Beryllium's fourth electron will have the fou | vedclass

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(C) The atomic number of Beryllium $(Be)$ is $4$. Its electronic configuration is $1s^2 2s^2$. The fourth electron enters the $2s$ orbital. For the $2

Vedclass · Accepted Answer

$n=2, l=0, m=0, s=-1/2$

Vedclass · Answer

(C) The atomic number of Beryllium $(Be)$ is $4$. Its electronic configuration is $1s^2 2s^2$. The fourth electron enters the $2s$ orbital. For the $2s$ orbital,the principal quantum number $n=2$ and the azimuthal quantum number $l=0$. Since $l=0$,the magnetic quantum number $m=0$. According to the Pauli exclusion principle,the second electron in the $2s$ orbital must have an opposite spin to the first,so $s=-1/2$. Thus,the four quantum numbers for the fourth electron are $n=2, l=0, m=0, s=-1/2$.

Beryllium's fourth electron will have the four quantum numbers :

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