JEE Main 2021 Chemistry Question Paper with Answer and Solution

(D) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$T^2 \propto R^3$ or $T \propto R^{3/2}$.
Taking the logarithmic derivative,we get $\frac{dT}{T} = \frac{3}{2} \frac{dR}{R}$.
The percentage change in radius is $\frac{dR}{R} \times 100 = \frac{1.02R - R}{R} \times 100 = 0.02 \times 100 = 2\%$.
Therefore,the percentage change in the time period is $\frac{dT}{T} \times 100 = \frac{3}{2} \times (\frac{dR}{R} \times 100) = \frac{3}{2} \times 2\% = 3\%$.
Thus,the percentage difference in the time periods is $3\%$.

(B) The magnetic field on the axis of a circular coil of radius $R$ at a distance $x$ from the center is given by $B = \frac{\mu_0 N i R^2}{2(R^2 + x^2)^{3/2}}$.
Since $B \propto (R^2 + x^2)^{-3/2}$,we have the ratio $\frac{B_1}{B_2} = \left( \frac{R^2 + x_2^2}{R^2 + x_1^2} \right)^{3/2}$.
Given $\frac{B_1}{B_2} = 8$,$x_1 = 0.05\,m$,and $x_2 = 0.2\,m$,we substitute these values:
$8 = \left( \frac{R^2 + (0.2)^2}{R^2 + (0.05)^2} \right)^{3/2}$.
Taking the power of $2/3$ on both sides:
$8^{2/3} = \frac{R^2 + 0.04}{R^2 + 0.0025} \Rightarrow 4 = \frac{R^2 + 0.04}{R^2 + 0.0025}$.
$4(R^2 + 0.0025) = R^2 + 0.04 \Rightarrow 4R^2 + 0.01 = R^2 + 0.04$.
$3R^2 = 0.03 \Rightarrow R^2 = 0.01 \Rightarrow R = 0.1\,m$.

(C) For an ideal gas, the equation of state is $PV = \mu RT$, which implies $P = \frac{\mu RT}{V}$.
Given the process equation $PT^2 = k$ (where $k$ is a constant).
Substituting $P$ in the process equation: $\left(\frac{\mu RT}{V}\right)T^2 = k$.
This simplifies to $\mu RT^3 = kV$.
Differentiating both sides with respect to $T$: $\mu R(3T^2 dT) = k dV$.
Dividing the differential equation by the original equation $\mu RT^3 = kV$:
$\frac{3\mu RT^2 dT}{\mu RT^3} = \frac{kdV}{kV}$.
This yields $\frac{3dT}{T} = \frac{dV}{V}$.
The coefficient of volume expansion $\gamma$ is defined as $\gamma = \frac{1}{V} \frac{dV}{dT}$.
From the derived relation, $\frac{1}{V} \frac{dV}{dT} = \frac{3}{T}$.
Therefore, $\gamma = 3/T$.

(C) Given: Diameter of lens $= 6 \ cm$,so radius $a = 3 \ cm$. Thickness $t = 3 \ mm = 0.3 \ cm$. Speed of light in lens $v = 2 \times 10^8 \ m/s$. Speed of light in vacuum $c = 3 \times 10^8 \ m/s$.
Refractive index $\mu = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$.
For a spherical surface of radius $R$,the relationship between radius of aperture $a$,thickness $t$,and radius of curvature $R$ is given by $R^2 = a^2 + (R - t)^2$.
$R^2 = a^2 + R^2 - 2Rt + t^2 \implies 2Rt = a^2 + t^2$.
$R = \frac{a^2 + t^2}{2t} = \frac{3^2 + (0.3)^2}{2 \times 0.3} = \frac{9 + 0.09}{0.6} = \frac{9.09}{0.6} = 15.15 \ cm$.
Using the lens maker's formula for a plano-convex lens: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} \right)$.
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{15.15} \right) = 0.5 \times \frac{1}{15.15} = \frac{1}{30.3}$.
Thus,$f \approx 30 \ cm$.

(C) Given the process equation: $PT^2 = K$ (where $K$ is a constant).
Using the ideal gas law $PV = \mu RT$,we can express pressure as $P = \frac{\mu RT}{V}$.
Substituting this into the process equation: $\left(\frac{\mu RT}{V}\right) T^2 = K$.
This simplifies to $\frac{\mu R T^3}{V} = K$,or $T^3 = \left(\frac{K}{\mu R}\right) V$.
Let $C = \frac{K}{\mu R}$,so $T^3 = CV$.
Differentiating both sides with respect to $T$: $3T^2 dT = C dV$.
Thus,$\frac{dV}{dT} = \frac{3T^2}{C} = \frac{3T^2}{(T^3/V)} = \frac{3V}{T}$.
The coefficient of volume expansion $\gamma$ is defined as $\gamma = \frac{1}{V} \frac{dV}{dT}$.
Substituting the expression for $\frac{dV}{dT}$: $\gamma = \frac{1}{V} \left(\frac{3V}{T}\right) = \frac{3}{T}$.

(D) The reaction sequence involves the diazotization of $2,4,6$-tribromoaniline using $NaNO_2/HCl$ at low temperature to form the corresponding diazonium salt.
Subsequent treatment with ethanol $(EtOH)$ and heating results in the reduction of the diazonium group,replacing it with a hydrogen atom.
Thus,the $-NH_2$ group is removed,yielding $1,3,5$-tribromobenzene as the major product $X$.

(B) The coefficient of linear expansion for metal $A$ is $\alpha_A$ and for metal $B$ is $\alpha_B$. Given that $\alpha_A > \alpha_B$.
When the temperature decreases (placed in a cold bath),both metals undergo contraction.
The change in length is given by $\Delta L = L_0 \alpha \Delta T$.
Since $\alpha_A > \alpha_B$,the contraction in metal $A$ is greater than the contraction in metal $B$.
Because metal $A$ is on the left side and contracts more than metal $B$ on the right,the strip will bend towards the left.

(B) The coefficient of linear expansion is denoted by $\alpha$. Given that $\alpha_A > \alpha_B$.
When the temperature decreases (placed in a cold bath),the change in length is given by $\Delta L = L_0 \alpha \Delta T$,where $\Delta T$ is negative.
Since $\alpha_A > \alpha_B$,the magnitude of contraction in metal $A$ is greater than that in metal $B$ $(|\Delta L_A| > |\Delta L_B|)$.
Because metal $A$ is on the left and it contracts more than metal $B$,the strip will bend towards the left side (the side of the metal that contracts more).

(C) The reaction involves the Friedel-Crafts acylation of $o$-cresol with $\gamma$-butyrolactone in the presence of a Lewis acid catalyst,$AlCl_3$.
$1$. The $AlCl_3$ coordinates with the carbonyl oxygen of the lactone,activating it for nucleophilic attack.
$2$. The electron-rich aromatic ring of $o$-cresol attacks the activated carbonyl carbon,leading to the ring-opening of the lactone.
$3$. This forms an intermediate ketone with a hydroxypropyl side chain.
$4$. Under the reaction conditions (heating,$\Delta$),this intermediate undergoes intramolecular cyclization (Friedel-Crafts alkylation) to form a bicyclic ketone.
$5$. The regioselectivity is governed by the directing effects of the $-OH$ and $-CH_3$ groups on the aromatic ring,leading to the formation of the product shown in option $C$.

(C) The reaction proceeds in two steps:
$1$. Electrophilic addition of $Cl_2$ across the double bond in the presence of $CCl_4$ yields a vicinal dichloride: $CH_3O-C_6H_4-CH_2-CH_2-CHCl-CH_2Cl$.
$2$. In the presence of anhydrous $AlCl_3$ (a Lewis acid),one of the chlorine atoms is abstracted to form a carbocation. The electron-rich benzene ring (activated by the $-OCH_3$ group) then undergoes intramolecular Friedel-Crafts alkylation.
$3$. The cyclization occurs such that the more stable carbocation intermediate leads to the formation of a six-membered ring,resulting in $6-methoxy-2-chlorotetralin$ (or a related isomer depending on the specific cyclization pathway). Based on the provided options,the product is $C$.

