JEE Main 2019 Chemistry Question Paper with Answer and Solution

(B) In acyclic compounds,the bond angles are generally free to rotate,and the carbon atoms can adopt geometries that minimize angle strain. Therefore,angle strain is not a factor that governs the stability of different conformations in acyclic compounds. Conversely,torsional strain,steric interactions,and electrostatic forces are significant factors that determine the relative stability of conformations in acyclic systems.

(A) In option $(A)$,the equilibrium constant is large,which suggests the reaction goes to completion,but this does not imply that no catalyst is required; catalysts are often used to increase the rate of reaction.
In option $(B)$,$\Delta n_g = 2 - (2 + 1) = -1$. Since $\Delta n_g < 0$,an increase in pressure will shift the reaction in the forward direction according to Le Chatelier's principle.
In option $(C)$,as the reaction is exothermic $(\Delta H < 0)$,an increase in temperature will decrease the equilibrium constant.
In option $(D)$,the equilibrium constant is a function of temperature only; therefore,the addition of an inert gas at constant volume does not affect the equilibrium constant.
Hence,option $(A)$ is the incorrect statement.

(D) The most probable velocity is given by the formula $V_{mp} = \sqrt{\frac{2RT}{M}}$,which implies $V_{mp} \propto \sqrt{\frac{T}{M}}$.
Calculating the relative values for the given gases:
$1$. For $N_2$ $(300 \ K)$: $\sqrt{\frac{300}{28}} \approx \sqrt{10.71} \approx 3.27$
$2$. For $O_2$ $(400 \ K)$: $\sqrt{\frac{400}{32}} = \sqrt{12.5} \approx 3.53$
$3$. For $H_2$ $(300 \ K)$: $\sqrt{\frac{300}{2}} = \sqrt{150} \approx 12.25$
Comparing these values,we get $V_{mp}(N_2, 300 \ K) < V_{mp}(O_2, 400 \ K) < V_{mp}(H_2, 300 \ K)$.
Thus,point $I$ corresponds to $N_2$ $(300 \ K)$,point $II$ corresponds to $O_2$ $(400 \ K)$,and point $III$ corresponds to $H_2$ $(300 \ K)$.

(B) $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$.
The formula for the $pH$ of such a salt solution is:
$pH = \frac{1}{2} [pK_w - pK_b - \log C]$
Given:
$C = 0.02 \ M = 2 \times 10^{-2} \ M$
$K_b = 10^{-5} \implies pK_b = 5$
$K_w = 10^{-14} \implies pK_w = 14$
Substituting the values:
$pH = \frac{1}{2} [14 - 5 - \log(2 \times 10^{-2})]$
$pH = \frac{1}{2} [9 - (\log 2 + \log 10^{-2})]$
$pH = \frac{1}{2} [9 - (0.301 - 2)]$
$pH = \frac{1}{2} [9 - (-1.699)]$
$pH = \frac{1}{2} [10.699]$
$pH = 5.3495 \approx 5.35$

(C) To find the amount of $O_2$ consumed per gram of reactant,we calculate the mass of $O_2$ required for $1 \ g$ of each reactant:
$(A)$ $C_3H_8$: Molar mass $= 3(12) + 8(1) = 44 \ g/mol$. $1 \ mol$ $C_3H_8$ reacts with $5 \ mol$ $O_2$ $(160 \ g)$. $O_2$ per gram $= 160/44 \approx 3.64 \ g$.
$(B)$ $P_4$: Molar mass $= 4(31) = 124 \ g/mol$. $1 \ mol$ $P_4$ reacts with $5 \ mol$ $O_2$ $(160 \ g)$. $O_2$ per gram $= 160/124 \approx 1.29 \ g$.
$(C)$ $Fe$: Molar mass $= 56 \ g/mol$. $4 \ mol$ $Fe$ $(224 \ g)$ react with $3 \ mol$ $O_2$ $(96 \ g)$. $O_2$ per gram $= 96/224 \approx 0.43 \ g$.
$(D)$ $Mg$: Molar mass $= 24 \ g/mol$. $2 \ mol$ $Mg$ $(48 \ g)$ react with $1 \ mol$ $O_2$ $(32 \ g)$. $O_2$ per gram $= 32/48 \approx 0.67 \ g$.
Comparing the values,the minimum amount of $O_2$ is consumed in reaction $(C)$.

(C) The slope of the tangent to the curve $y = \frac{x}{x^2 - 3}$ at $(\alpha, \beta)$ is given by $\frac{dy}{dx} = \frac{(x^2 - 3)(1) - x(2x)}{(x^2 - 3)^2} = \frac{-x^2 - 3}{(x^2 - 3)^2}$.
At point $(\alpha, \beta)$,the slope is $m = \frac{-\alpha^2 - 3}{(\alpha^2 - 3)^2}$.
The given line is $2x + 6y - 11 = 0$,which can be written as $y = -\frac{1}{3}x + \frac{11}{6}$. The slope of this line is $-\frac{1}{3}$.
Since the tangent is parallel to the line,$\frac{-\alpha^2 - 3}{(\alpha^2 - 3)^2} = -\frac{1}{3}$.
$3(\alpha^2 + 3) = (\alpha^2 - 3)^2 \Rightarrow 3\alpha^2 + 9 = \alpha^4 - 6\alpha^2 + 9 \Rightarrow \alpha^4 - 9\alpha^2 = 0$.
Since $\alpha \neq 0$,we have $\alpha^2 = 9$,so $\alpha = \pm 3$.
If $\alpha = 3$,$\beta = \frac{3}{3^2 - 3} = \frac{3}{6} = \frac{1}{2}$.
If $\alpha = -3$,$\beta = \frac{-3}{(-3)^2 - 3} = \frac{-3}{6} = -\frac{1}{2}$.
For $(\alpha, \beta) = (3, 1/2)$,$|6\alpha + 2\beta| = |6(3) + 2(1/2)| = |18 + 1| = 19$.
For $(\alpha, \beta) = (-3, -1/2)$,$|6\alpha + 2\beta| = |6(-3) + 2(-1/2)| = |-18 - 1| = 19$.

(A) In $(SiH_3)_3N$,the lone pair of electrons on the nitrogen atom is involved in $p\pi-d\pi$ back-bonding with the vacant $d$-orbitals of the silicon atoms.
This results in the molecule adopting a planar geometry ($sp^2$ hybridization).
Because the lone pair is delocalized into the $d$-orbitals of silicon,it is not readily available for donation,making $(SiH_3)_3N$ much less basic than $(CH_3)_3N$,where the lone pair is localized on the nitrogen atom.

(A) Photochemical smog is formed by the action of sunlight on nitrogen oxides ($NO$,$NO_2$) and hydrocarbons released from automobiles and factories.
These pollutants react in the presence of sunlight to produce secondary pollutants like ozone $(O_3)$,formaldehyde,acrolein,and peroxyacetyl nitrate $(PAN)$.
Therefore,the primary species responsible for its formation are $NO$,$NO_2$,$O_3$,and hydrocarbons.

(C) Step $1$: The reaction of $3$-bromocyclohexene with $KO^tBu$ (a strong,bulky base) undergoes an $E2$ elimination reaction to form $1,3$-cyclohexadiene.
Step $2$: The subsequent reaction with $O_3/Me_2S$ is a reductive ozonolysis of $1,3$-cyclohexadiene.
Step $3$: Reductive ozonolysis of $1,3$-cyclohexadiene cleaves both double bonds to yield two molecules of glyoxal $(OHC-CHO)$ and one molecule of succinaldehyde $(OHC-CH_2-CH_2-CHO)$.
However,based on the provided options,the product is a mixture of $OHC-CH_2-CH_2-CHO$ and $OHC-CHO$.

