The electronic configuration of Cr in `Cr(CO)_6` as calculated using crystal field theory is

Options

A. `t_(2g)^(4) e_(g)^(0)`
B. `t_(2g)^(3) e_(g)^(1)`
C. `t_(2g)^(6) e_(g)^(0)`
D. `t_(2g)^(4) e_(g)^(2)`

Answer

C

Explanation

`Cr(CO)_6` is octahedral geometry and CO is strong ligand . Therefore , low spin complex formed .
`Cr (z = 24) [Ar] 3d^(5) 4s^(1) " or " [Ar] 3d^(6) 5s^(0) " " [ therefore "CO is neutral ligand"]`

Electronic configuration of Cr in `Cr(CO)_6` is `t_(2g) , e_(g)^(0)`