WBJEE 2025 Chemistry Question Paper with Answer and Solution

(B) The relationship between the real coefficient of expansion $(\gamma_r)$,apparent coefficient of expansion $(\gamma_a)$,and the volume coefficient of expansion of the vessel $(\gamma_v)$ is given by: $\gamma_r = \gamma_a + \gamma_v$.
Since the volume coefficient of expansion $\gamma_v = 3\alpha$,where $\alpha$ is the linear coefficient of expansion,we have $\gamma_r = \gamma_a + 3\alpha$.
For the copper vessel: $\gamma_r = C + 3A$.
For the silver vessel: $\gamma_r = S + 3\alpha_{Ag}$,where $\alpha_{Ag}$ is the linear coefficient of expansion of silver.
Since the real coefficient of expansion of the liquid is constant,we equate the two expressions: $C + 3A = S + 3\alpha_{Ag}$.
Rearranging for $\alpha_{Ag}$: $3\alpha_{Ag} = C + 3A - S$.
Therefore,$\alpha_{Ag} = \frac{C + 3A - S}{3}$.

(C) The equation of the first circle is ${x^2 + y^2 - 4x - 6y - 12 = 0}$. Its centre ${C_1}$ is $(2, 3)$ and radius ${r_1} = \sqrt{2^2 + 3^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5$.
The equation of the second circle is ${x^2 + y^2 + 6x + 18y + 26 = 0}$. Its centre ${C_2}$ is $(-3, -9)$ and radius ${r_2} = \sqrt{(-3)^2 + (-9)^2 - 26} = \sqrt{9 + 81 - 26} = \sqrt{64} = 8$.
The distance between the centres ${C_1}$ and ${C_2}$ is ${d = \sqrt{(2 - (-3))^2 + (3 - (-9))^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13}$.
Since ${r_1 + r_2 = 5 + 8 = 13}$,which is equal to the distance between the centres ${d}$,the circles touch each other externally.
For two circles touching each other externally,the number of common tangents is $3$.

(B) Let the arithmetic progression be $a, a+d, a+2d, \dots, a+(n-1)d$. Given $a = 11$.
Sum of the first four terms: $a + (a+d) + (a+2d) + (a+3d) = 56$.
$4a + 6d = 56 \implies 4(11) + 6d = 56 \implies 44 + 6d = 56 \implies 6d = 12 \implies d = 2$.
Sum of the last four terms: $(a+(n-1)d) + (a+(n-2)d) + (a+(n-3)d) + (a+(n-4)d) = 112$.
$4a + (4n - 10)d = 112$.
Substitute $a = 11$ and $d = 2$: $4(11) + (4n - 10)(2) = 112$.
$44 + 8n - 20 = 112 \implies 8n + 24 = 112 \implies 8n = 88 \implies n = 11$.

(D) For pushing the body up the inclined plane,the minimum force $F_1$ must overcome both the component of gravity acting down the plane $(mg \sin \theta)$ and the maximum static friction force acting down the plane $(\mu mg \cos \theta)$. Thus,$F_1 = mg \sin \theta + \mu mg \cos \theta = mg(\sin \theta + \mu \cos \theta)$.
For preventing the body from sliding down the inclined plane,the minimum force $F_2$ acts up the plane. The component of gravity acts down the plane $(mg \sin \theta)$,and the maximum static friction force acts up the plane $(\mu mg \cos \theta)$. Thus,$F_2 = mg \sin \theta - \mu mg \cos \theta = mg(\sin \theta - \mu \cos \theta)$.
The ratio is given by $\frac{F_1}{F_2} = \frac{mg(\sin \theta + \mu \cos \theta)}{mg(\sin \theta - \mu \cos \theta)} = \frac{\sin \theta + \mu \cos \theta}{\sin \theta - \mu \cos \theta}$.
Dividing the numerator and denominator by $\cos \theta$,we get $\frac{F_1}{F_2} = \frac{\tan \theta + \mu}{\tan \theta - \mu}$.
Given $\tan \theta = 2 \mu$,substituting this value gives $\frac{F_1}{F_2} = \frac{2 \mu + \mu}{2 \mu - \mu} = \frac{3 \mu}{\mu} = 3$.

(C) For the circle $x^2+y^2-4x-6y-12=0$,the center $C_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$.
For the circle $x^2+y^2+6x+18y+26=0$,the center $C_2 = (-3, -9)$ and radius $r_2 = \sqrt{(-3)^2+(-9)^2-26} = \sqrt{9+81-26} = \sqrt{64} = 8$.
The distance between the centers $d = C_1C_2 = \sqrt{(2-(-3))^2 + (3-(-9))^2} = \sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13$.
Since $r_1 + r_2 = 5 + 8 = 13$,we have $d = r_1 + r_2$.
Because the distance between the centers is equal to the sum of the radii,the two circles touch each other externally.
When two circles touch each other externally,there are exactly $3$ common tangents.