(B) In the addition of chlorine water $(Cl_2/H_2O)$ to an alkene (halohydrin formation),the reaction proceeds via a cyclic chloronium ion intermediate.
The nucleophile $(H_2O)$ then attacks the more substituted carbon atom because it carries a greater partial positive charge in the transition state.
This results in the formation of $1$-chloropropan-$2$-ol as the major product.
$CH_3-CH=CH_2 + Cl_2 + H_2O \rightarrow CH_3-CH(OH)-CH_2-Cl + HCl$

(A) The thermal stability of alkaline earth metal carbonates increases as the size of the cation increases down the group.
As the size of the cation increases,the polarizing power of the cation decreases,which makes the $M-O$ bond stronger and the carbonate ion more stable.
Therefore,the order of thermal stability is $MgCO_3 < CaCO_3 < SrCO_3 < BaCO_3$.

(A) The electronic configuration of an element with atomic number $15$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^3$.
Since the valence shell is the $3rd$ shell $(n=3)$,the number of valence electrons is $2 + 3 = 5$.
For $p$-block elements,the group number is calculated as $10 + \text{valence electrons} = 10 + 5 = 15$.
The valency is calculated as $8 - \text{valence electrons} = 8 - 5 = 3$.
Thus,the group number is $15$,the number of valence electrons is $5$,and the valency is $3$.

(C) Amphoteric metals react with both acids and bases to liberate hydrogen gas.
Zinc $(Zn)$ is an amphoteric metal.
Reaction with acid: $Zn + 2HCl \rightarrow ZnCl_2 + H_2 \uparrow$
Reaction with base: $Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2 \uparrow$

(B) The probability of finding an electron in an atom is given by the square of the wave function,$\psi^2$.
In the given graph,the regions $a$ and $c$ represent the peaks of the $\psi^2$ curve,where the probability density is maximum.
Therefore,electrons are most likely to be found in regions $a$ and $c$.

(C) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
In the reaction $2CuBr \longrightarrow CuBr_2 + Cu$,the copper ion $Cu^{+}$ in $CuBr$ undergoes disproportionation.
The oxidation state of $Cu$ changes from $+1$ in $CuBr$ to $+2$ in $CuBr_2$ (oxidation) and to $0$ in $Cu$ (reduction).
Thus,the reaction is $2Cu^{+} \longrightarrow Cu^{2+} + Cu$.

(C) Molar mass is the mass of $1 \, mol$ of a substance.
For $AB_2$: Molar mass $= \frac{125 \times 10^{-3} \, kg}{5 \, mol} = 25 \times 10^{-3} \, kg \, mol^{-1}$.
So,$M_A + 2M_B = 25 \times 10^{-3} \quad (i)$.
For $A_2B_2$: Molar mass $= \frac{300 \times 10^{-3} \, kg}{10 \, mol} = 30 \times 10^{-3} \, kg \, mol^{-1}$.
So,$2M_A + 2M_B = 30 \times 10^{-3} \quad (ii)$.
Subtracting equation $(i)$ from $(ii)$:
$(2M_A + 2M_B) - (M_A + 2M_B) = (30 - 25) \times 10^{-3}$.
$M_A = 5 \times 10^{-3} \, kg \, mol^{-1}$.
Substituting $M_A$ in equation $(i)$:
$5 \times 10^{-3} + 2M_B = 25 \times 10^{-3}$.
$2M_B = 20 \times 10^{-3}$.
$M_B = 10 \times 10^{-3} \, kg \, mol^{-1}$.

(D) The minerals feldspar,zeolites,mica,and asbestos are all classified as silicate minerals.
The fundamental structural unit of all silicate minerals is the tetrahedral $(SiO_4)^{4-}$ anion,where one silicon atom is bonded to four oxygen atoms.

(D) Using Kirchhoff's equation: $\Delta H_2 = \Delta H_1 + \Delta C_P (T_2 - T_1)$.
Given: $\Delta H_1 = 24 \ cal \ g^{-1}$,$T_1 = 200 + 273 = 473 \ K$,$T_2 = 250 + 273 = 523 \ K$.
$\Delta C_P = C_P(vap) - C_P(solid) = 0.031 - 0.055 = -0.024 \ cal \ g^{-1} K^{-1}$.
$\Delta H_2 = 24 + (-0.024) \times (523 - 473)$.
$\Delta H_2 = 24 - 0.024 \times 50$.
$\Delta H_2 = 24 - 1.2 = 22.8 \ cal \ g^{-1}$.

(A) The process described is the Lassaigne's test modification or a standard qualitative analysis for phosphorus. $1$. The organic compound is fused with sodium peroxide $(Na_2O_2)$ to convert phosphorus into phosphate $(PO_4^{3-})$. $2$. Boiling with $HNO_3$ removes excess peroxide and ensures the phosphorus is in the phosphate form. $3$. Upon adding ammonium molybdate $(NH_4)_2MoO_4$ in the presence of $HNO_3$,a canary yellow precipitate of ammonium phosphomolybdate $(NH_4)_3PO_4 \cdot 12MoO_3$ is formed. $4$. Therefore,the element present is phosphorus.

(D) The dissociation of $Al(OH)_3$ is given by: $Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^-$
$K_{sp} = [Al^{3+}][OH^-]^3$
In $0.2 \ M \ NaOH$,the concentration of $OH^-$ is $0.2 \ M$ (assuming complete dissociation of $NaOH$ and neglecting the contribution from $Al(OH)_3$ due to the common ion effect).
Let $s$ be the molar solubility of $Al(OH)_3$.
Then $[Al^{3+}] = s$ and $[OH^-] = 0.2 + 3s \approx 0.2$.
Substituting these values into the $K_{sp}$ expression:
$2.4 \times 10^{-24} = s(0.2)^3$
$2.4 \times 10^{-24} = s(0.008)$
$s = \frac{2.4 \times 10^{-24}}{0.008} = 3 \times 10^{-22} \ M$.

(C) When but$-2-$ene $(CH_3-CH=CH-CH_3)$ reacts with alkaline $KMnO_4$ at an elevated temperature,the double bond undergoes oxidative cleavage.
Each $=CH-CH_3$ fragment is oxidized to $CH_3COOH$ (acetic acid).
Therefore,two molecules of acetic acid are formed.
$CH_3-CH=CH-CH_3$ $\xrightarrow[\Delta]{alk. KMnO_4} 2CH_3COOK$ $\xrightarrow{H^+} 2CH_3COOH$

(A) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} \cdot \Delta V$.
Given: $P_{ext} = 1 \, bar$, $V_1 = 1 \, L$, $V_2 = 10 \, L$.
$\Delta V = V_2 - V_1 = 10 \, L - 1 \, L = 9 \, L$.
$W = -1 \, bar \cdot 9 \, L = -9 \, bar \cdot L$.
Since $1 \, bar \cdot L = 100 \, J$, then $W = -9 \times 100 \, J = -900 \, J$.
Converting to $kJ$: $W = -900 \, J / 1000 = -0.9 \, kJ$.

(B) Given that the mean of the binomial distribution is $np = 8$ and the variance is $npq = 4$.
Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{4}{8} = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 8$,we get $n \times \frac{1}{2} = 8$,which implies $n = 16$.
The probability mass function is $P(X = r) = ^{16}C_r (\frac{1}{2})^r (\frac{1}{2})^{16-r} = ^{16}C_r (\frac{1}{2})^{16}$.
We need to find $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$.
$P(X \le 2) = \frac{^{16}C_0 + ^{16}C_1 + ^{16}C_2}{2^{16}}$.
Calculating the combinations: $^{16}C_0 = 1$,$^{16}C_1 = 16$,and $^{16}C_2 = \frac{16 \times 15}{2} = 120$.
Summing these values: $1 + 16 + 120 = 137$.
Thus,$P(X \le 2) = \frac{137}{2^{16}}$.
Comparing this with $\frac{k}{2^{16}}$,we get $k = 137$.

