Question

The electronic configuration of $Cr$ in $Cr ( CO )_{6}$ as calculated using crystal field theory is

• A. $t_{2 g}^{4} e_{g}^{0}$
• B. $t_{2 g}^{3} e_{g}^{1}$
• C. $t_{2 g}^{6} e_{g}^{0}$
• D. $t_{2 g}^{4} e_{g}^{2}$

Answer 1

Answer: (C)

$Cr ( CO )_{6}$ is octahedral geometry and $CO$ is strong ligand. Therefore, low spin complex formed.

$Cr (z=24)[ Ar ] \,3 d^{5} \,4 s^{1}$ or $[ Ar ] 3 d^{6} \,4 s^{\circ}$

$[\therefore \,CO$ is neutral ligand]



Electronic configuration of $Cr$ in $Cr ( CO )_{6}$ is $t_{2 g}^{6} e_{g}^{0}$