JEE Main 2022 Chemistry Question Paper with Answer and Solution

(A) First,calculate the solubility $S$ in $mol / L$:
$S = \frac{2.34 \times 10^{-3} \, g}{78 \, g/mol} \times \frac{1}{0.1 \, L} = \frac{0.03 \times 10^{-3}}{0.1} = 3 \times 10^{-4} \, mol / L$.
The dissociation of $CaF_{2}$ is $CaF_{2} \rightleftharpoons Ca^{2+} + 2F^-$.
$K_{sp} = [Ca^{2+}][F^-]^{2} = (S)(2S)^{2} = 4S^{3}$.
$K_{sp} = 4 \times (3 \times 10^{-4})^{3} = 4 \times 27 \times 10^{-12} = 108 \times 10^{-12} = 0.0108 \times 10^{-8} \, (mol / L)^{3}$.
Thus,$x = 0.0108$.

(A) In oxalic acid $(H_{2}C_{2}O_{4})$,the oxidation state of carbon is $+3$.
In carbon dioxide $(CO_{2})$,the oxidation state of carbon is $+4$.
The change in oxidation number of carbon is $|4 - 3| = 1$.
The balanced chemical equation is: $2KMnO_{4} + 5H_{2}C_{2}O_{4} + 3H_{2}SO_{4} \rightarrow K_{2}SO_{4} + 2MnSO_{4} + 10CO_{2} + 8H_{2}O$.

(D) The optical purity of an enantiomeric mixture is calculated using the formula: $\text{Optical Purity} = \frac{\text{observed rotation of mixture}}{\text{specific rotation of pure enantiomer}} \times 100$.
Given: $\text{Observed rotation} = +12.6^{\circ}$ and $\text{Specific rotation} = +30^{\circ}$.
Substituting the values: $\text{Optical Purity} = \frac{12.6}{30} \times 100 = 0.42 \times 100 = 42\%$.
Thus,the optical purity is $42\%$.

(D) Let the initial moles of benzene be $n$.
For reaction $I$,the yield is $60 \%$,so the moles of benzene sulphonic acid obtained $= 0.6 \times n = 0.6n$.
For reaction $II$,the yield is $50 \%$,so the moles of phenol obtained $= 0.5 \times (0.6n) = 0.3n$.
The overall yield of the complete reaction is calculated as: $\text{Overall yield} = \frac{\text{Final moles of product}}{\text{Initial moles of reactant}} \times 100$.
$\text{Overall yield} = \frac{0.3n}{n} \times 100 = 30 \%$.

(D) An atomic orbital is a mathematical function $(\Psi)$ that describes the probability of finding an electron in a region of space around the nucleus. It is fully characterized by three quantum numbers: principal quantum number $(n)$,azimuthal quantum number $(l)$,and magnetic quantum number $(m_l)$. Therefore,the statement that an atomic orbital is characterized by $n$ and $l$ only is incorrect.

(A) By definition,the enthalpy change is given by $\Delta H = \Delta U + \Delta(PV)$.
At constant pressure,this becomes $\Delta H = \Delta U + P \Delta V$.
Option $A$ is written as $\Delta H = \Delta U - P \Delta V$,which is incorrect because the correct sign is positive.
Option $B$ represents the first law of thermodynamics $(\Delta U = q + W)$.
Option $C$ represents the second law of thermodynamics for a spontaneous process.
Option $D$ is the Gibbs-Helmholtz equation.

(C) The electron gain enthalpy $(\Delta_{eg}H)$ values for elements in the same group generally become less negative as we move down the group.
$1$. For $Rb$ and $Cs$ (Group $1$): The values are approximately $-47 \ kJ/mol$ and $-46 \ kJ/mol$ respectively,which are nearly identical.
$2$. For $Ar$ and $Kr$ (Group $18$): Noble gases have positive electron gain enthalpy values due to stable electronic configurations. Both $Ar$ and $Kr$ have values around $+96 \ kJ/mol$ to $+99 \ kJ/mol$,which are considered nearly identical.
Therefore,the pairs $A$ $(Rb, Cs)$ and $C$ $(Ar, Kr)$ have nearly identical electron gain enthalpies.

(B) $Clark's$ method is used to remove temporary hardness of water by adding calculated amounts of lime,$Ca(OH)_2$.
The chemical reactions involved are:
$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3(s) + 2H_2O(l)$
$Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3(s) + Mg(OH)_2(s) + 2H_2O(l)$
As seen from the reactions,the insoluble metal salts formed are $CaCO_3$ and $Mg(OH)_2$.

(A) The low solubility of $LiF$ in water is due to its very high lattice enthalpy,which outweighs the hydration enthalpy. Therefore,the statement in option $A$ is incorrect.

(D) Statement $I$ is false because in polluted water,the value of dissolved oxygen is low,but the value of $BOD$ (Biological Oxygen Demand) is high due to the presence of organic matter that requires oxygen for decomposition.
Statement $II$ is true because eutrophication leads to the excessive growth of algae and weeds,which consume dissolved oxygen,thereby decreasing its concentration in water bodies.
Therefore,Statement $I$ is false and Statement $II$ is true.

(A) Step $1$: Analyze the structures in List-$I$.
$(A)$ is Tetrahydrofuran,which is a non-planar heterocyclic compound $(III)$.
$(B)$ is a bicyclo compound $(IV)$.
$(C)$ is a spiro compound $(I)$.
$(D)$ is Furan,which is an aromatic compound $(II)$.
Step $2$: Match the findings.
$A-III, B-IV, C-I, D-II$.
Therefore,the correct option is $A$.

(B) The reaction with $Hg(OAc)_2, H_2O$ followed by $NaBH_4$ is Oxymercuration-Demercuration,which follows Markovnikov's rule to give product $A$ ($3$,$3$-dimethylbutan$-2-$ol).
The reaction with $(BH_3)_2$ followed by $H_2O_2/OH^-$ is Hydroboration-Oxidation,which follows anti-Markovnikov's rule to give product $B$ ($3$,$3$-dimethylbutan$-1-$ol).
Therefore,$A$ is the Markovnikov product and $B$ is the anti-Markovnikov product.

(D) The balanced chemical equation is: $X + Y + 3 Z \rightarrow XYZ_3$.
Initial moles: $X = 1 \ mol$,$Y = 1 \ mol$,$Z = 0.05 \ mol$.
To find the Limiting Reagent ($L$.$R$.),divide the moles by their stoichiometric coefficients:
$X: 1/1 = 1$
$Y: 1/1 = 1$
$Z: 0.05/3 \approx 0.0167$
Since $Z$ has the smallest ratio,$Z$ is the Limiting Reagent.
Moles of $XYZ_3$ formed = $\frac{1}{3} \times \text{moles of } Z = \frac{0.05}{3} \ mol$.
Molar mass of $XYZ_3 = 10 + 20 + (3 \times 30) = 120 \ g/mol$.
Yield of $XYZ_3 = \frac{0.05}{3} \times 120 = 0.05 \times 40 = 2 \ g$.

(D) To determine the paramagnetic nature,we use Molecular Orbital Theory $(MOT)$ to find the number of unpaired electrons:
$1. B_{2} (10 \ e^-): (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^1 (\pi 2p_y)^1$ ($2$ unpaired electrons,paramagnetic)
$2. Li_{2} (6 \ e^-): (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2$ ($0$ unpaired electrons,diamagnetic)
$3. C_{2} (12 \ e^-): (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2$ ($0$ unpaired electrons,diamagnetic)
$4. C_{2}^{-} (13 \ e^-): (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^1$ ($1$ unpaired electron,paramagnetic)
$5. O_{2}^{2-} (18 \ e^-): (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^2$ ($0$ unpaired electrons,diamagnetic)
$6. O_{2}^{+} (15 \ e^-): (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$ ($1$ unpaired electron,paramagnetic)
$7. He_{2}^{+} (3 \ e^-): (\sigma 1s)^2 (\sigma^* 1s)^1$ ($1$ unpaired electron,paramagnetic)
Thus,the paramagnetic species are $B_{2}, C_{2}^{-}, O_{2}^{+},$ and $He_{2}^{+}$. The total count is $4$.

