JEE Main 2021 Chemistry Question Paper with Answer and Solution

(B) The molar mass of aniline $(C_6H_5NH_2)$ is $93 \ g/mol$ and the molar mass of acetanilide $(C_6H_5NHCOCH_3)$ is $135 \ g/mol$.
Reaction: $C_6H_5NH_2 + CH_3COCl \rightarrow C_6H_5NHCOCH_3 + HCl$.
$1 \ mol$ of aniline produces $1 \ mol$ of acetanilide.
Moles of aniline used $= \frac{1.86 \ g}{93 \ g/mol} = 0.02 \ mol$.
Theoretical yield of acetanilide $= 0.02 \ mol \times 135 \ g/mol = 2.70 \ g$.
Since $10 \ \%$ of the product is lost during purification,the yield is $90 \ \%$ of the theoretical yield.
Actual yield $= 2.70 \ g \times 0.90 = 2.43 \ g$.
$2.43 \ g = 243 \times 10^{-2} \ g$.

(A) For the reaction: $3 HC \equiv CH_{(g)} \rightleftharpoons C_6H_{6(\ell)}$
$\Delta G^{\circ} = \sum \Delta_f G^{\circ}(\text{products}) - \sum \Delta_f G^{\circ}(\text{reactants})$
$\Delta G^{\circ} = [1 \times (-1.24 \times 10^5)] - [3 \times (-2.04 \times 10^5)]$
$\Delta G^{\circ} = -1.24 \times 10^5 + 6.12 \times 10^5 = 4.88 \times 10^5 \, J \, mol^{-1}$
We know that $\Delta G^{\circ} = -RT \ln K = -2.303 RT \log K$
$4.88 \times 10^5 = -2.303 \times 8.314 \times 298 \times \log K$
$\log K = -\frac{4.88 \times 10^5}{2.303 \times 8.314 \times 298} \approx -85.52$
The magnitude of $\log K$ is $85.52$.
Given $\log K = x \times 10^{-1}$,so $85.52 = x \times 10^{-1} \Rightarrow x = 855.2 \approx 855$.

(C) . In $SF_{4}$,the central atom $S$ undergoes $sp^{3}d$ hybridization,resulting in a see-saw geometry. Due to the presence of a lone pair in the equatorial position,the axial $S-F$ bonds are longer than the equatorial $S-F$ bonds. In contrast,$BF_{4}^{-}$,$XeF_{4}$,and $SiF_{4}$ have symmetric geometries where all bond lengths are equal.

Species	Hybridization and Bond Length Characteristics
$BF_{4}^{-}$	$sp^{3}$ (Tetrahedral); All bond lengths are equal
$XeF_{4}$	$sp^{3}d^{2}$ (Square planar); All bond lengths are equal
$SF_{4}$	$sp^{3}d$ (See-saw); Axial bond lengths > Equatorial bond lengths
$SiF_{4}$	$sp^{3}$ (Tetrahedral); All bond lengths are equal

(B) The reaction involves the electrophilic addition of benzene to the alkene $CH_3-CH(NO_2)-CH=CH_2$ in the presence of an acid catalyst $(H_2SO_4)$.
$1$. The acid protonates the alkene to form a carbocation intermediate.
$2$. The most stable carbocation is formed,which is the secondary carbocation at the carbon adjacent to the $NO_2$ group due to resonance stabilization or inductive effects.
$3$. Benzene acts as a nucleophile and attacks the carbocation to form the final product,which is $2-nitro-3-phenylbutane$.

(A) The correct answer is $(A)$.
Water reacts with carbon and hydrocarbons at high temperatures to produce syngas $(CO + H_2)$.
$C + H_2O \xrightarrow{1270 \ K} CO + H_2$
$CH_4 + H_2O \xrightarrow{1270 \ K, \ Ni} CO + 3H_2$
$C_3H_8 + 3H_2O \xrightarrow{\Delta, \ Ni} 3CO + 7H_2$
Reaction with $CO_2$:
$CO_2 + H_2O \rightleftharpoons H_2CO_3$ (Carbonic acid).
No $CO$ is formed in this reaction.

(D) Statement $I$ is true because normal rain water has a $pH$ of approximately $5.6$ due to the dissolution of atmospheric $CO_2$ to form weak carbonic acid $(H_2CO_3)$.
Statement $II$ is true because when the $pH$ of rain water drops below $5.6$ due to the presence of pollutants like oxides of sulfur $(SO_x)$ and nitrogen $(NO_x)$,it is termed as acid rain.
Therefore,both statements are correct.

(A) For the testing of halogens,$HNO_3$ (Nitric acid) is added to the sodium extract.
This is done because if $CN^-$ or $S^{2-}$ ions are present in the extract,they would interfere with the test by forming $AgCN$ or $Ag_2S$ precipitates.
$HNO_3$ decomposes these ions into $HCN$ and $H_2S$ gases,respectively,which are then removed by boiling,ensuring that only halides react with $AgNO_3$.

(C) The dissociation of $Ca(OH)_2$ is represented as:
$Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$
If $s$ is the solubility,then $[Ca^{2+}] = s$ and $[OH^-] = 2s$.
The solubility product expression is:
$K_{sp} = [Ca^{2+}][OH^-]^2 = (s)(2s)^2 = 4s^3$
Given $K_{sp} = 5.5 \times 10^{-6}$,we have:
$4s^3 = 5.5 \times 10^{-6}$
$s^3 = \frac{5.5}{4} \times 10^{-6} = 1.375 \times 10^{-6}$
$s = (1.375 \times 10^{-6})^{1/3} \approx 1.11 \times 10^{-2} \text{ M}$

(A) The reaction of ethylene glycol $(HOCH_2-CH_2OH)$ with oxalic acid $((COOH)_2)$ at $110^{\circ}C$ forms a cyclic ester (ethylene oxalate).
Upon further heating to $210^{\circ}C$,this cyclic ester undergoes decarboxylation to produce ethene $(CH_2=CH_2)$ and carbon dioxide $(CO_2)$.
Therefore,the major product $X$ formed at $210^{\circ}C$ is ethene $(CH_2=CH_2)$.

(B) The photoelectric effect depends on the work function of the metal.
$Cs$ (Cesium) has the lowest ionization energy among the alkali metals,making it the most suitable for use in photoelectric cells.
While other alkali metals like $Na$ and $Rb$ can exhibit the photoelectric effect,$Cs$ is the standard choice for such applications due to its low work function.
Therefore,only $1$ metal $(Cs)$ is typically cited for this specific application in the context of this question.

(D) The energy $E$ required to ionize one mole of atoms is given by the formula:
$E = \frac{hcN_{A}}{\lambda \times 1000} \text{ (in } kJ \ mol^{-1} \text{)}$
Given:
$h = 6.63 \times 10^{-34} \ J \cdot s$
$c = 3.00 \times 10^{8} \ m \cdot s^{-1}$
$N_{A} = 6.02 \times 10^{23} \ mol^{-1}$
$\lambda = 663 \ nm = 663 \times 10^{-9} \ m$
Substituting the values:
$E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8} \times 6.02 \times 10^{23}}{663 \times 10^{-9} \times 1000}$
$E = \frac{6.63 \times 3.00 \times 6.02 \times 10^{-3}}{663 \times 10^{-6}}$
$E = \frac{119.799 \times 10^{-3}}{663 \times 10^{-6}} = \frac{119.799}{0.663} \approx 180.69 \ kJ \ mol^{-1}$
Rounding off to the nearest integer,we get $181 \ kJ \ mol^{-1}$.

(C) For an isothermal process of an ideal gas,$\Delta U = 0$.
According to the first law of thermodynamics,$\Delta U = q + W$,so $q = -W$.
Work done against a constant external pressure is given by $W = -P_{ext} \Delta V$.
Using the ideal gas law $PV = nRT$,we have $V = \frac{nRT}{P}$.
Thus,$W = -P_{ext} \left( \frac{nRT}{P_2} - \frac{nRT}{P_1} \right) = -nRT P_{ext} \left( \frac{1}{P_2} - \frac{1}{P_1} \right)$.
Given $n = 5 \, mol$,$T = 293 \, K$,$P_{ext} = 4.3 \, MPa$,$P_1 = 2.1 \, MPa$,$P_2 = 1.3 \, MPa$.
$W = -5 \times 8.314 \times 293 \times 4.3 \times \left( \frac{1}{1.3} - \frac{1}{2.1} \right) \, J$.
$W = -52504.97 \times \left( 0.7692 - 0.4762 \right) \approx -52504.97 \times 0.293 = -15383.96 \, J \approx -15.38 \, kJ$.
Since $q = -W$,$q = 15.38 \, kJ$ for $5 \, moles$.
Heat transferred per mole $q_{mol} = \frac{15.38}{5} = 3.076 \, kJ \, mol^{-1}$.
Rounding to the nearest integer,we get $3 \, kJ \, mol^{-1}$ (closest option is $4$).

