Question

The number of electrons having $(n\times l\times m)=0$ in $N{a}^{+}$ is :

• A. 2
• B. 4
• C. 6
• D. 7

Answer 1

Answer: (C)

6

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 $\left(s\right)\to wheres\Rightarrow \pm 1/2$
$N{a}^{+}=1s^{2}2s^2{p}^6$
As we know,
s have $l$ = 0 and $p_z$ have $m=0$
So, it have six electron with $(n\times l\times m)=0$.