The electronic configuration of Cr in Cr(CO)_ | vedclass

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Question

(C) $Cr(CO)_6$ has an octahedral geometry and $CO$ is a strong field ligand. Therefore,a low spin complex is formed.$Cr$ (atomic number $Z = 24$) has

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) $Cr(CO)_6$ has an octahedral geometry and $CO$ is a strong field ligand. Therefore,a low spin complex is formed.$Cr$ (atomic number $Z = 24$) has the ground state electronic configuration $[Ar] 3d^5 4s^1$.In $Cr(CO)_6$,the oxidation state of $Cr$ is $0$ because $CO$ is a neutral ligand. Thus,the $Cr$ atom retains its $6$ valence electrons ($3d^5 4s^1$ becomes $3d^6$ in the complex).Due to the strong field nature of $CO$,the $6$ electrons pair up in the lower energy $t_{2g}$ orbitals.Therefore,the electronic configuration of $Cr$ in $Cr(CO)_6$ is $t_{2g}^6 e_g^0$.

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

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