JEE Main 2022 Chemistry Question Paper with Answer and Solution

(A) The reaction involves a bimolecular nucleophilic substitution $(SN^{2})$ at the benzylic carbon. The nucleophile $PhS^-$ attacks the benzylic carbon from the side opposite to the leaving group $(I^-)$,resulting in an inversion of configuration. The fluorine atom on the benzene ring is not replaced because the reaction conditions favor $SN^{2}$ at the aliphatic benzylic position rather than nucleophilic aromatic substitution $(S_NAr)$ at the ring,especially given the specific stereochemical outcome required.

(A) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$.
Substituting $\Delta p = m \Delta v$,we get: $m \Delta v \Delta x = \frac{h}{4 \pi}$.
Given: $\Delta v = 2.4 \times 10^{-26} \, m \, s^{-1}$,$\Delta x = 10^{-7} \, m$,and $h = 6.626 \times 10^{-34} \, J \, s$.
$m \times (2.4 \times 10^{-26}) \times (10^{-7}) = \frac{6.626 \times 10^{-34}}{4 \times 3.14159}$.
$m \times 2.4 \times 10^{-33} = \frac{6.626 \times 10^{-34}}{12.566}$.
$m = \frac{6.626 \times 10^{-34}}{12.566 \times 2.4 \times 10^{-33}} = \frac{6.626}{30.1584} \times 10^{-1} \, kg$.
$m \approx 0.2197 \times 10^{-1} \, kg = 0.02197 \, kg$.
Converting to grams: $m = 0.02197 \times 1000 \, g = 21.97 \, g$.
The nearest integer is $22$.

(B) The reaction is: $2 NOCl_{(g)} \rightleftharpoons 2 NO_{(g)} + Cl_{2(g)}$
Initial concentration: $[NOCl] = 2.0 \ M$,$[NO] = 0 \ M$,$[Cl_2] = 0 \ M$
At equilibrium: $[NOCl] = (2 - x) \ M$,$[NO] = x \ M$,$[Cl_2] = \frac{x}{2} \ M$
Given that at equilibrium,$[NO] = 0.4 \ M$,so $x = 0.4 \ M$.
Therefore,$[NOCl]_{eq} = 2 - 0.4 = 1.6 \ M$ and $[Cl_2]_{eq} = \frac{0.4}{2} = 0.2 \ M$.
The equilibrium constant $K_c$ is given by:
$K_c = \frac{[NO]^2 [Cl_2]}{[NOCl]^2} = \frac{(0.4)^2 \times (0.2)}{(1.6)^2}$
$K_c = \frac{0.16 \times 0.2}{2.56} = \frac{0.032}{2.56} = 0.0125$
$K_c = 125 \times 10^{-4}$

(C) Let the atomic weight of $A$ be $A$ and the atomic weight of $B$ be $B$.
The molar mass of $A_2 B$ is $(2A + B) \ g/mol$.
The molar mass of $AB_3$ is $(A + 3B) \ g/mol$.
Given that $0.15 \ mol$ of $A_2 B$ and $0.15 \ mol$ of $AB_3$ have equal mass:
$0.15 \times (2A + B) = 0.15 \times (A + 3B)$
Dividing both sides by $0.15$:
$2A + B = A + 3B$
Rearranging the terms:
$2A - A = 3B - B$
$A = 2B$
Thus,the atomic weight of $A$ is $2$ times the atomic weight of $B$.

(C) To find the total number of stereoisomers of dimethylcyclopentane,we analyze the structural isomers:
$1$. $1,1$-dimethylcyclopentane: This molecule has a plane of symmetry and no chiral centers,so it has $0$ stereoisomers.
$2$. $1,2$-dimethylcyclopentane: This molecule has $2$ chiral centers. It exists as a cis-isomer (which is meso,$1$ isomer) and a trans-isomer (which is a pair of enantiomers,$2$ isomers). Total = $3$ stereoisomers.
$3$. $1,3$-dimethylcyclopentane: This molecule also has $2$ chiral centers. It exists as a cis-isomer (which is meso,$1$ isomer) and a trans-isomer (which is a pair of enantiomers,$2$ isomers). Total = $3$ stereoisomers.
Summing these up,the total number of stereoisomers is $0 + 3 + 3 = 6$.

(D)

$C$: $74\,\%$	$\frac{74}{12} = 6.16$; $\frac{6.16}{1.23} \approx 5$
$N$: $17.3\,\%$	$\frac{17.3}{14} = 1.23$; $\frac{1.23}{1.23} = 1$
$H$: $8.7\,\%$	$\frac{8.7}{1} = 8.7$; $\frac{8.7}{1.23} \approx 7$

Empirical formula $= C_{5}H_{7}N$
Empirical formula mass $= (5 \times 12) + (7 \times 1) + (1 \times 14) = 60 + 7 + 14 = 81\, g\, mol^{-1}$
Multiplying factor $(n) = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{162}{81} = 2$
Molecular formula $= n \times (\text{Empirical formula}) = 2 \times (C_{5}H_{7}N) = C_{10}H_{14}N_{2}$

(D) The principal quantum number $n$ represents the shell and takes positive integer values $n = 1, 2, 3, \dots$.
$(B)$ The azimuthal quantum number $l$ determines the subshell and takes values $l = 0, 1, 2, \dots, (n-1)$.
$(C)$ The magnetic quantum number $m_l$ determines the orientation of orbitals and takes $(2l+1)$ values ranging from $-l$ to $+l$.
$(D)$ The spin quantum number $m_s$ describes the electron spin,which can be $\pm 1/2$.
$(E)$ For $l = 5$,the number of orbitals is given by $2l + 1 = 2(5) + 1 = 11$.
Since all statements $(A), (B), (C), (D),$ and $(E)$ are correct,the correct option is $(D)$.

(A) The central atom $S$ in $SF_{4}$ has $5$ electron pairs ($4$ bond pairs and $1$ lone pair),resulting in $sp^{3}d$ hybridization.
According to $VSEPR$ theory,the lone pair occupies the equatorial position in a trigonal bipyramidal geometry to minimize repulsion.
In the equatorial position,the lone pair experiences repulsion from two axial bond pairs at $90^{\circ}$ and two equatorial bond pairs at $120^{\circ}$.
Therefore,there are two lone pair-bond pair repulsions at $90^{\circ}$.

(B) The Henderson-Hasselbalch equation for an acidic buffer is given by:
$pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$
First,calculate $pK_{a}$:
$pK_{a} = -\log(K_{a}) = -\log(1.3 \times 10^{-5}) = 5 - \log(1.3) \approx 5 - 0.1139 = 4.8861$
Substitute the values into the equation:
$4 = 4.8861 + \log \frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]}$
$\log \frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]} = 4 - 4.8861 = -0.8861$
Taking the antilog:
$\frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]} = 10^{-0.8861} \approx 0.13$

(B) $Cl_2O_7$ is an acidic oxide because it is a non-metal oxide with a high oxidation state.
$Na_2O$ is a basic oxide as it is an alkali metal oxide.
$Al_2O_3$ is an amphoteric oxide as it reacts with both acids and bases.
$N_2O$ is a neutral oxide.
Therefore,the correct matching is: $A-IV, B-II, C-I, D-III$.

(D) Isotopes are atoms of the same element that have the same atomic number (number of protons and electrons) but different mass numbers due to a different number of neutrons in their nuclei.
Since mass number is the sum of protons and neutrons,a difference in the number of neutrons leads to a difference in atomic mass.

(C) $SO_{3}$ and $SiO_{2}$ are acidic oxides.
$CaO$ is a basic oxide because it is an oxide of an alkaline earth metal.
$Al_{2}O_{3}$ is an amphoteric oxide.

(A) To determine the oxidation states of sulphur in the given oxoacids,we examine their structures:
$1$. $H_{2}S_{2}O_{3}$ (Thiosulphuric acid): The structure contains one central $S$ atom bonded to another $S$ atom. The central $S$ atom is in the $+6$ oxidation state,while the terminal $S$ atom is in the $-2$ oxidation state.
$2$. $H_{2}S_{2}O_{6}$ (Dithionic acid): Both $S$ atoms are in the $+5$ oxidation state.
$3$. $H_{2}S_{2}O_{7}$ (Pyrosulphuric acid): Both $S$ atoms are in the $+6$ oxidation state.
$4$. $H_{2}S_{2}O_{8}$ (Peroxodisulphuric acid): Both $S$ atoms are in the $+6$ oxidation state.
Thus,$H_{2}S_{2}O_{3}$ is the only acid among the options that contains sulphur in two different oxidation states ($-2$ and $+6$).

