TS EAMCET 2024 Chemistry Question Paper with Answer and Solution

(D) Given the equation of the pair of lines is $2x^2 + hxy + 6y^2 = 0$.
Comparing this with the general form $ax^2 + 2h'xy + by^2 = 0$,we have $a = 2$,$2h' = h$,and $b = 6$.
Let the slopes of the two lines be $m_1$ and $m_2$.
We know that $m_1 + m_2 = -\frac{2h'}{b} = -\frac{h}{6}$ and $m_1 m_2 = \frac{a}{b} = \frac{2}{6} = \frac{1}{3}$.
Given that one slope is thrice the other,let $m_1 = 3m_2$.
Substituting this into the product equation: $(3m_2) \cdot m_2 = \frac{1}{3}$ $\Rightarrow 3m_2^2 = \frac{1}{3}$ $\Rightarrow m_2^2 = \frac{1}{9}$ $\Rightarrow m_2 = \pm \frac{1}{3}$.
If $m_2 = \frac{1}{3}$,then $m_1 = 1$.
If $m_2 = -\frac{1}{3}$,then $m_1 = -1$.
Using the sum of slopes: $m_1 + m_2 = 4m_2 = -\frac{h}{6}$.
For $m_2 = \frac{1}{3}$,$4(\frac{1}{3}) = -\frac{h}{6}$ $\Rightarrow \frac{4}{3} = -\frac{h}{6}$ $\Rightarrow h = -8$.
For $m_2 = -\frac{1}{3}$,$4(-\frac{1}{3}) = -\frac{h}{6}$ $\Rightarrow -\frac{4}{3} = -\frac{h}{6}$ $\Rightarrow h = 8$.
Thus,$h = \pm 8$.

(A) Step $1$: Dehydration of propan-$1$-ol $(C_3H_7OH)$ with concentrated $H_2SO_4$ at $443 \ K$ yields propene $(X)$ as the major product via an elimination reaction: $CH_3CH_2CH_2OH \xrightarrow{Conc. H_2SO_4, 443 \ K} CH_3CH=CH_2 + H_2O$.
Step $2$: Addition of $HBr$ to propene in the presence of benzoyl peroxide $((C_6H_5CO)_2O_2)$ follows the anti-Markovnikov rule (peroxide effect or Kharasch effect),resulting in $1$-bromopropane $(Y)$ as the major product: $CH_3CH=CH_2 + HBr \xrightarrow{(C_6H_5CO)_2O_2} CH_3CH_2CH_2Br$.

(B) The controlled oxidation of methane under different conditions yields different products.
$1$. When methane is heated with $Cu$ at $523 \ K$ and $100 \ atm$ pressure, it gives methanol: $2CH_4 + O_2 \xrightarrow{Cu / 523 \ K / 100 \ atm} 2CH_3OH$.
$2$. When methane is heated with $Mo_2O_3$, it gives methanal: $CH_4 + O_2 \xrightarrow{Mo_2O_3 / \Delta} HCHO + H_2O$.
Therefore, $X = Cu$ and $Y = Mo_2O_3$.

(B) $1$. Oxidation of $2-\text{methylpropane}$ with $KMnO_4$ yields $2-\text{methylpropan-2-ol}$ $(X)$: $(CH_3)_3CH \xrightarrow{KMnO_4} (CH_3)_3COH$.
$2$. Dehydration of $2-\text{methylpropan-2-ol}$ with $20 \% \ H_3PO_4$ at $358 \ K$ gives $2-\text{methylpropene}$ $(Y)$: $(CH_3)_3COH \xrightarrow{H_3PO_4, \Delta} CH_2=C(CH_3)_2$.
$3$. Ozonolysis of $2-\text{methylpropene}$ followed by reductive workup with $Zn \mid H_2O$ yields acetone and formaldehyde: $CH_2=C(CH_3)_2 \xrightarrow[(ii) Zn \mid H_2O]{(i) O_3} (CH_3)_2C=O + CH_2=O$.
$4$. Thus,$A$ and $B$ are $(CH_3)_2C=O$ and $CH_2=O$.

(A) In $CH_3OH$ (methanol),the oxygen atom is $sp^3$ hybridized with two lone pairs. Due to the repulsion between lone pairs,the bond angle is compressed to less than the tetrahedral angle of $109^{\circ} 28'$.
In $CH_3OCH_3$ (dimethyl ether),the oxygen atom is also $sp^3$ hybridized,but the presence of two bulky methyl groups causes steric repulsion between them,which increases the bond angle to greater than $109^{\circ} 28'$.
Therefore,$X < 109^{\circ} 28'$ and $Y > 109^{\circ} 28'$.

(D) In an $LR$ series circuit,the total voltage $V$ is the phasor sum of the voltage across the inductor $(V_L)$ and the voltage across the resistor $(V_R)$.
The relationship is given by: $V^2 = V_L^2 + V_R^2$.
Given: $V = 10 \ V$ and $V_L = 6 \ V$.
Substituting the values into the equation:
$(10)^2 = (6)^2 + V_R^2$
$100 = 36 + V_R^2$
$V_R^2 = 100 - 36 = 64$
$V_R = \sqrt{64} = 8 \ V$.
Therefore,the potential difference across the resistor is $8 \ V$.

(C) The centripetal acceleration of an electron in the $n^{th}$ orbit is given by $a_c = \frac{v^2}{r}$. Since $v \propto \frac{1}{n}$ and $r \propto n^2$,we have $a_c \propto \frac{(1/n)^2}{n^2} = \frac{1}{n^4}$.
Given the ratio of accelerations for two successive orbits $n$ and $(n-1)$ is $\frac{a_n}{a_{n-1}} = \frac{(n-1)^4}{n^4} = \frac{16}{81}$.
Taking the fourth root,we get $\frac{n-1}{n} = \frac{2}{3}$,which implies $3n - 3 = 2n$,so $n = 3$.
The two orbits are $n = 3$ and $n = 2$.
The angular momentum of an electron in the $n^{th}$ orbit is $L_n = \frac{nh}{2\pi}$.
The change in angular momentum is $\Delta L = L_n - L_{n-1} = \frac{nh}{2\pi} - \frac{(n-1)h}{2\pi} = \frac{h}{2\pi}(n - n + 1) = \frac{h}{2\pi}$.

(D) Given: $C_1 = 10 \mu F$,$V_1 = 100 \text{ V}$,$C_2 = 30 \mu F$,$V_2 = 0 \text{ V}$.
When two capacitors are connected,the electrostatic energy lost during the redistribution of charge is given by the formula:
$\Delta U = \frac{1}{2} \left( \frac{C_1 C_2}{C_1 + C_2} \right) (V_1 - V_2)^2$
Substituting the values:
$\Delta U = \frac{1}{2} \left( \frac{10 \times 10^{-6} \times 30 \times 10^{-6}}{10 \times 10^{-6} + 30 \times 10^{-6}} \right) (100 - 0)^2$
$\Delta U = \frac{1}{2} \left( \frac{300 \times 10^{-12}}{40 \times 10^{-6}} \right) (10000)$
$\Delta U = \frac{1}{2} \times (7.5 \times 10^{-6}) \times 10^4$
$\Delta U = 3.75 \times 10^{-2} \text{ J}$.

(A) Let the vertical plate be $1$ and the horizontal plate be $2$. Both have area $A = L \times a$.
For the initial position:
The centre of mass of the vertical plate is at $y_1 = L/2$.
The centre of mass of the horizontal plate is at $y_2 = L + a/2$.
The combined centre of mass $y_{cm}$ is:
$y_{cm} = \frac{A_1 y_1 + A_2 y_2}{A_1 + A_2} = \frac{(La)(L/2) + (La)(L + a/2)}{2La} = \frac{L/2 + L + a/2}{2} = \frac{3L + a}{4}$.
When inverted,the horizontal plate is at the bottom. The new centre of mass $y'_{cm}$ from the surface is:
The centre of mass of the horizontal plate is at $y'_1 = a/2$.
The centre of mass of the vertical plate is at $y'_2 = a + L/2$.
$y'_{cm} = \frac{(La)(a/2) + (La)(a + L/2)}{2La} = \frac{a/2 + a + L/2}{2} = \frac{3a + L}{4}$.
The shift in the centre of mass is:
$\Delta y_{cm} = y_{cm} - y'_{cm} = \frac{3L + a}{4} - \frac{3a + L}{4} = \frac{2L - 2a}{4} = \frac{L - a}{2}$.

