JEE Main 2022 Chemistry Question Paper with Answer and Solution

(A) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_{g} RT$.
For the reaction $H_{2}F_{2(g)} \rightarrow H_{2(g)} + F_{2(g)}$,the change in the number of gaseous moles is $\Delta n_{g} = (1 + 1) - 1 = 1$.
The temperature is $T = 27 + 273 = 300 \ K$.
Given $\Delta U = -59.6 \ kJ \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1} = 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.
Substituting the values: $\Delta H = -59.6 + (1 \times 8.314 \times 10^{-3} \times 300) = -59.6 + 2.4942 = -57.1058 \ kJ \ mol^{-1}$.
The nearest integer value is $-57 \ kJ \ mol^{-1}$.

(B) The reaction of ethanol with $PCl_{3}$ produces ethyl chloride and phosphorous acid $(H_{3}PO_{3})$,which is compound $A$.
$3C_{2}H_{5}OH + PCl_{3} \rightarrow 3C_{2}H_{5}Cl + H_{3}PO_{3} (A)$
Further reaction of $H_{3}PO_{3}$ with $PCl_{3}$ yields pyrophosphorous acid $(H_{4}P_{2}O_{5})$,which is compound $B$.
$2H_{3}PO_{3} + PCl_{3} \rightarrow H_{4}P_{2}O_{5} (B) + PCl_{3} + ...$ (The reaction stoichiometry leads to $H_{4}P_{2}O_{5}$).
In the structure of pyrophosphorous acid $(H_{4}P_{2}O_{5})$,there are two $P-H$ bonds.
Protons attached directly to the phosphorus atom $(P-H)$ are non-ionisable.
Therefore,there are $2$ non-ionisable protons in $H_{4}P_{2}O_{5}$.

(D) Monochlorination of methylcyclohexane $(C_7H_{14})$ occurs at different positions:
$1$. At the methyl group: $1$ isomer (chloromethylcyclohexane).
$2$. At the $C_1$ position: $1$ isomer ($1$-chloro$-1-$methylcyclohexane).
$3$. At the $C_2$ position: $2$ chiral centers are present,leading to $2^2 = 4$ stereoisomers.
$4$. At the $C_3$ position: $2$ chiral centers are present,leading to $2^2 = 4$ stereoisomers.
$5$. At the $C_4$ position: $2$ stereoisomers (cis and trans$-1-$chloro$-4-$methylcyclohexane).
Total isomers = $1 + 1 + 4 + 4 + 2 = 12$.

(C) The reaction is: $CH_3CH_2MgBr + CH_3OH \rightarrow CH_3CH_3 + Mg(OCH_3)Br$.
The gas produced is ethane $(C_2H_6)$.
At $STP$,$22400 \ mL$ of gas corresponds to $1 \ mol$.
Therefore,$2.24 \ mL$ of gas corresponds to $n = \frac{2.24}{22400} = 10^{-4} \ mol$.
The molar mass of ethane $(C_2H_6)$ is $2 \times 12 + 6 \times 1 = 30 \ g/mol$.
The weight of gas produced is $W = n \times M = 10^{-4} \ mol \times 30 \ g/mol = 30 \times 10^{-4} \ g = 3 \times 10^{-3} \ g$.
Since $1 \ g = 1000 \ mg$,the weight is $3 \times 10^{-3} \times 1000 = 3 \ mg$.

(C) tautology is a statement that is always true for all possible truth values of its components.
$(A)$ $((\sim q) \wedge p) \wedge q = (\sim q \wedge q) \wedge p = F \wedge p = F$ (Contradiction)
$(B)$ $((\sim q) \wedge p) \wedge (p \wedge (\sim p)) = (\sim q \wedge p) \wedge F = F$ (Contradiction)
$(C)$ $((\sim q) \wedge p) \vee (p \vee (\sim p)) = ((\sim q) \wedge p) \vee T = T$ (Tautology)
$(D)$ $(p \wedge q) \wedge (\sim (p \wedge q)) = F$ (Contradiction)
Thus,option $C$ is a tautology.

(C) The angle of incidence $i$ is the angle between the incident ray and the normal. The given angle is with the interface,so $i = 90^{\circ} - 45^{\circ} = 45^{\circ}$.
The angle of deviation $D$ is given by $D = i - r$,where $r$ is the angle of refraction.
Given $D = 15^{\circ}$,we have $15^{\circ} = 45^{\circ} - r$,which gives $r = 30^{\circ}$.
Using Snell's law,$n_1 \sin i = n_2 \sin r$,where $n_1 = 1$ (for air) and $n_2 = \mu$ (refractive index of the medium).
$1 \cdot \sin 45^{\circ} = \mu \cdot \sin 30^{\circ}$
$\frac{1}{\sqrt{2}} = \mu \cdot \frac{1}{2}$
$\mu = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414$.

(C) The shapes of the given compounds are as follows:
$A$. $BrF_5$: It has $5$ bond pairs and $1$ lone pair,resulting in a square pyramidal geometry $(II)$.
$B$. $[CrF_6]^{3-}$: It has $6$ bond pairs and $0$ lone pairs,resulting in an octahedral geometry $(IV)$.
$C$. $O_3$: It has $2$ bond pairs and $1$ lone pair on the central atom,resulting in a bent shape $(I)$.
$D$. $PCl_5$: It has $5$ bond pairs and $0$ lone pairs,resulting in a trigonal bipyramidal geometry $(III)$.
Therefore,the correct matching is $A-II, B-IV, C-I, D-III$.

(D) Statement $I$ is incorrect because the covalent radius is defined as half of the internuclear distance between two bonded atoms in a molecule,not double the atomic radius.
Statement $II$ is correct because in an anion,the number of electrons increases while the nuclear charge remains the same,leading to increased electron-electron repulsion and a decrease in effective nuclear charge,which results in a larger ionic radius compared to the parent atom.

(A) $Urea$ acts as a stabilizer for $H_{2}O_{2}$ and prevents its decomposition.

(B) The reaction between $BeCl_{2}$ and $LiAlH_{4}$ is a common method for the preparation of beryllium hydride $(BeH_{2})$.
The balanced chemical equation is:
$2BeCl_{2} + LiAlH_{4} \rightarrow 2BeH_{2} + LiCl + AlCl_{3}$
Thus,the products formed are $BeH_{2}$,$LiCl$,and $AlCl_{3}$,which correspond to options $(B)$,$(D)$,and $(A)$.

(B) The preparation of borazine $(B_3N_3H_6)$ involves the reaction of diborane $(B_2H_6)$ with ammonia $(NH_3)$ at high temperatures.
The balanced chemical equation is:
$3 B_2H_6 + 6 NH_3 \xrightarrow{\Delta} 2 B_3N_3H_6 + 12 H_2$
Here,$X$ is $B_2H_6$ and $Y$ is $NH_3$.

(C) disproportionation reaction is one in which the same element in a given oxidation state is simultaneously oxidized and reduced.
$A$: $2 H_2O_2 \rightarrow 2 H_2O + O_2$: Oxygen in $H_2O_2$ is in $-1$ state. It goes to $-2$ in $H_2O$ (reduction) and $0$ in $O_2$ (oxidation). This is a disproportionation reaction.
$B$: $2 NO_2 + H_2O \rightarrow HNO_3 + HNO_2$: Nitrogen in $NO_2$ is in $+4$ state. It goes to $+5$ in $HNO_3$ (oxidation) and $+3$ in $HNO_2$ (reduction). This is a disproportionation reaction.
$C$: $MnO_4^- + 4 H^+ + 3 e^- \rightarrow MnO_2 + 2 H_2O$: Manganese in $MnO_4^-$ is in $+7$ state and goes to $+4$ in $MnO_2$. This is a simple reduction reaction,not disproportionation.
$D$: $3 MnO_4^{2-} + 4 H^+ \rightarrow 2 MnO_4^- + MnO_2 + 2 H_2O$: Manganese in $MnO_4^{2-}$ is in $+6$ state. It goes to $+7$ in $MnO_4^-$ (oxidation) and $+4$ in $MnO_2$ (reduction). This is a disproportionation reaction.

