JEE Main 2024 Chemistry Question Paper with Answer and Solution

(B) Given: Molarity $(M) = 0.2 \ M$,Volume $(V) = 500 \ mL = 0.5 \ L$,Density $(d) = 1.25 \ g/mL$,Molar mass of $CuSO_4 = 159.5 \ g/mol$.
Step $1$: Calculate moles of solute $(n)$:
$n = M \times V = 0.2 \times 0.5 = 0.1 \ mol$.
Step $2$: Calculate mass of solute $(x)$:
$x = n \times \text{Molar mass} = 0.1 \times 159.5 = 15.95 \ g$.
Step $3$: Calculate mass of solution:
$Mass_{soln} = V \times d = 500 \times 1.25 = 625 \ g$.
Step $4$: Calculate mass of solvent $(W_{solvent})$:
$W_{solvent} = Mass_{soln} - x = 625 - 15.95 = 609.05 \ g$.
Step $5$: Calculate molality $(m)$:
$m = \frac{n \times 1000}{W_{solvent} (in \ g)} = \frac{0.1 \times 1000}{609.05} \approx 0.16418 \ m$.
$m \approx 164 \times 10^{-3} \ m$.

(C) Ambidentate ligands are those which possess two different donor atoms but can coordinate to the central metal atom through only one of them at a time.
In the given list:
$1$. $NO_2^{-}$ (can coordinate through $N$ or $O$)
$2$. $SCN^{-}$ (can coordinate through $S$ or $N$)
$3$. $CN^{-}$ (can coordinate through $C$ or $N$)
Other ligands like $C_2O_4^{2-}$,$NH_3$,$SO_4^{2-}$,and $H_2O$ are not ambidentate.
Therefore,the total number of ambidentate ligands is $3$.

(A) Essential amino acids are those that cannot be synthesized by the human body and must be obtained from the diet.
From the given list:
$1$. Arginine (Essential for children/semi-essential)
$2$. Phenylalanine (Essential)
$3$. Histidine (Essential)
$4$. Valine (Essential)
Aspartic acid,Cysteine,and Proline are non-essential amino acids.
Note: Arginine and Histidine are often classified as essential in the context of human nutrition.
Counting the essential amino acids from the list: Arginine,Phenylalanine,Histidine,and Valine.
Total count = $4$.

(B) The colour of lanthanoid ions depends on the presence of unpaired $f$-electrons ($f-f$ transitions).
Electronic configurations of the given ions are:
$La^{3+} = [Xe] 4f^0$ (No unpaired electrons)
$Nd^{3+} = [Xe] 4f^3$ ($3$ unpaired electrons)
$Sm^{3+} = [Xe] 4f^5$ ($5$ unpaired electrons)
$Eu^{3+} = [Xe] 4f^6$ ($6$ unpaired electrons)
$Lu^{3+} = [Xe] 4f^{14}$ (No unpaired electrons)
Ions with $f^0$ or $f^{14}$ configurations are colourless because they lack unpaired electrons for $f-f$ transitions.
Thus,$La^{3+}$ and $Lu^{3+}$ are colourless.
The total number of colourless ions is $2$.

(A) . $Br_2$ water test is a test for unsaturation in which the reddish-orange colour of bromine water disappears.
$B$. Alcohols give a red colour with ceric ammonium nitrate.
$C$. Phenols give a violet colour with neutral ferric chloride.
$D$. Aldehydes and ketones give yellow,orange,or red coloured precipitates with $2,4-DNP$ ($2$,$4$-dinitrophenylhydrazine).

(C) The correct matching is as follows:
$A$. Leclanche cell: The anode reaction is $Zn \rightarrow Zn^{2+} + 2e^-$. Thus,$A-IV$.
$B$. Ni-Cd cell: This is a rechargeable cell. Thus,$B-III$.
$C$. Fuel cell: It converts the energy of combustion of fuels like $H_2$ directly into electrical energy. Thus,$C-I$.
$D$. Mercury cell: It does not involve any ion in the solution,so the cell potential remains constant during its life. It is used in hearing aids. Thus,$D-II$.
Therefore,the correct sequence is $A-IV, B-III, C-I, D-II$.