(C) The reaction proceeds in two steps:
$1$. Treatment with $t-BuOK$ (a strong base) causes an intramolecular elimination reaction (specifically an $E2$ mechanism) to form an $\alpha,\beta$-unsaturated ketone. The base abstracts an $\alpha$-hydrogen,and the resulting enolate displaces the chloride ion to form a vinyl ketone derivative.
$2$. Treatment with concentrated $H_2SO_4$ and heat induces a Friedel-Crafts type cyclization. The acid protonates the carbonyl oxygen,increasing the electrophilicity of the $\beta$-carbon,which then undergoes an intramolecular electrophilic aromatic substitution (cyclization) onto the benzene ring.
Given the structure,the cyclization occurs at the ortho position relative to the alkyl chain,leading to the formation of a six-membered ring fused to the benzene ring,resulting in $7$-isopropyl-$3,4$-dihydronaphthalen-$2(1H)$-one (or a related isomer depending on the specific substitution pattern). Based on the provided options,the product is represented by option $C$.

(C) The reaction involves the cleavage of ethers with excess $HI$.
$1$. The oxygen atoms are protonated by $H^+$.
$2$. The methoxy group attached to the benzene ring is cleaved to form a phenol and $CH_3I$.
$3$. The other ether linkage is cleaved by $I^-$ via an $S_N1$ or $S_N2$ mechanism (depending on the stability of the carbocation) to form an alcohol and an alkyl iodide.
$4$. In this specific case,the benzylic position is stabilized by the cyano group,allowing the formation of the iodo product at that position.
$5$. The final product is $3$-hydroxyphenyl-cyano-iodomethane.

(C) The reaction involves the acid-catalyzed hydration of the double bond in the presence of $dil. \ H_2SO_4$.
Specifically,this is an intramolecular cyclization reaction of geraniol (or a related terpene alcohol).
The protonation of the double bond leads to the formation of a carbocation,which then undergoes intramolecular nucleophilic attack by the other double bond to form a six-membered ring.
Finally,the addition of water to the resulting carbocation followed by deprotonation yields $\alpha$-terpineol as the major product.
The structure of $\alpha$-terpineol is given in option $C$.

(D) Given observations are $a, b, c$ with $b = a + c$.
The mean $\bar{x} = \frac{a + b + c}{3}$.
Substituting $a + c = b$,we get $\bar{x} = \frac{b + b}{3} = \frac{2b}{3}$.
The standard deviation of $a + 2, b + 2, c + 2$ is the same as the standard deviation of $a, b, c$,which is $d$.
Thus,$d^{2} = \frac{a^{2} + b^{2} + c^{2}}{3} - (\bar{x})^{2}$.
Substituting $\bar{x} = \frac{2b}{3}$,we have $d^{2} = \frac{a^{2} + b^{2} + c^{2}}{3} - \frac{4b^{2}}{9}$.
Multiplying by $9$,we get $9d^{2} = 3(a^{2} + b^{2} + c^{2}) - 4b^{2}$.
$9d^{2} = 3(a^{2} + c^{2}) + 3b^{2} - 4b^{2}$.
$9d^{2} = 3(a^{2} + c^{2}) - b^{2}$.
Therefore,$b^{2} = 3(a^{2} + c^{2}) - 9d^{2}$.

(D) Given that the ratio of intensities of two coherent sources is $\frac{I_1}{I_2} = 2x$.
We know that $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Substituting these into the expression $\frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$:
Numerator: $I_{\max} - I_{\min} = (I_1 + I_2 + 2\sqrt{I_1I_2}) - (I_1 + I_2 - 2\sqrt{I_1I_2}) = 4\sqrt{I_1I_2}$.
Denominator: $I_{\max} + I_{\min} = (I_1 + I_2 + 2\sqrt{I_1I_2}) + (I_1 + I_2 - 2\sqrt{I_1I_2}) = 2(I_1 + I_2)$.
Thus,the ratio is $\frac{4\sqrt{I_1I_2}}{2(I_1 + I_2)} = \frac{2\sqrt{I_1I_2}}{I_1 + I_2}$.
Dividing the numerator and denominator by $I_2$,we get $\frac{2\sqrt{I_1/I_2}}{I_1/I_2 + 1}$.
Substituting $\frac{I_1}{I_2} = 2x$,we get $\frac{2\sqrt{2x}}{2x + 1}$.

(C) Given integral: $I = \int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x$
We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2\sin x \cos x) = 1 - (\sin x - \cos x)^2$. However,it is more convenient to write $\sin 2x = (\sin x + \cos x)^2 - 1$.
Substituting this into the denominator:
$8 - \sin 2x = 8 - ((\sin x + \cos x)^2 - 1) = 8 - (\sin x + \cos x)^2 + 1 = 9 - (\sin x + \cos x)^2$.
Let $t = \sin x + \cos x$. Then $dt = (\cos x - \sin x) dx$.
The integral becomes:
$I = \int \frac{dt}{\sqrt{9 - t^2}} = \sin^{-1}\left(\frac{t}{3}\right) + c$.
Substituting $t$ back:
$I = \sin^{-1}\left(\frac{\sin x + \cos x}{3}\right) + c$.
Comparing this with $a \sin^{-1}\left(\frac{\sin x + \cos x}{b}\right) + c$,we get $a = 1$ and $b = 3$.
Thus,the ordered pair $(a, b)$ is $(1, 3)$.

(A) Let the point be $P(a, 6, 9)$ and its image be $Q(20, b, -a-9)$.
Let the line be $L: \frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9} = k$.
The midpoint $M$ of $PQ$ is $\left( \frac{a+20}{2}, \frac{6+b}{2}, \frac{9-a-9}{2} \right) = \left( \frac{a+20}{2}, \frac{6+b}{2}, -\frac{a}{2} \right)$.
Since $M$ lies on the line $L$,we have $\frac{\frac{a+20}{2}-3}{7} = \frac{\frac{6+b}{2}-2}{5} = \frac{-\frac{a}{2}-1}{-9} = k$.
Simplifying,we get $\frac{a+14}{14} = \frac{b+2}{10} = \frac{a+2}{18} = k$.
From $\frac{a+14}{14} = \frac{a+2}{18}$,we get $18(a+14) = 14(a+2) \Rightarrow 18a + 252 = 14a + 28 \Rightarrow 4a = -224 \Rightarrow a = -56$.
Substituting $a = -56$ into $\frac{a+14}{14} = k$,we get $k = \frac{-56+14}{14} = \frac{-42}{14} = -3$.
Now,$\frac{b+2}{10} = -3 \Rightarrow b+2 = -30 \Rightarrow b = -32$.
Thus,$|a+b| = |-56 - 32| = |-88| = 88$.

(B) Statement $I$ is false because while $CaCl_2 \cdot 6H_2O$ can be dehydrated to anhydrous $CaCl_2$ by heating,$MgCl_2 \cdot 8H_2O$ (or $6H_2O$) undergoes hydrolysis upon heating to form $MgO$ and $HCl$ due to the high polarizing power of $Mg^{2+}$.
Statement $II$ is false because $BeO$ is indeed amphoteric,but the oxides of other elements in Group $2$ $(MgO, CaO, SrO, BaO)$ are basic,not acidic.

(B) To determine aromaticity,we use $H$ückel's rule ($4n+2$ $\pi$ electrons,planar,cyclic,conjugated):
$A$. Heptafulvene: It is non-aromatic because it is not fully conjugated in a way that satisfies the $4n+2$ rule for the entire ring system.
$B$. Benzene: It is aromatic ($6$ $\pi$ electrons,$n=1$,planar,cyclic,fully conjugated).
$C$. Cyclopentadienyl anion: It is aromatic ($6$ $\pi$ electrons,$n=1$,planar,cyclic,fully conjugated).
$D$. Cyclopentadienyl cation: It is anti-aromatic ($4$ $\pi$ electrons,$n=1$,planar,cyclic,fully conjugated).
Therefore,compounds $B$ and $C$ are aromatic.

(D) Partially deactivated palladised charcoal $(H_2 / Pd / CaCO_3)$ is known as Lindlar catalyst.
It is used for the partial hydrogenation of alkynes to cis-alkenes.

(B) Statement $I$ is true: $H_{2}O_{2}$ acts as an oxidising agent in basic medium,e.g.,$2Fe^{2+} + H_{2}O_{2} \rightarrow 2Fe^{3+} + 2OH^{-}$. It also acts as a reducing agent in basic medium,e.g.,$2MnO_{4}^{-} + 3H_{2}O_{2} \rightarrow 2MnO_{2} + 3O_{2} + 2H_{2}O + 2OH^{-}$.
Statement $II$ is true: The basic principle of the hydrogen economy is the transportation and storage of energy in the form of liquid or gaseous dihydrogen. Energy is transmitted as dihydrogen rather than as electric power.

