MHT CET 2025 Chemistry Question Paper with Answer and Solution

(B) The total entropy change is given by the formula: $\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}$.
Given $\Delta S_{\text{sys}} = 32 \ JK^{-1}$.
The entropy change of the surroundings is given by $\Delta S_{\text{surr}} = -\frac{\Delta H}{T}$.
Given $\Delta H = -150 \ kJ = -150000 \ J$ and $T = 300 \ K$.
$\Delta S_{\text{surr}} = -\frac{-150000 \ J}{300 \ K} = 500 \ JK^{-1}$.
Therefore,$\Delta S_{\text{total}} = 32 \ JK^{-1} + 500 \ JK^{-1} = 532 \ JK^{-1}$.

(C) For any reversible process,the system is always in equilibrium with its surroundings.
By definition,the total entropy change of the universe $(\Delta S_{total})$ for any reversible process is zero.
$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} = 0$.

(B) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative,where $\Delta G = \Delta H - T\Delta S$.
For the reaction to be spontaneous at high temperature,the term $T\Delta S$ must be large and positive to make $\Delta G$ negative,even if $\Delta H$ is positive.
This requires $\Delta S > 0$ and $\Delta H > 0$.
Thus,when $\Delta H > 0$ and $\Delta S > 0$,the reaction becomes spontaneous at high temperatures where $T\Delta S > \Delta H$.

(A) For a reaction to be spontaneous,the Gibbs free energy change must be negative,i.e.,$\Delta G = \Delta H - T\Delta S < 0$.
Given that the reaction is spontaneous below equilibrium temperature $(T < T_{eq})$,we analyze the condition $\Delta H - T\Delta S < 0$.
If $\Delta H < 0$ and $\Delta S < 0$,then $\Delta G = \Delta H - T\Delta S$.
For $\Delta G < 0$,we require $|\Delta H| > |T\Delta S|$,which implies $T < \frac{\Delta H}{\Delta S}$.
Since both $\Delta H$ and $\Delta S$ are negative,the ratio $\frac{\Delta H}{\Delta S}$ is positive,representing the equilibrium temperature.
Thus,the reaction is spontaneous when $T$ is less than this equilibrium temperature.

(A) The relationship between Gibbs free energy,enthalpy,and entropy is given by the equation: $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$.
Given values are: $\Delta G^{\circ} = 28000 \ J$,$\Delta S^{\circ} = 120 \ J \ K^{-1}$,and $T = 298 \ K$.
Rearranging the formula to solve for $\Delta H^{\circ}$: $\Delta H^{\circ} = \Delta G^{\circ} + T\Delta S^{\circ}$.
Substituting the values: $\Delta H^{\circ} = 28000 \ J + (298 \ K \times 120 \ J \ K^{-1})$.
$\Delta H^{\circ} = 28000 \ J + 35760 \ J = 63760 \ J$.
Converting to $kJ$: $\Delta H^{\circ} = 63.76 \ kJ$.

(C) The Gibbs free energy equation is given by $\Delta G = \Delta H - T \Delta S$.
For $\Delta G = 0$,the equation becomes $\Delta H = T \Delta S$.
Given: $\Delta H = -210 \ kJ = -210000 \ J$ and $\Delta S = -150 \ J \ K^{-1}$.
Substituting the values: $-210000 \ J = T \times (-150 \ J \ K^{-1})$.
$T = \frac{-210000}{-150} \ K = 1400 \ K$.
Therefore,the temperature is $1400 \ K$.

(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 non-spontaneous,$\Delta G$ must be greater than $0$ $(\Delta G > 0)$.
If $\Delta H > 0$ (endothermic) and $\Delta S < 0$ (decrease in entropy),then $\Delta G = (\text{positive value}) - T(\text{negative value}) = \text{positive value} + T(\text{positive value})$.
Since $T$ is always positive in Kelvin,$\Delta G$ will always be positive regardless of the temperature.
Therefore,the reaction is non-spontaneous at all temperatures when $\Delta H > 0$ and $\Delta S < 0$.

(C) The spontaneity of a reaction is determined by the Gibbs free energy change,given by the equation: $\Delta G = \Delta H - T\Delta S$.
For a reaction to be spontaneous at all temperatures,$\Delta G$ must be negative $(\Delta G < 0)$ for all values of $T$.
If $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (increase in entropy),then $\Delta G$ will always be negative regardless of the temperature $T$,because both terms ($\Delta H$ and $-T\Delta S$) will contribute to a negative value.
Therefore,the correct condition is $\Delta H < 0$ and $\Delta S > 0$.

(A) The relationship between Gibbs free energy change $(\Delta G)$,enthalpy change $(\Delta H)$,and entropy change $(\Delta S)$ is given by the equation: $\Delta G = \Delta H - T \Delta S$.
For $\Delta G = 0$,the equation becomes: $0 = \Delta H - T \Delta S$,which implies $T = \frac{\Delta H}{\Delta S}$.
Given: $\Delta H = -225 \ kJ \ mol^{-1} = -225000 \ J \ mol^{-1}$ and $\Delta S = -150 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $T = \frac{-225000 \ J \ mol^{-1}}{-150 \ J \ K^{-1} \ mol^{-1}} = 1500 \ K$.
Therefore,the temperature at which $\Delta G$ is zero is $1500 \ K$.

(A) The given line is $x-y+1=0$. The curve is $x=y^2$.
Let a point on the curve be $P(y^2, y)$.
The slope of the tangent to the curve at $P$ must be equal to the slope of the given line for the shortest distance.
The slope of the line $x-y+1=0$ is $m=1$.
Differentiating $x=y^2$ with respect to $y$,we get $\frac{dx}{dy} = 2y$.
Thus,$\frac{dy}{dx} = \frac{1}{2y}$.
Setting $\frac{1}{2y} = 1$,we find $y = \frac{1}{2}$.
Then $x = y^2 = (\frac{1}{2})^2 = \frac{1}{4}$.
The point on the curve is $P(\frac{1}{4}, \frac{1}{2})$.
The shortest distance $d$ from point $(x_0, y_0)$ to the line $Ax+By+C=0$ is given by $d = \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$.
Here,$A=1, B=-1, C=1, x_0=\frac{1}{4}, y_0=\frac{1}{2}$.
$d = \frac{|1(\frac{1}{4}) - 1(\frac{1}{2}) + 1|}{\sqrt{1^2+(-1)^2}} = \frac{|\frac{1}{4} - \frac{2}{4} + \frac{4}{4}|}{\sqrt{2}} = \frac{3/4}{\sqrt{2}} = \frac{3}{4\sqrt{2}} = \frac{3\sqrt{2}}{8}$.

