JEE Main 2021 Chemistry Question Paper with Answer and Solution

(B) The number of angular nodes is given by the azimuthal quantum number $l$.
Since there are no angular nodes,$l = 0$,which corresponds to an '$s$' orbital.
The number of radial nodes is given by the formula $n - l - 1$.
Given that the number of radial nodes is $2$,we have $n - 0 - 1 = 2$.
Solving for $n$,we get $n = 3$.
Therefore,the orbital is $3s$.

(D) For temporary hardness,the reaction on heating is:
$Mg(HCO_3)_2 \xrightarrow{\Delta} Mg(OH)_2 \downarrow + 2CO_2 \uparrow$
Thus,Assertion $A$ is false because $Mg(HCO_3)_2$ converts to $Mg(OH)_2$,not $MgCO_3$.
Regarding the solubility products $(K_{sp})$:
$K_{sp}(MgCO_3) = 3.5 \times 10^{-8}$
$K_{sp}(Mg(OH)_2) = 1.8 \times 10^{-11}$
Since $3.5 \times 10^{-8} > 1.8 \times 10^{-11}$,the solubility product of $MgCO_3$ is greater than that of $Mg(OH)_2$.
Therefore,Reason $R$ is also false.
Both statements are false.

(C) The reaction between phosphorus trichloride $(PCl_3)$ and phosphonic acid $(H_3PO_3)$ is given by:
$PCl_3 + 3H_3PO_3 \rightarrow 3H_3PO_2Cl + H_3PO_3$ (Note: The standard reaction for the formation of pyrophosphorous acid is $PCl_3 + 3H_3PO_3 \rightarrow 3H_3PO_2Cl + H_3PO_3$ is not the intended path here).
Based on the provided image,the reaction is $PCl_3 + H_3PO_3 \rightarrow H_4P_2O_5 + 3HCl$.
The product $H_4P_2O_5$ is pyrophosphorous acid.
In the structure of pyrophosphorous acid $(H_4P_2O_5)$,there are two $P-OH$ groups.
Since only the hydrogen atoms attached to oxygen atoms (in $P-OH$ groups) are ionisable,there are $2$ ionisable hydrogens in the molecule.

(B) According to Molecular Orbital Theory,the bond order is calculated as $\text{Bond Order} = \frac{N_b - N_a}{2}$.
For a diatomic molecule with $15$ electrons (like $NO$),the molecular orbital configuration is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$.
Number of bonding electrons $(N_b)$ = $2+2+2+2+2 = 10$.
Number of antibonding electrons $(N_a)$ = $2+2+1 = 5$.
$\text{Bond Order} = \frac{10 - 5}{2} = \frac{5}{2} = 2.5$.
Thus,the total number of electrons in $AX$ is $15$.

(D) The Henderson-Hasselbalch equation for an acidic buffer is given by:
$pH = pKa + \log \frac{[Salt]}{[Acid]}$
Given values are:
$pH = 5.74$
$pKa = 4.74$
$[Acid] = 1.0 \ M$
Substituting these values into the equation:
$5.74 = 4.74 + \log \frac{[Salt]}{1.0}$
$5.74 - 4.74 = \log [Salt]$
$1 = \log [Salt]$
Taking the antilog on both sides:
$[Salt] = 10^1 = 10 \ M$
Therefore,the concentration of sodium acetate is $10 \ M$.

(C) The reaction is $C_{2}H_{6} \rightarrow C_{2}H_{4} + H_{2}$.
In $C_{2}H_{6}$ (ethane),there are $1$ $C-C$ bond and $6$ $C-H$ bonds.
In $C_{2}H_{4}$ (ethene),there are $1$ $C=C$ bond and $4$ $C-H$ bonds.
In $H_{2}$,there is $1$ $H-H$ bond.
Reaction enthalpy $\Delta_{r}H = \sum \text{Bond Enthalpies of Reactants} - \sum \text{Bond Enthalpies of Products}$.
$\Delta_{r}H = [1 \times \epsilon_{C-C} + 6 \times \epsilon_{C-H}] - [1 \times \epsilon_{C=C} + 4 \times \epsilon_{C-H} + 1 \times \epsilon_{H-H}]$.
Simplifying,$\Delta_{r}H = [\epsilon_{C-C} + 2 \times \epsilon_{C-H}] - [\epsilon_{C=C} + \epsilon_{H-H}]$.
Substituting the values: $\Delta_{r}H = [347 + 2 \times 414] - [611 + 436]$.
$\Delta_{r}H = [347 + 828] - [1047]$.
$\Delta_{r}H = 1175 - 1047 = 128 \, kJ \, mol^{-1}$.

(B) The dehydration reaction of $3-$hydroxypropanal is:
$HO-CH_2-CH_2-CHO \xrightarrow{\Delta} CH_2=CH-CHO + H_2O$
Let the mass of $3-$hydroxypropanal be $x \ g$.
Number of moles of $3-$hydroxypropanal $= \frac{x}{74} \ mol$.
Theoretical moles of acrolein produced $= \frac{x}{74} \ mol$.
Actual moles of acrolein produced $= \frac{7.8 \ g}{56 \ g/mol} \approx 0.1393 \ mol$.
Given percentage yield $= 64\% = 0.64$.
Actual yield $= \text{Theoretical yield} \times \text{Percentage yield}$.
$0.1393 = \left(\frac{x}{74}\right) \times 0.64$.
$x = \frac{0.1393 \times 74}{0.64} \approx 16.11 \ g$.
Rounding off to the nearest integer,we get $16 \ g$.

(D) The balanced chemical equation for the combustion of ethane $(C_2H_6)$ is:
$2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(l)$
Molar mass of ethane $(C_2H_6)$ = $(2 \times 12.0) + (6 \times 1.0) = 30 \ g/mol$.
Moles of ethane = $\frac{3 \ g}{30 \ g/mol} = 0.1 \ mol$.
From the stoichiometry of the reaction,$2 \ mol$ of $C_2H_6$ produces $6 \ mol$ of $H_2O$.
Therefore,$1 \ mol$ of $C_2H_6$ produces $3 \ mol$ of $H_2O$.
So,$0.1 \ mol$ of $C_2H_6$ produces $0.1 \times 3 = 0.3 \ mol$ of $H_2O$.
Number of molecules of water = $\text{moles} \times N_A = 0.3 \times 6.023 \times 10^{23} = 1.8069 \times 10^{23} = 18.069 \times 10^{22}$.
Rounding to the nearest integer,the value of $x$ is $18$.

(C) Greenhouse gases are gases in the atmosphere that trap heat.
Major greenhouse gases include $CO_2$ (Carbon dioxide),$H_2O$ (Water vapour),and $CH_4$ (Methane).
Therefore,$A, C$ and $D$ are greenhouse gases.

(A) For element $X$,the jump from $IE_1$ $(495 \ kJ/mol)$ to $IE_2$ $(4563 \ kJ/mol)$ is very large,indicating that the second electron is removed from a stable noble gas core. This corresponds to $Na$ $([Ne] 3s^1)$.
For element $Y$,the values $IE_1$ $(731 \ kJ/mol)$ and $IE_2$ $(1450 \ kJ/mol)$ are relatively close,which is characteristic of an alkaline earth metal like $Mg$ $([Ne] 3s^2)$,where both valence electrons are removed easily compared to the third electron.
Therefore,$X=Na$ and $Y=Mg$.

(C) The reaction shows the chlorination of cyclohexene to form a dichlorocyclohexene derivative.
Specifically,the reaction involves the substitution of hydrogen atoms at the allylic positions.
Free radical halogenation using $Cl_2$ in the presence of $UV$ light is the standard method for allylic substitution in alkenes.
Therefore,the correct reagent and condition is $A = Cl_2$ with $UV$ light.

(A) $FeI_3$ is unstable because $I^-$ is a strong reducing agent that reduces $Fe^{3+}$ to $Fe^{2+}$.
The reaction is: $2 FeI_3 \longrightarrow 2 FeI_2 + I_2$.
Therefore,$Fe$ forms dihalides with $F, Cl, Br, I$ $(FeF_2, FeCl_2, FeBr_2, FeI_2)$ and trihalides with $F, Cl, Br$ $(FeF_3, FeCl_3, FeBr_3)$.