(C) For a weak acid,the $pH$ is given by the formula: $pH = \frac{1}{2} (pK_{a} - \log C)$.
Given $K_{a} = 2 \times 10^{-5}$,so $pK_{a} = -\log(2 \times 10^{-5}) = 5 - \log 2 = 5 - 0.30 = 4.7$.
Concentration $C = 0.2 \ M = 2 \times 10^{-1} \ M$.
$pH = \frac{1}{2} (4.7 - \log(2 \times 10^{-1}))$.
$pH = \frac{1}{2} (4.7 - (\log 2 + \log 10^{-1}))$.
$pH = \frac{1}{2} (4.7 - (0.30 - 1)) = \frac{1}{2} (4.7 - (-0.7)) = \frac{1}{2} (5.4) = 2.7$.
$pH = 2.7 = 27 \times 10^{-1}$.
Thus,the value is $27$.

(B) The disproportionation reaction of manganate ion $(MnO_{4}^{2-})$ in acidic medium is given by:
$3MnO_{4}^{2-} + 4H^{+} \rightarrow 2MnO_{4}^{-} + MnO_{2} + 2H_{2}O$
Here,the manganese compounds formed are $MnO_{4}^{-}$ (where $Mn$ is in $+7$ oxidation state) and $MnO_{2}$ (where $Mn$ is in $+4$ oxidation state).
Given that the oxidation state of $Mn$ in $B$ is smaller than that of $A$,we identify $A$ as $MnO_{4}^{-}$ $(+7)$ and $B$ as $MnO_{2}$ $(+4)$.
In $MnO_{2}$,$Mn$ is in the $+4$ oxidation state,which corresponds to the electronic configuration $[Ar] 3d^{3}$.
The number of unpaired electrons $(n)$ in $Mn^{4+}$ is $3$.
The spin-only magnetic moment $(\mu)$ is calculated as:
$\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
The nearest integer value is $4$.

(A) Mass of organic compound $= 0.492 \ g$.
Moles of $CO_2 = \frac{0.7938 \ g}{44 \ g/mol} = 0.01804 \ mol$.
Moles of $C = 0.01804 \ mol$,so mass of $C = 0.01804 \times 12 = 0.2165 \ g$.
Moles of $H_2O = \frac{0.4428 \ g}{18 \ g/mol} = 0.0246 \ mol$.
Moles of $H = 2 \times 0.0246 = 0.0492 \ mol$,so mass of $H = 0.0492 \times 1 = 0.0492 \ g$.
Mass of $O = 0.492 - (0.2165 + 0.0492) = 0.492 - 0.2657 = 0.2263 \ g$.
Percentage of $O = \frac{0.2263}{0.492} \times 100 \approx 46 \%$.

(D) Zero overlap occurs when the orbitals do not have the proper symmetry or orientation to overlap effectively.
Assertion $A$ is false because zero overlap is not necessarily an out-of-phase overlap (which refers to destructive interference of wave functions). Zero overlap often occurs due to improper orientation or symmetry,such as the overlap between $s$ and $p_y$ orbitals along the $x$-axis.
Reason $R$ is true because zero overlap is indeed caused by the different orientation or direction of approach of the orbitals relative to each other,as shown in the provided figure where $p_x$ and $p_y$ orbitals do not overlap effectively.
Therefore,$A$ is false but $R$ is true.

(A) Metallic character is defined by the ease with which an element can lose electrons.
Across a period from left to right,the effective nuclear charge increases and atomic size decreases,making it harder to lose electrons.
Therefore,metallic character decreases across a period.
For the given elements: $Na$ (Group $1$),$Mg$ (Group $2$),$Be$ (Group $2$),$Si$ (Group $14$),and $P$ (Group $15$).
Comparing their positions: $Na$ is the most metallic,followed by $Mg$,then $Be$,then $Si$,and finally $P$ is the least metallic.
The correct decreasing order is $Na > Mg > Be > Si > P$.

(C) Clark's method is used to remove temporary hardness of water by adding calculated amounts of lime,$Ca(OH)_{2}$.
The chemical reactions are:
$Ca(HCO_{3})_{2} + Ca(OH)_{2} \rightarrow 2CaCO_{3} \downarrow + 2H_{2}O$
$Mg(HCO_{3})_{2} + 2Ca(OH)_{2} \rightarrow 2CaCO_{3} \downarrow + Mg(OH)_{2} \downarrow + 2H_{2}O$
Thus,the products formed are $CaCO_{3}$ and $Mg(OH)_{2}$.

(B) Statement $I$ is false because an alloy of $Li$ and $Mg$ is used to make armour plates,not aircraft plates.
Statement $II$ is false because calcium ions $(Ca^{2+})$,not magnesium ions,are primarily responsible for cell-membrane integrity and neuromuscular function.

(B) The reactivity towards electrophilic aromatic substitution (like nitration) depends on the electron density of the benzene ring.
$1$. $-NO_2$ (in $D$) is a strongly electron-withdrawing group (deactivating),making the ring least reactive.
$2$. $-Br$ (in $B$) is an electron-withdrawing group due to the inductive effect (deactivating),but less so than $-NO_2$.
$3$. $E$ (benzene) is the reference point.
$4$. $-CH_3$ groups are electron-donating (activating) due to hyperconjugation and inductive effects.
$5$. $p$-xylene $(A)$ has two $-CH_3$ groups,and mesitylene $(C)$ has three $-CH_3$ groups. More electron-donating groups increase the reactivity further.
Thus,the order of increasing reactivity is: $D < B < E < A < C$.

(A) Thin layer chromatography $(TLC)$ is a type of adsorption chromatography.
It involves the separation of substances of a mixture over a thin layer of an adsorbent coated on a glass plate.
$A$ thin layer (about $0.2 \ mm$ thick) of an adsorbent,such as silica gel or alumina,is spread over a glass plate of suitable size.
Since the separation is based on the differential adsorption of the components on the stationary phase (silica gel),both the Assertion $(A)$ and the Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(A) The reaction is an iodolactonization followed by dehydroiodination.
$1$. Treatment of the unsaturated carboxylic acid with $I_2/NaHCO_3$ leads to the formation of an iodolactone intermediate. The carboxylate oxygen attacks the iodonium ion formed across the double bond,resulting in a bicyclic lactone with an iodine atom at the bridgehead position.
$2$. Subsequent treatment with pyridine and heat $(\Delta)$ promotes an $E2$ elimination reaction,removing the iodine atom and a hydrogen atom from the adjacent carbon to form a double bond.
$3$. The final product is a bicyclic lactone containing a double bond in the six-membered ring,which corresponds to option $A$.

(D) The balanced chemical equation is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$
Initial moles of $H_2SO_4 = 2 \ L \times 0.2 \ M = 0.4 \ mol$.
Initial moles of $NaOH = 2 \ L \times 0.1 \ M = 0.2 \ mol$.
Since $NaOH$ is the limiting reagent,$0.2 \ mol$ of $NaOH$ will react with $0.1 \ mol$ of $H_2SO_4$ to produce $0.1 \ mol$ of $Na_2SO_4$.
Total volume of the solution = $2 \ L + 2 \ L = 4 \ L$.
Molarity of $Na_2SO_4 = \frac{0.1 \ mol}{4 \ L} = 0.025 \ M$.
$0.025 \ M = 25 \ mM$.

(C) The kinetic energy $K$ of the emitted electron is given by the de Broglie wavelength formula: $\lambda = \frac{h}{\sqrt{2mK}}$.
Rearranging for $K$: $K = \frac{h^2}{2m\lambda^2}$.
Substituting the given values: $K = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (3.3 \times 10^{-10})^2} \approx 2.215 \times 10^{-18} \ J$.
The minimum energy required to escape from the ground state of $H$ atom is the ionization energy: $E_{req} = 13.6 \ eV = 13.6 \times 1.602 \times 10^{-19} \ J \approx 2.179 \times 10^{-18} \ J$.
The total energy absorbed is $E_{abs} = E_{req} + K$.
The ratio is $\frac{E_{abs}}{E_{req}} = 1 + \frac{K}{E_{req}} = 1 + \frac{2.215 \times 10^{-18}}{2.179 \times 10^{-18}} \approx 1 + 1.016 = 2.016$.
Rounding to the nearest integer,the value is $2$.

