AP EAMCET 2023 Chemistry Question Paper with Answer and Solution

(D) The given equation is $y = \frac{1}{\sqrt{a}}\sin \omega t \pm \frac{1}{\sqrt{b}}\cos \omega t$.
We know that $\cos \omega t = \sin(\omega t + \frac{\pi}{2})$.
Substituting this,the equation becomes $y = \frac{1}{\sqrt{a}}\sin \omega t \pm \frac{1}{\sqrt{b}}\sin(\omega t + \frac{\pi}{2})$.
This represents the superposition of two simple harmonic motions with amplitudes $A_1 = \frac{1}{\sqrt{a}}$ and $A_2 = \frac{1}{\sqrt{b}}$,and a phase difference of $\phi = \frac{\pi}{2}$.
The resultant amplitude $A$ is given by $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$.
Since $\cos(\frac{\pi}{2}) = 0$,the resultant amplitude is $A = \sqrt{A_1^2 + A_2^2}$.
Substituting the values: $A = \sqrt{(\frac{1}{\sqrt{a}})^2 + (\frac{1}{\sqrt{b}})^2} = \sqrt{\frac{1}{a} + \frac{1}{b}} = \sqrt{\frac{a + b}{ab}}$.

(C) The frequency of a light wave is a characteristic property of the source of light and does not depend on the medium through which it propagates.
When light travels from one medium to another,its speed and wavelength change,but its frequency remains constant.
Therefore,the frequency of the light ray in the medium of refractive index $1.5$ will remain the same as its frequency in vacuum (or air),which is $6 \times 10^{14} \ Hz$.

(C) Bond length is inversely proportional to bond order. Higher bond order results in shorter bond length.
$1.$ In $CO$, the bond order is $3$ $(:C \equiv O:^+)$, so the bond length is shortest $(112.8 \ pm)$.
$2.$ In $CO_2$, the bond order is $2$ $(O=C=O)$, so the bond length is intermediate $(122 \ pm)$.
$3.$ In $CO_3^{2-}$, the carbon atom is $sp^2$ hybridized and exhibits resonance. The bond order is $1.33$, which is the lowest, resulting in the longest bond length $(136 \ pm)$.
Therefore, the correct order of increasing bond length is $CO < CO_2 < CO_3^{2-}$.

(A) Given: $m_{1} = m$,$m_{2} = 2m$,$u_{1} = 2 \ m/s$,$u_{2} = 0$,and coefficient of restitution $e = 0.5$.
Let $v_{1}$ and $v_{2}$ be the velocities after collision.
Applying the law of conservation of linear momentum:
$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$
$m(2) + 2m(0) = mv_{1} + 2mv_{2}$
$2 = v_{1} + 2v_{2} \quad ...(i)$
By the definition of the coefficient of restitution:
$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$
$0.5 = \frac{v_{2} - v_{1}}{2 - 0}$
$1 = v_{2} - v_{1} \quad ...(ii)$
Adding equations $(i)$ and $(ii)$:
$(v_{1} + 2v_{2}) + (v_{2} - v_{1}) = 2 + 1$
$3v_{2} = 3 \implies v_{2} = 1 \ m/s$
Substituting $v_{2} = 1$ in equation $(ii)$:
$1 = 1 - v_{1} \implies v_{1} = 0 \ m/s$
Thus,the velocities after collision are $0 \ m/s$ and $1 \ m/s$.

(D) The given equation is $y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$.
We can rewrite $\cos \omega t$ as $\sin(\omega t + \frac{\pi}{2})$.
Thus,$y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \sin(\omega t + \frac{\pi}{2})$.
This represents the superposition of two simple harmonic motions with amplitudes $A_1 = \frac{1}{\sqrt{a}}$ and $A_2 = \frac{1}{\sqrt{b}}$,and a phase difference of $\phi = \frac{\pi}{2}$.
The resultant amplitude $A$ is given by $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$.
Since $\cos(\frac{\pi}{2}) = 0$,the formula simplifies to $A = \sqrt{A_1^2 + A_2^2}$.
Substituting the values,$A = \sqrt{(\frac{1}{\sqrt{a}})^2 + (\frac{1}{\sqrt{b}})^2} = \sqrt{\frac{1}{a} + \frac{1}{b}}$.
Simplifying the expression,$A = \sqrt{\frac{a + b}{ab}}$.

(D) The bond length is inversely proportional to the bond order. Higher bond order implies a shorter bond length.
$1$. In $CO$,the bond order is $3.0$.
$2$. In $CO_2$,the bond order is $2.0$.
$3$. In $CO_3^{2-}$,the bond order is $1.33$ (due to resonance).
Thus,the bond order order is $CO > CO_2 > CO_3^{2-}$.
Therefore,the bond length order is $CO < CO_2 < CO_3^{2-}$.

(A) Given: $m_1 = m$,$m_2 = 2m$,$u_1 = 2 \, m/s$,$u_2 = 0$,and $e = 0.5$.
Applying the law of conservation of linear momentum:
$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
$m(2) + 2m(0) = m v_1 + 2m v_2$
$2m = m v_1 + 2m v_2 \implies v_1 + 2v_2 = 2$ $...(i)$
Using the definition of the coefficient of restitution:
$e = \frac{v_2 - v_1}{u_1 - u_2}$
$0.5 = \frac{v_2 - v_1}{2 - 0}$
$v_2 - v_1 = 1$ $...(ii)$
Adding equations $(i)$ and $(ii)$:
$(v_1 + 2v_2) + (v_2 - v_1) = 2 + 1$
$3v_2 = 3 \implies v_2 = 1 \, m/s$
Substituting $v_2 = 1$ into equation $(ii)$:
$1 - v_1 = 1 \implies v_1 = 0 \, m/s$
Thus,the velocities after collision are $0 \, m/s$ and $1 \, m/s$.

(B) The reaction involves the oxidation of alkyl groups attached to a benzene ring using alkaline $KMnO_4$ followed by acidic workup $(H_3O^+)$.
Alkaline $KMnO_4$ is a strong oxidizing agent that oxidizes any alkyl side chain attached to the benzene ring to a carboxylic acid group $(-COOH)$,provided that the benzylic carbon has at least one hydrogen atom.
In the given reactant,$1$-ethyl-$4$-methylbenzene,both the ethyl group $(-CH_2CH_3)$ and the methyl group $(-CH_3)$ are attached to the benzene ring at the $1$ and $4$ positions.
Both of these alkyl groups have benzylic hydrogens,so both will be oxidized to carboxylic acid groups.
Therefore,the product $X$ is benzene-$1,4$-dicarboxylic acid,also known as terephthalic acid.

(B) $BF_3$ and $CCl_4$ are non-polar molecules due to their symmetrical geometry,which leads to the cancellation of individual bond dipoles,resulting in a net dipole moment of $\mu = 0$. Therefore,the set containing $BF_3$ and $CCl_4$ as polar molecules is incorrectly matched.

(D) In the $XeOF_4$ molecule,the central atom $Xe$ is bonded to four $F$ atoms via single bonds and one $O$ atom via a double bond.
Thus,the total number of bond pairs is $4 + 2 = 6$.
Regarding lone pairs:
- Each of the four $F$ atoms has $3$ lone pairs $(4 \times 3 = 12)$.
- The $O$ atom has $2$ lone pairs.
- The $Xe$ atom has $1$ lone pair.
Total lone pairs $= 12 + 2 + 1 = 15$.