(A) The total number of electrons in $HeH^{+}$ is $2 + 1 - 1 = 2$.
The molecular orbital electronic configuration is $\sigma_{1s}^2$.
Bond Order $= \frac{1}{2} [N_b - N_a] = \frac{1}{2} [2 - 0] = 1$.

(A) The strength of a hydrogen bond $X-H \cdots Y$ depends on the electronegativity difference between $X$ and $H$.
As the electronegativity of $X$ decreases,the polarity of the $X-H$ bond decreases,resulting in a smaller partial positive charge $(\delta+)$ on the hydrogen atom.
This leads to a weaker electrostatic attraction between the hydrogen atom and the electronegative atom $Y$.
The electronegativity values are $C (2.5) < N (3.0) < O (3.5)$.
Therefore,the $C-H \cdots O$ hydrogen bond is the weakest among the given options.

(C)

Species	$PCl_{5(g)}$	$PCl_{3(g)}$	$Cl_{2(g)}$
Initial moles	$1$	$0$	$0$
Equilibrium moles	$1 - \alpha$	$\alpha$	$\alpha$

Total moles at equilibrium = $(1 - \alpha) + \alpha + \alpha = 1 + \alpha$.
Partial pressures: $P_{PCl_5} = \frac{1 - \alpha}{1 + \alpha} P$,$P_{PCl_3} = \frac{\alpha}{1 + \alpha} P$,$P_{Cl_2} = \frac{\alpha}{1 + \alpha} P$.
Equilibrium constant $K_P = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} = \frac{(\frac{\alpha}{1 + \alpha} P)(\frac{\alpha}{1 + \alpha} P)}{(\frac{1 - \alpha}{1 + \alpha} P)} = \frac{\alpha^2 P}{1 - \alpha^2}$.
Rearranging: $K_P(1 - \alpha^2) = \alpha^2 P \implies K_P = \alpha^2(P + K_P) \implies \alpha^2 = \frac{K_P}{K_P + P}$.
Therefore,$\alpha = \left( \frac{K_P}{K_P + P} \right)^{1/2}$.

(B, C) compound is optically active if it possesses at least one chiral center and lacks a plane of symmetry or center of symmetry.
$(A)$ Glycine $(H_2N-CH_2-COOH)$ has no chiral center.
$(B)$ Lactic acid $(CH_3-CH(OH)-COOH)$ has one chiral center at the central carbon atom,making it optically active.
$(C)$ Glyceraldehyde $(HOCH_2-CH(OH)-CHO)$ has one chiral center at the central carbon atom,making it optically active.
$(D)$ Tartaric acid $(HOOC-CH(OH)-CH(OH)-COOH)$ has two chiral centers,but the meso form has a plane of symmetry and is optically inactive. However,the chiral isomers of tartaric acid are optically active.
Given the options provided,$(B)$ and $(C)$ are clearly optically active.

(C) Kjeldahl's method is not applicable to compounds containing nitrogen in nitro groups $(-NO_2)$,azo groups $(-N=N-)$,or nitrogen present in the ring (like pyridine) because these nitrogen atoms cannot be converted into ammonium sulfate under the conditions of the Kjeldahl method.
In the given options,$C_6H_5-N=N-C_6H_5$ is an azo compound (azobenzene),which does not respond to Kjeldahl's method.

(B) $1$. Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
$2$. The formation of acetone $(CH_3)_2C=O$ indicates that the alkene must have a $(CH_3)_2C=$ group.
$3$. The molecular formula is $C_8H_{16}$. If one part is acetone $(C_3H_6O)$,the other part must be a $C_5$ aldehyde or ketone.
$4$. For the alkene to be optically active,it must contain a chiral center. In the structure $2,3$-dimethylhex-$2$-ene,the molecule is achiral. In $3,4$-dimethylhex-$2$-ene,the carbon at position $3$ is chiral because it is bonded to four different groups: $-H$,$-CH_3$,$-CH_2CH_3$,and $-C(CH_3)=C(CH_3)_2$.
$5$. The structure shown in option $B$ corresponds to $3,4$-dimethylhex-$2$-ene,which contains a chiral center (marked with an asterisk) and produces acetone upon ozonolysis.

