JEE Main 2023 Chemistry Question Paper with Answer and Solution

(A) The combustion of methane $(CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O)$ is an exothermic process.
In System $A$,the temperature rises because the system is adiabatic,meaning no heat can escape to the surroundings.
In System $B$,the temperature remains constant because the system is diathermic,allowing heat to be exchanged with the surroundings to maintain thermal equilibrium.
Therefore,System $A$ is an adiabatic system and System $B$ is a diathermic system.

(C) Acidic strength is directly proportional to the $-I$ effect and inversely proportional to the $+I$ effect.
$F$ and $Cl$ exert $-I$ effect,while the methyl group exerts $+I$ effect. Thus,$C$ is the least acidic.
Inductive effect is distance-dependent. In compound $A$,the $-I$ group $(Cl)$ is at the $\alpha$-position relative to the $-OH$ group,whereas in compound $B$,the $-I$ group $(F)$ is at the $\delta$-position.
Due to the closer proximity of the electron-withdrawing group in $A$,it exerts a stronger $-I$ effect on the acidic proton than the more distant group in $B$,despite $F$ being inherently more electronegative than $Cl$.
Therefore,the order of acidity is $A > B > C$. Assertion $A$ is true.
Reason $R$ is also true because $F$ is indeed a stronger electron-withdrawing group than $Cl$ due to higher electronegativity.
However,the reason for $A > B$ is the distance dependence of the inductive effect,not the inherent strength of the electron-withdrawing group. Thus,$R$ is not the correct explanation for $A$.

(B) The standard enthalpy of formation $(\Delta_{f}H^0)$ for ionic compounds like sodium halides is primarily determined by the lattice energy and the hydration/sublimation energies.
As the size of the halide ion increases from $F^-$ to $I^-$,the lattice energy decreases (becomes less negative).
Consequently,the overall enthalpy of formation becomes less negative (i.e.,increases algebraically) in the order $NaF < NaCl < NaBr < NaI$.
Therefore,the correct order of $\Delta_{f}H^0$ values is $NaF < NaCl < NaBr < NaI$.

(A) In the $[BF_4]^-$ ion,Boron forms $4$ covalent bonds with $4$ fluorine atoms. Therefore,the covalency of Boron is $4$.
To find the oxidation state of Boron $(x)$:
$x + 4 \times (-1) = -1$
$x - 4 = -1$
$x = +3$
Thus,the covalency is $4$ and the oxidation state is $+3$.

(D) In the Carius method,the mass of $Br$ is calculated from the mass of $AgBr$ formed.
Mole of $AgBr = \frac{0.376 \ g}{188 \ g \ mol^{-1}} = 0.002 \ mol$.
Since $1 \ mol$ of $AgBr$ contains $1 \ mol$ of $Br$,the mole of $Br = 0.002 \ mol$.
Mass of $Br = 0.002 \ mol \times 80 \ g \ mol^{-1} = 0.16 \ g$.
Percentage of $Br = \frac{\text{Mass of } Br}{\text{Mass of compound}} \times 100 = \frac{0.16 \ g}{0.400 \ g} \times 100 = 40 \%$.

(D) The balanced chemical equation for the reaction is:
$M_2CO_3 + 2HCl \rightarrow 2MCl + H_2O + CO_2$
From the stoichiometry of the reaction,$1 \ mol$ of $M_2CO_3$ produces $1 \ mol$ of $CO_2$.
Given that $0.01 \ mol$ of $CO_2$ is produced,the moles of $M_2CO_3$ reacted must be $0.01 \ mol$.
We know that $\text{moles} = \frac{\text{mass}}{\text{molar mass}}$.
Therefore,$0.01 \ mol = \frac{1 \ g}{\text{molar mass of } M_2CO_3}$.
$\text{Molar mass of } M_2CO_3 = \frac{1 \ g}{0.01 \ mol} = 100 \ g \ mol^{-1}$.

(D) The balanced chemical equation is:
$Cr_2O_7^{2-} + 14H^{+} + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$
Comparing this with the given equation:
$X = 14$
$Y = 2$
$Z = 7$
Therefore,the sum $(X + Y + Z) = 14 + 2 + 7 = 23$.

(D) The chemical formula of Borax is $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$.
Comparing this with the given formula $Na_2B_4O_x(OH)_y \cdot zH_2O$,we get $x = 5$,$y = 4$,and $z = 8$.
Therefore,$x + y + z = 5 + 4 + 8 = 17$.

(D) The reaction is: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$.
Initial moles of $CH_3COOH = 50 \ mL \times 0.1 \ M = 5 \ mmol$.
Initial moles of $NaOH = 20 \ mL \times 0.1 \ M = 2 \ mmol$.
After reaction,moles of $CH_3COONa$ (salt) $= 2 \ mmol$ and remaining moles of $CH_3COOH$ (acid) $= 5 - 2 = 3 \ mmol$.
Using the Henderson-Hasselbalch equation: $pH = pKa + \log_{10} (\frac{[salt]}{[acid]})$
$pH = 4.76 + \log_{10} (\frac{2}{3})$
$pH = 4.76 + (\log 2 - \log 3) = 4.76 + (0.30 - 0.48) = 4.76 - 0.18 = 4.58$.
Thus,$pH = 4.58 = 458 \times 10^{-2}$.

(D) The orbital angular momentum is calculated using the formula $L = \sqrt{l(l+1)} \frac{h}{2\pi}$.
For any $s$ orbital,including $3s$,the azimuthal quantum number $l$ is $0$.
Substituting $l = 0$ into the formula: $L = \sqrt{0(0+1)} \frac{h}{2\pi} = 0$.
Comparing this with $\frac{xh}{2\pi}$,we get $x = 0$.

(D) Photochemical smog is formed by the action of sunlight on nitrogen oxides and hydrocarbons. It requires a warm,dry,and sunny climate to occur. Srinagar in January experiences cold,winter conditions,which are unfavorable for the formation of photochemical smog. Therefore,the possibility of its formation is minimum at Srinagar in January.