(A) In acidic medium,the reduction reaction of the permanganate ion is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
In $MnO_4^-$,the oxidation state of $Mn$ is $+7$.
In $Mn^{2+}$,the oxidation state of $Mn$ is $+2$.
The change in the oxidation number of manganese is $|(+2) - (+7)| = 5$.

(C) The reaction between a chlorine radical $(\dot{Cl})$ and methane $(CH_4)$ is a propagation step in the free radical substitution mechanism.
The chlorine radical abstracts a hydrogen atom from methane to form a methyl radical $(\dot{CH}_3)$ and hydrogen chloride $(HCl)$.
The reaction is: $\dot{Cl} + CH_4 \rightarrow \dot{CH}_3 + HCl$.
Therefore,$A = \dot{CH}_3$ and $B = HCl$.

(D) $p-$nitrophenol exhibits intermolecular hydrogen bonding,while picric acid ($2,4,6-$trinitrophenol) exhibits intramolecular hydrogen bonding.
Due to these differences in hydrogen bonding,they possess different polarities and adsorption characteristics on a stationary phase like silica gel.
For a small quantity like $100 \ mg$,chromatographic techniques are highly effective.
Preparative $TLC$ is a suitable technique for separating small amounts of mixtures based on differences in their $R_f$ values.

(C) $[10]$ Annulene,although it follows the $(4n + 2) \pi$ electron rule,is non-aromatic due to its non-planar nature.
It is non-planar because of the steric repulsion between the two $C-H$ bonds present inside the ring.

(A) The given reaction is an example of Birch reduction,which involves the reduction of an aromatic ring to a non-conjugated $1,4-$cyclohexadiene using an alkali metal (like $Na$ or $Li$) in liquid ammonia $(NH_3)$ in the presence of an alcohol (like $CH_3CH_2OH$).
For an aromatic ring substituted with an electron-donating group (like $-OCH_3$),the substituent remains on the double bond of the resulting $1,4-$cyclohexadiene.
For an aromatic ring substituted with an electron-withdrawing group (like $-NO_2$),the substituent is located on the saturated carbon (the $sp^3$ hybridized carbon) of the resulting $1,4-$cyclohexadiene.
In the given reactant,$o$-nitroanisole,the $-NO_2$ group is a strong electron-withdrawing group,while the $-OCH_3$ group is an electron-donating group. The reduction occurs such that the $-NO_2$ group ends up on the saturated carbon atom of the $1,4-$cyclohexadiene ring. Thus,the correct product is the one where the $-NO_2$ group is at the $sp^3$ carbon and the $-OCH_3$ group is on the double bond.

(A) The stability of the diazonium cation is increased by electron-donating groups $(EDG)$ and decreased by electron-withdrawing groups $(EWG)$ attached to the benzene ring.
$(A)$ contains $-OCH_3$,which is an $EDG$ (via resonance),thus it is the most stable.
$(C)$ is the unsubstituted benzene diazonium salt.
$(D)$ contains $-CN$,which is an $EWG$.
$(B)$ contains $-NO_2$,which is a very strong $EWG$.
Therefore,the stability order is $(A) > (C) > (D) > (B)$.

(B) Concentration in $ppm = \frac{\text{Mass of solute (in } g)}{\text{Volume of solution (in } mL)} \times 10^{6}$.
Given $ppm = 48$,Volume $= 2 \, L = 2000 \, mL$.
Mass of $Mg = \frac{48 \times 2000}{10^{6}} = 96 \times 10^{-3} \, g = 0.096 \, g$.
Moles of $Mg = \frac{\text{Mass}}{\text{Atomic mass}} = \frac{0.096}{24} = 0.004 \, mol = 4 \times 10^{-3} \, mol$.
Number of $Mg$ atoms $= \text{Moles} \times N_{A} = 4 \times 10^{-3} \times 6.02 \times 10^{23} = 24.08 \times 10^{20}$.
Comparing with $x \times 10^{20}$,we get $x = 24.08$.
The nearest integer value of $x$ is $24$.

(B) Let the total mass of the mixture be $100 \, g$.
Mass of $H_2 = 40 \, g$,so moles of $H_2$ $(n_{H_2})$ $= \frac{40}{2} = 20 \, mol$.
Mass of $O_2 = 60 \, g$,so moles of $O_2$ $(n_{O_2})$ $= \frac{60}{32} = 1.875 \, mol$.
Mole fraction of $H_2$ $(x_{H_2})$ $= \frac{n_{H_2}}{n_{H_2} + n_{O_2}} = \frac{20}{20 + 1.875} = \frac{20}{21.875} \approx 0.9143$.
Partial pressure of $H_2$ $(P_{H_2})$ $= x_{H_2} \times P_{total} = 0.9143 \times 2.2 \, bar = 2.01146 \, bar$.
The nearest integer is $2 \, bar$.

(B) Given that the de Broglie wavelengths are equal: $\lambda_{e} = \lambda_{N}$.
Using the de Broglie relation $\lambda = \frac{h}{mv}$,we have $\frac{h}{m_{e} v_{e}} = \frac{h}{m_{N} v_{N}}$.
This simplifies to $m_{e} v_{e} = m_{N} v_{N}$,or $v_{e} = \frac{m_{N}}{m_{e}} v_{N}$.
Given $v_{e} = x \ v_{N}$,we have $x = \frac{m_{N}}{m_{e}}$.
Substituting the values: $x = \frac{1.6 \times 10^{-27} \ kg}{9.1 \times 10^{-31} \ kg}$.
$x = \frac{1.6}{9.1} \times 10^{4} = 0.175824 \times 10^{4} = 1758.24$.
The nearest integer value of $x$ is $1758$.

(D) The combustion reaction is: $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$; $\Delta H = -x \ kJ \ mol^{-1}$.
Heat released $(Q)$ is calculated as: $Q = C \times \Delta T = 20.0 \ kJ \ K^{-1} \times (300 \ K - 298 \ K) = 20.0 \ kJ \ K^{-1} \times 2 \ K = 40 \ kJ$.
This heat is released by $2.4 \ g$ of carbon $(C)$.
For $1 \ mole$ of $C$ $(12 \ g)$:
$Q = \frac{40 \ kJ}{2.4 \ g} \times 12 \ g = 200 \ kJ \ mol^{-1}$.
Since $\Delta n_g = 0$ for the reaction,$\Delta H = \Delta U = -200 \ kJ \ mol^{-1}$.
Therefore,$x = 200$.

(D) Initial moles of $HNO_3$ = $Molarity \times Volume(L) = 0.5 \times 0.8 = 0.4 \ mol$.
Moles of $HNO_3$ evaporated = $\frac{Mass}{Molar \ mass} = \frac{11.5 \ g}{63 \ g \ mol^{-1}} \approx 0.1825 \ mol$.
Moles of $HNO_3$ remaining = $0.4 - 0.1825 = 0.2175 \ mol$.
New volume of solution = $\frac{800 \ mL}{2} = 400 \ mL = 0.4 \ L$.
New molarity = $\frac{Moles \ remaining}{Volume(L)} = \frac{0.2175}{0.4} = 0.54375 \ M$.
Expressing in $x \times 10^{-2} \ M$,we get $54.375 \times 10^{-2} \ M$.
Rounding to the nearest integer,$x = 54$.

(C) When benzene $(C_{6}H_{6})$ reacts with excess chlorine $(Cl_{2})$ in the presence of sunlight ($UV$ light),an addition reaction occurs.
This reaction results in the formation of benzene hexachloride $(C_{6}H_{6}Cl_{6})$,also known as gammaxene or lindane.
The product $X$ is $C_{6}H_{6}Cl_{6}$.
In this molecule,each of the $6$ carbon atoms is bonded to one hydrogen atom and one chlorine atom.
Therefore,the total number of hydrogen atoms in $X$ is $6$.

