NEET 2022 Chemistry Question Paper with Answer and Solution

(C) The stability of carbocations is determined by the inductive effect and hyperconjugation.
$(CH_3)_3C^+$ is a tertiary $(3^\circ)$ carbocation.
It has $9 \, \alpha \text{-hydrogen}$ atoms,which provide maximum hyperconjugation,making it the most stable among the given options.

(D) The mixture of a weak acid $(CH_{3}COOH)$ and its salt with a strong base $(CH_{3}COONa)$ forms an acidic buffer solution.
The concentration of salt $[Salt] = 0.10 \ M$.
The concentration of acid $[Acid] = 0.01 \ M$.
The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation:
$pH = pK_{a} + \log \left( \frac{[Salt]}{[Acid]} \right)$
Substituting the given values:
$pH = 4.57 + \log \left( \frac{0.10}{0.01} \right)$
$pH = 4.57 + \log(10)$
Since $\log(10) = 1$,we get:
$pH = 4.57 + 1 = 5.57$

(C) $Li$ is used in electrochemical cells.
$Na$ is used as a coolant in fast breeder reactors.
$KOH$ is used as an absorbent for $CO_2$.
$Cs$ is used in photoelectric cells.
Therefore,the correct matching is $(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)$.

(B) In diamond,each carbon atom is bonded to four other carbon atoms,resulting in a $sp^{3}$ hybridisation.
In graphite,each carbon atom is bonded to three other carbon atoms,resulting in a $sp^{2}$ hybridisation.

(A) In a $p-V$ graph,the area under the curve represents the magnitude of the work done.
Comparing the given graphs,the area enclosed under the curve in option $A$ is the largest among the choices provided.
Therefore,the curve in option $A$ represents the maximum work done.

(C) To determine if a compound is aromatic,it must follow $H$ückel's rule:
$1$. The molecule must be cyclic.
$2$. The molecule must be planar.
$3$. The molecule must be fully conjugated.
$4$. The molecule must have $(4n + 2) \pi$ electrons,where $n = 0, 1, 2, ...$
Let's analyze each option:
- $A$: Cycloheptatrienyl cation has $6 \pi$ electrons $(n=1)$,is cyclic,planar,and conjugated. It is aromatic.
- $B$: Cyclopropenyl cation has $2 \pi$ electrons $(n=0)$,is cyclic,planar,and conjugated. It is aromatic.
- $C$: Cyclooctatetraene has $8 \pi$ electrons. It is not planar (it adopts a tub shape) and does not follow the $(4n + 2) \pi$ rule. It is non-aromatic.
- $D$: Cyclopentadienyl anion has $6 \pi$ electrons $(n=1)$,is cyclic,planar,and conjugated. It is aromatic.
Therefore,the compound that is not aromatic is cyclooctatetraene.

(C) Dalton's law of partial pressure states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases: $p = p_{1} + p_{2} + p_{3}$.
The partial pressure of a gas $i$ is given by $p_{i} = \chi_{i} p$,where $\chi_{i}$ is the mole fraction of the gas in the mixture.
Option $A$ is correct because $p = \sum n_{i} \frac{RT}{V} = (\sum n_{i}) \frac{RT}{V} = n_{total} \frac{RT}{V}$,which is the ideal gas law for the mixture.
Option $C$ represents Raoult's Law for solutions,not Dalton's Law for gaseous mixtures. Therefore,it is the incorrect equation.

(D) $1$. $MgH_2$ is an ionic hydride.
$2$. $GeH_4$ is an electron-precise hydride (has $8$ electrons in the valence shell,no lone pairs).
$3$. $B_2H_6$ is an electron-deficient hydride (has less than $8$ electrons in the valence shell).
$4$. $HF$ is an electron-rich hydride (has $8$ electrons in the valence shell with lone pairs).
Therefore,the correct matching is: $a-iv, b-i, c-ii, d-iii$.

(A) In $KO_{2}$,potassium $(K)$ is an alkali metal and always exhibits an oxidation state of $+1$.
The superoxide ion is $O_{2}^{-}$,where the oxidation state of oxygen is $-1/2$.
Therefore,the statement that the oxidation number of $K$ in $KO_{2}$ is $+4$ is incorrect.
The correct oxidation number of $K$ is $+1$.

(C) The electronic configuration of $O_{2}^{+}$ is $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, \pi 2p_{x}^{2} = \pi 2p_{y}^{2}, \pi^{*} 2p_{x}^{1}$.
Since $O_{2}^{+}$ ion has $15$ electrons,it contains one unpaired electron in the $\pi^{*}$ orbital.
Therefore,it is paramagnetic in nature,making the statement in option $C$ incorrect.

(A) The balanced chemical equation is:
$CaCO_{3(s)} + 2HCl_{(aq)} \rightarrow CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$
Number of moles of $HCl = \text{Molarity} \times \text{Volume (in L)} = 0.5 \times 0.050 = 0.025 \ \text{mol}$.
From the stoichiometry of the reaction,$1 \ \text{mole}$ of $CaCO_3$ reacts with $2 \ \text{moles}$ of $HCl$.
Therefore,moles of pure $CaCO_3$ required $= \frac{1}{2} \times 0.025 = 0.0125 \ \text{mol}$.
Mass of pure $CaCO_3 = \text{moles} \times \text{molar mass} = 0.0125 \times 100 \ \text{g/mol} = 1.25 \ \text{g}$.
Given that the sample is $95 \%$ pure,the mass of the impure sample is calculated as:
$\text{Mass of impure sample} = \frac{\text{Mass of pure } CaCO_3}{\text{Purity percentage}} \times 100 = \frac{1.25}{95} \times 100 \approx 1.3157 \ \text{g}$.
Rounding to the second decimal place,we get $1.32 \ \text{g}$.

(B) The Kjeldahl's method is not applicable to compounds containing nitrogen in nitro $(-NO_2)$ or azo $(-N=N-)$ groups,or nitrogen present in a ring (like pyridine),because the nitrogen in these compounds is not quantitatively converted to ammonium sulphate under the conditions of the Kjeldahl's method.
Among the given options:
$A$. Nitrobenzene contains a nitro group.
$B$. Aniline $(C_6H_5NH_2)$ contains an amino group,which is easily converted to ammonium sulphate.
$C$. Azobenzene contains an azo group.
$D$. Pyridine contains nitrogen in the ring.
Therefore,the Kjeldahl's method can be used for Aniline.

(C) The shapes of $d$-orbitals are determined by the angular part of the wave function,which depends on the azimuthal quantum number $l$. For all $d$-orbitals,$l = 2$. Therefore,$3d$,$4d$,and $5d$ orbitals have the same shapes.
Option $A$ is correct because $4d$ and $3d$ orbitals have identical shapes.
Option $B$ is correct because in a free atom,all five $d$-orbitals are degenerate (equal in energy).
Option $C$ is incorrect because $d_{x^{2}-y^{2}}$ and $d_{z^{2}}$ do not have similar shapes. $d_{x^{2}-y^{2}}$ has four lobes along the $x$ and $y$ axes,while $d_{z^{2}}$ has two lobes along the $z$-axis and a ring of electron density in the $xy$-plane.
Option $D$ is correct because as the principal quantum number $n$ increases,the size of the orbital increases. Thus,$5d$ orbitals are larger than $4d$ orbitals.