(B) The relationship between standard Gibbs free energy change $(\Delta G^o)$ and the equilibrium constant $(K)$ is given by the equation: $\Delta G^o = -RT \ln K$ or $\Delta G^o = -2.303 \, RT \log K$.
$1$. If $\Delta G^o < 0$,then $\ln K > 0$,which implies $K > 1$ (Spontaneous reaction).
$2$. If $\Delta G^o = 0$,then $\ln K = 0$,which implies $K = 1$ (Equilibrium).
$3$. If $\Delta G^o > 0$,then $\ln K < 0$,which implies $K < 1$ (Non-spontaneous reaction).
Therefore,the match $\Delta G^o < 0, K < 1$ is $INCORRECT$.

(A) The thermal decomposition of $LiNO_3$ is given by: $2LiNO_3 \xrightarrow{\Delta} Li_2O + 2NO_2 + \frac{1}{2}O_2$.
Thus,it produces $Li_2O$ (lithium oxide),not $LiNO_2$ (lithium nitrite).
Therefore,the statement in option $A$ is $INCORRECT$.

(C) In $diamond$, each carbon atom is $sp^3$ hybridized and forms four single $C-C$ sigma bonds.
The bond length of a pure $C-C$ single bond is $154 \ pm$.
In $graphite$ and fullerenes ($C_{60}$, $C_{70}$), the carbon atoms are $sp^2$ hybridized, resulting in resonance and partial double bond character.
Partial double bond character leads to a shorter bond length (approximately $141.5 \ pm$ in graphite).
Therefore, the $C-C$ bond length is maximum in diamond.

(B) reacts with $Ag_2O$ to give a precipitate,indicating it is a terminal alkyne.
$A$ undergoes hydration $(Hg^{2+}/H^{+})$ to form $B$,which is reduced by $NaBH_4$ to $C$.
$C$ gives turbidity with Lucas reagent $(conc. HCl / ZnCl_2)$ within $5 \text{ minutes}$,indicating $C$ is a secondary alcohol.
$CH_3-C \equiv CH (A) \xrightarrow{Ag_2O} CH_3-C \equiv CAg \downarrow (\text{ppt})$
$CH_3-C \equiv CH (A)$ $\xrightarrow{Hg^{2+}/H^{+}} CH_3-COCH_3 (B)$ $\xrightarrow{NaBH_4} CH_3-CH(OH)-CH_3 (C)$
$CH_3-CH(OH)-CH_3 (C) \xrightarrow{conc. HCl / ZnCl_2} CH_3-CH(Cl)-CH_3 + H_2O$ (Turbidity within $5 \text{ minutes}$).

(B) The relationship between $K_p$ and $K_c$ is given by the equation $K_p = K_c(RT)^{\Delta n_g}$,where $\Delta n_g$ is the change in the number of moles of gaseous products and reactants.
If $\Delta n_g = 0$,then $K_p = K_c$.
If $\Delta n_g \neq 0$,then $K_p \neq K_c$.
Let us calculate $\Delta n_g$ for each option:
$(A)$ $2NO_{(g)} \rightleftharpoons N_{2(g)} + O_{2(g)}$: $\Delta n_g = (1 + 1) - 2 = 0$.
$(B)$ $2C_{(s)} + O_{2(g)} \rightleftharpoons 2CO_{(g)}$: $\Delta n_g = 2 - 1 = 1 \neq 0$.
$(C)$ $NO_{2(g)} + SO_{2(g)} \rightleftharpoons NO_{(g)} + SO_{3(g)}$: $\Delta n_g = (1 + 1) - (1 + 1) = 0$.
$(D)$ $2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$: $\Delta n_g = (1 + 1) - 2 = 0$.
Since $\Delta n_g \neq 0$ only in option $(B)$,$K_p \neq K_c$ for this reaction.

(B) $NO_2$ and hydrocarbons are the primary precursors of photochemical smog. Nitrogen oxides $(NO_x)$ are released into the atmosphere by automobiles and factories,which then react with hydrocarbons in the presence of sunlight to form photochemical smog.

(D) The solubility product constant $K_{sp}$ for $Cd(OH)_2$ is given by $K_{sp} = 4s^3$,where $s$ is the molar solubility in water.
$K_{sp} = 4 \times (1.84 \times 10^{-5})^3$.
Given $pH = 12$,we have $pOH = 14 - 12 = 2$.
Therefore,$[OH^-] = 10^{-2} \ M$.
Using the expression $K_{sp} = [Cd^{2+}][OH^-]^2$,we substitute the values:
$4 \times (1.84 \times 10^{-5})^3 = [Cd^{2+}] \times (10^{-2})^2$.
$[Cd^{2+}] = \frac{4 \times (1.84)^3 \times 10^{-15}}{10^{-4}}$.
$[Cd^{2+}] = 4 \times 6.2295 \times 10^{-11} \approx 2.49 \times 10^{-10} \ M$.

(A) The combustion reaction of a hydrocarbon $C_xH_y$ is given by: $C_xH_y + (x + y/4)O_2 \to xCO_2 + (y/2)H_2O$.
Mass of $C$ in $88 \ g$ of $CO_2 = (12/44) \times 88 = 24 \ g$.
Mass of $H$ in $9 \ g$ of $H_2O = (2/18) \times 9 = 1 \ g$.
Total mass of hydrocarbon = Mass of $C$ + Mass of $H = 24 \ g + 1 \ g = 25 \ g$.
Thus,the hydrocarbon contains $24 \ g$ of carbon and $1 \ g$ of hydrogen.

(A) Electrical conductivity of an aqueous solution depends on the concentration of ions produced.
Greater the acidic strength (higher $K_a$ value),greater will be the degree of dissociation and hence higher electrical conductivity.
The $K_a$ values at $298 \ K$ are:
Formic acid $(HCOOH)$: $1.8 \times 10^{-4}$
Benzoic acid $(C_6H_5COOH)$: $6.5 \times 10^{-5}$
Acetic acid $(CH_3COOH)$: $1.8 \times 10^{-5}$
Since the order of acidic strength is Formic acid > Benzoic acid > Acetic acid,the order of electrical conductivity is $a > c > b$.

(A) The energy of an orbital in a multi-electron system depends on the effective nuclear charge $(Z_{eff})$.
As the nuclear charge $(Z)$ increases,the attraction between the nucleus and the electrons increases,which stabilizes the orbital and lowers its energy.
Among the given elements,the atomic numbers are: $H$ $(Z=1)$,$Li$ $(Z=3)$,$Na$ $(Z=11)$,and $K$ $(Z=19)$.
Since $K$ has the highest nuclear charge $(Z=19)$,the $2s$ orbital in $K$ experiences the strongest attraction,resulting in the lowest energy.

(C) The dihedral angle is the angle between the $C-H$ bonds on adjacent carbon atoms.
In the given Newman projection,the angle between the front $C-H'$ bond and the back $C-H''$ bond is calculated by considering the standard staggered angle of $120^\circ$ between adjacent bonds and the deviation shown.
From the geometry,the dihedral angle between $H'$ and $H''$ is the sum of the angle between the bonds in the staggered conformation $(120^\circ)$ and the given deviation angle $(29^\circ)$.
Dihedral angle $= 120^\circ + 29^\circ = 149^\circ$.

(A) $1$. Identify the longest carbon chain containing both the double and triple bonds. The chain has $7$ carbons,so the parent alkane is heptane.
$2$. Number the chain from the end that gives the lowest locants to the multiple bonds. Starting from the double bond end,the double bond is at $C-1$ and the triple bond is at $C-6$. If we start from the other end,the triple bond is at $C-1$ and the double bond is at $C-6$. According to $IUPAC$ rules,when both double and triple bonds are present,the double bond gets priority for the lowest number if they are at equivalent positions. Thus,the chain is numbered from the alkene end.
$3$. The substituents are two methyl groups at positions $3$ and $5$,and a propyl group at position $4$.
$4$. Combining these,the name is $3, 5-$dimethyl$-4-$propylhept$-1-$en$-6-$yne.

(D) The atomic number of beryllium $(Be)$ is $4$ and boron $(B)$ is $5$.
Since $B$ has a higher atomic number than $Be$,it has a greater nuclear charge.
The electronic configuration of $Be$ is $1s^2 2s^2$ (fully filled $s$-orbital),while $B$ is $1s^2 2s^2 2p^1$.
Due to the stable fully filled $2s$-orbital in $Be$,it requires more energy to remove an electron compared to $B$.
Therefore,$Be$ has a lesser nuclear charge and a greater first ionisation enthalpy compared to $B$.