(C) $I$. $NaH$ (sodium hydride) contains hydrogen in the $-1$ oxidation state,making it a strong reducing agent,not an oxidising agent. Thus,statement $I$ is false.
$II$. In pyridine,the nitrogen atom has a lone pair of electrons that is not involved in the aromatic sextet. This lone pair is available for donation,which makes pyridine basic in nature. Thus,statement $II$ is true.

(D) The structure of $C_{60}$ (Buckminsterfullerene) consists of $60$ carbon atoms arranged in a truncated icosahedron shape.
It contains $20$ six-membered rings (hexagons) and $12$ five-membered rings (pentagons).
Each carbon atom is $sp^2$ hybridized and forms three sigma bonds.
The five-membered rings are fused only to six-membered rings,while the six-membered rings are fused to both six and five-membered rings.
Therefore,the statement that it contains $12$ six-membered rings and $24$ five-membered rings is incorrect.

(B) $H_2O_2$ is used in the treatment of effluents (industrial waste).
$H_2O_2$ acts as both an oxidising agent $(O.A.)$ and a reducing agent $(R.A.)$ in both acidic and basic media.
$H_2O_2$ has a non-planar,open-book structure. Therefore,the two hydroxyl groups do not lie in the same plane.
$H_2O_2$ is miscible with water in all proportions due to hydrogen bonding.
Thus,statements $A, B,$ and $D$ are correct.

(C) The reaction of an unsaturated hydrocarbon $X$ with ozone (ozonolysis) produces compound $A$.
Compound $A$ gives a positive Tollen's test (forms a silver mirror with ammoniacal silver nitrate),which indicates that $A$ is an aldehyde or formic acid $(HCOOH)$.
Let us analyze the options:
$(A)$ $CH_3-C(CH_3)=C(CH_3)-CH_3$ on ozonolysis gives $CH_3-CO-CH_3$ (acetone),which does not give Tollen's test.
$(B)$ $CH_3-C(CH_3)=C(CH_2)_2$ on ozonolysis gives $CH_3-CO-CH_3$ and cyclobutanone,neither of which gives Tollen's test.
$(C)$ $CH_3-CH_2-C \equiv CH$ on ozonolysis gives $CH_3-CH_2-COOH$ and $HCOOH$. Formic acid $(HCOOH)$ gives a positive Tollen's test.
$(D)$ $CH_3-C \equiv C-CH_3$ on ozonolysis gives $2CH_3-COOH$ (acetic acid),which does not give Tollen's test.
Therefore,the correct hydrocarbon is $CH_3-CH_2-C \equiv CH$.

(B) The atomic number $33$ corresponds to Arsenic $(As)$,which is a metalloid.
The atomic number $53$ corresponds to Iodine $(I)$,which is a non-metal.
The atomic number $83$ corresponds to Bismuth $(Bi)$,which is a metal.
Therefore,$X$ is a metalloid,$Y$ is a non-metal,and $Z$ is a metal.

(C) The correct matches are as follows:
$(a)$ Lassaigne's test is used for the detection of nitrogen,sulphur,phosphorus,and halogens in organic compounds. Thus,$(a)-(iii)$.
$(b)$ $Cu(II)$ oxide is used in the detection of carbon in organic compounds (via the formation of $CO_2$). Thus,$(b)-(i)$.
$(c)$ Silver nitrate $(AgNO_3)$ is used specifically for the detection of halogens. Thus,$(c)-(iv)$.
$(d)$ The sodium fusion extract gives a black precipitate with acetic acid and lead acetate,which confirms the presence of sulphur $(PbS)$. Thus,$(d)-(ii)$.
Therefore,the correct sequence is $(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)$.

(C) $H_3PO_3$ is a dibasic acid ($n$-factor = $2$) and $H_3PO_2$ is a monobasic acid ($n$-factor = $1$).
For neutralization,equivalents of base = equivalents of acid.
$(M_1 \times n_1) \times V_1 = (M_2 \times n_2) \times V_2$
$1.$ For $H_3PO_3$:
$1 \times 1 \times V_{NaOH} = 1 \times 2 \times 50 \implies V_{NaOH} = 100\, mL$.
$2.$ For $H_3PO_2$:
$1 \times 1 \times V_{NaOH} = 2 \times 1 \times 100 \implies V_{NaOH} = 200\, mL$.
Therefore,the volumes are $100\, mL$ and $200\, mL$.

(B) In Duma's method,the pressure of dry nitrogen is calculated by subtracting the aqueous tension from the total pressure: $P_{N_2} = 758 \, mm - 14 \, mm = 744 \, mm \, Hg$.
Using the ideal gas equation at $STP$ $(P_1V_1/T_1 = P_2V_2/T_2)$:
$V_{STP} = \frac{P_{N_2} \times V_{obs} \times 273}{760 \times T} = \frac{744 \times 30 \times 273}{760 \times 287} \approx 27.935 \, mL$.
Since $22400 \, mL$ of $N_2$ at $STP$ weighs $28 \, g$,the mass of $N_2$ obtained is:
$Mass_{N_2} = \frac{28 \times 27.935}{22400} \approx 0.03492 \, g$.
Percentage of nitrogen = $\frac{Mass_{N_2}}{Mass_{compound}} \times 100 = \frac{0.03492}{0.1840} \times 100 \approx 18.98 \%$.
Rounding off to the nearest integer,we get $19 \% $.

(A) For a given principal quantum number $n = 5$,the possible values of azimuthal quantum number $\ell$ are $0, 1, 2, 3, 4$.
The magnetic quantum number $m_{l}$ ranges from $-\ell$ to $+\ell$ for each $\ell$.
We need to find the number of orbitals where $m_{l} = +2$:
$1$. For $\ell = 0$ $(5s)$,$m_{l} = 0$.
$2$. For $\ell = 1$ $(5p)$,$m_{l} = \{-1, 0, +1\}$.
$3$. For $\ell = 2$ $(5d)$,$m_{l} = \{-2, -1, 0, +1, +2\}$. Here,$m_{l} = +2$ is possible ($1$ orbital).
$4$. For $\ell = 3$ $(5f)$,$m_{l} = \{-3, -2, -1, 0, +1, +2, +3\}$. Here,$m_{l} = +2$ is possible ($1$ orbital).
$5$. For $\ell = 4$ $(5g)$,$m_{l} = \{-4, -3, -2, -1, 0, +1, +2, +3, +4\}$. Here,$m_{l} = +2$ is possible ($1$ orbital).
Total number of orbitals with $m_{l} = +2$ is $1 + 1 + 1 = 3$.

(B) $H_{2}SO_{3}$ is a dibasic acid with concentration $c = 0.588 \ M$.
Since $Ka_{1} \gg Ka_{2}$,the $pH$ is primarily determined by the first dissociation step:
$H_{2}SO_{3(aq)} \rightleftharpoons H^{+}_{(aq)} + HS{O_{3}}^{-}_{(aq)}$
$Ka_{1} = \frac{[H^{+}][HSO_{3}^{-}]}{[H_{2}SO_{3}]} = 1.7 \times 10^{-2}$
Let $x$ be the concentration of $H^{+}$ at equilibrium:
$\frac{x^{2}}{0.588 - x} = 0.017$
$x^{2} + 0.017x - 0.009996 = 0$
Using the quadratic formula $x = \frac{-b + \sqrt{b^{2} - 4ac}}{2a}$:
$x = \frac{-0.017 + \sqrt{(0.017)^{2} - 4(1)(-0.009996)}}{2} \approx 0.09186 \ M$
$pH = -\log[H^{+}] = -\log(0.09186) \approx 1.036$
Rounding to the nearest integer,the $pH$ is $1$.