(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) Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these into the given equation:
$\frac{2 \cos A}{2R \sin A} + \frac{\cos B}{2R \sin B} + \frac{2 \cos C}{2R \sin C} = \frac{2R \sin A}{(2R \sin B)(2R \sin C)} + \frac{2R \sin B}{(2R \sin C)(2R \sin A)}$
$\frac{\cot A}{R} + \frac{\cot B}{2R} + \frac{\cot C}{R} = \frac{\sin A}{2R \sin B \sin C} + \frac{\sin B}{2R \sin C \sin A}$
Multiply by $2R$:
$2 \cot A + \cot B + 2 \cot C = \frac{\sin A}{\sin B \sin C} + \frac{\sin B}{\sin C \sin A}$
Using $\sin A = \sin(B+C) = \sin B \cos C + \cos B \sin C$ and $\sin B = \sin(A+C) = \sin A \cos C + \cos A \sin C$:
$2 \cot A + \cot B + 2 \cot C = \frac{\sin B \cos C + \cos B \sin C}{\sin B \sin C} + \frac{\sin A \cos C + \cos A \sin C}{\sin C \sin A}$
$2 \cot A + \cot B + 2 \cot C = \cot C + \cot B + \cot C + \cot A$
$2 \cot A + \cot B + 2 \cot C = \cot A + \cot B + 2 \cot C$
$2 \cot A = \cot A$
$\cot A = 0$
Since $A$ is an angle of a triangle,$A = \frac{\pi}{2}$.

(C) Let the first line be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = \lambda$. Then any point on this line is $(2\lambda+1, 3\lambda-1, 4\lambda+1)$.
Let the second line be $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1} = \mu$. Then any point on this line is $(\mu+3, 2\mu+k, \mu)$.
If the lines intersect,there exist $\lambda$ and $\mu$ such that:
$2\lambda+1 = \mu+3 \implies 2\lambda - \mu = 2$ (Equation $1$)
$4\lambda+1 = \mu \implies 4\lambda - \mu = -1$ (Equation $2$)
Subtracting Equation $1$ from Equation $2$: $(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2 \implies 2\lambda = -3 \implies \lambda = -\frac{3}{2}$.
Substituting $\lambda = -\frac{3}{2}$ into Equation $1$: $2(-\frac{3}{2}) - \mu = 2 \implies -3 - \mu = 2 \implies \mu = -5$.
Now,equate the $y$-coordinates: $3\lambda - 1 = 2\mu + k$.
Substitute $\lambda = -\frac{3}{2}$ and $\mu = -5$: $3(-\frac{3}{2}) - 1 = 2(-5) + k$.
$-\frac{9}{2} - 1 = -10 + k \implies -\frac{11}{2} = -10 + k$.
$k = 10 - \frac{11}{2} = \frac{20-11}{2} = \frac{9}{2}$.

(D) Given the equation $2 \sin^2 x + 5 \sin x - 3 = 0$.
Let $u = \sin x$. Then the equation becomes $2u^2 + 5u - 3 = 0$.
Factoring the quadratic: $2u^2 + 6u - u - 3 = 0 \implies 2u(u + 3) - 1(u + 3) = 0 \implies (2u - 1)(u + 3) = 0$.
This gives $u = \frac{1}{2}$ or $u = -3$.
Since the range of $\sin x$ is $[-1, 1]$,$u = -3$ is rejected.
Thus,$\sin x = \frac{1}{2}$.
In the interval $[0, 3\pi]$,we look for solutions for $\sin x = \frac{1}{2}$.
In $[0, 2\pi]$,the solutions are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.
In the interval $(2\pi, 3\pi]$,we add $2\pi$ to the first solution: $x = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6}$.
So,the values are $\frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}$.
The total number of values is $3$.

(D) Let the given expression be $E = (2 \bar{a}-\bar{b}) \cdot ((\bar{a} \times \bar{b}) \times (\bar{a}+2 \bar{b}))$.
Using the vector triple product identity $(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{v} \cdot \vec{w}) \vec{u}$,we have:
$(\bar{a} \times \bar{b}) \times (\bar{a}+2 \bar{b}) = (\bar{a} \cdot (\bar{a}+2 \bar{b})) \bar{b} - (\bar{b} \cdot (\bar{a}+2 \bar{b})) \bar{a}$.
Since $\bar{a}$ and $\bar{b}$ are unit vectors,$\bar{a} \cdot \bar{a} = 1$ and $\bar{b} \cdot \bar{b} = 1$.
Also,$\bar{a} \cdot \bar{b} = \frac{1}{7\sqrt{10}} (3 \times 2 + 0 \times 3 + 1 \times (-6)) = 0$.
Thus,$(\bar{a} \cdot (\bar{a}+2 \bar{b})) = \bar{a} \cdot \bar{a} + 2(\bar{a} \cdot \bar{b}) = 1 + 0 = 1$.
And $(\bar{b} \cdot (\bar{a}+2 \bar{b})) = \bar{b} \cdot \bar{a} + 2(\bar{b} \cdot \bar{b}) = 0 + 2(1) = 2$.
So,$(\bar{a} \times \bar{b}) \times (\bar{a}+2 \bar{b}) = 1 \bar{b} - 2 \bar{a} = - (2 \bar{a} - \bar{b})$.
Now,$E = (2 \bar{a}-\bar{b}) \cdot (-(2 \bar{a}-\bar{b})) = - |2 \bar{a}-\bar{b}|^2$.
$|2 \bar{a}-\bar{b}|^2 = (2 \bar{a}-\bar{b}) \cdot (2 \bar{a}-\bar{b}) = 4|\bar{a}|^2 + |\bar{b}|^2 - 4(\bar{a} \cdot \bar{b}) = 4(1) + 1 - 4(0) = 5$.
Therefore,$E = -5$.