(B) The reaction of $MeMgBr$ (a Grignard reagent) with hydrocarbons occurs if the hydrocarbon has an acidic hydrogen atom. $MeMgBr$ acts as a base and abstracts the most acidic proton to form methane $(CH_4)$.
Among the given options,$1,3$-cyclopentadiene contains a methylene group $(-CH_2-)$ between two double bonds. The hydrogen atoms on this carbon are significantly more acidic than those in other hydrocarbons because the resulting cyclopentadienyl anion is aromatic ($6\pi$ electrons,$H$ückel's rule),which provides high stability.
Therefore,$1,3$-cyclopentadiene reacts readily with $MeMgBr$ to form methane and the cyclopentadienyl Grignard reagent.

(C) For solution $A$: $pH=6.0$,so $[H^{+}]_A = 10^{-6} \ M$.
For solution $B$: $pH=4.0$,so $[H^{+}]_B = 10^{-4} \ M$.
Let the volume of each solution be $1 \ L$.
Total moles of $H^{+}$ = $(1 \ L \times 10^{-6} \ M) + (1 \ L \times 10^{-4} \ M) = 10^{-6} + 10^{-4} = 1.01 \times 10^{-4} \ \text{moles}$.
Total volume of the mixture = $1 \ L + 1 \ L = 2 \ L$.
Resultant $[H^{+}] = \frac{1.01 \times 10^{-4} \ \text{moles}}{2 \ L} = 5.05 \times 10^{-5} \ M$.
$pH = -\log(5.05 \times 10^{-5}) = 5 - \log(5.05) \approx 5 - 0.703 = 4.297$.
Since $4.297$ lies between $4$ and $5$,the correct range is between $4$ and $5$.

(A) The solubility of a sparingly soluble salt like $AgCl$ is affected by the common ion effect and the salt effect.
$1$. In $H_2O$,the solubility is $S = \sqrt{K_{sp}}$.
$2$. In $1 \ M \ NaCl$ and $1 \ M \ CaCl_2$,the common ion $Cl^-$ is present. Since $CaCl_2$ provides $2 \ M \ Cl^-$ ions and $NaCl$ provides $1 \ M \ Cl^-$ ions,the solubility is lowest in $CaCl_2$ due to the strongest common ion effect.
$3$. In $1 \ M \ NaNO_3$,the solubility is slightly higher than in pure water due to the salt effect (or diverse ion effect),which increases the ionic strength of the solution.
$4$. Therefore,the order of increasing solubility is: $CaCl_2 < NaCl < H_2O < NaNO_3$.

(D) To push the body up the inclined plane,the applied force $F_1$ must overcome both the component of gravitational force $mg \sin \theta$ and the maximum static friction force $f_{max} = \mu mg \cos \theta$ acting downwards. Thus,$F_1 = mg \sin \theta + \mu mg \cos \theta$.
To prevent the body from sliding down the inclined plane,the applied force $F_2$ acts upwards along with the maximum static friction force $f_{max} = \mu mg \cos \theta$ to balance the gravitational component $mg \sin \theta$ acting downwards. Thus,$F_2 = mg \sin \theta - \mu mg \cos \theta$.
Given $\tan \theta = 2\mu$,we can write $\sin \theta = 2\mu \cos \theta$. Substituting this into the expressions for $F_1$ and $F_2$:
$F_1 = mg(2\mu \cos \theta) + \mu mg \cos \theta = 3\mu mg \cos \theta$
$F_2 = mg(2\mu \cos \theta) - \mu mg \cos \theta = \mu mg \cos \theta$
Therefore,the ratio is:
$\frac{F_1}{F_2} = \frac{3\mu mg \cos \theta}{\mu mg \cos \theta} = 3$.

(A) In the structure of diborane $(B_2H_6)$,there are two boron atoms.
Each boron atom is bonded to two terminal hydrogen atoms via normal covalent bonds ($2 \times 2 = 4$ terminal hydrogens).
The two boron atoms are linked by two bridging hydrogen atoms,which form $3c-2e$ (three-center two-electron) bonds.
Therefore,there are $4$ terminal hydrogens and $2$ bridging hydrogens.

(D) The reduction of nitrogen gas $(N_2)$ to ammonia $(NH_3)$ involves the following balanced half-reaction:
$N_2 + 6e^{-} + 6H^{+} \rightarrow 2NH_3$
In this reaction,the oxidation state of nitrogen changes from $0$ in $N_2$ to $-3$ in $NH_3$.
Since there are two nitrogen atoms,the total change in oxidation state is $2 \times 3 = 6$.
Therefore,$6$ electrons are required for the reduction process.