(C) To determine the maximum 'lone pair - lone pair' (lp-lp) repulsions,we analyze the number of lone pairs on the central atom for each molecule:
$1$. $IF_{5}$: Iodine has $7$ valence electrons. It forms $5$ bonds with $F$ atoms,leaving $1$ lone pair. Number of lp-lp repulsions = $0$.
$2$. $SF_{4}$: Sulfur has $6$ valence electrons. It forms $4$ bonds with $F$ atoms,leaving $1$ lone pair. Number of lp-lp repulsions = $0$.
$3$. $XeF_{2}$: Xenon has $8$ valence electrons. It forms $2$ bonds with $F$ atoms,leaving $3$ lone pairs. These $3$ lone pairs occupy the equatorial positions of the trigonal bipyramidal geometry,resulting in $3$ lp-lp repulsions at $90^\circ$ angles.
$4$. $ClF_{3}$: Chlorine has $7$ valence electrons. It forms $3$ bonds with $F$ atoms,leaving $2$ lone pairs. Number of lp-lp repulsions = $1$.
Comparing these,$XeF_{2}$ has the maximum number of lone pair - lone pair repulsions.

(D) $IUPAC$ nomenclature for elements with atomic number $Z > 100$ is based on the digits of the atomic number.
For $119$:
$1$ = un
$1$ = un
$9$ = enn
Suffix = -ium
Therefore,the name is $Ununennium$ $(Uue)$.

(C) In diborane $(B_2H_6)$,each boron atom undergoes $sp^3$ hybridization,not $sp^2$.
Diborane has a non-planar structure where the two bridging hydrogen atoms lie in a plane perpendicular to the plane containing the four terminal hydrogen atoms and the two boron atoms.
Therefore,the statement that both boron atoms are $sp^2$ hybridized is incorrect.

(D) According to Bohr's atomic model, the radius $r$ is given by $r \propto \frac{n^2}{Z}$.
For the $3^{rd}$ orbit of $Li^{2+}$, $n_1 = 3$ and $Z_1 = 3$.
For the $2^{nd}$ orbit of $He^{+}$, $n_2 = 2$ and $Z_2 = 2$.
Using the ratio formula: $\frac{(r_3)_{Li^{2+}}}{(r_2)_{He^{+}}} = \frac{n_1^2}{n_2^2} \times \frac{Z_2}{Z_1}$.
Substituting the values: $\frac{(r_3)_{Li^{2+}}}{105.8 \, pm} = \frac{3^2}{2^2} \times \frac{2}{3} = \frac{9}{4} \times \frac{2}{3} = \frac{3}{2} = 1.5$.
Therefore, $(r_3)_{Li^{2+}} = 105.8 \, pm \times 1.5 = 158.7 \, pm$.

(C) Given:
Volume $V = 10.0 \, L$
Mass of $O_2$ $(W)$ = $64 \, g$
Molar mass of $O_2$ $(M)$ = $32 \, g \, mol^{-1}$
Temperature $T = 27^{\circ} C = 27 + 273 = 300 \, K$
Gas constant $R = 0.0831 \, L \, bar \, K^{-1} \, mol^{-1}$
Step $1$: Calculate the number of moles $(n)$:
$n = \frac{W}{M} = \frac{64 \, g}{32 \, g \, mol^{-1}} = 2 \, mol$
Step $2$: Use the ideal gas equation $PV = nRT$ to find pressure $(P)$:
$P = \frac{nRT}{V}$
$P = \frac{2 \, mol \times 0.0831 \, L \, bar \, K^{-1} \, mol^{-1} \times 300 \, K}{10.0 \, L}$
$P = \frac{49.86}{10} \, bar = 4.986 \, bar \approx 4.9 \, bar$

(D) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond and the addition of oxygen atoms to each carbon atom of the double bond.
Formaldehyde is $HCHO$ and $2-$methylpropanal is $(CH_{3})_{2}CHCHO$.
By joining the carbonyl carbons of these products by a double bond,we get the structure of the original alkene $X$:
$CH_{3}-CH(CH_{3})-CH=O + O=CH_{2} \xrightarrow{Zn/H_{2}O} CH_{3}-CH(CH_{3})-CH=CH_{2} + H_{2}O$
The structure $CH_{3}-CH(CH_{3})-CH=CH_{2}$ corresponds to $3-$methylbut$-1-$ene.
Therefore,the correct option is $D$.

(D) In neutral or faintly alkaline medium,the reaction is:
$2KMnO_4 + H_2O + KI \rightarrow 2MnO_2 + 2KOH + KIO_3$
In $KMnO_4$,the oxidation state of $Mn$ is $+7$.
In $MnO_2$,the oxidation state of $Mn$ is $+4$.
Therefore,the change in oxidation state of manganese is from $+7$ to $+4$.

(D) $1$. Identify the longest carbon chain containing the principal functional group ($-OH$ group). The chain has $6$ carbons,so the parent alkane is hexane.
$2$. Number the chain from the end that gives the lowest locant to the principal functional group $(-OH)$. Numbering from right to left gives the $-OH$ group at position $3$.
$3$. Identify the substituents: $Br$ at position $1$,$CH_3$ at position $4$,and $Cl$ at position $5$.
$4$. Arrange the substituents in alphabetical order: $1-$bromo,$5-$chloro,$4-$methyl.
$5$. Combining these,the correct $IUPAC$ name is $1-$bromo$-5-$chloro$-4-$methylhexan$-3-$ol.

(D) The equilibrium constant expression for the reaction $3 O_{2(g)} \rightleftharpoons 2 O_{3(g)}$ is given by $K_c = \frac{[O_3]^2}{[O_2]^3}$.
Given $K_c = 3.0 \times 10^{-59}$ and $[O_2] = 0.040 \ M = 4 \times 10^{-2} \ M$.
Substituting the values into the expression:
$3.0 \times 10^{-59} = \frac{[O_3]^2}{(4 \times 10^{-2})^3}$
$[O_3]^2 = 3.0 \times 10^{-59} \times (64 \times 10^{-6})$
$[O_3]^2 = 192 \times 10^{-65} = 1.92 \times 10^{-63}$
$[O_3] = \sqrt{1.92 \times 10^{-63}} = \sqrt{19.2 \times 10^{-64}} \approx 4.38 \times 10^{-32} \ M$.

(A) The presence of particulate matter in polluted air catalyses the oxidation of sulphur dioxide to sulphur trioxide: $2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)}$.
This reaction is also promoted by ozone $(O_3)$ and hydrogen peroxide $(H_2O_2)$:
$SO_{2(g)} + O_{3(g)} \rightarrow SO_{3(g)} + O_{2(g)}$
$SO_{2(g)} + H_2O_{2(l)} \rightarrow H_2SO_{4(aq)}$
Thus,particulate matter,ozone,and hydrogen peroxide all enhance the pollution caused by oxides of sulphur.

(A) $1$. Determine the hybridization and geometry of each species:
- $CO_2$: $sp$ hybridization,linear,bond angle = $180^{\circ}$.
- $NH_4^{+}$: $sp^3$ hybridization,tetrahedral,bond angle = $109.5^{\circ}$.
- $NH_3$: $sp^3$ hybridization,trigonal pyramidal,one lone pair,bond angle = $107^{\circ}$.
- $H_2O$: $sp^3$ hybridization,bent,two lone pairs,bond angle = $104.5^{\circ}$.
$2$. Comparing the bond angles: $104.5^{\circ} (H_2O) < 107^{\circ} (NH_3) < 109.5^{\circ} (NH_4^{+}) < 180^{\circ} (CO_2)$.
$3$. The correct order is $H_2O < NH_3 < NH_4^{+} < CO_2$.