(D) The total energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
For the hydrogen atom $(1H^1)$,the atomic number $Z_H = 1$. For the $2^{\text{nd}}$ orbit,$n = 2$. Thus,$E_{H} = -13.6 \times \frac{1^2}{2^2} = -13.6 \times \frac{1}{4} \text{ eV}$.
For the helium ion $(He^{+})$,the atomic number $Z_{He} = 2$. For the $2^{\text{nd}}$ orbit,$n = 2$. Thus,$E_{He} = -13.6 \times \frac{2^2}{2^2} = -13.6 \times 1 \text{ eV}$.
The ratio of the total energy of the hydrogen atom to that of the helium ion is: $\frac{E_H}{E_{He}} = \frac{-13.6 \times (1/4)}{-13.6 \times 1} = \frac{1}{4}$.

(A) The circuit consists of two batteries connected in opposition and a resistor in series.
The net electromotive force $(EMF)$ of the circuit is $E_{net} = 200 \, V - 10 \, V = 190 \, V$.
The total resistance of the circuit is $R = 38 \, \Omega$.
According to Ohm's law, the current $i$ flowing through the circuit is given by:
$i = \frac{E_{net}}{R} = \frac{190 \, V}{38 \, \Omega} = 5 \, A$.

(C) The circuit consists of two batteries connected in series with a resistor of $38 \ \Omega$.
Looking at the polarities,the $10 \ V$ battery and the $200 \ V$ battery are connected in opposition (positive terminals facing each other).
Therefore,the net electromotive force $(EMF)$ in the circuit is $E_{net} = 200 \ V - 10 \ V = 190 \ V$.
The total resistance in the circuit is $R = 38 \ \Omega$.
According to Ohm's law,the current $I$ flowing through the circuit is given by $I = \frac{E_{net}}{R}$.
Substituting the values,we get $I = \frac{190 \ V}{38 \ \Omega} = 5 \ A$.
Thus,the current flowing through the circuit is $5 \ A$.

(A) The total e.m.f of the series combination is $E_{eq} = E + E = 2E$.
The total resistance of the circuit is $R_{total} = R + r_1 + r_2$.
The current flowing through the circuit is $I = \frac{2E}{R + r_1 + r_2}$.
The potential difference across cell $E_1$ is given by $V_1 = E - I r_1$.
Given that $V_1 = 0$,we have $E = I r_1$.
Substituting the value of $I$,we get $E = \left( \frac{2E}{R + r_1 + r_2} \right) r_1$.
Dividing both sides by $E$,we get $1 = \frac{2r_1}{R + r_1 + r_2}$.
Thus,$R + r_1 + r_2 = 2r_1$.
Solving for $R$,we get $R = 2r_1 - r_1 - r_2 = r_1 - r_2$.

(B) The work function $\Phi$ is related to the threshold wavelength $\lambda_0$ by the equation: $\Phi = \frac{hc}{\lambda_0}$.
This implies that $\lambda_0 = \frac{hc}{\Phi}$.
Let $\Phi_{Na} = 2.3 \ eV$ and $\Phi_{Cu} = 4.5 \ eV$.
The ratio of the threshold wavelengths is given by: $\frac{\lambda_{Na}}{\lambda_{Cu}} = \frac{hc / \Phi_{Na}}{hc / \Phi_{Cu}} = \frac{\Phi_{Cu}}{\Phi_{Na}}$.
Substituting the given values: $\frac{\lambda_{Na}}{\lambda_{Cu}} = \frac{4.5}{2.3} \approx 1.956$.
Rounding this value to the nearest option,we get $2: 1$.

(D) The induced e.m.f. $(e)$ in an inductor of self-inductance $L$ is given by the formula: $e = -L \frac{dI}{dt}$.
Here,the negative sign indicates Lenz's Law,which states that the induced e.m.f. opposes the change in current.
If we consider the magnitude of the induced e.m.f. $|e| = L \frac{dI}{dt}$,we can see that $|e|$ is directly proportional to $\frac{dI}{dt}$.
Since $L$ is a constant,the graph of $|e|$ versus $\frac{dI}{dt}$ will be a straight line passing through the origin with a slope equal to $L$.
Looking at the given options,graph $(D)$ represents a linear relationship between $e$ and $dI/dt$ passing through the origin.
Therefore,the correct graph is $(D)$.

(D) The magnetic field $B$ at the center of a circular loop of radius $R_1$ carrying current $I_1$ is given by $B = \frac{\mu_0 I_1}{2 R_1}$.
Since $R_1$ > $R_2$, we assume the smaller loop of radius $R_2$ is placed within the magnetic field of the larger loop.
The magnetic flux $\phi_2$ linked with the smaller loop is $\phi_2 = B \cdot A_2$, where $A_2 = \pi R_2^2$ is the area of the smaller loop.
Substituting the values, we get $\phi_2 = \left( \frac{\mu_0 I_1}{2 R_1} \right) (\pi R_2^2) = \left( \frac{\mu_0 \pi R_2^2}{2 R_1} \right) I_1$.
By definition, the mutual inductance $M$ is given by $\phi_2 = M I_1$, so $M = \frac{\mu_0 \pi R_2^2}{2 R_1}$.
Thus, $M$ is directly proportional to $\frac{R_2^2}{R_1}$.

(B) The self-inductance $L$ of a solenoid or a coil is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A$ is the area,and $l$ is the length. For a coil of a fixed radius and length,$L \propto N^2$.
Given $L_1 = 108 \ mH$ for $N_1 = 600$ turns.
We need to find $L_2$ for $N_2 = 500$ turns.
Using the ratio: $\frac{L_2}{L_1} = \left( \frac{N_2}{N_1} \right)^2$.
$\frac{L_2}{108} = \left( \frac{500}{600} \right)^2 = \left( \frac{5}{6} \right)^2 = \frac{25}{36}$.
$L_2 = 108 \times \frac{25}{36} = 3 \times 25 = 75 \ mH$.