(C) Temporary hardness in water is caused by the presence of magnesium or calcium bicarbonates.
For magnesium bicarbonate,the reaction upon boiling is:
$Mg(HCO_3)_{2(aq)} \xrightarrow{\Delta} Mg(OH)_{2(s)} + 2CO_{2(g)}$.
Here,$X$ is $Mg(HCO_3)_2$ and $Y$ is $Mg(OH)_2$.
Thus,the correct option is $C$.

(D) The first ionization energy $(IE_1)$ of an element is the energy required to remove the first electron,and the second ionization energy $(IE_2)$ is the energy required to remove the second electron.
For alkali metals like potassium $(K)$,the electronic configuration is $[Ar] 4s^1$.
After the removal of the first electron,it achieves the stable noble gas configuration of argon $([Ar])$.
Removing the second electron requires breaking this stable octet,which demands a very high amount of energy.
Therefore,the difference between $IE_2$ and $IE_1$ is exceptionally large for alkali metals compared to alkaline earth metals or transition metals.

(D) An equimolar mixture of $CO$ and $H_{2}$ is obtained when steam is passed over red hot coke. This mixture is commonly known as synthesis gas or $syn \ gas$.
The reaction is:
$H_{2}O_{(g)} + C_{(s)} \rightarrow CO_{(g)} + H_{2(g)}$

(D) The size of alkali metal ions increases down the group as follows: $Li ^{+} < Na ^{+} < K ^{+} < Rb ^{+} < Cs ^{+}$.
Hydration enthalpy is inversely proportional to the ionic size $(Hydration \ Enthalpy \ \propto \ \frac{1}{\text{ionic size}})$.
Therefore,the smaller the ion,the higher its hydration enthalpy.
Thus,the correct order of hydration enthalpies is $Li ^{+} > Na ^{+} > K ^{+} > Rb ^{+} > Cs ^{+}$.

(C) The electronic configuration of $O_{2}$ ($16$ electrons) is: $\sigma 1s^{2} \sigma^{*} 1s^{2} \sigma 2s^{2} \sigma^{*} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{2} = \pi 2p_{y}^{2} \pi^{*} 2p_{x}^{1} = \pi^{*} 2p_{y}^{1}$.
When an electron is added to $O_{2}$ to form $O_{2}^{-}$,the incoming electron enters the lowest energy vacant or half-filled molecular orbital,which is the $\pi^{*}$ antibonding orbital.
Thus,the electron enters the $\pi^{*} 2p_{x}$ or $\pi^{*} 2p_{y}$ orbital.
Therefore,option $(C)$ is correct.

(B) The electronic configuration of $NO$ ($15$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. It has one unpaired electron,so it is paramagnetic. Bond order $= (10-5)/2 = 2.5$.
After losing one electron,$NO^{+}$ ($14$ electrons) has the configuration $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. It has no unpaired electrons,so it is diamagnetic. Bond order $= (10-4)/2 = 3$.
Thus,in the process $NO \rightarrow NO^{+}$,the bond order increases from $2.5$ to $3$ and the character changes from paramagnetic to diamagnetic.

(D) The reaction proceeds in two steps:
$1$. Diazotization: The primary aromatic amine group $(-NH_2)$ reacts with $NaNO_2$ and $HCl$ at $0-5 \ ^\circ C$ to form a diazonium salt $(-N_2^+Cl^-)$.
$2$. Intramolecular Coupling: In the presence of aqueous $NaOH$,the phenolic $-OH$ group is deprotonated to form a phenoxide ion,which is a strong activating group. This phenoxide ion then undergoes an intramolecular electrophilic aromatic substitution (coupling) with the diazonium group to form a cyclic azo compound.

(D) The total work done on the particle is the sum of the work done by the electric field and the magnetic field.
Work done by the magnetic field on a moving charge is always zero because the magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ is always perpendicular to the velocity $\vec{v}$.
Thus,the total work done is equal to the work done by the electric field.
The work done by the electric field is given by $W = \vec{F}_e \cdot \vec{d} = q \vec{E} \cdot \vec{d}$.
The displacement vector is $\vec{d} = (1 - 0) \hat{i} + (1 - 0) \hat{j} = \hat{i} + \hat{j}$.
Substituting the values: $W = q(2 \hat{i} + 3 \hat{j}) \cdot (\hat{i} + \hat{j})$.
$W = q(2 \times 1 + 3 \times 1) = 5q$.
Therefore,the magnitude of the total work done is $5q$.

(A) First,we determine the set $S$ by solving the inequality $x^2+30 \leq 11x$.
$x^2-11x+30 \leq 0$
$(x-5)(x-6) \leq 0$
Thus,$S = [5, 6]$.
Next,we find the derivative of $f(x) = 3x^3-18x^2+27x-40$:
$f'(x) = 9x^2-36x+27 = 9(x^2-4x+3) = 9(x-1)(x-3)$.
For $x \in [5, 6]$,both $(x-1)$ and $(x-3)$ are positive,so $f'(x) > 0$.
Since $f'(x) > 0$ on the interval $[5, 6]$,the function $f(x)$ is strictly increasing on this interval.
Therefore,the maximum value occurs at the right endpoint $x=6$.
$f(6) = 3(6)^3 - 18(6)^2 + 27(6) - 40$
$f(6) = 3(216) - 18(36) + 162 - 40$
$f(6) = 648 - 648 + 162 - 40 = 122$.

(A) Given expression: $[\sim(\sim p \vee q) \vee (p \wedge r)] \wedge (\sim q \wedge r)$
Applying De Morgan's law: $[(p \wedge \sim q) \vee (p \wedge r)] \wedge (\sim q \wedge r)$
Applying Distributive law: $[p \wedge (\sim q \vee r)] \wedge (\sim q \wedge r)$
Applying Associative law: $p \wedge [(\sim q \vee r) \wedge (\sim q \wedge r)]$
Since $(\sim q \vee r) \wedge (\sim q \wedge r) = (\sim q \wedge r)$,the expression becomes: $p \wedge (\sim q \wedge r)$
Rearranging using Commutative law: $(p \wedge r) \wedge \sim q$

(C) Given expression: $[(\sim(\sim p \vee q)) \vee (p \wedge r)] \wedge (\sim q \wedge r)$
Using De Morgan's law,$\sim(\sim p \vee q) \equiv (p \wedge \sim q)$.
So,the expression becomes: $[(p \wedge \sim q) \vee (p \wedge r)] \wedge (\sim q \wedge r)$
Applying the Distributive law: $[p \wedge (\sim q \vee r)] \wedge (\sim q \wedge r)$
Using the Associative and Commutative laws: $p \wedge [(\sim q \vee r) \wedge (\sim q \wedge r)]$
Since $(\sim q \vee r) \wedge (\sim q \wedge r) \equiv (\sim q \wedge r)$ by the Absorption law:
The expression simplifies to: $p \wedge (\sim q \wedge r)$
Which is equivalent to: $(p \wedge r) \wedge \sim q$

(C) For a Binomial distribution,the mean is given by $np = 8$ and the variance is given by $npq = 4$.
Dividing the variance by the mean,we get $q = \frac{4}{8} = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 8$,we get $n \times \frac{1}{2} = 8$,so $n = 16$.
Now,$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.
Using the formula $P(X=r) = {}^{n}C_{r} p^{r} q^{n-r}$,we have:
$P(X \leq 2) = {}^{16}C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{16} + {}^{16}C_{1} (\frac{1}{2})^{1} (\frac{1}{2})^{15} + {}^{16}C_{2} (\frac{1}{2})^{2} (\frac{1}{2})^{14}$
$P(X \leq 2) = ({}^{16}C_{0} + {}^{16}C_{1} + {}^{16}C_{2}) (\frac{1}{2})^{16}$
Calculating the combinations:
${}^{16}C_{0} = 1$
${}^{16}C_{1} = 16$
${}^{16}C_{2} = \frac{16 \times 15}{2} = 120$
Sum $= 1 + 16 + 120 = 137$.
Thus,$P(X \leq 2) = \frac{137}{2^{16}}$.
Comparing this with $\frac{k}{2^{16}}$,we get $k = 137$.

