MHT CET 2022 Chemistry Question Paper with Answer and Solution

(D) For a conical pendulum,the forces acting on the mass $M$ are the tension $T$ in the string and its weight $Mg$.
Resolving the tension $T$ into horizontal and vertical components:
$T \sin \theta = M \omega^2 R$ (Centripetal force equation) ... $(i)$
$T \cos \theta = Mg$ (Vertical equilibrium) ... (ii)
From the geometry,the radius of the circular path is $R = L \sin \theta$.
Substituting $R$ into equation $(i)$:
$T \sin \theta = M \omega^2 (L \sin \theta)$
$T = M \omega^2 L$
Given the frequency $n = 2/\pi$ revolutions per second,the angular velocity is $\omega = 2 \pi n = 2 \pi (2/\pi) = 4 \text{ rad/s}$.
Substituting $\omega = 4$ into the expression for tension:
$T = M (4)^2 L = 16 ML$.

(C) The relationship between linear momentum $P$,mass $m$,and kinetic energy $E$ is given by the formula $P = \sqrt{2mE}$.
Since the kinetic energies $E$ are equal for both masses,we can write the ratio of their momenta as:
$\frac{P_1}{P_2} = \frac{\sqrt{2m_1E}}{\sqrt{2m_2E}} = \sqrt{\frac{m_1}{m_2}}$.
Given $m_1 = 4 \, kg$ and $m_2 = 1 \, kg$,we substitute these values into the equation:
$\frac{P_1}{P_2} = \sqrt{\frac{4}{1}} = \sqrt{4} = 2$.
Therefore,the ratio of the magnitudes of their linear momenta is $2:1$.

(D) soap bubble has two surfaces (inner and outer),so the change in surface area is $2 \times 4\pi (R_2^2 - R_1^2)$.
Given the initial diameter is $d$,the initial radius is $r_1 = d/2$. The final diameter is $D$,so the final radius is $r_2 = D/2$.
The change in surface area $\Delta A = 2 \times 4\pi (r_2^2 - r_1^2) = 8\pi \left( \frac{D^2}{4} - \frac{d^2}{4} \right)$.
Simplifying this,we get $\Delta A = 2\pi (D^2 - d^2)$.
The work done $W$ is given by $W = T \times \Delta A$.
Therefore,$W = T \times 2\pi (D^2 - d^2) = 2\pi (D^2 - d^2)T$.

(A) Beats are a phenomenon that occurs due to the superposition of two sound waves of slightly different frequencies traveling in the same direction.
When these waves overlap,they undergo the principle of superposition,which is a specific case of interference.
As the waves move in and out of phase,the resultant amplitude varies periodically,leading to the phenomenon known as beats.

(A) The total electric field between the plates of a parallel plate capacitor is given by $E = \frac{V}{d}$.
The electric field produced by one plate alone is half of the total electric field,which is $E_1 = \frac{E}{2} = \frac{V}{2d}$.
The charge on the plate is $Q = CV$.
The force $F$ exerted by one plate on the other is given by the product of the charge on one plate and the electric field due to the other plate:
$F = Q \times E_1 = (CV) \times \left( \frac{V}{2d} \right) = \frac{CV^2}{2d}$.

(C) Since the capacitors $C_1$ and $C_2$ are connected in series,the charge $Q$ on both capacitors must be the same.
Let $V_D$ be the potential at point $D$.
The charge on capacitor $C_1$ is $Q = C_1(V_A - V_D) = C_1(V_1 - V_D)$.
The charge on capacitor $C_2$ is $Q = C_2(V_D - V_B) = C_2(V_D - V_2)$.
Equating the charges: $C_1(V_1 - V_D) = C_2(V_D - V_2)$.
Expanding the terms: $C_1V_1 - C_1V_D = C_2V_D - C_2V_2$.
Rearranging to solve for $V_D$: $C_1V_1 + C_2V_2 = V_D(C_1 + C_2)$.
Therefore,$V_D = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}$.

(A) The capacitive reactance $X_C$ is given by the formula $X_C = \frac{1}{2\pi \nu C}$.
Given $X_C = \frac{1}{1000}\,\Omega$ and $C = 5\,\mu F = 5 \times 10^{-6}\,F$.
Substituting the values into the formula:
$\frac{1}{1000} = \frac{1}{2\pi \times \nu \times 5 \times 10^{-6}}$
$2\pi \times \nu \times 5 \times 10^{-6} = 1000$
$10\pi \times \nu \times 10^{-6} = 10^3$
$\nu = \frac{10^3}{10\pi \times 10^{-6}} = \frac{10^2}{\pi \times 10^{-6}} = \frac{10^8}{\pi}\,Hz$.
Converting to $MHz$ $(1\,MHz = 10^6\,Hz)$:
$\nu = \frac{10^8}{\pi} \times 10^{-6}\,MHz = \frac{100}{\pi}\,MHz$.

(A) In a $P-$ type semiconductor,the material is doped with trivalent impurity atoms.
This creates a large number of holes,which act as the majority charge carriers.
Thermal excitation at room temperature generates a small number of free electrons,which act as the minority charge carriers.
Therefore,a $P-$ type semiconductor has a large number of holes and few electrons.

(D) Let the distance of the object from the convex lens be $u$ and the distance of the real image be $v$. Both $u$ and $v$ are positive in magnitude. The total distance $D$ between the object and the image is $D = u + v$.
From the lens formula,$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. Using sign convention,$u$ is negative,so $\frac{1}{f} = \frac{1}{v} - \frac{1}{-u} = \frac{1}{v} + \frac{1}{u}$.
Thus,$v = \frac{uf}{u-f}$.
The total distance is $D = u + \frac{uf}{u-f} = \frac{u^2 - uf + uf}{u-f} = \frac{u^2}{u-f}$.
To find the minimum distance,we differentiate $D$ with respect to $u$ and set it to zero: $\frac{dD}{du} = \frac{(u-f)(2u) - u^2(1)}{(u-f)^2} = 0$.
$2u^2 - 2uf - u^2 = 0 \implies u^2 - 2uf = 0 \implies u = 2f$.
Substituting $u = 2f$ into the expression for $v$,we get $v = \frac{2f \cdot f}{2f - f} = 2f$.
Therefore,the minimum distance $D = u + v = 2f + 2f = 4f$.

(B) Given expression: $\frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} - 1$
Factor out $i^{584}$ from the numerator and $i^{574}$ from the denominator:
$= \frac{i^{584}(i^8 + i^6 + i^4 + i^2 + 1)}{i^{574}(i^8 + i^6 + i^4 + i^2 + 1)} - 1$
$= \frac{i^{584}}{i^{574}} - 1$
$= i^{584-574} - 1$
$= i^{10} - 1$
Since $i^2 = -1$,then $i^{10} = (i^2)^5 = (-1)^5 = -1$.
$= -1 - 1 = -2$.