(A) According to the $(n+\ell)$ rule,the energy of an orbital is determined by the sum of its principal quantum number $(n)$ and azimuthal quantum number $(\ell)$.
For $A$: $n+\ell = 3+0 = 3$ ($3s$ orbital)
For $B$: $n+\ell = 4+0 = 4$ ($4s$ orbital)
For $C$: $n+\ell = 3+1 = 4$ ($3p$ orbital)
For $D$: $n+\ell = 3+2 = 5$ ($3d$ orbital)
Rules for energy:
$1$. Higher $(n+\ell)$ value means higher energy.
$2$. If $(n+\ell)$ values are equal,the orbital with the higher $n$ value has higher energy.
Comparing the values:
$D$ $(n+\ell=5)$ has the highest energy.
Between $B$ and $C$,both have $(n+\ell)=4$. Since $B$ has $n=4$ and $C$ has $n=3$,$B$ has higher energy than $C$.
$A$ $(n+\ell=3)$ has the lowest energy.
Therefore,the decreasing order of energy is $D > B > C > A$.

(C) . $\Psi_{MO} = \Psi_{A} - \Psi_{B}$ represents the Anti-bonding molecular orbital $(III)$.
$B$. $\mu = Q \times r$ is the formula for Dipole moment $(I)$.
$C$. $\frac{N_{b} - N_{a}}{2}$ is the formula for Bond order $(IV)$.
$D$. $\Psi_{MO} = \Psi_{A} + \Psi_{B}$ represents the Bonding molecular orbital $(II)$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(A) The titration of a weak base $(NH_4OH)$ with a strong acid $(HCl)$ involves the addition of acid to the base.
Initially,the $pH$ is high (basic).
As the strong acid is added,the $pH$ decreases gradually.
Near the equivalence point,there is a sharp drop in $pH$.
Since the resulting salt $(NH_4Cl)$ is acidic due to the hydrolysis of the $NH_4^+$ ion,the equivalence point $pH$ is less than $7$.
Therefore,the correct plot shows a downward curve starting from a high $pH$ value.

(B) The elements are identified as:
$A: 3s^{2} \rightarrow Mg$
$B: 3s^{2} 3p^{1} \rightarrow Al$
$C: 3s^{2} 3p^{3} \rightarrow P$
$D: 3s^{2} 3p^{4} \rightarrow S$
Ionization enthalpy generally increases across a period from left to right.
However,there are exceptions due to stable electronic configurations:
$1$. $Mg$ $(3s^{2})$ has a fully filled $s$-orbital,making it more stable than $Al$ $(3s^{2} 3p^{1})$.
$2$. $P$ $(3s^{2} 3p^{3})$ has a half-filled $p$-orbital,making it more stable than $S$ $(3s^{2} 3p^{4})$.
Comparing the values:
$Al < Mg < S < P$
Substituting the labels:
$B < A < D < C$

(A) In group $1$,$Li^{+}$ has the maximum hydration enthalpy due to its smallest ionic size.
$Li$ shows a diagonal relationship with $Mg$ of group $2$.
Therefore,the element $B$ is $Mg$.

(B) Assertion $A$ is true: Boron cannot form $BF_{6}^{3-}$ because it lacks $d$-orbitals in its valence shell,which prevents the expansion of its octet.
Reason $R$ is true: The size of the Boron atom is indeed very small.
Conclusion: While both statements are true,the inability to form $BF_{6}^{3-}$ is primarily due to the absence of $d$-orbitals,not just the small size of the atom. Therefore,$R$ is not the correct explanation for $A$.

(D) In a neutral or weakly alkaline medium,the permanganate ion $(MnO_{4}^{-})$ acts as an oxidizing agent and gets reduced to manganese dioxide $(MnO_{2})$.
Thiosulphate $(S_{2}O_{3}^{2-})$ is oxidized to sulphate $(SO_{4}^{2-})$.
The balanced chemical equation for this reaction is:
$8 MnO_{4}^{-} + 3 S_{2}O_{3}^{2-} + H_{2}O \rightarrow 8 MnO_{2} + 6 SO_{4}^{2-} + 2 OH^{-}$

(A) Statement $I$: Fly ash and slag generated as waste from the steel industry are effectively utilized by the cement industry as raw materials.
Statement $II$: Fuel derived from plastic waste through processes like pyrolysis has a high octane rating and is lead-free,often referred to as green fuel.
Therefore,both Statement $I$ and Statement $II$ are correct.

(D) Chloroform and Aniline are separated by $(III)$ Distillation because they have different boiling points.
$(B)$ Benzoic acid and Naphthalene are separated by $(IV)$ Crystallisation based on differences in their solubilities in a suitable solvent.
$(C)$ Water and Aniline are separated by $(I)$ Steam distillation as Aniline is steam volatile and immiscible with water.
$(D)$ Naphthalene and Sodium chloride are separated by $(II)$ Sublimation because Naphthalene sublimes while Sodium chloride does not.

(A) The reaction is: $H_{2}SO_{4} + 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O$.
Number of milliequivalents of $H_{2}SO_{4} = \text{Molarity} \times \text{n-factor} \times \text{Volume (mL)} = 0.1 \times 2 \times 100 = 20 \ mEq$.
Number of milliequivalents of $NaOH = 0.1 \times 1 \times 50 = 5 \ mEq$.
Since $H_{2}SO_{4}$ is in excess,the remaining milliequivalents of $H_{2}SO_{4} = 20 - 5 = 15 \ mEq$.
Total volume of the solution $= 100 \ mL + 50 \ mL = 150 \ mL$.
Normality of the resulting solution $= \frac{\text{Remaining mEq}}{\text{Total Volume (mL)}} = \frac{15}{150} = 0.1 \ N = 1 \times 10^{-1} \ N$.
Thus,the nearest integer is $1$.

(A) For a real gas under high pressure,the compressibility factor $Z$ is given by the equation: $Z = 1 + \frac{Pb}{RT}$.
Given $Z = 2$,$P = 99 \ bar$,$T = 25 + 273 = 298 \ K$,and $R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$.
Substituting the values: $2 = 1 + \frac{99 \times b}{0.083 \times 298}$.
$1 = \frac{99 \times b}{24.734}$.
$b = \frac{24.734}{99} \approx 0.2498 \ L \ mol^{-1}$.
$b \approx 0.25 \ L \ mol^{-1} = 25 \times 10^{-2} \ L \ mol^{-1}$.
Thus,the nearest integer value is $25$.

(A) Let $x \ g$ be the mass of the gas burnt.
Moles of gas $= \frac{x}{280} \ mol$.
Heat released by the combustion $= \text{Heat capacity} \times \Delta T = 2.5 \ kJ \ K^{-1} \times (298.45 - 298.0) \ K = 2.5 \times 0.45 \ kJ = 1.125 \ kJ$.
Since the combustion occurs in a constant volume calorimeter,the heat released is equal to the internal energy change,$\Delta U$.
Given the enthalpy of combustion $\Delta H = 9 \ kJ \ mol^{-1}$,and assuming $\Delta H \approx \Delta U$ for this calculation,we have:
$\Delta U \text{ per mole} = 9 \ kJ \ mol^{-1}$.
Total heat released $= \text{moles} \times \Delta U \text{ per mole}$.
$1.125 \ kJ = (\frac{x}{280} \ mol) \times 9 \ kJ \ mol^{-1}$.
$x = \frac{1.125 \times 280}{9} = 35 \ g$.

(C) In a basic medium,potassium permanganate $(KMnO_4)$ reacts with hydrogen peroxide $(H_2O_2)$ to produce manganese dioxide $(MnO_2)$,oxygen gas $(O_2)$,water $(H_2O)$,and potassium hydroxide $(KOH)$.
The balanced chemical equation is:
$2KMnO_4 + 3H_2O_2 \longrightarrow 2MnO_2 + 3O_2 + 2H_2O + 2KOH$
In the product $MnO_2$,let the oxidation state of $Mn$ be $x$.
$x + 2(-2) = 0$
$x - 4 = 0$
$x = +4$
Therefore,the oxidation state of manganese in the product is $4$.