(D) . $NF_3$ has a trigonal pyramidal shape due to the presence of one lone pair on the $N$ atom.
$B$. The bond order of $N_2$ is $3$ and $O_2$ is $2$. Higher bond order implies shorter bond length,so $N_2$ has a shorter bond length than $O_2$. Statement $B$ is correct.
$C$. Isoelectronic species (e.g.,$N_2$ and $CO$) have identical bond orders. Statement $C$ is correct.
$D$. The dipole moment of $H_2O$ $(1.85 \ D)$ is higher than that of $H_2S$ $(0.95 \ D)$ due to the higher electronegativity of oxygen compared to sulfur. Statement $D$ is incorrect.
Therefore,statements $B$ and $C$ are correct.

(A) Statement $I$ is correct because according to Bohr's postulate,the angular momentum of an electron in a stationary orbit is an integral multiple of $\frac{h}{2\pi}$,i.e.,$mvr = \frac{nh}{2\pi}$.
Statement $II$ is correct because Bohr's model assumes electrons move in well-defined circular orbits with fixed velocity and position,which directly contradicts the Heisenberg uncertainty principle,which states that it is impossible to determine simultaneously the exact position and momentum of a subatomic particle.

(B) The reactivity of benzene derivatives towards electrophilic substitution depends on the electron density of the benzene ring. Substituents that increase electron density (electron-donating groups) activate the ring,while those that decrease electron density (electron-withdrawing groups) deactivate the ring.
$1$. $-NMe_2$ (in $e$) has a strong $+M$ effect,making it the most reactive.
$2$. $-OCH_3$ (in $d$) also has a $+M$ effect,but it is weaker than $-NMe_2$ due to the higher electronegativity of oxygen.
$3$. $-CH_3$ (in $a$) shows a $+H$ (hyperconjugation) effect,which is weaker than the $+M$ effect.
$4$. Benzene $(b)$ has no substituent.
$5$. $-CF_3$ (in $c$) is a strong electron-withdrawing group due to the $-I$ effect,making it the least reactive.
Therefore,the decreasing order of reactivity is: $e > d > a > b > c$.

(A) $ (A) $ The $pH$ of $1 \times 10^{-8} \ M \ HCl$ is not $8$ because the contribution of $H^+$ ions from water cannot be neglected. The $pH$ is approximately $6.98$.
$ (B) $ The conjugate base of $H_2PO_4^{-}$ is obtained by removing one $H^+$ ion,which gives $HPO_4^{2-}$. This is correct.
$ (C) $ The auto-ionization of water is an endothermic process,so $K_w$ increases with an increase in temperature. This is correct.
$ (D) $ According to the Henderson-Hasselbalch equation,$pH = pK_a + \log \frac{[Salt]}{[Acid]}$. At the half-neutralization point,$[Salt] = [Acid]$,so $pH = pK_a$. This is correct.

(B) The electrons in $Be$ and $Mg$ are strongly bound to the nucleus due to their small size and high ionization enthalpy.
Consequently,the energy required to excite these electrons to higher energy levels is very high,which cannot be provided by the heat of a Bunsen flame.
Therefore,$Be$ and $Mg$ do not impart any characteristic color to the flame.
Assertion $(A)$ is false because $BeCl_2$ and $MgCl_2$ do not produce a characteristic flame.
Reason $(R)$ is true because the excitation energy for $Be$ and $Mg$ is indeed very high.

(A) The water-gas shift reaction is represented by the equation: $CO(g) + H_2O(g) \xrightarrow{\text{Iron chromate}} CO_2(g) + H_2(g)$.
In this reaction,the oxidation state of carbon in $CO$ is $+2$,and in $CO_2$ it is $+4$.
Since the oxidation state of carbon increases from $+2$ to $+4$,$CO$ is oxidized to $CO_2$.

(A) For good quality cement,the ratio of silica $(SiO_2)$ to alumina $(Al_2O_3)$ should be between $2.5$ and $4.0$.

(B) In paper chromatography,a special quality paper known as chromatography paper is used.
This paper contains water trapped in its pores,which acts as the stationary phase.

(A) In an acidic medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $Mn^{2+}$.
The reaction is: $2KMnO_4 + 10KI + 8H_2SO_4 \rightarrow 6K_2SO_4 + 2MnSO_4 + 5I_2 + 8H_2O$.
The oxidation state of $Mn$ in $KMnO_4$ is $+7$.
The oxidation state of $Mn$ in $MnSO_4$ is $+2$.
The change in oxidation state is $|(+2) - (+7)| = 5$.

(A) The chemical formula for Chromyl chloride is $CrO_2Cl_2$.
Let the oxidation state of $Cr$ be $x$.
$x + 2(-2) + 2(-1) = 0$
$x - 4 - 2 = 0$
$x = +6$
Therefore,the oxidation state of chromium in $CrO_2Cl_2$ is $+6$.

(A) Isoelectronic species are those that have the same number of electrons.
$O^{2-} = 8 + 2 = 10 \text{ electrons}$
$F^{-} = 9 + 1 = 10 \text{ electrons}$
$Mg^{2+} = 12 - 2 = 10 \text{ electrons}$
$Na^{+} = 11 - 1 = 10 \text{ electrons}$
$Al^{3+} = 13 - 3 = 10 \text{ electrons}$
Other species:
$Al = 13 \text{ electrons}$
$O^{+} = 7 \text{ electrons}$
$Mg = 12 \text{ electrons}$
$F = 9 \text{ electrons}$
The isoelectronic species are $O^{2-}, F^{-}, Mg^{2+}, Na^{+}, \text{ and } Al^{3+}$.
Total count = $5$.

(A) The relationship between enthalpy change of fusion $(\Delta H_{fus})$,entropy change of fusion $(\Delta S_{fus})$,and melting point $(T_{mp})$ is given by the formula: $\Delta S_{fus} = \frac{\Delta H_{fus}}{T_{mp}}$.
Given: $\Delta H_{fus} = 30.4 \, kJ \, mol^{-1} = 30400 \, J \, mol^{-1}$ and $\Delta S_{fus} = 28.4 \, J \, K^{-1} \, mol^{-1}$.
Substituting the values into the equation: $28.4 = \frac{30400}{T_{mp}}$.
Solving for $T_{mp}$: $T_{mp} = \frac{30400}{28.4} \approx 1070.42 \, K$.
Rounding to the nearest integer,the melting point is $1070 \, K$.