(D) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{V - N}{2}$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of monovalent atoms bonded to it.
$1$. For $IF_7$: Central atom $I$ has $7$ valence electrons. It forms $7$ bonds with $F$ atoms. $\text{Lone pairs} = \frac{7 - 7}{2} = 0$.
$2$. For $IF_5$: Central atom $I$ has $7$ valence electrons. It forms $5$ bonds with $F$ atoms. $\text{Lone pairs} = \frac{7 - 5}{2} = 1$.
$3$. For $ClF_3$: Central atom $Cl$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms. $\text{Lone pairs} = \frac{7 - 3}{2} = 2$.
$4$. For $XeF_2$: Central atom $Xe$ has $8$ valence electrons. It forms $2$ bonds with $F$ atoms. $\text{Lone pairs} = \frac{8 - 2}{2} = 3$.
The number of lone pairs is: $IF_7 (0) < IF_5 (1) < ClF_3 (2) < XeF_2 (3)$.
Thus,the correct order is $IF_7 < IF_5 < ClF_3 < XeF_2$.

(A) $I$: $SiH_4$ has $Si$ in $sp^3$ hybridization with $0$ lone pairs,resulting in a tetrahedral geometry. This is correct.
$II$: $BeCl_2$ has $Be$ in $sp$ hybridization with $0$ lone pairs,resulting in a linear geometry. The table incorrectly states $sp^2$ and $1$ lone pair.
$III$: $SF_4$ has $S$ in $sp^3d$ hybridization with $1$ lone pair,resulting in a see-saw geometry. The table incorrectly states $dsp^3$ and square planar geometry.
$IV$: $SnCl_2$ has $Sn$ in $sp^2$ hybridization with $1$ lone pair,resulting in an angular (bent) geometry. The table incorrectly states $sp$ and linear geometry.
Therefore,only set $I$ is correct.

(D) The positively charged carbon (carbocation) is bonded to three other atoms and has no lone pairs. The number of hybrid orbitals = $3 \text{ (bond pairs)} + 0 \text{ (lone pairs)} = 3$,which corresponds to $sp^2$ hybridization.
The negatively charged carbon (carbanion) is bonded to three other atoms and has one lone pair. The number of hybrid orbitals = $3 \text{ (bond pairs)} + 1 \text{ (lone pair)} = 4$,which corresponds to $sp^3$ hybridization.
Therefore,the hybridization is $sp^2$ and $sp^3$ respectively.

(A) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{V - M - C + A}{2}$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$(A)$ For $XeF_2$: $V=8, M=2, C=0, A=0$. $\text{Lone pairs} = \frac{8-2}{2} = 3$. Thus,$A-III$.
$(B)$ For $XeO_3$: $V=8, M=0$ (Oxygen is divalent),$C=0, A=0$. $\text{Lone pairs} = \frac{8-0}{2} = 4$ valence electrons involved in bonding,but since $O$ is divalent,$3$ double bonds use $6$ electrons,leaving $1$ lone pair. Thus,$B-IV$.
$(C)$ For $XeF_4$: $V=8, M=4, C=0, A=0$. $\text{Lone pairs} = \frac{8-4}{2} = 2$. Thus,$C-I$.
$(D)$ For $PF_6^-$: $V=5, M=6, C=0, A=1$. $\text{Lone pairs} = \frac{5-6+1}{2} = 0$. Thus,$D-II$.
The correct matching is $A-III, B-IV, C-I, D-II$.

$ (A) $ The shapes of the given molecules/ions are determined by $VSEPR$ theory:
$A. I_3^-$: The central iodine atom has $3$ lone pairs and $2$ bond pairs, resulting in a linear geometry ($sp^3d$ hybridization). Thus, $A-4$.
$B. ClF_3$: The central chlorine atom has $2$ lone pairs and $3$ bond pairs, resulting in a $T$-shaped geometry ($sp^3d$ hybridization). Thus, $B-1$.
$C. H_2O$: The central oxygen atom has $2$ lone pairs and $2$ bond pairs, resulting in an angular or bent geometry ($sp^3$ hybridization). Thus, $C-3$.
$D. SF_4$: The central sulfur atom has $1$ lone pair and $4$ bond pairs, resulting in a see-saw geometry ($sp^3d$ hybridization). Thus, $D-2$.
Therefore, the correct matching is $A-4, B-1, C-3, D-2$.

(B) To determine the hybridization,we calculate the steric number $(SN)$ using the formula: $SN = \text{number of sigma bonds} + \text{number of lone pairs}$.
$1$. For $ClF_3$: The central atom $Cl$ has $3$ bond pairs and $2$ lone pairs. $SN = 3 + 2 = 5$,which corresponds to $sp^3d$ hybridization.
$2$. For $NH_3$: The central atom $N$ has $3$ bond pairs and $1$ lone pair. $SN = 3 + 1 = 4$,which corresponds to $sp^3$ hybridization.
$3$. For $SO_3$: The central atom $S$ has $3$ bond pairs and $0$ lone pairs. $SN = 3 + 0 = 3$,which corresponds to $sp^2$ hybridization.
Thus,the hybridizations are $sp^3d, sp^3, sp^2$ respectively.

(B) . $dsp^2$ hybridisation corresponds to a square planar shape.
$B$. $sp^3$ hybridisation corresponds to a tetrahedral shape.
$C$. $d^2sp^3$ hybridisation corresponds to an octahedral shape.
$D$. $sp^3d$ hybridisation corresponds to a trigonal bipyramidal shape.
Therefore,the correct matching is $A-I, B-II, C-III, D-IV$.

(B) $Sn$ and $Pb$ both belong to group $14$ of the periodic table.
Both $SnCl_2$ and $PbCl_2$ have a central atom with $2$ bond pairs and $1$ lone pair,resulting in a bent or angular molecular geometry.
Therefore,they are isostructural.

(A) In the gas phase, the water molecule $(H_2O)$ has a bent geometry.
Experimental data shows that the $O-H$ bond length is $95.7 \ pm$ and the bond angle is $104.5^{\circ}$.

(C) The bond order $(B.O.)$ is calculated as $B.O. = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For $C_2$ ($12$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2$. $B.O. = \frac{8 - 4}{2} = 2$. Thus,$x = 2$.
For $B_2$ ($10$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1, \pi 2p_y^1$. $B.O. = \frac{6 - 4}{2} = 1$. Since $x = 2$,$B.O. = \frac{1}{2} x = 1$.
For $O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1, \pi^* 2p_y^1$. $B.O. = \frac{10 - 6}{2} = 2$. Since $x = 2$,$B.O. = x = 2$.
Therefore,the bond orders for $B_2$ and $O_2$ are $\frac{1}{2} x$ and $x$,respectively.

(A) According to $VSEPR$ theory,the bond angle depends on the hybridization and the number of lone pairs on the central atom.1. $SO_2$ $(IV)$: The central atom $S$ is $sp^2$ hybridized with one lone pair. The bond angle is approximately $119.5^\circ$.2. $CH_4$ $(III)$: The central atom $C$ is $sp^3$ hybridized with no lone pairs. It has a perfect tetrahedral bond angle of $109.5^\circ$.3. $NH_3$ $(II)$: The central atom $N$ is $sp^3$ hybridized with one lone pair. Due to lone pair-bond pair repulsion,the bond angle reduces to approximately $107^\circ$.4. $H_2O$ $(I)$: The central atom $O$ is $sp^3$ hybridized with two lone pairs. Due to greater lone pair-lone pair repulsion,the bond angle reduces further to approximately $104.5^\circ$.Therefore,the correct order is: $SO_2$ $(IV)$ $> CH_4$ $(III)$ $> NH_3$ $(II)$ $> H_2O$ $(I)$.