(A) Given: Mass of balls $m_1 = m_2 = m = 1.2 \ kg$. Initial velocities $u_1 = 12 \ ms^{-1}$,$u_2 = 0$. Coefficient of restitution $e = \frac{1}{\sqrt{2}}$.
By the principle of conservation of linear momentum:
$m u_1 + m u_2 = m v_1 + m v_2$
$12 + 0 = v_1 + v_2 \Rightarrow v_1 + v_2 = 12 \quad ...(i)$
By the definition of coefficient of restitution:
$e = \frac{v_2 - v_1}{u_1 - u_2}$
$\frac{1}{\sqrt{2}} = \frac{v_2 - v_1}{12 - 0} \Rightarrow v_2 - v_1 = \frac{12}{\sqrt{2}} = 6\sqrt{2} \quad ...(ii)$
Adding equations $(i)$ and $(ii)$:
$2v_2 = 12 + 6\sqrt{2} \Rightarrow v_2 = 6 + 3\sqrt{2} \ ms^{-1}$
Subtracting equation $(ii)$ from $(i)$:
$2v_1 = 12 - 6\sqrt{2} \Rightarrow v_1 = 6 - 3\sqrt{2} \ ms^{-1}$
Initial kinetic energy $(KE)_i = \frac{1}{2} m u_1^2 = \frac{1}{2} m (12)^2 = 72m$
Final kinetic energy $(KE)_f = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 = \frac{1}{2} m [(6 - 3\sqrt{2})^2 + (6 + 3\sqrt{2})^2]$
$(KE)_f = \frac{1}{2} m [(36 + 18 - 36\sqrt{2}) + (36 + 18 + 36\sqrt{2})] = \frac{1}{2} m [54 + 54] = 54m$
Ratio $\frac{(KE)_f}{(KE)_i} = \frac{54m}{72m} = \frac{3}{4}$ or $3:4$.

(C) $[SiCl_6]^{2-}$ does not exist because six large $Cl^{-}$ ions cannot be accommodated around the small $Si^{4+}$ ion due to steric hindrance and inter-electronic repulsion between the chloride ions.
In contrast,$[SiF_6]^{2-}$ exists because the $F^{-}$ ion is small enough to fit around the $Si^{4+}$ ion.
$[GeCl_6]^{2-}$ and $[Sn(OH)_6]^{2-}$ are stable due to the larger size of the central $Ge^{4+}$ and $Sn^{4+}$ ions,which can accommodate six ligands.

(A) The ratio of $s$-electrons to $p$-electrons is determined by the electronic configuration of each species.
For $K^{+}$ $(Z=19)$: $1s^2 2s^2 2p^6 3s^2 3p^6$. Total $s$-electrons = $2+2+2 = 6$. Total $p$-electrons = $6+6 = 12$. Ratio = $6:12 = 1:2$.
For $Cr^{3+}$ $(Z=24)$: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^3$. Total $s$-electrons = $2+2+2 = 6$. Total $p$-electrons = $6+6 = 12$. Ratio = $6:12 = 1:2$.
Since both $K^{+}$ and $Cr^{3+}$ have the same ratio of $1:2$,the correct option is $A$.

(D) The given reactions are part of the copper extraction process:
$1. 2 Cu_2S + 3 O_2 \rightarrow 2 Cu_2O + 2 SO_2$ (where $X = Cu_2S$)
$2. 2 Cu_2O + Cu_2S \rightarrow 6 Cu + SO_2$ (where $Y = SO_2$)
In the $SO_2$ molecule,the sulfur atom is $sp^2$ hybridized with one lone pair of electrons.
Due to the presence of the lone pair,the geometry is bent or angular.

(A) Both $SnCl_2$ and $H_2O$ have a bent geometry.
In $SnCl_2$,the central atom $Sn$ has $2$ bond pairs and $1$ lone pair,resulting in $sp^2$ hybridization.
In $H_2O$,the central atom $O$ has $2$ bond pairs and $2$ lone pairs,resulting in $sp^3$ hybridization.
Since both have a bent geometry but different hybridization states,the correct pair is $SnCl_2$ and $H_2O$.

(A) The reaction is: $NH_2CONH_2 + 2 H_2O \rightarrow (NH_4)_2CO_3 \rightleftharpoons 2 NH_3 + H_2O + CO_2$.
Here,$[X]$ is ammonium carbonate $(NH_4)_2CO_3$ and $[Y]$ is carbon dioxide $(CO_2)$.
In the carbonate ion $(CO_3^{2-})$,the carbon atom is bonded to three oxygen atoms with one double bond and two single bonds,resulting in $sp^2$ hybridization.
In the carbon dioxide molecule $(O=C=O)$,the carbon atom is bonded to two oxygen atoms with two double bonds,resulting in $sp$ hybridization.
Therefore,the hybridization of carbon in $X$ and $Y$ is $sp^2$ and $sp$ respectively.

(D) To determine the shape of the molecules, we use the $VSEPR$ theory:
$A. SF_4$: Hybridisation is $sp^3d$ with one lone pair, resulting in a see-saw shape $(III)$.
$B. ClF_3$: Hybridisation is $sp^3d$ with two lone pairs, resulting in a $T$-shaped geometry $(I)$.
$C. BrF_5$: Hybridisation is $sp^3d^2$ with one lone pair, resulting in a square pyramidal shape $(IV)$.
$D. XeF_4$: Hybridisation is $sp^3d^2$ with two lone pairs, resulting in a square planar shape $(II)$.
Therefore, the correct matching is $A-III, B-I, C-IV, D-II$.

(A) For $BF_3$: The central atom $B$ has $3$ valence electrons and forms $3$ bonds with $F$ atoms. Steric number $= 3 + 0 = 3$,which corresponds to $sp^2$ hybridisation.
For $SnCl_2$: The central atom $Sn$ has $4$ valence electrons. It forms $2$ bonds with $Cl$ atoms and has $1$ lone pair. Steric number $= 2 + 1 = 3$,which corresponds to $sp^2$ hybridisation.
For $HgCl_2$: The central atom $Hg$ has $2$ valence electrons and forms $2$ bonds with $Cl$ atoms. Steric number $= 2 + 0 = 2$,which corresponds to $sp$ hybridisation.
Therefore,the correct sequence is $sp^2, sp^2, sp$.

(C) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{1}{2} (V - N)$,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 $ClF_3$: $V = 7, N = 3$. $\text{Lone pairs} = \frac{1}{2} (7 - 3) = 2$.
$2$. For $NF_3$: $V = 5, N = 3$. $\text{Lone pairs} = \frac{1}{2} (5 - 3) = 1$.
$3$. For $SF_4$: $V = 6, N = 4$. $\text{Lone pairs} = \frac{1}{2} (6 - 4) = 1$.
$4$. For $XeF_4$: $V = 8, N = 4$. $\text{Lone pairs} = \frac{1}{2} (8 - 4) = 2$.

Compound	Lone pairs
$ClF_3$	$2$
$NF_3$	$1$
$SF_4$	$1$
$XeF_4$	$2$

Therefore,the correct sequence is $2, 1, 1, 2$.

(B) In the planar structure of the nitric acid $(HNO_3)$ molecule,the nitrogen atom is $sp^2$ hybridized.
Based on the experimental structural data provided in the image:
The bond angle $H-O-N$ is approximately $102.2^{\circ}$.
The bond angle $O-N-O$ (between the two oxygen atoms bonded to nitrogen) is approximately $130.27^{\circ}$.
Therefore,the respective bond angles are $102^{\circ}$ and $130^{\circ}$.

(B) The bond order is calculated as: $\text{Bond order} = \frac{1}{2} [N_b - N_a]$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of anti-bonding electrons.
For $O_2$ ($16$ electrons): $\text{Bond order} = \frac{1}{2} [10 - 6] = 2.0$.
For $O_2^{-}$ ($17$ electrons): $\text{Bond order} = \frac{1}{2} [10 - 7] = 1.5$.
For $O_2^{+}$ ($15$ electrons): $\text{Bond order} = \frac{1}{2} [10 - 5] = 2.5$.
For $O_2^{2-}$ ($18$ electrons): $\text{Bond order} = \frac{1}{2} [10 - 8] = 1.0$.
For $O_2^{2+}$ ($14$ electrons): $\text{Bond order} = \frac{1}{2} [10 - 4] = 3.0$.
Calculating the differences:
$(A)$ $|2.5 - 1.5| = 1.0$
$(B)$ $|3.0 - 1.0| = 2.0$
$(C)$ $|3.0 - 2.0| = 1.0$
$(D)$ $|3.0 - 2.5| = 0.5$
The maximum difference is $2.0$ for the pair $O_2^{2-}$ and $O_2^{2+}$.

(D) The electronic configuration of $C_2$ ($12$ electrons) is: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2$.
Bond order $= \frac{1}{2}(N_b - N_a) = \frac{1}{2}(8 - 4) = 2$.
Since all electrons are paired,$C_2$ is diamagnetic.
$C_2^{-}$ has $13$ electrons,configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^1$. It has one unpaired electron,so it is paramagnetic.
In $C_2$,the two highest occupied molecular orbitals are $\pi$ orbitals,meaning $C_2$ consists of two $\pi$ bonds.
Therefore,the statement that $C_2$ consists of $1 \sigma$ and $1 \pi$ bond is incorrect.