(A) Hinsberg's reagent is $C_6H_5SO_2Cl$,which is known as benzene sulphonyl chloride.
It is used to distinguish between primary,secondary,and tertiary amines.

(C) $20\% \, w/w$ aqueous solution of $KI$ means $20 \, g$ of $KI$ is present in $100 \, g$ of solution.
Mass of solute $(KI) = 20 \, g$.
Mass of solvent (water) $= 100 \, g - 20 \, g = 80 \, g = 0.08 \, kg$.
Molar mass of $KI = 166 \, g \, mol^{-1}$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in } kg}$.
Moles of $KI = \frac{20 \, g}{166 \, g \, mol^{-1}} \approx 0.1205 \, mol$.
Molality $(m) = \frac{0.1205 \, mol}{0.08 \, kg} = 1.506 \, mol \, kg^{-1} \approx 1.51 \, m$.

(A) The given structure $[HN-CO-NH-CH_2]_n$ represents the repeating unit of Urea-Formaldehyde resin.
It is formed by the condensation polymerization of Urea $(NH_2-CO-NH_2)$ and Formaldehyde $(HCHO)$.
Thus,Formaldehyde is one of its constituents.

(C) $(I)$ Valence Bond Theory $(VBT)$ does not explain the color exhibited by transition metal complexes because the splitting of $d-$orbitals is explained by Crystal Field Theory $(CFT)$.
$(II)$ $VBT$ cannot predict the magnetic properties of transition metal complexes quantitatively; it only provides qualitative information.
$(III)$ $VBT$ does not distinguish between strong field and weak field ligands.
Therefore,statements $I$ and $III$ are correct.

(A) In this titration,a strong acid $(HCl)$ is added to a strong base $(NaOH)$.
Initially,the solution contains $NaOH$,so the $pH$ is high (basic).
As $HCl$ is added,the $pH$ decreases gradually.
Near the equivalence point,there is a sharp drop in $pH$ because the concentration of $OH^-$ ions decreases rapidly.
After the equivalence point,the solution becomes acidic due to excess $HCl$,and the $pH$ levels off at a low value.
Graph $(a)$ correctly represents this sigmoidal decrease in $pH$ as the volume of the acid titrant increases.

(D) The reaction of $p$-hydroxybenzophenone with $Br_2$ in $CCl_4$ is an electrophilic aromatic substitution reaction.
In $p$-hydroxybenzophenone,there are two benzene rings. One ring is attached to a carbonyl group $(-CO-)$,which is an electron-withdrawing group ($-R$ effect),deactivating the ring towards electrophilic aromatic substitution $(EAS)$.
The other ring is attached to a hydroxyl group $(-OH)$,which is a strong electron-donating group ($+R$ effect),activating the ring towards $EAS$.
The $-OH$ group is ortho/para directing. Since the para position is already occupied by the carbonyl group,the incoming electrophile $(Br^+)$ will attack the ortho position relative to the $-OH$ group.
Thus,the product formed is $3$-bromo-$4$-hydroxybenzophenone.

(B) For a complex to exhibit $cis-trans$ isomerism,it must have at least two identical ligands in different spatial arrangements.
$A$. $[Pt(en)Cl_2]$ is a square planar complex of the type $[M(AA)X_2]$. It shows $cis-trans$ isomerism.
$B$. $[Pt(en)_2Cl_2]^{2+}$ is an octahedral complex of the type $[M(AA)_2X_2]$. It shows $cis-trans$ isomerism.
$C$. $[Cr(en)_2(ox)]^+$ is an octahedral complex of the type $[M(AA)_3]$. It does not show $cis-trans$ isomerism.
$D$. $[Zn(en)Cl_2]$ is a tetrahedral complex. Tetrahedral complexes do not show geometric isomerism.
However,in the context of standard chemistry problems,$[Pt(en)_2Cl_2]^{2+}$ is the classic example of an octahedral complex exhibiting $trans$-isomerism as shown in the provided image.

(D) Amines are more nucleophilic than alcohols.
Therefore, the $-NH_2$ group reacts preferentially with ethyl formate $(HCOOC_2H_5)$ to form the formamide derivative $CH_3-CH(OH)-CH_2-CH_2-NH-CHO$.

(B) The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
$1. Sc^{3+} (Z=21): [Ar] 3d^0$. Number of unpaired electrons $(n) = 0$. $\mu = 0 \ B.M.$
$2. Ti^{3+} (Z=22): [Ar] 3d^1$. Number of unpaired electrons $(n) = 1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \ B.M. \approx 1.73 \ B.M.$
$3. Ti^{2+} (Z=22): [Ar] 3d^2$. Number of unpaired electrons $(n) = 2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \ B.M. \approx 2.83 \ B.M.$
$4. V^{2+} (Z=23): [Ar] 3d^3$. Number of unpaired electrons $(n) = 3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \ B.M. \approx 3.87 \ B.M.$
Comparing the values: $0 < 1.73 < 2.83 < 3.87$.
Therefore,the correct order is $Sc^{3+} < Ti^{3+} < Ti^{2+} < V^{2+}$.

(C) Amylopectin is a branched-chain polysaccharide that acts as a homopolymer of $\alpha-D$-glucose units.
It consists of a linear chain formed by $\alpha-D$-glucose units linked through $C_1-C_4$ glycosidic bonds.
Additionally,branching occurs at certain points where $\alpha-D$-glucose units are linked through $C_1-C_6$ glycosidic bonds.

(A) The Freundlich adsorption isotherm is given by $\frac{x}{m} = k p^{1/n}$.
According to the given equation $\frac{x}{m} = k p^{0.5}$,the extent of adsorption $\frac{x}{m}$ is directly proportional to the square root of pressure $p$.
Therefore,an increase in pressure $p$ leads to an increase in the extent of adsorption.
Physical adsorption is an exothermic process,which means it releases heat.
According to Le Chatelier's principle,for an exothermic process,a decrease in temperature $T$ favors the forward reaction,thereby increasing the extent of adsorption.
Thus,adsorption increases with an increase in $p$ and a decrease in $T$.

(A) For $t \le 1 \ hour$,the population grows as $N(t) = N_0 \exp(t)$. Thus,$\frac{N_0}{N} = \exp(-t)$. This is a decreasing exponential function.
For $t > 1 \ hour$,the population follows $\frac{dN}{dt} = -5N^2$. Integrating this from $t=1$ to $t$,we get $\int_{N_1}^{N} \frac{dN}{N^2} = \int_{1}^{t} -5 dt$,where $N_1 = N_0 \exp(1)$.
This gives $[-\frac{1}{N}]_{N_1}^{N} = -5(t-1)$,so $\frac{1}{N} - \frac{1}{N_1} = 5(t-1)$.
Multiplying by $N_0$,we get $\frac{N_0}{N} = \frac{N_0}{N_1} + 5N_0(t-1) = \exp(-1) + 5N_0(t-1)$.
This is a linear function with a positive slope for $t > 1$. Therefore,the plot of $\frac{N_0}{N}$ vs. $t$ decreases until $t=1$ and then increases linearly.