(A) According to Newton's law of universal gravitation,the force between the sun and a planet is given by $F = \frac{G M m}{r^2}$.
From this formula,we can conclude that $F \propto \frac{1}{r^2}$,which confirms statement $A$ is correct.
Statement $B$ is incorrect because the gravitational force is directly proportional to the product of the masses $(F \propto M m)$,not inversely proportional.
Statement $C$ is incorrect because the centripetal force is provided by the gravitational force,which is directed towards the sun,not away from it.
Statement $D$ represents Kepler's third law of planetary motion,which states that $T^2 \propto a^3$,where $T$ is the time period and $a$ is the semi-major axis. Thus,statement $D$ is correct.
Therefore,statements $A$ and $D$ are correct.

(D) The molar conductivity $\wedge_m$ is calculated as: $\wedge_m = \frac{k \times 1000}{C}$
Given $k = 5 \times 10^{-5} \ S \ cm^{-1}$ and $C = 0.0025 \ M$.
$\wedge_m = \frac{5 \times 10^{-5} \times 1000}{0.0025} = \frac{0.05}{0.0025} = 20 \ S \ cm^2 \ mol^{-1}$.
The degree of dissociation $\alpha$ is given by $\alpha = \frac{\wedge_m}{\wedge_m^\circ} = \frac{20}{400} = 0.05$.
The dissociation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{1 - \alpha}$.
Since $\alpha$ is very small,$1 - \alpha \approx 1$,so $K_a \approx C \alpha^2 = 0.0025 \times (0.05)^2 = 0.0025 \times 0.0025 = 6.25 \times 10^{-6} = 62.5 \times 10^{-7}$.
Rounding to the nearest integer,we get $63 \times 10^{-7}$.

(C) . For a zero-order reaction,$t_{1/2} = \frac{[A]_0}{2K}$. As the concentration decreases,the half-life decreases. (Correct statement)
$B$. If the order with respect to a reactant is zero,it will not affect the rate of reaction. (Correct statement)
$C$. Order can be fractional,but molecularity is always a whole number and cannot be fractional. (Incorrect statement)
$D$. For a zero-order reaction,the unit is $mol \ L^{-1} s^{-1}$,and for a second-order reaction,the unit is $mol^{-1} L s^{-1}$. (Correct statement)
Thus,only statement $C$ is incorrect. The number of incorrect statements is $1$.

(B) $1$. $X$ is a tetrose $(C_4H_8O_4)$ with two chiral carbons and a $-CHO$ group (positive Schiff's test).
$2$. Acetylation yields a triacetate,confirming three $-OH$ groups.
$3$. The $L$-isomer of a tetrose with two chiral centers and the given reactivity corresponds to $L$-threose.
$4$. Oxidation of $L$-threose with $HNO_3$ gives $L$-tartaric acid $(A)$.
$5$. Reduction of $L$-tartaric acid with $NaBH_4$ gives a chiral compound $(B)$.
$6$. Based on the Fischer projection for $L$-threose,the correct structure is represented in option $B$.

(D) Ethene undergoes addition polymerisation to form high density polythene in the presence of a catalyst such as triethylaluminium $(Al(C_2H_5)_3)$ and titanium tetrachloride $(TiCl_4)$,which is known as the Ziegler-Natta catalyst.
This process occurs at a temperature of $333 \ K$ to $343 \ K$ and under a pressure of $6-7 \ \text{atm}$.
High density polythene consists of linear molecules and is closely packed,which results in high density and high melting point.

(A) $1$. The formation of a dark brown ring with freshly prepared ferrous sulphate in an acidic medium is the characteristic test for the nitrate ion $(NO_3^{-})$,known as the Brown Ring Test: $2NO_3^{-} + 4H_2SO_4 + 6Fe^{2+} \rightarrow 6Fe^{3+} + 2NO + 4SO_4^{2-} + 4H_2O$ followed by $[Fe(H_2O)_6]^{2+} + NO \rightarrow [Fe(H_2O)_5(NO)]^{2+} + H_2O$ (brown ring).
$2$. The formation of a deep red colour with neutral $FeCl_3$ is the characteristic test for the acetate ion $(CH_3COO^{-})$,which forms ferric acetate: $CH_3COO^{-} + FeCl_3 \rightarrow [Fe_3(OH)_2(CH_3COO)_6]^+$. This red colour disappears on boiling,resulting in a brown-red precipitate of basic ferric acetate: $[Fe_3(OH)_2(CH_3COO)_6]^+ + 4H_2O \rightarrow 3Fe(OH)_2(CH_3COO) \downarrow + 3CH_3COOH + H^+$.
$3$. Therefore,the mixture contains $CH_3COO^{-}$ and $NO_3^{-}$.

(A) The reaction of $2$-substituted cyclohexanone with $KMnO_4$ involves oxidative cleavage of the ring.
$KMnO_4$ acts as a strong oxidizing agent that cleaves the $C-C$ bond adjacent to the carbonyl group,resulting in the formation of a keto-acid,$R-CO-(CH_2)_4-COOH$ (which is $A$).
Subsequently,the Wolff-Kishner reduction,using $NH_2NH_2$ and $KOH$ followed by $H_3O^+$,reduces the carbonyl group $(C=O)$ to a methylene group $(-CH_2-)$.
Therefore,the final product $B$ is $R-CH_2-(CH_2)_4-COOH$,which is $R-(CH_2)_5-COOH$.

(A) An ambidentate ligand is a ligand that can coordinate to a metal atom through more than one donor atom (e.g.,$NO_2^-$ can coordinate through $N$ or $O$,and $NCS^-$ can coordinate through $N$ or $S$).
Analyzing the options:
$A$: $C_2O_4^{2-}$ (oxalate) is a didentate ligand,ethylene diamine $(en)$ is a didentate ligand,and $H_2O$ is a monodentate ligand. None of these are ambidentate.
$B$: $NCS^-$ is an ambidentate ligand.
$C$: $NO_2^-$ is an ambidentate ligand.
$D$: $NO_2^-$ and $NCS^-$ are ambidentate ligands.
Therefore,the set that does not contain any ambidentate ligand is $A$.