(B) Let the mass of the lighter stone be $m_1 = m$ and its radius be $r_1 = r$. Let its tangential speed be $v_1$.
Let the mass of the heavier stone be $m_2 = 3m$ and its radius be $r_2 = \frac{r}{3}$. Let its tangential speed be $v_2$.
Given that $v_1 = n v_2$.
The centripetal force is given by $F = \frac{mv^2}{r}$.
Equating the centripetal forces for both stones: $F_1 = F_2$.
$\frac{m_1 v_1^2}{r_1} = \frac{m_2 v_2^2}{r_2}$.
Substituting the given values: $\frac{m (n v_2)^2}{r} = \frac{3m v_2^2}{r/3}$.
$\frac{m n^2 v_2^2}{r} = \frac{9m v_2^2}{r}$.
Canceling common terms $m, v_2^2,$ and $r$ from both sides,we get $n^2 = 9$.
Therefore,$n = 3$.

(C) At the mean position,the velocity of the mass $M$ is maximum,given by $v_{max} = \omega_1 A_1 = \sqrt{\frac{k}{M}} A_1$.
When mass $m$ is placed on $M$,the total mass becomes $(M+m)$.
Since the collision is perfectly inelastic and occurs at the mean position,the momentum is conserved.
$M v_{max} = (M+m) v'_{max}$.
So,$v'_{max} = \frac{M}{M+m} v_{max}$.
The new angular frequency is $\omega_2 = \sqrt{\frac{k}{M+m}}$.
Since $v'_{max} = \omega_2 A_2$,we have $\frac{M}{M+m} (\omega_1 A_1) = \omega_2 A_2$.
Substituting the values: $\frac{M}{M+m} \sqrt{\frac{k}{M}} A_1 = \sqrt{\frac{k}{M+m}} A_2$.
$\frac{A_1}{A_2} = \frac{M+m}{M} \sqrt{\frac{k}{M+m}} \sqrt{\frac{M}{k}} = \frac{M+m}{M} \sqrt{\frac{M}{M+m}} = \sqrt{\frac{M+m}{M}} = \left(\frac{M+m}{M}\right)^{\frac{1}{2}}$.

(D) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the axis.
For a square of side $L$,the distance of each corner from the centre is $r = \frac{\text{diagonal}}{2} = \frac{\sqrt{2}L}{2} = \frac{L}{\sqrt{2}}$.
Since there are four particles each of mass $M$,the total moment of inertia $I$ about the axis passing through the centre and perpendicular to the plane is:
$I = 4 \times M \times r^2 = 4 \times M \times (\frac{L}{\sqrt{2}})^2 = 4 \times M \times \frac{L^2}{2} = 2ML^2$.
The radius of gyration $k$ is defined by the relation $I = Mk_{total} k^2$,where $M_{total} = 4M$.
So,$2ML^2 = (4M)k^2$.
$k^2 = \frac{2ML^2}{4M} = \frac{L^2}{2}$.
$k = \frac{L}{\sqrt{2}}$.

(C) For cylinder $A$ (isobaric process,piston is free to move): The heat supplied is $dQ = n C_{p} dT_{A}$.
For cylinder $B$ (isochoric process,piston is fixed): The heat supplied is $dQ = n C_{v} dT_{B}$.
Since the heat supplied is the same in both cases,we have $n C_{p} dT_{A} = n C_{v} dT_{B}$.
Dividing both sides by $n C_{v}$,we get $\frac{C_{p}}{C_{v}} dT_{A} = dT_{B}$.
Given $\gamma = \frac{C_{p}}{C_{v}}$,we substitute this into the equation to get $dT_{B} = \gamma dT_{A}$.

(B) The structure of $Benzene-1,3-diol$ consists of a benzene ring with two hydroxyl $(-OH)$ groups attached at the $1$ and $3$ positions.
This compound is commonly known as $Resorcinol$.
$Catechol$ is $Benzene-1,2-diol$.
$Quinol$ (or $Hydroquinone$) is $Benzene-1,4-diol$.
$Pyrogallol$ is $Benzene-1,2,3-triol$.

(A) The structure of $Benzene-1,2-diol$ consists of a benzene ring with two hydroxyl $(-OH)$ groups attached at adjacent positions ($1$ and $2$).
This compound is commonly known as $Catechol$ (or $Pyrocatechol$).

(B) trihydric phenol contains three hydroxyl $(-OH)$ groups attached to the benzene ring.
$1$. Catechol is $1,2$-dihydroxybenzene (dihydric).
$2$. Pyrogallol is $1,2,3$-trihydroxybenzene (trihydric).
$3$. Resorcinol is $1,3$-dihydroxybenzene (dihydric).
$4$. Quinol (Hydroquinone) is $1,4$-dihydroxybenzene (dihydric).
Therefore,Pyrogallol is the correct answer.

(B) Mesityl oxide is an $\alpha,\beta$-unsaturated ketone with the chemical formula $(CH_3)_2C=CHCOCH_3$.
It is formed by the aldol condensation of two molecules of acetone followed by dehydration.
The structure consists of an isopropylidene group attached to an acetyl group.

(A) compound is optically inactive if it does not contain a chiral center (an asymmetric carbon atom bonded to four different groups).
$1.$ $2-$Chloro-$2-$methylbutane: The $C-2$ carbon is bonded to two identical methyl groups,so it is not chiral. Thus,it is optically inactive.
$2.$ $3-$Chlorohexane: The $C-3$ carbon is bonded to $-H$,$-Cl$,$-CH_2CH_3$,and $-CH_2CH_2CH_3$. Since all four groups are different,it is chiral and optically active.
$3.$ $2-$Chloro-$3-$methylbutane: The $C-2$ carbon is bonded to $-H$,$-Cl$,$-CH_3$,and $-CH(CH_3)_2$. Since all four groups are different,it is chiral and optically active.
$4.$ $2-$Chloropentane: The $C-2$ carbon is bonded to $-H$,$-Cl$,$-CH_3$,and $-CH_2CH_2CH_3$. Since all four groups are different,it is chiral and optically active.
Therefore,$2-$Chloro-$2-$methylbutane is the optically inactive compound.

