In an atom having 2 K,8 L,8 M and 2 N electro | vedclass

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(A) The electronic configuration of the atom is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$. We need to find the number of electrons with magnetic quantum number

Vedclass · Accepted Answer

$6$

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(A) The electronic configuration of the atom is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$. We need to find the number of electrons with magnetic quantum number $m=0$ and spin quantum number $s=+\frac{1}{2}$. - $1s^2$: One electron has $m=0, s=+\frac{1}{2}$. - $2s^2$: One electron has $m=0, s=+\frac{1}{2}$. - $2p^6$: The $2p_z$ orbital has $m=0$. It contains $2$ electrons,one of which has $s=+\frac{1}{2}$. - $3s^2$: One electron has $m=0, s=+\frac{1}{2}$. - $3p^6$: The $3p_z$ orbital has $m=0$. It contains $2$ electrons,one of which has $s=+\frac{1}{2}$. - $4s^2$: One electron has $m=0, s=+\frac{1}{2}$. Total number of electrons with $m=0$ and $s=+\frac{1}{2}$ is $1+1+1+1+1+1 = 6$.

In an atom having $2$ $K$,$8$ $L$,$8$ $M$ and $2$ $N$ electrons,the number of electrons with $m=0$ and $s=+\frac{1}{2}$ is $:-$

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