(C) In reaction $(I)$,the substrate has an electron-donating group $(-OMe)$ at the para position,which stabilizes the carbocation intermediate. This facilitates the $S_N1$ mechanism,which is a $1^{st}$ order reaction.
In reaction $(II)$,the substrate has an electron-withdrawing group $(-NO_2)$ at the para position,which destabilizes the carbocation intermediate,making the $S_N1$ pathway unfavorable. Instead,it proceeds via the $S_N2$ mechanism,which is a $2^{nd}$ order reaction.
Therefore,reaction $(I)$ is of $1^{st}$ order and reaction $(II)$ is of $2^{nd}$ order.

(A) The reactions given are:
$(i) 2 Cu_2S + 3 O_2 \rightarrow 2 Cu_2O + 2 SO_2$
$(ii) 2 Cu_2O + Cu_2S \rightarrow 6 Cu + SO_2$
These reactions represent the self-reduction process in the extraction of copper.
Due to the evolution of $SO_2$ gas during the solidification of the molten copper,the surface of the metal develops bubbles or blisters.
Therefore,the copper obtained is referred to as blister copper.

(A) Complexes of the type $[MA_3B_3]$ exhibit facial (fac) and meridional (mer) isomerism.
In the given options,$[Co(NH_3)_3(NO_2)_3]$ is of the type $[MA_3B_3]$,where $M = Co$,$A = NH_3$,and $B = NO_2$.
Therefore,it can exist as both facial and meridional isomers.

(B) $Fe_4[Fe(CN)_6]_3$ (Prussian Blue) is insoluble in water.
$K_3[Co(NO_2)_6]$ is very poorly soluble in water.
$(NH_4)_3[As(Mo_3O_{10})_4]$ (Ammonium arsenomolybdate) is insoluble in water.
$[Fe_3(OH)_2(OAc)_6]Cl$ is a basic iron$(III)$ acetate complex which is soluble in water.

(A) When $o$-phenylenediamine reacts with nitrous acid $(HNO_2)$,one of the amino groups is converted into a diazonium salt group $(-N_2^+)$.
Due to the proximity of the other amino group $(-NH_2)$ at the ortho position,an intramolecular nucleophilic attack occurs.
The lone pair on the nitrogen of the $-NH_2$ group attacks the diazonium nitrogen,leading to the formation of a cyclic compound known as benzotriazole.
The reaction proceeds as follows:
$o$-Phenylenediamine $\xrightarrow{HNO_2}$ $o$-amino-benzenediazonium ion $\xrightarrow{-H_2O, -H^+}$ Benzotriazole.

(D) For $[Cr(CN)_6]^{3-}$,the oxidation state of $Cr$ is $+3$. The electronic configuration is $3d^3$. Since $CN^-$ is a strong field ligand,the configuration is $t_{2g}^3 e_g^0$. The number of unpaired electrons $n = 3$.
$\mu_1 = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \ BM$.
For $[Cr(H_2O)_6]^{3+}$,the oxidation state of $Cr$ is $+3$. The electronic configuration is $3d^3$. Since $H_2O$ is a weak field ligand,the configuration is $t_{2g}^3 e_g^0$. The number of unpaired electrons $n = 3$.
$\mu_2 = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \ BM$.
The ratio $\frac{\mu_1}{\mu_2} = \frac{\sqrt{15}}{\sqrt{15}} = 1$.

(D) The density formula for a cubic unit cell is given by $d = \frac{Z \times M}{N_{A} \times a^3}$.
Given values:
$d = 3.0 \ g \ cm^{-3}$
$M = 12 \ g \ mol^{-1}$
$a = 300 \ pm = 300 \times 10^{-10} \ cm = 3 \times 10^{-8} \ cm$
$N_{A} = 6.02 \times 10^{23} \ mol^{-1}$
Rearranging for $Z$:
$Z = \frac{d \times N_{A} \times a^3}{M}$
Substituting the values:
$Z = \frac{3.0 \times 6.02 \times 10^{23} \times (3 \times 10^{-8})^3}{12}$
$Z = \frac{3.0 \times 6.02 \times 10^{23} \times 27 \times 10^{-24}}{12}$
$Z = \frac{3.0 \times 6.02 \times 27 \times 10^{-1}}{12}$
$Z = \frac{487.62 \times 0.1}{12} = \frac{48.762}{12} \approx 4.06$
The nearest integer is $4$.

(D) The reactant is $4$-hydroxybutanal,which contains both an aldehyde group $(-CHO)$ and a hydroxyl group $(-OH)$.
$1$ mole of $MeMgBr$ reacts with the acidic hydroxyl group $(-OH)$ to form an alkoxide,releasing methane gas.
Another $1$ mole of $MeMgBr$ performs a nucleophilic attack on the carbonyl carbon of the aldehyde group.
Thus,for $y = 1$ mole of the reactant,$x = 2$ moles of $MeMgBr$ are required.
The ratio $x / y = 2 / 1 = 2$.

(D) The balanced chemical equation for the reaction is: $Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe$
The enthalpy change of the reaction is calculated as: $\Delta H_{rxn}^{\ominus} = [\Delta H_{f}^{\ominus}(Al_2O_3) + 2\Delta H_{f}^{\ominus}(Fe)] - [\Delta H_{f}^{\ominus}(Fe_2O_3) + 2\Delta H_{f}^{\ominus}(Al)]$
Since $\Delta H_{f}^{\ominus}$ for elements in their standard state is $0$,we have: $\Delta H_{rxn}^{\ominus} = [-1700 + 0] - [-840 + 0] = -860 \ kJ \ mol^{-1}$
The molar mass of the mixture (for $1 \ mol$ of $Fe_2O_3$ and $2 \ mol$ of $Al$) is: $M_{mixture} = (1 \times 160) + (2 \times 27) = 160 + 54 = 214 \ g$
Heat evolved per gram of the mixture is: $\frac{860 \ kJ}{214 \ g} \approx 4.018 \ kJ \ g^{-1}$
Rounding to the nearest integer,we get $4$.

