Question

What is the minimum possible atomic number of the element containing four electrons in the ground state with magnetic quantum number $m_l$=$+1$ ?

• A. $18$
• B. $14$
• C. $12$
• D. $16$

Answer 1

The correct option is D $16$

We know, value of $m_l$ varies from $-l$ to $+l$.

For $m_l=$ $+1$, minimum value of $l=$ $1$.
$l=$ $1$ corresponds to $p$-orbital and each $p$-subshell has $2$ electrons with $m_l$ value $+1$.

So, we need two $p$-subshells in order to have $4$ electrons with $m_l=+1$.

$1s^{2}2s^{2}2p^{6}3s^{2}3p^4$ have $4$ electrons having $m_l=+1$.

Therefore minimum atomic number of element having $m_l=+1$ is $16$.