(B) According to Fajan's rules,covalent character increases with:
$1$. Smaller size of the cation.
$2$. Larger size of the anion.
$3$. Higher charge on the cation or anion.
Evaluating the options:
- $KF < KI$: $I^-$ is larger than $F^-$,so $KI$ is more covalent. (Correct)
- $LiF < KF$: $Li^+$ is smaller than $K^+$,so $LiF$ should be more covalent than $KF$. Thus,$LiF > KF$. (Incorrect)
- $SnCl_2 < SnCl_4$: $Sn^{4+}$ has a higher charge than $Sn^{2+}$,so $SnCl_4$ is more covalent. (Correct)
- $NaCl < CuCl$: $Cu^+$ has a pseudo-noble gas configuration ($18$ electrons in the valence shell) and is more polarizing than $Na^+$. (Correct)
Therefore,the statement $LiF < KF$ is incorrect.

(A) Let us analyze each pair:
$(A)$ Bond angle: The bond angles are $NO_2$ $(134^{\circ})$,$O_3$ $(116.8^{\circ})$,and $H_2O$ $(104.5^{\circ})$. Thus,the order $NO_2 > O_3 > H_2O$ is correct.
$(B)$ Dipole moment: The dipole moments are $H_2O$ $(1.85 \ D)$,$HF$ $(1.78 \ D)$,and $NH_3$ $(1.47 \ D)$. Thus,the order $H_2O > HF > NH_3$ is correct.
$(C)$ Bond length: $I_2$ has a single bond with a large atomic size,$F_2$ has a single bond with a smaller atomic size,and $N_2$ has a triple bond with a very small atomic size. Bond length order is $I_2 > F_2 > N_2$. This is correct.
Therefore,all three pairs $(A), (B),$ and $(C)$ are correctly matched.

(C) Dipole moment $\mu = q \times d$.
The dipole moments of the given molecules are as follows:

$HF = 1.82 \ D$	$H_2O = 1.85 \ D$	$NH_3 = 1.47 \ D$
$HCl = 1.08 \ D$	$H_2S = 0.95 \ D$	$NF_3 = 0.24 \ D$
$HBr = 0.83 \ D$	$CO_2 = 0.00 \ D$	$BF_3 = 0.00 \ D$

Evaluating the options:
Option $A$: $1.82 > 1.08 > 0.83$ (Correct order).
Option $B$: $1.85 > 0.95 > 0.00$ (Correct order).
Option $C$: $0.95 > 1.08 > 1.82$ (Incorrect order,as $HF$ has the highest dipole moment).
Option $D$: $1.47 > 0.24 > 0.00$ (Correct order).

(D) $o$-hydroxybenzaldehyde $(A)$ exhibits intramolecular $H$-bonding,which leads to the formation of a stable chelate ring within the molecule. This restricts the association between different molecules.
In contrast,$p$-hydroxybenzaldehyde $(B)$ exhibits intermolecular $H$-bonding,which allows for strong association between different molecules,resulting in a higher melting point.
Therefore,the lower melting point of $(A)$ is due to the presence of intramolecular $H$-bonding,while $(B)$ has intermolecular $H$-bonding.

(A) The reaction is $AO_{2(g)} + BO_{2(g)} \rightleftharpoons AO_{3(g)} + BO_{(g)}$.
Given $K_c = 16$ and initial concentrations $[AO_2] = [BO_2] = [AO_3] = [BO] = 1.0 \ M$.
Let the change in concentration be $x$.
At equilibrium:
$[AO_2] = 1 - x$
$[BO_2] = 1 - x$
$[AO_3] = 1 + x$
$[BO] = 1 + x$
$K_c = \frac{[AO_3][BO]}{[AO_2][BO_2]} = \frac{(1+x)(1+x)}{(1-x)(1-x)} = 16$
Taking the square root of both sides:
$\frac{1+x}{1-x} = 4$
$1 + x = 4 - 4x$
$5x = 3 \implies x = 0.6$
Equilibrium concentration of $[BO] = 1 + x = 1 + 0.6 = 1.6 \ mol \ L^{-1}$.

(A) The dissociation reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
Given $K_c = 2 \times 10^{-2} \ mol \ L^{-1}$ and volume $V = 1.0 \ L$.
Let the initial moles of $PCl_5$ be $x$.

Reaction	$PCl_{5(g)}$	$PCl_{3(g)} + Cl_{2(g)}$
Initial moles	$x$	$0, 0$
Equilibrium moles	$x - 0.2$	$0.2, 0.2$

Since $V = 1.0 \ L$,the concentrations at equilibrium are $[PCl_5] = (x - 0.2) \ M$,$[PCl_3] = 0.2 \ M$,and $[Cl_2] = 0.2 \ M$.
The equilibrium constant expression is $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}$.
Substituting the values: $2 \times 10^{-2} = \frac{0.2 \times 0.2}{x - 0.2}$.
$0.02 = \frac{0.04}{x - 0.2} \implies x - 0.2 = \frac{0.04}{0.02} = 2$.
$x = 2 + 0.2 = 2.2 \ mol$.

(C) Given reactions:
$(i) \ 2 A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)} ; K_1 = 16$
$(ii) \ 2 B_{(g)} + C_{(g)} \rightleftharpoons 2 D_{(g)} ; K_2 = 25$
We need the equilibrium constant $(K_3)$ for:
$(iii) \ A_{(g)} + \frac{1}{2} B_{(g)} \rightleftharpoons D_{(g)}$
From $(i)$,$K_1 = \frac{[B][C]}{[A]^2} = 16$.
From $(ii)$,$K_2 = \frac{[D]^2}{[C][B]^2} = 25$.
Multiplying $(i)$ and $(ii)$:
$K_1 \times K_2 = \frac{[B][C]}{[A]^2} \times \frac{[D]^2}{[C][B]^2} = \frac{[D]^2}{[A]^2 [B]} = 16 \times 25 = 400$.
Taking the square root of the expression:
$\sqrt{\frac{[D]^2}{[A]^2 [B]}} = \frac{[D]}{[A][B]^{1/2}} = \sqrt{400} = 20$.
Thus,$K_3 = 20$.

(A) Lauryl alcohol is $1$-dodecanol,which has a total of $12$ carbon atoms in its chain.
The chemical formula for lauryl alcohol is $CH_3(CH_2)_{10}CH_2OH$.
Comparing this with the given general formula $CH_3(CH_2)_nCH_2OH$,we find that $n=10$.

(A) The acidic oxides are $B_2O_3, SiO_2,$ and $CO_2$.
$CO$ is a neutral oxide.
$Al_2O_3$ and $PbO_2$ are amphoteric oxides.
$Tl_2O_3$ is a basic oxide.
Therefore,the total number of acidic oxides is $3$.

(B) Electron gain enthalpy becomes less negative (or smaller in magnitude) as we move down a group due to an increase in atomic size.
Among the given elements,$Cl$ and $S$ are in the $3^{rd}$ period,while $O$ and $I$ are in the $2^{nd}$ and $5^{th}$ periods respectively.
$Cl$ has the most negative electron gain enthalpy.
$Iodine$ $(I)$ has the least negative electron gain enthalpy among the halogens listed due to its large atomic size,which reduces the attraction for an incoming electron.
Comparing $O$ and $S$,$O$ has a smaller electron gain enthalpy than $S$ due to inter-electronic repulsion in its small $2p$ orbital.
However,comparing all four,$Iodine$ $(I)$ has the least electron gain enthalpy.

(B) For $P$ $(Z=15)$: $1s^2 2s^2 2p^6 3s^2 3p^3$. Total $s$-electrons = $2+2+2 = 6$. Total $p$-electrons = $6+3 = 9$. Ratio = $6/9 = 2/3$.
For $Ca$ $(Z=20)$: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$. Total $s$-electrons = $2+2+2+2 = 8$. Total $p$-electrons = $6+6 = 12$. Ratio = $8/12 = 2/3$.
Thus,the pair is $P$ and $Ca$.

(D) $1.$ $C$ and $N$ are in the same period. The atomic radius of $C$ is $\approx 77 \ pm$ and $N$ is $\approx 75 \ pm$. The difference is $\approx 2 \ pm$.
$2.$ $O$ and $F$ are in the same period. The atomic radius of $O$ is $\approx 73 \ pm$ and $F$ is $\approx 71 \ pm$. The difference is $\approx 2 \ pm$.
$3.$ $P$ and $S$ are in the same period. The atomic radius of $P$ is $\approx 110 \ pm$ and $S$ is $\approx 103 \ pm$. The difference is $\approx 7 \ pm$.
$4.$ $Li$ and $Be$ are in the same period. The atomic radius of $Li$ is $\approx 152 \ pm$ and $Be$ is $\approx 111 \ pm$. The difference is $\approx 41 \ pm$.
Comparing all pairs, the difference in atomic radii is maximum for the pair $(Li, Be)$.