(B) The reaction of $tert$-butyl bromide $((CH_3)_3CBr)$ with silver fluoride $(AgF)$ is a Swarts reaction,which is used for the synthesis of alkyl fluorides.
However,$tert$-butyl bromide is a tertiary alkyl halide. When heated with $AgF$,it undergoes an $E1$ elimination reaction rather than a nucleophilic substitution ($S_N1$ or $S_N2$) because the tertiary carbocation is highly stable and the fluoride ion acts as a base.
The major product formed is $2$-methylpropene (isobutylene) due to the elimination of $HBr$.

(C) An allylic halide is a compound in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom next to a carbon-carbon double bond $(C=C)$.
In $3-$bromopropene $(CH_2=CH-CH_2Br)$,the bromine atom is attached to a carbon atom that is adjacent to the double bond,which satisfies the definition of an allylic halide.
$1-$Bromopropene and $2-$bromopropene are vinylic halides because the halogen is attached directly to the double-bonded carbon atom.

(D) benzylic halide is a compound in which the halogen atom is bonded to a carbon atom that is directly attached to an aromatic ring (specifically,a $sp^3$ hybridized carbon atom adjacent to the benzene ring).
$1.$ Bromophenylmethane $(C_6H_5CH_2Br)$ is a benzylic halide.
$2.$ $1-$Bromo$-1-$phenylpropane $(C_6H_5CH(Br)CH_2CH_3)$ is a benzylic halide.
$3.$ $2-$Bromo$-2-$phenylpropane $(C_6H_5C(Br)(CH_3)_2)$ is a benzylic halide.
$4.$ $1-$Bromo$-2-$phenylbutane $(C_6H_5CH_2CH(Br)CH_2CH_3)$ has the bromine atom attached to a carbon that is not directly bonded to the benzene ring. The carbon attached to the benzene ring is a $CH_2$ group,and the bromine is on the second carbon of the alkyl chain. Therefore,it is not a benzylic halide.

(C) The reaction between sodium ethoxide $(CH_3CH_2ONa)$ and isopropyl chloride $((CH_3)_2CHCl)$ is a base-promoted elimination reaction.
Since sodium ethoxide is a strong base and isopropyl chloride is a secondary alkyl halide,the $E2$ elimination mechanism predominates over the $S_N2$ substitution mechanism.
The base abstracts a $\beta$-hydrogen from the isopropyl chloride,leading to the formation of propene $(CH_3-CH=CH_2)$,ethanol $(CH_3CH_2OH)$,and sodium chloride $(NaCl)$.
Therefore,the product '$X$' is propene.

(D) The reaction of an alkyl halide with alcoholic $KCN$ is a nucleophilic substitution reaction ($S_N2$ mechanism) that produces an alkyl cyanide (nitrile).
To obtain butanenitrile $(CH_3CH_2CH_2CH_2CN)$,which contains $4$ carbon atoms in the chain plus the nitrile carbon,we need a starting alkyl halide with $4$ carbon atoms.
$n-$Propylchloride $(CH_3CH_2CH_2Cl)$ has $3$ carbons,so it would yield butanenitrile $(CH_3CH_2CH_2CN)$ which is $3$ carbons in the chain plus the nitrile carbon,totaling $4$ carbons.
Wait,butanenitrile is $CH_3CH_2CH_2CN$. This contains $4$ carbon atoms in total.
Therefore,$n-$propylchloride $(CH_3CH_2CH_2Cl)$ reacts with $KCN$ to form $CH_3CH_2CH_2CN$ (butanenitrile).
Reaction: $CH_3CH_2CH_2Cl + KCN \text{ (alc.)} ightarrow CH_3CH_2CH_2CN + KCl$.

(A) The reaction of a mixture of alkyl halides with sodium metal in dry ether is known as the Wurtz reaction. When a mixture of chloroethane $(CH_3CH_2Cl)$ and $1-$chloropropane $(CH_3CH_2CH_2Cl)$ is treated with sodium in dry ether,the following products are formed due to the coupling of alkyl radicals:
$1$. Coupling of two chloroethane molecules: $CH_3CH_2-CH_2CH_3$ ($n-$butane).
$2$. Coupling of two $1-$chloropropane molecules: $CH_3CH_2CH_2-CH_2CH_2CH_3$ ($n-$hexane).
$3$. Cross-coupling of chloroethane and $1-$chloropropane: $CH_3CH_2-CH_2CH_2CH_3$ ($n-$pentane).
Propane is not formed in this reaction because it would require the loss of a carbon atom or a different mechanism not associated with the Wurtz coupling of these specific reactants.

(A) The reactivity of alkyl halides towards $S_N2$ reactions follows the order: $Primary > Secondary > Tertiary$.
This is due to the steric hindrance,which increases from primary to tertiary alkyl halides,making it difficult for the nucleophile to attack the electrophilic carbon.
Among the given options:
$(A)$ $n$-Butyl iodide is a primary $(1^{\circ})$ alkyl halide.
$(B)$ $sec$-Butyl iodide is a secondary $(2^{\circ})$ alkyl halide.
$(C)$ Isobutyl iodide is a primary $(1^{\circ})$ alkyl halide,but it has steric hindrance at the $\beta$-carbon.
$(D)$ $tert$-Butyl iodide is a tertiary $(3^{\circ})$ alkyl halide.
Since $n$-Butyl iodide is a primary alkyl halide with the least steric hindrance,it shows the highest reactivity for $S_N2$ reactions.

(D) An allylic halide is a compound in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom that is adjacent to a carbon-carbon double bond $(C=C-C-X)$.
In option $A$,$CH_2=CH-CH_2-X$,the halogen is attached to an allylic carbon.
In option $B$,$CH_3-CH=CH-CH_2-X$,the halogen is attached to an allylic carbon.
In option $C$,$CH_3-CH_2-CH=CH-CH_2-X$,the halogen is attached to an allylic carbon.
In option $D$,$CH_3-CH=CH-CH_2-CH_2-X$,the halogen is attached to a carbon atom that is two positions away from the double bond,making it a homoallylic halide,not an allylic halide.