(D) Mass of sugar in the first solution $= 200 \ g \times 0.25 = 50 \ g$.
Mass of sugar in the second solution $= 500 \ g \times 0.40 = 200 \ g$.
Total mass of sugar $= 50 \ g + 200 \ g = 250 \ g$.
Total mass of the resulting solution $= 200 \ g + 500 \ g = 700 \ g$.
Mass percentage of the resulting sugar solution $= (\frac{250 \ g}{700 \ g}) \times 100 \approx 35.71 \%$.
Rounding to the nearest integer,we get $36 \%$.

(D) The balanced chemical equation is: $KClO_3 + 6FeSO_4 + 3H_2SO_4 \rightarrow KCl + 3Fe_2(SO_4)_3 + 3H_2O$.
The rate of reaction $(ROR)$ is given by: $ROR = -\frac{1}{6} \frac{\Delta[FeSO_4]}{\Delta t} = \frac{1}{3} \frac{\Delta[Fe_2(SO_4)_3]}{\Delta t}$.
Therefore,the rate of production of $Fe_2(SO_4)_3$ is: $\frac{\Delta[Fe_2(SO_4)_3]}{\Delta t} = -\frac{3}{6} \frac{\Delta[FeSO_4]}{\Delta t} = -\frac{1}{2} \frac{\Delta[FeSO_4]}{\Delta t}$.
Given $\Delta[FeSO_4] = 8.8 \ M - 10 \ M = -1.2 \ M$ and $\Delta t = 30 \ \text{min} = 1800 \ s$.
Rate of production $= -\frac{1}{2} \times \frac{-1.2 \ M}{1800 \ s} = \frac{0.6}{1800} \ M \ s^{-1} = 0.000333 \ M \ s^{-1}$.
Converting to the required units: $0.000333 \times 10^6 \times 10^{-6} \ mol \ L^{-1} \ s^{-1} = 333 \times 10^{-6} \ mol \ L^{-1} \ s^{-1}$.

(D) For isotonic solutions, the osmotic pressure $(\pi)$ is equal: $\pi_{K_2SO_4} = \pi_{\text{glucose}}$.
Since $\pi = iCRT$, we have $i_{K_2SO_4} \times 0.004 \times RT = 1 \times 0.01 \times RT$.
Solving for the van't Hoff factor $(i)$: $i = \frac{0.01}{0.004} = 2.5$.
For the dissociation of $K_2SO_4$ $(K_2SO_4 \rightarrow 2K^+ + SO_4^{2-})$, the number of ions produced $(n)$ is $3$.
The degree of dissociation $(\alpha)$ is given by the formula: $i = 1 + (n - 1)\alpha$.
Substituting the values: $2.5 = 1 + (3 - 1)\alpha$.
$1.5 = 2\alpha$, which gives $\alpha = 0.75$.
The percentage dissociation is $\alpha \times 100 = 75\%$.

(D) The standard Gibbs free energy change $\Delta G^0$ is related to the standard electrode potential $E^0$ by the equation $\Delta G^0 = -nFE^0$.
For the reaction $Pb^{2+} + 2e^{-} \rightarrow Pb$,$\Delta G^0_1 = -2Fm$.
For the reaction $Pb^{4+} + 4e^{-} \rightarrow Pb$,$\Delta G^0_2 = -4Fn$.
We want the potential for $Pb^{2+} \rightarrow Pb^{4+} + 2e^{-}$,which is the reverse of the second reaction added to the first.
$\Delta G^0_3 = \Delta G^0_1 - \Delta G^0_2 = -2Fm - (-4Fn) = -2Fm + 4Fn$.
Since $\Delta G^0_3 = -2FE^0_{(Pb^{2+}/Pb^{4+})}$,we have $-2FE^0_{(Pb^{2+}/Pb^{4+})} = -2Fm + 4Fn$.
Dividing by $-2F$,we get $E^0_{(Pb^{2+}/Pb^{4+})} = m - 2n$.
Comparing this with $m - xn$,we find $x = 2$.

(C) Assertion $A$ is false. Wolff-Kishner reduction involves the use of a strong base (like $KOH$ or $EtO^-K^+$) at high temperatures. The substrate $3$-chloro-$2$-methylpentan-$3$-one contains a chlorine atom at the $\alpha$-position relative to the carbonyl group. In the presence of a strong base,this compound undergoes dehydrohalogenation (elimination of $HCl$) to form an $\alpha,\beta$-unsaturated ketone rather than the desired reduction product.
Reason $R$ is true. Wolff-Kishner reduction is indeed a standard method used to reduce carbonyl groups $(C=O)$ to methylene groups $(CH_2)$.

(B) The given reaction involves the treatment of a $\beta$-hydroxy aldehyde with Clemmensen reduction conditions $(Zn(Hg)/HCl, \Delta)$.
$1$. Under these acidic conditions,the aldehyde group $(-CHO)$ is reduced to a methyl group $(-CH_3)$.
$2$. Simultaneously,the $\beta$-hydroxy group $(-OH)$ undergoes dehydration due to the acidic medium and heat,leading to the formation of a stable conjugated alkene.
$3$. The reaction proceeds through the reduction of the aldehyde to a hydrocarbon chain,followed by the elimination of water to form the most stable alkene,which is $C_6H_5-CH=C(CH_3)-C_2H_5$.

(A) Freons are chlorofluorocarbons of methane and ethane.
Among the given options,$CCl_2F_2$ (dichlorodifluoromethane) is a well-known Freon,specifically known as Freon-$12$.
Therefore,the correct option is $A$.

(A) The rate law for the reaction is given by $r = k[A]^1[B]^1$.
For the first experiment: $0.10 = k(20)(0.5)$ $\Rightarrow 0.10 = 10k$ $\Rightarrow k = 0.01 \ L \ mol^{-1} \ s^{-1}$.
For the second experiment: $0.40 = k(x)(0.5)$. Substituting $k = 0.01$: $0.40 = 0.01(x)(0.5)$ $\Rightarrow 0.40 = 0.005x$ $\Rightarrow x = 80 \ mol \ L^{-1}$.
For the third experiment: $0.80 = k(40)(y)$. Substituting $k = 0.01$: $0.80 = 0.01(40)(y)$ $\Rightarrow 0.80 = 0.4y$ $\Rightarrow y = 2 \ mol \ L^{-1}$.
Therefore,the values are $x = 80$ and $y = 2$.