(B) Tritium $\left({ }_{1}^{3} H \right)$ is a radioactive isotope of hydrogen.
It undergoes $\beta^{-}$ decay to form Helium-$3$ according to the reaction: ${ }_{1}^{3} H \rightarrow { }_{2}^{3} He + { }_{-1}^{0} e + \bar{\nu}$.
This process involves the conversion of a neutron into a proton,emitting a low-energy $\beta^{-}$ particle.

(A) The electron gain enthalpy becomes less negative as we move down the group from $S$ to $Te$ due to an increase in atomic size.
However,$O$ has an exceptionally low electron gain enthalpy because of its small size,which leads to strong inter-electronic repulsions when an electron is added.
Therefore,the correct order is $S > Se > Te > O$.

(A) In the molecule $CH_2=C=CH-CH_3$:
Carbon-$1$ $(CH_2=)$: It is bonded to two $H$ atoms and one $C$ atom via a double bond. It has $3$ sigma bonds and $0$ lone pairs,so it is $sp^2$ hybridized.
Carbon-$2$ $(=C=)$: It is bonded to two $C$ atoms via double bonds. It has $2$ sigma bonds and $0$ lone pairs,so it is $sp$ hybridized.
Carbon-$3$ $(=CH-)$: It is bonded to one $C$ atom via a double bond,one $C$ atom via a single bond,and one $H$ atom. It has $3$ sigma bonds and $0$ lone pairs,so it is $sp^2$ hybridized.
Carbon-$4$ $(-CH_3)$: It is bonded to one $C$ atom and three $H$ atoms via single bonds. It has $4$ sigma bonds and $0$ lone pairs,so it is $sp^3$ hybridized.
Thus,the hybridization of carbon $1, 2, 3,$ and $4$ is $sp^2, sp, sp^2, sp^3$.

(B) $B_2O_3$ and $SiO_2$ are acidic oxides.
Generally,oxides of non-metals are acidic in nature.
$CaO$ and $BaO$ are basic oxides (metal oxides),and $N_2O$ is a neutral oxide.

(A) Calgon is sodium hexametaphosphate,with the formula $(NaPO_{3})_{6}$ or $Na_{6}P_{6}O_{18}$.
$1$. The $2^{nd}$ most abundant element in the Earth's crust by weight is Silicon $(Si)$,which is not present in Calgon. Thus,statement $A$ is false.
$2$. Calgon is a polymeric compound and is water-soluble. Statement $B$ is true.
$3$. It is commonly known as Graham's salt. Statement $C$ is true.
$4$. It removes $Ca^{2+}$ and $Mg^{2+}$ ions by forming soluble complex ions,not by precipitation. Statement $D$ is true.
Therefore,the statement that is $NOT$ true is $A$.

(A) The given reaction involves two steps:
$1$. Clemmensen reduction: The ketone group $(C=O)$ is reduced to a methylene group $(-CH_2-)$ using $Zn-Hg/HCl$. The reactant is $3-ethylhexan-3-one$ (or a similar ketone structure). The reduction converts the carbonyl group into a methylene group,resulting in an alkane chain.
$2$. Dehydrocyclization (Aromatization): The resulting alkane is treated with $Cr_2O_3$ at $773 \ K$ and $10-20 \ atm$. This is a standard industrial process for the aromatization of alkanes to form aromatic hydrocarbons.
Given the structure,the alkane formed is $3-ethylhexane$. Upon aromatization,it cyclizes to form ethylbenzene.

(A) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
$(a) \ Ne_2$ $(20 \ e^-)$: Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2, \sigma^* 2p_z^2$. $B.O. = \frac{10-10}{2} = 0$.
$(b) \ N_2$ $(14 \ e^-)$: $B.O. = \frac{10-4}{2} = 3$.
$(c) \ F_2$ $(18 \ e^-)$: $B.O. = \frac{10-8}{2} = 1$.
$(d) \ O_2$ $(16 \ e^-)$: $B.O. = \frac{10-6}{2} = 2$.
Thus,the correct matching is: $(a)$ $\rightarrow (iii), (b)$ $\rightarrow (iv), (c)$ $\rightarrow (i), (d)$ $\rightarrow (ii)$.

(B) Total mass of $Na^{+}$ required $= 70.0 \ mg/mL \times 50 \ mL = 3500 \ mg = 3.5 \ g$.
Molar mass of $NaNO_3 = 23 + 14 + (3 \times 16) = 85 \ g \ mol^{-1}$.
Moles of $Na^{+}$ $= \frac{3.5 \ g}{23 \ g \ mol^{-1}} \approx 0.15217 \ mol$.
Since $1 \ mol$ of $NaNO_3$ contains $1 \ mol$ of $Na^{+}$,moles of $NaNO_3 = 0.15217 \ mol$.
Mass of $NaNO_3 = 0.15217 \ mol \times 85 \ g \ mol^{-1} = 12.93445 \ g$.
Rounding off to the nearest integer,we get $13 \ g$.

(A) The reaction for the atomization of $SF_{6(g)}$ is: $SF_{6(g)} \rightarrow S_{(g)} + 6F_{(g)}$
The enthalpy of reaction $(\Delta_{r}H)$ is given by: $\Delta_{r}H = \Delta_{f}H(S, g) + 6 \times \Delta_{f}H(F, g) - \Delta_{f}H(SF_{6}, g)$
Substituting the given values: $\Delta_{r}H = 275 + 6 \times 80 - (-1100)$
$\Delta_{r}H = 275 + 480 + 1100 = 1855 \ kJ \ mol^{-1}$
Since there are $6$ $S-F$ bonds in $SF_{6}$,the average bond energy $(\epsilon_{S-F})$ is: $\epsilon_{S-F} = \frac{1855}{6} \approx 309.16 \ kJ \ mol^{-1}$
Rounding off to the nearest integer,we get $309 \ kJ \ mol^{-1}$.

(C) In mildly alkaline medium,the reaction between thiosulphate ion $(S_2O_3^{2-})$ and permanganate ion $(MnO_4^-)$ is as follows:
$8MnO_4^- + 3S_2O_3^{2-} + H_2O \rightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^-$
Here,the product $A$ is the sulphate ion $(SO_4^{2-})$.
To find the oxidation state of sulphur $(S)$ in $SO_4^{2-}$:
Let the oxidation state of $S$ be $x$.
$x + 4(-2) = -2$
$x - 8 = -2$
$x = +6$
Therefore,the oxidation state of sulphur in $A$ is $6$.

(C) Ammonium phosphate $(NH_4)_3PO_4$ is a salt of a weak acid $(H_3PO_4)$ and a weak base $(NH_4OH)$.
The formula for the $pH$ of a salt of a weak acid and a weak base is:
$pH = 7 + \frac{1}{2}(pK_a - pK_b)$
Given:
$pK_a = 5.23$
$pK_b = 4.75$
Substituting the values:
$pH = 7 + \frac{1}{2}(5.23 - 4.75)$
$pH = 7 + \frac{1}{2}(0.48)$
$pH = 7 + 0.24$
$pH = 7.24$
Rounding to the nearest integer,the $pH$ is approximately $7$.

(A) The reaction involves the electrophilic addition of $HBr$ to a conjugated diene,$2$-methylbuta-$1,3$-diene (isoprene).
When $HBr$ is added to a conjugated diene,both $1,2$-addition and $1,4$-addition products can be formed.
In the presence of excess $HBr$,the reaction proceeds further. However,the question asks for the major product of the initial addition step or the primary outcome of the reaction.
The $1,4$-addition product is generally more stable due to the formation of a more substituted alkene.
The reaction of $2$-methylbuta-$1,3$-diene with $HBr$ yields $1$-bromo-$3$-methylbut-$2$-ene as the major $1,4$-addition product.

(A) The $F^{-}$ ions make the enamel on teeth much harder by converting hydroxyapatite,$[3 Ca_{3}(PO_{4})_{2} \cdot Ca(OH)_{2}]$,which is the enamel on the surface of the teeth,into much harder fluoroapatite,$[3 Ca_{3}(PO_{4})_{2} \cdot CaF_{2}]$.
Thus,the correct formula for hydroxyapatite is $[3 Ca_{3}(PO_{4})_{2} \cdot Ca(OH)_{2}]$.

