The d-electron configurations of $Cr^{2+}$, $Mn^{2+}$, $Fe^{2+}$ and $Co^{2+}$ are $d^4$, $d^5$, $d^6$ and $d^

Question

The d-electron configurations of $Cr^{2+}$, $Mn^{2+}$, $Fe^{2+}$ and $Co^{2+}$ are $d^4$, $d^5$, $d^6$ and $d^7$ respectively. Which one of the following will exhibit minimum paramagnetic behaviour? (At. no. Cr = 24, Mn = 25, Fe = 26, Co = 27)

Accepted Answer

Paramagnetic behaviour depends on the number of unpaired electrons ($n$). The greater the number of unpaired electrons, the higher the paramagnetic behaviour. In all the given complexes, $H_2O$ is a weak field ligand, so they form high-spin complexes without pairing of electrons against Hund's rule.

Let's find the number of unpaired electrons for each:
1. $Fe^{2+}$ ($3d^6$): High spin configuration is $t_{2g}^4 e_g^2$. It has 4 unpaired electrons.
2. $Co^{2+}$ ($3d^7$): High spin configuration is $t_{2g}^5 e_g^2$. It has 3 unpaired electrons.
3. $Cr^{2+}$ ($3d^4$): High spin configuration is $t_{2g}^3 e_g^1$. It has 4 unpaired electrons.
4. $Mn^{2+}$ ($3d^5$): High spin configuration is $t_{2g}^3 e_g^2$. It has 5 unpaired electrons.

Since $[Co(H_2O)_6]^{2+}$ has the minimum number of unpaired electrons ($n=3$), it will exhibit the minimum paramagnetic behaviour.

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