(C) Since $\alpha$ and $\beta$ are the imaginary cube roots of unity,we can write $\alpha = \omega$ and $\beta = \omega^2$ (or vice versa).
Given expression: $\alpha^4 + \beta^{28} + \frac{1}{\alpha\beta}$.
Substituting the values: $\omega^4 + (\omega^2)^{28} + \frac{1}{\omega \cdot \omega^2}$.
Using the properties $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$:
$\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$.
$\beta^{28} = (\omega^2)^{28} = \omega^{56} = \omega^{54} \cdot \omega^2 = (\omega^3)^{18} \cdot \omega^2 = 1^{18} \cdot \omega^2 = \omega^2$.
$\frac{1}{\alpha\beta} = \frac{1}{\omega \cdot \omega^2} = \frac{1}{\omega^3} = \frac{1}{1} = 1$.
Therefore,$\alpha^4 + \beta^{28} + \frac{1}{\alpha\beta} = \omega + \omega^2 + 1 = 0$.

(B) Given $y = \tan^{-1}(\sec x - \tan x)$.
First,simplify the expression inside the inverse tangent function:
$\sec x - \tan x = \frac{1}{\cos x} - \frac{\sin x}{\cos x} = \frac{1 - \sin x}{\cos x}$.
Using the half-angle identities $1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$ and $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = (\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})$:
$\frac{1 - \sin x}{\cos x} = \frac{(\cos \frac{x}{2} - \sin \frac{x}{2})^2}{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})} = \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}}$.
Dividing numerator and denominator by $\cos \frac{x}{2}$,we get $\frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} = \tan(\frac{\pi}{4} - \frac{x}{2})$.
Thus,$y = \tan^{-1}(\tan(\frac{\pi}{4} - \frac{x}{2})) = \frac{\pi}{4} - \frac{x}{2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{4} - \frac{x}{2}) = 0 - \frac{1}{2} = -\frac{1}{2} = -0.5$.

(B) We evaluate the integral $I = \int_{0}^{2\pi} (\sin x + |\sin x|) \, dx$.
Since $|\sin x| = \sin x$ for $x \in [0, \pi]$ and $|\sin x| = -\sin x$ for $x \in [\pi, 2\pi]$,we split the integral:
$I = \int_{0}^{\pi} (\sin x + \sin x) \, dx + \int_{\pi}^{2\pi} (\sin x - \sin x) \, dx$
$I = \int_{0}^{\pi} 2\sin x \, dx + \int_{\pi}^{2\pi} 0 \, dx$
$I = 2 [-\cos x]_{0}^{\pi} + 0$
$I = -2 (\cos \pi - \cos 0)$
$I = -2 (-1 - 1) = -2(-2) = 4$.

(C) The impulse applied is $P = F \times \Delta t$.
Angular impulse about the center of mass is given by $J_{\theta} = P \times \frac{l}{2}$.
Since angular impulse equals the change in angular momentum,we have $P \times \frac{l}{2} = I \omega$,where $I = \frac{ml^2}{12}$ is the moment of inertia about the center of mass.
Thus,$P \times \frac{l}{2} = \frac{ml^2}{12} \omega$,which simplifies to $\omega = \frac{6P}{ml}$.
The rod rotates with a constant angular velocity $\omega$ after the impulse. The time taken to turn through a right angle ($\pi/2$ radians) is $t = \frac{\theta}{\omega} = \frac{\pi/2}{\omega}$.
Substituting $\omega$,we get $t = \frac{\pi/2}{6P/ml} = \frac{\pi ml}{12P}$.

(D) We need to find the negation of the expression $\sim s \vee (\sim r \wedge s)$.
Using De Morgan's Law,$\sim (p \vee q) \equiv \sim p \wedge \sim q$:
$\sim (\sim s \vee (\sim r \wedge s)) \equiv \sim (\sim s) \wedge \sim (\sim r \wedge s)$
Using the Double Negation Law,$\sim (\sim s) \equiv s$:
$\equiv s \wedge \sim (\sim r \wedge s)$
Applying De Morgan's Law again,$\sim (p \wedge q) \equiv \sim p \vee \sim q$:
$\equiv s \wedge (\sim (\sim r) \vee \sim s)$
Using the Double Negation Law,$\sim (\sim r) \equiv r$:
$\equiv s \wedge (r \vee \sim s)$
Using the Distributive Law,$p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$:
$\equiv (s \wedge r) \vee (s \wedge \sim s)$
Since $s \wedge \sim s \equiv F$ (Contradiction Law):
$\equiv (s \wedge r) \vee F$
Using the Identity Law,$p \vee F \equiv p$:
$\equiv s \wedge r$

(C) The capacitors $C_1$ and $C_2$ are connected in series between points $A$ and $B$.
Let $V_D$ be the potential at point $D$.
The charge $q$ on both capacitors must be the same because they are in series.
The potential difference across $C_1$ is $V_1 - V_D = \frac{q}{C_1}$.
The potential difference across $C_2$ is $V_D - V_2 = \frac{q}{C_2}$.
From these,we have $q = C_1(V_1 - V_D) = C_2(V_D - V_2)$.
Expanding this,$C_1 V_1 - C_1 V_D = C_2 V_D - C_2 V_2$.
Rearranging to solve for $V_D$,we get $C_1 V_1 + C_2 V_2 = V_D(C_1 + C_2)$.
Therefore,$V_D = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$.

(C) When no work is done,the process is isochoric (constant volume).
For an ideal monatomic gas,the molar heat capacity at constant volume is $C_V = \frac{3}{2} R$.
The amount of heat required is given by the formula $\Delta Q = n C_V \Delta T$.
Given: $n = 2 \, mol$,$\Delta T = 373 \, K - 273 \, K = 100 \, K$.
Substituting the values:
$\Delta Q = 2 \times \left( \frac{3}{2} R \right) \times 100$
$\Delta Q = 3 \times 100 \times R = 300 \, R$.

(B) The impedance $Z$ of the $RC$ series circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
As the angular frequency $\omega$ increases,the capacitive reactance $X_C = \frac{1}{\omega C}$ decreases.
Since $Z = \sqrt{R^2 + X_C^2}$,a decrease in $X_C$ leads to a decrease in the total impedance $Z$ of the circuit.
The rms current in the circuit is given by $I_{rms} = \frac{V_{rms}}{Z}$. As $Z$ decreases,$I_{rms}$ increases.
The power dissipated by the bulb is $P = I_{rms}^2 R$. Since $I_{rms}$ increases,the power $P$ increases,which means the bulb glows brighter.