(D) Photochemical smog is formed by the action of sunlight on unsaturated hydrocarbons and nitrogen oxides $(NO_x)$ in the atmosphere.
It is an oxidizing type of smog,unlike classical smog which is reducing.

(C) $1$. Identify the principal functional group: The aldehyde group $(-CHO)$ has the highest priority and is assigned position $1$.
$2$. Select the longest carbon chain containing the principal functional group and the double bond: The chain has $7$ carbons,making it a heptenal derivative.
$3$. Number the chain: Starting from the aldehyde carbon as $C-1$,the ketone group is at $C-3$,the double bond starts at $C-4$,the methyl group is at $C-4$,and the nitro group is at $C-6$.
$4$. Assemble the name: $4-$methyl$-6-$nitro$-3-$oxohept$-4-$enal.

(B) The ideal gas equation is $PV = nRT$,where $n = \frac{m}{M}$.
Given: $P = 1.5 \, bar$,$V = 416 \, L$,$m = 100 \, g$,$T = 27 + 273 = 300 \, K$,and $R = 0.083 \, L \, bar \, K^{-1} \, mol^{-1}$.
Substituting the values: $1.5 \times 416 = \frac{100}{M} \times 0.083 \times 300$.
$624 = \frac{2490}{M}$.
$M = \frac{2490}{624} \approx 3.99 \, g \, mol^{-1}$.
Rounding to the nearest integer,we get $M = 4 \, g \, mol^{-1}$.

(B) The combustion reaction of magnesium is: $Mg(s) + \frac{1}{2}O_{2}(g) \rightarrow MgO(s)$
The change in the number of gaseous moles is $\Delta n_{g} = n_{p(g)} - n_{r(g)} = 0 - \frac{1}{2} = -0.5 \ mol$.
The relationship between enthalpy change and internal energy change is given by: $\Delta H = \Delta U + \Delta n_{g}RT$.
Given $\Delta H = -601.70 \ kJ \ mol^{-1}$,$R = 8.3 \ J \ K^{-1} \ mol^{-1} = 8.3 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,and $T = 300 \ K$.
Substituting the values: $-601.70 = \Delta U + (-0.5) \times (8.3 \times 10^{-3}) \times 300$.
$-601.70 = \Delta U - 1.245$.
$\Delta U = -601.70 + 1.245 = -600.455 \ kJ \ mol^{-1}$.
The magnitude of the change in internal energy is $|\Delta U| = 600.455 \ kJ \ mol^{-1}$.
Rounding to the nearest integer,we get $600 \ kJ$.

(D) Moles of $CO_{2} = \frac{0.793}{44} \approx 0.01802 \, mol$. Since each $CO_{2}$ molecule contains one $C$ atom,moles of $C = 0.01802 \, mol$.
Weight of $C = 0.01802 \times 12 = 0.21624 \, g$.
Moles of $H_{2}O = \frac{0.442}{18} \approx 0.02456 \, mol$. Since each $H_{2}O$ molecule contains two $H$ atoms,moles of $H = 0.02456 \times 2 = 0.04912 \, mol$.
Weight of $H = 0.04912 \times 1 = 0.04912 \, g$.
Weight of $O = \text{Total weight} - (\text{Weight of } C + \text{Weight of } H)$
Weight of $O = 0.492 - (0.21624 + 0.04912) = 0.492 - 0.26536 = 0.22664 \, g$.
Percentage of $O = \frac{0.22664}{0.492} \times 100 \approx 46.06 \, \%$.
Rounding to the nearest integer,we get $46 \, \%$.

(A) The reaction of the given substrate with $Br_2$ in the presence of $h\nu$ (ultraviolet light) proceeds via a free-radical substitution mechanism.
In this specific molecule,the allylic position (the carbon adjacent to the double bond in the ethyl group) is the most reactive site for radical bromination.
Therefore,the bromine atom replaces one of the hydrogen atoms at the allylic position.
The major product contains $1$ bromine atom.

(A) The balanced chemical equation for the reaction between $KMnO_{4}$ and Mohr's salt $(FeSO_{4} \cdot (NH_{4})_{2}SO_{4} \cdot 6H_{2}O)$ in acidic medium is:
$MnO_{4}^{-} + 5Fe^{2+} + 8H^{+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_{2}O$
Using the equivalence concept:
$n_{factor} \times M_{1} \times V_{1} = n_{factor} \times M_{2} \times V_{2}$
For $KMnO_{4}$,$n_{factor} = 5$ (reduction of $Mn^{+7}$ to $Mn^{+2}$).
For Mohr's salt,$n_{factor} = 1$ (oxidation of $Fe^{+2}$ to $Fe^{+3}$).
$5 \times 0.01 \times V_{1} = 1 \times 0.05 \times 20.0$
$0.05 \times V_{1} = 1.0$
$V_{1} = \frac{1.0}{0.05} = 20.0\, mL$
The volume of $KMnO_{4}$ used is $20.0\, mL$.
The volume left in the $50\, mL$ burette is:
$V_{left} = 50.0 - 20.0 = 30.0\, mL$.

(A) The element $E$ is in period $4$ and group $16$. The group $16$ elements are oxygen family (chalcogens).
The element $E$ is Selenium $(Se)$,which has the atomic number $34$ and electronic configuration $[Ar] 3 d^{10} 4 s^{2} 4 p^{4}$.
The element just above $E$ in group $16$ is Sulfur $(S)$,which is in period $3$.
The valence shell electronic configuration of Sulfur $(S)$ is $3 s^{2} 3 p^{4}$.

(B) The reaction between dihydrogen $(H_{2})$ and copper$(II)$ oxide $(CuO)$ is a redox reaction where $H_{2}$ acts as a reducing agent.
$CuO(s) + H_{2}(g) \xrightarrow{\Delta} Cu(s) + H_{2}O(l)$
Thus,$CuO$ is reduced to metallic copper $(Cu)$.

(C) In Thin Layer Chromatography $(TLC)$,the separated components are often invisible to the naked eye. To visualize them,several techniques are used:
$1$. $I_{2}$ (Solid) chambers: Iodine vapors adsorb onto the components,making them visible.
$2$. $U.V.$ Light: Many organic compounds fluoresce under $U.V.$ light.
$3$. Spraying of an appropriate reagent: Chemical reagents react with the components to produce colored spots.
However,adding a visualization agent directly to the mobile phase is not a standard technique for spotting components on a $TLC$ plate,as it would interfere with the separation process itself.

(B) compound is aromatic if it is cyclic,planar,fully conjugated,and follows $H$ückel's rule ($4n+2$ $\pi$ electrons).
$A$ (Cyclopentadienyl anion): It has $6$ $\pi$ electrons ($4$ from double bonds + $2$ from the lone pair on the $sp^3$ carbon,which becomes $sp^2$ due to conjugation). It is cyclic,planar,and aromatic.
$B$ (Tropylium cation): It has $6$ $\pi$ electrons. It is cyclic,planar,and aromatic.
$C$ (Cyclooctatetraene derivative): It is non-planar (tub-shaped) to avoid anti-aromaticity,hence it is non-aromatic.
$D$ (Cycloheptatrienyl anion): It has $8$ $\pi$ electrons ($4n$ system),making it anti-aromatic.
Therefore,only $A$ and $B$ are aromatic.

(C) The reaction is an electrophilic addition of $HBr$ to the alkene $Ph-CH=CH-CH_2Br$.
$1$. The proton $(H^+)$ from $HBr$ attacks the double bond to form the most stable carbocation.
$2$. The carbocation formed at the benzylic position $(Ph-CH^+-CH_2-CH_2Br)$ is resonance-stabilized by the phenyl group.
$3$. The bromide ion $(Br^-)$ then attacks this stable carbocation to form the major product.
$4$. Thus,the major product is $Ph-CH(Br)-CH_2-CH_2Br$.

(A) In $Lassaigne's$ test for sulphur,the sodium fusion extract is treated with sodium nitroprusside.
The reaction is: $Na_2S + Na_2[Fe(CN)_5NO] \rightarrow Na_4[Fe(CN)_5(NOS)]$.
The complex $Na_4[Fe(CN)_5(NOS)]$ is responsible for the deep violet or purple colouration.