(A) In the $P_4$ molecule,the four phosphorus atoms are arranged at the corners of a regular tetrahedron. The bond angle between any two $P-P$ bonds is $60^{\circ}$.
In the cyclo $S_8$ molecule,the sulfur atoms are arranged in a puckered ring structure often referred to as a crown structure. The bond angle between $S-S-S$ bonds in this structure is approximately $107^{\circ}$.

(A) The bond length is inversely proportional to the bond order.
$1$. In $CO$,the bond order is $3$ $(C \equiv O)$.
$2$. In $CO_2$,the bond order is $2$ $(O=C=O)$.
$3$. In $CO_3^{2-}$,the bond order is $1.33$ due to resonance.
Since the bond order follows the order $CO > CO_2 > CO_3^{2-}$,the bond length follows the order $CO < CO_2 < CO_3^{2-}$.

(D) The bond angles for the given molecules are as follows:
$P_4$: $60^\circ$
$S_6$: $102^\circ$
$S_8$: $107^\circ$
$O_3$: $116^\circ$
Therefore,the increasing order of bond angles is $P_4 < S_6 < S_8 < O_3$.

(C) For multi-electron atoms,the energy of the orbitals increases with an increase in their $(n+l)$ value.
According to the $(n+l)$ rule:
For $5p$: $n=5, l=1$,$(n+l) = 6$
For $6s$: $n=6, l=0$,$(n+l) = 6$
For $4f$: $n=4, l=3$,$(n+l) = 7$
For $5d$: $n=5, l=2$,$(n+l) = 7$
When $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.
Comparing $5p$ and $6s$: $5p$ has lower $n$,so $5p < 6s$.
Comparing $4f$ and $5d$: $4f$ has lower $n$,so $4f < 5d$.
Thus,the correct increasing order of energy is: $5p < 6s < 4f < 5d$.

(A) Due to the inert pair effect,the stability of the $+1$ oxidation state increases down the group $13$ elements because the $ns^2$ electrons become more reluctant to participate in bonding.
Thus,the stability order for the $+1$ state is $Tl^{+} > In^{+} > Ga^{+}$.
Conversely,the stability of the $+3$ oxidation state decreases down the group.
Thus,the stability order for the $+3$ state is $Ga^{3+} > In^{3+} > Tl^{3+}$.
Therefore,the correct order is $Tl^{+} > In^{+} > Ga^{+}$ and $Ga^{3+} > In^{3+} > Tl^{3+}$.

(C) Covalent substances are generally poor conductors of electricity because they do not contain free electrons or ions to carry charge. Therefore,being a good conductor of electricity is not a property of covalent substances.

(C) Ionic bonds are electrostatic forces of attraction that are non-directional in nature,meaning they act in all directions equally. Thus,Assertion $(A)$ is correct.
Ionic compounds are polar in nature and are soluble in polar solvents (like water) due to ion-dipole interactions. They are generally insoluble in nonpolar solvents. Thus,Reasoning $(R)$ is incorrect.

(D) The structure of ozone $(O_3)$ consists of a central oxygen atom double-bonded to one terminal oxygen atom and single-bonded to another terminal oxygen atom.
Using the formal charge formula: $FC = V - L - \frac{1}{2}B$,where $V$ is the number of valence electrons,$L$ is the number of lone pair electrons,and $B$ is the number of bonding electrons.
For the central oxygen atom: $FC = 6 - 2 - \frac{1}{2}(6) = +1$.
For the terminal oxygen atom with a double bond: $FC = 6 - 4 - \frac{1}{2}(4) = 0$.
For the terminal oxygen atom with a single bond: $FC = 6 - 6 - \frac{1}{2}(2) = -1$.
Thus,the formal charges on the terminal oxygen atoms are $0$ and $-1$.

(B) The dipole moments of the given molecules are as follows:
$1. \ H_2O$: It has a bent geometry with a high dipole moment of $1.85 \ D$ $(A-IV)$.
$2. \ BF_3$: It has a trigonal planar geometry,making it a symmetric molecule with a net dipole moment of $0 \ D$ $(B-I)$.
$3. \ NH_3$: It has a trigonal pyramidal geometry with a dipole moment of $1.47 \ D$ $(C-III)$.
$4. \ NF_3$: It also has a trigonal pyramidal geometry,but due to the high electronegativity of fluorine,the bond dipoles oppose the lone pair dipole,resulting in a lower dipole moment of $0.23 \ D$ $(D-II)$.
Therefore,the correct matching is $A-IV, B-I, C-III, D-II$.

(A) The reaction is: $H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$
Initial moles: $H_2O = 1.0 \ mol$,$CO = 1.0 \ mol$,$H_2 = 0 \ mol$,$CO_2 = 0 \ mol$.
Given that $40 \%$ of water reacts,the moles of $H_2O$ reacted $= 0.4 \times 1.0 = 0.4 \ mol$.
According to the stoichiometry of the reaction,$0.4 \ mol$ of $CO$ will also react,and $0.4 \ mol$ of $H_2$ and $0.4 \ mol$ of $CO_2$ will be formed.
Moles at equilibrium:
$n(H_2O) = 1.0 - 0.4 = 0.6 \ mol$
$n(CO) = 1.0 - 0.4 = 0.6 \ mol$
$n(H_2) = 0.4 \ mol$
$n(CO_2) = 0.4 \ mol$
Since the volume of the flask is $1 \ L$,the concentrations are equal to the number of moles.
$[H_2O] = 0.6 \ M, [CO] = 0.6 \ M, [H_2] = 0.4 \ M, [CO_2] = 0.4 \ M$
$K_c = \frac{[H_2][CO_2]}{[H_2O][CO]} = \frac{0.4 \times 0.4}{0.6 \times 0.6} = \frac{0.16}{0.36} = 0.444$

(A) For the reaction $a A_{(g)} \rightleftharpoons b B_{(g)}$,the equilibrium constant is $K_c = \frac{[B]^b}{[A]^a}$.
For the reaction $2a A_{(g)} \rightleftharpoons 2b B_{(g)}$,the equilibrium constant is $K_c^{\prime} = \frac{[B]^{2b}}{[A]^{2a}}$.
Comparing the two expressions,we can see that $K_c^{\prime} = \left( \frac{[B]^b}{[A]^a} \right)^2$.
Therefore,$K_c^{\prime} = (K_c)^2$.

(B) The dissociation reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
Initial moles: $1, 0, 0$
Equilibrium moles: $(1-0.1), 0.1, 0.1$
Since the volume is $1 \ L$,the concentrations are $[PCl_{5}] = 0.9 \ M$,$[PCl_{3}] = 0.1 \ M$,and $[Cl_{2}] = 0.1 \ M$.
$K_{c} = \frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]} = \frac{0.1 \times 0.1}{0.9} = \frac{0.01}{0.9} = 0.0111$
For the reaction,$\Delta n_{g} = (1+1) - 1 = 1$.
Using the relation $K_{p} = K_{c}(RT)^{\Delta n_{g}}$:
$K_{p} = 0.0111 \times (0.082 \times 500)^{1}$
$K_{p} = 0.0111 \times 41 = 0.4551 \approx 0.455 \ atm$.

