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

Question

The electronic configuration of Cr in Cr ( CO )6 as calculated using crystal field theory is (A) t2 g4 eg0 (B) t2 g3 eg1 (C) t2 g6 eg0 (D) t2 g4 eg2

Tardigrade Admin · Accepted Answer

Cr ( CO )6 is octahedral geometry and CO is strong ligand. Therefore, low spin complex formed. Cr (z=24)[ Ar ] 3 d5 4 s1 or [ Ar ] 3 d6 4 s° [∴ CO is neutral ligand] <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/f79ceeda8228f921ec805f34fab1b456-.png" /> Electronic configuration of Cr in Cr ( CO )6 is t2 g6 eg0

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

$t_{2 g}^{4} e_{g}^{0}$

$t_{2 g}^{3} e_{g}^{1}$

$t_{2 g}^{6} e_{g}^{0}$

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

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

$C r (z = 24) [A r] 3 d^{5} 4 s^{1}$ or $[A r] 3 d^{6} 4 s^{\circ}$

$[∴ CO$ is neutral ligand]

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

Q. The electronic configuration of Cr in Cr(CO)6​ as calculated using crystal field theory is

t2g4​eg0​

t2g3​eg1​

t2g6​eg0​

t2g4​eg2​