The four quantum numbers of the outermost orb | vedclass

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(B) The electronic configuration of $K$ $(Z = 19)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1$. The outermost electron is in the $4s$ orbital. For the $4s

Vedclass · Accepted Answer

$n = 4, l = 0, m = 0, s = + \frac{1}{2}$

Vedclass · Answer

(B) The electronic configuration of $K$ $(Z = 19)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1$. The outermost electron is in the $4s$ orbital. For the $4s$ orbital,the principal quantum number $n = 4$. For an $s$-orbital,the azimuthal quantum number $l = 0$. Since $l = 0$,the magnetic quantum number $m = 0$. The spin quantum number $s$ for the single electron in the $4s$ orbital is $+ \frac{1}{2}$. Thus,the set of quantum numbers is $n = 4, l = 0, m = 0, s = + \frac{1}{2}$.

The four quantum numbers of the outermost orbital of $K$ (atomic number $= 19$) are:

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