtz-Fittig reaction: Used for the synthesis of alkyl benzenes,e.g.,$C_6H_5CH_2CH_3$ (Ethyl benzene).
$D.$ Finkelstein reaction: Used for the preparation of alkyl iodides from alkyl chlorides/bromides,e.g.,$CH_3CH_2I$ (Ethyl iodide).
Therefore,the correct sequence is $A-IV, B-III, C-I, D-II$.

(B) Statement $I$ is true: Fructose is a ketohexose and does not contain an aldehydic group. However,in the alkaline medium of Tollen's reagent,it undergoes tautomerization to form an enediol intermediate,which then rearranges to form aldoses like glucose and mannose. These aldoses contain an aldehydic group and thus reduce Tollen's reagent.
Statement $II$ is true: As shown in the reaction,in the presence of a base,fructose undergoes tautomerization to form an enediol intermediate,which further rearranges to give a mixture of glucose and mannose.

(C) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 3.95 \ BM$,we have $\sqrt{n(n+2)} \approx 3.95$,which implies $n(n+2) \approx 15.6$,so $n \approx 3$.
$Co(II)$ has a $3d^7$ configuration.
In an octahedral field,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.
For a $d^7$ system with $3$ unpaired electrons,the configuration is $t_{2g}^5 e_g^2$ (high spin).
This corresponds to $5$ electrons in $t_{2g}$ (with $1$ unpaired) and $2$ electrons in $e_g$ (with $2$ unpaired),totaling $3$ unpaired electrons.

(A) $Ma_3b_3$ type complexes show Facial - Meridional isomerism.
$(i)$ $[Co(NH_3)_3Cl_3] \Rightarrow Ma_3b_3$
$(ii)$ $[Co(NH_3)_4Cl_2]^{+} \Rightarrow Ma_4b_2$
$(iii)$ $[Co(en)_3]^{3+} \Rightarrow M(AA)_3$
$(iv)$ $[Co(en)_2Cl_2]^{+} \Rightarrow M(AA)_2b_2$
Here,$a = NH_3$,$b = Cl^{-}$,and $AA = en$ (ethylenediamine).
Thus,only $[Co(NH_3)_3Cl_3]$ corresponds to the $Ma_3b_3$ type.

(C) The reaction of $CH_3-CH_2-CHO$ with excess $HCHO$ in the presence of an alkali involves two steps:
$1$. Multiple cross-aldol condensation: The $\alpha$-hydrogens of propanal are replaced by hydroxymethyl groups $(CH_2OH)$ through reaction with $HCHO$. Since propanal has two $\alpha$-hydrogens,both are replaced to form $CH_3-C(CH_2OH)_2-CHO$.
$2$. Cross-Cannizzaro reaction: The remaining aldehyde group $(-CHO)$ is then reduced to a primary alcohol $(-CH_2OH)$ by the excess $HCHO$,which itself gets oxidized to formate.
The final product is $CH_3-C(CH_2OH)_3$.

(A) The reaction of phenol with bromine water is: $C_6H_5OH + 3Br_2 \rightarrow C_6H_2Br_3OH + 3HBr$
Molar mass of phenol $(C_6H_5OH) = (6 \times 12) + (6 \times 1) + 16 = 72 + 6 + 16 = 94 \ g \ mol^{-1}$.
Moles of phenol = $\frac{2 \ g}{94 \ g \ mol^{-1}} = 0.02128 \ mol$.
According to the stoichiometry,$1 \ mol$ of phenol reacts with $3 \ mol$ of $Br_2$.
Moles of $Br_2$ required = $3 \times 0.02128 = 0.06384 \ mol$.
Molar mass of $Br_2 = 2 \times 80 = 160 \ g \ mol^{-1}$.
Mass of $Br_2$ required = $0.06384 \ mol \times 160 \ g \ mol^{-1} = 10.2144 \ g \approx 10.22 \ g$.

(A) $I$. $V^{2+}$ is violet,$Cr^{3+}$ is violet,and $Mn^{3+}$ is violet. All three ions exhibit a violet colour in aqueous solution.
$II$. $Zn^{2+}$ is colourless,$V^{3+}$ is green,and $Fe^{3+}$ is yellow.
$III$. $Ti^{4+}$ is colourless,$V^{4+}$ is blue,and $Mn^{2+}$ is pink.
$IV$. $Sc^{3+}$ is colourless,$Ti^{3+}$ is purple,and $Cr^{2+}$ is blue.
Therefore,the set with the same colour is $V^{2+}, Cr^{3+}, Mn^{3+}$.