(C) Bleaching powder is represented by the formula $Ca(OH)_2 \cdot CaCl_2 \cdot Ca(OCl)_2 \cdot 2H_2O$.
It is a mixture of calcium chloride $(CaCl_2)$,calcium hypochlorite $(Ca(OCl)_2)$,and calcium hydroxide $(Ca(OH)_2)$.

(C) . Sodium laurylsulphate is an anionic detergent.
$b$. Cetyltrimethyl ammonium chloride is a cationic detergent.
$c$. Sodium stearate is a toilet soap.
$d$. Polyethyleneglycyl stearate is a non-ionic detergent.
Therefore,the correct matching is $(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)$.

(A) decomposition redox reaction involves the breakdown of a single compound into two or more products where the oxidation states of the elements change.
In the reaction $2Pb(NO_3)_{2(s)} \longrightarrow 2PbO_{(s)} + 4NO_{2(g)} + O_{2(g)}$,the oxidation state of $N$ changes from $+5$ to $+4$ and $O$ changes from $-2$ to $0$. Since a single reactant breaks down into multiple products with a change in oxidation states,it is a decomposition redox reaction.
Option $A$ is the correct answer.

(A) For element $X$: The difference between the $2^{nd}$ and $1^{st}$ ionization enthalpy is $3069 - 419 = 2650 \ kJ \ mol^{-1}$,which is very large. This indicates that the removal of the second electron requires breaking a stable noble gas configuration,characteristic of an alkali metal.
For element $Y$: The difference between the $2^{nd}$ and $1^{st}$ ionization enthalpy is $1145 - 590 = 555 \ kJ \ mol^{-1}$,which is relatively small. This indicates that the second electron is removed from the same valence shell,characteristic of an alkaline earth metal.
Therefore,$X$ is an alkali metal and $Y$ is an alkaline earth metal.

(B) $NH_3$: $sp^3$ hybridization with $1$ lone pair $(LP)$ results in a Trigonal pyramidal geometry $(a-iii)$.
$ClF_3$: $sp^3d$ hybridization with $2$ lone pairs $(LP)$ results in a $T$-shape geometry $(b-iv)$.
$PCl_5$: $sp^3d$ hybridization with $0$ lone pairs $(LP)$ results in a Trigonal bipyramidal geometry $(c-ii)$.
$BrF_5$: $sp^3d^2$ hybridization with $1$ lone pair $(LP)$ results in a Square pyramidal geometry $(d-i)$.
Therefore,the correct matching is $a-iii, b-iv, c-ii, d-i$.

(A) The chemical formulas for the given compounds are as follows:
$a$. Borax: $Na_2B_4O_7 \cdot 10H_2O$
$b$. Kernite: $Na_2B_4O_7 \cdot 4H_2O$
$c$. Orthoboric acid: $H_3BO_3$
$d$. Borax bead: $NaBO_2$
Therefore,the correct matching is $a-iv, b-ii, c-iii, d-i$.

(B) Atomic hydrogen welding is a process that uses an electric arc between two tungsten electrodes in an atmosphere of hydrogen gas. The hydrogen molecules dissociate into atoms in the arc,absorbing a large amount of heat. When these atoms recombine on the surface of the metal,they release this energy,producing temperatures up to $4000 \ K$,which is sufficient to weld metals with high melting points.

(B) When borax $(Na_2B_4O_7 \cdot 10H_2O)$ is heated,it first loses its water of crystallization and then melts to form a transparent liquid which solidifies into a glass-like mass consisting of sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O_3)$.
The balanced chemical equation is:
$Na_2B_4O_7 \stackrel{\Delta}{\longrightarrow} 2NaBO_2 + B_2O_3$
Thus,the product $X$ is $B_2O_3$.

(C) The electronic configurations of the elements are as follows:
$C (Z=6): 1s^2 2s^2 2p^2$
$N (Z=7): 1s^2 2s^2 2p^3$
$O (Z=8): 1s^2 2s^2 2p^4$
$F (Z=9): 1s^2 2s^2 2p^5$
Across a period,ionization enthalpy generally increases due to increasing effective nuclear charge.
However,nitrogen $(N)$ has a stable half-filled $2p^3$ configuration,which makes its ionization enthalpy higher than that of oxygen $(O)$.
Therefore,the correct order is $C < O < N < F$.

(C) For $CH_3COOH$ :
$[H^{+}] = C \cdot \alpha$
Given concentration $C = 0.01 \, M = 10^{-2} \, M$ and degree of ionisation $\alpha = 1 \% = 0.01 = 10^{-2}$.
$[H^{+}] = 10^{-2} \times 10^{-2} = 10^{-4} \, M$
$pH = -\log [H^{+}]$
$pH = -\log(10^{-4}) = 4$

(D) For an ideal gas,the internal energy $U$ is a function of temperature only,i.e.,$U = f(T)$.
Since the process is isothermal,the temperature remains constant $(T_2 = T_1 = 300 \ K)$.
Therefore,the change in internal energy $\Delta U = nC_v(T_2 - T_1) = 0$.

(A) The structure of the compound is $OHC-CH=CH-CH_2-COOCH_3$.
Numbering the carbons as indicated in the structure:
$C_1$ is the carbonyl carbon of the ester group $(-COOCH_3)$,which is bonded to one oxygen by a double bond and two other groups by single bonds. It has $3$ sigma bonds and $0$ lone pairs,so it is $sp^2$ hybridized.
$C_2$ is the methylene carbon $(-CH_2-)$ bonded to four atoms by single bonds. It has $4$ sigma bonds and $0$ lone pairs,so it is $sp^3$ hybridized.
Therefore,$C_1$ is $sp^2$ and $C_2$ is $sp^3$ hybridized.

(B) The formula for mass is: $Mass = Density \times Volume$.
Given: $Density = 2.15 \ g \ mL^{-1}$ and $Volume = 2.5 \ mL$.
$Mass = 2.15 \ g \ mL^{-1} \times 2.5 \ mL = 5.375 \ g$.
According to the rules of significant figures,the result of multiplication should have the same number of significant figures as the measurement with the fewest significant figures.
Here,$2.5$ has $2$ significant figures,so the result must be rounded to $2$ significant figures.
Therefore,$5.375 \ g$ rounded to $2$ significant figures is $5.4 \ g$.

(D) The correct matches are as follows :
$a$. Frenkel defect is shown by ionic solids where density remains constant $(iv)$.
$b$. Schottky defect is shown by ionic solids where density decreases $(iii)$.
$c$. Vacancy defect is shown by non-ionic solids where density decreases $(i)$.
$d$. Interstitial defect is shown by non-ionic solids where density increases $(ii)$.
Therefore,the correct sequence is $(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)$.

(D) The ease of liquefaction of a gas is directly proportional to its critical temperature $(T_C)$.
As the temperature is lowered from $500 \ K$,the gas with the highest critical temperature will reach its liquefaction point first.
Comparing the given critical temperatures: $NH_3$ $(405.5 \ K)$ > $CO_2$ $(304.1 \ K)$ > $N_2$ $(126 \ K)$ > $He$ $(5.3 \ K)$.
Since $NH_3$ has the highest critical temperature,it will liquefy first.