(A) The wavelength of absorbed light $(\lambda)$ is inversely proportional to the crystal field splitting energy $(\Delta_o)$,i.e.,$\lambda \propto \frac{1}{\Delta_o}$.
In $[Co(NH_3)_5(H_2O)]^{3+}$,the ligands are $NH_3$ and $H_2O$. In $[CoCl(NH_3)_5]^{2+}$,the ligands are $NH_3$ and $Cl^-$.
According to the spectrochemical series,$H_2O$ is a stronger field ligand than $Cl^-$. Therefore,the crystal field splitting energy $(\Delta_o)$ for $[Co(NH_3)_5(H_2O)]^{3+}$ is greater than that for $[CoCl(NH_3)_5]^{2+}$.
Since $\Delta_o$ is higher for $[Co(NH_3)_5(H_2O)]^{3+}$,it absorbs at a lower wavelength. Thus,Assertion $A$ is false.
The wavelength of absorbed light depends on the nature of the ligand (spectrochemical series),not just the oxidation state of the metal ion. Thus,Reason $R$ is also false.

(B) Assertion $A$: The reaction of glycine $(H_2N-CH_2-COOH)$ with $Cl_2$ in the presence of red phosphorus is the Hell-Volhard-Zelinsky $(HVZ)$ reaction. This reaction replaces an $\alpha$-hydrogen with a chlorine atom,resulting in $H_2N-CH(Cl)-COOH$ ($2$-chloro$-2-$aminoacetic acid). The central carbon atom is bonded to four different groups $(-H, -NH_2, -Cl, -COOH)$,making it a chiral carbon atom. Thus,Assertion $A$ is true.
Reason $R$: $A$ molecule with $2$ chiral carbons is not always optically active. For example,meso compounds contain chiral centers but are optically inactive due to an internal plane of symmetry. Thus,Reason $R$ is false.
Therefore,$A$ is true but $R$ is false.

(A) The reaction proceeds in three steps:
$1$. Reaction of butan$-2-$ol with $NaI$ and $H_3PO_4$ replaces the $-OH$ group with $-I$ to form $2-$iodobutane $(CH_3-CH_2-CH(I)-CH_3)$.
$2$. Reaction of $2-$iodobutane with $Mg$ in dry ether forms the Grignard reagent,sec-butylmagnesium iodide $(CH_3-CH_2-CH(MgI)-CH_3)$.
$3$. The Grignard reagent reacts with $D_2O$ (deuterium oxide) to replace the $-MgI$ group with a deuterium atom,resulting in $2-$deuteriobutane $(CH_3-CH_2-CH(D)-CH_3)$.

(B) Statement $I$ is false. The reaction of ethene in the presence of $AlEt_3$ and $TiCl_4$ (Ziegler-Natta catalyst) produces High Density Polyethene $(HDPE)$,not Low Density Polyethene $(LDP)$.
Statement $II$ is true. Caprolactam,when heated with water at $533-543 \ K$,undergoes ring-opening polymerization (a type of step-growth polymerization) to form Nylon-$6$.

(B) The reaction sequence is as follows:
$1$. The starting material is $2$-isopropyl-$5$-methylphenol (thymol).
$2$. Treatment with $NaNO_2/HCl$ performs electrophilic aromatic substitution (nitrosation) at the para-position relative to the $-OH$ group,forming $A$ (a nitroso compound).
$3$. Treatment of the nitroso compound with $NH_4SH$ (ammonium hydrosulfide) reduces the $-NO$ group to an $-NH_2$ group,yielding $B$ as $4$-amino-$2$-isopropyl-$5$-methylphenol.

(A) The reaction with $AgNO_3$ involves the formation of a carbocation intermediate. Compounds that form stable carbocations react readily to produce a precipitate of $AgBr$.
$A$. $1$-Bromocyclohexene: The bromine atom is attached to an $sp^2$ hybridized carbon atom of a double bond (vinylic halide). The resulting vinylic carbocation is highly unstable due to the high electronegativity of the $sp^2$ carbon and the inability to stabilize the positive charge through resonance. Therefore,it does not react with $AgNO_3$ to form a precipitate.
$B$. Benzyl bromide: Forms a resonance-stabilized benzyl carbocation,which reacts readily.
$C$. $3$-Bromocyclopropene: Forms a cyclopropenyl cation,which is aromatic ($2\pi$ electrons) and highly stable.
$D$. $3$-Phenyl-$3$-bromoprop-$1$-ene: Forms a resonance-stabilized carbocation (allylic and benzylic stabilization).
Thus,$1$-Bromocyclohexene is the correct answer.

(D) Mass of solute $(X) = 2 \ g$
Mass of solvent $(H_2O) = 1 \ mole = 18 \ g$
Total mass of solution = Mass of solute + Mass of solvent = $2 \ g + 18 \ g = 20 \ g$
Mass percent of $X = \frac{\text{Mass of solute}}{\text{Total mass of solution}} \times 100$
Mass percent of $X = \frac{2 \ g}{20 \ g} \times 100 = 10 \ \%$

(A) In metallurgy,sulphide ores are first converted into oxides by roasting because it is thermodynamically easier to reduce metal oxides than metal sulphides.
$2 ZnS + 3 O_2 \rightarrow 2 ZnO + 2 SO_2$
Metal oxides are easier to reduce using carbon because the formation of $CO_2$ is more favorable and $CO_2$ is a stable gas,whereas the reduction of sulphides with carbon is difficult and often leads to the formation of $CS_2$,which is less stable and harder to remove.
Therefore,both Statement-$I$ and Statement-$II$ are correct.

(D) For a dilute solution,the relative lowering of vapour pressure is given by: $\frac{P^0 - P_s}{P^0} = \frac{n}{N}$
Where $P^0 - P_s = 0.20 \ mm \ Hg$,$P^0 = 54.2 \ mm \ Hg$,$n$ is the moles of glucose,and $N$ is the moles of water.
$N = \frac{100 \ g}{18 \ g \ mol^{-1}} = 5.55 \ mol$.
Substituting the values: $\frac{0.20}{54.2} = \frac{w / 180}{100 / 18}$.
$\frac{0.20}{54.2} = \frac{w \times 18}{180 \times 100}$.
$w = \frac{0.20 \times 180 \times 100}{54.2 \times 18} = \frac{3600}{975.6} \approx 3.69 \ g$.