(D) To determine the hybridization and number of lone pairs for $XeOF_2$:
$1$. The central atom is Xenon $(Xe)$,which has $8$ valence electrons.
$2$. In $XeOF_2$,$Xe$ forms two single bonds with two Fluorine $(F)$ atoms and one double bond with one Oxygen $(O)$ atom.
$3$. The number of bonding electron pairs = $2$ (from $F$) + $2$ (from $O$) = $4$ electron pairs involved in bonding.
$4$. The number of lone pairs on $Xe$ = (Total valence electrons - electrons used in bonding) / $2$ = $(8 - 6) / 2 = 1$ lone pair.
$5$. The steric number = (Number of sigma bonds) + (Number of lone pairs) = $3$ (sigma bonds) + $1$ (lone pair) = $4$. Wait,let us re-evaluate: $Xe$ is bonded to $2$ $F$ atoms and $1$ $O$ atom. Total sigma bonds = $3$. Total lone pairs = $2$. Steric number = $3 + 2 = 5$.
$6$. $A$ steric number of $5$ corresponds to $sp^3d$ hybridization.
$7$. Thus,$XeOF_2$ has $2$ lone pairs and $sp^3d$ hybridization.

(C) The reduction of $O_2$ by four electrons is represented by the equation: $O_2 + 4e^- \rightarrow 2O^{2-}$.
Each oxygen atom gains $2e^-$ to form an oxide ion $(O^{2-})$.

(B) The balanced chemical equation is: $HgO + 4 I^{-} + H_2O \rightarrow HgI_4^{2-} + 2 OH^{-}$.
From the stoichiometry,$1 \ mol$ of $HgO$ produces $2 \ mol$ of $OH^{-}$.
Number of moles of $HgO = \frac{0.217 \ g}{217 \ g \ mol^{-1}} = 0.001 \ mol$.
Therefore,moles of $OH^{-}$ produced $= 2 \times 0.001 = 0.002 \ mol$.
For neutralization,$n_{H^{+}} = n_{OH^{-}}$,so $n_{HCl} = 0.002 \ mol$.
Using the formula $n = M \times V(L)$,we have $0.002 = 0.01 \times V(L)$.
$V(L) = \frac{0.002}{0.01} = 0.2 \ L = 200 \ mL$.

(B) The molar mass of water $(H_2O)$ is $18 \ g/mol$.
Number of moles of water $= \frac{0.36 \ g}{18 \ g/mol} = 0.02 \ mol$.
Number of water molecules $= 0.02 \times 6.023 \times 10^{23} = 1.205 \times 10^{22}$ molecules.
Since each molecule of water $(H_2O)$ contains $1$ oxygen atom,the number of oxygen atoms is equal to the number of water molecules.
Therefore,the number of oxygen atoms $= 1.205 \times 10^{22}$.

(D) Mass of gas initially $= 30.0 \ kg - 15.0 \ kg = 15.0 \ kg$.
Mass of gas remaining $= 24.2 \ kg - 15.0 \ kg = 9.2 \ kg$.
Mass of gas used $= 15.0 \ kg - 9.2 \ kg = 5.8 \ kg = 5800 \ g$.
Molar mass of butane $(C_4H_{10})$ $= 4 \times 12 + 10 \times 1 = 58 \ g/mol$.
Number of moles of gas used $(n)$ $= \frac{5800 \ g}{58 \ g/mol} = 100 \ mol$.
Using the ideal gas equation $PV = nRT$ at $1 \ atm$ and $27^{\circ}C$ $(300 \ K)$:
$V = \frac{nRT}{P} = \frac{100 \ mol \times 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} \times 300 \ K}{1 \ atm} = 2463 \ L$.
Since $1 \ m^3 = 1000 \ L$,the volume in cubic metres is $\frac{2463}{1000} \ m^3 = 2.463 \ m^3$.

(A) According to Graham's Law of diffusion: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$
Here,$r_1 = \frac{360}{30} = 12 \ cm^3/min$ (for hydrocarbon) and $r_2 = \frac{360}{60} = 6 \ cm^3/min$ (for $SO_2$).
$M_2$ (molecular weight of $SO_2$) = $64 \ g/mol$.
Substituting the values: $\frac{12}{6} = \sqrt{\frac{64}{M_1}}$
$2 = \sqrt{\frac{64}{M_1}}$
Squaring both sides: $4 = \frac{64}{M_1}$
$M_1 = \frac{64}{4} = 16 \ g/mol$.
The hydrocarbon with a molecular weight of $16 \ g/mol$ is $CH_4$.

(B) For the first circle $x^2+y^2-4x-6y-12=0$:
Center $C_1 = (2, 3)$,Radius $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$.
For the second circle $x^2+y^2+6x+18y+26=0$:
Center $C_2 = (-3, -9)$,Radius $r_2 = \sqrt{(-3)^2+(-9)^2-26} = \sqrt{9+81-26} = \sqrt{64} = 8$.
The distance between centers $C_1C_2 = \sqrt{(2 - (-3))^2 + (3 - (-9))^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
Since $r_1 + r_2 = 5 + 8 = 13$,we have $C_1C_2 = r_1 + r_2$.
This implies that the two circles touch each other externally.
When two circles touch each other externally,the number of common tangents is $3$.