(A) The work function $\phi$ is related to the threshold wavelength $\lambda$ by the equation: $\phi = \frac{hc}{\lambda}$.
Given $\phi = 6.63 \times 10^{-19} \ J$,$h = 6.63 \times 10^{-34} \ J \ s$,and $c = 3 \times 10^{8} \ m \ s^{-1}$.
Substituting the values: $6.63 \times 10^{-19} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda}$.
$\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{6.63 \times 10^{-19}} \ m$.
$\lambda = 3 \times 10^{-7} \ m$.
Converting to nanometers $(nm)$: $\lambda = 3 \times 10^{-7} \times 10^{9} \ nm = 300 \ nm$.

(A) The central atom $P$ in $PF_{5}$ has $5$ valence electrons and forms $5$ sigma bonds with $5$ $F$ atoms.
The steric number is calculated as: $\text{Steric Number} = \frac{1}{2} (V + M - C + A) = \frac{1}{2} (5 + 5 - 0 + 0) = 5$.
$A$ steric number of $5$ corresponds to $sp^{3}d$ hybridization.
Comparing $sp^{3}d$ with $sp^{x}d^{y}$,we get $x=3$ and $y=1$.
Therefore,the value of $y$ is $1$.

(C) For free expansion into vacuum,the external pressure $P_{ext} = 0$.
Since work done $w = -P_{ext} \Delta V$,we have $w = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
For an isothermal expansion of an ideal gas,$\Delta U = 0$.
Therefore,$0 = q + 0$,which implies $q = 0$.

(C) The dissociation of the salt is given by: $A_{2}X_{3(s)} \rightleftharpoons 2A^{3+}_{(aq)} + 3X^{2-}_{(aq)}$.
Let the solubility be $s \ mol \ L^{-1}$. Then $[A^{3+}] = 2s$ and $[X^{2-}] = 3s$.
The solubility product $K_{sp} = (2s)^2(3s)^3 = 108s^5 = 1.1 \times 10^{-23}$.
$s^5 = \frac{1.1 \times 10^{-23}}{108} \approx 1.018 \times 10^{-25}$.
$s \approx 10^{-5} \ mol \ L^{-1} = 10^{-5} \ mol \ (10^{-3} \ m^3)^{-1} = 0.01 \ mol \ m^{-3}$.
For a sparingly soluble salt,the molar conductivity $\Lambda_m \approx \Lambda_m^{\infty} = \frac{\kappa}{C}$,where $\kappa$ is the specific conductance and $C$ is the concentration in $mol \ m^{-3}$.
$\Lambda_m^{\infty} = \frac{3 \times 10^{-5} \ S \ m^{-1}}{0.01 \ mol \ m^{-3}} = 3 \times 10^{-3} \ S \ m^2 \ mol^{-1}$.
Comparing this with $x \times 10^{-3} \ S \ m^2 \ mol^{-1}$,we get $x = 3$.

(D) The molar mass of $AgBr = 108 + 80 = 188 \ g/mol$.
The mass of $Br$ in $0.40 \ g$ of $AgBr = \frac{80}{188} \times 0.40 \ g$.
Percentage of $Br = \frac{\text{mass of } Br}{\text{mass of organic compound}} \times 100$.
Percentage of $Br = \frac{(\frac{80}{188} \times 0.40)}{0.5} \times 100$.
Percentage of $Br = \frac{32}{188 \times 0.5} \times 100 = \frac{32}{94} \times 100 \approx 34.04 \ \%$.
The nearest integer is $34 \ \%$.

(B) First,calculate the value of the expression: $\frac{0.02858 \times 0.112}{0.5702} \approx 0.0056166$
In multiplication and division,the result should be reported to the same number of significant figures as the term with the fewest significant figures.
$0.02858$ has $4$ significant figures,$0.112$ has $3$ significant figures,and $0.5702$ has $4$ significant figures.
The least number of significant figures is $3$.
Rounding $0.0056166$ to $3$ significant figures gives $0.00562$ (Note: The provided options suggest $0.00561$ is the intended answer based on standard rounding of the specific calculation result $0.0056166$ which is closer to $0.00562$,but given the options,$0.00561$ is the closest choice).

(B) For the $2s$ orbital,the number of radial nodes is calculated using the formula: $\text{Radial nodes} = n - l - 1$.
For $2s$,$n = 2$ and $l = 0$,so $\text{Radial nodes} = 2 - 0 - 1 = 1$.
This means there is one point where the probability density $\psi^{2}(r)$ becomes zero.
Also,the probability density $\psi^{2}(r)$ is always non-negative,meaning it cannot go below the $r$-axis.
Among the given plots,option $B$ shows the probability density starting from a high value at the nucleus,touching the $r$-axis at one radial node,and then rising again before decaying,which correctly represents the $2s$ orbital.

(B) $1$. Calculate the total number of electrons in each species:
$CH_4$: $6 + 4(1) = 10$ electrons.
$NH_4^+$: $7 + 4(1) - 1 = 10$ electrons.
$BH_4^-$: $5 + 4(1) + 1 = 10$ electrons.
Since all species have $10$ electrons,they are isoelectronic.
$2$. Determine the geometry:
All three species have a central atom bonded to $4$ hydrogen atoms with no lone pairs on the central atom. According to $VSEPR$ theory,they all exhibit $sp^3$ hybridization and possess a tetrahedral geometry.

(A) The reaction is: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$.
Initial moles: $n(PCl_5) = 5.0 \, mol$,$n(Ar) = 4.0 \, mol$. Total initial moles = $9.0 \, mol$.
Initial pressure $(P_i)$ using $PV = nRT$: $P_i = \frac{9.0 \times 0.082 \times 610}{100} = 4.5 \, atm$.
Initial partial pressures: $P_{PCl_5} = \frac{5}{9} \times 4.5 = 2.5 \, atm$,$P_{Ar} = \frac{4}{9} \times 4.5 = 2.0 \, atm$.
At equilibrium:
$PCl_5 \rightleftharpoons PCl_3 + Cl_2$
Initial: $2.5 \, atm \quad 0 \quad 0$
Change: $-x \quad +x \quad +x$
Equilibrium: $(2.5 - x) \quad x \quad x$
Total pressure at equilibrium: $P_{total} = P_{PCl_5} + P_{PCl_3} + P_{Cl_2} + P_{Ar} = 6.0 \, atm$.
$(2.5 - x) + x + x + 2.0 = 6.0$
$4.5 + x = 6.0 \implies x = 1.5 \, atm$.
Equilibrium partial pressures: $P_{PCl_5} = 2.5 - 1.5 = 1.0 \, atm$,$P_{PCl_3} = 1.5 \, atm$,$P_{Cl_2} = 1.5 \, atm$.
$K_p = \frac{P_{PCl_3} \times P_{Cl_2}}{P_{PCl_5}} = \frac{1.5 \times 1.5}{1.0} = 2.25$.

(A) The electronic configuration of $N$ $(Z=7)$ is $1s^2 2s^2 2p^3$,which has a stable half-filled $p$-orbital.
The electronic configuration of $O$ $(Z=8)$ is $1s^2 2s^2 2p^4$.
Due to the half-filled stability of the $2p$ subshell in nitrogen,it requires more energy to remove an electron compared to oxygen.
In oxygen,the $2p^4$ configuration means that two electrons must occupy the same $2p$ orbital,leading to increased inter-electronic repulsion,which makes it easier to remove an electron.
Thus,both Assertion $A$ and Reason $R$ are correct,and $R$ is the correct explanation for $A$.

(A) The chemical formulas for the given substances are:
$1$. Baking soda: $NaHCO_3$ (contains bicarbonate anion,$HCO_3^-$).
$2$. Washing soda: $Na_2CO_3 \cdot 10H_2O$ (contains carbonate anion,$CO_3^{2-}$).
$3$. Caustic soda: $NaOH$ (contains hydroxide anion,$OH^-$).
Therefore,the carbonate anion $(CO_3^{2-})$ is present only in washing soda.

(C) For $BrF_{3}$ molecule:
Central atom is $Br$ (Bromine),which has $7$ valence electrons.
Number of bond pairs $= 3$ (three $Br-F$ bonds).
Number of lone pairs $= (7 - 3) / 2 = 2$.
Steric number $= 3$ (bond pairs) $+ 2$ (lone pairs) $= 5$,which corresponds to $sp^{3}d$ hybridization.
Due to the presence of $2$ lone pairs,the geometry is trigonal bipyramidal,but the shape is bent $T$-shape.

