The electronic configuration of $\mathrm{Cr}$ in $\mathrm{Cr}(\mathrm{CO})_6$ as calculated using crystal field theory is

• A

  $t_{2 g}^4 e_g^0$
• B

  $t_{2 g}^3 e_g^1$
• Correct answer
  C

  $t_{2 g}^6 e_g^0$
• D

  $t_{2 g}^4 e_g^2$

Solution

$\mathrm{Cr}(\mathrm{CO})_6$ is octahedral geometry and $\mathrm{CO}$ is strong ligand. Therefore, low spin complex formed. $$ \operatorname{Cr}(z=24)[\mathrm{Ar}] 3 d^5 4 s^1 \text { or }[\mathrm{Ar}] 3 d^6 4 s^{\circ} $$ $$ [\therefore \mathrm{CO} \text { is neutral ligand }] $$
$$ \text { Electronic configuration of } \mathrm{Cr} \text { in } \mathrm{Cr}(\mathrm{CO})_6 \text { is } t_{2 g}^6 e_g{ }^0 \text {. } $$

Hence, the correct answer is option: 'C'.