How many electrons are there in $f$ subshell of $N p (Z = 93)$ in ground state electronic configuration

18

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4

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7

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14

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Solution

The correct option is A $18$
Ground state electronic configuration of $_{93} N p = 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{6} 4 f^{14} 5 d^{10} 6 s^{2} 6 p^{6} 5 f^{4} 6 d^{1} 7 s^{2}$
Total number of electrons in $f$ subshell = $14 in 4f + 4 in 5f subshell = 18 e^{-}$

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