(A, B, C) An extensive property depends on the amount of matter,while an intensive property is independent of the amount of matter.
$1$. The product of an intensive property $(x)$ and an extensive property $(X)$ is extensive ($xX$ is extensive).
$2$. The ratio of an intensive property $(x)$ to an extensive property $(X)$ is intensive ($\frac{x}{X}$ is intensive).
$3$. The ratio of an extensive property $(X)$ to an intensive property $(x)$ is extensive ($\frac{X}{x}$ is extensive).
$4$. The ratio of the change in an extensive property $(dX)$ to the change in an intensive property $(dx)$ is extensive ($\frac{dX}{dx}$ is extensive).
Therefore,statements $A$,$B$,and $C$ are correct.

(C) According to the first law of thermodynamics,$Q = \Delta U + W$.
In an adiabatic process,$Q = 0$.
In free expansion,the external pressure $P_{ext} = 0$,so the work done $W = P_{ext} \Delta V = 0$.
Substituting these values into the first law: $0 = \Delta U + 0$,which implies $\Delta U = 0$.
For an ideal gas,internal energy $U$ is a function of temperature only $(U = f(T))$.
Since $\Delta U = 0$,the change in temperature $\Delta T = 0$,meaning the temperature remains constant.
Therefore,the process is isothermal.

(C) Given $f(x) = \alpha \log |x| + \beta x^2 + x$.
The derivative is $f'(x) = \frac{\alpha}{x} + 2\beta x + 1$.
Since $x = -1$ and $x = 2$ are extreme points,$f'(-1) = 0$ and $f'(2) = 0$.
For $x = -1$: $f'(-1) = \frac{\alpha}{-1} + 2\beta(-1) + 1 = 0 \Rightarrow -\alpha - 2\beta + 1 = 0 \Rightarrow \alpha + 2\beta = 1$ ...$(i)$
For $x = 2$: $f'(2) = \frac{\alpha}{2} + 2\beta(2) + 1 = 0 \Rightarrow \frac{\alpha}{2} + 4\beta + 1 = 0 \Rightarrow \alpha + 8\beta = -2$ ...(ii)
Subtracting $(i)$ from (ii): $(\alpha + 8\beta) - (\alpha + 2\beta) = -2 - 1 \Rightarrow 6\beta = -3 \Rightarrow \beta = -\frac{1}{2}$.
Substituting $\beta = -\frac{1}{2}$ into $(i)$: $\alpha + 2(-\frac{1}{2}) = 1 \Rightarrow \alpha - 1 = 1 \Rightarrow \alpha = 2$.

(D) Given the differential equation: $\frac{d p}{d t} = 0.5 p - 450 = 0.5(p - 900)$.
Separating the variables,we get: $\frac{d p}{p - 900} = 0.5 dt$.
Integrating both sides from $t=0$ to $t=T$ where $p(0)=850$ and $p(T)=0$:
$\int_{850}^{0} \frac{d p}{p - 900} = \int_{0}^{T} 0.5 dt$.
$[\ln |p - 900|]_{850}^{0} = 0.5 T$.
$\ln |0 - 900| - \ln |850 - 900| = 0.5 T$.
$\ln(900) - \ln(50) = 0.5 T$.
$\ln(\frac{900}{50}) = 0.5 T$.
$\ln(18) = 0.5 T$.
$T = 2 \ln(18)$.

(C) Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if their scalar triple product $[\vec{u}, \vec{v}, \vec{w}] = 0$.
Given vectors are $\vec{u} = \vec{a}+2 \vec{b}+3 \vec{c}$,$\vec{v} = 0\vec{a} + \lambda \vec{b}+4 \vec{c}$,and $\vec{w} = 0\vec{a} + 0\vec{b} + (2 \lambda-1) \vec{c}$.
The scalar triple product is given by the determinant of the coefficients of $\vec{a}, \vec{b}, \vec{c}$:
$\left|\begin{array}{ccc} 1 & 2 & 3 \\ 0 & \lambda & 4 \\ 0 & 0 & 2 \lambda-1 \end{array}\right| = 0$.
Expanding along the first column:
$1 \times [\lambda(2 \lambda-1) - 0] = 0$.
$\lambda(2 \lambda-1) = 0$.
This gives $\lambda = 0$ or $\lambda = \frac{1}{2}$.
Since the vectors are non-coplanar when the determinant is non-zero,the vectors are non-coplanar for all real values of $\lambda$ except $\lambda = 0$ and $\lambda = \frac{1}{2}$.