(A) haloalkyne is a compound in which a halogen atom is directly attached to a carbon atom that is part of a carbon-carbon triple bond $(C \equiv C)$.
In a carbon-carbon triple bond,the carbon atom is $sp$ hybridized.
Therefore,the halogen atom is bonded to an $sp$ hybridized carbon atom.

(B) Racemization occurs in $SN^1$ reactions when the substrate forms a stable,planar carbocation intermediate that is chiral.
Among the given options,$CH_3-CH_2-CH(Cl)-CH_3$ ($2$-chlorobutane) is a secondary alkyl halide that can form a planar carbocation $(CH_3-CH_2-CH^+-CH_3)$ upon the loss of the chloride ion.
This carbocation is prochiral and can be attacked by the nucleophile from either side,leading to a racemic mixture.
$CH_3-CH(Cl)-CH_3$ (isopropyl chloride) forms a less stable secondary carbocation and is less prone to $SN^1$ compared to $2-$chlorobutane.
$CH_3-CH_2-CH(Cl)-CH_2-CH_3$ ($3$-chloropentane) forms a carbocation that is achiral due to symmetry,so it does not result in a racemic mixture.
$(CH_3)_3C-CH_2-Cl$ is a primary alkyl halide and typically undergoes $SN^2$ reactions.

(D) When $2-$Bromobutane $(CH_3CH(Br)CH_2CH_3)$ is treated with an aqueous solution of sodium hydroxide $(NaOH)$,it undergoes a nucleophilic substitution reaction ($S_N1$ or $S_N2$ mechanism depending on conditions,but typically $S_N2$ for secondary halides in aqueous media).
The hydroxide ion $(OH^-)$ acts as a nucleophile and replaces the bromide ion $(Br^-)$.
The reaction is: $CH_3CH(Br)CH_2CH_3 + NaOH_{(aq)} \rightarrow CH_3CH(OH)CH_2CH_3 + NaBr$.
The product formed is Butan$-2-$ol.

(D) The boiling point of haloalkanes increases with an increase in molecular mass and the number of halogen atoms present in the molecule.
This is because the magnitude of van der Waals forces increases as the size and number of halogen atoms increase.
Comparing the given compounds: $CHBr_3$ (tribromomethane) has the highest molecular mass,followed by $CH_2Br_2$ (dibromomethane),$CH_3Br$ (bromomethane),and $CH_3Cl$ (chloromethane) has the lowest molecular mass.
Therefore,the correct decreasing order is $CHBr_3 > CH_2Br_2 > CH_3Br > CH_3Cl$.

(C) The reaction of alkyl halides with ammonia is a nucleophilic substitution reaction $(S_N2)$.
In this reaction,the halide ion $(X^-)$ acts as a leaving group.
The reactivity of alkyl halides depends on the strength of the $C-X$ bond.
The bond dissociation energy decreases as the size of the halogen atom increases,making the $C-I$ bond the weakest and the $C-Cl$ bond the strongest.
Therefore,the leaving group ability follows the order: $I^- > Br^- > Cl^-$.
Thus,the order of reactivity of alkyl halides with ammonia is $R-I > R-Br > R-Cl$.

(D) The reaction of haloalkanes $(R-X)$ with silver nitrite $(AgNO_2)$ leads to the formation of nitroalkanes $(R-NO_2)$ as the major product.
This occurs because $AgNO_2$ is a covalent compound,and the nitrogen atom is more nucleophilic than the oxygen atom.
In contrast,ionic nitrites like $KNO_2$ or $NaNO_2$ primarily yield alkyl nitrites $(R-ONO)$ because the oxygen atom is more nucleophilic in these ionic species.

(A) vinylic halide is a compound in which the halogen atom is bonded to a carbon atom that is part of a carbon-carbon double bond $(C=C)$.
In the option $CH_2=CH-Cl$,the chlorine atom is directly attached to a carbon atom involved in a double bond,which defines it as a vinylic halide.
Therefore,the correct option is $A$.

(C) Chlorobenzene undergoes electrophilic aromatic substitution (nitration) when treated with a mixture of conc. $HNO_3$ and conc. $H_2SO_4$.
Because the chlorine atom is ortho/para-directing due to the resonance effect,the nitration occurs at both the $ortho$ and $para$ positions.
Therefore,the reaction yields a mixture of $1-$chloro$-2-$nitrobenzene ($ortho$-isomer) and $1-$chloro$-4-$nitrobenzene ($para$-isomer).

(A) The Sandmeyer reaction is a chemical reaction used to synthesize aryl halides from aryl diazonium salts using copper salts as reagents or catalysts.
Specifically,the reaction of benzene diazonium chloride with $CuCl/HCl$ or $CuBr/HBr$ or $CuCN/KCN$ leads to the formation of chlorobenzene,bromobenzene,or benzonitrile,respectively.
Among the given options,$CuCN / KCN$ is the reagent used to introduce a cyano group via the Sandmeyer reaction.

(D) The reaction shown is the nucleophilic aromatic substitution of a chlorobenzene derivative with water (hydrolysis) to form a phenol derivative.
The product formed is $2,4,6-$trinitrophenol,commonly known as picric acid.
This reaction proceeds easily under mild conditions (warm water) because the three electron-withdrawing $-NO_2$ groups at the ortho and para positions significantly activate the benzene ring towards nucleophilic attack by $H_2O$.
Therefore,the substrate '$S$' must be $2,4,6-$trinitrochlorobenzene.

(D) The reaction involving the cleavage of the $C-X$ bond in haloarenes is Nucleophilic Aromatic Substitution $(S_NAr)$.
This reaction is facilitated by the presence of electron-withdrawing groups (like $-NO_2$) at the ortho and para positions,which stabilize the carbanion intermediate formed during the reaction.
Compound $I$ has one $-NO_2$ group at the para position.
Compound $II$ has two $-NO_2$ groups (one at ortho and one at para position).
Compound $III$ has three $-NO_2$ groups (two at ortho and one at para position).
As the number of electron-withdrawing groups increases,the reactivity towards nucleophilic substitution increases.
Therefore,the correct order of reactivity is $III > II > I$.