(B) The equilibrium constant $K_C$ for the reaction $A_{(g)} \rightleftharpoons B_{(g)}$ is given by $K_C = \frac{[B]}{[A]} = 10^{-1}$.
Since the temperature remains constant,the value of the equilibrium constant $K_C$ remains unchanged regardless of the addition of reactants or products.
Therefore,at the new equilibrium state,the ratio $\frac{[B]}{[A]}$ will still be equal to $10^{-1}$.
Consequently,the value of $\frac{[A]}{[B]} = \frac{1}{K_C} = \frac{1}{10^{-1}} = 10$.

(C) precipitate of $AgCl$ forms when the ionic product $[Ag^{+}][Cl^{-}]$ exceeds the solubility product constant $K_{sp} = 1 \times 10^{-10} \ M^2$.
When mixing $1 \ L$ of each solution,the final volume becomes $2 \ L$,so the concentrations are halved.
For $(iv)$ and $(v)$:
$[Ag^{+}] = \frac{10^{-3} \ M}{2} = 0.5 \times 10^{-3} \ M$
$[Cl^{-}] = \frac{10^{-5} \ M}{2} = 0.5 \times 10^{-5} \ M$
Ionic product = $(0.5 \times 10^{-3}) \times (0.5 \times 10^{-5}) = 0.25 \times 10^{-8} = 2.5 \times 10^{-9}$.
Since $2.5 \times 10^{-9} > 1 \times 10^{-10}$,a precipitate will form.

(D) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n_g}$.
Given values are: $K_c = 10 \ mol \ L^{-1}$,$T = 1000 \ K$,and $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.
The change in the number of moles of gaseous species is $\Delta n_g = (n_{products} - n_{reactants}) = (1 + 1) - 1 = 1$.
Substituting these values into the formula: $K_p = 10 \times (0.082 \times 1000)^1$.
$K_p = 10 \times 82 = 820 \ atm$.

(C) For the reaction: $A_2B_{6(g)} \rightleftharpoons A_2B_{4(g)} + B_{2(g)}$
Initial pressure: $P_{A_2B_6} = 4 \text{ atm}$
At equilibrium: $P_{A_2B_6} = 4 - x$,$P_{A_2B_4} = x$,$P_{B_2} = x$
$K_p = \frac{P_{A_2B_4} \cdot P_{B_2}}{P_{A_2B_6}} = \frac{x^2}{4-x} = 0.04$
$x^2 = 0.16 - 0.04x$
$x^2 + 0.04x - 0.16 = 0$
Solving the quadratic equation: $x = \frac{-0.04 + \sqrt{(0.04)^2 - 4(1)(-0.16)}}{2} \approx 0.38 \text{ atm}$
Equilibrium pressure of $A_2B_6 = 4 - x = 4 - 0.38 = 3.62 \text{ atm}$.

(D) According to Le Chatelier's principle,adding an inert gas like $He$ at constant volume does not change the partial pressures of the reacting species,and thus the equilibrium position remains unaffected.
Furthermore,for this reaction,the number of moles of gaseous reactants $(1 + 1 = 2)$ is equal to the number of moles of gaseous products $(2)$.
Therefore,even if the total pressure is changed,the equilibrium will not shift because the reaction quotient $Q_c$ remains equal to the equilibrium constant $K_c$.

(C) Addition of oxalic acid $(H_2C_2O_4)$ provides oxalate ions $(C_2O_4^{2-})$,which react with $Fe^{3+}$ ions to form a stable complex $[Fe(C_2O_4)_3]^{3-}$.
This reaction significantly decreases the concentration of free $Fe^{3+}$ ions in the solution.
According to Le Chatelier's principle,the equilibrium will shift to the left to compensate for the loss of $Fe^{3+}$ ions.
As the equilibrium shifts to the left,the concentration of the deep red complex $[Fe(SCN)]^{2+}$ decreases,leading to a decrease in the intensity of the deep red color.

(D) The rate of reaction is given by the expression: $-\frac{1}{2} \frac{d[A]}{dt} = \frac{1}{4} \frac{d[B]}{dt}$.
Given that the rate of increase of concentration of $B$ is $\frac{\Delta[B]}{\Delta t} = \frac{5 \times 10^{-3} \ mol \ L^{-1}}{10 \ s} = 5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
Substituting this into the rate expression: $-\frac{d[A]}{dt} = \frac{2}{4} \times \frac{d[B]}{dt} = \frac{1}{2} \times (5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1})$.
Therefore,the rate of disappearance of $A$ is $2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

(B) The rate of reaction is given by the expression: $-\frac{1}{3} \frac{d[X]}{dt} = \frac{1}{2} \frac{d[Y]}{dt} = \frac{d[Z]}{dt}$.
Given that the rate of disappearance of $X$ is $-\frac{d[X]}{dt} = 7.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
Substituting this into the relation: $\frac{1}{2} \frac{d[Y]}{dt} = \frac{1}{3} \left( -\frac{d[X]}{dt} \right)$.
Therefore,the rate of formation of $Y$ is $\frac{d[Y]}{dt} = \frac{2}{3} \times (7.2 \times 10^{-3}) = 4.8 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

(A) The rate of reaction is defined as the change in concentration of reactant over time: $Rate = -\frac{\Delta[R]}{\Delta t}$.
Given: $\Delta[R] = [R]_f - [R]_i = 0.02 \ mol \ L^{-1} - 0.03 \ mol \ L^{-1} = -0.01 \ mol \ L^{-1}$.
Time $\Delta t = 25 \ min = 25 \times 60 \ s = 1500 \ s$.
$Rate = -\frac{-0.01 \ mol \ L^{-1}}{1500 \ s} = \frac{0.01}{1500} \ mol \ L^{-1} \ s^{-1}$.
$Rate = 6.667 \times 10^{-6} \ mol \ L^{-1} \ s^{-1}$.

(C) Since the concentration of the reactant decreases from $0.6 \ M$ to $0.3 \ M$ (i.e.,halved) in $15 \ min$,the half-life $(t_{1/2})$ of the reaction is $15 \ min$.
For a first-order reaction,the time required for the concentration to change from $[A]_0$ to $[A]$ is given by $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}$.
Given $[A]_0 = 0.1 \ M$ and $[A] = 0.025 \ M$,the ratio $\frac{[A]_0}{[A]} = \frac{0.1}{0.025} = 4 = 2^2$.
This implies that the concentration reduces to $1/4$ of its initial value,which corresponds to two half-lives $(2 \times t_{1/2})$.
Therefore,$t = 2 \times 15 \ min = 30 \ min$.

(D) $Li-Mg$ alloy is used in making armour plates due to its high strength-to-weight ratio.
$Cu-Be$ alloy is used in high-strength springs because of its excellent fatigue resistance and mechanical properties.
$Mg-Al$ alloy (Magnalium) is used in aircraft construction due to its lightweight and high strength properties.
Therefore,all the given pairs are correctly matched.

(C) diagonal relationship is observed between certain elements of the second and third periods.
The pairs that exhibit a diagonal relationship are:
$1. Li \text{ and } Mg$
$2. Be \text{ and } Al$
$3. B \text{ and } Si$
Therefore,there are $3$ such pairs in the given list.
The pair $Al \text{ and } S$ does not exhibit a diagonal relationship.

