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(C) The electronic configuration of rubidium atom $(Z = 37)$ is $[Kr] 5s^1$.The valence electron is in the $5s$ orbital.For the $5s$ orbital,the princ

Vedclass · Accepted Answer

$5, 0, 0, +1/2$

Vedclass · Answer

(C) The electronic configuration of rubidium atom $(Z = 37)$ is $[Kr] 5s^1$.The valence electron is in the $5s$ orbital.For the $5s$ orbital,the principal quantum number $(n)$ is $5$.For an $s$-orbital,the azimuthal quantum number $(l)$ is $0$.Since $l = 0$,the magnetic quantum number $(m_l)$ is $0$.The spin quantum number $(m_s)$ can be either $+1/2$ or $-1/2$.Thus,the set of quantum numbers is $(n = 5, l = 0, m_l = 0, m_s = +1/2)$.Therefore,the correct option is $C$.

The correct set of four quantum numbers for the valence electron of rubidium atom $(Z = 37)$ is

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