(C) The energy of absorbed light $(E)$ is inversely proportional to the wavelength $(\lambda)$ of the absorbed light,i.e.,$E = \frac{hc}{\lambda}$.
The energy of absorption corresponds to the Crystal Field Splitting Energy ($CFSE$ or $\Delta_o$),which depends on the strength of the ligands.
The spectrochemical series order for the ligands is $Cl^- < H_2O < NH_3$.
Comparing the complexes:
$(I) [CoCl(NH_3)_5]^{2 }$ has $Cl^-$ as a ligand.
$(II) [Co(NH_3)_5H_2O]^{3 }$ has $H_2O$ as a ligand.
$(III) [Co(NH_3)_6]^{3 }$ has $NH_3$ as a ligand.
Since $NH_3$ is a stronger ligand than $H_2O$,and $H_2O$ is stronger than $Cl^-$,the order of $CFSE$ is $(III) > (II) > (I)$.
Since $E \propto \Delta_o$,the order of energy absorbed is $(III) > (II) > (I)$.
Since $\lambda \propto \frac{1}{E}$,the order of wavelength absorbed is $(I) > (II) > (III)$.

(B) The given reaction is a cross-Cannizzaro reaction between benzaldehyde $(C_6H_5CHO)$ and formaldehyde $(HCHO)$.
In a cross-Cannizzaro reaction,the more reactive carbonyl compound (formaldehyde) is oxidized to the corresponding acid (formate ion,which gives formic acid $HCOOH$ upon acidification),and the less reactive carbonyl compound (benzaldehyde) is reduced to the corresponding alcohol (benzyl alcohol,$C_6H_5CH_2OH$).
Therefore,the products are benzyl alcohol and formic acid.

(D) The correct matches are as follows:
$I$. Liquation is used for metals with low melting points like $Sn$.
$II$. Zone refining is used for semiconductors like $Ga$.
$III$. Mond process is specifically used for the refining of $Ni$.
$IV$. Van Arkel method is used for refining metals like $Zr$ and $Ti$.
Therefore,the correct matching is $I-c, II-d, III-b, IV-a$.

(D) The conversion of $N$-ethylphthalimide to ethylamine is the final step of the Gabriel phthalimide synthesis.
This reaction involves the hydrolysis of $N$-ethylphthalimide using an aqueous base (like $NaOH$) followed by acidification,or more commonly,treatment with hydrazine $(NH_2NH_2)$ to cleave the phthalimide ring and release the primary amine.
In the context of the provided options,$NH_2NH_2$ is the standard reagent used to liberate the primary amine from the $N$-substituted phthalimide.

(A) The reaction of $2$-bromo-$3$-methylbutane with methanol $(CH_3OH)$ proceeds via an $S_N1$ mechanism.
First,the $Br^-$ ion leaves to form a secondary carbocation $CH_3-CH(CH_3)-C^+H-CH_3$.
This carbocation undergoes a $1,2$-hydride shift to form a more stable tertiary carbocation $CH_3-C^+(CH_3)-CH_2-CH_3$.
Finally,the nucleophile $CH_3OH$ attacks the tertiary carbocation,followed by deprotonation,to yield $2$-methoxy-$2$-methylbutane $(CH_3-C(CH_3)(OCH_3)-CH_2-CH_3)$ as the major product.

(A) Nylon $-6,6$ is a condensation polymer formed by the reaction between hexamethylene diamine and adipic acid with the elimination of water molecules.
Buna $-S$,Teflon,and Neoprene are examples of addition polymers.

(A) Conductivity $(K)$ is defined as the conductance of a unit volume of solution. As the concentration of an electrolyte decreases,the number of ions per unit volume decreases,leading to a decrease in conductivity. Thus,$S_1$ is incorrect.
Molar conductivity $(\lambda_m)$ is defined as $\lambda_m = \frac{K}{C}$. As the concentration $(C)$ decreases,the volume of solution containing one mole of electrolyte increases significantly,which outweighs the decrease in conductivity $(K)$. Consequently,molar conductivity increases with a decrease in concentration. Thus,$S_2$ is correct.

(D) The rate of $S_N1$ reaction is directly proportional to the stability of the carbocation intermediate formed after the loss of the leaving group $(I^-)$.
$(A)$ The carbocation is a secondary benzylic carbocation.
$(B)$ The $-OCH_3$ group is at the meta position. It exerts a strong $-I$ effect and no resonance effect on the carbocation. This destabilizes the carbocation compared to $(A)$.
$(C)$ The $-CH_3$ group is at the para position. It provides stability through $+I$ and hyperconjugation effects.
$(D)$ The $-OCH_3$ group is at the para position. It provides significant stability through the $+M$ (resonance) effect,which outweighs its $-I$ effect.
Comparing the stability: The meta-methoxy substituted carbocation $(B)$ is the least stable due to the $-I$ effect. The unsubstituted benzylic carbocation $(A)$ is next. The para-methyl substituted carbocation $(C)$ is more stable due to hyperconjugation. The para-methoxy substituted carbocation $(D)$ is the most stable due to the strong $+M$ effect.
Therefore,the order of increasing rate is $B < A < C < D$.

(C) To determine which oxoacid does not contain a $S-S$ bond,we examine their structures:
$1$. $H_2S_2O_3$ (Thiosulphuric acid): Contains a $S-S$ bond $(HO-SO_2-SH)$.
$2$. $H_2S_2O_4$ (Dithionous acid): Contains a $S-S$ bond $(HO-SO-SO-OH)$.
$3$. $H_2S_2O_7$ (Pyrosulphuric acid or Oleum): Its structure is $HO-SO_2-O-SO_2-OH$. It contains a $S-O-S$ linkage,not a $S-S$ bond.
$4$. $H_2S_4O_6$ (Tetrathionic acid): Contains a chain of sulphur atoms ($S-S-S-S$ linkage).
Therefore,$H_2S_2O_7$ is the correct answer.

(D) The lowering of vapour pressure is given by the formula: $\Delta p = p^o \cdot X_{solute}$
Given:
$p^o = 35 \ mm \ Hg$
$w_{urea} = 0.60 \ g$,$M_{urea} = 60 \ g \ mol^{-1}$
$w_{water} = 360 \ g$,$M_{water} = 18 \ g \ mol^{-1}$
Calculate moles of solute $(n)$ and solvent $(N)$:
$n = \frac{0.60}{60} = 0.01 \ mol$
$N = \frac{360}{18} = 20 \ mol$
Calculate mole fraction of solute $(X_{solute})$:
$X_{solute} = \frac{n}{n + N} = \frac{0.01}{0.01 + 20} = \frac{0.01}{20.01} \approx 0.00049975$
Calculate lowering of vapour pressure $(\Delta p)$:
$\Delta p = 35 \times 0.00049975 \approx 0.01749 \ mm \ Hg$
Rounding to the nearest value,we get $0.017 \ mm \ Hg$.

(D) The given reaction is an intramolecular Friedel-Crafts alkylation.
$AlCl_3$ acts as a Lewis acid and reacts with the chlorine atom to form a carbocation at the secondary carbon position.
This carbocation then undergoes an electrophilic aromatic substitution on the benzene ring.
Since the oxygen atom is ortho/para directing,the cyclization occurs at the ortho position relative to the ether linkage,leading to the formation of a six-membered ring fused to the benzene ring.
Thus,the major product is the structure shown in option $D$.

(D) The first ionization enthalpy generally increases across a period from left to right due to an increase in effective nuclear charge.
For the $3d$ series elements $(Ti, Mn, Ni, Zn)$,the general trend is $Ti < Mn < Ni < Zn$.
$Ti$ $([Ar] 3d^2 4s^2)$ has the lowest ionization enthalpy,while $Zn$ $([Ar] 3d^{10} 4s^2)$ has the highest due to its stable,fully-filled $d$-orbital configuration.

