Question
Why is following electronic configuration not correct for ground state of Cr atom?
(Atomic number $=24$ ) $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^4$.

Answer

The electronic configuration of $\mathrm{Cr}(24)$ in ground state is,
$1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^1 3 d^5$ because half-filled orbitals are more stable.

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