(A) The boiling point of alkanes depends on the molecular mass and the extent of branching.
As the molecular mass increases,the boiling point increases.
For isomers,the boiling point decreases with an increase in branching due to a decrease in surface area.
Heptane $(C_7H_{16})$: $371.4 \ K$
Hexane $(C_6H_{14})$: $341.9 \ K$
$2-$Methylbutane $(C_5H_{12})$: $300.9 \ K$
Butane $(C_4H_{10})$: $272.4 \ K$
$2-$Methylpropane $(C_4H_{10})$: $261 \ K$
Comparing these values,the decreasing order is: $(a) > (e) > (c) > (b) > (d)$.

(D) The orbital is represented by $n\ell$,where $n$ is the principal quantum number and $\ell$ is the azimuthal quantum number.
For $\ell=0$,the orbital is $s$.
For $\ell=1$,the orbital is $p$.
For $\ell=2$,the orbital is $d$.
Matching the values:
$a. n=2, \ell=1 \rightarrow 2p$ $(iii)$
$b. n=3, \ell=2 \rightarrow 3d$ $(iv)$
$c. n=3, \ell=0 \rightarrow 3s$ $(ii)$
$d. n=2, \ell=0 \rightarrow 2s$ $(i)$
Therefore,the correct match is $(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)$.

(D) Calcination is a process of heating an ore in a limited supply or absence of air to remove volatile impurities or moisture.
$A$,$B$,and $C$ represent thermal decomposition reactions typical of calcination.
Option $D$ is a chemical reaction between a carbonate and an acid (displacement/double decomposition),not a calcination process.

(D) The energy of a single photon is given by $E = \frac{hc}{\lambda}$.
For one mole of photons,the energy is $E_{mol} = \frac{hcN_A}{\lambda}$.
$E_{mol} = \frac{6.626 \times 10^{-34} \, J \, s \times 3 \times 10^8 \, m \, s^{-1} \times 6.022 \times 10^{23} \, mol^{-1}}{300 \times 10^{-9} \, m} = 3.99 \times 10^5 \, J \, mol^{-1}$.
According to the photoelectric effect equation,$E_{photon} = \Phi + KE$,where $\Phi$ is the work function (minimum energy to remove an electron).
$\Phi = E_{mol} - KE = (3.99 \times 10^5 - 1.68 \times 10^5) \, J \, mol^{-1} = 2.31 \times 10^5 \, J \, mol^{-1}$.

(C) The synthesis of alkenes involves elimination reactions.
$(A)$ Treatment of alkynes with $Na$ in liquid $NH_3$ (Birch reduction) produces trans-alkenes.
$(B)$ Heating alkyl halides with alcoholic $KOH$ undergoes dehydrohalogenation to form alkenes.
$(C)$ Treating alkyl halides with aqueous $KOH$ results in nucleophilic substitution,producing alcohols $(R-X + KOH_{(aq)} \rightarrow R-OH + KX)$,not alkenes.
$(D)$ Treating vicinal dihalides with $Zn$ metal undergoes debromination/dehalogenation to form alkenes.
Therefore,option $C$ is the incorrect method for alkene synthesis.

(A) In the compound $Fe_{0.96}O$,the total charge must be zero.
Let the number of $Fe^{2+}$ ions be $x$.
Then the number of $Fe^{3+}$ ions is $(0.96 - x)$.
The charge balance equation is: $(x)(+2) + (0.96 - x)(+3) + 1(-2) = 0$.
$2x + 2.88 - 3x - 2 = 0$.
$-x + 0.88 = 0$,so $x = 0.88$.
The number of $Fe^{3+}$ ions is $0.96 - 0.88 = 0.08$.
The fraction of $Fe$ existing as $Fe(III)$ is $\frac{0.08}{0.96} = \frac{1}{12}$.

(A) Moles of dioxygen gas $(O_2)$ $= \frac{3.2 \ g}{32 \ g \ mol^{-1}} = 0.1 \ mol$.
Volume of dioxygen gas at $STP$ $(V_1)$ $= 0.1 \ mol \times 22.4 \ L \ mol^{-1} = 2.24 \ L$.
According to Boyle's Law,at constant temperature,$P_1V_1 = P_2V_2$.
Given $P_2 = \frac{1}{3}P_1$,so $P_1 \times 2.24 = (\frac{1}{3}P_1) \times V_2$.
$V_2 = 2.24 \times 3 = 6.72 \ L$.

(C) . Biochemical oxygen demand $(BOD)$ is a measure of the amount of dissolved oxygen required by bacteria to decompose organic matter in water. Thus,$a-iii$.
$b$. Photochemical smog is formed by the action of sunlight on nitrogen oxides and hydrocarbons,creating an oxidising atmosphere. Thus,$b-i$.
$c$. Classical smog occurs in cool,humid climates and is a mixture of smoke,fog,and sulfur dioxide,which acts as a reducing mixture. Thus,$c-iv$.
$d$. Ozone layer depletion is primarily caused by chlorofluorocarbons $(CFCs)$ which lead to the formation of polar stratospheric clouds that facilitate ozone destruction. Thus,$d-ii$.
Therefore,the correct matching is $a-iii, b-i, c-iv, d-ii$.

(B) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$
For the reaction $CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = 2 - 1 = 1$.
Given $K_p = 3.0$,$T = 1000 \ K$,and $R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$.
Substituting these values into the equation: $3.0 = K_c(0.083 \times 1000)^1$
$3.0 = K_c(83)$
$K_c = \frac{3.0}{83} \approx 3.6 \times 10^{-2}$

(D) The stability of carbocations is determined by the inductive effect and hyperconjugation.
$1$. $CH_3^+$ is a methyl carbocation (zero alkyl groups).
$2$. $CH_3CH_2^+$ is a primary $(1^{\circ})$ carbocation.
$3$. $(CH_3)_2CH^+$ is a secondary $(2^{\circ})$ carbocation.
$4$. $(CH_3)_3C^+$ is a tertiary $(3^{\circ})$ carbocation.
As the number of alkyl groups attached to the positively charged carbon increases,the stability increases due to the electron-donating inductive effect ($+I$ effect) and hyperconjugation.
Therefore,the tertiary carbocation $(CH_3)_3C^+$ is the most stable.

(C) The ascent of xylem sap in plants is primarily driven by the $Transpiration$ $Pull$ mechanism,which is explained by the $Cohesion-Tension$ theory.
$1$. $Cohesion$: The mutual attraction between water molecules due to hydrogen bonding.
$2$. $Adhesion$: The attraction of water molecules to the polar surfaces of the xylem vessel walls.
$3$. $Surface$ $Tension$: Water molecules are more attracted to each other in the liquid phase than to water in the gas phase.
These forces create a continuous water column that is pulled upward by the negative pressure generated by transpiration at the leaf surface. While root pressure contributes to the movement of water,it is not sufficient to push water to the heights of tall trees,making cohesion and adhesion the primary factors.

(A) The boiling points of group $16$ hydrides are influenced by both molecular mass and intermolecular hydrogen bonding.
For $H_2S$,$H_2Se$,and $H_2Te$,the boiling point increases with increasing molar mass due to stronger van der Waals forces.
However,$H_2O$ exhibits strong intermolecular hydrogen bonding,which results in a significantly higher boiling point than the other hydrides in the group.
The correct order of boiling points is $H_2S < H_2Se < H_2Te < H_2O$.
Statement $I$ is incorrect because it places $H_2O$ at the beginning of the sequence.
Statement $II$ is incorrect because it suggests a simple trend based on molar mass,ignoring the dominant effect of hydrogen bonding in $H_2O$.

