Question

Calculate the minimum and maximum number of electrons which may have magnetic quantum number, $m=+1$ and spin quantum number, $s=-\frac{1}{2}$ in chromium $\left(Cr\right)$ :

• A. $0,1$
• B. $1,2$
• C. $4,6$
• D. $2,3$

Answer 1

The correct option is D $2,3$

The electronic configuration of $Cr$ is
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^5$
Out of $6$ electrons in $2p$ and $3p$,there must be one electron with $m=+1$ and $s=-\frac{1}{2}$.
But in $3d-$ subshell an orbital having $m=+1$ may have spin quantum number of $-\frac{1}{2}$ or $+\frac{1}{2}.$
Therefore, minimum and maximum posible values are $2$ and $3$ respectively.