(C) Since $C_1$ and $C_2$ are connected in series,the charge on $C_1$ must be equal to the charge on $C_2$ because the junction $D$ is isolated.
Let $V_D$ be the potential at point $D$.
The charge on capacitor $C_1$ is $Q_1 = C_1(V_1 - V_D)$.
The charge on capacitor $C_2$ is $Q_2 = C_2(V_D - V_2)$.
Since $Q_1 = Q_2$,we have:
$C_1(V_1 - V_D) = C_2(V_D - V_2)$
$C_1V_1 - C_1V_D = C_2V_D - C_2V_2$
$C_1V_1 + C_2V_2 = V_D(C_1 + C_2)$
$V_D = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}$

(C) The process is isochoric because no work is done $(W = 0)$.
For an isochoric process,the heat supplied is equal to the change in internal energy: $\Delta Q = \Delta U = n C_v \Delta T$.
For a monoatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$.
Given: $n = 2\, mol$,$\Delta T = 373\, K - 273\, K = 100\, K$.
Substituting the values:
$\Delta Q = 2 \times \frac{3R}{2} \times 100 = 300\,R$.

(C) The capacitors $C_1$ and $C_2$ are connected in series between points $A$ and $B$.
Since they are in series,the charge $Q$ on both capacitors must be the same.
Let $V_D$ be the potential at point $D$.
The potential difference across $C_1$ is $(V_1 - V_D)$ and across $C_2$ is $(V_D - V_2)$.
Since $Q = C_1(V_1 - V_D) = C_2(V_D - V_2)$,
$C_1 V_1 - C_1 V_D = C_2 V_D - C_2 V_2$
$C_1 V_1 + C_2 V_2 = V_D(C_1 + C_2)$
$V_D = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$

(B) Lenticels are small,raised pores found in the bark of woody stems and roots of trees and shrubs.
These structures allow for the exchange of gases between the internal tissues of the plant and the external atmosphere.
While stomata are primarily found on leaves for gaseous exchange,lenticels serve a similar function in the woody parts of the plant where the epidermis has been replaced by bark (periderm).

(B) The melting point of a compound depends on the strength of intermolecular forces.
$p$-Nitrophenol exhibits strong intermolecular hydrogen bonding,which leads to a crystal lattice structure that is more stable and requires more energy to break.
In contrast,$o$-Nitrophenol exhibits intramolecular hydrogen bonding,which reduces the intermolecular forces.
$p$-Cresol and Phenol have weaker intermolecular forces compared to $p$-Nitrophenol.
Therefore,$p$-Nitrophenol has the highest melting point.

(A) The structure $CH_3-CH=CH-CH_2-OH$ corresponds to the unsaturated alcohol derived from crotonaldehyde.
This compound is commonly known as $Crotonyl \ alcohol$.

(A) The Gatterman-Koch reaction is a chemical reaction in which benzene or its derivatives are treated with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ or cuprous chloride $(CuCl)$ to produce benzaldehyde or substituted benzaldehydes. Therefore,the correct reagent is $CO, HCl / AlCl_3$ (anhydrous).

(C) The reaction between a Grignard reagent $(C_2H_5MgBr)$ and ammonia $(NH_3)$ is an acid-base reaction.
$NH_3$ acts as a proton donor,and the ethyl group $(C_2H_5^-)$ in the Grignard reagent acts as a base.
$C_2H_5MgBr + NH_3 \rightarrow C_2H_6 + Mg(NH_2)Br$.
Thus,the product $P$ is ethane $(C_2H_6)$.

(B) Cane sugar (sucrose) has the chemical formula $C_{12}H_{22}O_{11}$.
When treated with concentrated $H_2SO_4$,it undergoes dehydration.
The chemical reaction is: $C_{12}H_{22}O_{11} \xrightarrow{\text{conc. } H_2SO_4} 12C + 11H_2O$.
From the stoichiometry of the balanced equation,$1 \text{ mole}$ of sucrose yields $11 \text{ moles}$ of water.

(C) The molecular formula of citric acid is $C_6H_8O_7$.
From the structure of citric acid,we can count the number of carbon atoms per molecule.
There are $3$ carboxylic acid groups $(-COOH)$,each containing $1$ carbon atom,and a central carbon chain containing $3$ carbon atoms.
Total number of carbon atoms per molecule $= 3 + 3 = 6$.
Therefore,in $1$ mole of citric acid,there are $6$ moles of carbon atoms.
In $n$ moles of citric acid,the total number of moles of carbon atoms is $6n$.

(D) The reaction of cyclohexene with hot acidic $KMnO_4$ (oxidative cleavage) leads to the breaking of the double bond.
Since the double bond is part of a ring,the ring opens to form a dicarboxylic acid.
The product formed is hexane$-1,6-$dioic acid,commonly known as adipic acid.

(D) $VBT$ (Valence Bond Theory) explains the formation of covalent bonds through the overlapping of half-filled atomic orbitals.
It accounts for the delocalization of electrons between two nuclei and the partial ionic character of covalent bonds due to electronegativity differences.
The concept that the number of lone pairs and bonded pairs of electrons determines the shape of molecules is introduced by $VSEPR$ (Valence Shell Electron Pair Repulsion) Theory,not $VBT$.
Shielding effect is a concept related to atomic structure and effective nuclear charge,not a primary feature of $VBT$.

(B) The molecule $N_2$ (Dinitrogen) consists of two nitrogen atoms bonded together by a triple bond. The Lewis structure is represented as $:N \equiv N:$.

(A) The central atom in $XeF_4$ is Xenon $(Xe)$,which has $8$ valence electrons.
It forms $4$ single bonds with $4$ Fluorine $(F)$ atoms.
This leaves $4$ electrons on the Xenon atom,which form $2$ lone pairs.
The total number of electron pairs (steric number) is $4 \text{ (bond pairs)} + 2 \text{ (lone pairs)} = 6$.
$A$ steric number of $6$ corresponds to $sp^3d^2$ hybridization,which has an octahedral electron geometry.
Due to the presence of $2$ lone pairs,the molecular geometry is square planar.

(B) $CH_4$ molecule exhibits $sp^3$ hybridization.
In $sp^3$ hybridization,the geometry is tetrahedral.
The bond angle in a regular tetrahedral geometry is $109^{\circ} 28^{\prime}$.