(D) The ease of dehydration of alcohols under acidic conditions depends on the stability of the carbocation formed after the loss of the water molecule.
In the given options,the dehydration of $4$-hydroxyhexan-$2$-one (Option $D$) leads to the formation of a carbocation that is stabilized by hyperconjugation with $5$ $\alpha$-hydrogen atoms.
Additionally,the resulting alkene product is conjugated with the carbonyl group,which provides extra stability through resonance.
Therefore,$4$-hydroxyhexan-$2$-one undergoes dehydration most readily.

(C) $1$. The reaction of benzene with $CH_3CH_2CH_2Cl$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation. The primary carbocation formed $(CH_3CH_2CH_2^+)$ undergoes a $1,2-H^-$ shift to form a more stable secondary carbocation $(CH_3CH^+CH_3)$.
$2$. This secondary carbocation attacks the benzene ring to form isopropylbenzene (cumene) as the major product '$P$'.
$3$. Cumene undergoes oxidation with $O_2/OH^-, \Delta$ to form cumene hydroperoxide,which upon treatment with dilute $H_2SO_4$ yields phenol and acetone $(CH_3COCH_3)$ as the major product '$Q$'.
$4$. Thus,'$P$' is isopropylbenzene and '$Q$' is acetone.

(A) Compounds that are stronger acids than carbonic acid $(H_2CO_3)$ react with sodium bicarbonate $(NaHCO_3)$ to release carbon dioxide $(CO_2)$ gas,which causes effervescence.
$2,4,6-$Trinitrophenol (also known as picric acid) is a strong acid due to the electron-withdrawing effect of three nitro $(-NO_2)$ groups,which stabilize the phenoxide ion.
Since the $pK_a$ of picric acid is lower than the $pK_a$ of $H_2CO_3$,it reacts with $NaHCO_3$ to produce effervescence.
Other options like acetone,ethanol,and diethyl ether are not acidic enough to react with $NaHCO_3$.

(A, C, D) The given compound is ethyl acetoacetate $(CH_3COCH_2COOC_2H_5)$.
$(A)$ It contains active methylene group,so it exhibits keto-enol tautomerism.
$(B)$ The enol form has an acidic hydroxyl group,so it reacts with metallic sodium to release $H_2$ gas.
$(C)$ The enol form gives a characteristic reddish-violet coloration with neutral $FeCl_3$ solution.
$(D)$ It contains a keto group,so it reacts with $2,4-$dinitrophenyl hydrazine $(2,4-DNP)$ to form a precipitate.
Therefore,statements $(A)$,$(C)$,and $(D)$ are correct.

(A) The acidic strength of dicarboxylic acids depends on the distance between the two carboxyl groups. As the distance between the two $-COOH$ groups decreases,the inductive effect ($-I$ effect) of one carboxyl group on the other increases,which stabilizes the conjugate base and increases the acidity.
Comparing the structures:
$I$: Adipic acid,$HOOC-(CH_2)_4-COOH$ (longest chain between groups)
$IV$: Glutaric acid,$HOOC-(CH_2)_3-COOH$
$II$: Methylmalonic acid,$CH_3-CH(COOH)_2$ (groups are on adjacent carbons)
$III$: Oxalic acid,$HOOC-COOH$ (groups are directly attached)
Ordering by increasing distance between carboxyl groups: $III < II < IV < I$.
Therefore,the order of increasing acid strength is $I < IV < II < III$.

(B) The nitrogen dioxide molecule $(NO_2)$ has a total of $17$ valence electrons (odd number of electrons).
Due to the presence of an unpaired electron on the nitrogen atom,$NO_2$ exhibits paramagnetic behavior.
Other oxides like $SO_2$,$SiO_2$,and $CO_2$ have all electrons paired,making them diamagnetic.

(A) For zero order reaction: $t_{1/2} = \frac{a}{2k}$ $\Rightarrow 10 = \frac{100}{2k}$ $\Rightarrow k = 5 \ mol \ L^{-1} \ min^{-1}$. After $t = 20 \ min$,amount reacted $x = kt = 5 \times 20 = 100 \ mol$. Remaining reactant $(a-x) = 100 - 100 = 0 \ mol$.
For first order reaction: Number of half-lives $n = \frac{20}{10} = 2$. Remaining reactant $(a-x) = \frac{a}{2^n} = \frac{100}{2^2} = 25 \ mol$.
For second order reaction: $t_{1/2} = \frac{1}{ak}$ $\Rightarrow 10 = \frac{1}{100k}$ $\Rightarrow k = 10^{-3} \ L \ mol^{-1} \ min^{-1}$. Using integrated rate law $\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt$,we get $\frac{1}{[A]_t} = \frac{1}{100} + (10^{-3} \times 20) = 0.01 + 0.02 = 0.03$. Thus,$[A]_t = \frac{1}{0.03} = 33.33 \ mol$.