(C) Hinsberg's reagent is benzenesulfonyl chloride $(C_6H_5SO_2Cl)$.
It reacts with $1^{\circ}$ and $2^{\circ}$ amines because they possess at least one hydrogen atom attached to the nitrogen atom.
$(A)$ $C_6H_5NH_2$ is a $1^{\circ}$ amine (reacts).
$(2)$ $C_6H_5N(CH_3)_2$ is a $3^{\circ}$ amine (does not react).
$(C)$ $CH_3NH_2$ is a $1^{\circ}$ amine (reacts).
$(4)$ $N(CH_3)_3$ is a $3^{\circ}$ amine (does not react).
$(E)$ $(C_6H_5)_2NH$ is a $2^{\circ}$ amine (reacts).
Therefore,$A, C$ and $E$ react with Hinsberg's reagent.

(C) The starting material is $3$-nitrotoluene.
$1$. Bromination with $Br_2/Fe$ gives $2$-bromo-$5$-nitrotoluene.
$2$. Reduction with $Sn/HCl$ converts the $-NO_2$ group to $-NH_2$,yielding $5$-amino-$2$-bromotoluene.
$3$. Diazotization with $NaNO_2/HCl$ at $273 \ K$ converts the $-NH_2$ group to a diazonium salt,$-N_2^+Cl^-$.
$4$. Reduction with $H_3PO_2/H_2O$ removes the diazonium group,resulting in $2$-bromotoluene as the major product $(A)$.
The chemical formula of $2$-bromotoluene is $C_7H_7Br$.
Molar mass $= (7 \times 12) + (7 \times 1) + 80 = 84 + 7 + 80 = 171 \ g \ mol^{-1}$.

(D) The reaction is $2N_2O_{5(g)} \rightarrow 2N_2O_{4(g)} + O_{2(g)}$.
For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{P_0}{P_t}$,where $P_0$ is the initial pressure of $N_2O_5$ and $P_t$ is the pressure at time $t$.
Given $k = 4.606 \times 10^{-2} \ s^{-1}$ and $t = 100 \ s$.
$4.606 \times 10^{-2} = \frac{2.303}{100} \log \frac{0.6}{P_{N_2O_5}}$
$2 = \log \frac{0.6}{P_{N_2O_5}} \Rightarrow \frac{0.6}{P_{N_2O_5}} = 10^2 = 100$.
$P_{N_2O_5} = \frac{0.6}{100} = 0.006 \ atm$.
Let $x$ be the pressure of $N_2O_5$ decomposed. Then $0.6 - x = 0.006 \Rightarrow x = 0.594 \ atm$.
The total pressure $P_{total} = (0.6 - x) + x + \frac{x}{2} = 0.6 + \frac{x}{2}$.
$P_{total} = 0.6 + \frac{0.594}{2} = 0.6 + 0.297 = 0.897 \ atm$.
$P_{total} = 897 \times 10^{-3} \ atm$.
Rounding to the nearest integer,$X = 897$.

(A) The relationship between standard Gibbs free energy change and standard cell potential is given by $\Delta G^0 = -nFE^0_{cell}$.
For a cell to have the most negative $\Delta G^0$,it must have the most positive $E^0_{cell}$ value.
$E^0_{cell} = E^0_{cathode} - E^0_{anode}$.
$A) \ E^0_{cell} = 0.80 - (-0.76) = 1.56 \ V; \Delta G^0 = -2 \times F \times 1.56 = -3.12 \ F$
$B) \ E^0_{cell} = -2.37 - (-0.76) = -1.61 \ V; \Delta G^0 = -2 \times F \times (-1.61) = +3.22 \ F$
$C) \ E^0_{cell} = -2.37 - 0.80 = -3.17 \ V; \Delta G^0 = -2 \times F \times (-3.17) = +6.34 \ F$
$D) \ E^0_{cell} = 0.80 - 0.34 = 0.46 \ V; \Delta G^0 = -2 \times F \times 0.46 = -0.92 \ F$
Comparing the values,option $(A)$ gives the most negative $\Delta G^0$.