(C) To determine the number of non-planar species,we analyze the geometry of each:
$1$. $NO_{3}^{-}$: $sp^{2}$ hybridization,trigonal planar.
$2$. $H_{2}O_{2}$: Open book structure,non-planar.
$3$. $BF_{3}$: $sp^{2}$ hybridization,trigonal planar.
$4$. $PCl_{3}$: $sp^{3}$ hybridization,trigonal pyramidal,non-planar.
$5$. $XeF_{4}$: $sp^{3}d^{2}$ hybridization,square planar.
$6$. $SF_{4}$: $sp^{3}d$ hybridization,see-saw shape,non-planar.
$7$. $XeO_{3}$: $sp^{3}$ hybridization,trigonal pyramidal,non-planar.
$8$. $PH_{4}^{+}$: $sp^{3}$ hybridization,tetrahedral,non-planar.
$9$. $SO_{3}$: $sp^{2}$ hybridization,trigonal planar.
$10$. $[Al(OH)_{4}]^{-}$: $sp^{3}$ hybridization,tetrahedral,non-planar.
The non-planar species are: $H_{2}O_{2}, PCl_{3}, SF_{4}, XeO_{3}, PH_{4}^{+}, [Al(OH)_{4}]^{-}$.
Counting these,we get a total of $6$ non-planar species.

(D) The molar mass of toluene $(C_7H_8)$ is $7 \times 12 + 8 \times 1 = 92 \ g/mol$.
Number of moles of toluene = $\frac{5 \ g}{92 \ g/mol} = \frac{5}{92} \ mol$.
Since the reaction stoichiometry is $1:1$,the theoretical moles of benzaldehyde produced would be $\frac{5}{92} \ mol$.
Given the yield is $92 \%$,the actual moles of benzaldehyde produced = $\frac{5}{92} \times \frac{92}{100} = 5 \times 10^{-2} \ mol$.
The molar mass of benzaldehyde $(C_7H_6O)$ is $7 \times 12 + 6 \times 1 + 16 = 106 \ g/mol$.
Mass of benzaldehyde produced = $\text{moles} \times \text{molar mass} = (5 \times 10^{-2} \ mol) \times (106 \ g/mol) = 530 \times 10^{-2} \ g$.
Thus,the value is $530$.

(A) For a hydrogen-like atom or ion,the energy of an orbital depends only on the principal quantum number $n$. However,for multi-electron atoms like lithium $(Z=3)$,the energy of an orbital depends on both $n$ and the effective nuclear charge $(Z_{eff})$.
As the atomic number $(Z)$ increases,the effective nuclear charge increases,which leads to a stronger attraction between the nucleus and the electrons.
Consequently,the energy of the same orbital (e.g.,$2s$) decreases as the atomic number increases.
Therefore,the energy of the $2s$ orbital in hydrogen $(Z=1)$ is higher than that in lithium $(Z=3)$.
Both Assertion $A$ and Reason $R$ are correct,and $R$ correctly explains $A$.

(D) $1$. Ionization enthalpy decreases down a group. Thus,for alkali metals: $Li > Na > K$. Statement $A$ is correct.
$2$. In noble gases,ionization enthalpy decreases down the group. $Xe$ is below $He, Ne, Ar, Kr$,so it has a lower ionization enthalpy than them. However,$Rn$ is below $Xe$,so $Rn$ has the lowest. Thus,$Xe$ is not the lowest. Statement $B$ is correct.
$3$. Atomic number $37$ is $Rb$ (Group $1$) and atomic number $38$ is $Sr$ (Group $2$). Across a period,ionization enthalpy increases. Thus,$Sr > Rb$. Statement $C$ is correct.
$4$. Atomic number $30$ is $Zn$ $([Ar] 3d^{10} 4s^2)$ and $Ga$ is $([Ar] 3d^{10} 4s^2 4p^1)$. Due to the stable $d^{10}$ configuration and higher effective nuclear charge,$Zn$ has a higher ionization enthalpy than $Ga$. Thus,the statement that $Ga$ is higher than $Zn$ is incorrect.

(C) Statement $I$ is true: $H_2O_2$ acts as an oxidizing agent in both acidic and basic media by gaining electrons ($H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O$ in acidic and $H_2O_2 + 2e^- \rightarrow 2OH^-$ in basic medium).
Statement $II$ is false: The density of $H_2O_2$ at $298 \ K$ is approximately $1.45 \ g/cm^3$,while the density of $D_2O$ is approximately $1.11 \ g/cm^3$. Therefore,the density of $H_2O_2$ is higher than that of $D_2O$.

(A) $BeCl_2$ (in vapor phase) and $Al_2Cl_6$ exist as $Cl^{-}$-bridged structures. Due to their covalent nature,they are soluble in organic solvents and act as Lewis acids because of their electron-deficient nature.
$Be(OH)_2$ and $Al(OH)_3$ are amphoteric in nature. They react with excess alkali to form soluble beryllate $[Be(OH)_4]^{2-}$ and aluminate $[Al(OH)_4]^{-}$ ions,respectively.
Therefore,both Statement $I$ and Statement $II$ are true.

(B) . Sulphate $(> 500 \ ppm)$ causes a laxative effect.
$B$. Nitrate $(> 50 \ ppm)$ causes methemoglobinemia (blue baby syndrome).
$C$. Lead $(> 50 \ ppb)$ causes kidney damage and affects $RBC$ production.
$D$. Fluoride $(> 2 \ ppm)$ causes brown mottling of teeth.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(D) $[6]-$Annulene (benzene) is aromatic as it is planar,cyclic,and has $6\pi$ electrons ($4n+2$ rule,$n=1$).
$[8]-$Annulene is non-aromatic because it is non-planar (tub-shaped) to avoid angle strain,despite having $8\pi$ electrons ($4n$ rule).
Cis$-10-$Annulene is non-aromatic because the internal hydrogen atoms at positions $1$ and $6$ experience steric hindrance,forcing the molecule out of planarity,which prevents continuous conjugation.
Thus,Assertion $A$ is incorrect because it labels cis$-10-$Annulene as aromatic.
Reason $R$ is correct because planarity is indeed a fundamental requirement for a molecule to be aromatic ($H$ückel's rule).
Therefore,$A$ is not correct but $R$ is correct.

(A) Mass of organic compound $= 0.45 \ g$
Mass of $AgBr$ obtained $= 0.36 \ g$
Moles of $AgBr = \frac{0.36 \ g}{188 \ g \ mol^{-1}} = 0.001915 \ mol$
Since $1 \ mol$ of $AgBr$ contains $1 \ mol$ of $Br$,moles of $Br = 0.001915 \ mol$
Mass of Bromine $= 0.001915 \ mol \times 80 \ g \ mol^{-1} = 0.1532 \ g$
Percentage of $Br = \frac{\text{Mass of } Br}{\text{Mass of organic compound}} \times 100$
Percentage of $Br = \frac{0.1532 \ g}{0.45 \ g} \times 100 = 34.04 \ \%$

(C) The balanced redox reaction in acidic medium is: $Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$.
According to the law of equivalence,the number of equivalents of $K_2Cr_2O_7$ equals the number of equivalents of $Fe^{2+}$.
$n.f$ of $K_2Cr_2O_7 = 6$ (as $Cr$ changes from $+6$ to $+3$,$2 \times 3 = 6$).
$n.f$ of $Fe^{2+} = 1$ (as $Fe$ changes from $+2$ to $+3$).
Equivalents of $K_2Cr_2O_7 = Molarity \times Volume \times n.f = 0.02 \times 20 \times 6 = 2.4$.
Equivalents of $Fe^{2+} = M \times 10 \times 1 = 10M$.
Equating both: $10M = 2.4$.
$M = 0.24 \, M$.
$M = 24 \times 10^{-2} \, M$.
Therefore,the value is $24$.

(C) To determine the paramagnetism,we check for the presence of unpaired electrons:
$Na_2O$: Diamagnetic (all electrons paired).
$KO_2$: Paramagnetic (contains $O_2^-$ ion with one unpaired electron).
$NO_2$: Paramagnetic (odd number of valence electrons,$17$ valence electrons).
$N_2O$: Diamagnetic (all electrons paired).
$ClO_2$: Paramagnetic (odd number of valence electrons,$19$ valence electrons).
$NO$: Paramagnetic (odd number of valence electrons,$11$ valence electrons).
$SO_2$: Diamagnetic (all electrons paired).
$Cl_2O$: Diamagnetic (all electrons paired).
The paramagnetic oxides are $KO_2$,$NO_2$,$ClO_2$,and $NO$.
Therefore,the total number of paramagnetic oxides is $4$.