(C) The rate of reaction $k$ is inversely proportional to the time taken $t$,so $k \propto 1/t$.
Using the Arrhenius equation: $\ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left[\frac{T_1 - T_2}{T_1 T_2}\right]$.
Given $t_1 = 4.0 \text{ min}$ at $T_1$ and $t_2 = 8.0 \text{ min}$ at $T_2$,we have $k_1 = 1/4$ and $k_2 = 1/8$.
Substituting these values: $\ln \left(\frac{1/8}{1/4}\right) = \frac{E_a}{R} \left[\frac{T_1 - T_2}{T_1 T_2}\right]$.
$\ln(0.5) = \frac{E_a}{R} \left[\frac{T_1 - T_2}{T_1 T_2}\right]$.
Since $\ln(0.5) = -\ln(2) \approx -0.693$,we get $-0.693 = \frac{E_a}{R} \left[\frac{T_1 - T_2}{T_1 T_2}\right]$.
Rearranging for $E_a$: $E_a = -0.693 R \left[\frac{T_1 T_2}{T_1 - T_2}\right] = 0.693 R \left[\frac{T_1 T_2}{T_2 - T_1}\right]$.
Note: The provided options suggest a specific sign convention; based on the derivation,the correct expression is $0.693 R \frac{T_1 T_2}{T_2 - T_1}$.

(A) The reaction of $[Cd(NH_3)_4](NO_3)_2$ with $KCN$ leads to the formation of the stable cyano-complex $K_2[Cd(CN)_4]$.
Thus,$P = K_2[Cd(CN)_4]$.
When $K_2[Cd(CN)_4]$ reacts with $H_2S$,it forms the yellow precipitate of cadmium sulfide $(CdS)$.
Thus,$Q = CdS$.
The correct option is $A$.

(C) The electronic configuration of $Eu$ $(Z=63)$ is $[Xe] 4f^7 6s^2$. It can lose two electrons to form $Eu^{2+}$ $([Xe] 4f^7)$,which is stable due to the half-filled $f$-orbital. It also exhibits a $+3$ oxidation state.
The electronic configuration of $Gd$ $(Z=64)$ is $[Xe] 4f^7 5d^1 6s^2$. It loses three electrons to form $Gd^{3+}$ $([Xe] 4f^7)$,which is highly stable due to the half-filled $f$-orbital. Thus,the common stable oxidation states are $+2$ for $Eu$ and $+3$ for $Gd$.

(A) The separation of lanthanoids is difficult due to their similar chemical properties. However,$Eu(II)$ shows similarity with alkaline earth metals like $Sr(II)$.
$Eu(II)$ can be precipitated as $EuSO_4$ in the presence of sulfate ions,similar to $SrSO_4$,while other trivalent lanthanoid ions like $Dy(III)$ or $Gd(III)$ remain in the solution.
Therefore,the pair $Eu(II)$ and $Dy(III)$ (or any other trivalent lanthanoid) can be separated by the precipitation method.

(B) According to Kohlrausch's law of independent migration of ions:
$\lambda_{m(NH_4OH)}^{\infty} = \lambda_{m(NH_4Cl)}^{\infty} + \lambda_{m(Ba(OH)_2)}^{\infty} / 2 - \lambda_{m(BaCl_2)}^{\infty} / 2$
$\lambda_{m(NH_4OH)}^{\infty} = 129.8 + 523.28 / 2 - 280.0 / 2$
$\lambda_{m(NH_4OH)}^{\infty} = 129.8 + 261.64 - 140.0$
$\lambda_{m(NH_4OH)}^{\infty} = 251.44 \ S \ cm^2 \ mol^{-1}$

(A) The reactivity of $S_N 1$ reaction is directly proportional to the stability of the carbocation intermediate formed after the departure of the leaving group $(Br^-)$.
In option $A$,the carbocation formed is the cyclopentadienyl cation,which is aromatic ($6 \pi$ electrons,planar,cyclic,conjugated). Aromatic compounds are exceptionally stable.
In option $B$,the carbocation is allylic,which is resonance-stabilized but not aromatic.
In option $C$,the carbocation is also allylic but less substituted than $B$.
In option $D$,the carbocation is also allylic but with different substitution.
Since the cyclopentadienyl cation is aromatic,it is the most stable carbocation among the choices,making the corresponding alkyl bromide the most reactive in an $S_N 1$ reaction.

(B) The rate of nucleophilic aromatic substitution is increased by the presence of electron-withdrawing groups $(EWG)$ on the benzene ring,as they stabilize the intermediate carbanion (Meisenheimer complex).
$II$ has a $-OCH_3$ group,which is an electron-donating group $(EDG)$ via resonance,thus it decreases the rate of nucleophilic substitution compared to chlorobenzene $(I)$.
$III$ has one $-NO_2$ group (a strong $EWG$),which increases the rate compared to $I$.
$IV$ has two $-NO_2$ groups,which further increase the rate compared to $III$.
Therefore,the increasing order of reactivity is: $II < I < III < IV$.