(B) For isoelectronic species,the size decreases as the nuclear charge (number of protons) increases.
$F^{-}$ has $9$ protons and $10$ electrons.
$Na^{+}$ has $11$ protons and $10$ electrons.
Since $Na^{+}$ has a higher nuclear charge than $F^{-}$,the electrons are pulled more strongly towards the nucleus,making $Na^{+}$ smaller than $F^{-}$.
Thus,in the pair $(F^{-}, Na^{+})$,the second ion is smaller than the first.

(D) The electronic configuration of $Na^{+}$ is $1s^2 2s^2 2p^6$ ($10$ electrons).
The electronic configuration of $F^{-}$ is $1s^2 2s^2 2p^6$ ($10$ electrons).
Since both have the same number of electrons,they are isoelectronic species,so $(R)$ is correct.
However,the ionic radius depends on the nuclear charge $(Z)$.
For $Na^{+}$,$Z = 11$,and for $F^{-}$,$Z = 9$.
As the nuclear charge increases,the attraction of the nucleus on the electrons increases,leading to a smaller ionic radius.
Therefore,the ionic radius of $Na^{+}$ $(102 \ pm)$ is smaller than that of $F^{-}$ $(133 \ pm)$.
Thus,$(A)$ is incorrect.

(B) Noble gases have large positive electron gain enthalpies because the added electron has to enter the next higher principal quantum level,which leads to a very unstable electronic configuration.
Therefore,the set containing noble gases,$Kr, Xe, Rn$,will have positive electron gain enthalpies.
Thus,option $(B)$ is correct.

(C) Across a period from left to right,the electronegativity of the element increases,which leads to an increase in the acidic strength of its oxides.
The elements $Al, Si, P$ and $S$ belong to the same period (Period $3$).
The order of electronegativity is $Al < Si < P < S$.
Thus,the correct order of acidic strength is $Al_2O_3 < SiO_2 < P_2O_3 < SO_2$.

(A) Ionization enthalpy generally increases across a period due to an increase in effective nuclear charge. However,there are exceptions due to stable electronic configurations.
The electronic configuration of $Mg$ $([Ne] 3s^2)$ is more stable than that of $Al$ $([Ne] 3s^2 3p^1)$ because $Mg$ has a completely filled $s$-orbital.
Consequently,the first ionization enthalpy of $Al$ is lower than that of $Mg$.
The values are: $Na (496) < Al (575) < Mg (737) < Si (786)$.
Therefore,the value for $Al$ is $575 \ kJ \ mol^{-1}$.

(D) The relationship between current $i$ and time $t$ is given by $i = 12t + 9t^2$.
We know that the charge $q$ flowing through a conductor is the integral of current with respect to time: $q = \int_{t_1}^{t_2} i \ dt$.
Given $t_1 = 2 \ s$ and $t_2 = 10 \ s$,we substitute the expression for $i$:
$q = \int_{2}^{10} (12t + 9t^2) \ dt$.
Integrating the terms:
$q = [12 \cdot \frac{t^2}{2} + 9 \cdot \frac{t^3}{3}]_{2}^{10} = [6t^2 + 3t^3]_{2}^{10}$.
Evaluating at the limits:
$q = [6(10)^2 + 3(10)^3] - [6(2)^2 + 3(2)^3]$.
$q = [600 + 3000] - [24 + 24] = 3600 - 48 = 3552 \ C$.

(D) Given: Length $l = 1.5 \ m$,Area $A = 3 \times 10^{-5} \ m^2$,Resistance $R = 15 \ \Omega$,Electric field $E = 21 \ Vm^{-1}$.
We know that current density $J = \sigma E$,where $\sigma$ is the conductivity.
Since $\sigma = \frac{1}{\rho} = \frac{l}{RA}$,we can write $J = \frac{l}{RA} \times E$.
Substituting the values:
$J = \frac{1.5 \times 21}{15 \times 3 \times 10^{-5}}$
$J = \frac{31.5}{45 \times 10^{-5}}$
$J = 0.7 \times 10^5 \ Am^{-2}$.

(C) Beryllium forms an alloy with copper known as beryllium copper or $Cu-Be$ alloy.
This alloy is widely used in the manufacturing of high-strength springs,electrical connectors,and non-sparking tools due to its excellent mechanical properties and conductivity.

(A) The de Broglie wavelength $\lambda$ of a particle is given by $\lambda = \frac{h}{\sqrt{2mK}}$,where $m$ is the mass and $K$ is the kinetic energy.
From this,we have $\lambda \propto \frac{1}{\sqrt{mK}}$,which implies $K \propto \frac{1}{m\lambda^2}$.
Let $m_p$ and $m_\alpha$ be the masses of the proton and alpha particle respectively. We know $m_\alpha = 4m_p$.
Given $\lambda_p = 2\lambda_\alpha$.
The ratio of kinetic energies is $\frac{K_p}{K_\alpha} = \frac{m_\alpha}{m_p} \times \left( \frac{\lambda_\alpha}{\lambda_p} \right)^2$.
Substituting the values: $\frac{K_p}{K_\alpha} = \left( \frac{4m_p}{m_p} \right) \times \left( \frac{\lambda_\alpha}{2\lambda_\alpha} \right)^2 = 4 \times \left( \frac{1}{2} \right)^2 = 4 \times \frac{1}{4} = 1$.
Thus,the ratio is $1:1$.

(A) Given: $\lambda^{\circ}(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}$ and $\lambda^{\circ}(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$.
Degree of dissociation $\alpha = 11.0 \% = 0.11$.
The molar conductivity at infinite dilution is calculated as: $\lambda_{m}^{\circ}(HCOOH) = \lambda^{\circ}(H^{+}) + \lambda^{\circ}(HCOO^{-}) = 349.6 + 54.6 = 404.2 \ S \ cm^2 \ mol^{-1}$.
The molar conductivity $\lambda_{m}$ at a given concentration is related to the degree of dissociation by the formula: $\lambda_{m} = \alpha \times \lambda_{m}^{\circ}$.
Substituting the values: $\lambda_{m} = 0.11 \times 404.2 = 44.46 \ S \ cm^2 \ mol^{-1}$.

(B) The maximum induced electromotive force $(emf)$ in a rotating coil is given by the formula: $e_{\max} = N B A \omega$.
Given:
Number of turns $(N)$ = $50$
Magnetic field $(B)$ = $2 \times 10^{-2} ~T$
Area $(A)$ = $200 ~cm^2 = 200 \times 10^{-4} ~m^2 = 2 \times 10^{-2} ~m^2$
Angular speed $(\omega)$ = $40 ~rad/s$
Substituting these values into the formula:
$e_{\max} = 50 \times (2 \times 10^{-2}) \times (2 \times 10^{-2}) \times 40$
$e_{\max} = 50 \times 4 \times 10^{-4} \times 40$
$e_{\max} = 200 \times 40 \times 10^{-4}$
$e_{\max} = 8000 \times 10^{-4} = 0.8 ~V$
Thus,the maximum induced $emf$ is $0.8 ~V$.

(D) For an electromagnetic wave,the relationship between the peak electric field $(E_0)$ and the peak magnetic field $(B_0)$ is given by $E_0 = c B_0$,where $c$ is the speed of light in vacuum.
Given:
$B_0 = 30 \times 10^{-9} \ T$
$c = 3 \times 10^8 \ m/s$
Substituting these values into the formula:
$E_0 = (3 \times 10^8 \ m/s) \times (30 \times 10^{-9} \ T)$
$E_0 = 90 \times 10^{-1} \ Vm^{-1}$
$E_0 = 9 \ Vm^{-1}$
Therefore,the peak value of the electric field is $9 \ Vm^{-1}$.

(C) Let the two charges be $q_1 = -10 \mu C$ at $x_1 = 0$ and $q_2 = +5 \mu C$ at $x_2 = \sqrt{2} \ m$.
For the electric field to be zero,the point must be closer to the charge with the smaller magnitude. Since $|q_2| < |q_1|$,the point $P$ must lie on the $X$-axis to the right of $q_2$.
Let the distance of point $P$ from $q_2$ be $x$. The position of $P$ is $x_P = \sqrt{2} + x$.
The electric field due to $q_1$ at $P$ is $E_1 = \frac{k|q_1|}{(\sqrt{2}+x)^2}$ (directed towards the origin).
The electric field due to $q_2$ at $P$ is $E_2 = \frac{k|q_2|}{x^2}$ (directed away from the origin).
Setting $E_1 = E_2$:
$\frac{10}{x^2 + 2\sqrt{2}x + 2} = \frac{5}{x^2}$
$2x^2 = x^2 + 2\sqrt{2}x + 2$
$x^2 - 2\sqrt{2}x - 2 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{2\sqrt{2} \pm \sqrt{8 - 4(1)(-2)}}{2} = \frac{2\sqrt{2} \pm \sqrt{16}}{2} = \sqrt{2} \pm 2$.
Since $x$ must be positive,$x = \sqrt{2} + 2$.
The coordinate of point $P$ is $x_P = \sqrt{2} + x = \sqrt{2} + (\sqrt{2} + 2) = 2\sqrt{2} + 2 = 2(\sqrt{2} + 1) \ m$.