(B) For an ideal gas,the relationship between molar heat capacities is $C_{p,m} - C_{v,m} = R$.
Given $C_{p,m} = 20.785 \ J \ K^{-1} \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,we calculate $C_{v,m}$ as:
$C_{v,m} = 20.785 - 8.314 = 12.471 \ J \ K^{-1} \ mol^{-1}$.
The change in internal energy $\Delta U$ is given by the formula $\Delta U = n C_{v,m} \Delta T$.
Here,$\Delta U = 5000 \ J$ and $\Delta T = 500 \ K - 300 \ K = 200 \ K$.
Substituting the values: $5000 = n \times 12.471 \times 200$.
$n = \frac{5000}{12.471 \times 200} = \frac{25}{12.471} \approx 2.0046$.
The nearest integer is $2$.

(B) The bond order for each species is calculated as follows:
$CN^{-}$ ($14$ electrons): Bond order $= \frac{10-4}{2} = 3$
$NO^{+}$ ($14$ electrons): Bond order $= \frac{10-4}{2} = 3$
$O_{2}$ ($16$ electrons): Bond order $= \frac{10-6}{2} = 2$
$O_{2}^{+}$ ($15$ electrons): Bond order $= \frac{10-5}{2} = 2.5$
$O_{2}^{2+}$ ($14$ electrons): Bond order $= \frac{10-4}{2} = 3$
Thus,$CN^{-}, NO^{+},$ and $O_{2}^{2+}$ have an identical bond order of $3$.
The total number of such species is $3$.

(D) The disproportionation reaction of manganate $(MnO_{4}^{2-})$ in acidic medium is given by:
$3 MnO_{4}^{2-} + 4 H^{+} \longrightarrow 2 MnO_{4}^{-} + MnO_{2} + 2 H_{2}O$
In this reaction,manganese is oxidized to $Mn^{+7}$ (in $MnO_{4}^{-}$) and reduced to $Mn^{+4}$ (in $MnO_{2}$).
The product with the higher oxidation state is $MnO_{4}^{-}$,where the oxidation state of $Mn$ is $+7$.
The electronic configuration of $Mn^{+7}$ is $[Ar] 3d^{0}$.
Since there are no unpaired electrons $(n = 0)$,the spin-only magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{0(0+2)} = 0 \, B.M.$

(C) An asymmetric carbon atom (chiral center) is a carbon atom bonded to four different groups.
$1$. $2$-chlorobutane $(CH_3-CHCl-CH_2-CH_3)$: The $C_2$ carbon is bonded to $-H, -Cl, -CH_3, -CH_2CH_3$. It is asymmetric.
$2$. Glycine $(NH_2-CH_2-COOH)$: No carbon is bonded to four different groups.
$3$. Alanine $(CH_3-CH(NH_2)-COOH)$: The $C_2$ carbon is bonded to $-H, -NH_2, -CH_3, -COOH$. It is asymmetric.
$4$. Styrene $(C_6H_5-CH=CH_2)$: No $sp^3$ hybridized carbon is bonded to four different groups.
$5$. Menthol: It contains multiple chiral centers (asymmetric carbons).
Thus,compounds $1, 3,$ and $5$ possess asymmetric carbon atoms. The total number is $3$.

(C) The fermentation of glucose $(C_6H_{12}O_6)$ by the enzyme zymase produces ethanol $(CH_3CH_2OH)$ and carbon dioxide $(CO_2)$.
$C_6H_{12}O_6 \xrightarrow{\text{Zymase}} 2CH_3CH_2OH + 2CO_2$
Here,$A$ is $CH_3CH_2OH$.
Ethanol reacts with $NaOI$ (sodium hypoiodite) in a haloform reaction to produce iodoform $(CHI_3)$ and sodium formate $(HCOONa)$.
$CH_3CH_2OH + 4NaOI \rightarrow CHI_3 + HCOONa + 3NaOH + H_2O$
Here,$B$ is sodium formate $(HCOONa)$.
The product $B$ $(HCOONa)$ contains $1$ carbon atom.

(B) The depression in freezing point $\Delta T_f$ is given by $\Delta T_f = K_f \cdot m$,where $m$ is the molality.
Since the mass of solvent $(1 \ kg)$ and the mass of solute $(1 \ g)$ are the same for both,the molality $m$ is inversely proportional to the molar mass $M$ $(m = \frac{n}{W_{solvent}} = \frac{mass}{M \cdot W_{solvent}})$.
Therefore,$\frac{\Delta T_x}{\Delta T_y} = \frac{M_y}{M_x}$.
Given $\frac{\Delta T_x}{\Delta T_y} = \frac{1}{4}$,we have $\frac{1}{4} = \frac{M_y}{M_x}$.
This implies $\frac{M_x}{M_y} = \frac{4}{1}$,so $M_x : M_y = 4 : 1$.

(A) The formula for molar conductivity is $\Lambda_{m} = \kappa \times \frac{1000}{M}$,where $M$ is the molarity of the solution.
Since the conductivities $\kappa$ are the same,$\Lambda_{m} \propto \frac{1}{M}$.
Calculate the molarity for both solutions:
$M_1 = \frac{n_1}{V_1} = \frac{10 \ mol}{20 \ mL} = 0.5 \ mol/mL$.
$M_2 = \frac{n_2}{V_2} = \frac{20 \ mol}{80 \ mL} = 0.25 \ mol/mL$.
Now,find the ratio:
$\frac{\Lambda_{m2}}{\Lambda_{m1}} = \frac{M_1}{M_2} = \frac{0.5}{0.25} = 2$.
Therefore,$\Lambda_{m2} = 2 \Lambda_{m1}$.

(C) Micelle formation is a spontaneous process driven by the hydrophobic effect.
During micelle formation,the hydrophobic tails of surfactant molecules aggregate,which leads to the release of water molecules previously ordered around the tails into the bulk solvent.
This release of water molecules results in a net increase in the entropy of the system,so $\Delta S > 0$.
Additionally,the process is endothermic,meaning $\Delta H > 0$.
Therefore,statements $B$ and $C$ are correct.

(B) Statement $-I$ is incorrect because cast iron is obtained by melting pig iron with scrap iron and coke using a hot air blast.
Statement $-II$ is incorrect because pig iron contains about $4 \%$ carbon,whereas cast iron contains about $3 \%$ carbon. Therefore,pig iron has a higher carbon content than cast iron.

(A) The energy of absorption is directly proportional to the crystal field splitting energy $(\Delta_o)$.
According to the spectrochemical series,the field strength of the ligands is $H_2O < NH_3 < en$.
Since the metal ion $(Ni^{2+})$ is the same in all complexes,the splitting energy depends on the ligand strength.
Therefore,the order of splitting energy (and thus energy of absorption) is $[Ni(H_2O)_6]^{2+} < [Ni(NH_3)_6]^{2+} < [Ni(en)_3]^{2+}$,which corresponds to $C < B < A$.

(B) The reaction involves the self-aldol condensation of isobutyraldehyde ($2$-methylpropanal) in the presence of a dilute base $(\text{OH}^-)$.
$1$. The base abstracts an $\alpha$-hydrogen from isobutyraldehyde to form an enolate ion.
$2$. This enolate ion acts as a nucleophile and attacks the carbonyl carbon of another molecule of isobutyraldehyde.
$3$. The resulting alkoxide is protonated to form the $\beta$-hydroxy aldehyde,which is $3-$hydroxy$-2,4-$dimethylpentanal.
Therefore,the major product is $3-$hydroxy$-2,4-$dimethylpentanal.