(D) Borax $(Na_2B_4O_7 \cdot 10H_2O)$ dissolves in water to form orthoboric acid $(H_3BO_3)$ and sodium hydroxide $(NaOH)$.
$Na_2B_4O_7 + 7H_2O \rightarrow 2NaOH + 4H_3BO_3$.
Since $NaOH$ is a strong base and $H_3BO_3$ is a very weak acid,the resulting aqueous solution is strongly basic.

(D) Acid rain is primarily caused by the oxidation of $SO_{2}$ in the atmosphere in the presence of moisture and oxygen.
The overall chemical reaction is: $2 SO_{2} + O_{2} + 2 H_{2}O \rightarrow 2 H_{2}SO_{4}$.

(D) The stability of a carbocation is increased by the presence of electron-donating groups that can stabilize the positive charge through resonance.
In the given options,the methoxy group $(-OCH_3)$ acts as a strong electron-donating group due to the $+M$ (mesomeric) effect.
Among the options,the carbocation $CH_3O-CH=CH-CH=CH^+$ is the most stable because the positive charge is delocalized over a larger conjugated system,and the $+M$ effect of the $-OCH_3$ group is effectively transmitted through the extended conjugation to stabilize the carbocation.

(B) In the Friedel-Crafts alkylation of benzene with $n$-propyl chloride $(CH_3CH_2CH_2Cl)$ in the presence of anhydrous $AlCl_3$,the primary carbocation $(CH_3CH_2CH_2^+)$ is initially formed.
This primary carbocation is unstable and undergoes a $1,2$-hydride shift to form a more stable secondary carbocation,which is the isopropyl carbocation $(CH_3-\stackrel{\oplus}{C}H-CH_3)$.

(D) The number of moles of water $(n)$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.90 \ g}{18 \ g \ mol^{-1}} = 0.05 \ mol$.
The temperature $(T)$ is $27 + 273 = 300 \ K$.
The pressure $(P)$ in $atm$ is $\frac{32.0 \ Torr}{760 \ Torr \ atm^{-1}} = \frac{32}{760} \ atm$.
Using the ideal gas equation $PV = nRT$,we find the volume $(V)$:
$V = \frac{nRT}{P} = \frac{0.05 \times 0.082 \times 300}{32 / 760} = \frac{0.05 \times 0.082 \times 300 \times 760}{32} = 29.21 \ L$.
The nearest integer is $29$.

(D) $1$. Calculate the number of moles of $N_{2}O$: $n = \frac{2.2 \, g}{44 \, g \, mol^{-1}} = 0.05 \, mol$.
$2$. Calculate the change in enthalpy $(\Delta H)$ at constant pressure: $\Delta H = n C_{p} \Delta T = 0.05 \, mol \times 100 \, J \, K^{-1} \, mol^{-1} \times (270 - 310) \, K = 5 \times (-40) = -200 \, J$.
$3$. Calculate the work done $(w)$ during the process: $w = -P_{ext} \Delta V = -1 \, atm \times (167.75 - 217.1) \, mL = -1 \times (-49.35) \, mL \cdot atm = 49.35 \, mL \cdot atm$.
$4$. Convert $w$ to Joules: $1 \, L \cdot atm = 101.3 \, J$,so $w = 49.35 \times 10^{-3} \, L \times 101.3 \, J \, L^{-1} \approx 5 \, J$.
$5$. Use the relation $\Delta H = \Delta U + P \Delta V$,which implies $\Delta U = \Delta H - P \Delta V = \Delta H + w$.
$6$. $\Delta U = -200 \, J + 5 \, J = -195 \, J$.
$7$. Since $\Delta U = -x \, J$,we have $-195 = -x$,so $x = 195$.

(A) Step $1$: Oxidation of cyclohexanol with $K_2Cr_2O_7$ gives cyclohexanone.
Step $2$: Reaction of cyclohexanone with phenylmagnesium bromide $(C_6H_5MgBr)$ followed by hydrolysis gives $1-$phenylcyclohexanol.
Step $3$: Acid-catalyzed dehydration of $1-$phenylcyclohexanol with $H^+$,heat gives $1-$phenylcyclohexene as the major product '$X$'.
Step $4$: In $1-$phenylcyclohexene,the phenyl ring has $6$ $sp^{2}$ carbons and the double bond in the cyclohexene ring involves $2$ $sp^{2}$ carbons.
Total $sp^{2}$ hybridised carbon atoms = $6 + 2 = 8$.

(C) $1$. Calculate the molar mass of $Fe_{3}O_{4} = (3 \times 56) + (4 \times 16) = 168 + 64 = 232 \ g \ mol^{-1}$.
$2$. Calculate the moles of $Fe_{3}O_{4} = \frac{4.640 \times 10^{3} \ g}{232 \ g \ mol^{-1}} = 20 \ mol$.
$3$. Calculate the molar mass of $CO = 12 + 16 = 28 \ g \ mol^{-1}$.
$4$. Calculate the moles of $CO = \frac{2.520 \times 10^{3} \ g}{28 \ g \ mol^{-1}} = 90 \ mol$.
$5$. According to the stoichiometry,$1 \ mol$ of $Fe_{3}O_{4}$ requires $4 \ mol$ of $CO$. Thus,$20 \ mol$ of $Fe_{3}O_{4}$ requires $20 \times 4 = 80 \ mol$ of $CO$.
$6$. Since we have $90 \ mol$ of $CO$ (which is more than $80 \ mol$),$Fe_{3}O_{4}$ is the limiting reagent.
$7$. From the equation,$1 \ mol$ of $Fe_{3}O_{4}$ produces $3 \ mol$ of $Fe$. Therefore,$20 \ mol$ of $Fe_{3}O_{4}$ produces $20 \times 3 = 60 \ mol$ of $Fe$.
$8$. Mass of $Fe$ produced $= 60 \ mol \times 56 \ g \ mol^{-1} = 3360 \ g$.

(D) $(A)$ The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$ due to the extra stability of half-filled $d$-orbitals. This statement is correct.
$(B)$ The magnetic quantum number $m_l$ ranges from $-\ell$ to $+\ell$, including zero, so it can have negative values. This statement is correct.
$(C)$ According to the Aufbau principle, electrons fill orbitals in the order of increasing energy in the ground state. This statement is correct.
$(D)$ The total number of nodes in an orbital is given by $n-1$, where $n$ is the principal quantum number. The statement provided $n-2$ is incorrect.
Therefore, statements $(A)$, $(B)$, and $(C)$ are correct.

(A) According to $Fajan's$ rule,the covalent character of an ionic bond increases with the decrease in the size of the cation.
The order of the size of the alkali metal cations is: $Li^+ < Na^+ < K^+ < Cs^+$.
Since the size of the cation increases from $Li^+$ to $Cs^+$,the polarizing power decreases.
Therefore,the decreasing order of covalent character is: $LiCl > NaCl > KCl > CsCl$,which corresponds to $(A) > (B) > (C) > (D)$.

(D) The solubility of a sparingly soluble salt like $AgCl$ is governed by the common ion effect.
In the presence of common ions like $Cl^-$ (from $KCl$ or $HCl$) or $Ag^+$ (from $AgNO_3$),the solubility of $AgCl$ decreases due to the common ion effect.
In deionized water,there are no common ions present to suppress the dissociation of $AgCl$.
Therefore,the solubility of $AgCl$ is maximum in deionized water.

(C) The reducing action of $H_2O_2$ involves its oxidation to $O_2$ $(H_2O_2 \rightarrow O_2 + 2H^{+} + 2e^{-})$.
In option $A$,$H_2O_2$ acts as a reducing agent in an acidic medium.
In option $C$,$H_2O_2$ reduces $MnO_4^{-}$ to $MnO_2$ in a basic medium,as indicated by the production of $OH^{-}$ ions.
Therefore,option $C$ represents the reducing ability of $H_2O_2$ in a basic medium.

(A) $1$. Compound $'P'$ is phenol $(C_6H_5OH)$.
$2$. Nitration of phenol with dil. $HNO_3$ gives a mixture of $o$-nitrophenol and $p$-nitrophenol.
$3$. $o$-Nitrophenol shows intramolecular $H$-bonding and is steam volatile,while $p$-nitrophenol shows intermolecular $H$-bonding and is not steam volatile.
$4$. Reaction of phenol with conc. $HNO_3$ yields $2,4,6$-trinitrophenol,commonly known as picric acid.
$5$. The chemical formula of picric acid is $C_6H_3N_3O_7$.
$6$. The number of oxygen atoms in picric acid $('C')$ is $7$.