(A) Methemoglobinemia is caused by an excess of nitrate concentration in drinking water.
High levels of nitrate in drinking water can cause methemoglobinemia,which is also known as $blue \ baby \ syndrome$.
When consumed,nitrates are absorbed into the blood and some are converted into nitrites,which react with hemoglobin to create methemoglobin,reducing the oxygen-carrying capacity of the blood.

(C) Non-biodegradable wastes are substances that cannot be broken down by biological processes. Thermal power plants generate fly ash,which is a non-biodegradable waste product.

(D) Photochemical smog is formed by the action of sunlight on nitrogen oxides $(NO_x)$ and volatile organic compounds $(VOCs)$ in the atmosphere.
Its common components include Ozone $(O_3)$,Nitrogen dioxide $(NO_2)$,Peroxyacetyl nitrate $(PAN)$,and aldehydes (compounds containing the $-CHO$ group).

(C) Eutrophication is a process where water bodies receive excessive amounts of nutrients,such as nitrates and phosphates,which results in excessive plant and algal growth.
As these plants die and decompose,the process consumes a large amount of oxygen.
Therefore,eutrophication leads to a significant decrease in the level of $DO$ (dissolved oxygen) in the water.

(C) The first reaction is the reduction of benzoic acid $(C_6H_5COOH)$ to benzaldehyde $(C_6H_5CHO)$. This cannot be done directly with $SOCl_2$ and $H_2|Ni$ (which reduces acid chlorides to aldehydes,but the conditions are specific). The standard laboratory method to reduce a carboxylic acid to an aldehyde involves converting it to an ester first,followed by reduction with $DiBAL-H$.
Step $1$: $C_6H_5COOH + C_2H_5OH \xrightarrow{H^+} C_6H_5COOC_2H_5 + H_2O$ (Esterification).
Step $2$: $C_6H_5COOC_2H_5 \xrightarrow{(i) \ DiBAL-H, (ii) \ H_2O} C_6H_5CHO$ (Reduction of ester to aldehyde).
Thus,$X$ is $(i) \ C_2H_5OH|H^+, (ii) \ DiBAL-H, (iii) \ H_2O$.
Step $3$: The second reaction is an Aldol condensation between benzaldehyde $(C_6H_5CHO)$ and acetophenone (implied,though the prompt shows $C_6H_5COOH$ as a reactant,the product $Y$ is clearly the Claisen-Schmidt condensation product,benzalacetophenone or chalcone,$C_6H_5-CH=CH-CO-C_6H_5$).
Therefore,the correct option is $C$.

(B) The given reaction involves two different chemical transformations of benzamide $(C_6H_5CONH_2)$:
$1$. Reaction with $LiAlH_4$ followed by $H_2O$: This is a reduction reaction. $LiAlH_4$ reduces the amide group $(-CONH_2)$ to an amine group $(-CH_2NH_2)$. Thus,$Y$ is benzylamine $(C_6H_5CH_2NH_2)$.
$2$. Reaction with $Br_2$ and $NaOH$: This is the Hoffmann bromamide degradation reaction. It converts an amide $(-CONH_2)$ into a primary amine with one carbon atom less $(-NH_2)$. Thus,$X$ is aniline $(C_6H_5NH_2)$.
Therefore,$X$ is $C_6H_5NH_2$ and $Y$ is $C_6H_5CH_2NH_2$.

(B) $1$. The nitration of chlorobenzene with $conc. HNO_3$ and $conc. H_2SO_4$ yields $1$-chloro-$4$-nitrobenzene as the major product $(X)$ because the $-Cl$ group is ortho/para directing and the para-isomer is sterically favored.
$2$. The conversion of $1$-chloro-$4$-nitrobenzene to $4$-nitrophenol involves nucleophilic aromatic substitution.
$3$. The reaction requires strong nucleophilic conditions,which are provided by $(i) NaOH$ at $443 \ K$ followed by $(ii) H^+$ (acidification) to convert the phenoxide ion to the phenol.
$4$. Thus,$Y$ is $(i) NaOH, 443 \ K, (ii) H^+$ and $Z$ is $4$-nitrophenol.

(A) The reaction of $4$-bromotoluene with $Br_2$ in the presence of $UV$ light is a free radical substitution reaction. This reaction occurs at the benzylic position,not on the aromatic ring. Therefore,$X$ is $4$-bromobenzyl bromide $(Br-C_6H_4-CH_2Br)$.
When $4$-bromobenzyl bromide is treated with $OH^-$,it undergoes a nucleophilic substitution reaction $(S_N2)$ where the $Br$ atom on the side chain is replaced by an $OH$ group. Therefore,$Y$ is $4$-bromobenzyl alcohol $(Br-C_6H_4-CH_2OH)$.

(D) The boiling points depend on intermolecular forces.
$1$. $II$ ($n$-butanol) and $III$ (isobutanol) are alcohols and exhibit strong intermolecular hydrogen bonding,leading to higher boiling points compared to ethers and alkanes.
$2$. Between $II$ and $III$,$II$ is a straight-chain alcohol with a larger surface area,leading to stronger van der Waals forces than the branched $III$. Thus,$II > III$.
$3$. $I$ (diethyl ether) is polar due to the $C-O-C$ bond,giving it a higher boiling point than the non-polar alkane $IV$ ($n$-pentane).
$4$. Therefore,the order is $II > III > I > IV$.

(D) The Lucas test uses conc. $HCl$ and anhydrous $ZnCl_2$ to distinguish between primary,secondary,and tertiary alcohols.
Tertiary alcohols react instantly to produce turbidity due to the formation of an insoluble alkyl chloride.
Given the molecular formula $C_5H_{12}O$,the alcohol $X$ must be a tertiary alcohol,which is $2-$methyl$-2-$butanol.
Acid-catalysed hydration of $2-$methyl$-1-$butene follows Markovnikov's rule to yield $2-$methyl$-2-$butanol,which is a tertiary alcohol.
Therefore,option $D$ is the correct answer.

(A) Step $1$: Reaction of tert-butyl alcohol with $Na$ gives sodium tert-butoxide $(P)$:
$(CH_3)_3 COH Na \rightarrow (CH_3)_3 CONa \frac{1}{2} H_2$
Step $2$: Reaction of $P$ with methyl bromide $(CH_3 Br)$ via $S_N2$ mechanism gives tert-butyl methyl ether $(Q)$:
$(CH_3)_3 CONa CH_3 Br \rightarrow (CH_3)_3 COCH_3 NaBr$
Step $3$: Cleavage of ether $Q$ with $HI$ at high temperature $(\Delta)$:
Since the ether contains a tertiary alkyl group,the reaction proceeds via an $S_N1$ mechanism.
The protonated ether undergoes cleavage to form a stable tertiary carbocation $(CH_3)_3 C $,which then reacts with $I^-$ to form tert-butyl iodide $(R = (CH_3)_3 CI)$ and methanol $(S = CH_3 OH)$.

(C) $1$. Dehydration of ethanol: $CH_3CH_2OH \xrightarrow[443 \ K]{\text{Conc. } H_2SO_4} CH_2=CH_2$ ($X$ is ethene).
$2$. Ozonolysis of ethene: $CH_2=CH_2 \xrightarrow[(2) \ Zn/H_2O]{(1) \ O_3} 2HCHO$ ($Y$ is formaldehyde).
$3$. Cannizzaro reaction: Formaldehyde $(HCHO)$ has no $\alpha$-hydrogen,so it undergoes the Cannizzaro reaction in the presence of concentrated $NaOH$ to form methanol $(CH_3OH)$ and sodium formate $(HCOONa)$.
$4$. Thus,the conversion of $Y$ to $Z$ is the Cannizzaro reaction.

(B) The reaction involves the conversion of an alcohol to an alkyl iodide.
Using $H_2SO_4$ with $KI$ is not ideal because $H_2SO_4$ is an oxidizing agent that can oxidize $HI$ to $I_2$.
Phosphoric acid $(H_3PO_4)$ is a non-oxidizing acid and is preferred for the in-situ generation of $HI$ from $KI$.
The reaction is: $3 KI + H_3PO_4 \rightarrow 3 HI + K_3PO_4$.
This $HI$ then reacts with the alcohol via an $S_N2$ mechanism to give the alkyl iodide with a good yield.