(D) Interhalogen compounds are generally more reactive than halogens because the bond between two different halogens $(X-X')$ is weaker than the bond between two similar halogens $(X-X)$,except for $F_{2}$.
Since the $I-Cl$ bond is weaker than the $I-I$ bond,$ICl$ is more reactive than $I_{2}$.
Therefore,both Assertion $(A)$ and Reason $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(C) Thermosetting polymers are cross-linked polymers that undergo extensive cross-linking during molding,making them hard and infusible. Therefore,they cannot be reshaped or reused upon heating.

(B) Enantiomers are defined as non-superimposable mirror images of each other. Therefore,the statement that they are superimposable is incorrect.

(A) The given complex is $\left[ Ag(H_{2}O)_{2} \right] \left[ Ag(CN)_{2} \right]$.
In this complex,the cation is $\left[ Ag(H_{2}O)_{2} \right]^+$ and the anion is $\left[ Ag(CN)_{2} \right]^-$.
For the cation $\left[ Ag(H_{2}O)_{2} \right]^+$,the oxidation state of $Ag$ is $x + 2(0) = +1$,so $x = +1$.
The name is diaquasilver$(I)$.
For the anion $\left[ Ag(CN)_{2} \right]^-$,the oxidation state of $Ag$ is $x + 2(-1) = -1$,so $x = +1$.
The name is dicyanidoargentate$(I)$.
Combining these,the $IUPAC$ name is diaquasilver$(I)$ dicyanidoargentate$(I)$.

(B) According to the Hardy-Schulze rule,the coagulating power of an ion depends on its valency. The greater the valency of the flocculating ion,the greater is its power to cause precipitation.
Statement $I$ is correct: For a negative sol,the coagulating power of cations follows the order $Al^{3+} > Ba^{2+} > Na^{+}$ because the valency of the cation is directly proportional to the coagulating power.
Statement $II$ is incorrect: For a positive sol,the coagulating power depends on the valency of the anion. The order should be based on the valency of the anions $(Cl^- < SO_{4}^{2-} < PO_{4}^{3-})$. Therefore,the correct order of flocculating power for the salts should be $NaCl < Na_{2}SO_{4} < Na_{3}PO_{4}$.

(C) The reactions of carbonyl compounds are as follows:
$(a).$ Cyanohydrin is formed by the reaction of carbonyl compounds with $HCN$ $(iv)$.
$(b).$ Acetal is formed by the reaction of aldehydes with alcohol in the presence of dry $HCl$ $(iii)$.
$(c).$ Schiff's base is formed by the reaction of carbonyl compounds with primary amines $(RNH_2)$ $(ii)$.
$(d).$ Oxime is formed by the reaction of carbonyl compounds with hydroxylamine $(NH_2OH)$ $(i)$.
Therefore,the correct matching is $(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)$.

(C) The standard reduction potentials $(SRP)$ are given as: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$,$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$,$E^{\circ}_{Cu^{2+}/Cu} = 0.34 \ V$,and $E^{\circ}_{Ag^{+}/Ag} = 0.80 \ V$.
$A$ metal with a lower $SRP$ value can displace a metal with a higher $SRP$ value from its salt solution.
Comparing the values,the reactivity order is $Zn > Fe > Cu > Ag$.
In option $C$,$Ag$ is less reactive than $Cu$ $(E^{\circ}_{Ag^{+}/Ag} > E^{\circ}_{Cu^{2+}/Cu})$,therefore $Ag$ cannot displace $Cu$ from $CuSO_4$ solution.
Thus,the reaction $2CuSO_{4(aq)} + 2Ag_{(s)} \rightarrow 2Cu_{(s)} + Ag_2SO_{4(aq)}$ is not spontaneous and cannot occur.

(B) The correct answer is $B$.
$1$. Like chemical catalysts,enzymes reduce the activation energy of bio-processes. This is a correct statement.
$2$. Enzymes are polysaccharides. This is an incorrect statement because enzymes are proteinaceous in nature (they are globular proteins).
$3$. Enzymes are very specific for a particular reaction and substrate. This is a correct statement.
$4$. Enzymes are biocatalysts. This is a correct statement.

(B) Statement $I$ is correct: Primary aliphatic amines react with nitrous acid $(HNO_2)$ to form highly unstable aliphatic diazonium salts,which decompose immediately to evolve $N_2$ gas and form alcohols.
Statement $II$ is incorrect: Primary aromatic amines react with $HNO_2$ to form benzene diazonium salts,which are stable only at low temperatures $(273-278 \ K)$. Above $278 \ K$ (or $5 \ ^\circ C$),these salts decompose; they are certainly not stable above $300 \ K$.

(D) The boiling point order for compounds of comparable molecular masses is: Alcohols > Aldehydes/Ketones > Hydrocarbons.
Statement $I$ is correct: Aldehydes and ketones possess a polar carbonyl group,leading to dipole-dipole interactions,which are stronger than the weak London dispersion forces found in non-polar hydrocarbons.
Statement $II$ is correct: Alcohols exhibit intermolecular $H$-bonding,which is significantly stronger than the dipole-dipole interactions present in aldehydes and ketones,resulting in higher boiling points for alcohols.
Therefore,both statements are correct.

(B) For a zero order reaction,the rate is independent of the concentration of the reactant. Thus,a graph of $y=$ rate versus $x=$ concentration is a horizontal line.
For a first order reaction,the half-life $(t_{1/2})$ is independent of the initial concentration of the reactant. Thus,a graph of $y=t_{1/2}$ versus $x=$ concentration is a horizontal line.
Therefore,the correct option is $B$.

(A) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
$m = \frac{\text{Moles of solute}}{\text{Mass of solvent in } kg}$
Given $m = 1 \ mol/kg$ and $\text{moles of solute} = 0.5 \ mol$.
$1 = \frac{0.5}{\text{Mass of solvent in } kg}$
$\text{Mass of solvent in } kg = 0.5 \ kg = 500 \ g$.

(D) The reaction of a Grignard reagent $(RMgX)$ with carbon dioxide $(CO_2)$ is a nucleophilic addition reaction.
The nucleophilic alkyl group $(R^-)$ of the Grignard reagent attacks the electrophilic carbon atom of $CO_2$.
This results in the formation of an intermediate carboxylate salt,which is $RCOO^{-}Mg^{+}X$ (often written as $RCOOMgX$).
Subsequent acid hydrolysis $(H_3O^{+})$ of this intermediate yields the corresponding carboxylic acid $(RCOOH)$.

(D) For the reaction to be spontaneous,the cell potential $E_{cell}^{o}$ must be positive.
Reduction (Cathode): $2MnO_{4}^{-} + 16H^{+} + 10e^{-} \rightarrow 2Mn^{2+} + 8H_{2}O$ $(E^{o} = +1.510 \, V)$
Oxidation (Anode): $5H_{2}O \rightarrow \frac{5}{2} O_{2} + 10H^{+} + 10e^{-}$ $(E^{o} = +1.223 \, V)$
Overall reaction: $2MnO_{4}^{-} + 6H^{+} \rightarrow 2Mn^{2+} + \frac{5}{2} O_{2} + 3H_{2}O$
$E_{cell}^{o} = E_{cathode}^{o} - E_{anode}^{o}$
$E_{cell}^{o} = 1.510 \, V - 1.223 \, V = +0.287 \, V$
Since $E_{cell}^{o} > 0$,the reaction is spontaneous,and $MnO_{4}^{-}$ will liberate $O_{2}$ from water in the presence of an acid.

