Q17 / 75 · JEE Main 2026 (24 January Shift 1) · Chemistry
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
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
- Correct answerABoth Statement I and Statement II are true
- BStatement I is false but Statement II is true
- CBoth Statement I and Statement II are false
- DStatement I is true but Statement II is false
Solution
Statement I: Analysis of paramagnetic species.
1. $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand (WFL), so configuration is $t_{2g}^4 e_g^2$. Unpaired electrons ($n$) = 4. Paramagnetic.
2. $[TiF_6]^{3-}$: $Ti^{3+}$ is $3d^1$. $n = 1$. Paramagnetic.
3. $V_2O_5$: $V^{5+}$ is $3d^0$. $n = 0$. Diamagnetic.
4. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ 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.
1. $K_4[Fe(CN)_6]$: $Fe^{2+}$ ($3d^6$) with SFL $\rightarrow t_{2g}^6 e_g^0$, $n = 0$.
2. $K_3[Fe(CN)_6]$: $Fe^{3+}$ ($3d^5$) with SFL $\rightarrow t_{2g}^5 e_g^0$, $n = 1$.
3. $[Fe(H_2O)_6]SO_4 \cdot H_2O$: $Fe^{2+}$ ($3d^6$) with WFL $\rightarrow t_{2g}^4 e_g^2$, $n = 4$.
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.
1. $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand (WFL), so configuration is $t_{2g}^4 e_g^2$. Unpaired electrons ($n$) = 4. Paramagnetic.
2. $[TiF_6]^{3-}$: $Ti^{3+}$ is $3d^1$. $n = 1$. Paramagnetic.
3. $V_2O_5$: $V^{5+}$ is $3d^0$. $n = 0$. Diamagnetic.
4. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ 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.
1. $K_4[Fe(CN)_6]$: $Fe^{2+}$ ($3d^6$) with SFL $\rightarrow t_{2g}^6 e_g^0$, $n = 0$.
2. $K_3[Fe(CN)_6]$: $Fe^{3+}$ ($3d^5$) with SFL $\rightarrow t_{2g}^5 e_g^0$, $n = 1$.
3. $[Fe(H_2O)_6]SO_4 \cdot H_2O$: $Fe^{2+}$ ($3d^6$) with WFL $\rightarrow t_{2g}^4 e_g^2$, $n = 4$.
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.
Hence, the correct answer is option: 'A'.