(D) In $CH_4$,the central atom $C$ has $4$ bond pairs and $0$ lone pairs,resulting in a tetrahedral geometry with a bond angle of $109^{\circ} 28'$.
In $NH_3$,the central atom $N$ has $3$ bond pairs and $1$ lone pair,which causes repulsion and reduces the bond angle to $107^{\circ}$.
In $H_2O$,the central atom $O$ has $2$ bond pairs and $2$ lone pairs,leading to greater repulsion and a further reduction in the bond angle to $104.5^{\circ}$.
Therefore,the decreasing order of bond angle is $CH_4 > NH_3 > H_2O$.

(C) The $SO_2$ molecule has a bent geometry due to the presence of one lone pair on the sulfur atom.
The sulfur atom undergoes $sp^2$ hybridization.
In an ideal $sp^2$ hybridized system,the bond angle is $120^{\circ}$.
However,due to the lone pair-bond pair repulsion,the bond angle is slightly compressed to approximately $119.5^{\circ}$.

(C) The central oxygen atom in $H_2O$ has $6$ valence electrons.
It forms two single covalent bonds with two hydrogen atoms,resulting in two bond pairs.
The remaining $4$ electrons form two lone pairs on the oxygen atom.
Thus,$H_2O$ contains two lone pairs and two bond pairs of electrons.

(B) In acetylene $(HC \equiv CH)$,each carbon atom undergoes $sp$ hybridization.
The $C-H$ sigma bond is formed by the head-on overlap of the $sp$ hybrid orbital of the carbon atom and the $1s$ orbital of the hydrogen atom.
Therefore,the bond is formed by $sp - s$ overlap.

(A) An odd electron molecule is a molecule that contains an unpaired electron in its valence shell.
$1$. $CO$: Total valence electrons = $4 + 6 = 10$ (even). It is not an odd electron molecule.
$2$. $NO$: Total valence electrons = $5 + 6 = 11$ (odd). It is an odd electron molecule.
$3$. $NO_2$: Total valence electrons = $5 + 6 + 6 = 17$ (odd). It is an odd electron molecule.
$4$. $ClO_2$: Total valence electrons = $7 + 6 + 6 = 19$ (odd). It is an odd electron molecule.
Therefore,$CO$ is the molecule that is $NOT$ an odd electron molecule.

(C) To determine the geometry,we calculate the hybridization of the central atom:
$1$. For $SF_6$,the central atom is $S$ (Sulfur). It has $6$ valence electrons and forms $6$ bonds with $F$ atoms. The steric number is $6 + 0 = 6$,which corresponds to $sp^3d^2$ hybridization and an octahedral geometry.
$2$. $TeF_4$ has a steric number of $5$ ($4$ bond pairs + $1$ lone pair),resulting in a see-saw shape.
$3$. $SeCl_2$ has a steric number of $4$ ($2$ bond pairs + $2$ lone pairs),resulting in a bent shape.
$4$. $SeF_4$ has a steric number of $5$ ($4$ bond pairs + $1$ lone pair),resulting in a see-saw shape.
Therefore,$SF_6$ is the only octahedral molecule among the given options.

(A) To determine the shape of the molecules,we look at their hybridization and the presence of lone pairs:
$1$. $SO_2$: The sulfur atom has $sp^2$ hybridization with one lone pair. Due to the lone pair-bond pair repulsion,the molecule adopts a bent (or angular) shape.
$2$. $BeBr_2$: The beryllium atom has $sp$ hybridization with no lone pairs,resulting in a linear shape.
$3$. $CO_2$: The carbon atom has $sp$ hybridization with no lone pairs,resulting in a linear shape.
$4$. $BF_3$: The boron atom has $sp^2$ hybridization with no lone pairs,resulting in a trigonal planar shape.
Therefore,$SO_2$ is the molecule with a bent shape.

(D) To determine the hybridisation of the central atom,we use the formula: $\text{Steric Number} = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$A$) For $NH_3$: $\text{Steric Number} = \frac{1}{2} [5 + 3] = 4$,which corresponds to $sp^3$ hybridisation.
$B$) For $H_2O$: $\text{Steric Number} = \frac{1}{2} [6 + 2] = 4$,which corresponds to $sp^3$ hybridisation.
$C$) For $CH_4$: $\text{Steric Number} = \frac{1}{2} [4 + 4] = 4$,which corresponds to $sp^3$ hybridisation.
$D$) For $BF_3$: $\text{Steric Number} = \frac{1}{2} [3 + 3] = 3$,which corresponds to $sp^2$ hybridisation.
Thus,the central atom in $BF_3$ does not undergo $sp^3$ hybridisation.

(B) To determine the shape of the molecules,we look at their hybridization and the presence of lone pairs:
$1$. $HBr$: It is a diatomic molecule,so it is linear.
$2$. $H_2S$: The central sulfur atom has $6$ valence electrons. It forms $2$ bonds with hydrogen atoms and has $2$ lone pairs. According to $VSEPR$ theory,the steric number is $4$ ($2$ bond pairs + $2$ lone pairs),resulting in $sp^3$ hybridization and a bent (angular) shape.
$3$. $BeBr_2$: The central beryllium atom has $2$ valence electrons and forms $2$ bonds with bromine atoms. It has no lone pairs,resulting in $sp$ hybridization and a linear shape.
$4$. $CO_2$: The central carbon atom forms $2$ double bonds with oxygen atoms. It has no lone pairs,resulting in $sp$ hybridization and a linear shape.
Therefore,$H_2S$ is the only molecule that is not linear.

(D) In $BeCl_2$,the central $Be$ atom undergoes $sp$ hybridisation.
In $BF_3$,the central $B$ atom undergoes $sp^2$ hybridisation.
In $C_2H_4$,each carbon atom undergoes $sp^2$ hybridisation.
In $H_2O$,the central oxygen atom has two bond pairs and two lone pairs,resulting in a steric number of $4$,which corresponds to $sp^3$ hybridisation.

(B) In the $H_2O$ molecule,the oxygen atom undergoes $sp^3$ hybridization.
Due to the presence of two lone pairs on the oxygen atom,the bond angle is compressed from the ideal tetrahedral angle of $109^{\circ} 28^{\prime}$ to $104^{\circ} 28^{\prime}$ according to the $VSEPR$ theory.

(A) According to Molecular Orbital Theory,the energy of molecular orbitals increases as follows:
$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$.
For the $F_2$ molecule,the order of the lower energy orbitals is $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s$.

(C) According to Molecular Orbital Theory,the electronic configuration of $O_2$ ($16$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Bond Order = $\frac{N_b - N_a}{2} = \frac{10 - 6}{2} = \frac{4}{2} = 2.0$.