(B) The Arrhenius equation is given by: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303 \ R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$
Given:
$K_1 = 2.5 \times 10^{-4} \ dm^3 \ mol^{-1} \ s^{-1}$,$T_1 = 327 + 273 = 600 \ K$
$K_2 = 1.0 \ dm^3 \ mol^{-1} \ s^{-1}$,$T_2 = 527 + 273 = 800 \ K$
$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
Substituting the values:
$\log \left( \frac{1}{2.5 \times 10^{-4}} \right) = \frac{E_a}{2.303 \times 8.314} \left( \frac{800 - 600}{600 \times 800} \right)$
$\log (4000) = \frac{E_a}{19.147} \left( \frac{200}{480000} \right)$
$3.602 = \frac{E_a}{19.147} \times 4.167 \times 10^{-4}$
$E_a = \frac{3.602 \times 19.147}{4.167 \times 10^{-4}} \approx 165500 \ J \ mol^{-1} = 165.5 \ kJ \ mol^{-1} \approx 166 \ kJ \ mol^{-1}$

(C) The $R_f$ (retention factor) value is defined as the ratio of the distance traveled by the solute to the distance traveled by the solvent front.
Since the solute cannot travel further than the solvent front,the $R_f$ value is always $\le 1$.
$A$ higher $R_f$ value indicates that the substance has a greater affinity for the mobile phase and a lower affinity for the stationary phase (i.e.,lower adsorption).
Therefore,the statement "Higher $R_f$ value means higher adsorption" is incorrect.

(C) Ruby is a gemstone composed of aluminium oxide $(Al_2O_3)$ in which some $Al^{3+}$ ions are replaced by $Cr^{3+}$ ions in octahedral sites.
Statement $C$ is incorrect because ruby is a variety of corundum $(Al_2O_3)$,not beryl.
Beryl is the mineral associated with emeralds.
Therefore,the statement claiming $Cr^{3+}$ occupies sites in beryl is false.

(B) In the froth flotation process,collectors like pine oil,fatty acids,and xanthates are added to the suspension of powdered ore in water to make the mineral particles preferentially wetted by oil. Froth stabilizers like cresols and aniline are added to stabilize the froth. Therefore,the statement that aniline is a froth stabilizer is correct. Zincite $(ZnO)$ is an oxide ore,not a carbonate ore. Titanium is refined using the Van Arkel method,not zone refining. Sodium cyanide $(NaCN)$ is used in the leaching process for the extraction of silver.

(A) Nucleophilicity is the ability of a species to donate an electron pair to an electrophile.
For charged species,the nucleophilicity is generally higher than for neutral species.
Among the given species,$H_2O$ is a neutral molecule and is the weakest nucleophile.
Comparing the anions,$CH_3SO_3^-$ is a very weak nucleophile because the negative charge is delocalized over three oxygen atoms,making it a stable conjugate base of a strong acid.
$CH_3CO_2^-$ has the negative charge delocalized over two oxygen atoms,making it a stronger nucleophile than $CH_3SO_3^-$.
$OH^-$ is the strongest nucleophile among these as the charge is localized on a single oxygen atom.
Thus,the increasing order is $(b) < (c) < (a) < (d)$.

(D) High density polythene is prepared using Ziegler-Natta catalyst $(III)$.
$(b)$ Polyacrylonitrile is prepared by addition polymerization using peroxide catalyst $(I)$.
$(c)$ Novolac is a linear polymer of phenol and formaldehyde prepared using acid or base catalyst $(IV)$.
$(d)$ Nylon-$6$ is prepared by heating caprolactam at high temperature and pressure $(II)$.
Therefore,the correct match is $a \to III, b \to I, c \to IV, d \to II$.

(A) $1$. The reaction of acetophenone $(PhCOCH_3)$ with $NaOCl$ is a haloform reaction,which converts the methyl ketone group into a carboxylate salt,followed by acidification to give benzoic acid $(PhCOOH)$ as product $X$.
$2$. Benzoic acid $(PhCOOH)$ reacts with $SOCl_2$ to form benzoyl chloride $(PhCOCl)$.
$3$. Benzoyl chloride then reacts with aniline $(PhNH_2)$ via a nucleophilic acyl substitution reaction to form $N$-phenylbenzamide $(PhCONHPh)$ as the major product $Y$.

(D) The electronic configuration of $U$ $(Z=92)$ is $[Rn] 5f^3 6d^1 7s^2$. The highest oxidation state shown by $U$ is $+6$.
The electronic configuration of $Pu$ $(Z=94)$ is $[Rn] 5f^6 7s^2$. The highest oxidation state shown by $Pu$ is $+7$.

(A) $1$. Compound $A$ $(C_9H_{10}O)$ gives a positive iodoform test,which indicates the presence of a $CH_3CO-$ group attached to a carbon or hydrogen atom.
$2$. Oxidation of $A$ with $KMnO_4/KOH$ yields acid $B$ $(C_8H_6O_4)$.
$3$. The anhydride of $B$ is used to prepare phenolphthalein,which identifies $B$ as phthalic acid (benzene$-1,2-$dicarboxylic acid).
$4$. For $A$ to oxidize to phthalic acid,it must have two alkyl groups at the ortho positions of the benzene ring,one of which is an acetyl group $(CH_3CO-)$ and the other is a methyl group $(CH_3-)$.
$5$. Therefore,$A$ is $2$-methylacetophenone.

(B) The noble gases are $He$,$Ne$,$Ar$,$Kr$,$Xe$,and $Rn$.
All noble gases except $Rn$ (Radon) are present in the atmosphere.
$Rn$ is a radioactive element formed by the radioactive decay of radium $(^{226}_{88}Ra \rightarrow ^{222}_{86}Rn + ^{4}_{2}He)$.
Therefore,$Rn$ does not occur in the atmosphere.

(A) $1$. For $[Fe(H_2O)_6]Cl_2$: The central metal ion is $Fe^{2+}$,which has a $d^6$ configuration. $H_2O$ is a weak field ligand,so the configuration is $t_{2g}^4 e_g^2$. The $CFSE$ is calculated as: $CFSE = [-0.4 \times 4 + 0.6 \times 2] \, \Delta_o = [-1.6 + 1.2] \, \Delta_o = -0.4 \, \Delta_o$.
$2$. For $K_2[NiCl_4]$: The central metal ion is $Ni^{2+}$,which has a $d^8$ configuration. It forms a tetrahedral complex $[NiCl_4]^{2-}$. The configuration is $e^4 t_2^4$. The $CFSE$ for a tetrahedral complex is calculated as: $CFSE = [-0.6 \times n_e + 0.4 \times n_{t2}] \, \Delta_t = [-0.6 \times 4 + 0.4 \times 4] \, \Delta_t = [-2.4 + 1.6] \, \Delta_t = -0.8 \, \Delta_t$.
$3$. Therefore,the values are $-0.4 \, \Delta_o$ and $-0.8 \, \Delta_t$.

(A) In electrophoresis,colloidal particles move towards the oppositely charged electrode and get neutralized,leading to precipitation. Thus,option $(A)$ is correct.
Brownian motion decreases as viscosity increases.
Colloidal medicines are more effective due to their large surface area.
Addition of alum to water is a standard process for purification (coagulation of impurities).

(B) The molar conductivity $(\Lambda_m)$ of strong electrolytes follows the Kohlrausch equation: $\Lambda_m = \Lambda_m^0 - A\sqrt{C}$.
Both $NaCl$ and $KCl$ are strong electrolytes.
The limiting molar conductivity $(\Lambda_m^0)$ for $KCl$ is higher than that of $NaCl$ because the ionic mobility of $K^+$ is greater than that of $Na^+$ due to smaller hydration of $K^+$ ions.
Since both are $1:1$ electrolytes,the slope $A$ is approximately the same for both,meaning the lines are parallel.
Therefore,the graph where the $KCl$ line is above the $NaCl$ line and they are parallel is correct,which corresponds to option $(B)$.

(C) The elevation in boiling point is given by the formula: $\Delta T_b = K_b \times m$,where $K_b$ is the ebullioscopic constant and $m$ is the molality of the solution.
Since the mass of the solute $(1 \ g)$ and the mass of the solvent $(100 \ g)$ are the same for both solvents $A$ and $B$,the molality $m$ is the same for both solutions $(m_A = m_B)$.
Therefore,the ratio of the elevation in boiling points is directly proportional to the ratio of their ebullioscopic constants:
$\frac{\Delta T_b (A)}{\Delta T_b (B)} = \frac{K_{b(A)}}{K_{b(B)}}$.
Given that the ratio of the ebullioscopic constants is $\frac{K_{b(A)}}{K_{b(B)}} = \frac{1}{5}$,the ratio of the elevation in boiling points is $1:5$.

