What is the number of electrons in a Na atom | vedclass

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(A) The atomic number of $Na$ is $11$. The electronic configuration is $1s^2 2s^2 2p^6 3s^1$.We need to find electrons where $l + m = 0$.For $s$-orbit

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$7$

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(A) The atomic number of $Na$ is $11$. The electronic configuration is $1s^2 2s^2 2p^6 3s^1$.We need to find electrons where $l + m = 0$.For $s$-orbitals,$l = 0$ and $m = 0$. Thus,$l + m = 0 + 0 = 0$.For $p$-orbitals,$l = 1$ and $m = -1, 0, +1$.- For $m = -1$,$l + m = 1 + (-1) = 0$.- For $m = 0$,$l + m = 1 + 0 = 1$.- For $m = +1$,$l + m = 1 + 1 = 2$.Counting electrons with $l + m = 0$:- $1s^2$: Both electrons have $l=0, m=0$,so $l+m=0$. ($2$ electrons)- $2s^2$: Both electrons have $l=0, m=0$,so $l+m=0$. ($2$ electrons)- $2p^6$: Only the $2p_y$ orbital (where $m=-1$) satisfies $l+m=0$. There are $2$ electrons in this orbital. ($2$ electrons)- $3s^1$: The electron has $l=0, m=0$,so $l+m=0$. ($1$ electron)Total electrons = $2 + 2 + 2 + 1 = 7$.

What is the number of electrons in a $Na$ atom for which $l + m = 0$?

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