(C) All the given species are isoelectronic,having $18$ electrons.
In an isoelectronic series,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic numbers are: $K (19)$,$Ca (20)$,$Sc (21)$,and $Ti (22)$.
Since $Ti^{4+}$ has the highest nuclear charge $(Z = 22)$,it exerts the strongest attraction on the electrons,resulting in the smallest ionic radius.
Thus,the order of ionic radii is: $K^{+} > Ca^{2+} > Sc^{3+} > Ti^{4+}$.

(B) Metalloids are elements that exhibit properties intermediate between metals and non-metals. In the given list,the metalloids are $B$ (Boron),$Sb$ (Antimony),$As$ (Arsenic),$Ge$ (Germanium),and $Si$ (Silicon).
Thus,the total number of metalloids is $5$.
Note: $Si$ is classified as a metalloid in most standard chemistry textbooks.

(C) Fluorine $(F)$ is the element with the highest electronegativity,as it is located at the extreme right of the second period in the periodic table.
Francium $(Fr)$ is the most electropositive or least electronegative element,as it is located at the bottom left of the periodic table.

(C) $Na, Mg, Al$ and $Si$ belong to the $3^{rd}$ period of the periodic table.
Generally,the first ionization enthalpy increases across a period due to an increase in effective nuclear charge.
However,$Mg$ $(3s^2)$ has a fully-filled $s$-orbital,making it more stable than $Al$ $(3s^2 3p^1)$.
Therefore,the first ionization enthalpy of $Al$ is lower than that of $Mg$ but higher than that of $Na$.
Given values: $Na = 496 \ kJ \ mol^{-1}$,$Mg = 737 \ kJ \ mol^{-1}$,$Si = 786 \ kJ \ mol^{-1}$.
The value for $Al$ must be between $496$ and $737 \ kJ \ mol^{-1}$.
Thus,the correct value is $575 \ kJ \ mol^{-1}$.

(C) As we move down the group $14$ in the periodic table,the number of electron shells increases.
This increase in the number of shells leads to an increase in the atomic size and covalent radii.
Therefore,the correct order of covalent radii for $Si$,$Ge$,and $Sn$ is $Si < Ge < Sn$.

(B) $1$. For the formation of $X$: Benzoic acid reacts with $B_2H_6$ followed by hydrolysis $(H_3O^+)$ to undergo reduction,yielding benzyl alcohol $(C_6H_5CH_2OH)$ as $X$.
$2$. For the formation of $Y$: Benzoic acid reacts with $SOCl_2$ to form benzoyl chloride $(C_6H_5COCl)$,which then undergoes Rosenmund reduction using $H_2/Pd-BaSO_4$ to yield benzaldehyde $(C_6H_5CHO)$ as $Y$.
$3$. Therefore,$X$ is benzyl alcohol and $Y$ is benzaldehyde.

(A) The preparation of $1$-ethyl cyclohexanol involves the acid-catalyzed hydration of ethylidene cyclohexane.
$1$. Ethylidene cyclohexane $(C_8H_{14})$ reacts with $H_3O^+$ (acid catalyst).
$2$. The double bond is protonated to form a stable tertiary carbocation at the $1$-position of the cyclohexane ring.
$3$. Water then attacks the carbocation,followed by deprotonation to yield $1$-ethyl cyclohexanol.
Thus,the correct reactants are ethylidene cyclohexane and $H_3O^+$.

(D) $LiAlH_4$ acts as a source of hydride ion $(H^-)$,which attacks the electrophilic carbonyl carbon of cyclohexanone to form an alkoxide intermediate.
In the second step,$D_2O$ acts as a source of deuterium $(D^+)$,which protonates (deuterates) the alkoxide oxygen to form the final product,which is a cyclohexanol derivative with an $-OD$ group and a hydrogen atom attached to the alpha carbon.

(B) The dehydration of alcohols to alkenes requires specific conditions depending on the degree of substitution of the alcohol.
Primary $(1^{\circ})$ alcohols,like butan$-1-$ol in option $(B)$,are the least reactive towards dehydration and typically require concentrated acid (e.g.,$95\% \ H_2SO_4$) and high temperatures (around $443 \ K$) to proceed.
Option $(B)$ suggests the dehydration of a primary alcohol at a very low temperature $(300 \ K)$ with a lower concentration of acid $(75\% \ H_2SO_4)$,which is insufficient to drive the elimination reaction.
Secondary and tertiary alcohols are more reactive and can undergo dehydration under milder conditions,making options $(C)$ and $(D)$ feasible.

(C) The conversion of toluene $(C_6H_5CH_3)$ to benzoic acid $(C_6H_5CO_2H)$ requires strong oxidation,which is achieved using alkaline $KMnO_4$ followed by acidic workup $(H_3O^{+})$.
Thus,$X = (i) \ KMnO_4 / OH^{-}, \Delta, (ii) \ H_3O^{+}$.
The reduction of benzoic acid to benzyl alcohol $(C_6H_5CH_2OH)$ is selectively performed by diborane $(B_2H_6)$ followed by acidic workup $(H_3O^{+})$,as $B_2H_6$ does not reduce the aromatic ring.
Thus,$Y = (i) \ B_2H_6, (ii) \ H_3O^{+}$.
Therefore,the correct option is $C$.

(D) The reaction of ethanol with phosphorus pentachloride $(PCl_5)$ is a standard method for the preparation of alkyl halides.
The chemical equation is:
$CH_3CH_2OH + PCl_5 \rightarrow CH_3CH_2Cl + HCl + POCl_3$
Here,the products formed are chloroethane $(CH_3CH_2Cl)$,hydrogen chloride $(HCl)$,and phosphorus oxychloride $(POCl_3)$.

(C) When the vapours of a primary alcohol are passed over heated copper at $573 \ K$,dehydrogenation takes place and an aldehyde is formed: $R-CH_2OH \xrightarrow{Cu/573 \ K} R-CHO + H_2$.
In the case of tertiary alcohols,dehydration takes place to form an alkene: $(CH_3)_3C-OH \xrightarrow{Cu/573 \ K} (CH_3)_2C=CH_2 + H_2O$.
Thus,$X$ is an aldehyde $(RCHO)$ and $Y$ is an alkene $((CH_3)_2C=CH_2)$.

(B) The presence of electron-withdrawing groups like $-NO_2$ at ortho and para positions increases the reactivity of haloarenes towards nucleophilic substitution. As the number of $-NO_2$ groups increases,the reaction conditions become milder.
$1.$ $1$-$Chloro$-$4$-$nitrobenzene$ requires $(i) \ NaOH, 443 \ K, (ii) \ H^{+}$ to form $4$-$nitrophenol$.
$2.$ $1$-$Chloro$-$2,4$-$dinitrobenzene$ requires $(i) \ NaOH, 368 \ K, (ii) \ H^{+}$ to form $2,4$-$dinitrophenol$.
$3.$ $1$-$Chloro$-$2,4,6$-$trinitrobenzene$ reacts with just $\text{Warm } H_2O$ to form $2,4,6$-$trinitrophenol$ (picric acid).

(B) $1$. Reaction with $Zn$ dust and heat: $p-Cresol$ $(p-methylphenol)$ reacts with $Zn$ dust to undergo reduction,where the $-OH$ group is removed and replaced by a hydrogen atom,resulting in the formation of $Toluene$ $(X)$.
$2$. Reaction with $(CH_3CO)_2O$ followed by $H^+$: $p-Cresol$ reacts with acetic anhydride $(CH_3CO)_2O$ to undergo acetylation of the phenolic $-OH$ group,forming $p-acetoxy-toluene$ $(Y)$.