(C) The reaction is $2 \ NO_{(g)} + O_{2(g)} \rightleftarrows 2 \ NO_{2(g)}$.
Initial moles: $NO = 2, O_2 = 1, NO_2 = 0$.
At equilibrium,moles of $O_2 = 1 - x = 0.6$,so $x = 0.4$.
Equilibrium moles: $NO = 2 - 2(0.4) = 1.2$,$O_2 = 0.6$,$NO_2 = 2(0.4) = 0.8$.
Total moles at equilibrium = $1.2 + 0.6 + 0.8 = 2.6$.
Mole fractions: $X_{NO} = \frac{1.2}{2.6}$,$X_{O_2} = \frac{0.6}{2.6}$,$X_{NO_2} = \frac{0.8}{2.6}$.
Partial pressures $(P_{total} = 1 \ atm)$: $P_{NO} = \frac{1.2}{2.6} \ atm$,$P_{O_2} = \frac{0.6}{2.6} \ atm$,$P_{NO_2} = \frac{0.8}{2.6} \ atm$.
$K_p = \frac{(P_{NO_2})^2}{(P_{NO})^2 \times P_{O_2}} = \frac{(\frac{0.8}{2.6})^2}{(\frac{1.2}{2.6})^2 \times (\frac{0.6}{2.6})} = \frac{0.8^2 \times 2.6}{1.2^2 \times 0.6} = \frac{0.64 \times 2.6}{1.44 \times 0.6} = \frac{1.664}{0.864} \approx 1.926$.
The nearest integer is $2$.

(A) Step $1$: Calculate the pressure of dry $N_2$ gas.
$P_{N_2} = P_{total} - P_{H_2O} = 759 \ mm \ Hg - 14.2 \ mm \ Hg = 744.8 \ mm \ Hg$.
Step $2$: Convert units to standard values.
$P = \frac{744.8}{760} \ atm$,$V = \frac{22.78}{1000} \ L$,$T = 280 \ K$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.
Step $3$: Calculate moles of $N_2$ using the ideal gas equation $PV = nRT$.
$n_{N_2} = \frac{PV}{RT} = \frac{744.8 \times 22.78}{760 \times 0.082 \times 280 \times 1000} \approx 0.000971 \ mol$.
Step $4$: Calculate mass of $N_2$ and percentage.
$Mass_{N_2} = 0.000971 \times 28 \ g/mol = 0.027188 \ g$.
$\%N = \frac{Mass_{N_2}}{Mass_{sample}} \times 100 = \frac{0.027188}{0.125} \times 100 \approx 21.75\%$.
The nearest integer is $22$.

(A) The reaction is: $IO_{4}^{-} + H_{2}O_{2} \rightarrow IO_{3}^{-} + O_{2} + H_{2}O$.
In $KIO_{4}$,the oxidation state of $I$ is calculated as: $x + 4(-2) = -1$,so $x = +7$.
In the product $IO_{3}^{-}$,the oxidation state of $I$ is: $x + 3(-2) = -1$,so $x = +5$.
Thus,the oxidation number of $I$ changes from $+7$ to $+5$.

(B) State variables are properties whose values depend only on the state of the system and not on the path taken to reach that state.
In the given list:
$1$. Internal energy $(U)$ is a state function.
$2$. Volume $(V)$ is a state function.
$3$. Enthalpy $(H)$ is a state function.
$4$. Heat $(q)$ is a path function,not a state function.
Therefore,there are $3$ state variables: Internal energy,Volume,and Enthalpy.

(B) An odd-electron molecule is one that has an odd number of valence electrons,such as $NO$ (Nitric oxide,$5+6=11$ valence electrons).
An expanded octet molecule is one where the central atom has more than $8$ electrons in its valence shell,such as $H_{2}SO_{4}$ (Sulfuric acid).
In $H_{2}SO_{4}$,the central sulfur atom is bonded to four oxygen atoms (two double bonds and two single bonds),resulting in $12$ electrons in its valence shell.
Thus,the pair containing an odd-electron molecule $(NO)$ and an expanded octet molecule $(H_{2}SO_{4})$ is $NO$ and $H_{2}SO_{4}$.

(C) The balanced chemical equation is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Calculate the number of moles of reactants:
$n(N_{2}) = \frac{20 \ g}{28 \ g/mol} \approx 0.714 \ mol$
$n(H_{2}) = \frac{5 \ g}{2 \ g/mol} = 2.5 \ mol$
Determine the limiting reagent by dividing moles by the stoichiometric coefficient:
For $N_{2}: \frac{0.714}{1} = 0.714$
For $H_{2}: \frac{2.5}{3} \approx 0.833$
Since $0.714 < 0.833$,$N_{2}$ is the limiting reagent.
The amount of $NH_{3}$ produced depends on the limiting reagent:
$n(NH_{3}) = 2 \times n(N_{2}) = 2 \times 0.714 = 1.428 \ mol \approx 1.42 \ mol$.

(C) The general trend for the first ionization enthalpy across the third period is: $Na < Al < Mg < Si$.
Given values are:
$IE(Na) = 496 \ kJ \ mol^{-1}$
$IE(Mg) = 737 \ kJ \ mol^{-1}$
$IE(Si) = 786 \ kJ \ mol^{-1}$
Based on the periodic trend,the ionization enthalpy of $Al$ should be greater than $Na$ but less than $Mg$.
Therefore,$496 < IE(Al) < 737$.
Among the given options,only $577 \ kJ \ mol^{-1}$ falls within this range.
Thus,the correct option is $(C)$.

(D) Zinc reacts with an excess of aqueous alkali (such as $NaOH$) to evolve hydrogen gas and form the soluble tetrahydroxozincate$(II)$ complex ion.
The chemical equation is:
$Zn(s) + 2OH^{-}(aq) + 2H_2O(l) \rightarrow [Zn(OH)_4]^{2-}(aq) + H_2(g) \uparrow$
In this reaction,the zinc metal is oxidized to the $+2$ oxidation state,forming the complex ion $[Zn(OH)_4]^{2-}$,which is the stable species in an aqueous alkaline medium.

(C) When heated,lithium nitrate decomposes to form lithium oxide $(Li_2O)$,nitrogen dioxide $(NO_2)$,and oxygen $(O_2)$:
$4LiNO_3 \longrightarrow 2Li_2O + 4NO_2 + O_2$
In contrast,sodium nitrate decomposes to form sodium nitrite $(NaNO_2)$ and oxygen $(O_2)$:
$2NaNO_3 \longrightarrow 2NaNO_2 + O_2$
Therefore,the products are $Li_2O$ and $NaNO_2$ respectively.
Note: At very high temperatures (above $800 \ ^{\circ}C$),$NaNO_2$ can further decompose to $Na_2O$,but the standard decomposition product for $NaNO_3$ is $NaNO_2$.

(B) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{1}{2} (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of bonding electrons (or atoms attached).
$1$. For $SCl_2$: Sulfur $(S)$ has $6$ valence electrons. It forms $2$ bonds with $Cl$ atoms. Lone pairs = $\frac{1}{2} (6 - 2) = 2$.
$2$. For $O_3$: The central Oxygen atom has $6$ valence electrons. It forms $2$ bonds with one oxygen atom and $1$ coordinate bond with another. The central atom has $1$ lone pair.
$3$. For $ClF_3$: Chlorine $(Cl)$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms. Lone pairs = $\frac{1}{2} (7 - 3) = 2$.
$4$. For $SF_6$: Sulfur $(S)$ has $6$ valence electrons. It forms $6$ bonds with $F$ atoms. Lone pairs = $\frac{1}{2} (6 - 6) = 0$.
Thus,the number of lone pairs are $2, 1, 2$ and $0$ respectively. The correct option is $B$.

(A) Transition metal ions are colourless if they do not have unpaired electrons to undergo $d-d$ transitions. This occurs in ions with $d^{0}$ or $d^{10}$ electronic configurations.
$A$) $Sc^{3+}$ $(3d^{0})$ and $Zn^{2+}$ $(3d^{10})$: Both have no unpaired electrons,so they are colourless.
$B$) $Ti^{4+}$ $(3d^{0})$ is colourless,but $Cu^{2+}$ $(3d^{9})$ has one unpaired electron and is coloured.
$C$) $V^{2+}$ $(3d^{3})$ and $Ti^{3+}$ $(3d^{1})$: Both have unpaired electrons and are coloured.
$D$) $Zn^{2+}$ $(3d^{10})$ is colourless,but $Mn^{2+}$ $(3d^{5})$ has five unpaired electrons and is coloured.
Therefore,the correct pair is $Sc^{3+}, Zn^{2+}$.