(A) The reactions are as follows:
$A$. Phenol reacts with $Zn$ dust to form benzene: $C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$ $(II)$.
$B$. Phenol is oxidized by $Na_2Cr_2O_7/H_2SO_4$ to form benzoquinone $(I)$.
$C$. Reimer-Tiemann reaction of phenol with $CHCl_3/NaOH$ followed by $H^+$ gives salicylaldehyde $(IV)$.
$D$. Kolbe's reaction of phenol with $NaOH$ followed by $CO_2/H^+$ gives salicylic acid $(III)$.
Therefore,the correct matching is $A-II, B-I, C-IV, D-III$.

(D) For ether $A$ (phenyl $n$-propyl ether): The $C-O$ bond between the phenyl group and oxygen cannot be cleaved because the $C-O$ bond has partial double bond character due to resonance. Therefore,cleavage occurs at the alkyl side,yielding phenol and $n$-propyl bromide.
For ether $B$ ($tert$-butyl $n$-propyl ether): The cleavage occurs to form the more stable carbocation. The $tert$-butyl group forms a $3^{\circ}$ carbocation,which is much more stable than the $n$-propyl carbocation. Thus,the $C-O$ bond between the $tert$-butyl group and oxygen is cleaved,yielding $tert$-butyl bromide and $n$-propanol.
Therefore,the bromides formed are $n$-propyl bromide and $tert$-butyl bromide.

(A) The reaction sequence is as follows:
$C_2H_5NH_2$ $\xrightarrow{NaNO_2/HCl} [C_2H_5N_2^+Cl^-] (X)$ $\xrightarrow{H_2O} C_2H_5OH (Y)$ $\xrightarrow{Cu, 573 \ K} CH_3CHO (Z)$
$Z$ is acetaldehyde $(CH_3CHO)$.
$I$. $Z$ is an aldehyde. (Correct)
$II$. $Z$ has $\alpha$-hydrogen atoms,so it does not undergo Cannizzaro reaction. (Incorrect)
$III$. $Z$ contains a $CH_3CO-$ group,so it gives a positive iodoform test. (Correct)
$IV$. $Z$ is an aldehyde,so it gives a positive test with Tollens' reagent. (Incorrect)
Thus,statements $I$ and $III$ are correct.

(A) The acidity of phenols is determined by the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups (like $-NO_2$) increase acidity (lower $pK_{a}$),while electron-donating groups (like $-CH_3$) decrease acidity (higher $pK_{a}$).
$1$. $p$-Nitrophenol has a $-NO_2$ group at the para position,which exerts a strong $-R$ (resonance) and $-I$ (inductive) effect,making it the most acidic $(pK_{a} = 7.1)$.
$2$. $m$-Nitrophenol has a $-NO_2$ group at the meta position,which exerts only a $-I$ effect,making it less acidic than $p$-nitrophenol $(pK_{a} = 8.3)$.
$3$. $p$-Cresol has a $-CH_3$ group at the para position,which is an electron-donating group ($+I$ and hyperconjugation),making it the least acidic $(pK_{a} = 10.2)$.
Thus,$X = m$-nitrophenol $(8.3)$,$Y = p$-nitrophenol $(7.1)$,and $Z = p$-cresol $(10.2)$.

(A) The reaction of alcohols with $SOCl_2$ proceeds via an $S_N2$ mechanism or $S_Ni$ mechanism to replace the $-OH$ group with $-Cl$.
In phenol,the lone pair of electrons on the oxygen atom participates in resonance with the benzene ring,giving the $C-O$ bond a partial double bond character.
This makes the $C-O$ bond stronger and shorter,preventing the cleavage of the $C-O$ bond required to replace the $-OH$ group with $-Cl$.
Therefore,chlorobenzene is not formed when phenol reacts with $SOCl_2$.
Both Assertion $(A)$ and Reason $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(C) The $pK_a$ value is inversely proportional to the acidic strength of the compound.
Acidic strength order:
$IV$ ($p$-Nitrobenzoic acid) > $II$ ($p$-Nitrophenol) > $I$ (Phenol) > $III$ (Ethanol).
$p$-Nitrobenzoic acid is a carboxylic acid,which is significantly more acidic than phenols and alcohols.
$p$-Nitrophenol is more acidic than phenol due to the electron-withdrawing effect of the $-NO_2$ group.
Phenol is more acidic than ethanol due to the resonance stabilization of the phenoxide ion.
Therefore,the increasing order of $pK_a$ values is the reverse of the acidic strength order:
$IV < II < I < III$.

(B) The reaction sequence is as follows:
$1$. Benzaldehyde $(C_6H_5CHO)$ is oxidized by $KMnO_4/H^+$ to form Benzoic acid $(C_6H_5COOH)$,which is '$X$'.
$2$. Benzoic acid reacts with $SOCl_2$ to form Benzoyl chloride $(C_6H_5COCl)$.
$3$. Benzoyl chloride undergoes Rosenmund reduction $(H_2, Pd/BaSO_4)$ to form Benzaldehyde $(C_6H_5CHO)$,which is '$Y$'.
$4$. Benzaldehyde undergoes Clemmensen reduction $(Zn-Hg/Conc. HCl)$ to form Toluene $(C_6H_5CH_3)$,which is '$Z$'.
Therefore,'$Z$' is Toluene.

(C) Assertion $(A)$ is correct: Aldehydes are more reactive than ketones towards nucleophilic addition reactions due to both electronic and steric factors.
Reason $(R)$ is incorrect: In aldehydes,the carbonyl carbon is more electrophilic than in ketones because aldehydes have only one electron-donating alkyl group attached to the carbonyl carbon,whereas ketones have two.
Additionally,aldehydes have less steric hindrance compared to ketones,making them more susceptible to nucleophilic attack.

(C) The reaction of an aldehyde or ketone with a primary amine $(RNH_2)$ leads to the formation of an imine,which is also known as a Schiff base.
In the given reaction,benzaldehyde reacts with aniline $(C_6H_5NH_2)$ to form a Schiff base.
Therefore,$X = C_6H_5NH_2$ and $Y = \text{Schiff base}$.
Other options are incorrect because:
- $NH_2OH$ forms an oxime.
- $NH_2NH_2$ forms a hydrazone.
- $NH_2CONHNH_2$ forms a semicarbazone.

(A) The reaction sequence is as follows:
$1$. Glucose reacts with $HCN$ to form a cyanohydrin $(X)$. The aldehyde group $(-CHO)$ is converted into a cyanohydrin group $(-CH(OH)CN)$.
$2$. The cyanohydrin $(X)$ undergoes acid-catalyzed hydrolysis $(H^+/H_2O)$ to convert the cyano group $(-CN)$ into a carboxylic acid group $(-COOH)$.
$3$. The starting material is glucose,which has $6$ carbon atoms. The addition of $HCN$ adds one carbon atom,resulting in a chain of $7$ carbon atoms.
$4$. The final product $Y$ is a heptanoic acid derivative with hydroxyl groups on carbons $2, 3, 4, 5, 6,$ and $7$. Thus,the $IUPAC$ name is $2,3,4,5,6,7$-hexahydroxyheptanoic acid.

(D) Step $1$: Reaction of acetone with $C_2H_5MgBr$ followed by hydrolysis gives $2-$methylbutan$-2-$ol $(X)$: $(CH_3)_2C=O + C_2H_5MgBr \rightarrow (CH_3)_2C(OH)C_2H_5$.
Step $2$: Reaction of $X$ with $SOCl_2$ gives $2-$chloro$-2-$methylbutane,which upon treatment with $CH_3ONa$ (base) undergoes dehydrohalogenation to form $2-$methylbut$-2-$ene $(Y)$ as the major product.
Step $3$: Addition of $HBr$ to $2-$methylbut$-2-$ene in the presence of peroxide follows anti-Markovnikov's rule to give $1-$bromo$-2-$methylbutane $(Z)$.

(C) $1$. The reaction of propanone $(CH_3)_2CO$ with methylmagnesium bromide $(CH_3MgBr)$ in the presence of ether forms an addition product $A$ (an alkoxide complex).
$2$. Acidic hydrolysis $(H_3O^+)$ of $A$ yields $B$,which is $2-$methylpropan$-2-$ol $(CH_3)_3C-OH$ (a tertiary alcohol).
$3$. When tertiary alcohols are passed over heated copper $(Cu)$ at $573 \ K$,they undergo dehydration to form alkenes.
$4$. Therefore,$2-$methylpropan$-2-$ol undergoes dehydration to form $2-$methylprop$-1-$ene $(CH_3)_2C=CH_2$ as the final product $C$.

(C) Statement $A$ is incorrect: Benzene diazonium chloride is stable only at low temperatures $(273-278 \ K)$. At $285 \ K$,it is unstable and decomposes.
Statement $B$ is correct: The $-NHCOCH_3$ group in $N$-phenylethanamide is electron-withdrawing due to resonance with the carbonyl group,making the ring less electron-rich compared to the $-NH_2$ group in aniline. Thus,it is less reactive towards electrophilic substitution like nitration.
Statement $C$ is incorrect: Hinsberg reagent is benzenesulphonyl chloride $(C_6H_5SO_2Cl)$,not $p-CH_3C_6H_4COCl$.
Therefore,only statement $B$ is correct.