(B) In the presence of ozone $(O_{3})$,oxidising smog increases during the daytime.
Automobiles and factories produce the main components of photochemical smog (oxidising smog) due to the action of sunlight on unsaturated hydrocarbons and nitrogen oxides.
Ozone is a strong oxidising agent and reacts with unburnt hydrocarbons in polluted air to produce various chemicals,thereby increasing the intensity of oxidising smog.

(B) The keto-enol tautomerism in acetone $[CH_{3}COCH_{3}]$ results in a very small amount of enol form $(< 0.1 \%)$ because the enol form is less stable than the keto form.
In acetylacetone $[CH_{3}COCH_{2}COCH_{3}]$,the enol form is significantly more stable due to the formation of a six-membered ring stabilized by intramolecular hydrogen bonding.
Therefore,both the assertion $A$ and the reason $R$ are true,and $R$ provides the correct explanation for $A$.

(D) The hybridization of oxygen in the water molecule $(H_2O)$ is $sp^3$.
According to $VSEPR$ theory,the electron geometry of the water molecule is tetrahedral,and the bond angle should ideally be $109^{\circ} 28^{\prime}$.
However,the water molecule contains two lone pairs on the oxygen atom.
According to $VSEPR$ theory,the order of repulsion is: lone pair $-$ lone pair $>$ lone pair $-$ bond pair $>$ bond pair $-$ bond pair.
Due to the strong lone pair $-$ lone pair repulsion,the bond pairs are pushed closer together,which decreases the $H-O-H$ bond angle from the ideal tetrahedral angle to $104.5^{\circ}$.
Therefore,both Assertion $A$ and Reason $R$ are true,and $R$ is the correct explanation of $A$.

(D) In chromatography technique,the separation and purification of a compound depend on the differential distribution of the substance between the stationary phase and the mobile phase. This process relies on factors such as the solubility of the compound in the solvent,the mobility of the solvent system,and the length of the column or $TLC$ plate. However,the purification process is independent of the physical state (solid,liquid,or gas) of the pure compound itself.

(C) The relationship between standard Gibbs free energy change and the equilibrium constant is given by $\Delta G^{\circ} = -RT \ln K_{eq}$.
Given $\Delta G^{\circ} = -9.478 \ kJ \ mol^{-1} = -9478 \ J \ mol^{-1}$.
Substituting the values: $-9478 = -8.314 \times 495 \times \ln K_{eq}$.
$\ln K_{eq} = \frac{9478}{8.314 \times 495} \approx 2.303$.
Since $\ln 10 = 2.303$,we have $K_{eq} = 10$.
For the reaction $A_{(g)} \rightleftharpoons B_{(g)}$:
At $t = 0$,moles of $A = 22$,moles of $B = 0$.
At equilibrium,moles of $A = 22 - x$,moles of $B = x$.
$K_{eq} = \frac{[B]}{[A]} = \frac{x}{22 - x} = 10$.
$x = 220 - 10x \implies 11x = 220 \implies x = 20$.
The amount of $B$ at equilibrium is $20 \ mmol$.

(C) Mass of carbon in $420 \ g$ of $CO_2 = \frac{12}{44} \times 420 = 114.545 \ g$.
Percentage of carbon $= \frac{114.545}{750} \times 100 = 15.27 \% \approx 15.3 \%$.
Mass of hydrogen in $210 \ g$ of $H_2O = \frac{2}{18} \times 210 = 23.33 \ g$.
Percentage of hydrogen $= \frac{23.33}{750} \times 100 = 3.11 \%$.
Rounding to the nearest integer,the percentage of hydrogen is $3 \%$.

(D) The given redox reaction is: $2 MnO_{4}^{-} + b C_{2}O_{4}^{2-} + c H^{+} \rightarrow x Mn^{2+} + y CO_{2} + z H_{2}O$
Step $1$: Write the reduction half-reaction:
$MnO_{4}^{-} + 8 H^{+} + 5 e^{-} \rightarrow Mn^{2+} + 4 H_{2}O$
Step $2$: Write the oxidation half-reaction:
$C_{2}O_{4}^{2-} \rightarrow 2 CO_{2} + 2 e^{-}$
Step $3$: Equalize the number of electrons by multiplying the reduction half-reaction by $2$ and the oxidation half-reaction by $5$:
$2 MnO_{4}^{-} + 16 H^{+} + 10 e^{-} \rightarrow 2 Mn^{2+} + 8 H_{2}O$
$5 C_{2}O_{4}^{2-} \rightarrow 10 CO_{2} + 10 e^{-}$
Step $4$: Add the two half-reactions:
$2 MnO_{4}^{-} + 5 C_{2}O_{4}^{2-} + 16 H^{+} \rightarrow 2 Mn^{2+} + 10 CO_{2} + 8 H_{2}O$
Comparing this with the given equation,the coefficient $c$ corresponding to $H^{+}$ is $16$.

(B) Energy incident $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{248 \times 10^{-9}} \, J$
Converting to $eV$: $E = \frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{248 \times 10^{-9} \times 1.6 \times 10^{-19}} \approx 5.0 \, eV$
Using Einstein's photoelectric equation: $E = \phi + K.E.$
$5.0 \, eV = 3.0 \, eV + K.E. \implies K.E. = 2.0 \, eV$
$K.E. = 2.0 \times 1.6 \times 10^{-19} \, J = 3.2 \times 10^{-19} \, J$
De-Broglie wavelength $\lambda = \frac{h}{\sqrt{2mK.E.}}$
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 3.2 \times 10^{-19}}}$
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{58.24 \times 10^{-50}}} = \frac{6.63 \times 10^{-34}}{7.63 \times 10^{-25}} \approx 0.868 \times 10^{-9} \, m$
$\lambda \approx 8.68 \times 10^{-10} \, m = 8.68 \, \mathring{A}$
Rounding to the nearest integer,we get $9 \, \mathring{A}$.

(B) For $A_2X$: The dissociation is $A_2X \rightleftharpoons 2A^+ + X^{2-}$.
$K_{sp} = [2S_1]^2 [S_1] = 4S_1^3$.
$4.0 \times 10^{-12} = 4S_1^3$ $\Rightarrow S_1^3 = 10^{-12}$ $\Rightarrow S_1 = 10^{-4} \text{ mol L}^{-1}$.
For $MX$: The dissociation is $MX \rightleftharpoons M^{2+} + X^{2-}$.
$K_{sp} = S_2^2$.
$4.0 \times 10^{-12} = S_2^2 \Rightarrow S_2 = 2.0 \times 10^{-6} \text{ mol L}^{-1}$.
Ratio $\frac{S(A_2X)}{S(MX)} = \frac{10^{-4}}{2.0 \times 10^{-6}} = \frac{100}{2} = 50$.

(A) For a compound to be aromatic,it must follow $H$ückel's rule: it must be cyclic,planar,fully conjugated,and contain $(4n + 2) \pi$ electrons.
$A$. The cyclopentadienyl anion $(C_5H_5^-)$ has $6 \pi$ electrons (two from each of the two double bonds and two from the lone pair on the negatively charged carbon). It is cyclic,planar,and fully conjugated. Thus,it is aromatic.
$B$. Cyclopentadiene is not fully conjugated because it contains an $sp^3$ hybridized carbon atom.
$C$. The cyclopentadienyl dianion has $8 \pi$ electrons,which follows the $4n$ rule,making it anti-aromatic.
$D$. The cyclopentadienyl cation has $4 \pi$ electrons,which follows the $4n$ rule,making it anti-aromatic.
Therefore,the cyclopentadienyl anion is the aromatic compound.

(C) The total number of electron pairs around the central atom is the sum of the number of bond pairs and lone pairs.
Total electron pairs = $3$ (bond pairs) + $2$ (lone pairs) = $5$.
According to $VSEPR$ theory,a total of $5$ electron pairs corresponds to $sp^3d$ hybridization with a trigonal bipyramidal electron geometry.
When there are $3$ bond pairs and $2$ lone pairs,the lone pairs occupy the equatorial positions to minimize repulsion.
This results in a $T$-shaped molecular geometry.

(C) The electron gain enthalpy is the energy released when an electron is added to a neutral gaseous atom.
For halogens,the expected trend based on size is $F > Cl > Br > I$.
However,due to the very small size of the fluorine atom,the inter-electronic repulsions are high,which makes the incoming electron experience less attraction compared to chlorine.
Therefore,the absolute value of electron gain enthalpy for chlorine is higher than that of fluorine.
The correct order of the absolute value of electron gain enthalpy is $Cl > F > Br > I$.