(D) To determine the hybridization,we analyze the electronic configuration of the central metal ion and the nature of the ligand:
$A. K_2[Ni(CN)_4]$: $Ni^{2+}$ is $[Ar] 3d^8$. $CN^-$ is a strong field ligand $(SFL)$,causing pairing of electrons in $3d$ orbitals,resulting in $dsp^2$ hybridization.
$B. [Ni(CO)_4]$: $Ni$ is $[Ar] 3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing,resulting in $sp^3$ hybridization.
$C. [Co(NH_3)_6]Cl_3$: $Co^{3+}$ is $[Ar] 3d^6$. $NH_3$ is a strong field ligand for $Co^{3+}$,causing pairing,resulting in $d^2sp^3$ hybridization.
$D. Na_3[CoF_6]$: $Co^{3+}$ is $[Ar] 3d^6$. $F^-$ is a weak field ligand $(WFL)$,so no pairing occurs,resulting in $sp^3d^2$ hybridization.
Matching the results: $A-III, B-I, C-IV, D-II$.

(B) $EDTA^{4-}$ is a hexadentate ligand,meaning it forms $6$ coordinate bonds with the central metal ion.
In the complex $[Ca(EDTA)]^{2-}$,the $Ca^{2+}$ ion is surrounded by $6$ donor atoms from the $EDTA^{4-}$ ligand.
Therefore,the coordination number is $6$,which corresponds to an octahedral geometry.

(A) $Glucose$ is soluble in water due to the presence of multiple hydroxyl $(-OH)$ groups which facilitate extensive hydrogen bonding with water molecules. The aldehyde group does not primarily account for its high solubility.
$Glucose$ exists in equilibrium between open-chain and cyclic (pyranose) isomeric forms in its aqueous solution.
$Glucose$ contains $6$ carbon atoms,making it a hexose,and possesses an aldehyde functional group,making it an aldose. Thus,it is an aldohexose.
$Sucrose$ is a disaccharide composed of one unit of $\alpha-D-glucose$ and one unit of $\beta-D-fructose$ linked by a glycosidic bond.
Therefore,the statement that $Glucose$ is soluble in water because of the aldehyde group is incorrect.

(A) The reaction involves the participation of the phenyl ring in a Neighboring Group Participation $(NGP)$ mechanism.
$1$. The $\pi$-electrons of the phenyl ring assist in the departure of the leaving group $(Br^{-})$.
$2$. This leads to the formation of a phenonium ion intermediate.
$3$. The nucleophile $(CN^{-})$ then attacks the more electrophilic carbon of the phenonium ion.
$4$. This process results in the substitution of the $Br$ atom by the $CN$ group,leading to the product shown in option $A$.

(B) The iodoform test is given by compounds containing the $CH_3CO-$ group or compounds that can be oxidized to this group.
Treatment of $CH_3-CH_2-CCl_2-CH_3$ with aqueous $KOH$ leads to the hydrolysis of the gem-dichloride to a gem-diol $(CH_3-CH_2-C(OH)_2-CH_3)$,which is unstable and loses a water molecule to form butan$-2-$one $(CH_3-CH_2-CO-CH_3)$.
Butan$-2-$one contains the $CH_3CO-$ group,which reacts with potassium hypoiodite $(KOI)$ to form iodoform $(CHI_3)$,a yellow precipitate.
Thus,the correct compound is $CH_3-CH_2-CCl_2-CH_3$.