(B) Statement $I$: In the titration of a strong acid with a weak base,the $pH$ at the equivalence point is in the acidic range $(pH < 7)$. Methyl orange has a $pH$ transition range of $3.2$ to $4.4$,which falls within the steep portion of the titration curve. Thus,it is a suitable indicator. Statement $I$ is true.
Statement $II$: In the titration of a weak acid (acetic acid) with a strong base $(NaOH)$,the $pH$ at the equivalence point is in the basic range $(pH > 7)$. Phenolphthalein has a $pH$ transition range of $8.2$ to $10.0$,which covers the steep portion of this titration curve. Thus,it is a suitable indicator. Statement $II$ is false.

(A) Pure demineralised (de-ionized) water,free from all soluble mineral salts,is obtained by passing water successively through a cation exchange resin (in the $H^{+}$ form) and an anion exchange resin (in the $OH^{-}$ form).

(B) When $CO_{2}$ is passed through slaked lime $(Ca(OH)_{2})$,it first forms a white precipitate of calcium carbonate $(CaCO_{3})$:
$Ca(OH)_{2} + CO_{2} \longrightarrow CaCO_{3} \downarrow + H_{2}O$
When excess $CO_{2}$ is passed,the calcium carbonate reacts further to form soluble calcium bicarbonate $(Ca(HCO_{3})_{2})$:
$CaCO_{3} + CO_{2} + H_{2}O \longrightarrow Ca(HCO_{3})_{2}$

(C) The velocity of an electron in Bohr's model is given by the relation $V \propto \frac{Z}{n}$.
Here,$Z$ is the atomic number (representing the magnitude of positive charge on the nucleus) and $n$ is the principal quantum number.
Statement $I$ claims that velocity increases with a decrease in positive charge $(Z)$. However,since $V \propto Z$,the velocity actually decreases as the positive charge on the nucleus decreases. Thus,Statement $I$ is false.
Statement $II$ claims that velocity increases with a decrease in the principal quantum number $(n)$. Since $V \propto \frac{1}{n}$,the velocity increases as $n$ decreases. Thus,Statement $II$ is true.

(B) The bromination of alkanes is highly selective.
The stability of the free radical intermediate formed during the reaction determines the major product.
In isobutane,the tertiary radical $((CH_{3})_{3}C^{\bullet})$ is more stable than the primary radical $((CH_{3})_{2}CH-CH_{2}^{\bullet})$.
Thus,$2-$bromo$-2-$methylpropane is formed as the major product.

(A) According to the $VSEPR$ theory,for an $AB_{3}$ molecule to have a $T$-shaped geometry,the central atom $A$ must have $3$ bond pairs and $2$ lone pairs of electrons.
This corresponds to a steric number of $5$,which implies $sp^{3}d$ hybridization.
As shown in the structure,there are $2$ lone pairs on the central atom $A$.

(A) The mixture consists of a weak base $(NH_{3})$ and its salt $(NH_{4}Cl)$,forming a basic buffer.
First,calculate the moles of each component:
$n(NH_{4}Cl) = 0.0504 \ M \times 5.0 \ mL = 0.252 \ mmol$
$n(NH_{3}) = 0.0210 \ M \times 2.0 \ mL = 0.042 \ mmol$
Using the Henderson-Hasselbalch equation for a basic buffer:
$[OH^{-}] = K_{b} \times \frac{[Base]}{[Salt]} = K_{b} \times \frac{n(NH_{3})}{n(NH_{4}Cl)}$
$[OH^{-}] = 1.8 \times 10^{-5} \times \frac{0.042}{0.252}$
$[OH^{-}] = 1.8 \times 10^{-5} \times \frac{1}{6} = 0.3 \times 10^{-5} = 3 \times 10^{-6} \ M$
Comparing this with $x \times 10^{-6} \ M$,we get $x = 3$.

(A) Mohr's salt is $(NH_4)_2Fe(SO_4)_2 \cdot 6H_2O$. The number of water molecules is $6$.
Potash alum is $KAl(SO_4)_2 \cdot 12H_2O$. The number of water molecules is $12$.
The ratio of the number of water molecules in Mohr's salt to potash alum is $\frac{6}{12} = \frac{1}{2} = 0.5$.
Expressing $0.5$ in the form $.... \times 10^{-1}$,we get $5 \times 10^{-1}$.

(A) The Born-Haber cycle equation is given by:
$\Delta_{f} H^{\ominus}_{KCl} = \Delta_{sub} H^{\ominus}_{(K)} + \Delta_{ionization} H^{\ominus}_{(K)} + \frac{1}{2} \Delta_{bond} H^{\ominus}_{(Cl_2)} + \Delta_{electron \ gain} H^{\ominus}_{(Cl)} + \Delta_{lattice} H^{\ominus}_{(KCl)}$
Substituting the given values:
$-436.7 = 89.2 + 419.0 + \frac{1}{2}(243.0) + (-348.6) + \Delta_{lattice} H^{\ominus}_{(KCl)}$
$-436.7 = 89.2 + 419.0 + 121.5 - 348.6 + \Delta_{lattice} H^{\ominus}_{(KCl)}$
$-436.7 = 281.1 + \Delta_{lattice} H^{\ominus}_{(KCl)}$
$\Delta_{lattice} H^{\ominus}_{(KCl)} = -436.7 - 281.1 = -717.8 \ kJ \ mol^{-1}$
The magnitude of lattice enthalpy is the absolute value,which is $717.8 \ kJ \ mol^{-1}$.
Rounding to the nearest integer,we get $718 \ kJ \ mol^{-1}$.

(D) Photochemical smog is an oxidizing mixture of pollutants like ozone $(O_3)$,nitric oxide $(NO)$,acrolein,formaldehyde,and peroxyacetyl nitrate $(PAN)$.
These oxidizing agents react with the double bonds in rubber,leading to its degradation and cracking.
Therefore,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation for $(A)$.

(C) For $London$ dispersion forces,the interaction energy $E$ is inversely proportional to the sixth power of the distance $r$ between the particles.
Mathematically,$E \propto \frac{1}{r^{6}}$,which can be written as $E \propto r^{-6}$.
Therefore,comparing this to $E \propto r^{x}$,we get $x = -6$.

(A) The electronic configuration of $O_{2}^{-}$ ($17$ electrons) is: $(\sigma_{1s})^{2}(\sigma_{1s}^{*})^{2}(\sigma_{2s})^{2}(\sigma_{2s}^{*})^{2}(\sigma_{2p_{z}})^{2}(\pi_{2p_{x}}^{2} = \pi_{2p_{y}}^{2})(\pi_{2p_{x}}^{*2} = \pi_{2p_{y}}^{*1})$.
$\text{Bond order} = \frac{N_b - N_a}{2} = \frac{10 - 7}{2} = 1.5$.
Since there is one unpaired electron in the $\pi^{*}$ orbital,the ion is paramagnetic.

(A) Assertion $A$ is true: $BaCO_3$ is insoluble in water due to high lattice energy and is thermally stable.
Reason $R$ is true: As the size of the alkaline earth metal cation increases down the group $(Be^{2+} < Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+})$,the polarising power of the cation decreases,which leads to an increase in the thermal stability of the corresponding carbonates.
Since $Ba^{2+}$ is the largest cation in the group,$BaCO_3$ is the most thermally stable carbonate among them.
Therefore,$R$ is the correct explanation for $A$.

(D) Assertion $(A)$ is true because heavy water $(D_2O)$ is commonly used as a tracer in studying reaction mechanisms.
Reason $(R)$ is false because the $O-H$ bond is weaker than the $O-D$ bond due to the kinetic isotope effect. Consequently,the rate of cleavage of the $O-H$ bond is faster than that of the $O-D$ bond,not slower.
Therefore,$(A)$ is true but $(R)$ is false.

(C) To determine if a compound is aromatic,it must satisfy $H$ückel's rule: it must be cyclic,planar,fully conjugated,and have $(4n + 2) \pi$ electrons.
$A$. Cycloheptatrienyl anion: It has $8 \pi$ electrons (anti-aromatic) or is non-planar,but in many contexts,it is considered non-aromatic due to lack of planarity or anti-aromaticity.
$B$. Cyclopentadienyl anion: It has $6 \pi$ electrons $(n=1)$,is cyclic,planar,and fully conjugated. It is aromatic.
$C$. Cyclooctatetraene: It has $8 \pi$ electrons ($4n$ system). It is non-planar (tub-shaped) to avoid anti-aromaticity,making it non-aromatic.
$D$. Phenol: It is a benzene derivative,which is aromatic.
Comparing the options,Cyclooctatetraene is the most classic example of a non-aromatic compound due to its non-planar structure.