(D) For $[Fe(H_2O)_6]^{3+}$:
$Fe^{3+}$ has $d^5$ configuration. $H_2O$ is a weak field ligand,so no pairing occurs.
Number of unpaired electrons $(n)$ = $5$.
$\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
For $[Fe(CN)_6]^{3-}$:
$Fe^{3+}$ has $d^5$ configuration. $CN^-$ is a strong field ligand,so pairing occurs.
Number of unpaired electrons $(n)$ = $1$.
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \ BM$.
Thus,the magnetic moments are $5.92 \ BM$ and $1.732 \ BM$ respectively.

(A) $Mg(NH_4)PO_4$ is a white precipitate formed during the test for magnesium ions.
$K_3[Co(NO_2)_6]$ (Potassium cobaltinitrite) is a yellow precipitate.
$MnO(OH)_2$ (or $MnO_2 \cdot H_2O$) is a brown precipitate.
$Fe_4[Fe(CN)_6]_3$ (Ferric ferrocyanide) is known as Prussian blue.
Therefore,the correct matching is $A-II, B-III, C-I, D-IV$.

(B) The complex $[NiCl_2Br_2]^{2-}$ involves a $3d$ metal ion with weak field ligands,resulting in a tetrahedral geometry,which is paramagnetic and does not exhibit geometrical isomerism.
The complex $[PtCl_2Br_2]^{2-}$ involves a $5d$ metal ion,which forces pairing of electrons,resulting in a square planar geometry,which is diamagnetic and exhibits geometrical isomerism (cis and trans forms).
Therefore,the properties that change are:
$1.$ Geometry (Tetrahedral to Square Planar)
$2.$ Geometrical isomerism (Absent to Present)
$3.$ Magnetic properties (Paramagnetic to Diamagnetic)
Thus,$A, B,$ and $D$ are expected to change.

(A) Benedict's solution is used to detect reducing sugars. Reducing sugars contain a free aldehyde or ketone group that can reduce $Cu^{2+}$ ions to $Cu^+$ ions,forming a red/orange precipitate of $Cu_2O$.
Among the given compounds:
$1$. Glucose: Reducing sugar (gives test).
$2$. Maltose: Reducing sugar (gives test).
$3$. Sucrose: Non-reducing sugar (does not give test).
$4$. Ribose: Reducing sugar (gives test).
$5$. $2$-deoxyribose: Reducing sugar (gives test).
$6$. Amylose: Non-reducing polysaccharide (does not give test).
$7$. Lactose: Reducing sugar (gives test).
Therefore,only $2$ compounds (Sucrose and Amylose) will not produce an orange-red precipitate with Benedict's solution.

(B) The modern adsorption theory of heterogeneous catalysis involves the following steps:
$1.$ Diffusion of reactants to the surface of the catalyst.
$2.$ Adsorption of reactant molecules on the surface of the catalyst.
$3.$ Occurrence of a chemical reaction on the catalyst's surface through the formation of an intermediate.
$4.$ Desorption of reaction products from the catalyst surface.
$5.$ Diffusion of reaction products away from the catalyst surface.
Evaluating the given statements:
$A.$ Incorrect: Reactants diffuse to the catalyst surface,not the other way around.
$B.$ Correct: Reactants are adsorbed on the catalyst surface.
$C.$ Correct: The reaction occurs via an intermediate on the surface.
$D.$ Correct: It combines the intermediate compound formation theory and the adsorption theory.
$E.$ Correct: It explains the action of catalysts,promoters,and poisons.
Thus,statements $B, C, D,$ and $E$ are correct. The total number of correct statements is $4$.

(A) The reaction of $3$-chloro-$1$-butene with $HCl$ proceeds via the formation of a carbocation intermediate.
First,the $H^+$ ion from $HCl$ adds to the double bond to form a carbocation.
This carbocation can undergo a $1,2$-hydride shift to form a more stable carbocation.
$1$. The initial carbocation reacts with $Cl^-$ to form $2,3$-dichlorobutane. This product has two chiral centers,leading to $3$ stereoisomers ($d, l$ and $meso$ forms).
$2$. The rearranged carbocation reacts with $Cl^-$ to form $2,2$-dichlorobutane,which is an achiral product ($1$ isomer).
Total number of possible isomeric products $= 3 + 1 = 4$.

(A) $A.$ $E_{cell}$ is an intensive property because it does not depend on the amount of matter present in the system.
$B.$ $A$ negative $E^{\Theta}$ value indicates that the species is a better reducing agent than $H_2$ gas.
$C.$ According to Faraday's laws,the amount of electricity required is determined by the number of moles of electrons involved in the balanced half-reaction,which is defined by the stoichiometry.
$D.$ This is the definition of Faraday's first law of electrolysis,which states that the mass of substance deposited or liberated is directly proportional to the quantity of electricity passed.
Therefore,all four statements are correct.

(B) The reaction involves the formation of a Grignard reagent from the alkyl bromide part of the molecule: $CH_3COCH_2CH(CH_3)CH_2Br + Mg \rightarrow CH_3COCH_2CH(CH_3)CH_2MgBr$.
This Grignard reagent acts as a nucleophile and attacks the carbonyl group of another molecule of the same reactant (intermolecular reaction).
The nucleophilic carbon of the Grignard reagent attacks the electrophilic carbonyl carbon of another molecule.
After hydrolysis with $H_2O$,the final product '$A$' is formed,which is a dimer containing both a ketone and an alcohol group.

(D) The extent of adsorption of a gas on a solid adsorbent like charcoal is directly proportional to its critical temperature $(T_c)$.
Higher critical temperature implies that the gas is more easily liquefiable and has stronger van der Waals forces of attraction.
Given critical temperatures are:
$A = 5.3 \ K$
$B = 33.2 \ K$
$C = 126.0 \ K$
$D = 154.3 \ K$
Therefore,the order of adsorption is $D > C > B > A$.

