The electronic configuration of an element is $1 s^{2} 2 s^{2} 2 p^{6}$ $3 s^{2} 3 p^{6} 3 d^{5}, 4

Question

The electronic configuration of an element is $1 s^{2} 2 s^{2} 2 p^{6}$ $3 s^{2} 3 p^{6} 3 d^{5}, 4 s^{1}$. This represents its\

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(b) : Since $3 d^{5} 4 s^{1}$ system has lower energy (because of half-filled $d$ orbitals) as compared to $3 d^{4} 4 s^{2}$, so $3 d^{5} 4 s^{1}$ represents the ground state of the atom.\