Be's 4^{th} electron will have four quantum n | vedclass

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(C) The atomic number of $Be$ is $4$. The electronic configuration of $Be$ is $1s^2 2s^2$. The $4^{th}$ electron enters the $2s$ orbital. For the $2s$

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$2, 0, 0, -1/2$

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(C) The atomic number of $Be$ is $4$. The electronic configuration of $Be$ is $1s^2 2s^2$. The $4^{th}$ electron enters the $2s$ orbital. For the $2s$ orbital,the principal quantum number $n = 2$. For an $s$ orbital,the azimuthal quantum number $l = 0$. Since $l = 0$,the magnetic quantum number $m = 0$. Assuming the first electron in the $2s$ orbital has $s = +1/2$,the second electron (the $4^{th}$ electron of $Be$) will have $s = -1/2$ according to the Pauli exclusion principle.

$Be$'s $4^{th}$ electron will have four quantum numbers $(n, l, m, s)$:

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