JEE Main 2025 Chemistry Question Paper with Answer and Solution

(D) $1$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $[Ar] 3d^6$. $NH_3$ is a Strong Field Ligand $(SFL)$,causing pairing. Hybridization is $d^2 sp^3$.
$2$. $SF_6$: Sulphur has $6$ valence electrons and forms $6$ bonds. Hybridization is $sp^3 d^2$.
$3$. $[CrF_6]^{3-}$: $Cr^{3+}$ is $[Ar] 3d^3$. Hybridization is $d^2 sp^3$.
$4$. $[CoF_6]^{3-}$: $Co^{3+}$ is $[Ar] 3d^6$. $F^-$ is a Weak Field Ligand $(WFL)$,no pairing. Hybridization is $sp^3 d^2$.
$5$. $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $[Ar] 3d^4$. $CN^-$ is $SFL$. Hybridization is $d^2 sp^3$.
$6$. $[MnCl_6]^{3-}$: $Mn^{3+}$ is $[Ar] 3d^4$. $Cl^-$ is $WFL$. Hybridization is $sp^3 d^2$.
Species with $sp^3 d^2$ hybridization are $SF_6, [CoF_6]^{3-}$,and $[MnCl_6]^{3-}$.
Total count is $3$.

(B) The given compound is $CH_3COCH_2CH_2CH_2CH_2CHO$ (heptane$-2,6-$dione derivative or similar keto-aldehyde).
Intramolecular aldol condensation occurs between the $\alpha$-hydrogen of the ketone and the carbonyl carbon of the aldehyde.
The enolate formed at the $\alpha$-position of the ketone attacks the aldehyde carbonyl carbon to form a five-membered ring.
Dehydration of the resulting $\beta$-hydroxy ketone under heating $( \Delta )$ yields the conjugated enone.
The major product is $1-\text{acetylcyclopentene}$.

(A) $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state. Configuration: $[Ar] 3d^8 4s^2$. Due to strong field ligand $CO$,electrons pair up. Hybridization: $sp^3$. Shape: Tetrahedral. Magnetic moment: $0 \ BM$ (Diamagnetic). Matches $III$.
$(B)$ $[Ni(CN)_4]^{2-}$: $Ni$ is in $+2$ oxidation state. Configuration: $[Ar] 3d^8$. Due to strong field ligand $CN^-$,electrons pair up. Hybridization: $dsp^2$. Shape: Square planar. Magnetic moment: $0 \ BM$ (Diamagnetic). Matches $II$.
$(C)$ $[NiCl_4]^{2-}$: $Ni$ is in $+2$ oxidation state. Configuration: $[Ar] 3d^8$. Weak field ligand $Cl^-$. Hybridization: $sp^3$. Shape: Tetrahedral. Two unpaired electrons,$\mu = \sqrt{n(n+2)} = \sqrt{2(4)} = \sqrt{8} \approx 2.8 \ BM$. Matches $I$.
$(D)$ $[MnBr_4]^{2-}$: $Mn$ is in $+2$ oxidation state. Configuration: $[Ar] 3d^5$. Weak field ligand $Br^-$. Hybridization: $sp^3$. Shape: Tetrahedral. Five unpaired electrons,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.9 \ BM$. Matches $IV$.
Therefore,the correct matching is $A-III, B-II, C-I, D-IV$.

(D) The acidic strength of hydrides of group $16$ elements increases down the group due to the decrease in bond dissociation enthalpy. Thus,the order is $H_2Se < H_2Te$. Therefore,Statement $I$ is false.
The bond dissociation enthalpy decreases down the group as the bond length increases. Thus,$H_2Se$ has a higher bond dissociation enthalpy $(276 \ kJ/mol)$ compared to $H_2Te$ $(238 \ kJ/mol)$. Therefore,Statement $II$ is true.

(A) The Nernst equation for the given half-cell reaction is: $E = E^{o} - \frac{0.059}{n} \log Q$.
Here,$n = 6$ and $Q = \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}] [H^{+}]^{14}}$.
Setting $E = 0$:
$0 = 1.33 - \frac{0.059}{6} \log \left( \frac{10^{-6}}{[H^{+}]^{14}} \right)$.
$1.33 = \frac{0.059}{6} (\log 10^{-6} - \log [H^{+}]^{14})$.
$1.33 = \frac{0.059}{6} (-6 + 14 pH)$.
$\frac{1.33 \times 6}{0.059} = -6 + 14 pH$.
$135.25 = -6 + 14 pH$.
$141.25 = 14 pH$.
$pH = \frac{141.25}{14} \approx 10.09$.
The nearest integer value is $10$.

(A) The correct answer is $A$.
Ascorbic acid is chemically known as $Vitamin \ C$,which is an essential water-soluble vitamin required for the synthesis of collagen and as an antioxidant.