(B) The reaction between an aryl halide $(Ar-X)$ and an alkyl halide $(R-X)$ in the presence of sodium metal $(Na)$ and dry ether is known as the $Wurtz-Fittig$ reaction.
The general equation is: $Ar-X + 2Na + R-X \xrightarrow{\text{dry ether}} Ar-R + 2NaX$.
This reaction is used to prepare alkyl-substituted aromatic compounds.

(B) The Fittig reaction is a coupling reaction where two aryl halides react with sodium metal in the presence of dry ether to form a diaryl compound.
Option $B$ represents the reaction of two molecules of chlorobenzene $(C_6H_5Cl)$ with sodium to form biphenyl $(C_6H_5-C_6H_5)$,which is the definition of the Fittig reaction.
Option $A$ is the Wurtz reaction.
Option $C$ is the Wurtz-Fittig reaction.
Option $D$ is the Wurtz reaction.

(B) When arene diazonium chloride $(Ar-N_2^+Cl^-)$ is treated with fluoroboric acid $(HBF_4)$,it forms arene diazonium fluoroborate $(Ar-N_2^+BF_4^-)$ as a precipitate.
Upon heating,this salt decomposes to yield fluoroarene $(Ar-F)$,boron trifluoride $(BF_3)$,and nitrogen gas $(N_2)$.
The reaction is: $Ar-N_2^+Cl^- + HBF_4 \rightarrow Ar-N_2^+BF_4^- + HCl$.
Then,$Ar-N_2^+BF_4^- \xrightarrow{\Delta} Ar-F + BF_3 + N_2$.

(C) The reaction of an aryl halide with an alkyl halide in the presence of sodium metal and dry ether is known as the $Wurtz-Fittig$ reaction.
In this reaction,the alkyl group from the alkyl halide replaces the halogen atom on the aromatic ring.
The general equation is: $Ar-X + R-X + 2Na \xrightarrow{\text{dry ether}} Ar-R + 2NaX$.

(D) The reaction involves an aryl halide (bromobenzene) and an alkyl halide (bromomethane) reacting with sodium metal in the presence of dry ether to form an alkylbenzene (toluene). This specific reaction is known as the $Wurtz-Fittig$ reaction. The general equation is: $C_6H_5Br + CH_3Br + 2Na \xrightarrow{\text{dry ether}} C_6H_5CH_3 + 2NaBr$.

(D) Nucleophilic aromatic substitution in aryl halides is facilitated by the presence of electron-withdrawing groups $(EWG)$ at the ortho and para positions relative to the halogen atom.
These groups stabilize the carbanion intermediate (Meisenheimer complex) formed during the reaction through resonance and inductive effects.
As the number of electron-withdrawing nitro groups $(-NO_2)$ increases,the electron density on the benzene ring decreases,making the carbon atom attached to the chlorine more electrophilic and thus more susceptible to nucleophilic attack.
Therefore,$2,4,6-$trinitrochlorobenzene has the highest number of electron-withdrawing groups,making it the most reactive towards nucleophilic substitution.

(C) The reaction of two molecules of an aryl halide with sodium metal in the presence of dry ether to form a diaryl (biphenyl) is known as the $Fittig$ reaction.
$2Ar-X + 2Na \xrightarrow{\text{dry ether}} Ar-Ar + 2NaX$
Here,$Ar$ represents an aryl group and $X$ represents a halogen atom.

(B) The ease of breaking the $C-X$ bond in haloarenes depends on the electron density at the carbon atom attached to the halogen.
Electron-withdrawing groups (like $-NO_2$) at the ortho and para positions increase the reactivity of the $C-X$ bond towards nucleophilic substitution by stabilizing the carbanion intermediate.
In $m-$Nitrochlorobenzene,the $-NO_2$ group is at the meta position,which does not exert a strong electron-withdrawing effect on the carbon attached to the chlorine through resonance.
Therefore,$m-$Nitrochlorobenzene has the least activation for nucleophilic substitution,making the $C-X$ bond the most difficult to break among the given options.

(B) The reaction is a Friedel-Crafts acylation of $A$ with acetyl chloride in the presence of anhydrous $AlCl_3$.
The products formed are $1-\text{chloroacetophenone}$ (ortho-substituted) and $4-\text{chloroacetophenone}$ (para-substituted).
Since the products contain a chlorine atom attached to the benzene ring,the starting material $A$ must be $Chlorobenzene$.
Chlorobenzene is an ortho/para directing group due to the $+M$ effect of the chlorine atom,which explains the formation of the observed products.

(C) Chlorobenzene undergoes electrophilic aromatic substitution when treated with a nitrating mixture (concentrated $HNO_3$ and concentrated $H_2SO_4$).
Since the chlorine atom is ortho/para-directing due to the resonance effect,the nitration occurs at both the ortho and para positions.
Therefore,the reaction yields a mixture of $1-$chloro$-2-$nitrobenzene (ortho-isomer) and $1-$chloro$-4-$nitrobenzene (para-isomer).

(D) The Fittig reaction involves the coupling of two aryl halide molecules in the presence of sodium metal and dry ether to form a diaryl compound.
Option $A$ represents the Wurtz-Fittig reaction.
Options $B$ and $C$ represent the Wurtz reaction.
Option $D$ represents the Fittig reaction: $2C_6H_5Br + 2Na \xrightarrow[\text{dry ether}]{} C_6H_5-C_6H_5 + 2NaBr$.

(A) benzylic halide is a compound in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom,which is directly attached to an aromatic ring.
In the structure $C_6H_5-CH_2-X$,the halogen atom $X$ is attached to a carbon atom that is bonded to a benzene ring $(C_6H_5)$.
This carbon atom is $sp^3$ hybridized,making it a benzylic halide.
Therefore,the correct option is $A$.

(D) haloarene is a compound in which a halogen atom is directly attached to an aromatic ring (benzene ring).
In option $D$,the halogen atom $X$ is directly bonded to the carbon atom of the benzene ring,which makes it an aryl halide or haloarene.
In option $A$,the halogen is attached to a side chain carbon,making it a benzyl halide.
In options $B$ and $C$,the compounds are aliphatic cyclic halides,not aromatic.

