JEE Main 2026 — Coordination Compounds Question with Solution

Statement I: Analysis of paramagnetic species.
1. $[Co{F}_6{]}^{3-}$: $C{o}^{3+}$ 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. $[Ti{F}_6{]}^{3-}$: $T{i}^{3+}$ is $3d^1$. $n=1$. Paramagnetic.
3. $V_2{O}_5$: $V^{5+}$ is $3d^0$. $n=0$. Diamagnetic.
4. $[Fe(CN{)}_6{]}^{3-}$: $F{e}^{3+}$ is $3d^5$. $C{N}^{-}$ 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. $K_4[Fe(CN{)}_6]$: $F{e}^{2+}$ ($3d^6$) with SFL $\to t_{2g}^6{e}_g^0$, $n=0$.
2. $K_3[Fe(CN{)}_6]$: $F{e}^{3+}$ ($3d^5$) with SFL $\to t_{2g}^5{e}_g^0$, $n=1$.
3. $[Fe(H_{2}O{)}_6]S{O}_4\cdot H_{2}O$: $F{e}^{2+}$ ($3d^6$) with WFL $\to t_{2g}^4{e}_g^2$, $n=4$.
4. $[Fe(H_{2}O{)}_6]C{l}_3$: $F{e}^{3+}$ ($3d^5$) with WFL $\to t_{2g}^3{e}_g^2$, $n=5$.
Order of $n$: $0<1<4<5$. Statement II is true.
Both statements are true.