(A) When $p$-cresol ($4$-methylphenol) is heated with zinc dust,it undergoes a reduction reaction.
In this reaction,the phenolic $-OH$ group is removed and replaced by a hydrogen atom.
This process converts the phenol derivative into the corresponding hydrocarbon.
Thus,$p$-cresol reacts with zinc dust to form toluene $(C_6H_5CH_3)$ and zinc oxide $(ZnO)$.
The reaction is: $CH_3-C_6H_4-OH + Zn \rightarrow CH_3-C_6H_5 + ZnO$.

(C) When phenol is treated with concentrated nitric acid $(Conc. HNO_3)$,it undergoes electrophilic aromatic substitution to form $2,4,6$-trinitrophenol,which is commonly known as picric acid.

(D) The reaction of phenol with $CHCl_3$ and aqueous $NaOH$ followed by acidification is the Reimer-Tiemann reaction,which yields $2$-hydroxybenzaldehyde (salicylaldehyde) as product $X$.
The reaction of phenol with $NaOH$ followed by $CO_2$ and then acidification is the Kolbe-Schmitt reaction,which yields $2$-hydroxybenzoic acid (salicylic acid) as product $Y$.
Therefore,$X$ is $2$-hydroxybenzaldehyde and $Y$ is $2$-hydroxybenzoic acid.

(B) The reaction involves a tertiary alkyl halide,$(CH_3)_3C-Cl$,and a strong base,$C_2H_5ONa$ (sodium ethoxide).
Since the alkyl halide is sterically hindered (tertiary),the $S_N2$ pathway is unfavorable.
Instead,the base acts as a proton acceptor,leading to an $E2$ elimination reaction.
The base abstracts a proton from one of the $\beta$-carbon atoms,resulting in the formation of an alkene.
The major product is $2$-methylpropene,which is $(CH_3)_2C=CH_2$.

(C) $1$. Reaction with $Br_2$ in $CH_3COOH$: Anisole $(C_6H_5OCH_3)$ undergoes electrophilic aromatic substitution. The $-OCH_3$ group is ortho/para directing. Due to steric hindrance,the para-isomer is the major product $(X)$,which is $p$-bromoanisole.
$2$. Reaction with $HI$ and $\Delta$: Anisole reacts with $HI$ to undergo cleavage of the $C-O$ bond. Since the phenyl group is attached to the oxygen,the $C(phenyl)-O$ bond has partial double bond character and is difficult to break. Thus,the $C(methyl)-O$ bond breaks,resulting in the formation of phenol $(C_6H_5OH)$ and methyl iodide $(CH_3I)$ as $Y$.

(B) The reaction of toluene with $Cl_2$ in the presence of $hv$ (photochemical chlorination) leads to the formation of benzal chloride $(C_6H_5CHCl_2)$ as the intermediate $X$,which upon hydrolysis at $373 \ K$ gives benzaldehyde.
The reaction of toluene with chromyl chloride $(CrO_2Cl_2)$ in $CS_2$ is the Etard reaction. This reaction proceeds through the formation of a brown chromium complex intermediate $Y$,which is $C_6H_5CH[OCr(OH)Cl_2]_2$. This complex upon acidic hydrolysis $(H_3O^+)$ yields benzaldehyde.
Therefore,$X = C_6H_5CHCl_2$ and $Y = C_6H_5CH[OCr(OH)Cl_2]_2$.

(A) The reaction of $p$-nitrobenzaldehyde with ethanol in the presence of dry $HCl(g)$ is an acetal formation reaction.
Aldehydes react with alcohols in the presence of dry $HCl$ to form hemiacetals,which further react with another molecule of alcohol to form acetals.
Step $1$: $p$-Nitrobenzaldehyde reacts with $C_2H_5OH$ in the presence of $HCl(g)$ to form the diethyl acetal,$X$,which is $p-NO_2-C_6H_4-CH(OC_2H_5)_2$.
Step $2$: The acetal $X$ on hydrolysis with dilute $HCl$ (acidic hydrolysis) regenerates the original aldehyde,$Y$,which is $p-nitrobenzaldehyde$ $(p-NO_2-C_6H_4-CHO)$.
Therefore,$X$ is the diethyl acetal and $Y$ is the original aldehyde.

(C) The reaction between benzaldehyde $(C_6H_5CHO)$ and acetophenone $(C_6H_5COCH_3)$ in the presence of a base $(OH^-)$ is a Claisen-Schmidt condensation reaction.
This reaction produces an $\alpha,\beta$-unsaturated ketone,which is $X = C_6H_5CH=CHCOC_6H_5$ (benzalacetophenone).
Next,$NaBH_4$ is a selective reducing agent that reduces the carbonyl group $(C=O)$ to a hydroxyl group $(-OH)$ without affecting the carbon-carbon double bond $(C=C)$.
Therefore,the reduction of $C_6H_5CH=CHCOC_6H_5$ with $NaBH_4$ yields $Y = C_6H_5CH=CHCH(OH)C_6H_5$ ($1$,$3$-diphenylprop$-2-$en$-1-$ol).
Thus,the correct option is $C$.

(D) The reaction involves $CH_3CHO$ (acetaldehyde) and $C_6H_5COCH_3$ (acetophenone) in the presence of $NaOH$ and heat. Both compounds contain $\alpha$-hydrogens,so they undergo self-aldol condensation and cross-aldol condensation.
$1$. Self-aldol condensation of $CH_3CHO$ gives one product.
$2$. Self-aldol condensation of $C_6H_5COCH_3$ gives one product.
$3$. Cross-aldol condensation of $CH_3CHO$ (as donor) and $C_6H_5COCH_3$ (as acceptor) gives one product.
$4$. Cross-aldol condensation of $C_6H_5COCH_3$ (as donor) and $CH_3CHO$ (as acceptor) gives one product.
Thus,a total of $4$ products are obtained.

(A) hemiacetal is a functional group characterized by a carbon atom bonded to one $-OH$ group,one $-OR$ group,and two other carbon-containing groups (or hydrogen atoms).
In the given options,the structure in option $A$ shows a carbon atom bonded to an $-OH$ group and an ether oxygen atom within a ring,which fits the definition of a cyclic hemiacetal.

(C) The given reaction is the Cannizzaro reaction. Benzaldehyde $(C_6H_5CHO)$ does not have an $\alpha$-hydrogen atom,so it undergoes a self-oxidation and reduction (disproportionation) reaction in the presence of a concentrated base like $NaOH$.
One molecule of benzaldehyde is reduced to benzyl alcohol $(C_6H_5CH_2OH)$,and another molecule is oxidized to sodium benzoate $(C_6H_5COONa)$.
Thus,the products '$X$' and '$Y$' are benzyl alcohol and sodium benzoate.

(A) Isobutyraldehyde is an aldehyde,which can be oxidized to the corresponding carboxylic acid (isobutyric acid) using oxidizing agents.
$HNO_3$ is a strong oxidizing agent.
$2[Ag(NH_3)_2]^+$ (Tollen's reagent) is a mild oxidizing agent that specifically oxidizes aldehydes to carboxylic acids.
$NH_2NH_2 / OH^-$ (Wolff-Kishner reduction) reduces aldehydes to alkanes.
$NaOH$ typically causes aldol condensation or Cannizzaro reaction (if no $\alpha$-hydrogen is present) but does not oxidize aldehydes to acids.
Therefore,reagents $I$ and $III$ are the correct oxidizing agents.