(B) Lewis base is a chemical species that has the capability to donate an electron pair.
In $NF_{3}$,the central atom $N$ has one lone pair.
In $SF_{4}$,the central atom $S$ has one lone pair.
In $ClF_{3}$,the central atom $Cl$ has two lone pairs.
Therefore,$NF_{3}$,$SF_{4}$,and $ClF_{3}$ can act as Lewis bases.
In $PCl_{5}$,the central atom $P$ is in its maximum oxidation state $(+5)$ and has no lone pairs available for donation. Thus,it cannot act as a Lewis base.

(B) Reducing smog,also known as classical smog,occurs in cool,humid climates. It is a mixture of smoke,fog,and sulfur dioxide $(SO_2)$.

(B) The reaction of $1\text{-methyl-1-vinylcyclohexane}$ with $HBr$ proceeds via an electrophilic addition mechanism.
$1$. Protonation of the double bond occurs to form the most stable carbocation. Initially,a secondary carbocation is formed at the terminal carbon of the vinyl group.
$2$. This secondary carbocation undergoes a $1,2\text{-methyl shift}$ from the cyclohexane ring to the adjacent carbon to form a more stable tertiary carbocation at the ring position.
$3$. Finally,the bromide ion $(Br^-)$ attacks this tertiary carbocation to form the major product,which is $1\text{-bromo-1-isopropylcyclohexane}$.

(C) Heavy water $(D_2O)$ is used as a moderator in nuclear reactors to slow down fast neutrons.
It is obtained as a by-product in the fertilizer industry during the production of hydrogen.
It is used in the study of reaction mechanisms by using $D$ as a tracer.
The dielectric constant of ordinary water $(H_2O)$ is $78.39$,while that of heavy water $(D_2O)$ is $78.06$ at $298 \ K$.
Therefore,the statement that heavy water has a higher dielectric constant than water is $INCORRECT$.

(B) The conductivity of ions in an aqueous solution depends on their hydrated ionic size.
As the size of the gaseous ion decreases,the extent of hydration increases,which leads to a larger hydrated ionic radius.
Larger hydrated ions experience greater resistance (viscous drag) while moving through the solvent,resulting in lower ionic mobility and lower conductivity.
The order of gaseous ionic size is: $Cs^{+} > Rb^{+} > K^{+} > Na^{+}$.
Consequently,the order of hydrated ionic size is: $Cs^{+} < Rb^{+} < K^{+} < Na^{+}$.
Therefore,the order of ionic conductivity in water is: $Cs^{+} > Rb^{+} > K^{+} > Na^{+}$.

(C) $R_f$ (Retardation factor) is defined as the ratio of the distance travelled by the substance to the distance travelled by the solvent front from the reference line.
$R_f = \frac{\text{Distance travelled by the substance from reference line (cm)}}{\text{Distance travelled by the solvent from reference line (cm)}}$
Since $R_f$ is a ratio of two identical units (length),it is a dimensionless quantity. Thus,Statement-$I$ is false.
The $R_f$ value of a compound depends on the nature of the solvent and the stationary phase used. Therefore,it is not constant for all solvents. Thus,Statement-$II$ is false.

(C) The reaction of white phosphorus $(P_4)$ with alkali $(NaOH)$ is a disproportionation reaction:
$P_4 + 3 OH^{-} + 3 H_2O \rightarrow PH_3 + 3 H_2PO_2^{-}$
Here,the product '$A$' is the hypophosphite ion $(H_2PO_2^{-})$.
The reaction of $1 \ mol$ of $H_2PO_2^{-}$ with excess $AgNO_3$ is a redox reaction:
$H_2PO_2^{-} + 4 Ag^{+} + 2 H_2O \rightarrow 4 Ag + H_3PO_4 + 3 H^{+}$
From the stoichiometry of the balanced equation,$1 \ mol$ of $H_2PO_2^{-}$ reacts to produce $4 \ mol$ of $Ag$.

(C) The dissociation equilibrium of the weak acid $HA$ is given by: $HA \rightleftharpoons H^{+} + A^{-}$.
Initial concentrations: $[HA] = 0.01 \ M$,$[H^{+}] = 0.1 \ M$ (from $HCl$),$[A^{-}] = 0$.
At equilibrium,let the concentration of $A^{-}$ be $x \ M$. Then $[HA] = (0.01 - x) \approx 0.01 \ M$ and $[H^{+}] = (0.1 + x) \approx 0.1 \ M$ (since $\alpha << 1$).
The acid dissociation constant expression is $K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}$.
Substituting the values: $2.0 \times 10^{-6} = \frac{0.1 \times x}{0.01}$.
Solving for $x$: $x = \frac{2.0 \times 10^{-6} \times 0.01}{0.1} = 2.0 \times 10^{-7} \ M$.
The degree of dissociation $\alpha = \frac{x}{C} = \frac{2.0 \times 10^{-7}}{0.01} = 2.0 \times 10^{-5}$.
Thus,the value is $2 \times 10^{-5}$.

(B) For a given orbital,the magnetic quantum number $m_{\ell}$ ranges from $-\ell$ to $+\ell$.
Given $m_{\ell} = -3$,the azimuthal quantum number $\ell$ must be at least $3$.
Since $n = 4$ and $\ell = 3$,this corresponds to a $4f$ orbital.
The formula for the number of radial nodes is $n - \ell - 1$.
Substituting the values: $4 - 3 - 1 = 0$.

(B) The chemical reaction for the nitration of benzene is: $C_6H_6 + HNO_3 \xrightarrow{H_2SO_4} C_6H_5NO_2 + H_2O$
Molar mass of benzene $(C_6H_6)$ $= 6 \times 12 + 6 \times 1 = 78 \ g/mol$.
Molar mass of nitrobenzene $(C_6H_5NO_2)$ $= 6 \times 12 + 5 \times 1 + 14 + 2 \times 16 = 123 \ g/mol$.
According to the stoichiometry,$1 \ mole$ of benzene produces $1 \ mole$ of nitrobenzene.
So,$78 \ g$ of benzene should theoretically produce $123 \ g$ of nitrobenzene.
Therefore,$3.9 \ g$ of benzene should produce: $\frac{123}{78} \times 3.9 = 6.15 \ g$ of nitrobenzene.
The actual amount of nitrobenzene formed is $4.92 \ g$.
Percentage yield $= \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{4.92}{6.15} \times 100 = 80 \%$.

(C) The given reaction is: $3 CaO + 2 Al \rightarrow Al_{2}O_{3} + 3 Ca$
The standard enthalpy of reaction is calculated using the formula: $\Delta_{r}H^{\circ} = \Sigma \Delta_{f}H^{\circ}(\text{products}) - \Sigma \Delta_{f}H^{\circ}(\text{reactants})$
Since the standard enthalpy of formation for elements in their standard state ($Ca$ and $Al$) is $0 \ kJ \ mol^{-1}$,we have:
$\Delta_{r}H^{\circ} = [1 \times \Delta_{f}H^{\circ}(Al_{2}O_{3}) + 3 \times \Delta_{f}H^{\circ}(Ca)] - [3 \times \Delta_{f}H^{\circ}(CaO) + 2 \times \Delta_{f}H^{\circ}(Al)]$
$\Delta_{r}H^{\circ} = [1 \times (-1675) + 3 \times 0] - [3 \times (-635) + 2 \times 0]$
$\Delta_{r}H^{\circ} = -1675 + 1905 = +230 \ kJ \ mol^{-1}$

(C) The balanced redox reaction in acidic medium is: $6Fe^{2+} + Cr_{2}O_{7}^{2-} + 14H^+ \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_{2}O$.
According to the law of equivalence,the number of equivalents of $Fe^{2+}$ equals the number of equivalents of $Cr_{2}O_{7}^{2-}$.
$n_{eq} (Fe^{2+}) = n_{eq} (Cr_{2}O_{7}^{2-})$.
$(Molarity \times Volume \times n-factor)_{Fe^{2+}} = (Molarity \times Volume \times n-factor)_{Cr_{2}O_{7}^{2-}}$.
For $Fe^{2+} \rightarrow Fe^{3+}$,the $n-factor = 1$.
For $Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+}$,the $n-factor = 6$.
$\left(\frac{15 \times M_{Fe^{2+}}}{1000}\right) \times 1 = \left(\frac{20 \times 0.03}{1000}\right) \times 6$.
$15 \times M_{Fe^{2+}} = 20 \times 0.03 \times 6$.
$15 \times M_{Fe^{2+}} = 3.6$.
$M_{Fe^{2+}} = \frac{3.6}{15} = 0.24 \ M$.
$0.24 \ M = 24 \times 10^{-2} \ M$.
Thus,the value is $24$.