(A) The basicity of amines is inversely proportional to their $pK_b$ values. $A$ lower $pK_b$ value indicates a stronger base.
$1$. $N$-Ethylethanamine (diethylamine) is an aliphatic secondary amine,which is much more basic than aromatic amines due to the absence of resonance stabilization of the lone pair on nitrogen. Thus,it has the lowest $pK_b$ value: $C-IV$ $(3.00)$.
$2$. Among the aromatic amines,$N,N$-Dimethylaniline $(A)$ is more basic than $N$-Methylaniline $(D)$ and Aniline $(B)$ due to the electron-donating $+I$ effect of the two methyl groups,which increases electron density on the nitrogen atom. Thus,$A$ has the lowest $pK_b$ among the aromatics: $A-II$ $(8.92)$.
$3$. $N$-Methylaniline $(D)$ is more basic than Aniline $(B)$ due to the $+I$ effect of one methyl group. Thus,$D-I$ $(9.30)$.
$4$. Aniline $(B)$ is the least basic among these due to the resonance of the lone pair with the benzene ring. Thus,$B-III$ $(9.38)$.
Therefore,the correct matching is $A-II, B-III, C-IV, D-I$.

(B) $1$. The starting material is $4$-amino-$3$-bromoethylbenzene. Treatment with $NaNO_2/HCl$ at $273 \ K$ converts the $-NH_2$ group into a diazonium salt $(-N_2^+Cl^-)$,forming compound $A$.
$2$. Treatment of the diazonium salt with $H_3PO_2/H_2O$ reduces the diazonium group to a hydrogen atom,resulting in $1$-bromo-$2$-ethylbenzene (compound $B$).
$3$. Finally,oxidation of the ethyl group $(-CH_2CH_3)$ with alkaline $KMnO_4$ followed by acidic workup $(H_3O^+)$ converts the alkyl side chain into a carboxylic acid group $(-COOH)$. The product $C$ is $2$-bromobenzoic acid.

(B) The conversion of benzamide $(C_6H_5CONH_2)$ to aniline $(C_6H_5NH_2)$ involves the removal of a carbonyl group,which is achieved by the Hoffmann bromamide degradation reaction using $NaOH + Br_2$. Thus,$X = NaOH + Br_2$.
The conversion of benzamide $(C_6H_5CONH_2)$ to benzylamine $(C_6H_5CH_2NH_2)$ involves the reduction of the amide group to an amine group,which is achieved using a strong reducing agent like $LiAlH_4$ followed by hydrolysis. Thus,$Y = (i) \ LiAlH_4 \ (ii) \ H_2O$.
Therefore,$X = NaOH + Br_2$ and $Y = (i) \ LiAlH_4 \ (ii) \ H_2O$.

(B) The conversion of cyanobenzene (benzonitrile) to a Schiff's base involves two main steps:
$1$. Reduction of cyanobenzene to benzaldehyde using $DIBAL-H$ followed by hydrolysis $(H_2O)$.
$2$. Reaction of benzaldehyde with an amine (like aniline) to form a Schiff's base $(R-CH=N-R')$.
Given the options,the primary reagents for the reduction step are $DIBAL-H$ and $H_2O$.

(A) Cellulose is a linear polysaccharide composed of $ \beta $-$D$-glucose units joined through $1,4$-glycosidic linkages.
These glucose units are linked by glycosidic bonds between the $C_1$ carbon of one unit and the $C_4$ carbon of the adjacent unit,where the anomeric $C_1$ carbon maintains a $ \beta $-configuration.

(B) Glucose $(CHO(CHOH)_4CH_2OH)$ on oxidation with mild oxidizing agents like $Br_2$ water gets converted to gluconic acid $(COOH(CHOH)_4CH_2OH)$. Here,the aldehyde group $(-CHO)$ is oxidized to a carboxylic acid group $(-COOH)$.
Gluconic acid $(COOH(CHOH)_4CH_2OH)$ on oxidation with strong oxidizing agents like $HNO_3$ gets converted to saccharic acid $(COOH(CHOH)_4COOH)$. Here,the primary alcoholic group $(-CH_2OH)$ is oxidized to a carboxylic acid group $(-COOH)$.
Thus,the functional groups involved in the respective conversions are $-CHO$ and $-CH_2OH$.

(D) Lactose,i.e.,milk sugar,is composed of a $\beta-D$-galactose unit and an $\alpha-D$-glucose unit joined by a $\beta$-glycosidic linkage between $C-1$ of galactose and $C-4$ of the glucose unit. Thus,Statement $I$ is incorrect because it mentions $\alpha-D$-galactose instead of $\beta-D$-galactose.
Sucrose,i.e.,cane sugar,is composed of $\alpha-D$-glucose and $\beta-D$-fructose units. These units are joined by an $\alpha,\beta$-glycosidic linkage between $C-1$ of glucose and $C-2$ of fructose. Thus,Statement $II$ is correct.

(A) The reaction of formic acid $(HCOOH)$ with concentrated $H_2SO_4$ is a dehydration reaction:
$HCOOH \xrightarrow{Conc. H_2SO_4, 373 \ K} H_2O + CO$
Here,$X$ is $H_2O$ (water) and $Y$ is $CO$ (carbon monoxide).
In $H_2O$ $(X)$,there are two $O-H$ single bonds,so it has $2 \sigma$ bonds and $0 \pi$ bonds.
In $CO$ $(Y)$,the structure is $C \equiv O$,which consists of $1 \sigma$ bond and $2 \pi$ bonds.
Therefore,for $X$ $(H_2O)$,the count is $2 \sigma, 0 \pi$ and for $Y$ $(CO)$,the count is $1 \sigma, 2 \pi$.
Thus,the correct option is $A$.

(C) The reaction sequence is as follows:
$1$. Benzoic acid reacts with $NH_3$ followed by heating to form benzamide $(X)$.
$2$. Benzamide undergoes Hofmann bromamide degradation with $Br_2/NaOH$ to form aniline $(Y)$.
$3$. Aniline reacts with $CHCl_3/KOH$ (carbylamine reaction) to form phenyl isocyanide $(Z)$,which is characterized by a foul smell.

(B) The reaction shown is a Fries rearrangement.
When phenyl benzoate is treated with $AlCl_3$ (a Lewis acid),it undergoes rearrangement to form a mixture of ortho-hydroxybenzophenone and para-hydroxybenzophenone.
The product '$X$' is ortho-hydroxybenzophenone,which contains a phenolic hydroxyl group $(-OH)$ and a ketone carbonyl group $(-C=O)$.
Therefore,the functional groups present are $-OH$ and $-C=O$ (ketone).

(C) The complete hydrolysis of $XeF_6$ is given by the reaction:
$XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$
Here,the oxide of xenon '$O$' is $XeO_3$.
In the structure of $XeO_3$,the xenon atom is bonded to three oxygen atoms by double bonds $(Xe=O)$.
Each $Xe=O$ bond consists of one $\sigma$ bond and one $\pi$ bond.
Therefore,there are $3$ $\sigma$ bonds and $3$ $\pi$ bonds.
Total number of $\sigma$ and $\pi$ bonds $= 3 + 3 = 6$.

(A) For a first order reaction,$k = \frac{2.303}{t} \log \frac{a}{a-x}$.
For $t_{1/4}$,$x = \frac{a}{4}$,so $t_{1/4} = \frac{2.303}{k} \log \frac{a}{3a/4} = \frac{2.303}{k} \log \frac{4}{3}$.
For $t_{1/10}$,$x = \frac{a}{10}$,so $t_{1/10} = \frac{2.303}{k} \log \frac{a}{9a/10} = \frac{2.303}{k} \log \frac{10}{9}$.
Taking the ratio: $\frac{t_{1/4}}{t_{1/10}} = \frac{\log(4/3)}{\log(10/9)} = \frac{\log 4 - \log 3}{\log 10 - \log 9} = \frac{2 \log 2 - \log 3}{1 - 2 \log 3}$.
Using $\log 2 = 0.3$ and $\log 3 = 0.477$: $\frac{t_{1/4}}{t_{1/10}} = \frac{2(0.3) - 0.477}{1 - 2(0.477)} = \frac{0.6 - 0.477}{1 - 0.954} = \frac{0.123}{0.046} \approx 2.67$.
Thus,$\frac{t_{1/4}}{t_{1/10}} \times 100 \approx 267$. The closest option is $272$.

(B) For the reaction $N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$,the rate of reaction is given by:
Rate $= -\frac{d[N_2O_5]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt} = 2 \frac{d[O_2]}{dt}$
Substituting the given rate expressions:
$K_1[N_2O_5] = \frac{1}{2} K_2[N_2O_5] = 2K_3[N_2O_5]$
Dividing by $[N_2O_5]$ gives $K_1 = \frac{1}{2} K_2 = 2K_3$.
Multiplying the entire relation by $2$,we get $2K_1 = K_2 = 4K_3$.