(B) The balanced chemical equation is: $3Pb(NO_3)_2 + Cr_2(SO_4)_3 \rightarrow 3PbSO_4 + 2Cr(NO_3)_3$
Calculate the initial moles of reactants:
Moles of $Pb(NO_3)_2 = 0.15 \ M \times 0.035 \ L = 0.00525 \ mol = 5.25 \times 10^{-3} \ mol$
Moles of $Cr_2(SO_4)_3 = 0.12 \ M \times 0.020 \ L = 0.0024 \ mol = 2.4 \times 10^{-3} \ mol$
Determine the limiting reagent:
For $Pb(NO_3)_2$,moles required per stoichiometric coefficient $= (5.25 \times 10^{-3}) / 3 = 1.75 \times 10^{-3}$
For $Cr_2(SO_4)_3$,moles required per stoichiometric coefficient $= (2.4 \times 10^{-3}) / 1 = 2.4 \times 10^{-3}$
Since $1.75 \times 10^{-3} < 2.4 \times 10^{-3}$,$Pb(NO_3)_2$ is the limiting reagent.
Calculate moles of $PbSO_4$ formed:
According to the stoichiometry,$3 \ mol$ of $Pb(NO_3)_2$ produces $3 \ mol$ of $PbSO_4$.
Therefore,$5.25 \times 10^{-3} \ mol$ of $Pb(NO_3)_2$ produces $5.25 \times 10^{-3} \ mol$ of $PbSO_4$.
$5.25 \times 10^{-3} \ mol = 525 \times 10^{-5} \ mol$.

(A) The chemical reaction is: $Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)$.
Moles of $Fe = \frac{50 \ g}{55.85 \ g \ mol^{-1}} \approx 0.89526 \ mol$.
Since $1 \ mol$ of $Fe$ produces $1 \ mol$ of $H_2(g)$,moles of $H_2(g)$ produced $(n)$ $= 0.89526 \ mol$.
Work done $(w)$ $= -P_{ext} \Delta V = -\Delta n_g RT$.
Given $T = 25^{\circ} C = 298 \ K$,$R = 8.314 \ J \ mol^{-1} \ K^{-1}$,and $P_{ext} = 1 \ bar$.
$w = -nRT = -0.89526 \ mol \times 8.314 \ J \ mol^{-1} \ K^{-1} \times 298 \ K$.
$w = -2218.059 \ J$.
The magnitude of work done by the gas is $2218 \ J$.

(D) The pairs $Li-Mg$,$B-Si$,and $Be-Al$ exhibit a diagonal relationship in the periodic table.
$Li$ and $Na$ belong to the same group $(Group \ 1)$ and do not show a diagonal relationship,as they are placed vertically relative to each other.

(C) Alkanes are generally inert towards oxidation. However,alkanes containing a tertiary hydrogen atom can be oxidized to the corresponding tertiary alcohol using a strong oxidizing agent like $KMnO_4$.
Statement-$I$ is correct because $2-$methylbutane contains a tertiary hydrogen atom at the $C-2$ position,which can be oxidized to $2-$methylbutan$-2-$ol.
Statement-$II$ is incorrect because $n-$alkanes lack tertiary hydrogen atoms and are not easily oxidized to alcohols by $KMnO_4$ under normal conditions.

(D) To determine the shape of the given species,we analyze their structures based on $VSEPR$ theory:
$1$. $NO_2$: The nitrogen atom has one unpaired electron and two oxygen atoms attached,resulting in a bent shape due to the lone electron and repulsion.
$2$. $Cl_2O$: The central oxygen atom has two lone pairs and two bonded chlorine atoms,resulting in a bent shape (similar to $H_2O$).
$3$. $O_3$: The central oxygen atom has one lone pair and is bonded to two other oxygen atoms,resulting in a bent shape.
$4$. $N_3^-$ (Azide ion): The central nitrogen atom is $sp$ hybridized and bonded to two other nitrogen atoms with no lone pairs on the central atom,resulting in a linear shape.
Thus,the linear species is $N_3^-$.

(C) In the Solvay process,$NH_{3}$ is recovered by treating the ammonium chloride $(NH_{4}Cl)$ solution with calcium hydroxide $(Ca(OH)_{2})$.
The chemical reaction is: $2NH_{4}Cl + Ca(OH)_{2} \rightarrow 2NH_{3} + CaCl_{2} + 2H_{2}O$.
Here,$CaCl_{2}$ (calcium chloride) is obtained as a by-product.

(D) Eutrophication is the process in which nutrient-enriched water bodies support a dense plant population,which kills animal life by depriving it of oxygen and results in a loss of biodiversity.
$A$. Excess usage of fertilisers leads to nutrient enrichment (e.g.,nitrates and phosphates),causing eutrophication.
$B$. Excess usage of detergents also adds phosphates to water bodies,causing eutrophication.
$C$. Dense plant population is a result/symptom of eutrophication,not a cause.
$D$. Lack of nutrients is the opposite of the condition required for eutrophication.
Since the question asks for the incorrect reason$(s)$,both $C$ and $D$ are incorrect as causes. However,looking at the provided options,$D$ is the most fundamentally incorrect statement regarding the cause of eutrophication.

(A) In the electrophilic addition of $HBr$ to but$-1-$ene,the proton $(H^+)$ attacks the double bond to form a carbocation intermediate.
Carbocation $A$ is a primary $(1^\circ)$ carbocation $(CH_3-CH_2-CH_2-CH_2^+)$,while carbocation $B$ is a secondary $(2^\circ)$ carbocation $(CH_3-CH_2-CH^+-CH_3)$.
Secondary carbocations are more stable than primary carbocations due to the inductive effect and hyperconjugation.
Because carbocation $B$ is more stable,its formation has a lower activation energy,meaning it is formed at a faster rate.

(A) . Freezing of water to ice at $0^{\circ} C$ involves a phase change from liquid to solid,which decreases disorder,so $\Delta S < 0$.
$B$. Freezing of water to ice at $-10^{\circ} C$ is a spontaneous process where liquid water turns to solid ice,decreasing entropy,so $\Delta S < 0$.
$C$. In the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the number of moles of gas decreases from $4$ to $2$,leading to a decrease in entropy,so $\Delta S < 0$.
$D$. Adsorption of $CO_{(g)}$ on a lead surface restricts the movement of gas molecules,decreasing entropy,so $\Delta S < 0$.
$E$. Dissolution of $NaCl$ in water increases the disorder of the system as the crystal lattice breaks down into ions,so $\Delta S > 0$.
Therefore,entropy decreases in processes $A, B, C,$ and $D$.

(A) The structure of mesityl oxide is $(CH_3)_2C=CH-CO-CH_3$.
Expanding the structure to identify $C-C$ sigma bonds:
$H_3C-C(CH_3)=CH-C(=O)-CH_3$.
The $C-C$ sigma bonds are:
$1$. $C_1-C_2$ (between $CH_3$ and $C$)
$2$. $C_2-C_3$ (between $C$ and $CH_3$)
$3$. $C_2-C_4$ (between $C$ and $CH$)
$4$. $C_4-C_5$ (between $CH$ and $C=O$)
$5$. $C_5-C_6$ (between $C=O$ and $CH_3$)
Total number of $C-C$ sigma bonds = $5$.

(B) The electronic configuration of $Fe$ $(Z = 26)$ is $[Ar] 3d^6 4s^2$.
In the $3d$ subshell,there are $6$ electrons. According to Hund's rule,the distribution is:

$\uparrow \downarrow$

$\uparrow$

$\uparrow$

$\uparrow$

$\uparrow$

Number of unpaired electrons $(n) = 4$.
The spin-only magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n + 2)} \,BM$.
$\mu = \sqrt{4(4 + 2)} = \sqrt{24} \,BM$.
Using $\sqrt{24} \approx 4.899 \,BM$.
$\mu = 48.99 \times 10^{-1} \,BM$.
Rounding off to the nearest integer,we get $49$.

(D) At $STP$,the pressure $P = 1 \, bar$ and temperature $T = 273.15 \, K$.
Using the ideal gas equation $PV = nRT$,where $n = \frac{N}{N_A}$:
$1 \times (20 \times 10^{-3} \, L) = \frac{N}{6.023 \times 10^{23}} \times 0.083 \times 273.15$
$N = \frac{0.020 \times 6.023 \times 10^{23}}{0.083 \times 273.15} \approx 5.31 \times 10^{20} \, \text{molecules of } Cl_2$.
Since each $Cl_2$ molecule contains $2$ chlorine atoms:
Number of $Cl$ atoms $= 2 \times 5.31 \times 10^{20} = 10.62 \times 10^{20} = 1.062 \times 10^{21}$.
Rounding to the nearest integer,we get $1 \times 10^{21}$.

(C) For the reaction $N_{2}O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the change in the number of moles of gaseous species is $\Delta n_{g} = 2 - 1 = 1$.
The relationship between $K_{P}$ and $K_{C}$ is given by $K_{P} = K_{C} \cdot (RT)^{\Delta n_{g}}$.
Substituting the given values: $600.1 = 20.4 \times (0.0831 \times T)^{1}$.
Solving for $T$: $T = \frac{600.1}{20.4 \times 0.0831} = \frac{600.1}{1.69524} \approx 353.99 \, K$.
Rounding off to the nearest integer,we get $T = 354 \, K$.