(D) First,calculate the number of moles of each gas:
$n_{CH_4} = \frac{6.4 \ g}{16 \ g \ mol^{-1}} = 0.4 \ mol$
$n_{CO_2} = \frac{8.8 \ g}{44 \ g \ mol^{-1}} = 0.2 \ mol$
Total moles,$n = 0.4 + 0.2 = 0.6 \ mol$
Convert volume to $m^3$:
$V = 10 \ L = 10 \times 10^{-3} \ m^3 = 0.01 \ m^3$
Convert temperature to Kelvin:
$T = 27 + 273 = 300 \ K$
Using the ideal gas equation $PV = nRT$:
$P = \frac{nRT}{V} = \frac{0.6 \ mol \times 8.314 \ J \ mol^{-1} K^{-1} \times 300 \ K}{0.01 \ m^3}$
$P = 149652 \ Pa = 149.652 \ kPa$
Rounding to the nearest integer,we get $150 \ kPa$.

(B) The ions $Na^{+}$,$Mg^{2+}$,and $Al^{3+}$ are isoelectronic species,all having $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic numbers are $Na$ $(Z=11)$,$Mg$ $(Z=12)$,and $Al$ $(Z=13)$.
Since the nuclear charge increases from $Na^{+}$ to $Al^{3+}$,the ionic radius decreases in the order $Na^{+} > Mg^{2+} > Al^{3+}$.
Given the radius of $Na^{+}$ is $1.02 \ \mathring{A}$,the values for $Mg^{2+}$ and $Al^{3+}$ must be smaller than $1.02 \ \mathring{A}$ and follow the decreasing trend.
Among the options,$0.72 \ \mathring{A}$ and $0.54 \ \mathring{A}$ fit this trend correctly.

(A) Methane $(CH_4)$ contributes to global warming and is involved in the formation of photochemical smog.
Paddy fields are a significant source of methane emission due to anaerobic decomposition.
Methane has a higher global warming potential compared to $CO_2$,making it a stronger greenhouse gas.
Methane is not a component of reducing smog (which typically consists of $SO_2$ and particulate matter).
Therefore,statements $A, B,$ and $C$ are correct.

(C) $Ca(OCl)_2$ is used as a bleaching agent $(iii)$.
$CaSO_4 \cdot \frac{1}{2} H_2O$ is known as Plaster of Paris $(iv)$.
$CaO$ is a major component in the manufacturing of cement $(ii)$.
$CaCO_3$ is used as an antacid $(i)$.
Therefore,the correct matching is $a-iii, b-iv, c-ii, d-i$.

(D) The molecular formula $C_3H_6O$ represents compounds such as propanal $(CH_3CH_2CHO)$ and propanone $(CH_3COCH_3)$.
Since these compounds have the same molecular formula but different functional groups (aldehyde and ketone),they exhibit functional group isomerism.

(C) Step $1$: Oxidation of the primary alcohol group with $CrO_3$ yields a carboxylic acid group. The phenolic $-OH$ group remains unaffected under these conditions.
Step $2$: Treatment with $SOCl_2/\Delta$ converts the carboxylic acid into an acid chloride $(-COCl)$.
Step $3$: Heating $(\Delta)$ induces a Friedel-Crafts acylation reaction. The acid chloride group undergoes intramolecular cyclization with the ortho-position of the phenol ring to form a cyclic ketone (indanone derivative).
The final product is a $5$-hydroxy-indanone derivative.

(D) In the actinide series,the ionic radius decreases from left to right due to actinoid contraction,which is similar to lanthanoid contraction.
Since $Bk$ (atomic number $97$) comes after $Np$ (atomic number $93$) in the actinide series,the size of the $Bk^{3+}$ ion is smaller than that of the $Np^{3+}$ ion.
Therefore,Assertion $A$ is true.
However,the decrease in size in the actinide series is due to actinoid contraction,not lanthanoid contraction. Thus,Reason $R$ is false.

(B) Vitamins are classified into two groups: water-soluble and fat-soluble.
Water-soluble vitamins (like Vitamin $B$ complex and Vitamin $C$) are excreted in urine and cannot be stored in the body for long,except for Vitamin $B_{12}$.
Fat-soluble vitamins (like Vitamin $A$,$D$,$E$,and $K$) are stored in the liver and adipose tissues.
Therefore,the pair of Vitamin $A$ and Vitamin $D$ is stored in our body for a longer duration.

(C) $DIBAL-H$ (Diisobutylaluminum hydride) is a selective reducing agent that reduces esters to aldehydes at low temperatures $(-78^{\circ}C)$.
In the given reaction,$DIBAL-H$ reduces the cyclic ester (lactone) to a lactol,which then opens to form a hydroxy-aldehyde.
$DIBAL-H$ does not reduce the carbon-carbon double bond $(C=C)$ present in the molecule.
Therefore,the reaction converts the lactone group into a hydroxy-aldehyde group while keeping the rest of the structure intact,which matches option $C$.

(C) Haber's process is used for $NH_{3}$ synthesis.
$(b)$ Ostwald's process is used for $HNO_{3}$ synthesis.
$(c)$ Contact process is used for $H_{2}SO_{4}$ synthesis.
$(d)$ Hall-Heroult process is used for Aluminium extraction.

(C) The reaction of aniline with $NaNO_2$ and $HCl$ at $273-278 \ K$ is a diazotization reaction,which produces benzene diazonium chloride as the intermediate $X$.
Benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$ is then hydrolyzed by heating with water $(H_2O/\Delta)$ to form phenol as the major product.
Therefore,$X$ is benzene diazonium chloride and $A$ is $H_2O/\Delta$.

(D) The $E^{\circ}$ value for $Ce^{4+} / Ce^{3+}$ is $+1.74 \, V$. This high positive value indicates that $Ce^{4+}$ has a strong tendency to gain an electron to form $Ce^{3+}$.
The most stable oxidation state of lanthanide series elements is $+3$. Therefore,$Ce^{3+}$ is more stable than $Ce^{4+}$.
Thus,Statement $I$ is correct and Statement $II$ is incorrect.

(B) Antihistamines act as both antacids and antiallergic agents.
They interfere with the natural action of histamine by competing with histamine for binding sites of receptors where histamine exerts its effect.
For example,drugs like cimetidine are designed to prevent the interaction of histamine with the receptors present in the stomach wall,thereby acting as an antacid.

(A) In the first reaction,$1$-methylcyclohexanol undergoes acid-catalyzed dehydration using $20\% \ H_3PO_4$ at $358 \ K$. This follows an $E_1$ mechanism,where the more stable,highly substituted alkene (Saytzeff product) is formed as the major product,which is $1-$methylcyclohexene.
In the second reaction,$1$-chlorocyclohexane reacts with potassium tert-butoxide $(CH_3)_3COK$,which is a bulky base. This reaction proceeds via an $E_2$ mechanism. Due to the steric hindrance of the bulky base,it abstracts a proton from the less hindered position,leading to the formation of the less substituted alkene (Hofmann product) as the major product,which is methylenecyclohexane.

(C) The Hoffmann Bromamide degradation reaction involves the conversion of an amide $(RCONH_2)$ to a primary amine $(RNH_2)$ using $Br_2$ and $NaOH$ (or $KOH$).
$A$. $C_6H_5CH_2CONH_2 + Br_2 + 4NaOH \rightarrow C_6H_5CH_2NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$. This is a Hoffmann Bromamide degradation.
$B$. $C_6H_5CN$ $\xrightarrow{KOH, H_2O} C_6H_5CONH_2$ $\xrightarrow{Br_2, NaOH} C_6H_5NH_2$. This involves Hoffmann Bromamide degradation in the second step.
$C$. $C_6H_5CH_2COCH_3$ $\xrightarrow{Br_2, NaOH} C_6H_5CH_2COOH$ $\xrightarrow{NH_3, \Delta} C_6H_5CH_2CONH_2$ $\xrightarrow{LiAlH_4} C_6H_5CH_2CH_2NH_2$. The first step is a haloform reaction,and the final step is a reduction. This reaction does not involve Hoffmann Bromamide degradation.
$D$. $C_6H_5COCl$ $\xrightarrow{NH_3} C_6H_5CONH_2$ $\xrightarrow{Br_2, NaOH} C_6H_5NH_2$. This involves Hoffmann Bromamide degradation in the second step.