(D) The balanced chemical equation for the reaction in a neutral or faintly alkaline medium is:
$8MnO_4^- + 3S_2O_3^{2-} + H_2O \rightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^-$
In $KMnO_4$,the oxidation state of $Mn$ is $+7$.
In $MnO_2$,the oxidation state of $Mn$ is $+4$.
The change in the oxidation state of $Mn$ is $|(+4) - (+7)| = 3$.

(D) Herbicides are chemical substances used to destroy unwanted plants. $Sodium \ chlorate$ $(NaClO_3)$ and $sodium \ arsenite$ $(Na_3AsO_3)$ are both well-known examples of herbicides. $Aldrin$ and $Dieldrin$ are classified as insecticides.

(C) Ozonolysis of a benzene ring with fixed double bonds involves the cleavage of the $C=C$ bonds to form carbonyl compounds.
For $1,2$-dimethylbenzene ($o$-xylene),the two resonance structures have different positions of the double bonds relative to the methyl groups.
In the first structure,the double bonds are between $C_1-C_2$,$C_3-C_4$,and $C_5-C_6$. Ozonolysis yields $CH_3-CO-CO-CH_3$ and $2 \ CHO-CHO$.
In the second structure,the double bonds are between $C_2-C_3$,$C_4-C_5$,and $C_6-C_1$. Ozonolysis yields $2 \ CH_3-CO-CHO$ and $CHO-CHO$.
Since the products are different,the pair of $o$-xylene structures gives different products upon ozonolysis.

(A) According to Heisenberg's uncertainty principle:
$\Delta x \times \Delta p_{x} \geq \frac{h}{4 \pi}$
Given $\Delta x = 2 a_{0} = 2 \times 52.9 \times 10^{-12} \ m = 105.8 \times 10^{-12} \ m$.
For minimum uncertainty,$\Delta x \times m \Delta v = \frac{h}{4 \pi}$.
$\Delta v = \frac{h}{4 \pi \times m \times \Delta x}$
Substituting the values:
$\Delta v = \frac{6.63 \times 10^{-34}}{4 \times 3.14159 \times 9.1 \times 10^{-31} \times 105.8 \times 10^{-12}}$
$\Delta v \approx 548273 \ m \ s^{-1} = 548.273 \ km \ s^{-1}$.
Rounding to the nearest integer,the value is $548 \ km \ s^{-1}$.

(B) The reaction is: $HNO_3 + NaOH \rightarrow NaNO_3 + H_2O$
Initial moles of $HNO_3 = 600 \; mL \times 0.2 \; M = 120 \; m \; mol = 0.12 \; mol$
Initial moles of $NaOH = 400 \; mL \times 0.1 \; M = 40 \; m \; mol = 0.04 \; mol$
Since $NaOH$ is the limiting reagent,the moles of water formed $= 0.04 \; mol$.
Heat released $(q)$ $= \text{moles of water} \times \Delta_{neut}H = 0.04 \; mol \times 57 \times 10^3 \; J \; mol^{-1} = 2280 \; J$.
Total volume of solution $= 600 \; mL + 400 \; mL = 1000 \; mL$.
Assuming density of solution $\approx 1 \; g \; mL^{-1}$,mass of solution $(m)$ $= 1000 \; g$.
Using $q = m \times S \times \Delta T$:
$2280 \; J = 1000 \; g \times 4.2 \; J \; K^{-1} \; g^{-1} \times \Delta T$
$\Delta T = \frac{2280}{4200} \; K = 0.54286 \; K = 0.54286 \; ^{\circ}C$
Expressing in terms of $10^{-2} \; ^{\circ}C$:
$\Delta T = 54.286 \times 10^{-2} \; ^{\circ}C \approx 54 \times 10^{-2} \; ^{\circ}C$.

(B) The dissociation of $PbS$ is given by: $PbS(s) \rightleftharpoons Pb^{2+}(aq) + S^{2-}(aq)$.
The solubility product expression is $K_{sp} = [Pb^{2+}][S^{2-}] = S \times S = S^{2}$.
Given $K_{sp} = 8 \times 10^{-28}$.
Therefore,$S = \sqrt{8 \times 10^{-28}} = \sqrt{4 \times 2 \times 10^{-28}} = 2 \sqrt{2} \times 10^{-14} \ mol \ L^{-1}$.
Using the given value $\sqrt{2} = 1.41$,we get $S = 2 \times 1.41 \times 10^{-14} = 2.82 \times 10^{-14} \ mol \ L^{-1}$.
To express this in the form $x \times 10^{-16} \ mol \ L^{-1}$,we write $2.82 \times 10^{-14} = 282 \times 10^{-16}$.
Thus,the value of $x$ is $282$.

(D) The chemical reaction is: $CH_3-C \equiv CH + 2Br_2 \rightarrow CH_3-CBr_2-CHBr_2$
$1$. Molar mass of $Br_2 = 2 \times 80 = 160 \; g/mol$.
$2$. Molar mass of $1,1,2,2-$tetrabromopropane $(C_3H_4Br_4)$ $= (3 \times 12) + (4 \times 1) + (4 \times 80) = 36 + 4 + 320 = 360 \; g/mol$.
$3$. According to the stoichiometry,$2 \; mol$ of $Br_2$ produces $1 \; mol$ of $1,1,2,2-$tetrabromopropane.
$4$. Moles of $Br_2$ used $= \frac{1 \; g}{160 \; g/mol} = 0.00625 \; mol$.
$5$. Theoretical moles of $1,1,2,2-$tetrabromopropane $= \frac{0.00625}{2} = 0.003125 \; mol$.
$6$. Theoretical mass of $1,1,2,2-$tetrabromopropane $= 0.003125 \; mol \times 360 \; g/mol = 1.125 \; g$.
$7$. Actual yield $(27\%)$ $= 1.125 \; g \times 0.27 = 0.30375 \; g$.
$8$. Expressing in $\times 10^{-1} \; g$: $0.30375 = 3.0375 \times 10^{-1} \; g$.
$9$. Nearest integer is $3$.

(C) The balanced chemical equation is:
$4 HNO_{3} + 3 KCl \rightarrow Cl_{2} + NOCl + 2 H_{2}O + 3 KNO_{3}$
Calculate the molar masses:
$M(HNO_{3}) = 1 + 14 + (3 \times 16) = 63 \ g/mol$
$M(KNO_{3}) = 39 + 14 + (3 \times 16) = 101 \ g/mol$
From the stoichiometry,$3 \ moles$ of $KNO_{3}$ are produced by $4 \ moles$ of $HNO_{3}$.
Number of moles of $KNO_{3}$ produced = $\frac{110.0 \ g}{101 \ g/mol} \approx 1.089 \ mol$.
Moles of $HNO_{3}$ required = $\frac{4}{3} \times (\text{moles of } KNO_{3}) = \frac{4}{3} \times \frac{110}{101} \approx 1.452 \ mol$.
Mass of $HNO_{3}$ required = $1.452 \ mol \times 63 \ g/mol = 91.5 \ g$.

(B) The energy of an orbital is determined by the $(n + l)$ rule.
$A$ $\Rightarrow 3d$ $\Rightarrow n + l = 3 + 2 = 5$
$B$ $\Rightarrow 4p$ $\Rightarrow n + l = 4 + 1 = 5$
$C$ $\Rightarrow 4d$ $\Rightarrow n + l = 4 + 2 = 6$
$D$ $\Rightarrow 3p$ $\Rightarrow n + l = 3 + 1 = 4$
According to the $(n + l)$ rule,the energy increases as the $(n + l)$ value increases.
If the $(n + l)$ values are the same,the orbital with the lower $n$ value has lower energy.
Comparing the values:
$D (n+l = 4) < A (n+l = 5, n=3) < B (n+l = 5, n=4) < C (n+l = 6)$
Therefore,the correct order of increasing energy is $D < A < B < C$.