(D) For $1,2-$dimethylcyclopropane,there are two chiral centers at $C1$ and $C2$.
$1,2-$dimethylcyclopropane exists in three stereoisomeric forms:
$1$. $cis-1,2-$dimethylcyclopropane: This molecule has a plane of symmetry and is achiral (meso compound).
$2$. $trans-1,2-$dimethylcyclopropane: This exists as a pair of enantiomers (dextrorotatory and levorotatory forms).
Thus,there are a total of $3$ stereoisomers ($1$ meso form + $2$ enantiomers).

(B) Molecular mass of $BaSO_4 = 137 + 32 + (4 \times 16) = 233 \, g/mol$.
Since $233 \, g$ of $BaSO_4$ contains $32 \, g$ of sulphur,
$1.44 \, g$ of $BaSO_4$ contains $\frac{32}{233} \times 1.44 \, g$ of sulphur.
Percentage of sulphur = $\frac{\text{Mass of sulphur}}{\text{Mass of organic compound}} \times 100$
Percentage of sulphur = $\frac{32 \times 1.44}{233 \times 0.471} \times 100 = 41.98 \, \%$.
Rounding to the nearest integer,we get $42 \, \%$.

(B) The reaction is $A + B \rightleftharpoons C + D$ with $K_{eq} = 100$.
Initial concentrations are $[A] = 1 \ M, [B] = 1 \ M, [C] = 1 \ M, [D] = 1 \ M$.
Calculate the reaction quotient $Q_{c} = \frac{[C][D]}{[A][B]} = \frac{1 \times 1}{1 \times 1} = 1$.
Since $Q_{c} < K_{eq}$,the reaction proceeds in the forward direction.
Let $x$ be the change in concentration at equilibrium:
$[A] = 1-x, [B] = 1-x, [C] = 1+x, [D] = 1+x$.
$K_{eq} = \frac{(1+x)(1+x)}{(1-x)(1-x)} = 100$.
Taking the square root: $\frac{1+x}{1-x} = 10$.
$1+x = 10 - 10x$ $\Rightarrow 11x = 9$ $\Rightarrow x = \frac{9}{11} \approx 0.818$.
Equilibrium concentration of $D = 1 + x = 1 + \frac{9}{11} = \frac{20}{11} \approx 1.8181 \ M$.
Expressing in terms of $10^{-2} \ M$: $1.8181 \times 10^{0} = 181.81 \times 10^{-2} \ M$.
Rounding to the nearest integer,we get $182 \times 10^{-2} \ M$.

(B) The evaporation reaction is: $H_2O(\ell) \longrightarrow H_2O(g)$.
Given $\Delta H = 41 \ kJ \ mol^{-1}$,$T = 373 \ K$,and $R = 8.3 \ J \ mol^{-1} \ K^{-1} = 8.3 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1}$.
The change in the number of moles of gas is $\Delta n_g = 1 - 0 = 1$.
Using the relation $\Delta H = \Delta U + \Delta n_g RT$:
$\Delta U = \Delta H - \Delta n_g RT$
$\Delta U = 41 \ kJ \ mol^{-1} - (1 \ mol) \times (8.3 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1}) \times (373 \ K)$
$\Delta U = 41 - 3.0959 = 37.9041 \ kJ \ mol^{-1} \approx 38 \ kJ \ mol^{-1}$.

(B) $v$: speed of electron having maximum kinetic energy.
From Einstein's photoelectric equation:
$E = \phi + K.E._{\max}$
$\frac{hc}{\lambda} = h \nu_{0} + \frac{1}{2} m_{e} v^{2}$
$\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{500 \times 10^{-9}} = 6.63 \times 10^{-34} \times 4.3 \times 10^{14} + \frac{1}{2} \times 9.0 \times 10^{-31} \times v^{2}$
$3.978 \times 10^{-19} = 2.8509 \times 10^{-19} + 4.5 \times 10^{-31} \times v^{2}$
$1.1271 \times 10^{-19} = 4.5 \times 10^{-31} \times v^{2}$
$v^{2} = \frac{1.1271 \times 10^{-19}}{4.5 \times 10^{-31}} \approx 0.2504 \times 10^{12} = 25.04 \times 10^{10}$
$v \approx 5.004 \times 10^{5} \ ms^{-1}$
Rounding to the nearest integer,the value is $5$.

(B) The dissociation of $Na_{3}PO_{4}$ is given by: $Na_{3}PO_{4} \rightarrow 3Na^{+} + PO_{4}^{3-}$.
Number of moles of $Na$ atoms = $\frac{3.45 \, g}{23.0 \, g \, mol^{-1}} = 0.15 \, mol$.
Since $1 \, mol$ of $Na_{3}PO_{4}$ contains $3 \, mol$ of $Na$,the moles of $Na_{3}PO_{4}$ = $\frac{1}{3} \times 0.15 \, mol = 0.05 \, mol$.
Volume of solution = $100 \, mL = 0.1 \, L$.
Molarity $(M)$ = $\frac{\text{moles of solute}}{\text{Volume of solution in } L} = \frac{0.05 \, mol}{0.1 \, L} = 0.5 \, mol \, L^{-1}$.
Expressing in terms of $10^{-2}$: $0.5 = 50 \times 10^{-2} \, mol \, L^{-1}$.
Thus,the nearest integer is $50$.

(A) For the reaction $[PtCl_4]^{2-} + H_2O \rightleftharpoons [Pt(H_2O)Cl_3]^- + Cl^-$,the rate of the forward reaction is $r_f = k_f [[PtCl_4]^{2-}]$ and the rate of the backward reaction is $r_b = k_b [[Pt(H_2O)Cl_3]^-] [Cl^-]$.
At equilibrium,the rate of the forward reaction equals the rate of the backward reaction $(r_f = r_b)$.
Therefore,$k_f [[PtCl_4]^{2-}] = k_b [[Pt(H_2O)Cl_3]^-] [Cl^-]$.
The equilibrium constant $K_c$ is given by $K_c = \frac{k_f}{k_b} = \frac{[[Pt(H_2O)Cl_3]^-] [Cl^-]}{[[PtCl_4]^{2-}]}$.
From the given rate expression,$k_f = 4.8 \times 10^{-5} \ s^{-1}$ and $k_b = 2.4 \times 10^{-3} \ M^{-1} s^{-1}$.
$K_c = \frac{4.8 \times 10^{-5}}{2.4 \times 10^{-3}} = 0.02$.
The nearest integer to $0.02$ is $0$.

(A) To determine the number of lone pairs on the central atom $(Xe)$ in each species:
$1$. $XeF_{2}$: $Xe$ has $8$ valence electrons. $2$ are used for bonding with $F$ atoms. Remaining electrons = $8-2 = 6$,which form $3$ lone pairs. Thus,$(a) - (iv)$.
$2$. $XeO_{2}F_{2}$: $Xe$ has $8$ valence electrons. $2$ are used for $F$ atoms and $4$ are used for $2$ double-bonded $O$ atoms. Remaining electrons = $8-2-4 = 2$,which form $1$ lone pair. Thus,$(b) - (ii)$.
$3$. $XeO_{3}F_{2}$: $Xe$ has $8$ valence electrons. $2$ are used for $F$ atoms and $6$ are used for $3$ double-bonded $O$ atoms. Remaining electrons = $8-2-6 = 0$. Thus,$(c) - (i)$.
$4$. $XeF_{4}$: $Xe$ has $8$ valence electrons. $4$ are used for bonding with $F$ atoms. Remaining electrons = $8-4 = 4$,which form $2$ lone pairs. Thus,$(d) - (iii)$.
Therefore,the correct matching is $a-iv, b-ii, c-i, d-iii$.