(B) Statement $I$ is correct: The $-NO_2$ group is a strong electron-withdrawing group $(EWG)$ due to both $-I$ and $-M$ effects. It stabilizes the phenoxide ion,thereby increasing the acidic strength of nitrophenols compared to phenol.
Statement $II$ is incorrect: The acidic strength of nitrophenols depends on the position of the $-NO_2$ group. The $-M$ effect is operative at ortho and para positions,while only the $-I$ effect is operative at the meta position. Therefore,their acidic strengths are different. The order of acidic strength is $p$-nitrophenol > $o$-nitrophenol > $m$-nitrophenol > phenol.

(A) $(i)$ Assertion $(A)$ is correct because in point defects like Schottky or metal deficiency defects,the ionic solid maintains electrical neutrality despite the absence of ions.
$(ii)$ Reason $(R)$ is correct because Frenkel defect occurs when a cation leaves its lattice site and occupies an interstitial site,which preserves the overall electrical neutrality of the crystal.
$(iii)$ However,the Reason $(R)$ describes the Frenkel defect,whereas the Assertion $(A)$ refers to a general property of point defects (specifically vacancy-type defects). Therefore,$(R)$ is not the correct explanation for $(A)$.

(A) The correct matches are as follows:
$(a)$ Antacids: Cimetidine $(iii)$
$(b)$ Antihistamines: Seldane $(iv)$
$(c)$ Analgesics: Morphine $(ii)$
$(d)$ Antimicrobials: Salvarsan $(i)$
Therefore,the correct sequence is $(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)$.

(D) Chlorobenzene can be synthesized by the electrophilic aromatic substitution of benzene with chlorine in the presence of a Lewis acid catalyst like anhydrous $FeCl_{3}$.
The reaction is:
$C_{6}H_{6} + Cl_{2} \xrightarrow{\text{anhydrous } FeCl_{3}} C_{6}H_{5}Cl + HCl$
Alternatively,it can also be prepared from aniline via the Sandmeyer reaction (using $NaNO_{2}, HCl$ followed by $CuCl$),but option $D$ is a direct and standard method for the synthesis of chlorobenzene from benzene.

(A) The electronic configuration of $Gd$ $(Z=64)$ is $[Xe] 4f^7 5d^1 6s^2$.
Upon formation of $Gd^{2+}$,the configuration becomes $[Xe] 4f^7 5d^1$.
The third ionisation enthalpy involves the removal of the $5d^1$ electron.
After the removal of this electron,the remaining $4f^7$ configuration is half-filled,which is highly stable due to high exchange energy.
Therefore,the energy required for the third ionisation is relatively low.

(A) The reaction between acetone $(CH_3COCH_3)$ and $2-$pentanone $(CH_3COCH_2CH_2CH_3)$ in the presence of dilute $NaOH$ and heating is a cross-aldol condensation reaction.
Acetone has two equivalent $\alpha-$hydrogens,while $2-$pentanone has two types of $\alpha-$hydrogens (at $C-1$ and $C-3$).
Self-aldol products of acetone and $2-$pentanone can also form.
The product shown in option $A$ requires a structure that cannot be derived from the condensation of these two specific ketones under the given conditions,as it would require a different carbon skeleton or starting materials.
Therefore,the structure in option $A$ is not formed.

(D) The formula for density is $d = \frac{Z \times M}{N_A \times a^3}$.
For an $fcc$ unit cell,the number of atoms per unit cell $Z = 4$.
The edge length $a = 3.608 \times 10^{-8} \, cm$.
Avogadro's number $N_A = 6.022 \times 10^{23} \, mol^{-1}$.
Substituting the values: $8.92 = \frac{4 \times M}{6.022 \times 10^{23} \times (3.608 \times 10^{-8})^3}$.
$8.92 = \frac{4 \times M}{6.022 \times 10^{23} \times 46.96 \times 10^{-24}}$.
$M = \frac{8.92 \times 6.022 \times 10^{23} \times 46.96 \times 10^{-24}}{4}$.
$M = 63.1 \, g \, mol^{-1}$.
Thus,the atomic mass of copper is $63.1 \, u$.

(B) Statement $I$ is correct because the Lucas test uses a mixture of conc. $HCl$ and anhydrous $ZnCl_{2}$ (Lucas Reagent) to distinguish between $1^{\circ}$,$2^{\circ}$,and $3^{\circ}$ alcohols based on their reactivity.
Statement $II$ is incorrect because $3^{\circ}$ alcohols are the most reactive and produce turbidity immediately at room temperature,whereas $1^{\circ}$ alcohols are the least reactive and do not produce turbidity at room temperature.
The reactivity order towards Lucas reagent is $3^{\circ} > 2^{\circ} > 1^{\circ}$ alcohol.

(A) For a first order reaction,the rate constant $K$ is given by the formula:
$K = \frac{2.303}{t} \log \left(\frac{[A]_0}{[A]_t}\right)$
Given:
Initial concentration $[A]_0 = 0.1 \, M$
Concentration after time $t$,$[A]_t = 0.001 \, M$
Time $t = 5 \, \min$
Substituting the values:
$K = \frac{2.303}{5} \log \left(\frac{0.1}{0.001}\right)$
$K = \frac{2.303}{5} \log(100)$
Since $\log(100) = 2$:
$K = \frac{2.303 \times 2}{5} = \frac{4.606}{5} = 0.9212 \, \min^{-1}$

(B) The cell reaction is: $Ni_{(s)} + 2Ag^{+}(0.001 \ M) \rightarrow Ni^{2+}(0.001 \ M) + 2Ag_{(s)}$
Using the Nernst equation: $E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} \log \frac{[Ni^{2+}]}{[Ag^{+}]^2}$
Here,$n = 2$,$[Ni^{2+}] = 10^{-3} \ M$,and $[Ag^{+}] = 10^{-3} \ M$.
$E_{cell} = 10.5 - \frac{0.059}{2} \log \frac{10^{-3}}{(10^{-3})^2}$
$E_{cell} = 10.5 - \frac{0.059}{2} \log \frac{10^{-3}}{10^{-6}}$
$E_{cell} = 10.5 - \frac{0.059}{2} \log 10^3$
$E_{cell} = 10.5 - \frac{0.059}{2} \times 3$
$E_{cell} = 10.5 - 0.0885 = 10.4115 \ V$

(C) The reaction sequence is as follows:
$1$. Reduction of benzonitrile $(C_6H_5CN)$ with $LiAlH_4$ followed by $H_2O$ gives benzylamine $(C_6H_5CH_2NH_2)$.
$2$. Treatment of benzylamine with $NaNO_2 + HCl$ at low temperature $(0-5 \ ^\circ C)$ forms unstable benzyl diazonium chloride $(C_6H_5CH_2N_2^+Cl^-)$.
$3$. Benzyl diazonium chloride rapidly decomposes to form a benzyl carbocation $(C_6H_5CH_2^+)$,which reacts with water $(H_2O)$ to yield benzyl alcohol $(C_6H_5CH_2OH)$.

(A) The correct matches are:
$a$. Haematite is $Fe_{2}O_{3}$ $(iii)$.
$b$. Magnetite is $Fe_{3}O_{4}$ $(i)$.
$c$. Calamine is $ZnCO_{3}$ $(ii)$.
$d$. Kaolinite is $[Al_{2}(OH)_{4}Si_{2}O_{5}]$ $(iv)$.
Therefore,the correct sequence is $a-iii, b-i, c-ii, d-iv$.