(D) Given equations:
$C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} + 400 \; kJ \; (I)$
$C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)} + 100 \; kJ \; (II)$
Mass of coal $= 0.6 \; kg = 600 \; g$
Pure Carbon $= 600 \times \frac{60}{100} = 360 \; g$
Moles of pure Carbon $= \frac{360 \; g}{12 \; g/mol} = 30 \; mol$
Carbon converted into $CO = 30 \times 0.60 = 18 \; mol$
Carbon converted into $CO_2 = 30 - 18 = 12 \; mol$
Energy from $CO$ formation $= 18 \; mol \times 100 \; kJ/mol = 1800 \; kJ$
Energy from $CO_2$ formation $= 12 \; mol \times 400 \; kJ/mol = 4800 \; kJ$
Total heat generated $= 1800 \; kJ + 4800 \; kJ = 6600 \; kJ$

(B) The total moles of $H^{+}$ ions from $HCl$ is $n_{1} = M \times V = 0.01 \times 0.2 = 0.002 \ mol$.
The total moles of $H^{+}$ ions from $H_{2}SO_{4}$ is $n_{2} = M \times V \times 2 = 0.01 \times 0.4 \times 2 = 0.008 \ mol$.
Total moles of $H^{+} = 0.002 + 0.008 = 0.01 \ mol$.
Total volume of the mixture = $200 \ mL + 400 \ mL = 600 \ mL = 0.6 \ L$.
Concentration of $[H^{+}] = \frac{0.01 \ mol}{0.6 \ L} = \frac{1}{60} \ M$.
$pH = -\log[H^{+}] = -\log(\frac{1}{60}) = \log(60) = \log(6 \times 10) = \log(6) + \log(10) \approx 0.778 + 1 = 1.778 \approx 1.78$.

(C) $SnH_4$ (stannane) is a molecular hydride formed by the reaction of $Sn$ with hydrogen,which is true.
$SnH_4$ has a tetrahedral geometry with $sp^3$ hybridization,which means it is a non-planar molecule.
Therefore,Statement $I$ is true and Statement $II$ is false.

(B) Gypsum $(CaSO_{4} \cdot 2 H_{2}O)$ is added to Portland cement to slow down the process of setting,thereby increasing the setting time.

(D) The target alcohol is $1,1-diphenyl-1-ethanol$ (or $1-phenylethanol$ derivative with two phenyl groups attached to the same carbon).
Reaction of a ketone with a Grignard reagent $(RMgX)$ followed by acidic hydrolysis yields a tertiary alcohol.
Specifically,the reaction of acetophenone $(Ph-CO-CH_3)$ with one mole of phenyl magnesium bromide $(PhMgBr)$ followed by acidic hydrolysis $(H^+)$ proceeds as follows:
$Ph-CO-CH_3 + PhMgBr \rightarrow Ph-C(OMgBr)(Ph)-CH_3$
$Ph-C(OMgBr)(Ph)-CH_3 + H_2O/H^+ \rightarrow Ph-C(OH)(Ph)-CH_3 + Mg(OH)Br$
This matches the structure of the product shown in the image. Therefore,the correct reactant is acetophenone $(Ph-CO-CH_3)$.

(D) The reaction between stearic acid $(CH_3(CH_2)_{16}COOH)$ and polyethylene glycol $(OH(CH_2CH_2O)_nCH_2CH_2OH)$ involves the elimination of a water molecule to form an ester linkage.
The chemical reaction is:
$CH_3(CH_2)_{16}COOH + OH(CH_2CH_2O)_nCH_2CH_2OH \rightarrow CH_3(CH_2)_{16}CO(OCH_2CH_2)_{n+1}OH + H_2O$
This product does not contain any ionic groups and is therefore classified as a non-ionic detergent.

(A) sugar is considered a reducing sugar if it possesses a free $-OH$ group at the anomeric carbon,which allows it to exist in equilibrium with its open-chain aldehyde or ketone form. This form can then act as a reducing agent.
In the provided options,the structure in option $A$ shows a cyclic monosaccharide with a free $-OH$ group at the anomeric carbon,making it a reducing sugar.

(C) Aniline is a Lewis base and reacts with the Lewis acid $AlCl_{3}$ to form an acid-base adduct (salt),$C_{6}H_{5}NH_{2} \cdot AlCl_{3}$.
In this adduct,the nitrogen atom carries a positive charge,which makes the $-NH_{2}$ group strongly electron-withdrawing (deactivating) and meta-directing.
However,because the ring is so strongly deactivated,it does not undergo Friedel-Crafts alkylation with $CH_{3}Cl$.
Therefore,the reaction does not yield $o-$ or $p-$methylaniline,nor does it yield $m-$methylaniline; it simply does not proceed.
Thus,Assertion $(A)$ is true,but Reason $(R)$ is false because the reaction does not yield $m-$methylaniline.

(D) Given reaction: $Cu_{(s)} + 2 Ag^{+}_{(aq)} \rightleftharpoons Cu^{2+}_{(aq)} + 2 Ag_{(s)}$ with $K_{eq} = 2 \times 10^{15}$.
The target reaction is $\frac{1}{2} Cu^{2+}_{(aq)} + Ag_{(s)} \rightleftharpoons \frac{1}{2} Cu_{(s)} + Ag^{+}_{(aq)}$.
This target reaction is the reverse of the original reaction multiplied by a factor of $\frac{1}{2}$.
Therefore,the new equilibrium constant $K_{eq}^{\prime} = (K_{eq})^{-1/2} = \frac{1}{\sqrt{K_{eq}}}$.
$K_{eq}^{\prime} = \frac{1}{\sqrt{2 \times 10^{15}}} = \frac{1}{\sqrt{20 \times 10^{14}}} = \frac{1}{\sqrt{20} \times 10^7} = \frac{1}{4.472 \times 10^7} \approx 0.2236 \times 10^{-7} = 2.236 \times 10^{-8}$.
Comparing this with $x \times 10^{-8}$,we get $x = 2.236$.
The nearest integer value of $x$ is $2$.

(C) The chemical formula for the reduction of $Fe_{3}O_{4}$ to $Fe$ is:
$Fe_{3}O_{4} + 8e^{-} \rightarrow 3Fe + 4O^{2-}$
In this reaction,$8$ moles of electrons are required to produce $3$ moles of iron $(Fe)$.
Therefore,the charge required to produce $1$ mole of iron is $8/3 \, F$.
$8/3 \approx 2.67 \, F$.
Rounding to the nearest integer,we get $3 \, F$.

(A) The half-life of a reaction of order $n$ is related to the initial concentration $[A_0]$ as $t_{\frac{1}{2}} \propto \frac{1}{[A_0]^{n-1}}$.
Given:
$t_{\frac{1}{2}, 1} = 100 \ s$ at $[A_0]_1 = 0.5 \ mol \ L^{-1}$
$t_{\frac{1}{2}, 2} = 50 \ s$ at $[A_0]_2 = 1.0 \ mol \ L^{-1}$
Taking the ratio:
$\frac{t_{\frac{1}{2}, 1}}{t_{\frac{1}{2}, 2}} = \left( \frac{[A_0]_2}{[A_0]_1} \right)^{n-1}$
$\frac{100}{50} = \left( \frac{1.0}{0.5} \right)^{n-1}$
$2 = (2)^{n-1}$
$2^1 = 2^{n-1}$
Comparing the exponents:
$n - 1 = 1$
$n = 2$
The order of the reaction is $2$.

(C) For $[Co(H_2O)_6]Cl_2$:
$Co^{2+}$ has a $3d^7$ configuration. In an octahedral field,the electrons are arranged as $t_{2g}^5 e_g^2$,resulting in $3$ unpaired electrons.
The spin-only magnetic moment $\mu = \sqrt{n(n+2)} \ BM = \sqrt{3(3+2)} = \sqrt{15} \ BM$.
For $[Cr(H_2O)_6]Cl_3$:
$Cr^{3+}$ has a $3d^3$ configuration. The electrons are arranged as $t_{2g}^3 e_g^0$,resulting in $3$ unpaired electrons.
The spin-only magnetic moment $\mu = \sqrt{n(n+2)} \ BM = \sqrt{3(3+2)} = \sqrt{15} \ BM$.
Difference in spin-only magnetic moment $= \sqrt{15} - \sqrt{15} = 0$.

(B) Statement $I$ is false because $KI$ is a strong electrolyte,and for strong electrolytes,molar conductivity increases only slightly with dilution.
Statement $II$ is false because carbonic acid $(H_2CO_3)$ is a weak electrolyte,and for weak electrolytes,molar conductivity increases sharply with dilution due to an increase in the degree of dissociation.
Therefore,both statements are false.

(C) Assertion $(A)$ is correct because dissolved substances (crystalloids) can pass through parchment paper,while colloidal particles cannot,allowing for purification via dialysis.
Reason $(R)$ is incorrect because particles in a true solution (crystalloids) can easily pass through parchment paper,whereas colloidal particles are too large to pass through it.
Therefore,$(A)$ is correct but $(R)$ is incorrect.