(A) For the reaction $A_2 + B_2 \rightarrow 2 AB$,the rate of reaction $R$ is defined as $R = k[A_2]^x [B_2]^y$.
The rate of formation of $AB$ is given as $\frac{d[AB]}{dt} = 2R$.
From the table:
$1.25 \times 10^{-4} = k(0.1)^x(0.1)^y \dots (i)$
$2.5 \times 10^{-4} = k(0.2)^x(0.1)^y \dots (ii)$
$5.0 \times 10^{-4} = k(0.2)^x(0.2)^y \dots (iii)$
Dividing $(ii)$ by $(i)$,we get $2^x = 2$,so $x = 1$.
Dividing $(iii)$ by $(ii)$,we get $2^y = 2$,so $y = 1$.
Thus,the rate law is $R = k[A_2][B_2]$.
Using values from $(i)$: $1.25 \times 10^{-4} = k(0.1)(0.1) = k(0.01)$.
$k = \frac{1.25 \times 10^{-4}}{0.01} = 1.25 \times 10^{-2} \ L \ mol^{-1} s^{-1}$.

(C) For a first order reaction,the integrated rate equation is given by:
$\log \frac{a}{a-x} = \frac{kt}{2.303}$
Comparing this with the equation of a straight line $y = mx + c$:
Here,$y = \log \frac{a}{a-x}$,$x = t$,and the slope $m = \frac{k}{2.303}$.
Given that the slope $m = 2 \times 10^{-3} \ min^{-1}$,we have:
$\frac{k}{2.303} = 2 \times 10^{-3} \ min^{-1}$
Therefore,$k = 2.303 \times 2 \times 10^{-3} \ min^{-1} = 4.606 \times 10^{-3} \ min^{-1}$.

(A) According to the Arrhenius equation: $\ln k = \ln A - \frac{E_a}{RT}$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = \ln k$ and $x = \frac{1}{T}$:
The slope $(m) = -\frac{E_a}{R}$.
Given the slope $(m) = -10^3 \ K$,we have:
$-\frac{E_a}{R} = -10^3 \ K$
$E_a = 10^3 \ K \times R$
$E_a = 10^3 \ K \times 8.314 \ J \ mol^{-1} \ K^{-1} = 8314 \ J \ mol^{-1}$.
To convert to $kJ \ mol^{-1}$,divide by $1000$:
$E_a = \frac{8314}{1000} \ kJ \ mol^{-1} = 8.314 \ kJ \ mol^{-1}$.

(C) Chloramphenicol is a chlorine-containing antibiotic produced by microorganisms. It is a broad-spectrum antibiotic that is very effective for the treatment of typhoid fever. It acts as a bacteriostatic drug,meaning it inhibits the growth of bacteria rather than killing them. Therefore,the statement that it is a bactericidal drug is incorrect.

(C) $\rightarrow (III), (B)$ $\rightarrow (II), (C)$ $\rightarrow (I), (D)$ $\rightarrow (IV)$
$(A)$ Antacid: Substances which neutralize excess acid in the stomach are called antacids. Example: Ranitidine.
$(B)$ Antihistamine: Substances which reduce the release of histamine are called antihistamines. Example: Seldane.
$(C)$ Tranquilizer: Substances that are used to treat nervous system disorders. Example: Serotonin.
$(D)$ Antibiotic: Substances which inhibit or kill the growth of bacteria. Example: Chloramphenicol.

(A) Vitamin $B_6$ is also known as Pyridoxine.
$B_2$ is known as Riboflavin.
$B_1$ is known as Thiamine.
$B_{12}$ is known as Cobalamin.

(B) Aspirin (acetylsalicylic acid) is a non-narcotic analgesic that exhibits the following properties:
$1$. Anti-inflammatory $(A)$: It reduces inflammation.
$2$. Antipyretic $(C)$: It reduces fever.
$3$. Anticoagulant $(D)$: It prevents blood clotting.
It does not act as an antidepressant $(B)$ or a hypnotic $(E)$.
Therefore,the correct effects are $A, C, D$.

(D) $\rightarrow (II), (B)$ $\rightarrow (I), (C)$ $\rightarrow (IV), (D)$ $\rightarrow (III)$
The electronic configuration of elements is as follows:
$(A)$ $Ra$ $(Z=88)$: The last electron enters the $7s$ orbital. Thus,it is an $s$-block element.
$(B)$ $Uuq$ ($Z=114$,Flerovium): The last electron enters the $7p$ orbital. Thus,it is a $p$-block element.
$(C)$ $Ds$ ($Z=110$,Darmstadtium): The last electron enters the $6d$ orbital. Thus,it is a $d$-block element.
$(D)$ $Fm$ ($Z=100$,Fermium): The last electron enters the $5f$ orbital. Thus,it is an $f$-block element.

(C) The general electronic configuration $(n-1)d^{1-10}ns^2$ represents the transition elements (d-block elements).
Let us analyze the configurations of the elements in option $C$:
${}_{108}Hs \Rightarrow [Rn] 5f^{14} 6d^6 7s^2$
${}_{80}Hg \Rightarrow [Xe] 4f^{14} 5d^{10} 6s^2$
${}_{74}W \Rightarrow [Xe] 4f^{14} 5d^4 6s^2$
All these elements belong to the d-block and follow the general configuration $(n-1)d^{1-10}ns^2$.

(A) The reaction of the complex with $AgNO_3$ precipitates $Cl^-$ ions present outside the coordination sphere as $AgCl$.
Since $1$ mole of the complex yields $3$ moles of $AgCl$,there must be $3$ chloride ions outside the coordination sphere.
The formula $[Co(H_2O)_6]Cl_3$ dissociates as:
$[Co(H_2O)_6]Cl_3(aq) \rightarrow [Co(H_2O)_6]^{3+}(aq) + 3Cl^-(aq)$
The $3Cl^-$ ions react with $AgNO_3$ to form $3$ moles of $AgCl$:
$3Ag^+(aq) + 3Cl^-(aq) \rightarrow 3AgCl(s)$
Thus,the correct formula is $[Co(H_2O)_6]Cl_3$.

(A) Geometrical isomerism in coordination complexes occurs when ligands can be arranged in different spatial positions relative to each other.
For the complex $\left[ Co(Cl)_2(en)_2 \right]^{+}$,the two $Cl^-$ ligands can be either adjacent to each other (cis-isomer) or opposite to each other (trans-isomer).
Therefore,$\left[ Co(Cl)_2(en)_2 \right]^{+}$ exhibits geometrical isomerism.

(A) In $[Fe(H_2O)_6]^{3+}$,the oxidation state of $Fe$ is $+3$.
$Fe$ $(Z=26)$: $[Ar] 3d^6 4s^2$.
$Fe^{3+}$: $[Ar] 3d^5$.
Since $H_2O$ is a weak field ligand,pairing of electrons does not occur.
According to Crystal Field Theory $(CFT)$,the five $d$-electrons occupy the $t_{2g}$ and $e_g$ orbitals singly,resulting in the electronic configuration $t_{2g}^3 e_g^2$.

(A) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. The lower the number of unpaired electrons,the lower the magnetic moment.

Ion	Electronic Configuration	Number of Unpaired Electrons $(n)$
$[Ti(H_2O)_6]^{3+}$	$3d^1$	$1$
$[Mn(H_2O)_6]^{2+}$	$3d^5$	$5$
$[Ni(H_2O)_6]^{2+}$	$3d^8$	$2$
$[Co(H_2O)_6]^{2+}$	$3d^7$	$3$

Since $[Ti(H_2O)_6]^{3+}$ has the lowest number of unpaired electrons $(n=1)$,it has the lowest spin-only magnetic moment.

(B) The crystal field splitting energy $(\Delta_o)$ depends on the strength of the ligand.
Stronger ligands cause greater splitting.
According to the spectrochemical series,the order of field strength for the given ligands is: $F^- < H_2O < NH_3 < CN^-$.
Therefore,the order of crystal field splitting energy is:
$[CoF_6]^{3-} < [Co(H_2O)_6]^{3 } < [Co(NH_3)_6]^{3 } < [Co(CN)_6]^{3-}$.
This corresponds to $IV < I < II < III$.

(C) The spin-only magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $\mu = 2.86 \ BM$,we have $\sqrt{n(n+2)} \approx 2.86$,which implies $n(n+2) \approx 8$,so $n = 2$.
In $[NiCl_4]^{2-}$,$Ni$ is in the $+2$ oxidation state ($3d^8$ configuration).
For $Ni^{2+}$ $(3d^8)$,the electrons are arranged as $t_{2g}^6 e_g^2$ in a tetrahedral field,resulting in $n = 2$ unpaired electrons.
Thus,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$,which matches the given value.