(B) The base present only in $RNA$ is Uracil.
$A$ nucleotide consists of a nitrogenous base,a ribose sugar,and a phosphate group.
$1$. Uracil contains $2$ oxygen atoms.
$2$. Ribose sugar contains $5$ oxygen atoms.
$3$. The phosphate group $(PO_4^{3-})$ contains $4$ oxygen atoms.
When these combine to form a nucleotide (Uridylic acid),water molecules are removed during the formation of the glycosidic bond and the ester bond.
- One water molecule is removed between the base and the sugar.
- One water molecule is removed between the sugar and the phosphate group.
Total oxygen atoms = (Oxygen in Uracil + Oxygen in Ribose + Oxygen in Phosphate) - (Oxygen in $2$ water molecules)
Total oxygen atoms = $(2 + 5 + 4) - 2 = 11 - 2 = 9$.
Therefore,the number of oxygen atoms in the nucleotide is $9$.

(D) Statement $I$ is incorrect because the leaching of gold with cyanide ion requires the presence of air or $O_2$ as an oxidizing agent to convert $Au$ to $Au^+$,forming the complex $[Au(CN)_2]^-$. It does not form $Au(III)$ complex.
Statement $II$ is correct because in the displacement reaction,zinc acts as a reducing agent and is oxidized to $[Zn(CN)_4]^{2-}$ while gold is reduced from $Au^+$ to $Au$.

(B) When an aqueous solution of ammonium chloride $(NH_4Cl)$ is treated with sodium nitrite $(NaNO_2)$,ammonium nitrite $(NH_4NO_2)$ is formed as an intermediate.
$NH_4Cl(aq) + NaNO_2(aq) \rightarrow NH_4NO_2(aq) + NaCl(aq)$
Ammonium nitrite is unstable and decomposes upon heating to produce nitrogen gas $(N_2)$ and water $(H_2O)$.
$NH_4NO_2(aq) \rightarrow N_2(g) + 2H_2O(l)$
Thus,the gas produced is $N_2$.

(A) Fluorine is the most electronegative element and has the smallest atomic size among all halogens.
Due to its small size and high electronegativity,it cannot accommodate multiple oxygen atoms or exhibit positive oxidation states,thus it forms only one oxoacid,which is hypofluorous acid $(HOF)$.
Therefore,both Assertion $A$ and Reason $R$ are correct,and $R$ is the correct explanation of $A$.

(C) The standard electrode potential $(E^{\circ})$ for $M^{2+}/M$ in the $3d$ series depends on the sum of enthalpy of sublimation,ionization enthalpy,and hydration enthalpy.
$Cr^{2+}/Cr \rightarrow -0.90 \, V$
$Fe^{2+}/Fe \rightarrow -0.44 \, V$
$Cu^{2+}/Cu \rightarrow +0.34 \, V$
$Zn^{2+}/Zn \rightarrow -0.76 \, V$
Among the given options,$Cu$ has the highest positive value $(+0.34 \, V)$,which is due to its high enthalpy of atomization and low hydration enthalpy,making it the only metal in the $3d$ series with a positive $E^{\circ}$ value for the $M^{2+}/M$ couple.

(C) The electronic configuration of $Tb$ $(Z=65)$ is $[Xe] 4f^9 6s^2$. Therefore,$Tb^{4+}$ has the configuration $[Xe] 4f^7$,which is half-filled.
The electronic configuration of $Yb$ $(Z=70)$ is $[Xe] 4f^{14} 6s^2$. Therefore,$Yb^{2+}$ has the configuration $[Xe] 4f^{14}$,which is completely filled.
Thus,$Tb^{4+}$ and $Yb^{2+}$ correspond to half-filled and completely filled $f$ orbitals,respectively.

(B) The magnetic moment $\mu$ is given by $\sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$(A)$ $[FeF_{6}]^{3-}$: $Fe^{3+}$ is $3d^{5}$. $F^-$ is a weak field ligand,so electrons remain unpaired. $n = 5$.
$(B)$ $[Fe(CN)_{6}]^{3-}$: $Fe^{3+}$ is $3d^{5}$. $CN^-$ is a strong field ligand,causing pairing. $n = 1$.
$(C)$ $[MnCl_{6}]^{3-}$: $Mn^{3+}$ is $3d^{4}$. High spin complex means $n = 4$.
$(D)$ $[Mn(CN)_{6}]^{3-}$: $Mn^{3+}$ is $3d^{4}$. $CN^-$ is a strong field ligand,causing pairing. $n = 2$.
Comparing $n$ values: $B(n=1) < D(n=2) < C(n=4) < A(n=5)$.
Therefore,the increasing order of magnetic moments is $B < D < C < A$.

(D) Step $1$: Electrophilic aromatic substitution $(Br_2, Fe)$ on ethylbenzene gives $p$-bromoethylbenzene as the major product due to the ortho/para directing nature of the ethyl group.
Step $2$: Benzylic halogenation $(Cl_2, \Delta)$ occurs at the benzylic position of the ethyl group,yielding $1-(4-bromophenyl)-1-chloroethane$.
Step $3$: Dehydrohalogenation using $alc. KOH$ (elimination reaction) removes $HCl$ from the side chain to form the alkene,resulting in $1-bromo-4-ethenylbenzene$ (also known as $p$-bromostyrene).

(A) The correct matches are:
$A$. Phenol to Salicylaldehyde is the Reimer-Tiemann reaction,which uses $CHCl_3/NaOH$ $(IV)$.
$B$. Phenol to Benzene is the reduction of phenol using $Zn$ dust $(III)$.
$C$. Phenol to $p$-Benzoquinone is the oxidation of phenol using $Na_2Cr_2O_7/H_2SO_4$ $(II)$.
$D$. Phenol to $p$-Bromophenol is the bromination of phenol in a non-polar solvent like $CS_2$ $(I)$.
Therefore,the correct matching is $A-IV, B-III, C-II, D-I$.

(D) The six isomers of diaminobenzoic acid are decarboxylated to form phenylenediamine isomers.
$1$. The three isomers that yield $m$-phenylenediamine $(A)$ are $2,4$-diaminobenzoic acid,$2,6$-diaminobenzoic acid,and $3,5$-diaminobenzoic acid.
$2$. The two isomers that yield $o$-phenylenediamine $(B)$ are $2,3$-diaminobenzoic acid and $3,4$-diaminobenzoic acid.
$3$. The one isomer that yields $p$-phenylenediamine $(C)$ is $2,5$-diaminobenzoic acid.
$p$-phenylenediamine $(C)$ has a melting point of $142^{\circ} C$.

(C) Buna-$N$ is a synthetic rubber.
It is obtained by the copolymerization of $1, 3-$butadiene and acrylonitrile in the presence of a peroxide catalyst.
The $N$ in Buna-$N$ stands for acrylonitrile.

(C) Maltose is a disaccharide composed of two $\alpha$-$D$-glucose units.
These units are linked by an $\alpha$-glycosidic linkage between $C_1$ of one glucose unit and $C_4$ of the other.
Since one of the glucose units has a free hemiacetal group at $C_1$,it can act as a reducing agent,making maltose a reducing sugar.
Therefore,Statement $I$ is true.
Statement $II$ is incorrect because maltose consists of two $\alpha$-$D$-glucose units (not $\alpha$ and $\beta$) linked at $C_1$ and $C_4$ (not $C_1$ and $C_6$),and it is a reducing sugar (not non-reducing).

(A) The correct matches are as follows:

$A$. Antipyretic	$III$. Reduces fever
$B$. Analgesic	$I$. Reduces pain
$C$. Tranquilizer	$II$. Reduces stress
$D$. Antacid	$IV$. Reduces acidity

Thus,the correct sequence is $A-III, B-I, C-II, D-IV$.

(D) $CO_3^{2-}$ reacts with dil. $H_2SO_4$ to evolve $CO_2$ gas,which turns lime water milky $(IV)$.
$S^{2-}$ reacts with dil. $H_2SO_4$ to evolve $H_2S$ gas,which turns lead acetate paper black $(I)$.
$SO_3^{2-}$ reacts with dil. $H_2SO_4$ to evolve $SO_2$ gas,which turns acidified potassium dichromate solution green $(II)$.
$NO_2^{-}$ reacts with dil. $H_2SO_4$ to evolve $NO_2$ gas (brown fumes),which turns acidified $KI$ solution containing starch blue $(III)$.
Therefore,the correct matching is $A-IV, B-I, C-II, D-III$.