(B) The bond length of the $C-F$ bond in a $CH_3F$ (fluoromethane) molecule is experimentally determined to be approximately $139 \ pm$.

(A) To find the number of electrons in each species:
$(a)$ $O^{2-}$: Oxygen has atomic number $8$. $O^{2-}$ has $8 + 2 = 10$ electrons. Thus,$(a)-(iii)$.
$(b)$ $Li^{2+}$: Lithium has atomic number $3$. $Li^{2+}$ has $3 - 2 = 1$ electron. Thus,$(b)-(iv)$.
$(c)$ $He$: Helium has atomic number $2$. It has $2$ electrons. Thus,$(c)-(ii)$.
$(d)$ $Ca^{2+}$: Calcium has atomic number $20$. $Ca^{2+}$ has $20 - 2 = 18$ electrons. Thus,$(d)-(i)$.
Therefore,the correct match is $a-iii, b-iv, c-ii, d-i$.

(D) In benzene $(C_6H_6)$, all carbon-carbon bonds are equivalent due to resonance.
The bond order is $1.5$, which results in a bond length of approximately $139 \ pm$ (or $1.39 \ \mathring{A}$), intermediate between a single bond $(154 \ pm)$ and a double bond $(134 \ pm)$.

(D) The bond order is calculated using the formula: $\text{Bond Order} = \frac{1}{2} (N_b - N_a)$.
For $F_2$ ($18$ electrons): $\text{Bond Order} = 1$.
For $H_2$ ($2$ electrons): $\text{Bond Order} = 1$.
For $Li_2$ ($6$ electrons): $\text{Bond Order} = 1$.
For $N_2$ ($14$ electrons): $\text{Bond Order} = 3$.
Since $3 > 1$,the molecule $N_2$ has a bond order greater than $1$.

(C) Lower members of alcohols (like methanol and ethanol) are highly soluble in water due to hydrogen bonding and are also soluble in organic solvents. Therefore,the statement that they are insoluble in water and organic solvents is incorrect. Thus,option $C$ is the correct answer.

(A) $LiAlH_4$ is a strong reducing agent.
It reduces esters $(R-COOR')$ to primary alcohols.
The ester bond is cleaved to form two alcohols: the alcohol derived from the acyl part $(R-CH_2OH)$ and the alcohol derived from the alkoxy part $(R'-OH)$.
Therefore,the reaction is: $R-COOR' \xrightarrow{LiAlH_4} R-CH_2OH + R'-OH$.
Thus,$A = R-CH_2OH$ and $B = R'-OH$.

(C) The conversion of an aldehyde $(R-CHO)$ to a primary alcohol $(R-CH_2-OH)$ is a reduction reaction.
$LiAlH_4$ (Lithium aluminium hydride) is a strong reducing agent that effectively reduces aldehydes to primary alcohols.
The reaction is: $R-CHO \xrightarrow{LiAlH_4 / H_3O^+} R-CH_2-OH$.

(C) The reaction of an alkyl halide with aqueous $KOH$ is a nucleophilic substitution reaction.
$C_2H_5Br + \text{aq. } KOH \rightarrow C_2H_5OH + KBr$
Here,the hydroxide ion $(OH^-)$ acts as a nucleophile and replaces the bromide ion $(Br^-)$ to form ethanol $(C_2H_5OH)$.

(D) Carboxylic acids $(R-COOH)$ are reduced to primary alcohols $(R-CH_2OH)$ using strong reducing agents like lithium aluminum hydride $(LiAlH_4)$.
The reaction is: $R-COOH \xrightarrow[\text{Reduction}]{LiAlH_4} R-CH_2OH$.

(C) In aryl halides,the halogen atom is attached to an $sp^2$ hybridized carbon atom,making the $C-X$ bond shorter and stronger than in alkyl halides.
Nucleophilic substitution in aryl halides is facilitated by the presence of strong electron-withdrawing groups (like $-NO_2$) at the $ortho$ and $para$ positions due to the stabilization of the carbanion intermediate.
In $o$-nitrochlorobenzene,$p$-nitrochlorobenzene,and $2, 4, 6$-trinitrochlorobenzene,the $-NO_2$ groups are at positions that allow for resonance stabilization of the intermediate.
However,in $m$-nitrochlorobenzene,the $-NO_2$ group at the $meta$ position does not provide resonance stabilization to the carbanion intermediate formed during nucleophilic attack.
Therefore,$m$-nitrochlorobenzene has the most difficulty in breaking the $C-X$ bond compared to the other options.

(C) The reaction of a Grignard reagent $(CH_3MgBr)$ with formaldehyde $(HCHO)$ followed by acid hydrolysis $(H_3O^+)$ yields a primary alcohol $(CH_3CH_2OH)$.
The reaction mechanism is as follows:
$HCHO + CH_3MgBr \rightarrow CH_3CH_2OMgBr$
$CH_3CH_2OMgBr + H_2O \rightarrow CH_3CH_2OH + Mg(OH)Br$
Thus,the compound $A$ is formaldehyde $(HCHO)$.

(A) Alcohols are polar molecules due to the presence of the $-OH$ group,which creates a dipole moment. Therefore,the statement that they are non-polar molecules is incorrect. Thus,option $(A)$ is the correct answer.

(D) The reactivity of alcohols with halo acids depends on the stability of the carbocation intermediate formed during the reaction.
Since the stability of carbocations follows the order $3^{\circ} > 2^{\circ} > 1^{\circ}$,the reactivity of alcohols towards halo acids also follows the order $3^{\circ} > 2^{\circ} > 1^{\circ}$.
Therefore,the statement $D$ is false.

(B) The Lucas reagent is a solution of anhydrous zinc chloride $(ZnCl_2)$ in concentrated hydrochloric acid $(HCl)$.
It is used to classify alcohols based on their reactivity.

(B) The preparation of phenol from aniline involves two main steps:
$1$. Aniline is treated with $NaNO_2$ and $HCl$ at $0-5 \ ^\circ C$ to form benzene diazonium chloride.
$2$. Benzene diazonium chloride is then hydrolyzed by boiling with water to produce phenol.
Therefore,the intermediate product is benzene diazonium chloride.

(A) When aniline $(C_6H_5NH_2)$ is treated with $NaNO_2 + HCl$ at $0-5 \ ^\circ C$,it undergoes a diazotization reaction to form benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$.
This intermediate is then hydrolyzed by heating with water to produce phenol $(C_6H_5OH)$,nitrogen gas $(N_2)$,and hydrochloric acid $(HCl)$.
The question specifically asks for the product formed prior to hydrolysis,which is benzene diazonium chloride.