(C) The $\alpha-$helix and $\beta-$pleated sheet are common types of secondary structures of proteins.
These structures are stabilized primarily by hydrogen bonds between the carbonyl oxygen and the amide hydrogen of the polypeptide backbone.

(B) The hydration of carbonyl compounds involves the nucleophilic addition of water to the carbonyl carbon.
For formaldehyde $(HCHO)$,the $K_{eq}$ for hydration is approximately $2280$. This high value is due to the lack of steric hindrance (small substituents) and the high electrophilicity of the carbonyl carbon.
For trichloroacetaldehyde ($CCl_3CHO$,also known as chloral),the $K_{eq}$ for hydration is approximately $2000$. This high value is primarily due to the strong electron-withdrawing $-I$ effect of the three chlorine atoms,which makes the carbonyl carbon highly electrophilic and susceptible to nucleophilic attack by water.
Both statements are scientifically accurate based on standard chemical literature.
Therefore,both Statement $I$ and Statement $II$ are true.

(B) The reaction sequence is as follows:
$1$. $PbS$ (Galena) reacts with $HNO_3$ to form lead$(II)$ nitrate: $3PbS + 8HNO_3 \rightarrow 3Pb(NO_3)_2 + 2NO + 3S + 4H_2O$. Thus,$A = PbS$.
$2$. $Pb(NO_3)_2$ reacts with $H_2SO_4$ to form lead$(II)$ sulfate precipitate: $Pb(NO_3)_2 + H_2SO_4 \rightarrow PbSO_4(s) + 2HNO_3$. Thus,$B = PbSO_4$.
$3$. $PbSO_4$ is dissolved in ammonium acetate and acetic acid to form lead$(II)$ acetate,which then reacts with $K_2CrO_4$ to form a yellow precipitate of lead$(II)$ chromate: $PbSO_4 + 2CH_3COONH_4 \rightarrow Pb(CH_3COO)_2 + (NH_4)_2SO_4$ and $Pb(CH_3COO)_2 + K_2CrO_4 \rightarrow PbCrO_4(s) + 2CH_3COOK$. Thus,$C = PbCrO_4$.
Therefore,$A = PbS$,$B = PbSO_4$,and $C = PbCrO_4$.

(D) Statement $(I)$ is true: The boiling point of alcohols and phenols increases with an increase in the number of carbon atoms due to an increase in van der Waals forces of attraction as the molecular weight increases $(B.P. \propto M.W.)$.
Statement $(II)$ is true: Alcohols and phenols exhibit intermolecular hydrogen bonding,which is significantly stronger than the dipole-dipole interactions found in ethers and haloalkanes of comparable molecular mass. Therefore,they have higher boiling points.

(A) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{P^{\circ} - P}{P^{\circ}} = X_{\text{solute}}$.
Since $P^{\circ} - P \propto X_{\text{solute}}$,we have $10 \propto 0.2$.
For a decrease of $20 \ mm \ Hg$,the new mole fraction of the solute $(X'_{\text{solute}})$ will be: $20 \propto X'_{\text{solute}}$.
Thus,$X'_{\text{solute}} = \frac{20}{10} \times 0.2 = 0.4$.
The mole fraction of the solvent is $X_{\text{solvent}} = 1 - X'_{\text{solute}} = 1 - 0.4 = 0.6$.

(A) The rate of solvolysis (which proceeds via an $S_{N}1$ mechanism) is directly proportional to the stability of the intermediate carbocation formed after the departure of the leaving group (bromide ion).
$1$. Compound $C$ forms a diphenylmethyl carbocation,which is highly stabilized by resonance with two phenyl rings.
$2$. Compound $B$ forms a benzylic carbocation (specifically,a $1-$tetralyl type),which is stabilized by resonance with one benzene ring.
$3$. Compound $A$ forms a tertiary carbocation in a cyclohexane ring,which is stabilized by inductive effects and hyperconjugation.
$4$. Compound $D$ forms a secondary carbocation,which is the least stable among the given options.
Thus,the stability order of the carbocations is $C > B > A > D$. Consequently,the ascending order of the rate of solvolysis is $D < A < B < C$.

