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Question

(A) The atomic number of rubidium $(Rb)$ is $37$.The electronic configuration of $Rb$ is: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 5s^1$.Th

Vedclass · Accepted Answer

$5, 0, 0, +\frac{1}{2}$

Vedclass · Answer

(A) The atomic number of rubidium $(Rb)$ is $37$.The electronic configuration of $Rb$ is: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 5s^1$.The valence electron is in the $5s$ orbital.For the $5s^1$ electron:Principal quantum number $(n) = 5$.Azimuthal quantum number $(l)$ for $s$-orbital $= 0$.Magnetic quantum number $(m_l) = 0$.Spin quantum number $(m_s) = +\frac{1}{2}$ (or $-\frac{1}{2}$).Thus,the set of quantum numbers is $(5, 0, 0, +\frac{1}{2})$.

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

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