The 'd' electron configuration 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?

Options

A. `[Cr(H_2O)_6]^(+2)`
B. `[Mn(H_2O)_6]^(+2)`
C. `Fe(H_2O)_6]^(+2)`
D. `[Co(H_2O)_6]^(+2)`

Answer

D

Explanation

In `[Mn(H_2O)_6]^(+2),Mn^(+2)=3d^5`
So, the number of unpaired electron=5.
In `[C(H_2O)_6]^(+2),Fe^(+2)=3d^6`
So, the number of unpaired electron=4.
In `[Co(H_2O)_6]^(+2), Co^(+2)=3d^7`
So, the number of unpaired electron =3
In `[Cr(H_2O)_6]^(+2), Cr^(+2)=3d^4`
So, the number of unpaired electron =4.
All these show `sp^3d^2` hybridisation forming outer orbital complexes. Hence, minimum paramagnetic behaviour is shown by `[Co(H_2O)_6]^(+2)`