(A) In complexes where the metal is in a low oxidation state,the metal center is electron-rich. To stabilize this state,the ligands must be able to accept electron density from the metal through back-bonding. Therefore,ligands with good $\pi$-accepting character (such as $CO$,$NO^+$,$CN^-$) are required to form stable complexes with metals in low oxidation states.

(D) The reaction involves the treatment of a ketone with a base $(NaOH)$ and an alkyl halide $(CH_3CH_2Br)$. This is an alkylation reaction of the enolate ion formed from the ketone.
$1$. The base $(NaOH)$ abstracts an $\alpha$-hydrogen from the ketone $(R-CO-CH_3)$ to form an enolate ion $(R-CO-CH_2^-)$.
$2$. The enolate ion then acts as a nucleophile and attacks the alkyl halide $(CH_3CH_2Br)$ via an $S_N2$ mechanism.
$3$. The alkyl group $(CH_3CH_2-)$ gets attached to the $\alpha$-carbon of the ketone.
$4$. The resulting product is $R-CO-CH_2-CH_2-CH_3$.
Comparing this with the given options,option $D$ represents the structure $R-CO-CH_2-CH_2-CH_3$.

(B) The reaction of $2$-methylbut-$2$-ene with $Br_2$ in the presence of $CH_3OH$ (a polar protic solvent) proceeds via the formation of a bromonium ion intermediate.
$CH_3OH$ acts as a nucleophile and attacks the more substituted carbon atom of the bromonium ion to form an ether,$A$ ($3$-bromo-$2$-methoxy-$2$-methylbutane).
Reaction: $CH_3-C(CH_3)=CH-CH_3 + Br_2 + CH_3OH \rightarrow CH_3-C(OCH_3)(CH_3)-CH(Br)-CH_3$ $(A)$.
When $A$ reacts with $HI$,the methoxy group $(-OCH_3)$ is protonated by $H^+$ to form an oxonium ion,which is a good leaving group.
$I^-$ then attacks the tertiary carbon atom,displacing $CH_3OH$ to form the final product $B$ ($3$-bromo-$2$-iodo-$2$-methylbutane).
Therefore,the major product $B$ is $CH_3-C(I)(CH_3)-CH(Br)-CH_3$.

(A) The correct matches are as follows:
$A$. $C_6H_5COCl + H_2 \xrightarrow{Pd-BaSO_4} C_6H_5CHO$ is the Rosenmund reaction $(IV)$.
$B$. $CH_3CN + SnCl_2 + HCl \rightarrow CH_3CHO$ is the Stephen reaction $(III)$.
$C$. $C_6H_5CH_3 + CrO_2Cl_2 \rightarrow C_6H_5CHO$ is the Etard reaction $(II)$.
$D$. $C_6H_6 + CO + HCl \xrightarrow{AlCl_3} C_6H_5CHO$ is the Gattermann-Koch reaction $(I)$.
Therefore,the correct sequence is $A-IV, B-III, C-II, D-I$.

(A) The correct matches are as follows:
$(A)$ Neoprene is a polymer of chloroprene $(2-chloro-1,3-butadiene)$. Thus,$A-II$.
$(B)$ Teflon is a polymer of tetrafluoroethene $(CF_2=CF_2)$. Thus,$B-III$.
$(C)$ Acrilan (polyacrylonitrile) is a polymer of acrylonitrile $(CH_2=CH-CN)$. Thus,$C-I$.
$(D)$ Natural rubber is a polymer of isoprene $(2-methyl-1,3-butadiene)$. Thus,$D-IV$.
Therefore,the correct sequence is $A-II, B-III, C-I, D-IV$.

(A) $1$. The compound $A$ contains $N$ and $Cl$ and forms an acidic solution in water,suggesting it is an amine hydrochloride salt $(R-NH_3^+Cl^-)$.
$2$. The molecular weight of $C_6H_5NH_3Cl$ is $77 + 14 + 3 + 35.5 = 129.5$,which matches the given $131 \pm 2$.
$3$. Treatment with $NaOH$ releases the free amine: $C_6H_5NH_3Cl + NaOH \rightarrow C_6H_5NH_2 + NaCl + H_2O$. The liquid $C_6H_5NH_2$ (aniline) contains $N$ but no $Cl$.
$4$. Aniline reacts with nitrous acid $(HNO_2)$ at $0-5^{\circ}C$ to form benzenediazonium chloride,which then couples with phenol in basic medium to form $p$-hydroxyazobenzene,an orange dye.
$5$. Thus,$A$ is $C_6H_5NH_3Cl$ (anilinium chloride).

(A) The reactions of glucose are as follows:
$A$. Glucose reacts with $HI$ to form $n$-hexane,which indicates the presence of a straight chain of six carbon atoms. Thus,$A-IV$.
$B$. Glucose reacts with $Br_{2}$ water (a mild oxidizing agent) to form gluconic acid,which indicates the presence of an aldehyde group. Thus,$B-I$.
$C$. Glucose reacts with acetic anhydride to form glucose pentacetate,which indicates the presence of five $-OH$ groups. Thus,$C-II$.
$D$. Glucose reacts with $HNO_{3}$ (a strong oxidizing agent) to form saccharic acid (glucaric acid),which indicates the presence of both a primary alcohol and an aldehyde group. Thus,$D-III$.
Therefore,the correct matching is $A-IV, B-I, C-II, D-III$.

(C) Rosin is added to soaps during manufacturing. It forms sodium rosinate,which is a substance that enhances the lathering property of the soap.

(D) The reaction between $Fe^{3+}$ ions and potassium ferrocyanide,$K_4[Fe(CN)_6]$,leads to the formation of ferric ferrocyanide,which is known as Prussian blue.
The balanced chemical equation is:
$4Fe^{3+} + 3[Fe(CN)_6]^{4-} \longrightarrow Fe_4[Fe(CN)_6]_3$ (Prussian blue).

(D) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is given by: $\frac{P^{0} - P_{s}}{P^{0}} = \frac{n_{A}}{n_{A} + n_{solvent}}$.
Given that the vapour pressure of the solution $P_{s}$ is half of the pure water $P^{0}$,so $P_{s} = \frac{P^{0}}{2}$.
Substituting the values: $\frac{P^{0} - P^{0}/2}{P^{0}} = \frac{n_{A}}{n_{A} + n_{solvent}} \implies \frac{1}{2} = \frac{n_{A}}{n_{A} + n_{solvent}}$.
This simplifies to $n_{A} + n_{solvent} = 2n_{A}$,which means $n_{A} = n_{solvent}$.
The number of moles of water $(n_{solvent})$ is $\frac{100 \ g}{18 \ g/mol} = 5.55 \ mol$.
Therefore,$n_{A} = 5.55 \ mol$.
Rounding to the nearest integer,we get $6$.

(B) For a first-order reaction,the half-life $t_{1/2}$ is given as $70 \ min$.
Converting time to seconds: $t_{1/2} = 70 \times 60 \ s = 4200 \ s$.
The rate constant $k$ is calculated as:
$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{4200} \ s^{-1}$.
$k = \frac{693}{4200000} \ s^{-1} = \frac{693}{42} \times 10^{-6} \ s^{-1}$.
$k = 165 \times 10^{-6} \ s^{-1}$.
Thus,the value of $x$ is $165$.

(C) The principal ores of iron are those from which iron is commercially extracted. From the given list,the iron ores are:
$1$. Siderite $(FeCO_3)$
$2$. Haematite $(Fe_2O_3)$
$3$. Magnetite $(Fe_3O_4)$
$4$. Limonite $(Fe_2O_3 \cdot 3H_2O)$
Thus,there are $4$ principal ores of iron.

(D) Fehling's reagent contains a complex of $Cu^{2+}$ ions,specifically copper$(II)$ tartrate.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
In a $3d^9$ configuration,there is $1$ unpaired electron.
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.
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ B.M.$
The nearest integer to $1.73$ is $2$.