(D) The preparation of phenol involves specific industrial and laboratory methods.
$A$. Aniline can be converted to phenol via benzene diazonium chloride (diazotization followed by hydrolysis). This is a standard method.
$B$. Cumene oxidation is the most common industrial method for producing phenol.
$C$. The Dow process involves the reaction of chlorobenzene with $NaOH$ at high temperature and pressure to form sodium phenoxide,which upon acidification yields phenol. This is a standard method.
$D$. Benzene reacts with concentrated $H_2SO_4$ to form benzene sulfonic acid,not phenol directly. Phenol is obtained from benzene sulfonic acid by fusion with $NaOH$ followed by acidification. Therefore,the reaction as written is incorrect.

(B) The process is known as the cumene process for the industrial production of phenol.
Step $1$: Cumene (isopropylbenzene) is oxidized by air in the presence of cobalt naphthenate to form cumene hydroperoxide.
Step $2$: Cumene hydroperoxide is then treated with dilute acid (like $dil. H_2SO_4$) to yield phenol and acetone as the final products.
The reaction is: $C_6H_5CH(CH_3)_2 + O_2$ $\xrightarrow{Co-naphthenate} C_6H_5C(CH_3)_2OOH$ $\xrightarrow{dil. H_2SO_4} C_6H_5OH + CH_3COCH_3$.

(A) The air oxidation of cumene (isopropylbenzene) in the presence of cobalt naphthenate as a catalyst leads to the formation of cumene hydroperoxide.
This is the first step in the industrial preparation of phenol and acetone from cumene.
The reaction is:
$C_6H_5-CH(CH_3)_2 + O_2 \xrightarrow{\text{Co-naphthenate}} C_6H_5-C(CH_3)_2OOH$
Thus,the product formed is cumene hydroperoxide.

(A) The Reimer-Tiemann reaction involves the treatment of phenol with chloroform $(CHCl_3)$ in the presence of an aqueous base like $KOH$ or $NaOH$.
This reaction results in the introduction of a formyl group $(-CHO)$ at the ortho position of the phenol ring.
The major product formed is $2-$hydroxybenzaldehyde,commonly known as Salicylaldehyde,along with a small amount of the para-isomer.

(B) When phenol $(C_6H_5OH)$ vapours are passed over heated zinc dust,it undergoes reduction to form benzene $(C_6H_6)$ and zinc oxide $(ZnO)$.
The chemical reaction is as follows:
$C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$

(A) The reaction of phenol with concentrated $H_2SO_4$ at $373 \ K$ is an electrophilic aromatic substitution reaction known as sulphonation.
At higher temperatures $(373 \ K)$,the $-SO_3H$ group is predominantly directed to the para-position due to steric hindrance at the ortho-position,resulting in the formation of $p-$phenol sulphonic acid as the major product.

(D) The reaction described is the $Kolbe-Schmitt$ reaction.
$1$. Phenol reacts with $NaOH$ to form sodium phenoxide.
$2$. Sodium phenoxide reacts with $CO_2$ at high pressure and temperature $(140 \ ^\circ C)$ to form sodium salicylate.
$3$. Subsequent acid hydrolysis $(H^+)$ converts the intermediate into salicylic acid.
Thus,the final product is salicylic acid.

(A) The preparation of ethers from alcohols by the action of concentrated $H_2SO_4$ is a dehydration reaction.
When $2 \ moles$ of alcohol are heated with concentrated $H_2SO_4$ at $413 \ K$ $(140^{\circ}C)$,an ether is formed.
The reaction is as follows:
$2R-OH \xrightarrow{Conc. H_2SO_4, 413 \ K} R-O-R + H_2O$
Therefore,the correct temperature is $413 \ K$.

(D) The chemical structure of propylene glycol (propane-$1,2$-diol) is $CH_3-CH(OH)-CH_2OH$.
As shown in the structure,there are two hydroxyl $(-OH)$ groups attached to the carbon chain.
Therefore,the number of hydroxyl groups present in propylene glycol is $2$.

(C) Ethylene glycol $(HO-CH_2-CH_2-OH)$ is used as an antifreeze agent in automobile radiators because it lowers the freezing point of water,preventing it from freezing in cold weather.

(D) When an aldehyde reacts with one equivalent of a monohydric alcohol in the presence of dry $HCl$ gas,it forms a hemiacetal.
However,when the aldehyde is treated with an excess of monohydric alcohol,the hemiacetal further reacts to form an acetal.
The reaction is: $R-CHO + 2R'-OH \xrightarrow{dry \ HCl} R-CH(OR')_2 + H_2O$.
Thus,the final product formed is an acetal.

(D) Let us analyze the given reactions:
$A$) Rosenmund reduction uses $H_2/Pd-BaSO_4$ to reduce acid chlorides to aldehydes. This is correct.
$B$) Stephen reaction uses $SnCl_2/HCl$ to reduce nitriles to imines,which are then hydrolyzed to aldehydes. This is correct.
$C$) Etard reaction uses $CrO_2Cl_2$ to oxidize toluene to benzaldehyde. This is correct.
$D$) Gatterman-Koch reaction uses $CO + HCl$ in the presence of anhydrous $AlCl_3$ to form benzaldehyde from benzene. The reagent given in the option $(CrO_3/(CH_3CO)_2O)$ is for the Etard-like oxidation of toluene to benzaldehyde diacetate. Thus,option $D$ is incorrectly matched.

(A) When a mixture of two different aldehydes (ethanal and propanal),both containing at least one $\alpha$-hydrogen,reacts with aqueous $NaOH$ and is warmed,it undergoes a cross-aldol condensation reaction.
Each aldehyde can undergo self-aldol condensation,and they can also undergo cross-aldol condensation with each other.
$1$. Self-aldol of ethanal: $CH_3CHO + CH_3CHO \rightarrow CH_3CH=CHCHO$ (after dehydration).
$2$. Self-aldol of propanal: $CH_3CH_2CHO + CH_3CH_2CHO \rightarrow CH_3CH_2CH=C(CH_3)CHO$ (after dehydration).
$3$. Cross-aldol (ethanal as donor,propanal as acceptor): $CH_3CHO + CH_3CH_2CHO \rightarrow CH_3CH=C(CH_3)CHO$ (after dehydration).
$4$. Cross-aldol (propanal as donor,ethanal as acceptor): $CH_3CH_2CHO + CH_3CHO \rightarrow CH_3CH_2CH=CHCHO$ (after dehydration).
Thus,a total of $4$ different aldol condensation products are formed.