(A) Aldehydes that do not have $\alpha$-hydrogens undergo the Cannizzaro reaction.
$1$. Methanal $(HCHO)$: It has no $\alpha$-carbon,hence no $\alpha$-hydrogen.
$2$. Trichloro ethanal $(CCl_3CHO)$: The carbon adjacent to the carbonyl group is bonded to three chlorine atoms,so there is no $\alpha$-hydrogen.
$3$. Phenyl ethanal $(C_6H_5CH_2CHO)$: It has two $\alpha$-hydrogens on the carbon attached to the phenyl ring.
$4$. $2-$Methoxy propanal $(CH_3OCH(CH_3)CHO)$: It has one $\alpha$-hydrogen on the carbon at the $2-$position.
Thus,only Methanal and Trichloro ethanal undergo the Cannizzaro reaction. The total number is $2$.

(D) The given reaction is the oxidation of an aldehyde (propanal) by Fehling's solution.
Fehling's solution is an alkaline solution of copper$(II)$ ions complexed with tartrate ions.
Aldehydes are oxidized to the corresponding carboxylate anions $(RCOO^{-})$ in the presence of base,while $Cu^{2+}$ ions are reduced to red-brown precipitate of copper$(I)$ oxide $(Cu_2O)$.
The balanced equation is:
$CH_3CH_2CHO + 2Cu^{2+} + 5OH^{-} \rightarrow CH_3CH_2COO^{-} + Cu_2O + 3H_2O$
Comparing this with the given reaction,$X = CH_3CH_2COO^{-}$ and $Y = Cu_2O$.

(C) The reaction of aniline with $CHCl_3$ and $KOH$ (carbylamine reaction) produces phenyl isocyanide $(C_6H_5NC)$ as $X$.
The reaction of aniline with $NaNO_2/HCl$ followed by $CuCN/KCN$ (Sandmeyer reaction) produces phenyl cyanide $(C_6H_5CN)$ as $Y$.
Therefore,$X$ is phenyl isocyanide and $Y$ is phenyl cyanide.

(A) The boiling points of isomeric amines follow the order: Primary $(1^{\circ})$ > Secondary $(2^{\circ})$ > Tertiary $(3^{\circ})$.
This is because primary amines $(R-NH_2)$ have two hydrogen atoms available for intermolecular hydrogen bonding,secondary amines $(R_2-NH)$ have one,and tertiary amines $(R_3-N)$ have none.
$I.$ $N$-Ethylethanamine (Secondary amine,$2^{\circ}$)
$II.$ Butanamine (Primary amine,$1^{\circ}$)
$III.$ $N,N$-Dimethylethanamine (Tertiary amine,$3^{\circ}$)
Therefore,the increasing order of boiling points is $III < I < II$.

(A) The conversion of aniline to benzene nitrile involves two steps:
$1$. Diazotization: Aniline reacts with $NaNO_2$ and $HCl$ at $273-278 \ K$ to form benzene diazonium chloride.
$2$. Sandmeyer reaction: Benzene diazonium chloride reacts with $CuCN / KCN$ to form benzene nitrile (cyanobenzene).

(B) secondary amine reacts with Hinsberg reagent (benzene sulphonyl chloride) to form a sulfonamide that has no acidic hydrogen,making it insoluble in alkali.
Thus,'$X$' is a secondary amine,$C_6H_5NHCH_3$ ($N$-methylaniline).
The reaction of $C_6H_5NHCH_3$ with ethanoyl chloride $(CH_3COCl)$ is an acetylation reaction:
$C_6H_5NHCH_3 + CH_3COCl \rightarrow C_6H_5N(CH_3)COCH_3 + HCl$
The product formed is $N$-methyl-$N$-phenylacetamide,which corresponds to option $B$.

(C) Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines.
It cannot be used for the preparation of aromatic primary amines (like Phenylamine or Aniline) because the aryl halides do not undergo nucleophilic substitution reaction with the potassium phthalimide salt easily due to the partial double bond character of the $C-X$ bond in aryl halides.

(A) $(i)$ Aniline is highly reactive towards electrophilic substitution. When treated with $Br_2$ in $H_2O$,it undergoes rapid poly-substitution to form $2, 4, 6-$tribromoaniline as the major product $(X)$.
$(ii)$ To control the reactivity of the $-NH_2$ group,it is first acetylated using acetic anhydride $(CH_3CO)_2O$ to form acetanilide. The $-NHCOCH_3$ group is less activating than the $-NH_2$ group,which restricts the bromination to the $p-$position,yielding $p-$bromoacetanilide as the major product $(Y)$.

(C) The reduction of nitrobenzene depends on the medium used:
$(i)$ In neutral medium,using $Zn$ dust and $NH_4Cl$ solution,nitrobenzene is reduced to phenylhydroxylamine $(C_6H_5NHOH)$. Thus,$X$ is phenylhydroxylamine.
(ii) In alkaline medium,using $Zn$ and $KOH/C_2H_5OH$,the reduction proceeds further to form hydrazobenzene $(C_6H_5NH-NHC_6H_5)$. Thus,$Y$ is hydrazobenzene.
Therefore,the correct option is $C$.

(A) The reaction between benzene diazonium chloride $(C_6H_5N_2Cl)$ and aniline $(C_6H_5NH_2)$ in the presence of a mild acid $(H^+)$ is a coupling reaction.
This is a type of electrophilic aromatic substitution reaction where the diazonium cation acts as an electrophile.
The electrophile attacks the electron-rich ring of aniline,primarily at the para-position due to steric hindrance at the ortho-position.
The product formed is $p$-aminoazobenzene $(C_6H_5-N=N-C_6H_4-NH_2)$.

(D) Primary aliphatic amines $(R-NH_2)$ react with nitrous acid $(HNO_2)$ to form unstable diazonium salts,which decompose in the presence of water to release nitrogen gas $(N_2 \uparrow)$.
$C_2 H_5 NH_2 + HNO_2$ $\xrightarrow{0-5^{\circ}C} [C_2 H_5 N_2^+ Cl^-]$ $\xrightarrow{H_2 O} C_2 H_5 OH + N_2 \uparrow + HCl$
Secondary amines form $N$-nitrosoamines (yellow oily liquids),and tertiary amines form salts with nitrous acid without evolving nitrogen gas.
Therefore,$C_2 H_5 NH_2$ (a primary amine) is the correct answer.

(B) $1$. The reaction of aniline with acetic anhydride in the presence of pyridine is an acylation reaction,which converts the $-NH_2$ group into an acetamido group $(-NHCOCH_3)$. The product $X$ is acetanilide.
$2$. Acetanilide undergoes electrophilic aromatic substitution (bromination) with $Br_2$ in $CH_3COOH$. The $-NHCOCH_3$ group is ortho/para directing. Due to steric hindrance,the para-isomer is the major product. The product $Y$ is $p$-bromoacetanilide.