(B) In the roasting process,metal sulphide $(MS)$ ores are converted into metal oxides,and sulphur is removed in the form of $SO_{2}$ gas.
$2 MS + 3 O_{2} \xrightarrow{\Delta} 2 MO + 2 SO_{2} \uparrow$

(A) Hypophosphorous acid $(H_{3}PO_{2})$: $3(+1) + x + 2(-2) = 0 \Rightarrow x = +1$.
$(b)$ Orthophosphoric acid $(H_{3}PO_{4})$: $3(+1) + x + 4(-2) = 0 \Rightarrow x = +5$.
$(c)$ Hypophosphoric acid $(H_{4}P_{2}O_{6})$: $4(+1) + 2x + 6(-2) = 0$ $\Rightarrow 2x = 8$ $\Rightarrow x = +4$.
$(d)$ Orthophosphorous acid $(H_{3}PO_{3})$: $3(+1) + x + 3(-2) = 0 \Rightarrow x = +3$.
Thus,the correct match is $(a)-(v), (b)-(i), (c)-(ii), (d)-(iii)$.

(D) In group $15$ $(N, P, As, Sb, Bi)$,the metallic character increases down the group.
$Bi$ is the most metallic element in this group.
The reducing power of hydrides $(MH_3)$ increases down the group from $NH_3$ to $BiH_3$ because the $M-H$ bond dissociation enthalpy decreases as the size of the central atom increases.
Therefore,$BiH_3$ is the strongest reducing agent among the group $15$ hydrides,and the element is $Bi$.

(D) For the dissociation reaction: $AB_2 \rightleftharpoons A^{2+} + 2B^-$,the number of ions produced per formula unit is $n = 3$.
Given the degree of dissociation $\alpha = 10\% = 0.1$.
The van't Hoff factor $i$ is calculated as $i = 1 + (n-1)\alpha = 1 + (3-1)0.1 = 1 + 0.2 = 1.2$.
The elevation in boiling point is given by $\Delta T_b = i \cdot K_b \cdot m$.
Substituting the values: $\Delta T_b = 1.2 \times 0.5 \text{ K kg mol}^{-1} \times 10.0 \text{ mol kg}^{-1} = 6 \text{ K}$ (or $6^\circ C$).
The boiling point of the solution is $T_b = T_b^\circ + \Delta T_b = 100^\circ C + 6^\circ C = 106^\circ C$.

(C) The complex $trans-CoCl_{3} \cdot 4NH_{3}$ can be written as $trans-[Co(NH_{3})_{4}Cl_{2}]Cl$.
In this complex,there are $4$ neutral $NH_{3}$ ligands in the coordination sphere.
Ethylene diamine $(en)$ is a bidentate ligand,meaning one molecule of $en$ occupies two coordination sites.
To replace $4$ monodentate $NH_{3}$ ligands,we require $2$ molecules of the bidentate ligand $en$ ($2 \times 2 = 4$ coordination sites).
Therefore,the number of equivalents of ethylene diamine required is $2$.

(C) Molality $(m)$ is defined as the number of moles of solute per $1000 \ g$ of solvent.
Given $6.50$ molal $KOH$ solution means $6.50 \ moles$ of $KOH$ are present in $1000 \ g$ of water.
Molar mass of $KOH = 39.0 + 16.0 + 1.0 = 56.0 \ g \ mol^{-1}$.
Mass of solute $(KOH)$ $= 6.50 \ mol \times 56.0 \ g \ mol^{-1} = 364.0 \ g$.
Total mass of solution $= \text{Mass of solvent} + \text{Mass of solute} = 1000 \ g + 364.0 \ g = 1364.0 \ g$.
Density of solution $= 1.89 \ g \ cm^{-3}$.
Volume of solution $= \frac{\text{Mass}}{\text{Density}} = \frac{1364.0 \ g}{1.89 \ g \ cm^{-3}} \approx 721.69 \ cm^3 = 0.72169 \ L$.
Molarity $(M)$ $= \frac{\text{moles of solute}}{\text{Volume of solution in } L} = \frac{6.50 \ mol}{0.72169 \ L} \approx 9.006 \ mol \ L^{-1}$.
Rounding to the nearest integer,the molarity is $9 \ mol \ dm^{-3}$.

(A) For $bcc$ lattice,the relation between edge length $a_1$ and atomic radius $r$ is $\sqrt{3} a_1 = 4 r$,so $r = \frac{\sqrt{3}}{4} a_1$.
Given $a_1 = 27 \mathring{A}$,$r = \frac{\sqrt{3}}{4} \times 27$.
For $fcc$ lattice,the relation between edge length $a_2$ and atomic radius $r$ is $\sqrt{2} a_2 = 4 r$,so $a_2 = \frac{4 r}{\sqrt{2}} = 2 \sqrt{2} r$.
Substituting the value of $r$: $a_2 = 2 \sqrt{2} \times \frac{\sqrt{3}}{4} \times 27 = \frac{\sqrt{6}}{2} \times 27$.
Using $\sqrt{6} = \sqrt{2} \times \sqrt{3} = 1.41 \times 1.73 = 2.4393$.
$a_2 = \frac{2.4393}{2} \times 27 = 1.21965 \times 27 = 32.93$.
Rounding off to the nearest integer,we get $33 \mathring{A}$.

(A) Given:
$T_1 = 300 \ K, K_1 = 1.0 \times 10^{-3} \ s^{-1}$
$T_2 = 200 \ K, K_2 = ?$
$E_a = 11.488 \ kJ \ mol^{-1} = 11488 \ J \ mol^{-1}$
$R = 8.314 \ J \ mol^{-1} K^{-1}$
Using the Arrhenius equation:
$\ln \left(\frac{K_1}{K_2}\right) = \frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$
$\ln \left(\frac{1.0 \times 10^{-3}}{K_2}\right) = \frac{11488}{8.314} \left(\frac{1}{200} - \frac{1}{300}\right)$
$\ln \left(\frac{1.0 \times 10^{-3}}{K_2}\right) = \frac{11488}{8.314} \left(\frac{300 - 200}{60000}\right) = \frac{11488}{8.314} \times \frac{100}{60000} = \frac{11488}{8.314 \times 600} \approx \frac{11488}{4988.4} \approx 2.303$
$\frac{1.0 \times 10^{-3}}{K_2} = e^{2.303} \approx 10$
$K_2 = \frac{1.0 \times 10^{-3}}{10} = 1.0 \times 10^{-4} = 10 \times 10^{-5} \ s^{-1}$
Thus,the value is $10$.

(C) Some drugs do not bind to the enzyme's active site. Instead,they bind to a different site on the enzyme called the allosteric site.
This binding of an inhibitor at the allosteric site changes the shape of the active site in such a way that the substrate cannot recognize or bind to it.
Such an inhibitor is known as a non-competitive inhibitor.
Therefore,the statement that an allosteric inhibitor competes with the enzyme's active site is incorrect,as it binds to a different site.

(B) The reaction of a ketone with ethylene glycol in the presence of an acid catalyst $(H^+)$ is a standard method for the protection of the carbonyl group as a cyclic acetal.
In the given molecule,ethyl acetoacetate,there are two carbonyl groups: a ketone and an ester.
Ketones are more reactive towards nucleophilic addition than esters.
Therefore,the ethylene glycol selectively reacts with the ketone group to form a cyclic acetal,while the ester group remains unaffected.
The product $A$ is the cyclic acetal of ethyl acetoacetate.

(A) The thermal decomposition of potassium permanganate $(KMnO_4)$ at $573 \ K$ is given by the reaction:
$2 KMnO_4 \xrightarrow{573 \ K} K_2MnO_4 + MnO_2 + O_2$
Thus,Statement-$I$ is true.
Regarding Statement-$II$:
- The permanganate ion $(MnO_4^-)$ has $Mn$ in the $+7$ oxidation state ($d^0$ configuration),making it diamagnetic.
- The manganate ion $(MnO_4^{2-})$ has $Mn$ in the $+6$ oxidation state ($d^1$ configuration),making it paramagnetic.
Both ions are tetrahedral,but they are not both paramagnetic.
Thus,Statement-$II$ is false.

(A) Tyrosine is an $\alpha$-amino acid with the general formula $R-CH(NH_2)-COOH$.
In tyrosine,the side chain $R$ is a $p$-hydroxybenzyl group,which is $-CH_2-C_6H_4-OH$.
Combining these,the structure is $HO-C_6H_4-CH_2-CH(NH_2)-COOH$.
Comparing this with the given options,option $A$ represents the correct structure.