(B) Back bonding in boron trihalides involves the donation of a lone pair from the halide $p$-orbital to the empty $2p$-orbital of boron.
This is a $(p\pi-p\pi)$ back bonding.
The strength of back bonding depends on the effective overlap between the orbitals.
For $BF_{3}$,the overlap is $(2p\pi-2p\pi)$,which is the most effective due to similar size.
For $BCl_{3}$,it is $(2p\pi-3p\pi)$,for $BBr_{3}$ it is $(2p\pi-4p\pi)$,and for $BI_{3}$ it is $(2p\pi-5p\pi)$.
As the size of the halide increases,the energy gap increases and the overlap efficiency decreases.
Therefore,the order of back bonding strength is $BF_{3} > BCl_{3} > BBr_{3} > BI_{3}$.

(A) The bond dissociation energy of $D_2$ is greater than $H_2$ due to the stronger $D-D$ bond compared to the $H-H$ bond.
Consequently,$D_2$ reacts slower than $H_2$ in chemical reactions.

(D) Blood is a negatively charged colloidal sol.
According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the valency of the active ion (ion with charge opposite to that of the sol).
For a negatively charged sol,a positively charged ion is required for coagulation.
The coagulating power increases with the increase in the magnitude of the charge on the cation.
Comparing the cations: $Na^+$,$Fe^{2+}$,$Mg^{2+}$,and $Fe^{3+}$.
The $Fe^{3+}$ ion has the highest charge $(+3)$,therefore it has the maximum coagulating power for the negatively charged blood sol.
Thus,$FeCl_3$ is the most suitable salt for the clotting of blood.

(A) Nucleophilic aromatic substitution reactions are facilitated by the presence of electron-withdrawing groups (like $-NO_2$) on the benzene ring.
These groups stabilize the intermediate carbanion through the $-I$ and $-M$ effects.
As the number of $-NO_2$ groups increases,the electron density on the ring decreases,making it more susceptible to nucleophilic attack.
The order of reactivity is:
$(i)$ Chlorobenzene (no $-NO_2$ group)
(ii) $p$-Nitrochlorobenzene (one $-NO_2$ group)
(iii) $2,4$-Dinitrochlorobenzene (two $-NO_2$ groups)
(iv) $2,4,6$-Trinitrochlorobenzene (three $-NO_2$ groups)
Therefore,the increasing order of tendency towards nucleophilic substitution is $(i) < (ii) < (iii) < (iv)$.

(C) $R-COCl \to R-CHO$ is the Rosenmund reduction,which uses $H_2 / Pd-BaSO_4$ as the reagent. Thus,$(a)-(ii)$.
$(b)$ $R-CH_2-COOH \to R-CH(Cl)-COOH$ is the Hell-Volhard-Zelinsky $(HVZ)$ reaction,which uses $Cl_2 / \text{Red } P, H_2O$ as the reagent. Thus,$(b)-(iv)$.
$(c)$ $R-CONH_2 \to R-NH_2$ is the Hoffmann bromamide degradation,which uses $Br_2 / NaOH$ as the reagent. Thus,$(c)-(i)$.
$(d)$ $R-CO-CH_3 \to R-CH_2-CH_3$ is the Clemmensen reduction,which uses $Zn(Hg) / \text{Conc. } HCl$ as the reagent. Thus,$(d)-(iii)$.
Therefore,the correct match is $(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)$.

(B) $(i)$ $[FeCl_{4}]^{2-}$: $Fe^{2+}$ is $d^{6}$ in a tetrahedral field. Configuration is $e^{3} t_{2}^{3}$,so $n = 4$ unpaired electrons. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
$(ii)$ $[Co(C_{2}O_{4})_{3}]^{3-}$: $Co^{3+}$ is $d^{6}$ in an octahedral field with a strong field ligand ($C_{2}O_{4}^{2-}$ is often considered strong enough to cause pairing in $Co^{3+}$). Configuration is $t_{2g}^{6} e_{g}^{0}$,so $n = 0$ unpaired electrons. $\mu = 0 \ BM$.
$(iii)$ $MnO_{4}^{2-}$: $Mn$ is in $+6$ oxidation state. $Mn^{6+}$ is $3d^{1}$. $n = 1$ unpaired electron. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

(D) Buna-$S$ is a copolymer formed by the polymerization of $1,3$-butadiene and styrene.
In the name Buna-$S$,'Bu' stands for $1,3$-butadiene,'na' stands for sodium (catalyst),and '$S$' stands for styrene.

(B) The reaction involves the reduction of a carbonyl group $(C=O)$ to a methylene group $(-CH_{2}-)$ in the presence of a carbon-carbon double bond $(C=C)$.
Wolff-Kishner reduction,which uses hydrazine $(NH_{2}NH_{2})$ followed by a strong base like sodium ethoxide $(C_{2}H_{5}O^{-}Na^{+})$ and heat,is specifically used to reduce carbonyl groups to alkanes without affecting the $C=C$ double bond.
$NaBH_{4}$ reduces carbonyls but is typically used for alcohols,and $Ni/H_{2}$ would reduce both the carbonyl and the $C=C$ double bond.

(B) Valium $-$ Tranquilizer: $(a-iv)$
$(b)$ Morphine $-$ Analgesic: $(b-iii)$
$(c)$ Norethindrone $-$ Antifertility drug: $(c-i)$
$(d)$ Vitamin $B_{12}$ $-$ Pernicious anaemia: $(d-ii)$
Therefore,the correct matching is $a-iv, b-iii, c-i, d-ii$.

(D) The ores are identified as follows:
$a$. Aluminium: Kaolinite $(Al_{2}(OH)_{4} \cdot Si_{2}O_{5})$
$b$. Copper: Malachite $(Cu(OH)_{2} \cdot CuCO_{3})$
$c$. Iron: Siderite $(FeCO_{3})$
$d$. Zinc: Calamine $(ZnCO_{3})$
Therefore,the correct matching is $a-iii, b-iv, c-i, d-ii$.

(C) In the $3d$ transition series,density generally increases from left to right as atomic mass increases and atomic volume decreases.
However,$Zn$ has a lower density than $Cr$ because of its larger atomic size due to fully filled $d$-orbitals.
The density values (in $g/cm^3$) are: $Zn (7.13) < Cr (7.19) < Fe (7.87) < Co (8.90) < Cu (8.96)$.
Therefore,the correct order is $Zn < Cr < Fe < Co < Cu$.

(A, D) $(I)$ In $VOSO_4$,$V$ is in $+4$ oxidation state. It acts as an oxidizing agent,not a reducing agent.
$(II)$ $Cr_2O_3$ is an amphoteric oxide.
$(III)$ In $RuO_4$,$Ru$ is in $+8$ oxidation state. It acts as an oxidizing agent.
$(IV)$ The red colour of ruby is due to the presence of $Cr^{3+}$ ions in $Al_2O_3$ lattice,not $Co^{3+}$.
Therefore,both $(A)$ and $(D)$ are incorrect statements.

(C) The formation of a coloured dye (azo dye) occurs through the coupling reaction of a diazonium salt with an aromatic compound like $\beta$-Naphthol in an alkaline medium $(NaOH)$.
Primary aromatic amines,such as aniline $(C_6H_5NH_2)$,react with $NaNO_2$ and $HCl$ at $0-5 \ ^\circ C$ to form stable benzenediazonium chloride.
This diazonium salt then undergoes an electrophilic substitution (coupling) reaction with $\beta$-Naphthol to produce an orange-red azo dye.
Benzylamine is an aliphatic amine and does not form a stable diazonium salt under these conditions.
$N,N$-Dimethylaniline and $N$-Methylaniline are secondary and tertiary amines,respectively,and do not form the same type of stable diazonium salt required for this specific coupling reaction to produce the characteristic dye.

(A) Gabriel phthalimide synthesis is used for the preparation of primary $(1^{\circ})$ aliphatic amines.
$1$. $(CH_3)_2CH-CH_2-NH_2$ is a primary aliphatic amine. It can be synthesized by this method.
$2$. $CH_3CH_2NH_2$ is a primary aliphatic amine. It can be synthesized by this method.
$3$. $C_6H_5-CH_2-NH_2$ (benzylamine) is a primary amine where the $-NH_2$ group is attached to an aliphatic carbon. It can be synthesized by this method.
$4$. $C_6H_5-NH_2$ (aniline) is an aromatic amine. Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Thus,there are $3$ such amines that can be synthesized by this method.

(B) Sulphur exists in various allotropic forms like $\alpha$-sulphur (rhombic) and $\beta$-sulphur (monoclinic),which consist of $S_8$ puckered ring structures. These forms are diamagnetic because all electrons are paired in the $S_8$ molecules.
The $S_2$ form of sulphur is analogous to $O_2$ and is paramagnetic in the vapour state at high temperatures due to the presence of two unpaired electrons in its antibonding $\pi^*$ molecular orbitals.
Therefore,only the $S_2$ form shows paramagnetism.
The number of such allotropic forms is $1$.

