Question

Total no. of electron in $Na$ having $n\times \ell =0$ is :

• A. $5$
• B. $2$
• C. $6$
• D. $8$

Answer 1

The correct option is A $5$

As we know,

principal quantum number $\Rightarrow$ denoted by $\left(n\right)\to wheren=1,2,3$
azimuthal quantum number $\Rightarrow$ denoted by $\left(\ell \right)\to where\ell =0ton+1$
magnetic quantum number $\Rightarrow$ denoted by $\left(m\right)\to wherem\Rightarrow -\ell to\ell$
spin quantum number $\Rightarrow$ denoted by $\left(s\right)\to wheres\Rightarrow \pm 1/2$
$Na=1s^{2}2s^2{p}^{6}3s^1$

As we know,
$s$ has $l=0$
So, it has 5 electrons with $(n\times l)=0$.