(D) Given:
Mass of solute $(w_2)$ $= 2.5 \times 10^{-3} \ kg = 2.5 \ g$
Mass of solvent $(w_1)$ $= 75 \times 10^{-3} \ kg = 75 \ g$
$K_b = 0.52 \ K \ kg \ mol^{-1}$
Boiling point of solution $(T_b)$ $= 373.535 \ K$
Boiling point of pure water $(T_b^o)$ $= 373.15 \ K$
Elevation in boiling point $(\Delta T_b)$ $= T_b - T_b^o = 373.535 - 373.15 = 0.385 \ K$
Formula: $\Delta T_b = K_b \times \frac{w_2 \times 1000}{M_2 \times w_1}$
$0.385 = 0.52 \times \frac{2.5 \times 1000}{M_2 \times 75}$
$M_2 = \frac{0.52 \times 2500}{0.385 \times 75} = \frac{1300}{28.875} \approx 45.02 \ g \ mol^{-1}$
Rounding to the nearest integer,the molar mass is $45 \ g \ mol^{-1}$.

(C) The cell reaction is: $\frac{1}{2} H_{2(g)} + Ag^{+}_{(aq)} \rightarrow H^{+}_{(aq)} + Ag_{(s)}$
Here,the number of electrons involved in the balanced reaction is $n = 1$.
The standard Gibbs free energy change is given by: $\Delta_r G^0 = -n F E_{Cell}^0$
Substituting the values: $\Delta_r G^0 = -1 \times 96500 \ C \ mol^{-1} \times 0.5332 \ V$
$\Delta_r G^0 = -51453.8 \ J \ mol^{-1} = -51.45 \ kJ \ mol^{-1}$
Rounding to the nearest integer,the value is $-51 \ kJ \ mol^{-1}$.

(C) The Arrhenius equation is given by: $\log_{10} \frac{K_2}{K_1} = \frac{E_a}{2.303 \, R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)$
Given: $T_1 = 300 \, K$,$T_2 = 309 \, K$,$\frac{K_2}{K_1} = 2$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$,$\log 2 = 0.30$.
Substituting the values:
$0.3 = \frac{E_a}{2.303 \times 8.3} \left(\frac{9}{300 \times 309}\right)$
$E_a = \frac{0.3 \times 2.303 \times 8.3 \times 300 \times 309}{9}$
$E_a = 59065.04 \, J \, mol^{-1}$
$E_a \approx 59 \, kJ \, mol^{-1}$

(C) The Freundlich adsorption isotherm is given by $\frac{x}{m} = k \, P^{\frac{1}{n}}$.
Taking logarithm on both sides,we get $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$.
Comparing this with the equation of a straight line $y = mx + c$,the slope is $\frac{1}{n}$ and the intercept is $\log k$.
From the graph,the slope is $1$,so $\frac{1}{n} = 1 \Rightarrow n = 1$.
The intercept is $\log k = 0.602$. Since $\log 4 \approx 0.602$,we have $k = 4$.
Now,substitute the values into the equation: $\frac{x}{m} = 4 \times (0.03)^1$.
$\frac{x}{m} = 0.12 \, g$.
To express this in the form $...... \times 10^{-2} \, g$,we write $0.12 = 12 \times 10^{-2} \, g$.

(A) Molality is defined as the number of moles of solute per kilogram of solvent $(mol \ kg^{-1})$.
Since both the number of moles of solute and the mass of the solvent are independent of temperature,molality does not change with temperature.
Therefore,Assertion $A$ is true.
Reason $R$ correctly explains that mass remains unaffected by temperature,which is the reason why molality is temperature-independent.
Thus,both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

(A) . Lyophilic colloids act as protective colloids because they form a protective layer around lyophobic particles.
$B$. Emulsions are colloidal systems where both the dispersed phase and the dispersion medium are liquids (liquid-liquid colloid).
$C$. The reaction of $FeCl_3$ with hot water leads to the formation of a positively charged sol of hydrated ferric oxide $(Fe_2O_3 \cdot xH_2O)$.
$D$. The reaction of $FeCl_3$ with $NaOH$ (in excess) leads to the formation of a negatively charged sol due to the preferential adsorption of $OH^-$ ions on the surface of the precipitate.
Therefore,the correct matching is $A-II, B-I, C-IV, D-III$.

(A) . Concentration of gold ore is done by leaching with $NaCN$ solution (cyanidation).
$B$. Leaching of alumina $(Al_2O_3)$ is performed using $NaOH$ solution (Bayer's process).
$C$. Aniline is used as a froth stabiliser in the froth flotation process.
$D$. Blister copper is obtained after the Bessemerization process,where $SO_2$ gas escapes,causing bubbles on the surface of the solidified copper.

(D) When white phosphorus $(P_4)$ is heated with concentrated $NaOH$ solution,it undergoes a disproportionation reaction.
The chemical equation for this reaction is:
$P_4 + 3 NaOH + 3 H_2O \rightarrow 3 NaH_2PO_2 + PH_3$
Here,$PH_3$ (phosphine) and $NaH_2PO_2$ (sodium hypophosphite) are the main products formed.

(C) The crystal field stabilization energy $(CFSE)$ depends on the nature of the ligand and the oxidation state of the metal ion.
$CN^-$ is a strong field ligand,which causes a large splitting of $d$-orbitals $(\Delta_o)$.
In $[Co(CN)_6]^{3-}$,$Co$ is in the $+3$ oxidation state ($d^6$ configuration).
Since $CN^-$ is a strong field ligand,it leads to a low-spin $t_{2g}^6 e_g^0$ configuration,resulting in a very high $CFSE$ value compared to the other complexes listed,which involve weak field ligands like $H_2O$ or have lower oxidation states.

(C) Let us analyze each reaction:
$(A)$ Benzoic acid reacts with $SOCl_2$ to form benzoyl chloride,which on Rosenmund reduction $(H_2/Pd/BaSO_4)$ yields benzaldehyde.
$(B)$ Benzyl alcohol on oxidation with $CrO_3/H_2SO_4$ (Jones reagent) yields benzoic acid,not benzaldehyde.
$(C)$ Methyl benzoate does not react with $NaBH_4$ to form benzaldehyde.
$(D)$ Toluene reacts with $CrO_3$ in the presence of acetic anhydride to form a gem-diacetate intermediate,which on hydrolysis yields benzaldehyde (Etard reaction).
Therefore,reactions $(A)$ and $(D)$ yield benzaldehyde.

(D) In the Hofmann bromamide degradation reaction,an amide reacts with bromine in an alkaline medium to form a primary amine.
The mechanism involves the formation of an $N$-bromamide intermediate,which loses a proton to form an electron-deficient nitrene-like species (or nitrenoid intermediate).
During the rearrangement,the alkyl or aryl group $(R)$ migrates from the carbonyl carbon to the electron-deficient nitrogen atom to form an isocyanate $(R-N=C=O)$.
Statement-$I$ is incorrect because not only alkyl groups but also aryl groups can migrate.
Statement-$II$ is correct because the migration occurs specifically to the electron-deficient nitrogen atom.
Therefore,Statement-$I$ is incorrect and Statement-$II$ is correct.

(A) . Bakelite is a thermosetting polymer used for making electrical switches $(A-II)$.
$B$. Glyptal is used in the manufacture of paints and lacquers $(B-III)$.
$C$. $PVC$ (Polyvinyl chloride) is used in the manufacture of water pipes,raincoats,and handbags $(C-IV)$.
$D$. Polystyrene is used in the manufacture of radio and television cabinets $(D-I)$.
Therefore,the correct matching is $A-II, B-III, C-IV, D-I$.

(A) $1$. Compound $A$ has the molecular formula $C_{4}H_{8}O_{4}$.
$2$. It gives a positive test with $[Ag(NH_{3})_{2}]^{+}$ (Tollens' reagent),indicating the presence of an aldehyde group.
$3$. Treatment with acetic anhydride yields a triacetate,suggesting the presence of three hydroxyl $(-OH)$ groups.
$4$. The $L$-isomer configuration means the $-OH$ group on the chiral carbon furthest from the aldehyde group is on the left in the Fischer projection.
$5$. Bromine water oxidizes the aldehyde group to a carboxylic acid,forming an aldonic acid. For the $L$-isomer of erythrose (where both $-OH$ groups are on the left),this product is optically active.
$6$. Concentrated $HNO_{3}$ oxidizes both the aldehyde and the primary alcohol group to carboxylic acids,forming an aldaric acid. For the $L$-isomer of erythrose,the resulting tartaric acid is meso (optically inactive due to an internal plane of symmetry).
$7$. Comparing these properties,compound $A$ is $L$-erythrose,which corresponds to the structure in option $A$.

