Correct set of four quantum numbers for the o | vedclass

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Question

(A) The atomic number of rubidium $(Rb)$ is $37$. The electronic configuration of $Rb$ is $[Kr] 5s^1$. The outermost electron is in the $5s$ orbital.

Vedclass · Accepted Answer

$n=5, l=0, m_l=0, m_s=+1/2$

Vedclass · Answer

(A) The atomic number of rubidium $(Rb)$ is $37$. The electronic configuration of $Rb$ is $[Kr] 5s^1$. The outermost electron is in the $5s$ orbital. For the $5s$ orbital: Principal quantum number $(n) = 5$. Azimuthal quantum number $(l) = 0$ (for $s$-orbital). Magnetic quantum number $(m_l) = 0$. Spin quantum number $(m_s) = +1/2$ (or $-1/2$). Thus,the set of quantum numbers is $(5, 0, 0, +1/2)$.

Correct set of four quantum numbers for the outermost electron of rubidium $(Z=37)$ is :

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