(B) The melting point of a compound is significantly influenced by the nature of hydrogen bonding.
$o-$Nitrophenol exhibits intramolecular hydrogen bonding,which restricts intermolecular association,leading to a lower melting point.
In $p-$Cresol,the presence of the non-polar $-CH_3$ group does not significantly enhance intermolecular forces compared to the nitro group.
$p-$Nitrophenol exhibits strong intermolecular hydrogen bonding,which leads to the association of molecules in the solid state,resulting in a much higher melting point $(114 \ ^\circ C)$ compared to the others.

(C) The given reaction is the hydrogenation of an acid chloride $(R-CO-Cl)$ to an aldehyde $(R-CHO)$ using hydrogen gas $(H_2)$ in the presence of a palladium catalyst supported on barium sulfate $(Pd-BaSO_4)$.
This specific reaction is known as the Rosenmund reduction.

(D) The Rosenmund reduction is a hydrogenation process used to convert acid chlorides into aldehydes.
It employs $H_2$ gas in the presence of a poisoned catalyst,specifically $Pd$ supported on $BaSO_4$ (palladium on barium sulfate).
The $BaSO_4$ acts as a poison to prevent further reduction of the aldehyde to an alcohol.

(C) The reagent $DIBAL-H$ (Diisobutylaluminium hydride) is a selective reducing agent that reduces nitriles $(-CN)$ to imines,which upon acidic hydrolysis $(H_3O^{+})$ yield aldehydes.
It does not affect the carbon-carbon double bond $(C=C)$.
The reaction is:
$CH_3-CH=CH-CH_2-C \equiv N \xrightarrow[H_3O^{+}]{\text{DIBAL-H}} CH_3-CH=CH-CH_2-CHO$
The product formed is $Pent-3-enal$.

(C) Step $1$: Reaction of methyl magnesium bromide with cadmium chloride $(CdCl_2)$ produces dimethyl cadmium $(A)$.
$2CH_3MgBr + CdCl_2 \rightarrow (CH_3)_2Cd + 2Mg(Br)Cl$
Step $2$: Dimethyl cadmium reacts with acetyl chloride $(CH_3COCl)$ to form propanone $(B)$.
$2CH_3COCl + (CH_3)_2Cd \rightarrow 2CH_3COCH_3 + CdCl_2$
Thus,the product $B$ is propanone.

(B) The given reaction involves the dehydrogenation of a secondary alcohol $(R-CH(OH)-R)$ in the presence of copper $(Cu)$ catalyst at $573 \ K$.
Secondary alcohols undergo dehydrogenation to form ketones as the product $P$.
The reaction is: $R-CH(OH)-R \xrightarrow{Cu, 573 \ K} R-CO-R + H_2$.
Therefore,the product $P$ is $R-CO-R$.

(B) Aldehydes $(R-CHO)$ undergo oxidation when treated with oxidizing agents like dilute nitric acid $(HNO_3)$.
This reaction converts the aldehyde group into a carboxylic acid group.
The chemical equation is: $R-CHO \xrightarrow{\text{Dil. } HNO_3} R-COOH$ (Carboxylic acid).

(D) The given reaction is the iodoform reaction,which is a characteristic test for methyl ketones or methyl carbinols.
In this reaction,acetone $(CH_3COCH_3)$ reacts with sodium hypoiodite $(NaOI)$ to form iodoform $(CHI_3)$ as a yellow precipitate along with sodium acetate $(CH_3COONa)$ and sodium hydroxide $(NaOH)$.
Thus,the product $Y$ is $CHI_3$ (iodoform).

(A) The given reaction is the iodoform test,which is a characteristic reaction for methyl ketones or compounds containing the $CH_3CO-$ group.
Acetone $(CH_3COCH_3)$ reacts with iodine $(I_2)$ in the presence of sodium hydroxide $(NaOH)$ to form sodium hypoiodite $(NaOI)$,which acts as the oxidizing and iodinating agent.
The balanced chemical equation for the reaction is:
$CH_3COCH_3 + 3NaOI \stackrel{\Delta}{\longrightarrow} CHI_3 \downarrow + CH_3COONa + 2NaOH$
Thus,the compound $X$ is $NaOI$ (which is generated in situ from $I_2$ and $NaOH$).

(C) Compounds that exhibit the iodoform test must contain either a methyl ketone group $(CH_3-CO-R)$ or a methyl carbinol group $(CH_3-CH(OH)-R)$.
Acetaldehyde $(CH_3CHO)$ contains the $CH_3-CO-$ group.
sec-Butyl alcohol $(CH_3-CH(OH)-CH_2CH_3)$ contains the $CH_3-CH(OH)-$ group.
Acetophenone $(C_6H_5-CO-CH_3)$ contains the $CH_3-CO-$ group.
n-Butyl alcohol $(CH_3-CH_2-CH_2-CH_2-OH)$ does not contain either of these structural units.
Therefore,$n$-Butyl alcohol does not show the iodoform test.

(B) The Gatterman-Koch formylation of an arene involves the treatment of benzene or a substituted benzene with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ under high pressure.
This reaction yields benzaldehyde or a substituted benzaldehyde.
The reaction is catalyzed by anhydrous aluminium chloride $(AlCl_3)$ or cuprous chloride $(CuCl)$.
Therefore,the correct reagent is $CO, HCl / AlCl_3$ (anhydrous).

(B) The reaction of a primary aliphatic amine $(R-NH_2)$ with nitrous acid $(HNO_2)$,which is generated in situ by the reaction of $NaNO_2$ and $HCl$ at $273-278 \ K$,forms an alkyl diazonium salt $(R-N_2^+ Cl^-)$ as an intermediate $(X)$.
This alkyl diazonium salt is highly unstable and decomposes rapidly in the presence of water to form an alcohol $(R-OH)$,nitrogen gas $(N_2)$,and hydrochloric acid $(HCl)$.
Therefore,the intermediate $X$ is $R-N_2^+ Cl^-$.

(C) The given reaction is the reaction of a secondary amine with benzenesulfonyl chloride (Hinsberg reagent).
In this reaction,the secondary amine $(C_2H_5)_2NH$ reacts with benzenesulfonyl chloride to form $N,N$-diethylbenzenesulfonamide.
The nitrogen atom in the secondary amine has one hydrogen atom,which is replaced by the sulfonyl group,releasing $HCl$.
Therefore,the compound $X$ is diethylamine,which is $(C_2H_5)_2NH$.

