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The d electron configurations of $Cr^{2+}$, $Mn^{2+}$, $Fe^{2+}$ and $Ni^{2+}$ are $3d^4$, $3d^5$, $3d^6$ and $3d^8$ respectively. Which one of the following aqua complexes will exhibit the minimum paramagnetic behaviour?
(At. No. Cr = 24, Mn = 25, Fe = 26, Ni = 28)
(At. No. Cr = 24, Mn = 25, Fe = 26, Ni = 28)
A
$[Fe(H_2O)_6]^{2+}$
B
$[Ni(H_2O)_6]^{2+}$
C
$[Cr(H_2O)_6]^{2+}$
D
$[Mn(H_2O)_6]^{2+}$
Detailed Solution
Lesser the number of unpaired electrons, smaller will be the paramagnetic behaviour ($\mu = \sqrt{n(n + 2)}$ BM).
$H_2O$ is a weak field ligand, so no pairing of electrons takes place and the complexes are high spin; the number of unpaired electrons is the same as in the free ion.
$Cr^{2+}$ ($3d^4$) = 4 unpaired electrons
$Mn^{2+}$ ($3d^5$) = 5 unpaired electrons
$Fe^{2+}$ ($3d^6$) = 4 unpaired electrons
$Ni^{2+}$ ($3d^8$) = 2 unpaired electrons
As $Ni^{2+}$ has the minimum number of unpaired electrons, $[Ni(H_2O)_6]^{2+}$ is the least paramagnetic.
$H_2O$ is a weak field ligand, so no pairing of electrons takes place and the complexes are high spin; the number of unpaired electrons is the same as in the free ion.
$Cr^{2+}$ ($3d^4$) = 4 unpaired electrons
$Mn^{2+}$ ($3d^5$) = 5 unpaired electrons
$Fe^{2+}$ ($3d^6$) = 4 unpaired electrons
$Ni^{2+}$ ($3d^8$) = 2 unpaired electrons
As $Ni^{2+}$ has the minimum number of unpaired electrons, $[Ni(H_2O)_6]^{2+}$ is the least paramagnetic.