(A) The acidic strength of phenols depends on the nature of the substituents attached to the benzene ring. Electron-withdrawing groups $(EWG)$ increase acidity,while electron-donating groups $(EDG)$ decrease it.
$A$: $p$-Nitrophenol ($-NO_2$ is a strong $EWG$ at para position,strong $-I$ and $-M$ effect).
$B$: $m$-Nitrophenol ($-NO_2$ is an $EWG$ at meta position,only $-I$ effect).
$C$: $m$-Methoxyphenol ($-OCH_3$ is an $EDG$ by $+M$ effect but acts as an $EWG$ by $-I$ effect at meta position).
$D$: $p$-Methoxyphenol ($-OCH_3$ is a strong $EDG$ by $+M$ effect at para position).
Comparing the effects:
$A$ is the most acidic due to strong $-M$ effect of $-NO_2$ at para position.
$B$ is next,as $-NO_2$ at meta position only shows $-I$ effect.
$C$ is less acidic than $B$ because $-OCH_3$ at meta position acts as an $EWG$ ($-I$ effect) but is less effective than $-NO_2$.
$D$ is the least acidic due to the strong $+M$ electron-donating effect of $-OCH_3$ at para position.
Therefore,the decreasing order of acidic strength is $A > B > C > D$.

(A) $1$. The reaction of $CH_{3}CH_{2}CN$ (propanenitrile) with $CH_{3}MgBr$ (Grignard reagent) followed by acid hydrolysis $(H_{3}O^{+})$ yields a ketone,$B$,which is butan$-2-$one $(CH_{3}CH_{2}COCH_{3})$.
$2$. The reaction is: $CH_{3}CH_{2}CN + CH_{3}MgBr$ $\rightarrow CH_{3}CH_{2}C(NMgBr)CH_{3}$ $\xrightarrow{H_{3}O^{+}} CH_{3}CH_{2}COCH_{3}$ $(B)$.
$3$. The subsequent step is the Clemmensen reduction,which uses $Zn-Hg/HCl$ to reduce the carbonyl group $(C=O)$ of the ketone to a methylene group $(-CH_{2}-)$.
$4$. Therefore,$CH_{3}CH_{2}COCH_{3} \xrightarrow[HCl]{Zn-Hg} CH_{3}CH_{2}CH_{2}CH_{3}$ ($C$,butane).

(B) $Nylon\,6,6$ is used for $Bristles\,of\,brushes$ $(A-I)$.
$Low\,density\,polythene$ is used for $Toys$ $(B-II)$.
$High\,density\,polythene$ is used for $Buckets$ $(C-III)$.
$Teflon$ is used for $Non-stick\,utensils$ $(D-IV)$.
Therefore,the correct matching is $A-I, B-II, C-III, D-IV$.

(B) Sucrose is a disaccharide composed of $\alpha$-$D$-glucose and $\beta$-$D$-fructose units.
These two monosaccharide units are joined together by a glycosidic linkage between $C_{1}$ of $\alpha$-$D$-glucose and $C_{2}$ of $\beta$-$D$-fructose.
This linkage is known as an $\alpha, \beta-1,2$-glycosidic bond.

(B) Enzymes have an active site where the substrate binds to undergo a chemical reaction.
Some drugs do not bind to the active site but instead bind to a different site on the enzyme surface,which is known as the $allosteric \ site$.
Binding at this site changes the shape of the active site,making it difficult for the substrate to bind,thereby inhibiting the enzyme's activity.

(C) For a first order reaction,the integrated rate equation is $\ln(P/P_0) = -kt$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln(P/P_0)$ and $x = t$,the slope $m = -k$.
From the given graph,the slope is $-3.465 \times 10^4 \ s^{-1}$.
Therefore,$-k = -3.465 \times 10^4 \ s^{-1}$,which gives $k = 3.465 \times 10^4 \ s^{-1}$.
Given $k = x \times 10^4 \ s^{-1}$,we have $x = 3.465 \approx 3$ (rounding to the nearest integer as per options).

(A) The standard electrode potential $E^{0}_{M^{3+}/M^{2+}}$ is negative only for the $Cr^{3+}/Cr^{2+}$ pair $(E^{0} = -0.41 \, V)$.
The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^{3}$.
The number of unpaired electrons $(n)$ in $Cr^{3+}$ is $3$.
The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \, B.M.$
$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, B.M.$
The nearest integer value is $4$.

(A) The reaction between a monohydric alcohol and methylmagnesium iodide is: $ROH + CH_3MgI \rightarrow ROMgI + CH_4(g)$.
According to the stoichiometry,$1 \ mole$ of $ROH$ produces $1 \ mole$ of $CH_4$ gas.
At $NTP$ conditions,$1 \ mole$ of gas occupies $22.4 \ L$.
Given volume of $CH_4 = 3.1 \ L$.
Moles of $CH_4 = \frac{3.1}{22.4} \approx 0.1384 \ mol$.
Since moles of $ROH = \text{moles of } CH_4$,we have $0.1384 = \frac{\text{mass}}{\text{Molar Mass}} = \frac{4.5}{M.M}$.
$M.M = \frac{4.5}{0.1384} \approx 32.51 \ g/mol$.
The nearest integer is $33$.

(C) Concentration of $HCOOH = 0.5 \, mL \, L^{-1}$.
Mass of $HCOOH = 0.5 \, mL \times 1.05 \, g \, mL^{-1} = 0.525 \, g \, L^{-1}$.
Molar mass of $HCOOH = 46 \, g \, mol^{-1}$.
Molality $(m) = \frac{0.525 \, g}{46 \, g \, mol^{-1} \times 1 \, kg} \approx 0.01141 \, mol \, kg^{-1}$.
Using the formula $\Delta T_{f} = i K_{f} m$,we have $i = \frac{\Delta T_{f}}{K_{f} m}$.
$i = \frac{0.0405}{1.86 \times 0.01141} \approx \frac{0.0405}{0.02122} \approx 1.908$.
Thus,the Van't Hoff factor is nearly $1.9$.

(A) The reactions are matched with their respective catalysts as follows:
$A. N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$ (Haber process) uses $Fe_xO_y + K_2O + Al_2O_3$ as a catalyst $(I)$.
$B. CO_{(g)} + 3H_{2(g)} \rightarrow CH_{4(g)} + H_2O_{(g)}$ uses $Ni$ as a catalyst $(II)$.
$C. CO_{(g)} + H_{2(g)} \rightarrow HCHO_{(g)}$ uses $Cu$ as a catalyst $(III)$.
$D. CO_{(g)} + 2H_{2(g)} \rightarrow CH_3OH_{(g)}$ uses $Cu/ZnO - Cr_2O_3$ as a catalyst $(IV)$.
Therefore,the correct matching is $A-I, B-II, C-III, D-IV$.

(D) During the extraction of copper from copper pyrites $(CuFeS_{2})$,the ore is roasted to remove sulfur as $SO_{2}$ and convert iron and copper into oxides.
In the smelting process,silica $(SiO_{2})$ is added as a flux to remove the iron oxide $(FeO)$ impurity.
$FeO$ reacts with $SiO_{2}$ to form iron silicate $(FeSiO_{3})$,which is removed as slag:
$FeO + SiO_{2} \rightarrow FeSiO_{3}$ (slag).
Therefore,only $FeO$ is removed as slag.

(C) When bromine reacts with an excess of fluorine,the reaction is as follows:
$Br_2 + 5F_2 (\text{excess}) \longrightarrow 2BrF_5$
Thus,the interhalogen compound formed is $BrF_5$ (bromine pentafluoride).

(C) The reaction proceeds in two steps:
$1$. The reaction of the secondary alcohol with $p$-toluenesulfonyl chloride $(TsCl)$ in the presence of pyridine converts the $-OH$ group into a good leaving group,the tosylate $(-OTs)$ group,with retention of configuration at the chiral center.
$2$. The subsequent reaction with $NaCN$ in $DMF$ is an $S_N2$ reaction. The nucleophile $CN^{-}$ attacks the chiral carbon from the side opposite to the $-OTs$ group,resulting in an inversion of configuration at that center.
Starting with the given stereoisomer,the $-OH$ is on a wedge. After forming the $-OTs$ group (still on a wedge),the $S_N2$ attack by $CN^{-}$ forces it to the dashed position (inversion). Thus,the correct product has the $CN$ group on a dash while the other chiral center remains unchanged.