(C) The reagent $Sn / HCl$ is a common reducing agent used to reduce nitro compounds $(R-NO_2)$ to primary amines $(R-NH_2)$.
The balanced chemical equation for the reduction of a nitroalkane is:
$R-NO_2 + 6[H] \xrightarrow{Sn / HCl} R-NH_2 + 2H_2O$.

(C) The basic strength of amines in the aqueous phase depends on the inductive effect,solvation effect,and steric hindrance.
For aliphatic amines where $R = CH_3$,the order of basicity is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$.
Since $pK_b = -\log(K_b)$,a higher basic strength corresponds to a lower $pK_b$ value.
Therefore,the order of $pK_b$ values is the reverse of the basicity order: $NH_3 > (CH_3)_3N > CH_3NH_2 > (CH_3)_2NH$.
However,in general terms for alkyl amines,the order of basicity is $R_2NH > RNH_2 > R_3N > NH_3$.
Thus,the correct order of $pK_b$ values is $NH_3 > R_3N > RNH_2 > R_2NH$.

(B) The reaction of ethylamine $(CH_3CH_2NH_2)$ with ethyl bromide $(C_2H_5Br)$ is an example of ammonolysis/alkylation of amines.
Step $1$: Ethylamine reacts with $1 \ mole$ of ethyl bromide to form diethylamine $(CH_3CH_2NHCH_2CH_3)$.
$CH_3CH_2NH_2 + C_2H_5Br \rightarrow (CH_3CH_2)_2NH + HBr$
Step $2$: Diethylamine further reacts with another $1 \ mole$ of ethyl bromide to form $N,N$-diethylethanamine (triethylamine,$(CH_3CH_2)_3N$).
$(CH_3CH_2)_2NH + C_2H_5Br \rightarrow (CH_3CH_2)_3N + HBr$
Total moles of ethyl bromide required = $1 + 1 = 2 \ moles$.

(C) The $pK_b$ value is inversely proportional to the basic strength of the amine. $A$ lower $pK_b$ value indicates a stronger base.
In the aqueous phase,the basicity of ethyl-substituted amines follows the order: $(C_2H_5)_2NH > C_2H_5NH_2 > (C_2H_5)_3N > NH_3$.
However,comparing the given options: $CH_3NH_2$ $(pK_b = 3.36)$,$CH_3CH_2NH_2$ $(pK_b = 3.29)$,$(CH_3CH_2)_3N$ $(pK_b = 3.25)$,and $(CH_3CH_2)_2NH$ $(pK_b = 3.00)$.
Thus,diethylamine $(CH_3CH_2)_2NH$ has the lowest $pK_b$ value,making it the strongest base among the choices.

(D) The Gabriel phthalimide synthesis is a method used for the preparation of primary aliphatic amines. The reaction involves the following steps:
$1$. Phthalimide reacts with ethanolic $KOH$ to form potassium phthalimide.
$2$. Potassium phthalimide reacts with an alkyl halide $(R-X)$ to form $N$-alkylphthalimide.
$3$. $N$-alkylphthalimide undergoes alkaline hydrolysis (using $NaOH_{\text{(aq.)}}$) to yield the corresponding primary amine $(R-NH_2)$ and sodium phthalate.

(A) The conversion of aniline to phenol involves two steps:
$1$. Diazotization of aniline using $NaNO_2 / HCl$ at $0-5 \ ^{\circ}C$ to form benzene diazonium chloride.
$2$. Hydrolysis of benzene diazonium chloride by warming with water to yield phenol.
Therefore,$NaNO_2 / HCl$ is the correct reagent for the initial step of this conversion.

(D) Primary aliphatic amines $(R-NH_2)$ react with $NaNO_2$ and $HCl$ to form unstable diazonium salts,which decompose to give alcohols $(R-OH)$.
Secondary amines $(R_2NH)$ react with $NaNO_2$ and $HCl$ to form $N$-nitrosoamines,which are yellow oily compounds,not alcohols.
$(CH_3)_2NH$ is a secondary amine,therefore it does not form alcohol.

(C) The Hofmann elimination reaction involves the conversion of an amine into an alkene.
First,the amine is converted into a quaternary ammonium iodide by reaction with excess methyl iodide $(CH_3I)$.
Then,the quaternary ammonium salt is treated with moist silver oxide $(Ag_2O + H_2O)$,which replaces the iodide ion with a hydroxide ion $(OH^-)$.
Finally,heating the quaternary ammonium hydroxide leads to the elimination reaction,producing an alkene,a tertiary amine,and water.

(A) The reagent used to distinguish between primary,secondary,and tertiary amines is Hinsberg reagent,which is Benzenesulfonyl chloride $(C_6H_5SO_2Cl)$.
$1$. Primary $(1^{\circ})$ amines react with Benzenesulfonyl chloride to form $N$-alkylbenzenesulfonamide,which is soluble in alkali due to the presence of an acidic hydrogen atom attached to the nitrogen.
$2$. Secondary $(2^{\circ})$ amines react to form $N$,$N$-dialkylbenzenesulfonamide,which is insoluble in alkali because it lacks an acidic hydrogen atom.
$3$. Tertiary $(3^{\circ})$ amines do not react with Benzenesulfonyl chloride.

(D) Hofmann's exhaustive methylation involves the reaction of an amine with an excess of an alkyl halide.
The reaction proceeds through successive alkylations until a quaternary ammonium salt is formed.
For example,with ethylamine and ethyl iodide:
$CH_3CH_2NH_2 + 3CH_3CH_2I \rightarrow (CH_3CH_2)_4N^{+}I^{-} + 3HI$.
Thus,the final product is a quaternary ammonium halide.

(A) The Hinsberg reagent is $C_6H_5SO_2Cl$,which is known as benzene sulphonyl chloride.
It is used to distinguish between primary,secondary,and tertiary amines.

(C) The correct answer is $C$.
$A$. $R-NH_2 + 3 R-X \rightarrow R-N^+(R)_3 X^-$ is known as Hofmann exhaustive alkylation.
$B$. $R-CONH_2 + Br_2 + 4 KOH \rightarrow R-NH_2 + K_2CO_3 + 2 KBr + 2 H_2O$ is known as Hofmann bromamide degradation.
$C$. The reduction of amides $(R-CONH_2)$ to amines $(R-CH_2-NH_2)$ using $LiAlH_4$ is a standard reduction of amides,not Mendius reduction. Mendius reduction specifically refers to the reduction of nitriles $(R-CN)$ to primary amines $(R-CH_2-NH_2)$ using $Na/EtOH$ or $H_2/Ni$.
$D$. The reaction of quaternary ammonium salts with moist $Ag_2O$ followed by heating is known as Hofmann elimination.