(A) The starting material is methyl $4-$cyanobenzoate. It contains two functional groups: a nitrile $(-CN)$ group and an ester $(-COOCH_3)$ group. Both groups react with Grignard reagents $(CH_3MgBr)$.
$1$. The ester group reacts with two equivalents of $CH_3MgBr$ to form a tertiary alcohol after acidic workup $(H_3O^+)$. The ester $-COOCH_3$ is converted to $-C(OH)(CH_3)_2$.
$2$. The nitrile group reacts with one equivalent of $CH_3MgBr$ to form an imine intermediate,which upon hydrolysis $(H_3O^+)$ yields a ketone. The $-CN$ group is converted to $-COCH_3$.
Therefore,the final product is $4-$($1$-hydroxy$-1-$methylethyl)acetophenone,which corresponds to the structure shown in option $A$.

(C) Statement $I$ is true: For transition elements,the stability of higher oxidation states increases down the group (e.g.,$W(VI)$ is more stable than $Cr(VI)$),whereas in $p-$block elements,the stability of higher oxidation states decreases down the group due to the inert pair effect.
Statement $II$ is true: Copper has a positive standard reduction potential $(E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V)$,which is higher than that of hydrogen $(E^{\circ}_{H^{+}/H_2} = 0.00 \ V)$. Therefore,copper cannot displace hydrogen from acids.

(A) The silver mirror test (Tollens' test) is given by aldehydes and $\alpha$-hydroxy ketones.
Among the given compounds:
$1$. Formic acid $(HCOOH)$ contains an aldehydic hydrogen atom attached to the carbonyl group,which allows it to be oxidized to $CO_2$ and $H_2O$ by Tollens' reagent,thus giving a positive silver mirror test.
$2$. Formaldehyde $(HCHO)$ is an aldehyde and gives a positive silver mirror test.
$3$. Benzaldehyde $(C_6H_5CHO)$ is an aromatic aldehyde and gives a positive silver mirror test.
$4$. Acetone $(CH_3COCH_3)$ is a ketone and does not give a silver mirror test.
Therefore,compounds $A$,$B$,and $C$ will give a silver mirror test.

(D) For a weak electrolyte $HA \rightleftharpoons H^{+} + A^{-}$,the dissociation constant $K_{a}$ is given by $K_{a} = \frac{\alpha^2 C}{1 - \alpha}$.
Since $\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}}$,we substitute this into the expression:
$K_{a} = \frac{(\Lambda_{m} / \Lambda_{m}^{\circ})^2 C}{1 - (\Lambda_{m} / \Lambda_{m}^{\circ})} = \frac{\Lambda_{m}^2 C}{\Lambda_{m}^{\circ}(\Lambda_{m}^{\circ} - \Lambda_{m})}$.
Rearranging this gives: $K_{a} \Lambda_{m}^{\circ}(\Lambda_{m}^{\circ} - \Lambda_{m}) = \Lambda_{m}^2 C$.
$K_{a} (\Lambda_{m}^{\circ})^2 - K_{a} \Lambda_{m} \Lambda_{m}^{\circ} = \Lambda_{m}^2 C$.
Therefore,the correct equation is $\Lambda_{m}^2 C + K_{a} \Lambda_{m} \Lambda_{m}^{\circ} - K_{a} (\Lambda_{m}^{\circ})^2 = 0$.

(B) The atomic number of Einsteinium $(Es)$ is $99$.
The electronic configuration is based on the noble gas Radon ($Rn$,atomic number $86$).
The remaining $13$ electrons are filled in the $5f$,$6d$,and $7s$ orbitals.
The correct configuration is $[Rn] 5f^{11} 6d^0 7s^2$.

(B) The second ionization enthalpy $(IE_2)$ involves removing an electron from the $M^{+}$ ion. For $Cr$ $(Z=24)$,the electronic configuration is $[Ar] 3d^5 4s^1$. The $Cr^{+}$ ion has the configuration $[Ar] 3d^5$. Removing the second electron from this stable half-filled $d^5$ subshell requires a very high amount of energy.
Thus,$M = Cr$.
The $Cr^{+}$ ion has the configuration $[Ar] 3d^5$,which means it has $n = 5$ unpaired electrons.
The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ BM$.
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
Rounding to the nearest integer,we get $6 \ BM$.