(C) The reaction is: $C_6H_5COCl + (C_6H_5)_2NH \rightarrow C_6H_5CON(C_6H_5)_2 + HCl$
Molar mass of $C_6H_5COCl = 77 + 12 + 16 + 35.5 = 140.5 \, g/mol$.
Molar mass of $C_6H_5CON(C_6H_5)_2 = 77 + 12 + 16 + 14 + 2(77) = 273 \, g/mol$.
Moles of $C_6H_5COCl = \frac{0.140 \, g}{140.5 \, g/mol} \approx 0.001 \, mol$.
Since the stoichiometry is $1:1$,theoretical moles of product $= 0.001 \, mol$.
Theoretical mass of product $= 0.001 \, mol \times 273 \, g/mol = 0.273 \, g$.
Actual mass of product formed $= 0.210 \, g$.
Percentage yield $= \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{0.210}{0.273} \times 100 \approx 76.92 \%$.
Rounding off to the nearest integer,we get $77 \%$.

(A) To find the oxidation state of nitrogen $(N)$ in each species,we use the rule that the oxidation state of oxygen $(O)$ is $-2$.
$1$. In $NO_{3}^{-}$: $x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5$.
$2$. In $NO_{2}$: $x + 2(-2) = 0 \implies x - 4 = 0 \implies x = +4$.
$3$. In $NO$: $x + (-2) = 0 \implies x = +2$.
$4$. In $N_{2}O$: $2x + (-2) = 0 \implies 2x = 2 \implies x = +1$.
Comparing the values: $+5 (NO_{3}^{-}) > +4 (NO_{2}) > +2 (NO) > +1 (N_{2}O)$.
Thus,the correct order is $NO_{3}^{-} > NO_{2} > NO > N_{2}O$.

(D) $H_{2}O_{2}$ acts as an oxidizing agent in basic medium:
$Mn^{2+} + H_{2}O_{2} \rightarrow Mn^{4+} + 2OH^{-}$
$H_{2}O_{2}$ acts as a reducing agent in basic medium:
$I_{2} + H_{2}O_{2} + 2OH^{-} \rightarrow 2I^{-} + 2H_{2}O + O_{2}$
$H_{2}O_{2}$ acts as an oxidizing agent in acidic medium:
$PbS_{(s)} + 4H_{2}O_{2(aq)} \rightarrow PbSO_{4(s)} + 4H_{2}O_{(\ell)}$
Thus,reactions $A$ and $B$ occur in basic medium.

(C) The first ionization energy $(IE_1)$ of $3^{rd}$ period elements follows the order: $Na < Al < Mg < Si < S < P < Cl < Ar$.
Magnesium $(Mg)$ has a higher $IE_1$ than $Na$ $(Z)$ but lower than $Cl$ and $Ar$ ($X$ and $Y$).
Therefore,$X$ and $Y$ are $Ar$ and $Cl$,and $Z$ is $Na$.

(B) Statement $-I$ is false because Bohr's theory is applicable only to single-electron species (e.g.,$H$,$He^{+}$,$Li^{2+}$). Since $Li^{+}$ has $2$ electrons,Bohr's theory cannot explain its spectrum.
Statement $-II$ is true because the splitting of spectral lines in a magnetic field is known as the Zeeman effect,which Bohr's theory could not explain.

(A) . $Be$ is used in the windows of $X$-ray tubes.
$b$. $Mg$ is used in incendiary bombs and signals.
$c$. $Ca$ is used in the extraction of metals (as a reducing agent).
$d$. $Ra$ is used in the treatment of cancer (radiotherapy).
Therefore,the correct matching is $a-iv, b-iii, c-ii, d-i$.

(D) Statement $I$ is true: Thermal power plants generate large amounts of fly ash,which is a non-biodegradable waste.
Statement $II$ is true: Biodegradable detergents,especially those containing phosphates,can lead to eutrophication in water bodies,which depletes dissolved oxygen and harms aquatic life.

(C) To determine the hybridisation of a carbon atom,we count the number of sigma $(\sigma)$ bonds and lone pairs (if any) attached to it.
$1$. Carbon $a$ is part of a methyl group $(-CH_3)$. It is bonded to three hydrogen atoms and one carbon atom,forming four single bonds. Thus,it has four $\sigma$ bonds and is $sp^{3}$ hybridised.
$2$. Carbon $b$ is part of a double bond $(C=C)$. It is bonded to one carbon atom (via a double bond,which counts as one $\sigma$ bond),one hydrogen atom,and one oxygen atom. It has three $\sigma$ bonds and is $sp^{2}$ hybridised.
$3$. Carbon $c$ is part of the benzene ring. It is bonded to two other carbon atoms in the ring and one oxygen atom. It is involved in a double bond within the aromatic ring,giving it three $\sigma$ bonds. Thus,it is $sp^{2}$ hybridised.
Therefore,the hybridisation of $a, b$ and $c$ are $sp^{3}, sp^{2}$ and $sp^{2}$ respectively.

(C) To determine the number of lone pairs $(l.p.)$ on the central atom,we use the formula: $l.p. = \frac{1}{2} (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of monovalent atoms bonded to it.
$1$. $SF_{4}$: $S$ has $6$ valence electrons,$4$ bonded atoms. $l.p. = \frac{1}{2}(6 - 4) = 1$.
$2$. $BF_{4}^{-}$: $B$ has $3$ valence electrons,$4$ bonded atoms,plus $1$ negative charge. $l.p. = \frac{1}{2}(3 + 1 - 4) = 0$.
$3$. $ClF_{3}$: $Cl$ has $7$ valence electrons,$3$ bonded atoms. $l.p. = \frac{1}{2}(7 - 3) = 2$.
$4$. $AsF_{3}$: $As$ has $5$ valence electrons,$3$ bonded atoms. $l.p. = \frac{1}{2}(5 - 3) = 1$.
$5$. $PCl_{5}$: $P$ has $5$ valence electrons,$5$ bonded atoms. $l.p. = \frac{1}{2}(5 - 5) = 0$.
$6$. $BrF_{5}$: $Br$ has $7$ valence electrons,$5$ bonded atoms. $l.p. = \frac{1}{2}(7 - 5) = 1$.
$7$. $XeF_{4}$: $Xe$ has $8$ valence electrons,$4$ bonded atoms. $l.p. = \frac{1}{2}(8 - 4) = 2$.
$8$. $SF_{6}$: $S$ has $6$ valence electrons,$6$ bonded atoms. $l.p. = \frac{1}{2}(6 - 6) = 0$.
The species with two lone pairs are $ClF_{3}$ and $XeF_{4}$.
Therefore,the total number of such species is $2$.

(B) The relationship between standard Gibbs free energy change and equilibrium constant is given by $\Delta_{r} G^{\circ} = -RT \ln K_{p}$.
Given $\Delta_{r} G^{\circ} = 25200 \ J \ mol^{-1}$,$R = 8.3 \ J \ mol^{-1} \ K^{-1}$,and $T = 400 \ K$.
Substituting the values: $25200 = -8.3 \times 400 \times \ln K_{p}$.
$\ln K_{p} = -\frac{25200}{3320} \approx -7.59$.
Converting to base $10$: $\log_{10} K_{p} = \frac{-7.59}{2.3} \approx -3.3$.
$K_{p} = 10^{-3.3} = 5.01 \times 10^{-4}$.
For the reaction $2A_{(g)} \rightleftharpoons A_{2(g)}$,$\Delta n = 1 - 2 = -1$.
Using $K_{p} = K_{c}(RT)^{\Delta n}$,we have $K_{c} = K_{p}(RT)^{-\Delta n} = K_{p}(RT)^{1}$.
$K_{c} = 5.01 \times 10^{-4} \times (8.3 \times 400) = 5.01 \times 10^{-4} \times 3320 = 1.6632$.
Expressing in terms of $10^{-2}$: $1.6632 = 166.32 \times 10^{-2}$.
Rounding to the nearest integer,we get $166$.