(B) The reaction of $3$-chloroanisole with $NaNH_2$ in liquid $NH_3$ proceeds via the benzyne mechanism.
$1$. The amide ion $(NH_2^-)$ abstracts a proton from the ortho position relative to the chlorine atom,leading to the formation of a benzyne intermediate.
$2$. The nucleophilic attack by $NH_2^-$ on the benzyne intermediate can occur at two positions.
$3$. Attack at the meta position relative to the $-OCH_3$ group leads to a carbanion that is stabilized by the inductive effect $(-I)$ of the methoxy group.
$4$. Attack at the ortho position leads to a less stable carbanion due to steric hindrance and electronic factors.
$5$. Therefore,the major product formed is $m-$anisidine ($3$-methoxyaniline).

(C) The reaction proceeds in two steps:
$1$. The Grignard reagent $CH_3MgBr$ acts as a nucleophile and attacks the carbonyl carbon of $5-chloropentan-2-one$. This forms an alkoxide intermediate.
$2$. Upon workup with $H_3O^+$,the alkoxide oxygen is protonated to form an alcohol. However,in the presence of the internal chloride group,the alkoxide oxygen performs an intramolecular nucleophilic substitution $(S_N2)$ reaction,displacing the chloride ion to form a five-membered cyclic ether,$2,2-dimethyltetrahydrofuran$.

(C) The reaction is an electrophilic aromatic substitution (bromination) of $N$-phenylbenzamide using $Br_2$ and $FeBr_3$.
$N$-phenylbenzamide consists of two phenyl rings attached to an amide group.
The nitrogen atom of the amide group is directly attached to one of the phenyl rings. The lone pair on the nitrogen atom is delocalized into this ring,making it an activated ring towards electrophilic substitution.
Conversely,the other phenyl ring is attached to the carbonyl group $(C=O)$,which is an electron-withdrawing group,making that ring deactivated.
Therefore,the electrophilic substitution occurs on the activated ring at the ortho and para positions. Due to steric hindrance,the para-substituted product is the major product.
Thus,the major product is $N$-($4$-bromophenyl)benzamide.

(C) Number of $A$ atoms at corners $= 8 \times \frac{1}{8} = 1$.
Number of $B$ atoms at body center $= 1 \times 1 = 1$.
Number of $C$ atoms at edge centers $= 12 \times \frac{1}{4} = 3$.
Therefore,the ratio of $A:B:C$ is $1:1:3$.
Hence,the formula of the compound is $ABC_3$.

(D) The formula for depression in freezing point is $\Delta T_f = K_f \times m$,where $m$ is molality.
For $PQ_2$: $0.8 = \frac{5.1 \times 1000 \times 1}{51 \times M_{PQ_2}}$ $\Rightarrow M_{PQ_2} = \frac{5100}{51 \times 0.8} = 125 \ g/mol$.
For $PQ_3$: $0.625 = \frac{5.1 \times 1000 \times 1}{51 \times M_{PQ_3}}$ $\Rightarrow M_{PQ_3} = \frac{5100}{51 \times 0.625} = 160 \ g/mol$.
Let atomic masses be $P$ and $Q$.
$P + 2Q = 125$ $(i)$
$P + 3Q = 160$ (ii)
Subtracting $(i)$ from (ii): $Q = 35$.
Substituting $Q$ in $(i)$: $P + 2(35) = 125 \Rightarrow P = 125 - 70 = 55$.
Thus,atomic mass of $P = 55$ and $Q = 35$.

(B) The given nuclear reaction is: $_{5}B^{10} + _{2}He^{4} \rightarrow _{b}^{a}X + _{0}n^{1}$
By balancing the atomic numbers (sum of protons):
$5 + 2 = b + 0$
$b = 7$
By balancing the mass numbers (sum of protons and neutrons):
$10 + 4 = a + 1$
$14 = a + 1$
$a = 13$
Therefore,the product $X$ is $_{7}N^{13}$.

(B) $As_{2}S_{3}$ is a negatively charged sol. According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the valency of the active ion (the ion carrying a charge opposite to that of the colloidal particles).
For a negatively charged sol,the coagulating power increases with the increase in the magnitude of the positive charge on the cation.
The charges on the given ions are: $Na^{+}$ $(+1)$,$Ba^{2+}$ $(+2)$,and $Al^{3+}$ $(+3)$.
Therefore,the order of coagulating power is $Na^{+} < Ba^{2+} < Al^{3+}$.