(B) The molecular formula of glucose is $C_6H_{12}O_6$. The molar mass of glucose is $6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \ g/mol$.
The mass of carbon in $180 \ g$ of glucose is $6 \times 12 = 72 \ g$.
Given that the solution contains $10.8 \ \%$ carbon by weight,the mass of carbon in $250 \ g$ of solution is $\frac{10.8}{100} \times 250 = 27 \ g$.
Since $72 \ g$ of carbon is present in $180 \ g$ of glucose,the mass of glucose containing $27 \ g$ of carbon is $\frac{180}{72} \times 27 = 67.5 \ g$.
The mass of the solvent (water) is $250 \ g - 67.5 \ g = 182.5 \ g = 0.1825 \ kg$.
The number of moles of glucose is $\frac{67.5 \ g}{180 \ g/mol} = 0.375 \ mol$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in } kg} = \frac{0.375 \ mol}{0.1825 \ kg} \approx 2.055 \ m \approx 2.06 \ m$.

(A) $O_2, Cu^{2+}$ and $Fe^{3+}$ contain unpaired electrons,hence they are paramagnetic.
Paramagnetic substances are weakly attracted by an external magnetic field and are magnetized in the direction of the field.
$NaCl$ and $H_2O$ have all electrons paired,hence they are diamagnetic.
Diamagnetic substances are weakly repelled by an external magnetic field and are weakly magnetized in the direction opposite to the applied magnetic field.
Therefore,both Statement $I$ and Statement $II$ are correct.

(C) The extent of adsorption of a gas on a solid adsorbent depends on its critical temperature $(T_c)$.
Gases with higher critical temperatures are more easily liquefied and thus more readily adsorbed on the surface of activated charcoal.
$SO_{2}$ has a higher critical temperature $(430 K)$ compared to $CH_{4}$ $(190 K)$.
Therefore,$SO_{2}$ is adsorbed more efficiently than $CH_{4}$,making Assertion $A$ correct.
Reason $R$ states that gases with lower critical temperatures are readily adsorbed,which is incorrect.
Thus,$A$ is correct but $R$ is not correct.

(B) For a dilute aqueous solution,the elevation in boiling point is given by $\Delta T_B = K_B \times m$,where $m$ is the molality.
Since the boiling points are equal,the elevation in boiling point is the same,so the molalities must be equal: $m_A = m_B$.
For $A$: $2 \ g$ of solute in $98 \ g$ of solvent. Molality $m_A = \frac{2 / M_A}{98 / 1000} = \frac{2000}{98 M_A}$.
For $B$: $8 \ g$ of solute in $92 \ g$ of solvent. Molality $m_B = \frac{8 / M_B}{92 / 1000} = \frac{8000}{92 M_B}$.
Equating $m_A = m_B$: $\frac{2000}{98 M_A} = \frac{8000}{92 M_B}$.
$\frac{1}{98 M_A} = \frac{4}{92 M_B}$.
$M_B = \frac{4 \times 98}{92} M_A = \frac{392}{92} M_A \approx 4.26 M_A$.
Given the standard approximation often used in such problems where the mass of solvent is taken as $100 \ g$ (assuming dilute solution),the ratio becomes $\frac{2/M_A}{100} = \frac{8/M_B}{100}$,which leads to $M_B = 4M_A$.

(A) Refining is the process of purifying crude metal.
$A$. Liquation is a refining method used for metals with low melting points (e.g.,$Sn$).
$B$. Calcination is a process of concentration of ore (converting carbonate ore to oxide),not refining.
$C$. Electrolysis is a common refining method (e.g.,for $Cu$,$Zn$).
$D$. Leaching is a process of concentration of ore (chemical extraction),not refining.
$E$. Distillation is a refining method used for volatile metals (e.g.,$Zn$,$Hg$).
Therefore,$B$ (Calcination) and $D$ (Leaching) are not used for refining.

(D) The chemical formulas for the given oxoacids of phosphorus are as follows:
$1$. Pyrophosphorous acid: $H_4P_2O_5$ ($5$ oxygen atoms)
$2$. Hypophosphoric acid: $H_4P_2O_6$ ($6$ oxygen atoms)
$3$. Phosphoric acid: $H_3PO_4$ ($4$ oxygen atoms)
$4$. Pyrophosphoric acid: $H_4P_2O_7$ ($7$ oxygen atoms)
Comparing the number of oxygen atoms,Pyrophosphoric acid $(H_4P_2O_7)$ contains the highest number of oxygen atoms $(7)$.

(B) Statement $I$ is false because neutral $KMnO_{4}$ oxidises $I^{-}$ to $IO_{3}^{-}$ (iodate ion),not $I_{2}$.
Statement $II$ is false because the manganate ion $(MnO_{4}^{2-})$ involves $d\pi-p\pi$ bonding,not $p\pi-p\pi$ bonding,although it is paramagnetic due to one unpaired electron.

(C) The structure of $Mn_2O_7$ consists of two $MnO_4$ tetrahedra sharing a common oxygen atom at one corner.
In this structure,each $Mn$ atom is bonded to three terminal oxygen atoms via double bonds $(Mn=O)$ and one bridging oxygen atom via a single bond $(Mn-O-Mn)$.
Therefore,there are $3$ $Mn=O$ bonds around each $Mn$ atom.
Total number of $Mn=O$ bonds = $3 + 3 = 6$.

(A) The reaction proceeds in two steps:
$1$. The reaction of $4$-hydroxybenzyl alcohol with $HCl$ under heating $(\Delta)$ leads to the substitution of the alcoholic $-OH$ group with $-Cl$,forming $4$-(chloromethyl)phenol as product $A$.
$2$. The reaction of $A$ with $NaI$ in acetone (Finkelstein reaction) proceeds via an $S_N2$ mechanism,where the $-Cl$ atom is replaced by an $-I$ atom to form $4$-(iodomethyl)phenol as product $B$.

(B) The correct matches are as follows:
$A$. Phenol-formaldehyde resin (linear) is known as Novolac $(II)$.
$B$. Copolymer of $1,3$-butadiene and styrene is known as Buna-$S$ $(III)$.
$C$. Polyester of glycol and phthalic acid is known as Glyptal $(I)$.
$D$. Polyester of glycol and terephthalic acid is known as Dacron $(IV)$.
Therefore,the correct matching is $A-II, B-III, C-I, D-IV$.

(A) The reaction described is the basis of the Bial's test or a variation of Seliwanoff's test conditions.
Aldopentoses,when heated with concentrated acids (like $HCl$),undergo dehydration to form furfural.
Furfural then reacts with resorcinol to produce a coloured condensation product.
While ketoses react rapidly (Seliwanoff's test),aldopentoses react more slowly under these conditions to yield furfural derivatives.
Therefore,the sugar '$X$' is an aldopentose.

(C) The correct matches are as follows:
$A$. Morphine is a narcotic analgesic,so $A-III$.
$B$. Chloroxylenol is used as an antiseptic,so $B-IV$.
$C$. Phenelzine is an anti-depressant,so $C-I$.
$D$. Saccharin is $550$ times sweeter than cane sugar,so $D-II$.
Therefore,the correct matching is $A-III, B-IV, C-I, D-II$.

(C) . Benzenesulphonyl chloride $\rightarrow$ $III$. Hinsberg reagent
$B$. Hoffmann bromamide reaction $\rightarrow$ $IV$. Known reaction of isocyanates
$R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + 2NaBr + Na_2CO_3 + 2H_2O$
Intermediate: $R-N=C=O$ (isocyanate)
$C$. Carbylamine reaction $\rightarrow$ $I$. Test for primary amine
$R-NH_2 + CHCl_3 + 3KOH \rightarrow RNC + 3KCl + 3H_2O$
$D$. Hoffmann orientation $\rightarrow$ $II$. Anti Saytzeff (Formation of less substituted alkene as major product)
Therefore,the correct matching is $A-III, B-IV, C-I, D-II$.

(C) Let the rate law be $Rate = k[P_{NO}]^x [P_{H_2}]^y$.
Comparing experiments $1$ and $2$ where $[P_{H_2}]$ is constant:
$\frac{Rate_1}{Rate_2} = (\frac{P_{NO,1}}{P_{NO,2}})^x$
$\frac{0.135}{0.033} \approx 4.09 \approx 4$
$(\frac{40.0}{20.1})^x \approx 2^x$
$4 = 2^x \implies x = 2$.
Thus,the order of the reaction with respect to $NO$ is $2$.