(A) In pure water,the solubility product constant $K_{sp}$ is given by $K_{sp} = S^{2} = (8.0 \times 10^{-4})^{2} = 64 \times 10^{-8}$.
In $0.01 \ M \ H_{2}SO_{4}$,the dissociation is $H_{2}SO_{4} \rightarrow 2H^{+} + SO_{4}^{2-}$.
The concentration of $SO_{4}^{2-}$ from $H_{2}SO_{4}$ is $0.01 \ M$.
Let the solubility of $CdSO_{4}$ in this solution be $x \ mol \ L^{-1}$.
$CdSO_{4} \rightleftharpoons Cd^{2+} + SO_{4}^{2-}$.
The equilibrium concentrations are $[Cd^{2+}] = x$ and $[SO_{4}^{2-}] = (x + 0.01) \ M$.
Since $x \ll 0.01$,we can approximate $(x + 0.01) \approx 0.01$.
$K_{sp} = [Cd^{2+}][SO_{4}^{2-}] = x(0.01) = 64 \times 10^{-8}$.
$x = \frac{64 \times 10^{-8}}{0.01} = 64 \times 10^{-6} \ mol \ L^{-1}$.
Thus,the value is $64$.

(A) The reaction is: $C_6H_5COOH + Br_2 \xrightarrow{FeBr_3} C_6H_4(Br)COOH + HBr$
Molar mass of benzoic acid $(C_7H_6O_2)$ $= (7 \times 12) + (6 \times 1) + (2 \times 16) = 84 + 6 + 32 = 122 \,g/mol$.
Molar mass of $m$-bromobenzoic acid $(C_7H_5BrO_2)$ $= 122 - 1 + 80 = 201 \,g/mol$.
Moles of benzoic acid used $= \frac{6.1 \,g}{122 \,g/mol} = 0.05 \,mol$.
According to the stoichiometry,$1 \,mol$ of benzoic acid produces $1 \,mol$ of $m$-bromobenzoic acid.
Theoretical yield of $m$-bromobenzoic acid $= 0.05 \,mol \times 201 \,g/mol = 10.05 \,g$.
Actual yield $= 7.8 \,g$.
Percentage yield $= \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{7.8}{10.05} \times 100 \approx 77.61 \,\%$.
Rounding off to the nearest integer,we get $78 \,\%$.

(A) When a vicinal or geminal dihalide reacts with magnesium metal in the presence of dry ether $(Et_2O)$,it forms a Grignard reagent.
In the given reaction,$CH_3-CH_2-CH(Br)-CH_2-CH(Br)-CH_3$ reacts with excess $Mg$ in $Et_2O$ to form the corresponding di-Grignard reagent:
$CH_3-CH_2-CH(MgBr)-CH_2-CH(MgBr)-CH_3$.
This is the first step of the reaction.

(B) In general,from left to right in a period,the first ionization enthalpy increases due to an increase in effective nuclear charge.
However,due to the extra stability of half-filled $(p^3)$ and fully-filled $(s^2)$ electronic configurations,the ionization enthalpy of certain elements is higher than their neighbors.
The electronic configurations are: $Mg$ $([Ne] 3s^2)$,$Al$ $([Ne] 3s^2 3p^1)$,$Si$ $([Ne] 3s^2 3p^2)$,$P$ $([Ne] 3s^2 3p^3)$,and $S$ $([Ne] 3s^2 3p^4)$.
Due to the stable $s^2$ configuration of $Mg$,its ionization enthalpy is higher than $Al$.
Due to the stable half-filled $p^3$ configuration of $P$,its ionization enthalpy is higher than $S$.
Thus,the correct increasing order is $Al < Mg < Si < S < P$.

(C) The reaction shows the oxidation of an alkyl group (specifically a methyl group attached to a benzene ring) to a carboxylic acid group $(-COOH)$.
Alkaline $KMnO_4$ followed by acidic workup $(H^+)$ is a strong oxidizing agent that converts alkyl side chains on aromatic rings into carboxylic acid groups,regardless of the length of the alkyl chain,provided there is at least one benzylic hydrogen atom.
Therefore,the reagent $A$ is alkaline $KMnO_4$ followed by $H^+$.

(A) The reduction of $Al_2O_3 \rightarrow Al$ cannot be carried out using coke (carbon) because aluminum has a very high affinity for oxygen and is more stable than $CO_2$.
It is instead carried out by the electrolytic reduction of its fused salts (Hall-Heroult process).
$ZnO$,$Fe_2O_3$,and $Cu_2O$ can be reduced to their respective metals using carbon (coke) as a reducing agent.

(C) The secondary structure of proteins includes two main types:
$(a)$ $\alpha-$Helix
$(b)$ $\beta-$pleated sheet
In the $\alpha-$Helix structure,the polypeptide chain is coiled due to the presence of intramolecular hydrogen bonding between the $C=O$ group of one amino acid and the $N-H$ group of another amino acid.
Similarly,in $\beta-$pleated sheets,the polypeptide chains are held together by intermolecular hydrogen bonding.

(C) Urea-formaldehyde resin is widely used in the manufacture of wood laminates and adhesives for bonding wood.

(C) In the reaction of alkyl halides with ammonia,the initial product is an alkylammonium salt,which is an acidic species (e.g.,$[R-NH_3]^+ X^-$).
$NaOH$ is added to the reaction mixture to neutralize this acidic salt,thereby liberating the free amine $(R-NH_2)$ which can then further react with more alkyl halide to form secondary and tertiary amines.
Thus,the purpose of $NaOH$ is to remove acidic impurities (the hydrogen halide salt) from the reaction medium.

(C) The basicity of an amine depends on the availability of the lone pair of electrons on the nitrogen atom for protonation.
$1$. In $(CH_3CO)\ddot{N}HC_2H_5$ (an amide),the lone pair on nitrogen is delocalized due to conjugation with the carbonyl group $(C=O)$,making it less available.
$2$. In $(C_2H_5)_3\ddot{N}$ (a tertiary amine),the lone pair is localized and available for donation.
$3$. In $(CH_3CO)_2\ddot{N}H$ (an imide),the lone pair on nitrogen is strongly delocalized due to conjugation with two electron-withdrawing carbonyl groups. This makes the lone pair least available for protonation,rendering it the least basic.
$4$. In $(C_2H_5)_2\ddot{N}H$ (a secondary amine),the lone pair is localized and available for donation.
Therefore,$(CH_3CO)_2\ddot{N}H$ is the least basic compound.

(B) Colloidal solutions exhibit colligative properties due to the presence of particles,although the values are small compared to true solutions.
An ordinary filter paper has large pores that allow colloidal particles to pass through easily; therefore,it cannot stop the flow of colloidal particles.
According to the Hardy-Schulze rule,the flocculating power of an ion increases with the increase in the magnitude of its charge. Thus,$Al^{3+}$ has a much higher flocculating power than $Na^{+}$.
Colloidal particles exhibit Brownian motion due to the unbalanced bombardment of the particles by the molecules of the dispersion medium.
Therefore,statement $B$ is incorrect.

(D) $_{58}Ce \rightarrow [Xe] 4f^1 5d^1 6s^2$. In the complex,$Ce^{4+} \rightarrow [Xe] 4f^0$. There are $0$ unpaired electrons,so $\mu_m = 0 \ B.M$.
$(b)$ $_{64}Gd^{3+} \rightarrow [Xe] 4f^7$. There are $7$ unpaired electrons,so $\mu_m = \sqrt{7(7+2)} = \sqrt{63} \ B.M$.
$(c)$ $_{63}Eu^{3+} \rightarrow [Xe] 4f^6$. There are $6$ unpaired electrons,so $\mu_m = \sqrt{6(6+2)} = \sqrt{48} \ B.M$.
Comparing the magnetic moments: $0 < \sqrt{48} < \sqrt{63}$.
Thus,the increasing order is $(a) < (c) < (b)$.

(D) The reaction of a nitrile $(-CN)$ with a Grignard reagent $(PhMgBr)$ followed by acid hydrolysis $(H_3O^+)$ is a standard method for the preparation of ketones.
$1$. The nucleophilic phenyl group $(Ph^-)$ from $PhMgBr$ attacks the electrophilic carbon of the nitrile group $(-CN)$ to form an imine magnesium salt intermediate.
$2$. Subsequent acid hydrolysis $(H_3O^+)$ converts the imine intermediate into a ketone $(C=O)$.
$3$. The other substituent on the benzene ring,$-\text{CH}(\text{CH}_3)\text{OCH}_3$,remains unaffected by these reagents.
$4$. Therefore,the final product $X$ is a ketone where the $-CN$ group is replaced by a $-COC_6H_5$ group,while the side chain $-\text{CH}(\text{CH}_3)\text{OCH}_3$ remains unchanged.