(B) The complexes are:
$A: [Ni(H_{2}O)_{2}(en)_{2}]^{2+}$
$B: [Ni(H_{2}O)_{4}(en)]^{2+}$
$C: [Ni(en)_{3}]^{2+}$
$en$ (ethylenediamine) is a stronger field ligand $(SFL)$ compared to $H_{2}O$.
According to the spectrochemical series,the crystal field splitting energy $(\Delta_{0})$ increases as the number of strong field ligands increases.
In complex $C$,there are $3$ $en$ ligands.
In complex $A$,there are $2$ $en$ ligands.
In complex $B$,there is $1$ $en$ ligand.
Therefore,the order of crystal field splitting energy $(\Delta_{0})$ is $C > A > B$.
Since the energy absorbed $(E = h\nu = \Delta_{0})$ is directly proportional to the splitting energy,the order of energy absorbed is $C > A > B$.

(A) According to Henry's Law, the solubility of a gas in a liquid is inversely proportional to the Henry's Law constant $(K_{H})$.
Mathematically, $S \propto 1 / K_{H}$.
Therefore, a higher value of $K_{H}$ indicates lower solubility, and a lower value of $K_{H}$ indicates higher solubility.
Comparing the given $K_{H}$ values:
$Ar (40.3) > CO_{2} (1.67) > CH_{4} (0.413) > HCHO (1.83 \times 10^{-5})$.
Thus, the order of solubility in water is:
$Ar < CO_{2} < CH_{4} < HCHO$.

(B) The large-scale industrial preparation of nitric acid is carried out by the $Ostwald$ process.
In this process,ammonia is catalytically oxidized by atmospheric oxygen to form nitric oxide $(NO)$.
The reaction is: $4NH_3(g) + 5O_2(g) \xrightarrow[Pt/Rh \ \text{catalyst}]{500 \ K, 9 \ bar} 4NO(g) + 6H_2O(g)$.

(B) The reaction sequence is as follows:
$1$. Benzaldehyde reacts with $HCN$ to form benzaldehyde cyanohydrin: $C_6H_5CHO + HCN \rightarrow C_6H_5CH(OH)CN$.
$2$. Acidic hydrolysis of the cyanohydrin yields mandelic acid: $C_6H_5CH(OH)CN + H_3O^+ \rightarrow C_6H_5CH(OH)COOH$.
$3$. Heating mandelic acid with $NaOH$ and $CaO$ (sodalime) leads to decarboxylation. However,the provided reaction sequence and options suggest a reduction pathway. Given the options,the final product is $C_6H_5CH_2OH$ (Benzyl alcohol).

(D) The correct matches are:
$a$. Gabriel synthesis is used for the preparation of $1^{\circ}$ amines $(iii)$.
$b$. Kolbe synthesis (Kolbe-Schmitt reaction) is used for the preparation of salicylic acid $(iv)$.
$c$. Williamson synthesis is used for the preparation of ethers $(ii)$.
$d$. Etard reaction is used for the oxidation of toluene to benzaldehyde $(i)$.
Therefore,the correct matching is $a-iii, b-iv, c-ii, d-i$.

(C) The reactivity of alkyl halides towards $S_{N}2$ reactions is primarily governed by steric hindrance. The order of reactivity is $Primary > Secondary > Tertiary$.
$(I)$ $CH_{3}CH_{2}CH_{2}CH_{2}Cl$ is a primary alkyl halide with minimal steric hindrance.
$(III)$ $(CH_{3})_{2}CHCH_{2}Cl$ is a primary alkyl halide but has more steric hindrance than $(I)$ due to the branching at the $\beta$-carbon.
$(II)$ $CH_{3}CH_{2}CH(Cl)CH_{3}$ is a secondary alkyl halide.
$(IV)$ $(CH_{3})_{3}CCl$ is a tertiary alkyl halide,which is the least reactive towards $S_{N}2$ due to maximum steric hindrance.
Therefore,the correct order is $I > III > II > IV$.

(B) Thermoplastic polymers are those that soften on heating and harden on cooling,and can be reshaped.
$Polythene$ is a classic example of a thermoplastic polymer.
$Bakelite$,$Urea-formaldehyde$ resin,and $Melamine$ polymer are thermosetting polymers,which undergo permanent chemical change upon heating and cannot be reshaped.

(C) For a cell to be feasible,$E^{\circ}_{cell}$ must be positive.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
Here,$Co^{3+}$ is reduced $(cathode)$ and $Al$ is oxidized $(anode)$.
$E^{\circ}_{cathode} = E^{\circ}_{Co^{3+}/Co^{2+}} = 1.81 \, V$
$E^{\circ}_{anode} = E^{\circ}_{Al^{3+}/Al} = -1.66 \, V$
$E^{\circ}_{cell} = 1.81 \, V - (-1.66 \, V) = 1.81 + 1.66 = 3.47 \, V$.

(B) Physical adsorption is an exothermic process. According to Le Chatelier's principle,an increase in temperature decreases the extent of adsorption $(\frac{x}{m})$ at a constant pressure.
From the given graph,at any constant pressure $p$,the extent of adsorption follows the order: $(x/m)_{T_1} > (x/m)_{T_2} > (x/m)_{T_3}$.
Since adsorption decreases with an increase in temperature,the corresponding temperatures must follow the inverse order: $T_1 < T_2 < T_3$,or $T_3 > T_2 > T_1$.

(B) For a first-order reaction,the number of half-lives $(n)$ is calculated as $n = \frac{t}{t_{1/2}} = \frac{8000}{2000} = 4$.
After $n$ half-lives,the remaining concentration $[A]_t$ is given by $[A]_t = \frac{[A]_0}{2^n}$.
Substituting the values: $0.02 = \frac{[A]_0}{2^4}$.
$0.02 = \frac{[A]_0}{16}$.
$[A]_0 = 0.02 \times 16 = 0.32 \, M$.

(A) The standard electrode potential $(E^{\circ})$ of a species is determined by the overall energy change involved in the process: $\frac{1}{2} F_2(g) + e^- \rightarrow F^-(aq)$.
This process involves three steps: $(1)$ Atomization (bond dissociation),$(2)$ Electron gain,and $(3)$ Hydration.
Fluorine has a very low bond dissociation enthalpy due to inter-electronic repulsions between lone pairs in the small $F_2$ molecule.
Additionally,the small size of the $F^-$ ion results in a very high hydration enthalpy,which makes the overall process highly exothermic.
Therefore,both the low bond dissociation enthalpy and the high hydration enthalpy contribute to the strong oxidizing power of fluorine.

(A) $[Co(NH_3)_5NO_2]Cl_2$ and $[Co(NH_3)_5ONO]Cl_2$ exhibit linkage isomerism due to the ambidentate ligand $NO_2^-$. Thus,$a-i$.
$(b)$ $[Cr(NH_3)_6][Co(CN)_6]$ and $[Cr(CN)_6][Co(NH_3)_6]$ exhibit coordination isomerism due to the exchange of ligands between cationic and anionic entities. Thus,$b-ii$.
$(c)$ $[Co(NH_3)_5(SO_4)]Br$ and $[Co(NH_3)_5Br]SO_4$ exhibit ionisation isomerism as they produce different ions in solution. Thus,$c-iii$.
$(d)$ $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$ exhibit solvate (hydrate) isomerism. Thus,$d-iv$.
Therefore,the correct match is $a-i, b-ii, c-iii, d-iv$.