(A) The complex $CoCl_3(NH_3)_4$ acts as a $1:1$ electrolyte,which means it dissociates into two ions in solution.
This indicates the formula is $[Co(NH_3)_4Cl_2]Cl$.
The primary valency of the central metal ion $(Co)$ corresponds to its oxidation state.
In $[Co(NH_3)_4Cl_2]Cl$,let the oxidation state of $Co$ be $x$.
$x + 4(0) + 2(-1) + 1(-1) = 0$
$x - 3 = 0$
$x = +3$.
Therefore,the primary valency is $3$.

(C) . $Cd_{(s)} + 2 Ni(OH)_{3(s)} \rightarrow CdO_{(s)} + 2 Ni(OH)_{2(s)} + H_2O_{(l)}$ represents the discharging of a nickel-cadmium secondary battery $(II)$.
$B$. $Zn(Hg) + HgO_{(s)} \rightarrow ZnO_{(s)} + Hg_{(l)}$ is the reaction in a mercury cell,which is a primary battery $(I)$.
$C$. $2 PbSO_{4(s)} + 2 H_2O_{(l)} \rightarrow Pb_{(s)} + PbO_{2(s)} + 2 H_2SO_{4(aq)}$ represents the charging of a lead storage battery $(IV)$.
$D$. $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$ is the reaction in a hydrogen-oxygen fuel cell $(III)$.
Therefore,the correct matching is $A-II, B-I, C-IV, D-III$.

(A) . $4NH_{3(g)} + 5O_{2(g)} \xrightarrow{Pt_{(s)}} 4NO_{(g)} + 6H_2O_{(g)}$ (Ostwald process uses $Pt_{(s)}$ as a catalyst).
$B$. $N_{2(g)} + 3H_{2(g)} \xrightarrow{Fe_{(s)}} 2NH_{3(g)}$ (Haber process uses $Fe_{(s)}$ as a catalyst).
$C$. $C_{12}H_{22}O_{11(aq)} + H_2O_{(l)} \xrightarrow{H^+} C_6H_{12}O_6 + C_6H_{12}O_6$ (Inversion of cane sugar uses $H_2SO_{4(l)}$ as a catalyst).
$D$. $2SO_{2(g)} + O_{2(g)} \xrightarrow{NO_{(g)}} 2SO_{3(g)}$ (Lead chamber process uses $NO_{(g)}$ as a catalyst).
Therefore,the correct matching is $A-III, B-IV, C-II, D-I$.

(C) Leaching is a process of concentrating ore by dissolving it in a suitable solvent in which the ore is soluble but the impurities are not.
The reaction $Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4]$ represents the Bayer's process,which is a classic example of leaching bauxite ore $(Al_2O_3)$ using a sodium hydroxide solution.

(C) The reactions are as follows:
$A$. $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + Cr_2O_3 + 4H_2O$. Thus,$A-II$.
$B$. $2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 5Cl_2 + 8H_2O$. Thus,$B-IV$.
$C$. $2Al + 2NaOH + 6H_2O \rightarrow 2Na[Al(OH)_4] + 3H_2$. Thus,$C-I$.
$D$. $2NaNO_3 \xrightarrow{\Delta} 2NaNO_2 + O_2$. Thus,$D-III$.
Therefore,the correct matching is $A-II, B-IV, C-I, D-III$.

(A) The tendency to liberate $H_2$ gas from mineral acids depends on the standard reduction potential of the metal. Metals with a negative reduction potential (above $H$ in the electrochemical series) can displace $H_2$.
Copper $(Cu)$ has a positive standard reduction potential $(E^0 = +0.34 \ V)$,meaning it lies below $H$ in the reactivity series.
Therefore,$Cu$ cannot displace $H_2$ from mineral acids,while $Mn$,$Ni$,and $Zn$ have negative reduction potentials and can liberate $H_2$.

(C) The $pK_{a}$ value is inversely proportional to the acidity of the compound. The acidity depends on the stability of the conjugate base formed after the removal of the proton.
$1$. In $A$,the conjugate base is stabilized by resonance with the carbonyl group,but it also experiences cross-conjugation with the $-OH$ group.
$2$. In $B$,the conjugate base is stabilized by resonance with the carbonyl group.
$3$. In $C$,the conjugate base is stabilized by resonance with the carbonyl group $AND$ the phenyl ring,making it highly stable.
$4$. In $D$,the conjugate base is stabilized by resonance with the carboxylate group,but it is less stable than $C$ due to the nature of the resonance.
Comparing the stability of the conjugate bases,the benzylic proton in $C$ $(C_6H_5-CH_2-CO-CH_3)$ forms the most stable conjugate base due to extended conjugation with both the phenyl ring and the carbonyl group. Therefore,it is the most acidic and has the lowest $pK_{a}$ value.

(B) $1$. The reaction of benzene with isopropyl chloride in the presence of $AlCl_3$ (Friedel-Crafts alkylation) gives cumene (isopropylbenzene).
$2$. Oxidation of cumene with $O_2$ followed by treatment with $H^+/H_2O$ yields phenol $(P)$ and acetone.
$3$. Phenol $(P)$ reacts with $Na_2Cr_2O_7/H_2SO_4$ (strong oxidizing agent) to form $p$-benzoquinone $(A)$.
$4$. Phenol $(P)$ reacts with $Br_2$ in $CS_2$ (a non-polar solvent) to give $p$-bromophenol $(B)$ as the major product due to the low polarity of the medium,which prevents further bromination.

(B) The given reaction is an elimination reaction of a quaternary ammonium salt $(CH_{3}CH_{2}CH_{2}CH(N^{+}(CH_{3})_{3})CH_{3})$ using a base $(C_{2}H_{5}ONa)$.
This is an example of the Hofmann elimination reaction.
In Hofmann elimination,the less substituted alkene is formed as the major product due to steric hindrance caused by the bulky leaving group $(N^{+}(CH_{3})_{3})$.
Therefore,the terminal alkene $CH_{3}CH_{2}CH_{2}CH=CH_{2}$ is the major product $(A)$,and the internal alkene $CH_{3}CH_{2}CH=CHCH_{3}$ is the minor product $(B)$.

(C) Terylene,also known as $Dacron$,is a polyester fiber.
It is prepared by the condensation polymerization of Ethane-$1, 2$-diol (ethylene glycol) and Benzene-$1, 4$-dicarboxylic acid (terephthalic acid).
The reaction involves the elimination of water molecules to form an ester linkage between the monomers.

(A) To convert the cyclic hemiacetal structure into a Haworth projection (pyranose form),we follow these steps:
$1$. Identify the carbon atoms in the chain. The hemiacetal is formed between the $C_1$ (anomeric carbon) and the $C_5$ hydroxyl group.
$2$. In the Haworth projection,the oxygen atom is placed at the top right corner of the ring.
$3$. Groups on the right side of the Fischer projection are placed pointing downwards,and groups on the left side are placed pointing upwards.
$4$. For the given structure,the $C_2-OH$ is on the left (up),$C_3-OH$ is on the right (down),$C_4-OH$ is on the left (up),and $C_5-CH_2OH$ is pointing upwards.
$5$. Comparing this configuration to the given options,the structure in option $(A)$ correctly represents the spatial arrangement of the hydroxyl groups and the hydroxymethyl group.

(C) Enzyme inhibitors are classified into two main types:
$1$. Competitive inhibitors: These drugs compete with the natural substrate for the active site of the enzyme.
$2$. Non-competitive inhibitors: These drugs bind to a different site known as the allosteric site,which changes the shape of the active site so that the substrate cannot bind.
Therefore,statement $A$ is correct because it identifies the two types.
Statement $B$ is correct because competitive inhibitors bind to the active site,while non-competitive inhibitors bind to the allosteric site.
Statements $C$ and $D$ are incorrect because they swap the definitions of competitive and non-competitive inhibitors.
Thus,only $A$ and $B$ are correct.

$(A)$ In the kinetic study of the reaction between $I^-$ and $H_2O_2$, starch is used as an indicator for $I_2$.
Freshly prepared starch solution is required because old solutions may undergo bacterial degradation.
$Na_2S_2O_3$ is added to react with the produced $I_2$ to prevent the blue colour from appearing until all $Na_2S_2O_3$ is consumed.
Therefore, the concentration of $Na_2S_2O_3$ must be less than that of $KI$ to ensure the reaction proceeds and the blue colour appears.
The time is recorded immediately after the appearance of the blue colour, which indicates the consumption of $Na_2S_2O_3$.
Thus, statements $(A)$, $(B)$, and $(C)$ are correct.