The differnece in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is

Options

A. `Mn^(2+)`
B. `Co^(2+)`
C. `Fe^(2+)`
D. `Ni^(2+)`

Answer

B

Explanation

For Co electronic configuration of valence shell is `4s^(2)3d^(7)`
For `Co^(2+): 4s^(0)3d^(7)`
In case of high spin complex, electron configuration will be (without pairing) `3d^(7)[t_(2)g^(5)eg^(2)]` and unpaired `e^(-)=3`
In case of low spin complex, electron configuration will be (considering pairing) `3d^(7)(t_(2)g^(6)eg^(1))` and number of unpaired `e^(-)=1`