(A) : $[CH_{3}(CH_{2})_{15}N(CH_{3})_{3}]^{+} Br^{-}$ is a cationic detergent used in hair conditioners.
$B$: $CH_{3}(CH_{2})_{11}-C_{6}H_{4}-SO_{3}^{-}Na^{+}$ is an anionic detergent used in toothpastes.
$C$: $C_{17}H_{35}COO^{-}Na^{+} + Na_{2}CO_{3} +$ Rosinate is used as laundry soap.
$D$: $CH_{3}(CH_{2})_{16}COO(CH_{2}CH_{2}O)_{n}CH_{2}CH_{2}OH$ is a non-ionic detergent formed from stearic acid and polyethylene glycol,used as liquid dishwashing detergent.
Therefore,the correct matching is $A-IV, B-II, C-III, D-I$.

(A) Let the number of $Fe^{2+}$ ions be $x$ and the number of $Fe^{3+}$ ions be $y$.
Total number of $Fe$ ions is $x + y = 0.93$.
Since the crystal is electrically neutral,the total positive charge must equal the total negative charge ($-2$ for $O^{2-}$).
$2x + 3y = 2$.
Substituting $y = 0.93 - x$ into the charge equation:
$2x + 3(0.93 - x) = 2$
$2x + 2.79 - 3x = 2$
$-x = -0.79$
$x = 0.79$.
Thus,the number of $Fe^{2+}$ ions is $0.79$.
The percentage of $Fe^{2+}$ ions is $\frac{0.79}{0.93} \times 100 \approx 84.94\%$.
The nearest integer is $85\%$.

(A) The elevation in boiling point is given by the formula $\Delta T_b = K_b \cdot m$,where $K_b$ is the ebullioscopic constant and $m$ is the molality of the solution.
Since the mass of the solute and the mass of the solvent are the same for both solutions,the molality $m$ is identical for both solvents $A$ and $B$.
Given the ratio of ebullioscopic constants $\frac{(K_b)_A}{(K_b)_B} = \frac{1}{8}$.
Therefore,the ratio of the elevation in boiling points is $\frac{(\Delta T_b)_A}{(\Delta T_b)_B} = \frac{(K_b)_A \cdot m}{(K_b)_B \cdot m} = \frac{(K_b)_A}{(K_b)_B} = \frac{1}{8}$.
Comparing this to $\frac{x}{y}$,we get $x = 1$ and $y = 8$.
The value of $y$ is $8$.

(B) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions.
$\lambda_{m}^{\infty}(AgI) = \lambda_{m}^{\infty}(Ag^+) + \lambda_{m}^{\infty}(I^-)$
We can express this using the given values:
$\lambda_{m}^{\infty}(AgI) = \lambda_{m}^{\infty}(AgNO_3) + \lambda_{m}^{\infty}(NaI) - \lambda_{m}^{\infty}(NaNO_3)$
Substituting the given values:
$\lambda_{m}^{\infty}(AgI) = 13.3 + 12.7 - 12.0$
$\lambda_{m}^{\infty}(AgI) = 26.0 - 12.0$
$\lambda_{m}^{\infty}(AgI) = 14.0 \, mS \, m^2 \, mol^{-1}$

(B) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_A}{RT}$.
Comparing this with the given equation $\ln k = 33.24 - \frac{2.0 \times 10^{4}}{T}$,we get:
$\frac{E_A}{R} = 2.0 \times 10^{4} \, K$.
Therefore,$E_A = 2.0 \times 10^{4} \times R$.
Substituting $R = 8.3 \, J \, K^{-1} \, mol^{-1}$:
$E_A = 2.0 \times 10^{4} \times 8.3 \, J \, mol^{-1} = 16.6 \times 10^{4} \, J \, mol^{-1}$.
Converting to $kJ \, mol^{-1}$:
$E_A = \frac{16.6 \times 10^{4}}{1000} \, kJ \, mol^{-1} = 166 \, kJ \, mol^{-1}$.

(C) $Cu(II)$ $(d^9)$ complexes have one unpaired electron,so they are paramagnetic. This is correct.
$(B)$ $Cu(I)$ $(d^{10})$ complexes have no unpaired electrons,so they are generally colourless due to the absence of $d-d$ transitions. This is correct.
$(C)$ $Cu(I)$ is unstable in aqueous solution and undergoes disproportionation to $Cu(0)$ and $Cu(II)$,meaning it is easily oxidized to $Cu(II)$. This is correct.
$(D)$ Fehling solution consists of two parts: Fehling $A$ ($CuSO_4$ solution) and Fehling $B$ (alkaline sodium potassium tartrate). The active species is the $Cu(II)$ tartrate complex. This is correct.
Therefore,all $4$ statements are correct.

(D) The balanced chemical equation for the reaction is: $2 \ KMnO_4 + 5 \ H_2C_2O_4 + 3 \ H_2SO_4 \rightarrow K_2SO_4 + 2 \ MnSO_4 + 10 \ CO_2 + 8 \ H_2O$.
In this reaction,the manganese product is $MnSO_4$,where manganese exists as the $Mn^{2+}$ ion.
The electronic configuration of $Mn^{2+}$ is $[Ar] \ 3d^5$.
It contains $5$ unpaired electrons.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \, B.M.$,where $n$ is the number of unpaired electrons.
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, B.M.$
Rounding to the nearest integer,we get $6 \, B.M.$

(C) The correct matches are as follows:
$A$. Negatively charged sol: $CdS$ sol $(II)$
$B$. Macromolecular colloid: Starch $(III)$
$C$. Positively charged sol: $Fe_{2}O_{3} \cdot xH_{2}O$ $(I)$
$D$. Cheese: $A$ gel $(IV)$
Therefore,the correct sequence is $A-II, B-III, C-I, D-IV$.

(D) In the extraction of copper,$FeO$ is an impurity present in the ore,which acts as a gangue.
$SiO_{2}$ is added as a flux to remove the $FeO$ impurity.
$FeSiO_{3}$ is the fusible material formed,which is known as slag.
Therefore,$FeO$ is gangue and $FeSiO_{3}$ is slag.

(B) To determine the number of compounds having an $N-N$ bond,we examine the structures of the given nitrogen oxides:
$1$. $N_2O$: The structure is $N \equiv N^+ - O^-$. There is no $N-N$ bond (it has an $N=N$ bond).
$2$. $N_2O_3$: The structure is $O_2N-NO$. It contains an $N-N$ bond.
$3$. $N_2O_4$: The structure is $O_2N-NO_2$. It contains an $N-N$ bond.
$4$. $N_2O_5$: The structure is $O_2N-O-NO_2$. It contains an $N-O-N$ linkage,not an $N-N$ bond.
Thus,the compounds $N_2O_3$ and $N_2O_4$ have an $N-N$ bond.
The total number of such compounds is $2$.

(C) The reaction proceeds via an acid-catalyzed dehydration of a secondary alcohol.
$1$. Protonation of the $-OH$ group occurs to form a good leaving group $(-OH_2^+)$.
$2$. Loss of a water molecule generates a secondary carbocation.
$3$. $A$ $1,2$-methyl shift occurs to rearrange the secondary carbocation into a more stable tertiary carbocation.
$4$. Finally,deprotonation (loss of $H^+$) from the adjacent carbon leads to the formation of the most stable,highly substituted alkene (Zaitsev product).
Thus,the major product is $1,2,5$-trimethylcyclohex-$1$-ene.

(C) The reaction sequence is as follows:
$1$. $4-$Bromotoluene reacts with $Cl_{2}$ in the presence of heat (free radical substitution) to form $4-$bromobenzyl chloride $(Br-C_{6}H_{4}-CH_{2}Cl)$.
$2$. This product then reacts with $CN^{-}$ (nucleophilic substitution) to form $4-$bromophenylacetonitrile $(Br-C_{6}H_{4}-CH_{2}CN)$.
$3$. Finally,acid-catalyzed hydrolysis of the nitrile group yields $4-$bromophenylacetic acid $(Br-C_{6}H_{4}-CH_{2}COOH)$.
Therefore,the starting material '$A$' is $4-$bromotoluene.

