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

Excited state

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Ground state

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Cationic form

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Anionic form

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Solution

The correct option is B Ground state
$3 d^{5} 4 s^{1}$ system is more stable than $3 d^{4} 4 s^{2}$ system because of the half filled $d$ -orbital. Hence former is the ground state configuration.

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Similar questions

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5} 4 s^{1}

{1 s}^{2} {2 s}^{2} {2 p}^{6} {3 s}^{2} {3 p}^{6} {3 d}^{6} {4 s}^{1}

{1 s}^{2} {2 s}^{2} {2 p}^{6} {3 s}^{2} {3 p}^{1}

{1 s}^{2} {2 s}^{2} {2 p}^{6} {3 s}^{2} {3 p}^{5}

{1 s}^{2} {2 s}^{2} {2 p}^{6} {3 s}^{2} {3 p}^{6} {4 s}^{1}

{(I E)}_{1}

{(I E)}_{4}

{1 s}^{2} {2 s}^{2} {2 p}^{6} {3 s}^{2} {3 p}^{6} {4 s}^{2}

An element having electronic configuration 1s², 2s² 2p⁶, 3s² 3p⁶, 4s¹ forms

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