JEE Main 2026 — Chemistry Inorganic Chemistry

2026mcqhard

Given below are two statements:

Statement I: The number of paramagnetic species among $\left[\mathrm{CoF}_{6}\right]^{3-},\left[\mathrm{TiF}_{6}\right]^{3-}$ , $\mathrm{V}_{2} \mathrm{O}_{5}$ and $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ is 3.

Statement II: $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]<\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]<\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}<\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$ is the correct order in terms of number of unpaired electron(s) present in the complexes.

In the light of the above statements, choose the correct answer from the options given below

Official previous-year question

Held on 24 Jan 2026 · Verified 6 Jul 2026.

Options

A
Both Statement I and Statement II are true
B
Statement I is false but Statement II is true
C
Both Statement I and Statement II are false
D
Statement I is true but Statement II is false

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Solution

Statement I: Analysis of paramagnetic species.

$[CoF_6]^{3-}$ : $Co^{3+}$ is $3d^6$ . $F^-$ is a weak field ligand (WFL), so configuration is $t_{2g}^4 e_g^2$ . Unpaired electrons ( $n$ ) = 4. Paramagnetic.
$[TiF_6]^{3-}$ : $Ti^{3+}$ is $3d^1$ . $n = 1$ . Paramagnetic.
$V_2O_5$ : $V^{5+}$ is $3d^0$ . $n = 0$ . Diamagnetic.
$[Fe(CN)_6]^{3-}$ : $Fe^{3+}$ is $3d^5$ . $CN^-$ is a strong field ligand (SFL), so configuration is $t_{2g}^5 e_g^0$ . $n = 1$ . Paramagnetic.

Total paramagnetic species = 3. Statement I is true.

Statement II: Analysis of unpaired electrons.

$K_4[Fe(CN)_6]$ : $Fe^{2+}$ ( $3d^6$ ) with SFL $\rightarrow t_{2g}^6 e_g^0$ , $n = 0$ .
$K_3[Fe(CN)_6]$ : $Fe^{3+}$ ( $3d^5$ ) with SFL $\rightarrow t_{2g}^5 e_g^0$ , $n = 1$ .
$[Fe(H_2O)_6]SO_4 \cdot H_2O$ : $Fe^{2+}$ ( $3d^6$ ) with WFL $\rightarrow t_{2g}^4 e_g^2$ , $n = 4$ .
$[Fe(H_2O)_6]Cl_3$ : $Fe^{3+}$ ( $3d^5$ ) with WFL $\rightarrow t_{2g}^3 e_g^2$ , $n = 5$ .

Order of $n$ : $0 < 1 < 4 < 5$ . Statement II is true.

Both statements are true.

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Solution

Statement I: Analysis of paramagnetic species.

$[CoF_6]^{3-}$ : $Co^{3+}$ is $3d^6$ . $F^-$ is a weak field ligand (WFL), so configuration is $t_{2g}^4 e_g^2$ . Unpaired electrons ( $n$ ) = 4. Paramagnetic.
$[TiF_6]^{3-}$ : $Ti^{3+}$ is $3d^1$ . $n = 1$ . Paramagnetic.
$V_2O_5$ : $V^{5+}$ is $3d^0$ . $n = 0$ . Diamagnetic.
$[Fe(CN)_6]^{3-}$ : $Fe^{3+}$ is $3d^5$ . $CN^-$ is a strong field ligand (SFL), so configuration is $t_{2g}^5 e_g^0$ . $n = 1$ . Paramagnetic.

Total paramagnetic species = 3. Statement I is true.

Statement II: Analysis of unpaired electrons.

$K_4[Fe(CN)_6]$ : $Fe^{2+}$ ( $3d^6$ ) with SFL $\rightarrow t_{2g}^6 e_g^0$ , $n = 0$ .
$K_3[Fe(CN)_6]$ : $Fe^{3+}$ ( $3d^5$ ) with SFL $\rightarrow t_{2g}^5 e_g^0$ , $n = 1$ .
$[Fe(H_2O)_6]SO_4 \cdot H_2O$ : $Fe^{2+}$ ( $3d^6$ ) with WFL $\rightarrow t_{2g}^4 e_g^2$ , $n = 4$ .
$[Fe(H_2O)_6]Cl_3$ : $Fe^{3+}$ ( $3d^5$ ) with WFL $\rightarrow t_{2g}^3 e_g^2$ , $n = 5$ .

Order of $n$ : $0 < 1 < 4 < 5$ . Statement II is true.

Both statements are true.

JEE Main 2026 — Chemistry Inorganic Chemistry

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JEE Main 2026 — Chemistry Inorganic Chemistry

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