(D) Let us analyze each reaction:
$A$. Rosenmund reduction: Acid chloride is reduced to aldehyde using $H_2$ and $Pd/BaSO_4$. This is correct.
$B$. Nitriles are reduced to aldehydes using $DIBAL-H$ followed by hydrolysis. This is correct.
$C$. Esters are reduced to aldehydes using $DIBAL-H$ at low temperature followed by hydrolysis. This is correct.
$D$. Oxidation of primary alcohols with strong oxidizing agents like $Na_2Cr_2O_7/H_2SO_4$ (Jones reagent) typically proceeds further to carboxylic acids. Therefore,$4$-methylbenzyl alcohol will be oxidized to $4$-methylbenzoic acid,not $4$-methylbenzaldehyde. Thus,this reaction does not represent the correct product.

(C) The reaction sequence described is the Gabriel phthalimide synthesis. \\
$1$. Compound $A$ is phthalic acid $(C_{6}H_{4}(COOH)_{2})$. \\
$2$. Reaction of phthalic acid with $NH_{3}$ followed by heating gives phthalimide $(C_{8}H_{5}NO_{2})$,which is compound $C$. The intermediate $B$ is phthalamide $(C_{6}H_{4}(CONH_{2})_{2})$. \\
$3$. Phthalimide $(C)$ reacts with ethanolic $KOH$ to form potassium phthalimide. \\
$4$. Potassium phthalimide reacts with alkyl chloride $(R-Cl)$ via $S_{N}2$ mechanism to form $N$-alkylphthalimide. \\
$5$. Hydrolysis of $N$-alkylphthalimide with alkali gives a primary amine $(RNH_{2})$ and phthalic acid. \\
Thus,compound $A$ is phthalic acid.

(A) Melamine polymer,also known as melamine-formaldehyde resin,is formed by the condensation polymerization of melamine and formaldehyde $(HCHO)$.
The reaction involves the formation of a resinous material through the elimination of water molecules.

(A) Denaturation of proteins involves the disruption of the $3D$ structure of the protein,such as the $secondary$,$tertiary$,and $quaternary$ structures,due to physical or chemical changes like temperature or $pH$ variations.
However,the $primary$ structure,which refers to the specific sequence of amino acids held together by peptide bonds,remains intact because these covalent bonds are not affected by the denaturation process.

(B) Drugs that bind to the receptor site and inhibit its natural function are known as $Antagonists$. These are useful when blocking a message is required.

(D) The Nernst equation is given by: $E_{cell} = E^{0}_{cell} - \frac{0.06}{n} \log Q$
First,calculate the standard cell potential: $E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode} = 0.008 \ V - (-0.763 \ V) = 0.771 \ V$
Substitute the given values into the Nernst equation: $0.801 = 0.771 - \frac{0.06}{n} \log(10^{-2})$
$0.801 - 0.771 = -\frac{0.06}{n} \times (-2)$
$0.03 = \frac{0.12}{n}$
$n = \frac{0.12}{0.03} = 4$
Thus,the number of electrons involved in the reaction is $4$.

(D) The half-life for a reaction of order $n$ is given by the relation $t_{1/2} \propto \frac{1}{P_0^{n-1}}$,where $P_0$ is the initial pressure.
Given: $(t_{1/2})_1 = 240 \ s = 4 \ min$ at $P_1 = 500 \ Torr$.
Given: $(t_{1/2})_2 = 4 \ min$ at $P_2 = 250 \ Torr$.
Since the half-life remains constant $(4 \ min)$ despite the change in initial pressure,the half-life is independent of the initial pressure.
For $t_{1/2}$ to be independent of $P_0$,the exponent $(n-1)$ must be $0$.
Therefore,$n-1 = 0$,which implies $n = 1$.
The reaction is of the $1^{st}$ order.

(B) The energy of absorbed light is inversely proportional to the wavelength: $\Delta_0 \propto \frac{1}{\lambda}$.
To absorb light with the shortest wavelength,the complex must have the largest crystal field splitting energy $(\Delta_0)$.
Among the ligands provided $(Cl^- < H_2O < NH_3 < CN^-)$,the cyanide ion $(CN^-)$ is the strongest field ligand $(SFL)$.
Therefore,$[Co(CN)_6]^{3-}$ has the largest $\Delta_0$ and absorbs light of the shortest wavelength.
In $[Co(CN)_6]^{3-}$,the oxidation state of $Co$ is $+3$,which corresponds to a $d^6$ configuration.
Since $CN^-$ is a strong field ligand,it causes pairing of electrons,resulting in a low-spin $t_{2g}^6 e_g^0$ configuration.
The number of unpaired electrons $(n)$ is $0$,so the spin-only magnetic moment $\mu = \sqrt{n(n+2)} = 0 \ B.M.$

(B) The standard electrode potential $E^{\circ}$ for the reduction of $H^{+}$ to $H_{2}$ is $0.00 \ V$.
$A$ metal ion can liberate $H_{2}$ from a dilute mineral acid if its standard reduction potential $E^{\circ}(M^{n+}/M)$ is negative.
Among the given ions,$Co^{3+}$ has a high positive reduction potential $(E^{\circ}(Co^{3+}/Co^{2+}) = +1.82 \ V)$,meaning it is a strong oxidizing agent and will not liberate $H_{2}$ from acid; instead,it is reduced to $Co^{2+}$.
The electronic configuration of $Co^{3+}$ $(Z=27)$ is $[Ar] 3d^{6}$.
In the $3d^{6}$ configuration,the number of unpaired electrons $(n)$ is $4$.
The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ B.M.$
$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.899 \ B.M.$
The nearest integer value is $5$.

(A) The reaction is an intramolecular Claisen condensation (Dieckmann condensation).
$1$. The ethoxide ion $(EtO^-)$ abstracts an $\alpha$-hydrogen from the ketone group,forming an enolate.
$2$. This enolate attacks the ester carbonyl carbon,leading to the formation of a bicyclic compound (product $A$).
$3$. The resulting product $A$ is a bicyclo[$4.3$.$0$]nonane derivative with two hydroxyl and ketone groups.
$4$. By examining the structure of the product,we identify two chiral centers (marked with asterisks in the structure),which are the bridgehead carbons where the ring fusion occurs and substituents are attached.
Therefore,the number of chiral carbons in product $A$ is $2$.

(C) Since the half-life is independent of the initial concentration,it is a $1^{st}$ order reaction.
$k = \frac{0.693}{T_{1/2}} = \frac{0.693}{200} = 3.465 \times 10^{-3} \, s^{-1}$.
For $80\%$ decomposition,the remaining concentration is $20\%$ of the initial concentration $(A = 0.2 A_{0})$.
The rate equation is $t = \frac{2.303}{k} \log \frac{A_{0}}{A}$.
$t = \frac{2.303}{3.465 \times 10^{-3}} \log \frac{A_{0}}{0.2 A_{0}} = \frac{2.303}{3.465 \times 10^{-3}} \log 5$.
Using $\log 5 = 0.70$ and $\frac{2.303}{k} = \frac{200}{0.693} \approx 288.66$.
$t = 288.66 \times 0.70 \approx 202.06 \times 2.303 \approx 466.67 \, s \approx 467 \, s$.

(A) The colour of a colloidal solution depends on the size and shape of the dispersed particles.
As the particle size of gold sol increases,the wavelength of light scattered by the particles changes,which results in a change in the observed colour (from red to purple to blue and finally to golden).
Therefore,Assertion $A$ is true.
Reason $R$ correctly explains that the colour of a colloidal solution depends on the wavelength of light scattered by the dispersed particles,which is a direct consequence of the particle size.
Thus,both $A$ and $R$ are true,and $R$ is the correct explanation of $A$.

(B) Gallium $(Ga)$ is a metal with an exceptionally low melting point $(29.76^\circ C)$.
In the periodic table,it is located in Group $13$ and is adjacent to the metalloid Germanium $(Ge)$.

(A) The metal $Al$ is extracted from its oxide ore,Bauxite $(Al_{2}O_{3} \cdot 2H_{2}O)$.
Iron is typically extracted from haematite $(Fe_{2}O_{3})$ or magnetite $(Fe_{3}O_{4})$,but it is also found as iron pyrites $(FeS_{2})$.
Lead is extracted from galena $(PbS)$.
Zinc is extracted from zinc blende $(ZnS)$.
Therefore,$Al$ is the correct answer as it is not extracted from a sulphide ore.

