JEE Main 2026 — Coordination Compounds Question with Solution

To determine the magnetic nature of the given complexes, we analyze the oxidation state, electronic configuration, and the nature of the ligands for each central metal atom/ion:

1. $[{\text{MnBr}}_4{]}^{2-}$: Mn${}^{2+}$ has a $3d^5$ configuration. Br${}^{-}$ is a weak field ligand. The complex is tetrahedral with $5$ unpaired electrons. (Paramagnetic)

2. $[{\text{NiCl}}_4{]}^{2-}$: Ni${}^{2+}$ has a $3d^8$ configuration. Cl${}^{-}$ is a weak field ligand. The complex is tetrahedral with $2$ unpaired electrons. (Paramagnetic)

3. $[{\text{Ni(CN)}}_4{]}^{2-}$: Ni${}^{2+}$ has a $3d^8$ configuration. CN${}^{-}$ is a strong field ligand. The complex is square planar ($ds{p}^2$) with all electrons paired. (Diamagnetic)

4. $[{\text{Ni(CO)}}_4]$: Ni${}^0$ has a $3d^{8}4s^2$ configuration, which becomes $3d^{10}$ in the presence of the strong field ligand CO. The complex is tetrahedral with all electrons paired. (Diamagnetic)

5. $[{\text{CoF}}_6{]}^{3-}$: Co${}^{3+}$ has a $3d^6$ configuration. F${}^{-}$ is a weak field ligand. The complex is octahedral and high spin ($t_{2g}^4{e}_g^2$) with $4$ unpaired electrons. (Paramagnetic)

6. $[{\text{Fe(CN)}}_6{]}^{4-}$: Fe${}^{2+}$ has a $3d^6$ configuration. CN${}^{-}$ is a strong field ligand. The complex is octahedral and low spin ($t_{2g}^6{e}_g^0$) with all electrons paired. (Diamagnetic)

7. $[{\text{Mn(CN)}}_6{]}^{3-}$: Mn${}^{3+}$ has a $3d^4$ configuration. CN${}^{-}$ is a strong field ligand. The complex is octahedral and low spin ($t_{2g}^4{e}_g^0$) with $2$ unpaired electrons. (Paramagnetic)

8. $[{\text{Ti(CN)}}_6{]}^{3-}$: Ti${}^{3+}$ has a $3d^1$ configuration. The complex is octahedral ($t_{2g}^1{e}_g^0$) with $1$ unpaired electron. (Paramagnetic)

9. $[{\text{Cu(H}}_2{\text{O)}}_6{]}^{2+}$: Cu${}^{2+}$ has a $3d^9$ configuration. The complex is octahedral ($t_{2g}^6{e}_g^3$) with $1$ unpaired electron. (Paramagnetic)

10. $[{\text{Co(C}}_2{\text{O}}_4{\text{)}}_3{]}^{3-}$: Co${}^{3+}$ has a $3d^6$ configuration. Oxalate (${\text{C}}_2{\text{O}}_4^{2-}$) acts as a strong field ligand for Co${}^{3+}$. The complex is octahedral and low spin ($t_{2g}^6{e}_g^0$) with all electrons paired. (Diamagnetic)

Therefore, the paramagnetic complexes are $[{\text{MnBr}}_4{]}^{2-}$, $[{\text{NiCl}}_4{]}^{2-}$, $[{\text{CoF}}_6{]}^{3-}$, $[{\text{Mn(CN)}}_6{]}^{3-}$, $[{\text{Ti(CN)}}_6{]}^{3-}$, and $[{\text{Cu(H}}_2{\text{O)}}_6{]}^{2+}$.

The total number of paramagnetic complexes is $6$.

Answer: $6$