(C) $1$. Isobutyraldehyde $((CH_{3})_{2}CHCHO)$ reacts with formaldehyde $(HCHO)$ in the presence of $K_{2}CO_{3}$ (aldol condensation followed by Cannizzaro-like step or simply aldol addition) to form compound '$A$',which is $3$-hydroxy-$2,2$-dimethylpropanal $(HO-CH_{2}-C(CH_{3})_{2}-CHO)$.
$2$. Compound '$A$' reacts with $KCN$ (cyanohydrin formation) to yield compound '$B$',which is $2,4$-dihydroxy-$3,3$-dimethylbutanenitrile $(HO-CH_{2}-C(CH_{3})_{2}-CH(OH)-CN)$.
$3$. Hydrolysis of '$B$' gives the corresponding carboxylic acid,$2,4$-dihydroxy-$3,3$-dimethylbutanoic acid $(HO-CH_{2}-C(CH_{3})_{2}-CH(OH)-COOH)$.
$4$. This hydroxy acid undergoes intramolecular cyclization (lactonization) to form a stable lactone,which is $3$-hydroxy-$4,4$-dimethyltetrahydrofuran-$2$-one (as shown in option $C$).

(C) In the nitration of aniline using $HNO_3$ and $H_2SO_4$,aniline is protonated to form the anilinium ion $(C_6H_5NH_3^+)$,which is meta-directing.
However,due to the presence of some unprotonated aniline,ortho and para products are also formed.
Experimental results show that $m$-nitroaniline $(47\%)$ and $p$-nitroaniline $(51\%)$ are the major products.
Thus,statement $(B)$ is correct.
In the reaction between $HNO_3$ and $H_2SO_4$,$H_2SO_4$ acts as an acid and $HNO_3$ acts as a base:
$HNO_3 + H_2SO_4$ $\rightarrow H_2NO_3^+ + HSO_4^-$ $\rightarrow NO_2^+ + H_2O + HSO_4^-$.
Thus,statement $(D)$ is correct.
Therefore,statements $(B)$ and $(D)$ are correct.

(C) Natural rubber is a linear polymer of isoprene $(2\text{-methyl-}1,3\text{-butadiene})$ and is also called $cis\text{-}1,4\text{-polyisoprene}.$
It possesses elastic properties due to its coiled structure.
The $cis\text{-polyisoprene}$ molecules consist of various chains held together by weak van der Waals interactions,not strong polar interactions.
Therefore,Assertion $(A)$ is true,but Reason $(R)$ is false.

(B) The hydrolysis of sucrose $(C_{12}H_{22}O_{11})$ in the presence of dilute acid yields glucose $(A)$ and fructose $(B)$.
$C_{12}H_{22}O_{11} + H_{2}O \xrightarrow{H^{+}} C_{6}H_{12}O_{6} (\text{Glucose}) + C_{6}H_{12}O_{6} (\text{Fructose})$
Glucose $(A)$ on oxidation with $HNO_{3}$ gives saccharic acid (glucaric acid).
Fructose $(B)$ is a laevorotatory sugar $([\alpha] = -92.4^{\circ})$.
Thus,the sugar $X$ is sucrose.

(C) Tegamet is the brand name of Cimetidine. The chemical structure of Cimetidine is shown in option $C$.

(B) The formula for osmotic pressure is $\pi = CRT$,where $C$ is the molar concentration.
Given: $\pi = 5.03 \times 10^{-3} \, bar$,$T = 300 \, K$,$R = 0.083 \, L \, bar \, K^{-1} \, mol^{-1}$,$V = 0.5 \, L$.
Calculating molarity $C$:
$C = \frac{\pi}{RT} = \frac{5.03 \times 10^{-3}}{0.083 \times 300} \approx 2.02 \times 10^{-4} \, mol \, L^{-1}$.
Moles of protein $= C \times V = 2.02 \times 10^{-4} \times 0.5 = 1.01 \times 10^{-4} \, mol$.
Molar mass of protein $(M) = \frac{\text{mass}}{\text{moles}} = \frac{2.5}{1.01 \times 10^{-4}} \approx 24752 \, g \, mol^{-1}$.
Molar mass of glycine $(C_2H_5NO_2) = (2 \times 12) + (5 \times 1) + 14 + (2 \times 16) = 75 \, g \, mol^{-1}$.
Number of glycine units $= \frac{24752}{75} \approx 330$.

(C) Given:
$(1) \ Sn^{2+} + 2e^{-}$ $\rightarrow Sn, \ E^{\circ}_{1} = -0.140 \ V, \ \Delta G^{\circ}_{1} = -nFE^{\circ}_{1} = -2 \times F \times (-0.140) = +0.280F$
$(2) \ Sn^{4+} + 4e^{-}$ $\rightarrow Sn, \ E^{\circ}_{2} = 0.010 \ V, \ \Delta G^{\circ}_{2} = -nFE^{\circ}_{2} = -4 \times F \times (0.010) = -0.040F$
We need the potential for $Sn^{4+} + 2e^{-} \rightarrow Sn^{2+}$.
This can be obtained by $(2) - (1)$:
$(Sn^{4+} + 4e^{-}) - (Sn^{2+} + 2e^{-}) \rightarrow Sn - Sn$
$Sn^{4+} + 2e^{-} \rightarrow Sn^{2+}$
$\Delta G^{\circ}_{3} = \Delta G^{\circ}_{2} - \Delta G^{\circ}_{1} = -0.040F - 0.280F = -0.320F$
Since $\Delta G^{\circ}_{3} = -nFE^{\circ}_{Sn^{4+}/Sn^{2+}}$,where $n=2$:
$-2FE^{\circ}_{Sn^{4+}/Sn^{2+}} = -0.320F$
$E^{\circ}_{Sn^{4+}/Sn^{2+}} = 0.160 \ V = 16 \times 10^{-2} \ V$
The value is $16$.

(C) The decay constant $\lambda$ is given by $\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{200} \approx 0.003465 \ day^{-1}$.
The activity $A$ at time $t$ is given by $A = A_0 e^{-\lambda t}$.
$\frac{A}{A_0} = e^{-\lambda t} = e^{-(0.693/200) \times 83} = e^{-0.2877}$.
Alternatively,using the formula $A = A_0 (1/2)^{t/t_{1/2}}$:
$\frac{A}{A_0} = (0.5)^{83/200} = (0.5)^{0.415}$.
Taking $\log$ on both sides: $\log(\frac{A}{A_0}) = 0.415 \times \log(0.5) = 0.415 \times (-0.301) \approx -0.1249$.
$\frac{A}{A_0} = \text{antilog}(-0.1249) = 10^{-0.1249} \approx 0.75$.
Percentage remaining = $0.75 \times 100 = 75 \ \%$.

(C) $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $d^6$,$CN^-$ is a strong field ligand,so $t_{2g}^6 e_g^0$. Diamagnetic.
$[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $d^5$,$CN^-$ is a strong field ligand,so $t_{2g}^5 e_g^0$. Paramagnetic ($1$ unpaired electron).
$[Ti(CN)_6]^{3-}$: $Ti^{3+}$ is $d^1$,so $t_{2g}^1 e_g^0$. Paramagnetic ($1$ unpaired electron).
$[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $d^8$,$CN^-$ is a strong field ligand,so $dsp^2$ hybridization. Diamagnetic.
$[Co(CN)_6]^{3-}$: $Co^{3+}$ is $d^6$,$CN^-$ is a strong field ligand,so $t_{2g}^6 e_g^0$. Diamagnetic.
Therefore,the number of paramagnetic complexes is $2$.

(D) $(a) \ [Co(NH_{3})_{4}Cl_{2}]Cl$ is of the type $[MA_{4}B_{2}]$,which exhibits $cis-trans$ isomerism.
$(b) \ [Co(NH_{3})_{5}Cl]Cl_{2}$ is of the type $[MA_{5}B]$,which does not exhibit $cis-trans$ isomerism.
$(c) \ [Co(NH_{3})_{6}]Cl_{3}$ is of the type $[MA_{6}]$,which does not exhibit $cis-trans$ isomerism.
$(d) \ [Co(NH_{3})_{5}Cl](NO_{3})_{2}$ or $[Co(NH_{3})_{5}(NO_{3})]Cl(NO_{3})$ are of the type $[MA_{5}B]$,which do not exhibit $cis-trans$ isomerism.
Therefore,only one complex,$[Co(NH_{3})_{4}Cl_{2}]Cl$,exhibits $cis-trans$ isomerism.