The electronic configuration of 24Cr is

The electronic configuration of $_{24} Cr$ is

A
$[Ar] (3 d)^{5} (4 s)^{1}$
B
$[Ar] (3 d)^{4} (4 s)^{2}$
C
$[Ar] (3 d)^{6} (4 s)^{0}$
D
$[Ar] (4 d)^{5} (4 s)^{1}$

Solution:

The electronic configuration of $_{24} Cr$ is $[Ar] (3 d)^{5} (4 s)^{'}$ where [Ar] represents the electronic configuration is of Ar atom.

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