(A) In $K_{3}[Cu(CN)_{4}]$,the oxidation state of $Cu$ is $+1$.
$Cu^{+}$ has the electronic configuration $[Ar] 3d^{10}$.
Since all $d$-orbitals are completely filled,there are no unpaired electrons.
Therefore,$K_{3}[Cu(CN)_{4}]$ is diamagnetic.

(B) Carbolic acid is the common name for phenol $(C_6H_5OH)$.
When benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ is treated with water (hydrolysis),it yields phenol,nitrogen gas $(N_2)$,and hydrochloric acid $(HCl)$.
The reaction is as follows:
$C_6H_5N_2^+Cl^- + H_2O \rightarrow C_6H_5OH + N_2 + HCl$

(A) $DiBAL-H$ (Diisobutylaluminium hydride) is a selective reducing agent that reduces esters $(-COOEt)$ and nitriles $(-CN)$ to aldehydes $(-CHO)$ at low temperatures (typically $-78 \ ^{\circ}C$).
In the given reaction,the starting material contains both an ester group $(-COOEt)$ and an aldehyde group $(-CHO)$.
$DiBAL-H$ will selectively reduce the ester group to an aldehyde group while leaving the existing aldehyde group unaffected.
Therefore,the ester group $EtO-CO-CH_2-$ is converted to $OHC-CH_2-$.
The final product is $OHC-CH_2-cyclopent-3-ene-CH_2CH_2CHO$.

(B) The starting material is $p$-nitroaniline. The goal is to convert the $-NO_2$ group to $-OH$ and the $-NH_2$ group to $-I$.
$1$. First,treat $p$-nitroaniline with $HNO_2$ ($NaNO_2 + HCl$ at $0-5^{\circ}C$) to form the diazonium salt at the $-NH_2$ position.
$2$. Then,treat with $KI$ to replace the diazonium group with $-I$.
$3$. Next,reduce the $-NO_2$ group to $-NH_2$ using $Fe/H^{+}$.
$4$. Treat with $HNO_2$ again to convert the new $-NH_2$ group into a diazonium salt.
$5$. Finally,heat with $H_2O$ to convert the diazonium group into an $-OH$ group.
This corresponds to the sequence: $HNO_2, KI, Fe/H^{+}, HNO_2, H_2O/\text{warm}$.

(C) Vulcanization is a process of heating raw rubber with sulphur to improve its physical properties,such as elasticity and strength.
Natural rubber is a polymer of $isoprene$ $(2-methyl-1,3-butadiene)$.
Therefore,vulcanization of rubber is carried out by heating a mixture of $isoprene$ (or natural rubber) and $sulphur$.

(C) Glycogen is a polysaccharide of glucose that serves as a form of energy storage in animals,fungi,and bacteria. Because its structure is similar to amylopectin (a component of starch) but more highly branched,it is commonly referred to as $ \text{animal starch} $.

(B) For elevation in boiling point: $\Delta T_{b} = K_{b} \times m_{1} = 3 \ K$ where $m_{1} = 1 \ m$. So,$K_{b} = 3 \ K \ kg \ mol^{-1}$.
For depression in freezing point: $\Delta T_{f} = K_{f} \times m_{2} = 6 \ K$ where $m_{2} = 2 \ m$. So,$K_{f} = 6 / 2 = 3 \ K \ kg \ mol^{-1}$.
The ratio $K_{b} / K_{f} = 3 / 3 = 1 / 1$.
Comparing this with $1 : X$,we get $X = 1$.

(B) From the stoichiometry of the reactions:
$2 Cu^{2+} \equiv I_{2} \equiv 2 S_{2}O_{3}^{2-}$.
Therefore,$n_{\text{eq.}}$ of $Cu^{2+} = n_{\text{eq.}}$ of $S_{2}O_{3}^{2-}$.
$n_{\text{eq.}}$ of $S_{2}O_{3}^{2-} = M \times V \times n_{\text{factor}} = 0.02 \times 20 \times 1 = 0.4 \, \text{mmol}$.
Since $n_{\text{eq.}}$ of $Cu^{2+} = 0.4 \, \text{mmol}$ and $n_{\text{factor}}$ for $Cu^{2+}$ is $1$ (change in oxidation state from $+2$ to $+1$ in $Cu_{2}I_{2}$),we have $n_{\text{mol}}$ of $Cu^{2+} = 0.4 \, \text{mmol}$.
$[Cu^{2+}] = \frac{n_{\text{mol}}}{V_{\text{solution}}} = \frac{0.4 \, \text{mmol}}{10 \, mL} = 0.04 \, M = 4 \times 10^{-2} \, M$.
Thus,the value is $4$.

(D) The oxidation states of $Mn$ in $MnF_{4}$,$MnF_{3}$,and $MnF_{2}$ are $+4$,$+3$,and $+2$ respectively.
The electronic configurations are:
$MnF_{4} (Mn^{4+}) = [Ar] 3d^{3}$
$MnF_{3} (Mn^{3+}) = [Ar] 3d^{4}$
$MnF_{2} (Mn^{2+}) = [Ar] 3d^{5}$
Among these,$Mn^{3+}$ is the strongest oxidizing agent because it can easily gain an electron to form the stable $d^{5}$ configuration $(Mn^{2+})$.
For $Mn^{3+}$ $(3d^{4})$,the number of unpaired electrons $(n)$ is $4$.
The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ B.M.$
$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89 \ B.M.$
The nearest integer value is $5$.

(C) Broad spectrum antibiotics are those that are effective against a wide range of Gram-positive and Gram-negative bacteria.
$Ofloxacin$ is a well-known broad spectrum antibiotic.
$Penicillin \ G$ is a narrow spectrum antibiotic.
$Terpineol$ is an antiseptic.
$Salvarsan$ is an antibacterial agent used for syphilis,but it is not classified as a broad spectrum antibiotic.
Therefore,only $1$ drug $(Ofloxacin)$ is a broad spectrum antibiotic.

(A) $2 SO_{2(g)} + O_{2(g)} \xrightarrow{V_{2}O_{5}} 2 SO_{3(g)}$: Contact process uses $V_{2}O_{5}$ as a catalyst.
$4 NH_{3(g)} + 5 O_{2(g)} \xrightarrow{Pt_{(s)}-Rh_{(s)}} 4 NO_{(g)} + 6 H_{2}O_{(g)}$: Ostwald's process uses $Pt_{(s)}-Rh_{(s)}$ as a catalyst.
$N_{2(g)} + 3 H_{2(g)} \xrightarrow{Fe_{(s)}} 2 NH_{3(g)}$: Haber's process uses $Fe_{(s)}$ as a catalyst.
$\text{Vegetable oil}(l) + H_{2(g)} \xrightarrow{Ni_{(s)}} \text{Vegetable ghee}(s)$: Hydrogenation uses $Ni_{(s)}$ as a catalyst.
Therefore,the correct matching is $A-I, B-II, C-III, D-IV$.

(A) Liquation is a process used for the purification of metals that have a lower melting point than the impurities present in them.
In this method,the crude metal is placed on the sloping hearth of a furnace and heated.
The metal melts and flows down,leaving the infusible impurities behind.

(B) The difference in the reaction products is observed due to the polarity of the solvent:
$(i)$ In a polar solvent like water,phenol ionizes to form the phenoxide ion $(C_6H_5O^-)$,which is much more reactive towards electrophilic aromatic substitution than phenol itself,leading to polybromination ($2$,$4$,$6$-tribromophenol).
$(ii)$ In a non-polar solvent like chloroform $(CHCl_3)$,phenol does not ionize significantly,and the reaction proceeds to form monobrominated products (o-bromophenol and p-bromophenol).

(D) The starting material is $4$-oxocyclohexanecarbaldehyde.
Step $1$: Treatment with Tollens' reagent $[Ag(NH_3)_2]^+ OH^-$ selectively oxidizes the aldehyde group to a carboxylic acid group while leaving the ketone group unaffected. Thus,$A$ is $4$-oxocyclohexanecarboxylic acid.
Step $2$: Treatment with $NaBH_4$ reduces the ketone group to a secondary alcohol but does not reduce the carboxylic acid group. Thus,$B$ is $4$-hydroxycyclohexanecarboxylic acid.