(C) In an $HCP$ structure,for every $1$ atom,there is $1$ octahedral void $(OV)$ and $2$ tetrahedral voids $(TV)$.
Therefore,the total number of voids per atom is $1 + 2 = 3$.
Number of moles of $Ga = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.581 \, g}{70 \, g/mol} = 0.0083 \, mol$.
Number of $Ga$ atoms $= \text{moles} \times N_A = 0.0083 \times 6.022 \times 10^{23} \approx 4.998 \times 10^{21}$ atoms.
Total number of voids $= 3 \times \text{Number of atoms} = 3 \times 4.998 \times 10^{21} = 14.994 \times 10^{21}$.
Rounding to the nearest integer,we get $15 \times 10^{21}$.

(C) Given: Concentration $(C) = 5.0 \, mmol \, dm^{-3} = 5.0 \times 10^{-3} \, mol \, L^{-1} = 5.0 \, mol \, m^{-3}$.
Conductance $(G) = 0.55 \, mS = 0.55 \times 10^{-3} \, S$.
Cell constant $(G^*) = 1.3 \, cm^{-1} = 130 \, m^{-1}$.
Conductivity $(\kappa) = G \times G^* = 0.55 \, mS \times 130 \, m^{-1} = 71.5 \, mS \, m^{-1} = 0.0715 \, S \, m^{-1}$.
Molar conductivity $(\lambda_m) = \frac{\kappa}{C} = \frac{0.0715 \, S \, m^{-1}}{5.0 \, mol \, m^{-3}} = 0.0143 \, S \, m^2 \, mol^{-1} = 14.3 \, mS \, m^2 \, mol^{-1}$.
Rounding to the nearest integer,the value is $14$.

(B) Given $(t_{1/2})_A = 54 \, min$ and $(t_{1/2})_B = 18 \, min$.
Starting with equimolar concentrations,let $[A]_0 = [B]_0 = x$.
For first order kinetics,the concentration at time $t$ is given by $[A]_t = [A]_0 \times (1/2)^n$,where $n = t / t_{1/2}$.
We want $[A]_t = 16 \times [B]_t$.
Substituting the expressions: $x \times (1/2)^{t/54} = 16 \times x \times (1/2)^{t/18}$.
$(1/2)^{t/54} = 2^4 \times (1/2)^{t/18}$.
$(1/2)^{t/54} = (1/2)^{-4} \times (1/2)^{t/18}$.
$(1/2)^{t/54} = (1/2)^{(t/18) - 4}$.
Equating the exponents: $t/54 = (t/18) - 4$.
$4 = t/18 - t/54 = (3t - t) / 54 = 2t / 54 = t / 27$.
$t = 4 \times 27 = 108 \, min$.

(A) Given: $P_{A}^{0} = 21 \ kPa$ and $P_{B}^{0} = 18 \ kPa$.
Number of moles of $A$ $(n_{A})$ = $1 \ mol$,Number of moles of $B$ $(n_{B})$ = $2 \ mol$.
Total moles = $1 + 2 = 3 \ mol$.
Mole fraction of $A$ $(X_{A})$ = $\frac{1}{3}$ and mole fraction of $B$ $(X_{B})$ = $\frac{2}{3}$.
According to Raoult's law for an ideal solution: $P_{T} = X_{A} P_{A}^{0} + X_{B} P_{B}^{0}$.
$P_{T} = (\frac{1}{3} \times 21) + (\frac{2}{3} \times 18)$.
$P_{T} = 7 + 12 = 19 \ kPa$.

(A) The energy of the absorbed photon corresponds to the octahedral splitting energy $\Delta_{0}$.
Using the formula $E = \frac{hc}{\lambda}$:
Given $\lambda = 498 \, nm = 498 \times 10^{-9} \, m$,$h = 6.626 \times 10^{-34} \, Js$,and $c = 3 \times 10^{8} \, ms^{-1}$.
$\Delta_{0} = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{498 \times 10^{-9}} \, J$
$\Delta_{0} = \frac{19.878 \times 10^{-26}}{498 \times 10^{-9}} \, J$
$\Delta_{0} \approx 0.039915 \times 10^{-17} \, J = 3.9915 \times 10^{-19} \, J$.
Rounding off to the nearest integer,we get $4 \times 10^{-19} \, J$.

(B) Fructose is a monosaccharide with the molecular formula $C_6H_{12}O_6$.
It contains a ketone group $(>C=O)$ at the $C-2$ position and six carbon atoms in its chain.
Therefore,it is classified as a ketohexose.

(A) Cation exchange resins contain acidic functional groups such as $-SO_3H$ or $-COOH$ which can release $H^+$ ions.
Anion exchange resins contain basic functional groups such as $-NH_2$ or $-NH_3^+OH^-$ which can release $OH^-$ ions.
Therefore,the correct pair for cation and anion exchange resins is $-SO_3H$ and $-NH_2$ respectively.

(D) The correct matches are as follows:
$(a).$ Haematite is $Fe_{2}O_{3}$ $(ii)$.
$(b).$ Bauxite is $Al_{2}O_{3} \cdot xH_{2}O$ $(i)$.
$(c).$ Magnetite is $Fe_{3}O_{4}$ $(iv)$.
$(d).$ Malachite is $CuCO_{3} \cdot Cu(OH)_{2}$ $(iii)$.
Therefore,the correct sequence is $(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)$.

(C) An ambident nucleophile is a species that has two or more nucleophilic sites through which it can attack.
$(A)$ $KCN$ and $AgCN$: $CN^-$ is an ambident nucleophile as it can attack through $C$ or $N$.
$(C)$ $AgNO_{2}$ and $KNO_{2}$: $NO_2^-$ is an ambident nucleophile as it can attack through $N$ or $O$.
$(B)$ $RCOOAg / RCOOK$ and $(D)$ $AgI / KI$ do not represent pairs of ambident nucleophiles in the context of typical organic reactions.
Therefore,the correct pairs are $(A)$ and $(C)$.

(A) $N_2O$ (nitrous oxide) and $NO$ (nitric oxide) are neutral oxides of nitrogen.
$N_2O_3$,$NO_2$,and $N_2O_5$ are acidic oxides of nitrogen.

(A) The correct matching is as follows:
$1$. $[Co(NH_3)_6][Cr(CN)_6]$ shows Coordination isomerism because both the cation and anion are complex ions,and ligands can be exchanged between them.
$2$. $[Co(NH_3)_3(NO_2)_3]$ shows Linkage isomerism due to the presence of the ambidentate ligand $NO_2^-$.
$3$. $[Cr(H_2O)_6]Cl_3$ shows Solvate (or Hydrate) isomerism as water molecules can act as ligands or be present as lattice water.
$4$. $cis-[CrCl_2(ox)_2]^{3-}$ shows Optical isomerism because the $cis$ isomer lacks a plane of symmetry.
Thus,the correct sequence is $(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)$.

(A) Primary,secondary,and tertiary amines can be distinguished using Hinsberg's reagent,which is $Para-Toluene$ sulphonyl chloride $(CH_3C_6H_4SO_2Cl)$.
$1.$ Primary amines react with $Para-Toluene$ sulphonyl chloride to form an $N-alkyl$ sulphonamide,which contains an acidic hydrogen atom attached to the nitrogen,making it soluble in $NaOH$.
$2.$ Secondary amines react to form an $N,N-dialkyl$ sulphonamide,which lacks an acidic hydrogen atom,rendering it insoluble in $NaOH$.
$3.$ Tertiary amines do not react with $Para-Toluene$ sulphonyl chloride because they lack a hydrogen atom on the nitrogen.

(A) The element with atomic number $24$ is Chromium $(Cr)$.
The electronic configuration of $Cr$ is $[Ar] \, 4s^{1} \, 3d^{5}$.
Chromium exhibits a range of common oxidation states from $+2$ to $+6$ in its various compounds.