(D) The reaction shown is the conversion of chlorobenzene to sodium phenoxide,which is the first step of the Dow process.
This nucleophilic substitution reaction of chlorobenzene with aqueous $NaOH$ requires harsh conditions due to the partial double bond character of the $C-Cl$ bond.
The required conditions are a temperature of $623 \ K$ and a pressure of $300 \ atm$.

(B) Step $1$: Hoffmann bromamide degradation of benzamide $(C_6H_5CONH_2)$ involves the reaction with $Br_2$ and $NaOH$ to form aniline $(C_6H_5NH_2)$ as product $A$.
Step $2$: Aniline $(C_6H_5NH_2)$ is a primary amine. When heated with $CHCl_3$ and $NaOH$ (or $KOH$),it undergoes the carbylamine reaction to form phenyl isocyanide $(C_6H_5NC)$ as product $B$.
Thus,$A$ is aniline and $B$ is phenyl isocyanide.

(D) Mesityl oxide is an $\alpha, \beta-$unsaturated ketone formed by the aldol condensation of acetone.
Its chemical structure is $(CH_3)_2C=CHCOCH_3$.
According to $IUPAC$ nomenclature,the longest carbon chain containing the ketone group has $5$ carbons.
The double bond is at position $3$ and the ketone group is at position $2$.
There is a methyl group at position $4$.
Therefore,the $IUPAC$ name is $4-$Methylpent$-3-$en$-2-$one.

(D) The process of cleavage of the $C-X$ bond by an ammonia molecule is known as ammonolysis.
In this reaction,the nucleophilic ammonia molecule attacks the alkyl halide,resulting in the substitution of the halogen atom by the amino group.
The reaction $C_{6}H_{5}CH_{2}Cl + NH_{3} \longrightarrow C_{6}H_{5}CH_{2}NH_{2} + HCl$ represents the nucleophilic substitution of a benzyl chloride by ammonia,which is a classic example of ammonolysis.

(A) colloidal system where a gas is dispersed in a solid medium is known as a $solid \ sol$. Examples include pumice stone and rubber foam.

(A) The electronic configuration of a divalent metal ion with atomic number $25$ $(Mn^{2 })$ is $[Ar] 3d^5 4s^0$.
The number of unpaired electrons $(n)$ is $5$.
The spin-only magnetic moment $(\mu)$ is calculated using the formula:
$\mu = \sqrt{n(n 2)} \ BM$
Substituting $n = 5$:
$\mu = \sqrt{5(5 2)} = \sqrt{35} \ BM \approx 5.92 \ BM$.

(A) In an Ellingham diagram,the point of intersection of two lines indicates that the $ \Delta G $ values for the two reactions are equal,meaning $ \Delta G = 0 $ for the overall coupled reaction.
$A$ sudden change in the slope of a line in the Ellingham diagram indicates a phase change (melting or boiling) of the metal or the metal oxide.

(A) $100$ molal aqueous solution means there are $100$ moles of solute in $1 \ kg$ $(1000 \ g)$ of water.
The number of moles of water $(n_{\text{solvent}})$ $= \frac{1000 \ g}{18 \ g/mol} = 55.56 \ mol$.
The mole fraction of solute $(x_{\text{solute}})$ $= \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$.
$x_{\text{solute}} = \frac{100}{100 + 55.56} = \frac{100}{155.56} \approx 0.6428$.
$0.6428 = 64.28 \times 10^{-2}$.
Rounding off to the nearest integer,we get $64 \times 10^{-2}$.

(A) For a $1^{st}$ order reaction,the rate constant $K$ is given by:
$K = \frac{2.303}{t} \log_{10} \left( \frac{[A]_0}{[A]_t} \right)$
Given $t = 570 \ s$ and $[A]_t = 32 \%$ of $[A]_0$,so $\frac{[A]_0}{[A]_t} = \frac{100}{32} = 3.125$.
$K = \frac{2.303}{570} \log_{10} (3.125)$
Since $\log_{10} (3.125) = \log_{10} (100/32) = \log_{10} 100 - \log_{10} 32 = 2 - 5 \log_{10} 2 = 2 - 5(0.301) = 2 - 1.505 = 0.495$.
$K = \frac{2.303 \times 0.495}{570} \approx \frac{1.14}{570} = 0.002 \ s^{-1} = 2 \times 10^{-3} \ s^{-1}$.
Rounding to the nearest integer,the value is $2$.

(D) According to Henry's Law,$P = K_H \cdot x$,where $x$ is the mole fraction of $O_2$ in water.
$20 \ kPa = (8.0 \times 10^{4} \ kPa) \cdot x$
$x = \frac{20}{8.0 \times 10^{4}} = 2.5 \times 10^{-4}$.
Since $x = \frac{n_{O_2}}{n_{O_2} + n_{H_2O}} \approx \frac{n_{O_2}}{n_{H_2O}}$ (as $n_{O_2} \ll n_{H_2O}$).
For $1 \ dm^{3}$ of water,mass $= 1000 \ g$,so $n_{H_2O} = \frac{1000}{18} = 55.55 \ mol$.
$n_{O_2} = x \cdot n_{H_2O} = (2.5 \times 10^{-4}) \times 55.55 = 1.38875 \times 10^{-2} \ mol$.
This is the molar solubility in $mol \ dm^{-3}$,which is $1388.75 \times 10^{-5} \ mol \ dm^{-3}$.
Rounding to the nearest integer,we get $1389$.

(A) The reaction of aniline with $NaNO_2$ and $HCl$ at $273-278 \ K$ is a diazotization reaction,which produces benzenediazonium chloride as the major product $(X)$.
Benzenediazonium chloride then undergoes an electrophilic aromatic substitution reaction (coupling reaction) with $N,N$-dimethylaniline. Since the $-N(CH_3)_2$ group is a strong activating group and is ortho/para-directing,the coupling occurs primarily at the para-position due to steric hindrance at the ortho-position.
Therefore,the major product $(Y)$ is $p$-(dimethylamino)azobenzene.

(C) $1$. The molecular formula of the reactant is $C_{8}H_{8}O$. Given the reaction with $C_{2}H_{5}MgBr$ (Grignard reagent) and subsequent hydrolysis,it is likely an aldehyde or ketone. Acetophenone $(C_{6}H_{5}COCH_{3})$ fits the formula $C_{8}H_{8}O$.
$2$. Reaction: $C_{6}H_{5}COCH_{3} + C_{2}H_{5}MgBr \rightarrow C_{6}H_{5}C(OH)(CH_{3})(C_{2}H_{5})$. This product $A$ is $2$-phenylbutan-$2$-ol.
$3$. Compound $A$ is a tertiary alcohol. Tertiary alcohols react instantly with Lucas reagent $(conc. \ HCl + ZnCl_{2})$ to form alkyl chlorides.
$4$. The reaction of $2$-phenylbutan-$2$-ol with $HCl$ replaces the $-OH$ group with $-Cl$ to form $2$-chloro-$2$-phenylbutane $(C_{10}H_{13}Cl)$.
$5$. Comparing this with the given options,the structure corresponding to $2$-chloro-$2$-phenylbutane is option $C$.

(D) The reagent consisting of $1$-naphthylamine and sulphanilic acid in acetic acid is known as the Griess reagent.
It is specifically used for the detection of nitrite ions $(NO_{2}^{-})$.
The test involves the diazotization of sulphanilic acid by nitrous acid $(HNO_{2})$,which is formed from $NO_{2}^{-}$ in the presence of acetic acid.
The resulting diazonium salt then couples with $1$-naphthylamine to form a red-colored azo dye,indicating the presence of $NO_{2}^{-}$.

(D) Sucrose is a non-reducing sugar because its glycosidic linkage involves the anomeric carbons of both glucose and fructose units.
Upon hydrolysis,it yields two reducing monosaccharides: glucose and fructose.
The reaction is:
$Sucrose + H_2O \rightarrow Glucose + Fructose$
Since both products contain a free hemiacetal or hemiketal group,they are reducing sugars.

(A) . Antacid: $ii$. Cimetidine
$b$. Artificial sweetener: $iv$. Alitame
$c$. Antifertility: $i$. Novestrol
$d$. Tranquilizers: $iii$. Valium
Therefore,the correct matching is $a-ii, b-iv, c-i, d-iii$.

(C) The reaction of a nitrile (cyclohexanecarbonitrile) with water in the presence of an acid $(H^+)$ leads to partial hydrolysis,which produces an amide as the major product $A$.
The reaction is as follows:
$R-CN + H_2O \xrightarrow{H^+} R-CONH_2$ (Partial Hydrolysis)
In this case,the nitrile group attached to the cyclohexane ring is converted to an amide group $(-CONH_2)$.
Therefore,product $A$ is cyclohexanecarboxamide.

