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 $2,3$

There are $6$ electrons in $2p$ and $3p$ each. Out of them one electron from $2p$ and one electron from $3p$ should have $m=+1$ and $s=-\frac{1}{2}$.
But in $3d$- subshell an orbital having $m=+1$ may have spin quantum no. -$\frac{1}{2}$ or $+\frac{1}{2}$.

Thereofore, minimum and maximum possible values are $2$ and $3$ respectively.