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(B) The atomic number of lithium $(Li)$ is $3$.The electronic configuration of $Li$ is $1s^2 2s^1$.The last electron enters the $2s$ orbital.For the $

Vedclass · Accepted Answer

$1s^2 2s^1, n=2, l=0, m_l=0, m_s=+1/2 	ext{ or } -1/2$

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(B) The atomic number of lithium $(Li)$ is $3$.The electronic configuration of $Li$ is $1s^2 2s^1$.The last electron enters the $2s$ orbital.For the $2s$ orbital,the principal quantum number $(n)$ is $2$.The azimuthal quantum number $(l)$ for an $s$-orbital is $0$.The magnetic quantum number $(m_l)$ for $l=0$ is $0$.The spin quantum number $(m_s)$ can be either $+1/2$ or $-1/2$.Therefore,the quantum numbers for the last electron are $n=2, l=0, m_l=0, m_s=\pm 1/2$.

The electronic configuration together with the quantum numbers of the last electron for lithium $(Li)$ is:

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