(B) The correct matches are:
$(a)$. Sucralose is an artificial sweetener. $(a)-(ii)$
$(b)$. Glyceryl ester of stearic acid is a soap. $(b)-(i)$
$(c)$. Sodium benzoate is a food preservative. $(c)-(iv)$
$(d)$. Bithionol is an antiseptic. $(d)-(iii)$
Therefore,the correct sequence is $(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)$.

(C) The hydrolysis of sucrose into glucose and fructose is catalyzed by the enzyme $Invertase$.
The fermentation of glucose into ethanol and carbon dioxide is catalyzed by the enzyme $Zymase$.
Therefore,the reactions are:
$C_{12}H_{22}O_{11} + H_2O \stackrel{\text{Invertase}}{\longrightarrow} C_6H_{12}O_6 + C_6H_{12}O_6$
$C_6H_{12}O_6 \stackrel{\text{Zymase}}{\longrightarrow} 2C_2H_5OH + 2CO_2$
Thus,enzyme $A$ is $Invertase$ and enzyme $B$ is $Zymase$.

(A) The reaction of $p-\text{methoxyphenyl diazonium chloride}$ with ethanol $(C_2H_5OH)$ is a reduction reaction where the diazonium group is replaced by a hydrogen atom to form anisole.
The reaction is as follows:
$CH_3O-C_6H_4-N_2^+Cl^- + CH_3CH_2OH \rightarrow CH_3O-C_6H_5 + N_2 + CH_3CHO + HCl$
Here,$(A)$ is $p-\text{methoxyphenyl diazonium chloride}$,$X$ is acetaldehyde $(CH_3CHO)$,and $Y$ is hydrogen chloride $(HCl)$.

(B) According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to the magnitude of its charge.
For a negative sol,the coagulating species must be a cation.
Comparing the given cations: $Na^{+}$ (charge $+1$) and $Ba^{2+}$ (charge $+2$).
Since $Ba^{2+}$ has a higher charge than $Na^{+}$,it will have a higher flocculating power.
Therefore,$Ba^{2+}$ is the correct species.

(D) The conductivity $\kappa$ is related to resistance $R$ and cell constant $G^{*}$ by the formula: $\kappa = \frac{1}{R} \cdot G^{*}$
For the same conductivity cell,$G^{*}$ is constant,therefore $\kappa \cdot R = G^{*} = \text{constant}$.
For the $KCl$ solution: $\kappa_{KCl} \cdot R_{KCl} = 0.14 \times 4.19 = 0.5866 \, S$.
For the $HCl$ solution: $\kappa_{HCl} \cdot R_{HCl} = 0.5866 \, S$.
$\kappa_{HCl} = \frac{0.5866}{1.03} \approx 0.5695 \, S m^{-1}$.
Converting to the required units: $0.5695 \, S m^{-1} = 56.95 \times 10^{-2} \, S m^{-1}$.
Rounding to the nearest integer,we get $57 \times 10^{-2} \, S m^{-1}$.

(D) The reaction of $FeCl_3$ with oxalic acid in the presence of $KOH$ leads to the formation of potassium trioxalatoferrate$(III)$:
$FeCl_3 + 3H_2C_2O_4 + 6KOH \rightarrow K_3[Fe(C_2O_4)_3] + 3KCl + 6H_2O$
The product $A$ is $K_3[Fe(C_2O_4)_3]$.
In this coordination complex,the oxalate ion $(C_2O_4^{2-})$ is a bidentate ligand.
Since there are $3$ oxalate ligands,the coordination number (secondary valency) of $Fe$ is $3 \times 2 = 6$.

(A) The given reaction is $2A + B_{2} \rightarrow 2AB$.
Since it is an elementary reaction,the rate law is given by $r_{1} = k[A]^{2}[B_{2}]$.
When the volume of the reaction vessel is reduced by a factor of $3$,the concentration of each reactant increases by a factor of $3$ (since $C = n/V$).
Let the new concentrations be $[A]' = 3[A]$ and $[B_{2}]' = 3[B_{2}]$.
The new rate $r_{2}$ is $r_{2} = k(3[A])^{2}(3[B_{2}])$.
$r_{2} = k \cdot 9[A]^{2} \cdot 3[B_{2}] = 27 \cdot k[A]^{2}[B_{2}] = 27 \cdot r_{1}$.
Thus,the rate of reaction increases by a factor of $27$.

(B) The dissociation of $K_{4}[Fe(CN)_{6}]$ is given by: $K_{4}[Fe(CN)_{6}] \rightleftharpoons 4K^{+} + [Fe(CN)_{6}]^{4-}$.
Initial concentration: $1 \, m$,$0$,$0$.
Final concentration: $(1 - 0.4) \, m$,$4 \times 0.4 \, m$,$0.4 \, m$.
Total effective molality $= 0.6 + 1.6 + 0.4 = 2.6 \, m$.
Since the boiling point is the same,the molality of the non-electrolytic solution must also be $2.6 \, m$.
For $18.1 \%$ by weight solution,$18.1 \, g$ of solute $A$ is present in $81.9 \, g$ of water.
Using the formula for molality: $m = \frac{w_A \times 1000}{M_A \times w_{solvent}}$.
$2.6 = \frac{18.1 \times 1000}{M_A \times 81.9}$.
$M_A = \frac{18100}{2.6 \times 81.9} \approx 85 \, u$.

(D) Molar mass of $KBr = 39.1 + 79.9 = 119 \ g/mol$.
$1 \ mole$ of $KBr$ is doped with $\frac{10^{-5}}{100} = 10^{-7} \ moles$ of $SrBr_2$.
Since $1 \ Sr^{2+}$ ion replaces $2 \ K^+$ ions to maintain electrical neutrality,it creates $1$ cationic vacancy.
Therefore,$1 \ mole$ of $KBr$ contains $10^{-7} \ moles$ of cationic vacancies.
Number of cationic vacancies in $1 \ g$ of $KBr = \frac{10^{-7} \times N_A}{\text{Molar mass of } KBr} = \frac{10^{-7} \times 6.023 \times 10^{23}}{119}$.
$= \frac{6.023 \times 10^{16}}{119} \approx 0.0506 \times 10^{16} = 5.06 \times 10^{14}$.
Rounding to the nearest integer,we get $5 \times 10^{14}$.

(A) The reaction described is the Hofmann bromamide degradation reaction.
In this reaction,an amide reacts with bromine in the presence of an aqueous or ethanolic solution of sodium hydroxide $(NaOH)$.
The overall reaction is: $R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$.
As shown in the reaction mechanism,the carbonyl carbon of the amide is released as the carbonate ion $(CO_3^{2-})$.

(B) $Mn_{3}O_{4}$ is a mixed oxide of $MnO$ and $Mn_{2}O_{3}$.
It exhibits ferrimagnetism due to the presence of unpaired electrons in the $d$-orbitals of manganese ions,which makes it magnetic.

(B) The reaction of an alkyl aryl ether with $HI$ involves the protonation of the ether oxygen atom followed by the nucleophilic attack of the iodide ion $(I^-)$ on the less sterically hindered alkyl group.
In $1-$methoxy naphthalene,the oxygen is attached to a methyl group and a naphthyl group.
The $I^-$ ion attacks the methyl group $(CH_3)$ via an $S_N2$ mechanism because the naphthyl group is bulky and the $C-O$ bond involving the aromatic ring has partial double bond character,making it resistant to nucleophilic substitution.
Thus,the products formed are $CH_3I$ (methyl iodide) and $1-$naphthol.

(A) Vitamin $K$ is essential for the synthesis of clotting factors in the liver.
Deficiency of vitamin $K$ leads to a reduction in these clotting factors,which results in an increase in blood clotting time.

(D) The reaction of an organic compound $A$ with benzene sulphonyl chloride (Hinsberg's reagent) to form a product $B$ that is soluble in $NaOH$ indicates that $A$ is a primary amine $(R-NH_2)$.
Primary amines react with benzene sulphonyl chloride to form $N$-alkylbenzene sulphonamide,which contains an acidic hydrogen atom attached to the nitrogen,making it soluble in $NaOH$.
Secondary amines form $N,N$-dialkylbenzene sulphonamides,which lack an acidic hydrogen and are insoluble in $NaOH$.
Tertiary amines do not react with benzene sulphonyl chloride.
Among the given options:
$A$: $C_6H_5-N(CH_3)_2$ is a tertiary amine.
$B$: $C_6H_5-NH-CH_2-CH_3$ is a secondary amine.
$C$: $C_6H_5-CH_2-NH-CH_3$ is a secondary amine.
$D$: $C_6H_5-CH(CH_3)-NH_2$ is a primary amine.
Therefore,the correct compound is $C_6H_5-CH(CH_3)-NH_2$.