(B) The $4f$ orbitals are deeply buried within the atom,shielded by outer shells,which limits their participation in chemical bonding.
In contrast,the $5f$ orbitals are less buried and extend further from the nucleus compared to $4f$ orbitals.
Because the $5f$ electrons experience less nuclear attraction and are more accessible,they can participate in bonding to a far greater extent than $4f$ electrons.
Therefore,both Assertion $A$ and Reason $R$ are true,and $R$ is the correct explanation of $A$.

(B) The reaction of $N,N$-dimethylaniline with $NaNO_2$ and $HX$ (nitrous acid) at low temperature $(0-5^{\circ}C)$ is an electrophilic aromatic substitution reaction.
$1$. The electrophile generated is the nitrosonium ion,$NO^{+}$.
$2$. Since the $-N(CH_3)_2$ group is a strong ortho/para directing group,the electrophile $NO^{+}$ attacks the para position to form $p$-nitroso-$N,N$-dimethylaniline.
$3$. Option $A$ is correct as $NO^{+}$ is the electrophile.
$4$. Option $B$ is incorrect because the product is $p$-nitroso-$N,N$-dimethylaniline,not an $N$-nitroso ammonium compound (which would form with secondary amines).
$5$. Option $C$ is correct as the reaction is performed at low temperatures to stabilize the nitrous acid and prevent side reactions.
$6$. Option $D$ is correct as the major product is the $p$-nitroso derivative.

(A) The $CFSE$ is calculated using the formula: $CFSE = (-0.4 n_{t_{2g}} + 0.6 n_{e_g}) \Delta_0$.
$A$. $[Cu(NH_3)_6]^{2+}$: $Cu^{2+}$ is $d^9$ $(t_{2g}^6 e_g^3)$. $CFSE = (-0.4 \times 6 + 0.6 \times 3) = -2.4 + 1.8 = -0.6 \Delta_0$. (Matches $I$)
$B$. $[Ti(H_2O)_6]^{3+}$: $Ti^{3+}$ is $d^1$ $(t_{2g}^1 e_g^0)$. $CFSE = (-0.4 \times 1 + 0.6 \times 0) = -0.4 \Delta_0$. (Matches $IV$)
$C$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $d^5$ (strong field,$t_{2g}^5 e_g^0$). $CFSE = (-0.4 \times 5 + 0.6 \times 0) = -2.0 \Delta_0$. (Matches $II$)
$D$. $[NiF_6]^{4-}$: $Ni^{2+}$ is $d^8$ $(t_{2g}^6 e_g^2)$. $CFSE = (-0.4 \times 6 + 0.6 \times 2) = -2.4 + 1.2 = -1.2 \Delta_0$. (Matches $III$)
Therefore,the correct matching is $A-I, B-IV, C-II, D-III$.

(D) . $2$-Chloro-$1,3$-butadiene is the monomer of Neoprene,which is a synthetic rubber $(A-II)$.
$B$. Nylon-$2$-nylon-$6$ is a copolymer of glycine and amino caproic acid,which is a biodegradable polymer $(B-I)$.
$C$. Polyacrylonitrile is formed by the polymerization of acrylonitrile,which is an addition polymer $(C-IV)$.
$D$. Dacron (also known as Terylene) is a polyester formed by the condensation of ethylene glycol and terephthalic acid $(D-III)$.
Therefore,the correct matching is $A-II, B-I, C-IV, D-III$.

(C) Statement $I$ is correct: According to Fajan's rule,the cation with higher charge $(Sb^{+5})$ has greater polarising power than the one with lower charge $(Sb^{+3})$,making $SbCl_5$ more covalent than $SbCl_3$.
Statement $II$ is incorrect: Generally,the lower oxides of halogens are more stable than the higher oxides. Higher oxides of halogens are often unstable and tend to decompose or explode.

(B) The Ellingham diagram plots $\Delta G^{\circ}$ versus $T$. The slope of the line is $-\Delta S$.
At the melting point of the metal $(Mg)$,the metal changes from solid to liquid state $(Mg_{(s)} \rightarrow Mg_{(l)})$.
This phase change involves a significant change in entropy $(\Delta S)$.
Since the entropy of the liquid state is higher than that of the solid state,the value of $\Delta S$ for the reaction $2Mg_{(l)} + O_2(g) \rightarrow 2MgO_{(s)}$ becomes more negative than for $2Mg_{(s)} + O_2(g) \rightarrow 2MgO_{(s)}$.
Consequently,the slope $(-\Delta S)$ increases sharply at the melting point $(\sim 1120^{\circ} C)$.
Thus,both Assertion $A$ and Reason $R$ are true,and $R$ is the correct explanation of $A$.

(A, B, C, D ARE ALL CORRECT (NOTE: THE PROVIDED OPTIONS DO NOT CONTAIN THE CORRECT COMBINATION A, B, C, D; HOWEVER, BASED ON THE QUESTION, ALL STATEMENTS ARE FACTUALLY CORRECT.)) In a lead storage battery,the anode is made of lead $(Pb)$ and the cathode is a grid of lead packed with lead dioxide $(PbO_2)$.
$A$ $\sim 38\%$ solution of sulphuric acid $(H_2SO_4)$ is used as the electrolyte.
During the charging process,the discharge reactions are reversed:
$1$. At the anode: $PbSO_{4(s)} + 2e^- \rightarrow Pb_{(s)} + SO_{4(aq)}^{2-}$
$2$. At the cathode: $PbSO_{4(s)} + 2H_2O_{(l)} \rightarrow PbO_{2(s)} + SO_{4(aq)}^{2-} + 4H_{(aq)}^+ + 2e^-$
Thus,statements $A$,$B$,$C$,and $D$ are all correct.

(C) The reaction is an intramolecular aldol condensation of heptane$-2,6-$dione.
$1$. The base $OEt^-$ abstracts an $\alpha$-hydrogen from one of the terminal methyl groups to form an enolate.
$2$. The enolate carbon attacks the other carbonyl carbon,forming a five-membered ring (intramolecular aldol addition).
$3$. This results in a $\beta$-hydroxy ketone intermediate.
$4$. Upon heating $(\Delta)$,the intermediate undergoes dehydration (elimination of water) to form the stable $\alpha,\beta$-unsaturated ketone,which is $3$-methylcyclopent-$2$-en-$1$-one derivative.