(B) Chlorophyll is a coordination compound of magnesium $(a-iv)$.
Vitamin $B_{12}$ (cyanocobalamin) is a coordination compound of cobalt $(b-iii)$.
Cisplatin is used as an anti-cancer drug and is a coordination compound of platinum $(c-ii)$.
Grubbs catalyst is a complex of ruthenium $(d-i)$.
Therefore,the correct matching is $a-iv, b-iii, c-ii, d-i$.

(B) Alcoholic potassium hydroxide is used for $\beta$-elimination reactions.
$(b)$ $Pd / BaSO_4$ is known as Lindlar's catalyst.
$(c)$ $BHC$ (Benzene hexachloride) is obtained by the addition reaction of chlorine to benzene.
$(d)$ Polyacetylene is a conducting polymer used as electrodes in batteries.
Therefore,the correct match is $a-iii, b-iv, c-ii, d-i$.

(A) $1$. For the square planar complex $trans-[NiBr_{2}(PPh_{3})_{2}]$,the two identical ligands ($Br$ or $PPh_{3}$) must be positioned opposite to each other (at $180^{\circ}$ angle). In the structure shown in option $A$,the $Br$ atoms are trans to each other,and the $PPh_{3}$ groups are trans to each other.
$2$. For the octahedral complex $meridional-[Co(NH_{3})_{3}(NO_{2})_{3}]$,the three identical ligands must lie on the same plane (meridian). In the structure shown in option $A$,the three $NH_{3}$ ligands are in a meridional arrangement,and the three $NO_{2}$ ligands are also in a meridional arrangement.
$3$. Therefore,the structure in option $A$ correctly represents both the trans-isomer of the nickel complex and the meridional-isomer of the cobalt complex.

(C) The reaction involves the oxidation of an alkyl group attached to a benzene ring using alkaline $KMnO_4$ followed by acidification $(H^+)$.
Alkaline $KMnO_4$ is a strong oxidizing agent that oxidizes the alkyl side chain (specifically,a methyl group attached to the benzene ring) to a carboxylic acid group $(-COOH)$.
In the given reactant,$1$-methoxy-$4$-methylbenzene,the methyl group $(-CH_3)$ at the para position is oxidized to a carboxylic acid group $(-COOH)$,while the methoxy group $(-OCH_3)$ remains unaffected.
Therefore,the product $X$ is $4$-methoxybenzoic acid.

(A) The correct matches are as follows:
$a$. Deacon's process uses $CuCl_2$ as a catalyst.
$b$. Contact process for the manufacture of $H_2SO_4$ uses $V_2O_5$ as a catalyst.
$c$. Cracking of hydrocarbons uses $ZSM-5$ as a catalyst.
$d$. Hydrogenation of vegetable oils uses $Ni$ as a catalyst.
Thus,the correct sequence is $a-ii, b-iv, c-i, d-iii$.

(B) In an $hcp$ structure,the number of atoms of element $A$ is $N = 6$.
The number of tetrahedral voids is $2N = 2 \times 6 = 12$.
Element $M$ occupies $2/3$ of the tetrahedral voids,so the number of $M$ atoms is $\frac{2}{3} \times 12 = 8$.
The ratio of $M:A$ is $8:6$,which simplifies to $4:3$.
Therefore,the formula of the compound is $M_4 A_3$.

(A) During the extraction of aluminium by the Hall-$H$éroult process,pure $Al_2O_3$ has a very high melting point.
To reduce the melting point of the reaction mixture and to increase its electrical conductivity,$Na_3AlF_6$ (Cryolite) is added.

(C) The rate law is given by $r = k [NO]^{m} [Cl_{2}]^{n}$.
From the data:
$0.18 = k (0.1)^{m} (0.1)^{n} \quad \dots (1)$
$0.35 = k (0.1)^{m} (0.2)^{n} \quad \dots (2)$
$1.40 = k (0.2)^{m} (0.2)^{n} \quad \dots (3)$
Dividing equation $(2)$ by $(1)$:
$\frac{0.35}{0.18} \approx 2 = (\frac{0.2}{0.1})^{n} = 2^{n} \implies n = 1$.
Dividing equation $(3)$ by $(2)$:
$\frac{1.40}{0.35} = 4 = (\frac{0.2}{0.1})^{m} = 2^{m} \implies m = 2$.
The overall order of the reaction is $m + n = 2 + 1 = 3$.

(B) The reaction of Benzylamine $(Ph-CH_2-NH_2)$ with bromomethane $(CH_3Br)$ is an exhaustive methylation reaction.
Benzylamine reacts with $3$ moles of bromomethane to form Benzyl trimethyl ammonium bromide $(Ph-CH_2-N(CH_3)_3^+Br^-)$.
The molar mass of Benzyl trimethyl ammonium bromide $(C_{10}H_{16}NBr)$ is calculated as:
$M = (10 \times 12) + (16 \times 1) + (14 \times 1) + (80 \times 1) = 120 + 16 + 14 + 80 = 230 \ g/mol$.
Given mass of product = $23 \ g$.
Number of moles of product = $\frac{23 \ g}{230 \ g/mol} = 0.1 \ mol$.
Since $1 \ mol$ of Benzylamine reacts with $3 \ mol$ of bromomethane to produce $1 \ mol$ of Benzyl trimethyl ammonium bromide,$0.1 \ mol$ of Benzylamine will react with $0.1 \times 3 = 0.3 \ mol$ of bromomethane.
Given that the number of moles of bromomethane consumed is $n \times 10^{-1} = 0.3$.
Therefore,$n = 3$.

(A) In the complex $K_3[Cr(C_2O_4)_3]$,the oxalate ion $(C_2O_4^{2-})$ has a charge of $-2$.
Let the oxidation state of $Cr$ be $x$.
$3(+1) + x + 3(-2) = 0$
$3 + x - 6 = 0$
$x = +3$.
The atomic number of $Cr$ is $24$,and its electronic configuration is $[Ar] 3d^5 4s^1$.
For $Cr^{3+}$,the configuration is $[Ar] 3d^3$.
Thus,there are $3$ unpaired electrons in the $3d$ orbitals.

(A) The freezing point depression is given by $\Delta T_f = T_f^{\circ} - T_f = 0^{\circ} C - (-3.885^{\circ} C) = 3.885 \ K$.
For a weak acid $HA$ undergoing dissociation,$HA \rightleftharpoons H^+ + A^-$,the van't Hoff factor $i = 1 + \alpha$.
Using the formula $\Delta T_f = i \times K_f \times m$:
$3.885 = (1 + \alpha) \times 1.85 \times 2$.
$3.885 = (1 + \alpha) \times 3.7$.
$1 + \alpha = \frac{3.885}{3.7} = 1.05$.
$\alpha = 1.05 - 1 = 0.05$.
Expressing $\alpha$ in terms of $10^{-3}$:
$\alpha = 0.05 = 50 \times 10^{-3}$.
Thus,the value is $50$.

(D) The reaction is $2 Fe^{3+} + 2 I^{-} \rightarrow 2 Fe^{2+} + I_{2}$.
First,calculate $E^{\circ}_{Fe^{3+}/Fe^{2+}}$ using the given Latimer diagram data:
$n_1 E_1^{\circ} + n_2 E_2^{\circ} = n_3 E_3^{\circ}$
$1 \times E^{\circ}_{Fe^{3+}/Fe^{2+}} + 2 \times E^{\circ}_{Fe^{2+}/Fe} = 3 \times E^{\circ}_{Fe^{3+}/Fe}$
$E^{\circ}_{Fe^{3+}/Fe^{2+}} + 2(-0.440) = 3(-0.036)$
$E^{\circ}_{Fe^{3+}/Fe^{2+}} = -0.108 + 0.880 = 0.772 \ V$.
Now,calculate the cell potential $E^{\circ}_{cell}$:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Fe^{3+}/Fe^{2+}} - E^{\circ}_{I_{2}/2I^{-}}$
$E^{\circ}_{cell} = 0.772 - 0.539 = 0.233 \ V$.
The standard Gibbs free energy change is $\Delta_{r} G^{\circ} = -n F E^{\circ}_{cell}$.
Here,$n = 2$ (moles of electrons transferred).
$\Delta_{r} G^{\circ} = -2 \times 96500 \times 0.233 = -44969 \ J \ mol^{-1} = -44.969 \ kJ \ mol^{-1}$.
The magnitude is approximately $45 \ kJ \ mol^{-1}$.