(B) In the structure of $CuSO_{4} \cdot 5H_{2}O,$ the copper ion $(Cu^{2+})$ is coordinated to $4$ water molecules directly,which defines its secondary valency as $4$.
The fifth water molecule is not directly coordinated to the $Cu^{2+}$ ion but is held in the crystal lattice by hydrogen bonding between the water molecule and the sulfate $(SO_{4}^{2-})$ group.
Therefore,the secondary valency is $4$ and the number of hydrogen-bonded water molecules is $1$.

(D) In the nitration of aniline using concentrated $HNO_3$ and $H_2SO_4$ at $288 \ K$,the amino group $(-NH_2)$ is protonated to form the anilinium ion $(-NH_3^+)$ in the acidic medium.
The $-NH_3^+$ group is meta-directing,which leads to a significant amount of meta-nitroaniline $(B)$.
However,some unprotonated aniline also reacts,leading to ortho $(A)$ and para $(C)$ products.
The experimental percentage yields are approximately:
$p$-nitroaniline $(C)$ $\approx 51\%$
$m$-nitroaniline $(B)$ $\approx 47\%$
$o$-nitroaniline $(A)$ $\approx 2\%$
Thus,the order of percentage yield is $C > B > A$.

(D) $CdS$ sol is a negatively charged sol.
$TiO_2$ sol is a positively charged sol.
Therefore,the charges are negative and positive,respectively.

(C) . Antifertility drug $\rightarrow$ $(iii)$. Norethindrone
$(b)$. Antibiotic $\rightarrow$ $(iv)$. Salvarsan
$(c)$. Tranquilizer $\rightarrow$ $(i)$. Meprobamate
$(d)$. Artificial sweetener $\rightarrow$ $(ii)$. Alitame
Therefore,the correct matching is $(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)$.

(D) The reaction is an Aldol condensation of cyclopentanone.
$1$. In the presence of $dil. NaOH$,two molecules of cyclopentanone undergo self-aldol condensation to form a $\beta$-hydroxy ketone,which is $X$ ($2$-($1$-hydroxycyclopentyl)cyclopentanone).
$2$. Upon heating with $H^+$,the $\beta$-hydroxy ketone undergoes dehydration to form an $\alpha,\beta$-unsaturated ketone,which is $Y$ ($2$-cyclopentylidenecyclopentanone).

(A) Statement $I$ is true. $C_2H_5OH$ can act as a nucleophile due to the lone pairs on the oxygen atom. $AgCN$ is a covalent compound where the nitrogen atom has a lone pair,making it an ambident nucleophile.
Statement $II$ is false. $KCN$ is an ionic compound that provides $CN^-$ ions,which act as carbon-centered nucleophiles (forming nitriles). However,$AgCN$ is covalent and primarily acts as a nitrogen-centered nucleophile (forming isonitriles) because the carbon atom is involved in a coordinate bond with silver.
Therefore,Statement $I$ is true and Statement $II$ is false.

(C) Mercury $\rightarrow$ Distillation refining $(ii)$
$(b)$ Copper $\rightarrow$ Electrolytic refining $(iii)$
$(c)$ Silicon $\rightarrow$ Zone refining $(iv)$
$(d)$ Nickel $\rightarrow$ Vapour phase refining $(i)$
Therefore,the correct matching is $a-ii, b-iii, c-iv, d-i$.

(D) Covalent or network solids possess a giant three-dimensional structure held together by strong covalent bonds,which results in a very high melting point.
Since all electrons are involved in bonding and are not free to move,these substances act as insulators in both their solid and molten states.

(D) For a first-order reaction,the time required for $n \%$ completion is given by $t = \frac{1}{k} \ln \frac{[A]_0}{[A]_t}$.
Given $t_{1/2} = 1 \, \text{min}$,we know $k = \frac{\ln 2}{t_{1/2}} = \frac{0.69}{1} = 0.69 \, \text{min}^{-1}$.
For $99.9 \, \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$.
$t_{99.9} = \frac{1}{k} \ln \frac{[A]_0}{0.001[A]_0} = \frac{1}{k} \ln 1000 = \frac{1}{k} \ln 10^3 = \frac{3 \ln 10}{k}$.
Substituting the values: $t_{99.9} = \frac{3 \times 2.3}{0.69} = \frac{6.9}{0.69} = 10 \, \text{min}$.

(D) According to Kohlrausch's law of independent migration of ions:
$\Lambda_{m}^{\infty}(BaSO_{4}) = \lambda_{m}^{\infty}(Ba^{2+}) + \lambda_{m}^{\infty}(SO_{4}^{2-})$
We can express this in terms of the given electrolytes:
$\Lambda_{m}^{\infty}(BaSO_{4}) = \Lambda_{m}^{\infty}(BaCl_{2}) + \Lambda_{m}^{\infty}(H_{2}SO_{4}) - 2\Lambda_{m}^{\infty}(HCl)$
Substituting the given values:
$\Lambda_{m}^{\infty}(BaSO_{4}) = 280 + 860 - 2(426)$
$= 1140 - 852$
$= 288 \, S \, cm^{2} \, mol^{-1}$

(C) The partial hydrolysis of $XeF_{6}$ is given by the reaction:
$XeF_{6} + 2H_{2}O \longrightarrow XeO_{2}F_{2} + 4HF$
Thus,compound $A$ is $XeF_{6}$.
The structure of $XeF_{6}$ is distorted octahedral with $1$ lone pair on the $Xe$ atom.
Each fluorine $(F)$ atom has $3$ lone pairs of electrons.
Total number of lone pairs in $XeF_{6} = 1 \text{ (on } Xe) + (6 \times 3 \text{ on } F) = 1 + 18 = 19$.

(B) The oxidation of an aldehyde to a carboxylate ion by Tollen's reagent is represented by the following balanced chemical equation:
$RCHO + 2[Ag(NH_{3})_{2}]^{+} + 3OH^{-} \rightarrow RCOO^{-} + 2Ag + 4NH_{3} + 2H_{2}O$
In this reaction,the aldehyde group $(-CHO)$ is oxidized to a carboxylate group $(-COO^{-})$,which involves the loss of $2$ electrons.
These $2$ electrons are transferred to the two silver ions $(Ag^{+})$ present in the Tollen's reagent,reducing them to metallic silver $(Ag^{0})$.
Therefore,the total number of electrons transferred to the Tollen's reagent per aldehyde group is $2$.

(C) The elevation in boiling point is given by $\Delta T_{b} = T_{b} - T_{b}^{0} = 100.52^{\circ} C - 100^{\circ} C = 0.52^{\circ} C$.
For dimerization,the van't Hoff factor is $i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.
Using the formula $\Delta T_{b} = i \times K_{b} \times m$,where $m = 2 \, M$:
$0.52 = (1 - \frac{\alpha}{2}) \times 0.52 \times 2$.
Dividing both sides by $0.52$,we get $1 = (1 - \frac{\alpha}{2}) \times 2$.
$1 = 2 - \alpha$.
$\alpha = 1$.
Thus,the percentage association is $1 \times 100 = 100 \, \%$.

(C) The most consistent and precise volume of $HCl$ obtained from the readings is $5.0 \, mL$.
At the equivalence point,the number of milliequivalents $(meq)$ of $Na_{2}CO_{3}$ equals the number of milliequivalents of $HCl$.
Let the molarity of $Na_{2}CO_{3}$ solution be $M$.
The reaction is: $Na_{2}CO_{3} + 2HCl \rightarrow 2NaCl + H_{2}O + CO_{2}$.
The n-factor for $Na_{2}CO_{3}$ is $2$ and for $HCl$ is $1$.
Using the formula: $M_{1} \times V_{1} \times n_{1} = M_{2} \times V_{2} \times n_{2}$
$M \times 10 \, mL \times 2 = 0.2 \, M \times 5.0 \, mL \times 1$
$20 \times M = 1.0$
$M = 0.05 \, mol/L$
To convert to $mM$ (millimolar),we multiply by $1000$:
$0.05 \times 1000 = 50 \, mM$.