(B) The reaction is a Friedel-Crafts acylation of a substituted benzene ring.
Since the products are $1-\text{chloroacetophenone}$ (ortho-isomer) and $4-\text{chloroacetophenone}$ (para-isomer),the starting material '$A$' must be chlorobenzene.
Chlorobenzene undergoes electrophilic aromatic substitution where the chlorine atom is ortho/para-directing,leading to the formation of these specific products.

(B) benzylic halide is a compound in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom,which is further attached to an aromatic ring.
In $Bromophenylmethane$ (also known as benzyl bromide),the bromine atom is attached to a $-CH_2-$ group,which is directly bonded to a benzene ring.
This structure fits the definition of a benzylic halide.
Therefore,the correct option is $B$.

(C) The reaction described is the $Etard$ reaction.
In this reaction,toluene reacts with chromyl chloride $(CrO_2Cl_2)$ in the presence of a non-polar solvent like $CS_2$ or $CCl_4$ to form a brown chromium complex.
This complex,upon subsequent hydrolysis,yields benzaldehyde $(C_6H_5CHO)$.

(D) The Wolff-Kishner reduction is a reaction used to convert carbonyl groups (aldehydes or ketones) into alkanes.
It involves the formation of a hydrazone intermediate by reacting the carbonyl compound with hydrazine $(H_2N-NH_2)$,followed by base-catalyzed decomposition (using $KOH$ in ethylene glycol) at high temperatures.
Option $D$ correctly represents this process: $RCOR' \xrightarrow[\text{ii) } KOH, HO(CH_2)_2OH]{\text{i) } H_2N-NH_2} RCH_2R'$.
Option $A$ is Rosenmund reduction,option $B$ is Stephen reduction,and option $C$ is Clemmensen reduction.

(B) The Fischer-Tropsch process is a collection of chemical reactions that converts a mixture of carbon monoxide and hydrogen into liquid hydrocarbons.
In this process,catalysts such as $Fe$ (iron) or $Co$ (cobalt) are typically used.
Specifically,for the synthesis of gasoline-range hydrocarbons,a cobalt-thorium $(Co-Th)$ catalyst is commonly employed to facilitate the hydrogenation of carbon monoxide.

(B) Ethylbenzene reacts with dilute nitric acid under specific conditions,but typically,dilute nitric acid is not a strong enough nitrating agent to perform electrophilic aromatic substitution on the benzene ring of ethylbenzene effectively. However,if the reaction is considered in the context of standard organic chemistry problems,the alkyl group $(-CH_2CH_3)$ is an ortho/para directing group. Therefore,the expected products of nitration would be a mixture of $o$-nitroethylbenzene and $p$-nitroethylbenzene. Given the options,both $A$ and $B$ are valid isomers,but usually,$p$-nitroethylbenzene is the major product due to steric hindrance at the ortho position. If the question implies a specific reaction condition leading to a single major product,$p$-nitroethylbenzene is the standard answer.

(D) The reaction of an alkylbenzene with an alkyl group having at least one benzylic hydrogen atom with alkaline $KMnO_4$ followed by acidic hydrolysis $(H_3O^+)$ results in the oxidation of the alkyl side chain to a carboxylic acid group.
In the given reactant,$C_6H_5-CH_2-CH_3$ (ethylbenzene),the benzylic carbon is the $CH_2$ group.
Upon oxidation,the entire alkyl side chain is converted into a carboxyl group $(-COOH)$ attached to the benzene ring.
Therefore,the product formed is benzoic acid,$C_6H_5-COOH$.

(C) Azo coupling reactions involve the reaction of a diazonium salt with an electron-rich aromatic compound (like a phenol or an aromatic amine) to form an azo compound $(Ar-N=N-Ar')$.
In this reaction,benzenediazonium chloride reacts with phenol in a basic medium to produce $p$-hydroxyazobenzene.
Therefore,$p$-hydroxyazobenzene is the product obtained via an azo coupling reaction.

(D) The reaction of $Chlorobenzene$ with $Cl_2$ in the presence of anhydrous $FeCl_3$ is an electrophilic aromatic substitution reaction (chlorination).
$Chlorine$ atom on the benzene ring is ortho/para-directing due to the resonance effect.
However,the para-isomer is the major product due to less steric hindrance compared to the ortho-isomer.
Therefore,$1,4-dichlorobenzene$ is the major product.

(B) The reaction is an electrophilic aromatic substitution,specifically a Friedel-Crafts alkylation.
Chlorobenzene reacts with chloromethane in the presence of anhydrous $AlCl_3$ to form $1-\text{chloro}-2-\text{methylbenzene}$ $(o-\text{chlorotoluene})$ and $1-\text{chloro}-4-\text{methylbenzene}$ $(p-\text{chlorotoluene})$.
Since the product is a mixture of $1-\text{chlorotoluene}$ (ortho-isomer) and $4-\text{chlorotoluene}$ (para-isomer),the reactant $A$ must be $Chlorobenzene$ because the chlorine atom on the benzene ring is ortho/para-directing.

(B) The reaction is a Friedel-Crafts acylation of an aromatic compound '$A$' with acetyl chloride in the presence of anhydrous $AlCl_3$.
Since the products formed are $1-$chloroacetophenone (ortho-substituted) and $4-$chloroacetophenone (para-substituted),the starting material '$A$' must be chlorobenzene.
The chlorine atom on the benzene ring is ortho/para-directing,which explains the formation of these specific isomers.

(C) The acidity of hydrides of group $16$ elements increases down the group.
As we move down the group from $O$ to $Te$,the atomic size of the central atom increases.
This leads to a decrease in the bond dissociation enthalpy of the $E-H$ bond (where $E = O, S, Se, Te$).
Consequently,the ease of releasing $H^+$ ions increases,making the hydrides more acidic.
Therefore,the correct order of acidity is $H_2O < H_2S < H_2Se < H_2Te$ or $H_2Te > H_2Se > H_2S > H_2O$.

(B) The decomposition of hydrogen peroxide $(H_2O_2)$ is a spontaneous reaction. Glycerol acts as a negative catalyst or an inhibitor in this reaction. It slows down the rate of decomposition of $H_2O_2$ by forming a complex or by stabilizing the reactant,thereby preventing its rapid breakdown into water and oxygen.