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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. nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)
A
$[Cr(H_2O)_6]^{2+}$
B
$[Mn(H_2O)_6]^{2+}$
C
$[Fe(H_2O)_6]^{2+}$
D
$[Co(H_2O)_6]^{2+}$
Detailed Solution
$H_2O$ is a weak field ligand, so all these octahedral complexes are high spin and the electrons are not forced to pair.
$Cr^{2+}$ ($d^4$): $t_{2g}^3e_g^1$, 4 unpaired electrons.
$Mn^{2+}$ ($d^5$): $t_{2g}^3e_g^2$, 5 unpaired electrons.
$Fe^{2+}$ ($d^6$): $t_{2g}^4e_g^2$, 4 unpaired electrons.
$Co^{2+}$ ($d^7$): $t_{2g}^5e_g^2$, 3 unpaired electrons.
Paramagnetism increases with the number of unpaired electrons ($\mu = \sqrt{n(n+2)}$ BM).
$[Co(H_2O)_6]^{2+}$ has the fewest unpaired electrons and shows minimum paramagnetic behaviour.
$Cr^{2+}$ ($d^4$): $t_{2g}^3e_g^1$, 4 unpaired electrons.
$Mn^{2+}$ ($d^5$): $t_{2g}^3e_g^2$, 5 unpaired electrons.
$Fe^{2+}$ ($d^6$): $t_{2g}^4e_g^2$, 4 unpaired electrons.
$Co^{2+}$ ($d^7$): $t_{2g}^5e_g^2$, 3 unpaired electrons.
Paramagnetism increases with the number of unpaired electrons ($\mu = \sqrt{n(n+2)}$ BM).
$[Co(H_2O)_6]^{2+}$ has the fewest unpaired electrons and shows minimum paramagnetic behaviour.