(B) Given:
Mass of solute $(C_{4}H_{10})$ = $10 \ g$
Molar mass of $C_{4}H_{10}$ = $(4 \times 12) + (10 \times 1) = 58 \ g/mol$
Moles of solute = $\frac{10}{58} \ mol$
Mass of solvent $(C_{6}H_{6})$ = $200 \ g = 0.2 \ kg$
Molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{10/58}{0.2} = \frac{10}{58 \times 0.2} = \frac{10}{11.6} \approx 0.862 \ m$
Freezing point depression constant $(K_{f})$ = $5.12 \ ^{\circ} C/m$
Depression in freezing point $(\Delta T_{f})$ = $K_{f} \times m = 5.12 \times 0.862 \approx 4.41 \ ^{\circ} C$
Freezing point of solution $(T_{f})$ = $T_{f}^{\circ} - \Delta T_{f} = 5.5 - 4.41 = 1.09 \ ^{\circ} C$
Rounding to the nearest integer,the temperature is $1 \ ^{\circ} C$.

(A) The reduction half-reaction is:
$MnO_4^- + 8H^{+} + 5e^- \rightarrow Mn^{2+} + 4H_2O$
Using the Nernst equation for the electrode potential:
$E = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-][H^{+}]^8}$
Case $I$: When $[H^{+}] = 1 \, M$
$E_1 = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-]}$
Case $II$: When $[H^{+}] = 10^{-4} \, M$
$E_2 = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-] \times (10^{-4})^8}$
$E_2 = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-]} + \frac{0.059}{5} \log (10^{-4})^8$
$E_2 = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-]} + \frac{0.059}{5} \times (-32)$
Calculating the magnitude of change $|E_1 - E_2|$:
$|E_1 - E_2| = \frac{0.059}{5} \times 32$
$|E_1 - E_2| = 0.0118 \times 32 = 0.3776 \, V$
Given the change is $x \times 10^{-4} \, V$:
$0.3776 \, V = 3776 \times 10^{-4} \, V$
Therefore,$x = 3776$.

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{3.33 \ h} = \frac{0.693}{10/3 \ h} = 0.2079 \ h^{-1}$.
The integrated rate law for a first order reaction is $\ln \frac{[A]_0}{[A]_t} = kt$,which can be written as $\log_{10} \frac{[A]_0}{[A]_t} = \frac{kt}{2.303}$.
Given that the fraction of sucrose remaining is $f = \frac{[A]_t}{[A]_0}$,we have $\frac{1}{f} = \frac{[A]_0}{[A]_t}$.
Substituting the values: $\log_{10} (\frac{1}{f}) = \frac{0.2079 \times 9}{2.303} = \frac{0.693 \times 3 \times 9}{10 \times 2.303} = \frac{0.693}{2.303} \times 2.7 = 0.3009 \times 2.7 = 0.81243$.
Thus,$\log_{10} (\frac{1}{f}) = 81.24 \times 10^{-2}$.
Rounding to the nearest integer,we get $81$.

(D) The carbylamine test is given by $1^{\circ}$ amines. When aniline $(C_6H_5NH_2)$ is treated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$,it undergoes a reaction to form phenyl isocyanide $(C_6H_5NC)$,which is characterized by an extremely unpleasant odor.
The chemical equation is:
$C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$

(A) Maltose is a disaccharide formed by two units of $\alpha-D-$glucopyranose joined by an $\alpha(1 \rightarrow 4)$ glycosidic linkage. In the $\alpha-$anomer of maltose,the hydroxyl group at the $C-1$ position of the second glucose unit is oriented downwards (axial position). The structure in image $983-$s843 represents the correct $\alpha-$anomer of maltose,where the $C-1$ hydroxyl group is in the $\alpha$ configuration.

(D) The synthesis of $4-$bromo$-2-$nitroethylbenzene from benzene involves the following steps:
$1$. Friedel-Crafts acylation of benzene with $CH_3COCl / AlCl_3$ to form acetophenone.
$2$. Clemmensen reduction of acetophenone using $Zn-Hg / HCl$ to form ethylbenzene.
$3$. Electrophilic aromatic bromination of ethylbenzene using $Br_2 / AlBr_3$ to form $4-$bromoethylbenzene (para-product is major due to steric hindrance at ortho position).
$4$. Nitration of $4-$bromoethylbenzene using $HNO_3 / H_2SO_4$ to form $4-$bromo$-2-$nitroethylbenzene.

(D) The acidic strength depends on the stability of the conjugate base formed after the loss of a proton $(H^+)$.
$1.$ Phenol $(I)$ is much less acidic than carboxylic acids $(II, III, IV)$.
$2.$ Among the carboxylic acids,electron-withdrawing groups $(-NO_2)$ increase acidity,while electron-donating groups $(-CH_3)$ decrease it.
$3.$ Compound $II$ ($p$-nitrobenzoic acid) has a strong $-I$ and $-M$ effect,making it the most acidic.
$4.$ Compound $III$ (benzoic acid) is the reference.
$5.$ Compound $IV$ ($p$-toluic acid) has a $+I$ and $+H$ effect,making it the least acidic among the carboxylic acids.
$6.$ Therefore,the order is $II > III > IV > I$.

(D) In the presence of strong acid $(H_2SO_4)$,aniline is protonated to form the anilinium ion $(C_6H_5NH_3^+)$.
The $-NH_3^+$ group is meta-directing due to its strong electron-withdrawing nature.
As a result,a significant amount of the $m$-nitroaniline (compound $(B)$) is formed,along with the $p$-nitroaniline (compound $(A)$) and $o$-nitroaniline (compound $(C)$).
Experimental data shows that $p$-nitroaniline $(51\%)$ is the major product,followed by $m$-nitroaniline $(47\%)$ and $o$-nitroaniline $(2\%)$.

(C) The correct order of bond dissociation enthalpy for halogens is $Cl_2 > Br_2 > F_2 > I_2$.
Due to the small size of the fluorine atom,the lone pairs of electrons on the two fluorine atoms in $F_2$ experience significant inter-electronic repulsion.
This makes the $F-F$ bond weaker than the $Cl-Cl$ and $Br-Br$ bonds,resulting in a lower bond dissociation enthalpy for $F_2$ compared to $Cl_2$ and $Br_2$.

(C) German silver is an alloy that does not contain silver.
It is composed of copper $(Cu)$,zinc $(Zn)$,and nickel $(Ni)$.
The typical composition is approximately $50\% \ Cu$,$30\% \ Zn$,and $20\% \ Ni$.

(A) To determine the spin-only magnetic moment $(\mu = \sqrt{n(n+2)} \ B.M.)$,we find the number of unpaired electrons $(n)$ for each complex:
$(i)$ $[FeF_{6}]^{3-}$: $Fe^{3+}$ is $3d^{5}$. $F^{-}$ is a weak field ligand $(WFL)$,so electrons remain unpaired. $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \ B.M.$
$(ii)$ $[Co(NH_{3})_{6}]^{3+}$: $Co^{3+}$ is $3d^{6}$. $NH_{3}$ is a strong field ligand $(SFL)$,causing pairing. $n = 0$,$\mu = 0 \ B.M.$
$(iii)$ $[NiCl_{4}]^{2-}$: $Ni^{2+}$ is $3d^{8}$. $Cl^{-}$ is a $WFL$. In tetrahedral geometry,$n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \ B.M.$
$(iv)$ $[Cu(NH_{3})_{4}]^{2+}$: $Cu^{2+}$ is $3d^{9}$. $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \ B.M.$
Comparing the values: $\sqrt{35} > \sqrt{8} > \sqrt{3} > 0$,which corresponds to $i > iii > iv > ii$.

(A) The viscosity of hydrophilic sols is generally higher than that of the dispersion medium $(H_2O)$.
Hydrophilic sols are more stable due to strong interaction between the dispersed phase and the dispersion medium,making them difficult to coagulate.
Hydrophilic sols are reversible in nature,meaning they can be reformed by simply mixing the dispersed phase with the dispersion medium after coagulation.
They do not require electrolytes for stability; in fact,the addition of electrolytes can lead to coagulation.
Therefore,the statement that their viscosity is of the order of that of $H_2O$ is false.