(A) Step $1$: Oxidation of benzylamine $(C_6H_5CH_2NH_2)$ to benzamide $(C_6H_5CONH_2)$ is not a standard single-step reaction. However,considering the options provided,the conversion of the side chain to an amide group typically involves oxidation. $KMnO_4 / H^{+} / \Delta$ is a strong oxidizing agent capable of oxidizing the benzylic carbon to a carboxylic acid,which can then be converted to an amide.
Step $2$: The conversion of benzamide $(C_6H_5CONH_2)$ to aniline $(C_6H_5NH_2)$ is the classic $Hoffmann$ bromamide degradation reaction,which uses $Br_2 / KOH$ (or $Br_2 / NaOH$).
Therefore,$X$ is $KMnO_4 / H^{+} / \Delta$ and $Y$ is $Br_2 / KOH$.

(C) The conversion of sucrose into glucose and fructose is catalyzed by the enzyme $\text{invertase}$,not $\text{zymase}$.
$\text{Zymase}$ is the enzyme responsible for the conversion of glucose or fructose into ethanol and carbon dioxide.
Therefore,the pair in option $C$ is incorrectly matched.

(B) dipeptide is formed by the condensation reaction between the carboxyl group $(-COOH)$ of one amino acid and the amino group $(-NH_2)$ of another amino acid.
In the dipeptide $Gly-Ala$,$Gly$ (Glycine) is the $N$-terminal amino acid and $Ala$ (Alanine) is the $C$-terminal amino acid.
Glycine structure: $H_2N-CH_2-COOH$
Alanine structure: $H_2N-CH(CH_3)-COOH$
When $Gly$ and $Ala$ combine,the $-COOH$ group of $Gly$ reacts with the $-NH_2$ group of $Ala$ to form a peptide bond $(-CO-NH-)$.
The resulting structure is $H_2N-CH_2-CO-NH-CH(CH_3)-COOH$.

(D) $S$-containing amino acids are Cysteine and Methionine.
Serine is an $OH$-containing amino acid.
Lysine is a basic amino acid containing an extra $-NH_2$ group.
Therefore,the correct options are $(B)$ and $(D)$.

(A) In $D-glucose$ $(X)$,the hydroxyl group at $C-5$ attacks the aldehyde group at $C-1$ to form a six-membered pyranose ring (hemiacetal).
In $D-fructose$ $(Y)$,the hydroxyl group at $C-5$ attacks the ketone group at $C-2$ to form a five-membered furanose ring (hemiketal).
Therefore,in both cases,the hydroxyl group at $C-5$ participates in the ring formation.

(C) To determine which amino acids contain an $-OH$ (hydroxyl) group in their side chain,let us examine the structures:
$1$. Lysine: Contains an amino group $(-NH_2)$ in its side chain.
$2$. Serine: Contains a hydroxymethyl group $(-CH_2OH)$ in its side chain,which includes an $-OH$ group.
$3$. Tyrosine: Contains a phenolic group $(-C_6H_4OH)$ in its side chain,which includes an $-OH$ group.
$4$. Valine: Contains an isopropyl group $(-CH(CH_3)_2)$ in its side chain,which does not contain an $-OH$ group.
Therefore,both Serine and Tyrosine contain an $-OH$ group. The correct option is $(C)$.

(D) Essential amino acids cannot be synthesized in our body,so we need to take them in our diet.
Amino acids are soluble in water but generally insoluble in organic solvents like ether.
Due to their crystalline nature,they are high melting solids because of strong dipole-dipole interactions.
In aqueous solutions,they exist as dipolar zwitter ions,as shown below:
$H_3N^+-CH(R)-COO^-$

(B) Fibrous proteins are long,thread-like structures that are insoluble in water. Examples include $Keratin$ (found in hair,wool,skin) and $Myosin$ (found in muscles).
Globular proteins have a spherical shape and are generally soluble in water. Examples include $Insulin$ and $Albumin$.
Therefore,$Keratin$ $(A)$ and $Myosin$ $(C)$ are fibrous proteins.

(D) nucleoside consists of a nitrogenous base attached to a pentose sugar.
In $RNA$,the pentose sugar is ribose,which has a hydroxyl $(-OH)$ group at both the $2'$ and $3'$ positions.
$RNA$ contains the nitrogenous bases Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Uracil $(U)$.
Thymine $(T)$ is found in $DNA$,not $RNA$.
Looking at the options:
Option $A$ has $T$ (Thymine),which is characteristic of $DNA$.
Option $B$ and $C$ show a deoxyribose sugar (missing the $-OH$ group at the $2'$ position),which is characteristic of $DNA$.
Option $D$ shows a ribose sugar (with $-OH$ groups at both $2'$ and $3'$ positions) attached to Uracil $(U)$,which is a characteristic nucleoside of $RNA$ (Uridine).

(D) According to Chargaff's rule,in a double-stranded $DNA$ molecule,the amount of adenine $(A)$ is equal to the amount of thymine $(T)$,and the amount of guanine $(G)$ is equal to the amount of cytosine $(C)$.
This is because $A$ pairs with $T$ via two hydrogen bonds,and $G$ pairs with $C$ via three hydrogen bonds.
Therefore,the Assertion $(A)$ is incorrect because hydrolysis of $DNA$ yields equal amounts of $A$ and $T$,and $G$ and $C$.
The Reason $(R)$ is correct as it describes the base-pairing mechanism.
Thus,the correct option is $D$.

(A) nucleoside is formed by the attachment of a nitrogenous base to the $1'$ position of a pentose sugar through an $N$-glycosidic linkage.
As shown in the structure,the nitrogenous base is attached to the $C-1'$ carbon atom of the sugar molecule.

(A) In the given structures:
$A$ is Adenine,$B$ is Uracil,$C$ is Cytosine,and $D$ is Thymine.
Adenine $(A)$,Guanine $(G)$,and Cytosine $(C)$ are present in both $DNA$ and $RNA$.
Thymine $(T)$ is present only in $DNA$,and Uracil $(U)$ is present only in $RNA$.
Therefore,the bases present in both are Adenine $(A)$ and Cytosine $(C)$.

(A) In $DNA$,the nitrogenous bases form specific hydrogen-bonded pairs.
Adenine $(A)$ pairs with Thymine $(T)$ through two hydrogen bonds $(A=T)$.
Guanine $(G)$ pairs with Cytosine $(C)$ through three hydrogen bonds $(G \equiv C)$.
Thus,the correct representation is $G \equiv C$ and $T=A$.

(C) nucleoside is composed of a pentose sugar and a nitrogenous base.
$A$ nucleotide is composed of a nucleoside and a phosphate group.
Option $A$ shows a nucleotide (cytidine monophosphate).
Option $B$ shows a sugar phosphate.
Option $C$ shows a nucleoside (uridine),which consists of a nitrogenous base (uracil) attached to a pentose sugar (ribose).
Option $D$ shows a pentose sugar (ribose).

(D) Deficiency of vitamin $B_6$ (pyridoxine) leads to neurological symptoms,including convulsions.

(A) Scurvy is caused by the deficiency of Vitamin $C$ (ascorbic acid).
Vitamin $C$ is abundantly found in citrus fruits and amla (Indian gooseberry).

(B) Beri-Beri is caused by the deficiency of Vitamin $B_1$ (thiamine).
Scurvy is caused by the deficiency of Vitamin $C$ (ascorbic acid).
Cheilosis is caused by the deficiency of Vitamin $B_2$ (riboflavin).
Rickets is caused by the deficiency of Vitamin $D$ (calciferol).
Therefore,the correct matching is $A-II, B-IV, C-I, D-V$.