(C) Denaturation is a process that causes the loss of the secondary and tertiary structures of proteins,leading to the loss of their biological activity.
$1$. It is caused by changes in physical conditions such as temperature or chemical conditions like $pH$.
$2$. During this process,the helical structure of the protein uncoils.
$3$. Option $C$ describes the primary structure of a protein (formation from amino acids linked by peptide bonds),which remains intact during denaturation.
Therefore,the statement in option $C$ is incorrect regarding the process of denaturation.

(A) Step $1$: Aniline reacts with acetic anhydride $(CH_3CO)_2O$ in the presence of pyridine to form acetanilide $(C_6H_5NHCOCH_3)$.
Step $2$: Acetanilide is then reduced by $LiAlH_4$ (a strong reducing agent) which reduces the amide group $(-NHCOCH_3)$ to an amine group $(-NHCH_2CH_3)$.
Step $3$: Hydrolysis $(H_2O)$ yields the final product,$N$-ethylaniline $(C_6H_5NHCH_2CH_3)$.

(A) Chlorine is an electron-withdrawing group due to its high electronegativity ($-I$ effect),which deactivates the benzene ring.
However,it has lone pairs of electrons that participate in resonance ($+R$ effect).
During electrophilic aromatic substitution,the resonance effect stabilizes the carbocation intermediate specifically at the ortho and para positions.
Since both statements are scientifically accurate and the reason correctly explains the dual nature of the chlorine substituent,the correct option is $(A)$.

(A) Nucleophilic addition-elimination reactions involve the addition of a nucleophile to the carbonyl carbon followed by the elimination of a water molecule.
In the reaction of $CH_3CHO$ with $NaHSO_3$,the bisulfite ion acts as a nucleophile and adds to the carbonyl carbon to form a bisulfite addition product.
This is a pure nucleophilic addition reaction,not an addition-elimination reaction,as no water molecule is eliminated.
Options $B$,$C$,and $D$ represent the formation of oximes,hydrazones,and imines respectively,which are classic examples of nucleophilic addition-elimination reactions.

(C) The decrease in atomic size across the actinoid series is more pronounced than in the lanthanoid series.
This is because the $5f$ orbitals have a poorer shielding effect compared to the $4f$ orbitals.
Due to the poor shielding by $5f$ electrons,the effective nuclear charge $(Z_{eff})$ increases more significantly across the actinoid series.
Since atomic size is inversely proportional to $Z_{eff}$ $(Size \propto \frac{1}{Z_{eff}})$,the size decreases more rapidly.

(C) The reactions of phenol are as follows:
$a$. Phenol reacts with $(i)$ $NaOH$,$(ii)$ $CO_2$,$(iii)$ $H^{+}$ to form Salicylic acid (Kolbe's reaction). Thus,$a-iv$.
$b$. Phenol reacts with $(i)$ Aqueous $NaOH + CHCl_3$,$(ii)$ $H^{+}$ to form Salicylaldehyde (Reimer-Tiemann reaction). Thus,$b-iii$.
$c$. Phenol reacts with $Zn$ dust upon heating to form Benzene. Thus,$c-ii$.
$d$. Phenol reacts with $Na_2Cr_2O_7$ and $H_2SO_4$ (chromic acid) to form Benzoquinone. Thus,$d-i$.
Therefore,the correct matching is $a-iv, b-iii, c-ii, d-i$.

(A) The rate of reaction is given by the expression: $\text{Rate} = -\frac{1}{4} \frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt} = \frac{1}{6} \frac{d[C]}{dt} = \frac{1}{9} \frac{d[D]}{dt}$.
Given,rate of disappearance of $A$ $(-\frac{d[A]}{dt})$ $= 4 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$.
Therefore,$\text{Rate of reaction} = \frac{1}{4} \times (4 \times 10^{-2}) = 1 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$.
Now,$\text{Rate of disappearance of } B = 3 \times \text{Rate of reaction} = 3 \times 1 \times 10^{-2} = 3 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$.
Amount of $B$ consumed in $10 \ s = \text{Rate of disappearance of } B \times \text{Time} = (3 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}) \times 10 \ s = 30 \times 10^{-2} \ mol \ L^{-1}$.

(D) Let us analyze each reaction pathway for the synthesis of benzaldehyde:
$A$. Etard reaction: Toluene is oxidized by $CrO_2Cl_2$ in $CS_2$ followed by hydrolysis to give benzaldehyde. This is a correct method.
$B$. Rosenmund reduction: Benzoyl chloride is reduced by $H_2$ in the presence of $Pd-BaSO_4$ (Lindlar catalyst) to give benzaldehyde. This is a correct method.
$C$. Toluene is oxidized by $CrO_3$ in the presence of acetic anhydride to form a gem-diacetate intermediate,which upon hydrolysis yields benzaldehyde. This is a correct method.
$D$. Reaction of benzonitrile $(C_6H_5CN)$ with $CH_3MgBr$ (a Grignard reagent) followed by hydrolysis yields acetophenone $(C_6H_5COCH_3)$,not benzaldehyde. Therefore,this is an incorrect method for the synthesis of benzaldehyde.

(A) In metal carbonyls,the metal-carbon bond is formed by a synergic bonding mechanism.
$1$. The ligand $(CO)$ donates a lone pair of electrons from its carbon atom into an empty orbital of the metal,forming a $\sigma$ bond.
$2$. The metal then donates a pair of electrons from its filled $d$-orbital into the empty antibonding $\pi^*$ orbital of the $CO$ ligand,forming a $\pi$ bond.
$3$. This back-donation strengthens the metal-carbon bond and weakens the carbon-oxygen bond.
$4$. Thus,both Assertion $(A)$ and Reason $(R)$ are correct,and $(R)$ correctly explains $(A)$.

(C) The preparation of phenol from chlorobenzene requires harsh conditions (high temperature and high pressure,e.g.,$623 \ K$ and $300 \ atm$) because the $C-Cl$ bond in chlorobenzene has partial double bond character due to resonance.
Under $STP$ conditions,chlorobenzene does not react with $NaOH$ to form phenol.
Therefore,the reaction sequence involving chlorobenzene and $NaOH$ under $STP$ conditions is incorrect for the preparation of phenol.

(B) The reaction sequence is as follows:
$1$. Styrene $(Ph-CH=CH_2)$ reacts with $HBr$ to form $1-$bromo$-1-$phenylethane $(Ph-CH(Br)-CH_3)$ via Markovnikov addition.
$2$. This alkyl bromide reacts with $Mg$ in dry ether to form the Grignard reagent $(Ph-CH(MgBr)-CH_3)$.
$3$. The Grignard reagent reacts with $CO_2$ followed by acidic workup $(H_3O^+)$ to form $2-$phenylpropanoic acid $(Ph-CH(CH_3)-COOH)$,which is product $A$.
$4$. $2-$phenylpropanoic acid $(A)$ reacts with $SOCl_2$ to form the corresponding acid chloride $(Ph-CH(CH_3)-COCl)$.
$5$. This acid chloride then reacts with methylamine $(CH_3NH_2)$ to form the amide,$N$-methyl$-2-$phenylpropanamide $(Ph-CH(CH_3)-CONHCH_3)$,which is product $B$.