(C) Given that the mixture is equimolar,the mole fraction of benzene $(X_B)$ and methyl benzene $(X_M)$ in the liquid phase are both $0.5$.
The partial vapour pressure of methyl benzene is $P_M = X_M \times P^{\circ}_M = 0.5 \times 24 \ Torr = 12 \ Torr$.
The partial vapour pressure of benzene is $P_B = X_B \times P^{\circ}_B = 0.5 \times 80 \ Torr = 40 \ Torr$.
The total vapour pressure is $P_{\text{total}} = P_M + P_B = 12 \ Torr + 40 \ Torr = 52 \ Torr$.
The mole fraction of methyl benzene in the vapour phase $(Y_M)$ is given by $Y_M = \frac{P_M}{P_{\text{total}}}$.
$Y_M = \frac{12}{52} \approx 0.2307$.
Rounding to the nearest integer as requested,$Y_M \approx 23 \times 10^{-2}$.

(D) The group-$IV$ cation forming a black precipitate with $H_2S$ is $Co^{2+}$.
$Co^{2+} + H_2S \rightarrow CoS \downarrow$ (Black precipitate,$A$)
$CoS + \text{aqua regia} \rightarrow Co^{2+} + NOCl + S + H_2O$ ($B$ is $Co^{2+}$)
$Co^{2+} + KNO_2 + CH_3COOH \rightarrow K_3[Co(NO_2)_6] + NO + H_2O$ ($C$ is $K_3[Co(NO_2)_6]$)
In $K_3[Co(NO_2)_6]$,the oxidation state of $Co$ is $+3$.
Electronic configuration of $Co^{3+}$ is $3d^6$.
Since $NO_2^-$ is a strong field ligand,it causes pairing of electrons,resulting in $d^2sp^3$ hybridization.
Number of unpaired electrons $(n)$ = $0$.
Magnetic moment = $\sqrt{n(n+2)} = \sqrt{0(0+2)} = 0 \ BM$.

(A) For the reaction $A_{(g)} \rightarrow 2B_{(g)} + C_{(g)}$,let the initial pressure of $A$ be $P_0$.
At $t = 0$: $P_A = P_0, P_B = 0, P_C = 0, P_{total} = P_0$.
At $t = 23 \ s$: $P_A = P_0 - x, P_B = 2x, P_C = x, P_{total} = P_0 + 2x = 200 \ torr$.
At $t = \infty$: $P_A = 0, P_B = 2P_0, P_C = P_0, P_{total} = 3P_0 = 300 \ torr$.
Thus,$P_0 = 100 \ torr$.
Substituting $P_0$ in $P_{total} = P_0 + 2x = 200$,we get $100 + 2x = 200$,so $x = 50 \ torr$.
The pressure of $A$ at $t = 23 \ s$ is $P_A = P_0 - x = 100 - 50 = 50 \ torr$.
For a first-order reaction,$k = \frac{2.303}{t} \log \frac{P_0}{P_A} = \frac{2.303}{23} \log \frac{100}{50} = \frac{2.303}{23} \log(2) = \frac{2.303 \times 0.301}{23} \approx 0.0301 \ s^{-1} = 3.01 \times 10^{-2} \ s^{-1}$.
The nearest integer is $3$.

(D) Compounds which cannot undergo Friedel-Crafts reaction are:
$1$. Nitrobenzene: The $-NO_2$ group is a strongly deactivating group,which makes the ring electron-deficient and unsuitable for electrophilic substitution.
$2$. Aniline: The $-NH_2$ group forms a complex with the Lewis acid catalyst $AlCl_3$ (e.g.,$C_6H_5NH_2 \cdot AlCl_3$),which deactivates the ring and prevents the reaction.
$3$. $m$-nitroaniline: It contains both the strongly deactivating $-NO_2$ group and the $-NH_2$ group that forms a complex with the catalyst.
$4$. $m$-dinitrobenzene: It contains two strongly deactivating $-NO_2$ groups,making the ring highly electron-deficient.
Thus,a total of $4$ compounds cannot undergo Friedel-Crafts reaction.