The d-electron configurations of Cr²⁺, Mn²⁺, Fe²⁺ and Ni²⁺ are 3d⁴, 3d⁵, 3d⁶ and 3d⁸ respectively. Which one o

Question

The d-electron configurations of Cr²⁺, Mn²⁺, Fe²⁺ and Ni²⁺ are 3d⁴, 3d⁵, 3d⁶ and 3d⁸ respectively. Which one of the following aqua complexes will exhibit the minimum paramagnetic behaviour? (At. No. Cr = 24, Mn = 25, Fe = 26, Ni = 28)

Accepted Answer

Paramagnetic behaviour is directly proportional to the number of unpaired electrons ($n$). The magnetic moment is given by $\mu = \sqrt{n(n+2)}$ BM. Water ($H_2O$) is a weak field ligand , so these complexes are 'high spin', meaning electrons will occupy higher energy orbitals before pairing up (following Hund's rule).

1.  **$[Cr(H_2O)_6]^{2+}$ ($3d^4$):** Configuration $t_{2g}^3 e_g^1$. Unpaired electrons ($n$) = 4.
2.  **$[Mn(H_2O)_6]^{2+}$ ($3d^5$):** Configuration $t_{2g}^3 e_g^2$. Unpaired electrons ($n$) = 5 (Maximum) .
3.  **$[Fe(H_2O)_6]^{2+}$ ($3d^6$):** Configuration $t_{2g}^4 e_g^2$. Pairing occurs in one $t_{2g}$ orbital. Unpaired electrons ($n$) = 4 .
4.  **$[Ni(H_2O)_6]^{2+}$ ($3d^8$):** Configuration $t_{2g}^6 e_g^2$. The $t_{2g}$ set is fully filled (6 electrons), leaving 2 unpaired electrons in the $e_g$ set. Unpaired electrons ($n$) = 2 .

Since $Ni^{2+}$ has the minimum number of unpaired electrons (2), it exhibits the minimum paramagnetic behaviour.

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