(A) Cyanobenzene $(C_6H_5CN)$ can be reduced to benzylamine $(C_6H_5CH_2NH_2)$ using strong reducing agents like $LiAlH_4$ or catalytic hydrogenation $(H_2/Ni)$.
Option $A$ involves $HCl/H_2O$ (hydrolysis) which converts the nitrile to a carboxylic acid,and $NaBH_4$ is not strong enough to reduce nitriles to amines.
Option $C$ involves the Stephen reduction $(SnCl_2/HCl)$ which typically reduces nitriles to imines,and $NaBH_4$ is not the standard reagent to complete this specific conversion to a primary amine.
Therefore,options $A$ and $C$ are not standard methods for this preparation.

(C) $C_{60}$ (Buckminsterfullerene) has a truncated icosahedron structure consisting of $20$ hexagons and $12$ pentagons.
White phosphorus $(P_4)$ exists as a tetrahedral molecule where each phosphorus atom is bonded to the other three,forming $4$ triangular faces.

(A) In the linear structure of $D$-glucose $(CHO-(CHOH)_4-CH_2OH)$,there are $4$ chiral carbons (stereo centers) at positions $C_2, C_3, C_4,$ and $C_5$.
In the cyclic structure of $\alpha-D$-glucose,the formation of the hemiacetal linkage creates a new chiral center at the anomeric carbon $(C_1)$.
Thus,the cyclic structure has $5$ stereo centers ($C_1, C_2, C_3, C_4,$ and $C_5$).

(B) The reaction starts with $2$-chloro-$3$-methylbutane.
Treatment with $EtONa$ (a strong base) and heat leads to an $E2$ elimination reaction.
The major product formed is the more substituted alkene,which is $2$-methylbut-$2$-ene (Saytzeff product).
Subsequent addition of $HBr$ to $2$-methylbut-$2$-ene follows Markovnikov's rule,where the $H^+$ adds to the carbon with more hydrogens and the $Br^-$ adds to the more substituted carbon.
This results in the formation of $2$-bromo-$2$-methylbutane as the major product '$Y$'.

(C) Given: Mole fraction of solvent $(X_{solvent})$ = $0.8$.
Since it is an aqueous solution,the solvent is water $(H_2O)$,with molar mass $M_{solvent} = 18 \ g \ mol^{-1}$.
Mole fraction of solute $(X_{solute})$ = $1 - X_{solvent} = 1 - 0.8 = 0.2$.
Let the total number of moles be $n_{total} = 1$. Then $n_{solute} = 0.2 \ mol$ and $n_{solvent} = 0.8 \ mol$.
Mass of solvent = $n_{solvent} \times M_{solvent} = 0.8 \ mol \times 18 \ g \ mol^{-1} = 14.4 \ g = 0.0144 \ kg$.
Molality $(m)$ = $\frac{n_{solute}}{\text{Mass of solvent in } kg} = \frac{0.2 \ mol}{0.0144 \ kg} = 13.88 \ mol \ kg^{-1}$.

(B) The reaction proceeds in two main steps:
$1$. Reimer-Tiemann reaction: $4$-chlorophenol reacts with $CHCl_3$ and aqueous $NaOH$ to form $5$-chloro$-2-$hydroxybenzaldehyde.
$2$. Cannizzaro reaction: The aldehyde formed reacts with $HCHO$ and concentrated $NaOH$ (cross-Cannizzaro reaction). The aldehyde is reduced to the corresponding alcohol ($5$-chloro$-2-$hydroxybenzyl alcohol),and $HCHO$ is oxidized to sodium formate $(HCOONa)$.
$3$. Acidification: Finally,treatment with $H_3O^+$ converts the phenoxide and formate salt into $5$-chloro$-2-$hydroxybenzyl alcohol and formic acid $(HCOOH)$.

(D) The Crystal Field Stabilization Energy $(CFSE)$ for an octahedral complex is given by $CFSE = (-0.4n_{t2g} + 0.6n_{eg})\Delta_o$.
For $[Fe(phen)_3]^{2+}$,the central metal is $Fe^{2+}$ ($d^6$ configuration). Since $phen$ is a strong field ligand,it forms a low-spin complex with $t_{2g}^6 e_g^0$ configuration.
$CFSE = (-0.4 \times 6 + 0.6 \times 0)\Delta_o = -2.4\Delta_o$.
Upon oxidation to $Fe^{3+}$ ($d^5$ configuration),the complex becomes $[Fe(phen)_3]^{3+}$ with $t_{2g}^5 e_g^0$ configuration.
$CFSE = (-0.4 \times 5 + 0.6 \times 0)\Delta_o = -2.0\Delta_o$.
However,the question implies a significant loss or change in stability. Let's check $[Co(phen)_3]^{2+}$ $(d^7)$: $t_{2g}^6 e_g^1$ $(CFSE = -1.8\Delta_o)$. Oxidation to $Co^{3+}$ $(d^6)$: $t_{2g}^6 e_g^0$ $(CFSE = -2.4\Delta_o)$.
Actually,for $Fe^{2+}$ $(d^6)$ to $Fe^{3+}$ $(d^5)$,the loss of an electron from the $t_{2g}$ orbital significantly reduces the stabilization energy. Thus,$[Fe(phen)_3]^{2+}$ is the correct answer.

(B) Peptization is the process of converting a freshly prepared precipitate into a colloidal solution by adding a suitable electrolyte,known as a peptizing agent.

(B) Thermosetting polymers are cross-linked polymers that undergo extensive cross-linking in molds during molding,which makes them hard and infusible.
Bakelite is a classic example of a thermosetting polymer,formed by the condensation reaction of phenol and formaldehyde.

(A) In an $fcc$ lattice,there are $8$ tetrahedral voids,each located on the body diagonals at a distance of $\frac{1}{4}$ of the body diagonal length from each corner.
Each body diagonal has a length of $\sqrt{3}a$.
The distance of each tetrahedral void from the nearest corner is $\frac{\sqrt{3}a}{4}$.
There are two tetrahedral voids on each body diagonal.
The distance between these two nearest tetrahedral voids is $\frac{\sqrt{3}a}{4} + \frac{\sqrt{3}a}{4} = \frac{\sqrt{3}a}{2}$ is incorrect; let us re-evaluate.
The distance between two nearest tetrahedral voids is the distance between the two voids located on the same body diagonal,which is $\frac{\sqrt{3}a}{2}$ is not correct. Actually,the distance between two nearest tetrahedral voids is $\frac{a}{2}$.

(B) The froth floatation process is a method used for the concentration of sulphide ores.
It is said that the idea of this process was inspired by a fisherwoman who noticed that sulphide minerals,being hydrophobic,float on water while gangue particles sink,similar to how oily substances float on water.
Thus,$X$ is a fisherwoman and $Y$ is the concentration of ores.

(D) The rate expression for the reaction $x A \longrightarrow y B$ is given by: $-\frac{1}{x} \frac{d[A]}{dt} = \frac{1}{y} \frac{d[B]}{dt}$.
Given the equation: $\log_{10} \left[ -\frac{d[A]}{dt} \right] = \log_{10} \left[ \frac{d[B]}{dt} \right] + 0.3010$.
Since $\log_{10}(2) \approx 0.3010$,we can write: $\log_{10} \left[ -\frac{d[A]}{dt} \right] = \log_{10} \left[ 2 \times \frac{d[B]}{dt} \right]$.
Taking antilog on both sides: $-\frac{d[A]}{dt} = 2 \frac{d[B]}{dt}$,which implies $-\frac{1}{2} \frac{d[A]}{dt} = \frac{d[B]}{dt}$.
Comparing this with the general rate expression,we get $x = 2$ and $y = 1$.
Thus,the reaction is $2 A \longrightarrow B$. For $A = C_2H_4$ and $B = C_4H_8$,the reaction is $2 C_2H_4 \longrightarrow C_4H_8$.