(B) For a zero order reaction: $A \rightarrow P$
$\text{Rate} = k[A]^0 = k$
Since the rate is constant,the concentration of the reactant $[A]_t$ decreases linearly with time according to the equation:
$[A]_t = [A]_0 - kt$
Comparing this to the equation of a straight line $y = mx + c$,where $y = [A]_t$,$x = t$,$m = -k$,and $c = [A]_0$,the graph of reactant concentration versus time is a straight line with a negative slope. Therefore,the graph in option $B$ represents a zero order reaction.

(A) Bronze is an alloy of $Cu$ and $Sn$. $(A-IV)$
Brass is an alloy of $Cu$ and $Zn$. $(B-III)$
$UK$ silver coin is an alloy of $Cu$ and $Ni$. $(C-I)$
Stainless steel is an alloy of $Fe, Cr, Ni,$ and $C$. $(D-II)$
Therefore,the correct matching is $A-IV, B-III, C-I, D-II$.

(C) To find the $d$-configuration of the central metal ion, we determine the oxidation state of the metal in each complex:
$(A)$ In $[FeO_4]^{2-}$, $Fe + 4(-2) = -2 \implies Fe = +6$. $Fe^{6+} = [Ar] 3d^2$.
$(B)$ In $[Mn(CN)_6]^{3-}$, $Mn + 6(-1) = -3 \implies Mn = +3$. $Mn^{3+} = [Ar] 3d^4$.
$(C)$ In $[Fe(CN)_6]^{3-}$, $Fe + 6(-1) = -3 \implies Fe = +3$. $Fe^{3+} = [Ar] 3d^5$.
$(D)$ In $[Cr(H_2O)_6]^{2+}$, $Cr + 6(0) = +2 \implies Cr = +2$. $Cr^{2+} = [Ar] 3d^4$.
$(E)$ In $[NiF_6]^{2-}$, $Ni + 6(-1) = -2 \implies Ni = +4$. $Ni^{4+} = [Ar] 3d^6$.
The complexes with $d^4$ configuration are $(B)$ and $(D)$.

(C) The reaction of chlorobenzene with $NaOH$ at $623 \ K$ and $300 \ atm$ followed by acidification with $H^+$ gives phenol as product $[A]$.
Phenol on oxidation with chromic acid $(Na_2Cr_2O_7 / H_2SO_4)$ yields $p$-benzoquinone as product $[B]$.

(A) According to Raoult's law for a binary solution of two volatile liquids $1$ and $2$:
$P_T = P_1^0 x_1 + P_2^0 x_2 = P_1^0 x_1 + P_2^0 (1 - x_1) = P_2^0 + x_1 (P_1^0 - P_2^0)$
Also,from Dalton's law,$P_1 = P_T y_1$,where $P_1 = P_1^0 x_1$.
So,$P_T y_1 = P_1^0 x_1 \implies P_T = \frac{P_1^0 x_1}{y_1}$.
Equating the two expressions for $P_T$:
$P_2^0 + x_1 (P_1^0 - P_2^0) = \frac{P_1^0 x_1}{y_1}$
Dividing both sides by $x_1 P_2^0$:
$\frac{P_2^0}{x_1} + \frac{P_1^0 - P_2^0}{P_2^0} = \frac{P_1^0}{P_2^0 y_1}$
Rearranging to the form $Y = mX + c$ where $Y = \frac{1}{x_1}$ and $X = \frac{1}{y_1}$:
$\frac{1}{x_1} = \left( \frac{P_1^0}{P_2^0} \right) \frac{1}{y_1} + \left( \frac{P_2^0 - P_1^0}{P_2^0} \right)$
Comparing with $Y = mX + c$,the slope is $\frac{P_1^0}{P_2^0}$ and the intercept is $\frac{P_2^0 - P_1^0}{P_2^0}$.

(C) The reaction of potassium dichromate $(K_2Cr_2O_7)$ with potassium hydroxide $(KOH)$ yields potassium chromate $(K_2CrO_4)$ and water:
$K_2Cr_2O_7 + 2KOH \rightarrow 2K_2CrO_4 + H_2O$
Thus,$[A] = K_2CrO_4$.
Next,the reaction of potassium chromate $(K_2CrO_4)$ with sulfuric acid $(H_2SO_4)$ yields potassium dichromate $(K_2Cr_2O_7)$,potassium sulfate $(K_2SO_4)$,and water:
$2K_2CrO_4 + H_2SO_4 \rightarrow K_2Cr_2O_7 + K_2SO_4 + H_2O$
Thus,$[B] = K_2Cr_2O_7$.