(C) Statement $I$ is true because dimethyl glyoxime $(DMG)$ is used as a specific reagent for the detection of $Ni^{2+}$ ions in an ammoniacal medium $(NH_{4}OH)$.
Statement $II$ is false because dimethyl glyoxime $(C_{4}H_{8}N_{2}O_{2})$ acts as a bidentate monoanionic ligand $(dmg^{-})$ after losing a proton,not as a neutral ligand.
The reaction is: $Ni^{2+}_{(aq)} + 2dmg^{-} \rightarrow [Ni(dmg)_{2}] \text{ (rosy red precipitate)}$.

(A) The reaction is the $OXO$ process or hydroformylation of alkenes.
In this reaction,an alkene reacts with $CO$ and $H_2$ in the presence of a rhodium $(Rh)$ or cobalt catalyst to form an aldehyde.
For the substrate $CH_3CH_2CH=CH_2$ (but$-1-$ene),the hydroformylation leads to two isomeric aldehydes:
$1$. $CH_3CH_2CH_2CH_2CHO$ (pentanal,linear product)
$2$. $CH_3CH_2CH(CH_3)CHO$ ($2$-methylbutanal,branched product)
Using a rhodium catalyst,the linear aldehyde is typically the major product due to less steric hindrance during the formation of the transition state.
Therefore,the major product is $CH_3CH_2CH_2CH_2CHO$.

(C) $Zone$ refining is used for the purification of $Indium$ $(In)$.
This method is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.

(C) The transition temperature for sulphur is $369 \ K$. Below $369 \ K$,$\alpha-$sulphur (rhombic) is stable,and above $369 \ K$,$\beta-$sulphur (monoclinic) is stable.
These two forms can change reversibly between themselves with slow heating or slow cooling,making Statement-$I$ true.
At room temperature,$\alpha-$sulphur (rhombic) is the most stable form,not monoclinic sulphur. Therefore,Statement-$II$ is false.

(B) The degree of dissociation $\alpha = 0.75$ and for $AB$,the number of ions $n = 2$.
The van't Hoff factor $i = 1 - \alpha + n\alpha = 1 - 0.75 + 2 \times 0.75 = 1.75$.
The elevation in boiling point is given by $\Delta T_{b} = i \times K_{b} \times m$.
Substituting the values: $2.5 = 1.75 \times 0.52 \times m$.
Calculating molality $m = \frac{2.5}{1.75 \times 0.52} = \frac{2.5}{0.91} \approx 2.747$.
Rounding off to the nearest integer,we get $3$.

(A) Let us analyze the structures of the given compounds:
$1$. Sulphanilic acid $(p-NH_2C_6H_4SO_3H)$: Contains a sulphonic acid group $(-SO_3H)$,not a carboxylic acid group $(-COOH)$.
$2$. Picric acid ($2,4,6-$trinitrophenol): Contains a phenolic hydroxyl group $(-OH)$ and nitro groups $(-NO_2)$,not a carboxylic acid group $(-COOH)$.
$3$. Aspirin (Acetylsalicylic acid): Contains a carboxylic acid group $(-COOH)$ and an ester group.
$4$. Ascorbic acid (Vitamin $C$): Contains hydroxyl groups and a cyclic ester (lactone),not a carboxylic acid group $(-COOH)$.
Thus,only Aspirin contains the $-COOH$ group.
The total number of such compounds is $1$.

(C) Given: $T_1 = 27 + 273 = 300 \, K$,$T_2 = 52 + 273 = 325 \, K$,$K_2 = 5K_1$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
Using the Arrhenius equation: $\ln \frac{K_2}{K_1} = \frac{E_a}{R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$.
$\ln(5) = \frac{E_a}{8.314} \left[ \frac{325 - 300}{300 \times 325} \right]$.
$1.6094 = \frac{E_a}{8.314} \left[ \frac{25}{97500} \right]$.
$1.6094 = \frac{E_a}{8.314} \times 0.0002564$.
$E_a = \frac{1.6094 \times 8.314}{0.0002564} \approx 52194 \, J \, mol^{-1} = 52.194 \, kJ \, mol^{-1}$.
Rounding to the nearest integer,we get $52 \, kJ \, mol^{-1}$.

(A) The atomic number of the element is $Z = 29$,which corresponds to Copper $(Cu)$.
The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
For the divalent ion $Cu^{2+}$,the electronic configuration is $[Ar] 3d^9$.
In the $3d$ orbital,there are $9$ electrons,which means there is $1$ unpaired electron.
The spin-only magnetic moment formula is $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Substituting $n = 1$: $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
Rounding $1.73$ to the nearest integer gives $2 \ BM$.

(A) The concordant volume of $NaOH$ is the average of the consistent readings $(iii, iv, v)$,which is $4.4\, mL$.
Using the principle of equivalence: $\text{equivalents of } NaOH = \text{equivalents of } H_2C_2O_4$.
$M_{NaOH} \times V_{NaOH} \times n_{factor} = M_{acid} \times V_{acid} \times n_{factor}$.
For $NaOH$,$n_{factor} = 1$. For oxalic acid $(H_2C_2O_4)$,$n_{factor} = 2$.
$M \times 4.4 \times 1 = 1.25 \times 10.0 \times 2$.
$M = \frac{25}{4.4} \approx 5.68\, M$.
Rounding to the nearest integer,we get $6\, M$.

(B) For the reduction of $NO_{3}^{-}$ to $NO$:
$NO_{3}^{-} + 4H^{+} + 3e^{-} \longrightarrow NO + 2H_{2}O$
$E_{1} = E_{NO_{3}^{-} / NO}^{\circ} - \frac{0.059}{3} \log \frac{P_{NO}}{[NO_{3}^{-}][H^{+}]^{4}}$
For the reduction of $NO_{3}^{-}$ to $NO_{2}$:
$NO_{3}^{-} + 2H^{+} + e^{-} \longrightarrow NO_{2} + H_{2}O$
$E_{2} = E_{NO_{3}^{-} / NO_{2}}^{\circ} - \frac{0.059}{1} \log \frac{P_{NO_{2}}}{[NO_{3}^{-}][H^{+}]^{2}}$
Given $[HNO_{3}] = y$,so $[H^{+}] = y$ and $[NO_{3}^{-}] = y$. Also $P_{NO} = P_{NO_{2}} = 1 \ bar$.
Equating $E_{1} = E_{2}$:
$0.96 - \frac{0.059}{3} \log \frac{1}{y \cdot y^{4}} = 0.79 - 0.059 \log \frac{1}{y \cdot y^{2}}$
$0.17 = \frac{0.059}{3} \log (y^{-5}) - 0.059 \log (y^{-3})$
$0.17 = -\frac{0.059}{3} \cdot 5 \log y + 0.059 \cdot 3 \log y$
$0.17 = 0.059 \log y (3 - \frac{5}{3}) = 0.059 \log y (\frac{4}{3})$
$\log y = \frac{0.17 \times 3}{0.059 \times 4} = \frac{0.51}{0.236} \approx 2.16$
$y = 10^{2.16}$,so $x = 2.16$.
$2x = 4.32$.

(C) For an $FCC$ unit cell,the number of atoms per unit cell $(Z)$ is $4$.
The formula for density $(d)$ is $d = \frac{Z \times M}{N_{A} \times a^{3}}$.
Given: $Z = 4$,$M = 63.54 \, g/mol = 0.06354 \, kg/mol$,$N_{A} = 6.022 \times 10^{23} \, mol^{-1}$,and $a = 3.596 \, \mathring{A} = 3.596 \times 10^{-10} \, m$.
Substituting the values: $d = \frac{4 \times 0.06354}{6.022 \times 10^{23} \times (3.596 \times 10^{-10})^{3}} \approx 9076.6 \, kg/m^{3}$.
Thus,the density is approximately $9077 \, kg/m^{3}$.

(D) $T \ell I_{3}$ is isomorphous to $CsI_{3}$,which means it contains the triiodide ion $I_{3}^{\ominus}$.
Thus,$T \ell I_{3}$ is formulated as $T \ell^{\oplus} I_{3}^{\ominus}$,where $T \ell$ is in the $+1$ oxidation state. So,Assertion $A$ is correct.
The electronic configuration of $T \ell$ $(Z=81)$ is $[Xe] 4f^{14} 5d^{10} 6s^{2} 6p^{1}$.
It contains fourteen $4f$ electrons. So,Reason $R$ is correct.
However,the presence of $14f$ electrons is a general feature of the $T \ell$ atom and is not the direct reason why $T \ell I_{3}$ exists as $T \ell^{\oplus} I_{3}^{\ominus}$ (which is due to the inert pair effect and the stability of the $T \ell^{\oplus}$ ion). Therefore,$R$ is not the correct explanation of $A$.

