The maximum number of electrons present in $N i$ for which $| m_{l} |$ = 1 is:

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Solution

The correct option is C 11
The electronic configuration of $N i \to$
$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{8}$
So,
In $2 p$ orbital there are 4 electrons having $| m_{l} | = 1$
In $3 p$ orbital there are 4 electrons having $| m_{l} | = 1$
In $3 d$ orbital there are 3 electrons having $| m_{l} | = 1$
$∴$ The total number of electrons for which $| m_{l} | = 1$ , will be: $4 + 4 + 3 = 11$

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