(C) The hydrolysis products of disaccharides are as follows:
$1$. Sucrose on hydrolysis yields $\alpha-D$-Glucose and $\beta-D$-Fructose. Thus,$(a) \rightarrow (ii)$.
$2$. Lactose on hydrolysis yields $\beta-D$-Galactose and $\beta-D$-Glucose. Thus,$(b) \rightarrow (i)$.
$3$. Maltose on hydrolysis yields $\alpha-D$-Glucose and $\alpha-D$-Glucose. Thus,$(c) \rightarrow (iii)$.
Therefore,the correct matching is $(a)$ $\rightarrow (ii), (b)$ $\rightarrow (i), (c)$ $\rightarrow (iii)$.

(D) The basic strength of amines depends on the availability of the lone pair on the nitrogen atom.
$A$ (Phenyl methanamine,$C_6H_5CH_2NH_2$): The lone pair is not in conjugation with the benzene ring,making it the most basic.
$B$ ($N,N$-Dimethylaniline,$C_6H_5N(CH_3)_2$): The lone pair is in conjugation with the benzene ring,but the two methyl groups provide inductive effect $(+I)$,increasing basicity compared to $C$ and $D$.
$C$ ($N$-Methylaniline,$C_6H_5NHCH_3$): The lone pair is in conjugation with the benzene ring,and one methyl group provides inductive effect $(+I)$.
$D$ (Benzenamine,$C_6H_5NH_2$): The lone pair is in conjugation with the benzene ring,and there is no inductive effect from alkyl groups.
Therefore,the order of basic strength is $A > B > C > D$.

(C) The Seliwanoff test is a chemical test which distinguishes between aldose and ketose sugars. It specifically identifies ketoses by producing a red color.
The Xanthoproteic test is used to detect the presence of proteins,specifically those containing amino acids with aromatic rings (like tyrosine,tryptophan,and phenylalanine),which produce a yellow color upon treatment with concentrated nitric acid.

(B) $2,4-DNP$ stands for $2,4-dinitrophenylhydrazine$.
This reagent is used to identify carbonyl compounds,which include both aldehydes and ketones.
Among the given options,aldehyde is a carbonyl compound.

(C) Ceric ammonium nitrate is used as a test for the identification of alcohols,which gives a characteristic red or yellow color.
$CHCl_3 / \text{alc. } KOH$ is used in the carbylamine test,which is specific for the identification of primary amines $(R-NH_2)$.

(A) The given reaction is an intramolecular aldol condensation of $o$-phenylenediacetaldehyde.
$1$. The base $(OH^-)$ abstracts an $\alpha$-hydrogen from one of the aldehyde groups to form an enolate ion.
$2$. This enolate ion performs a nucleophilic attack on the carbonyl carbon of the other aldehyde group,leading to the formation of a cyclic $\beta$-hydroxy aldehyde.
$3$. Upon heating in the presence of an acid or base (dehydration),the $\beta$-hydroxy aldehyde undergoes elimination of a water molecule to form an $\alpha,\beta$-unsaturated aldehyde.
$4$. The final product is $2$-formyl$-1,2,3,4-$tetrahydronaphthalene.

(B) The reaction proceeds in three steps:
$1$. Cross-Cannizzaro reaction: $p$-methoxybenzaldehyde reacts with $HCHO$ in the presence of $NaOH$ to form $p$-methoxybenzyl alcohol $(CH_3O-C_6H_4-CH_2OH)$ and sodium formate.
$2$. Williamson ether synthesis: The alcohol $p$-methoxybenzyl alcohol reacts with $CH_3CH_2Br$ in the presence of $NaH$ to form $p$-methoxybenzyl ethyl ether $(CH_3O-C_6H_4-CH_2-O-CH_2CH_3)$.
$3$. Cleavage with $HI$ and $\Delta$: The ether linkage and the methoxy group are cleaved by $HI$ upon heating. The methoxy group $(CH_3O-)$ is converted to $OH$ and $CH_3I$,and the benzyl ether $(R-CH_2-O-CH_2CH_3)$ is cleaved to form $p$-hydroxybenzyl iodide $(HO-C_6H_4-CH_2I)$ and ethanol (which further reacts with $HI$ to form $CH_3CH_2I$).
Thus,the final product $A$ is $4$-hydroxybenzyl iodide.

(C) . Benzenediazonium chloride reacting with $Cu_2Cl_2/HCl$ is the Sandmeyer reaction $(a \rightarrow ii)$.
$b$. Benzenediazonium chloride reacting with $Cu/HCl$ is the Gatterman reaction $(b \rightarrow iv)$.
$c$. The reaction of alkyl halides with $Na$ in ether to form alkanes is the Wurtz reaction $(c \rightarrow i)$.
$d$. The reaction of aryl halides with $Na$ in ether to form diaryls is the Fittig reaction $(d \rightarrow iii)$.
Therefore,the correct matching is $a$ $\rightarrow ii, b$ $\rightarrow iv, c$ $\rightarrow i, d$ $\rightarrow iii$.

(B) The reaction involves the treatment of a polyfunctional compound with $SOCl_2$.
$SOCl_2$ (thionyl chloride) is a reagent commonly used to convert alcohols into alkyl chlorides.
In the given substrate,there are three hydroxyl groups:
$1$. $A$ phenolic $-OH$ group.
$2$. $A$ secondary aliphatic $-OH$ group on the saturated ring.
$3$. $A$ primary benzylic $-CH_2OH$ group.
$SOCl_2$ reacts readily with aliphatic alcohols (both primary and secondary) to form alkyl chlorides.
However,it does not typically convert phenolic $-OH$ groups into aryl chlorides under standard conditions.
Therefore,both the secondary $-OH$ and the primary benzylic $-CH_2OH$ groups will be converted into $-Cl$ and $-CH_2Cl$ respectively.
The resulting major product $A$ will have a phenolic $-OH$ group intact,a secondary $-Cl$ group,and a $-CH_2Cl$ group.
This corresponds to the structure shown in option $B$.

(C) Siderite is $FeCO_{3}$,which is an ore of $Fe$.
$(b)$ Calamine is $ZnCO_{3}$,which is an ore of $Zn$.
$(c)$ Malachite is $Cu(OH)_{2} \cdot CuCO_{3}$,which is an ore of $Cu$.
$(d)$ Cryolite is $Na_{3}AlF_{6}$,which is an ore of $Al$.
Therefore,the correct matching is: $(a)$ $\rightarrow (iii), (b)$ $\rightarrow (v), (c)$ $\rightarrow (i), (d)$ $\rightarrow (iv)$.

(A) When $FeCl_3$ is added to excess of hot water,it undergoes hydrolysis to form hydrated ferric oxide,$Fe_2O_3 \cdot xH_2O$.
During this process,the colloidal particles preferentially adsorb $Fe^{3+}$ ions from the solution,resulting in the formation of a positively charged sol.
The representation is $Fe_2O_3 \cdot xH_2O / Fe^{3+}$.
Therefore,the nature of the charge is positive.

(A) Sodium carbonate $(Na_2CO_3)$ is prepared by the Solvay process.
$(b)$ Titanium $(Ti)$ is refined by the Van-Arkel process.
$(c)$ Chlorine $(Cl_2)$ is prepared by the Deacon process.
$(d)$ Sodium hydroxide $(NaOH)$ is prepared by the Castner-Kellner process.
Therefore,the correct matching is $(a)$ $\rightarrow (iv), (b)$ $\rightarrow (iii), (c)$ $\rightarrow (i), (d)$ $\rightarrow (ii)$.

(A) The cell reaction is: $Zn_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + 2Ag_{(s)}$
$E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode} = 0.80 - (-0.76) = 1.56 \,V$
Using the Nernst equation: $E_{cell} = E^{0}_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Ag^{+}]^{2}}$
Here,$n = 2$,$[Zn^{2+}] = 0.1$,and $[Ag^{+}] = 0.01$
$E_{cell} = 1.56 - \frac{0.059}{2} \log \frac{0.1}{(0.01)^{2}}$
$E_{cell} = 1.56 - 0.0295 \times \log(1000)$
$E_{cell} = 1.56 - 0.0295 \times 3 = 1.56 - 0.0885 = 1.4715 \,V$
$E_{cell} = 147.15 \times 10^{-2} \,V$
Rounding to the nearest integer,$x = 147$.