The set of quantum numbers for the 19^{th} el | vedclass

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(A) The atomic number of chromium $(Cr)$ is $24$. The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$. Filling the electrons according to the Auf

Vedclass · Accepted Answer

$n = 4, l = 0, m = 0, s = +1/2 	ext{ or } -1/2$

Vedclass · Answer

(A) The atomic number of chromium $(Cr)$ is $24$. The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$. Filling the electrons according to the Aufbau principle: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1, 3d^5$. The $19^{th}$ electron enters the $4s$ orbital. For the $4s$ orbital,the principal quantum number $n = 4$. Since it is an $s$-orbital,the azimuthal quantum number $l = 0$. Consequently,the magnetic quantum number $m = 0$. The spin quantum number $s$ can be $+1/2$ or $-1/2$. Thus,the correct set is $n = 4, l = 0, m = 0, s = +1/2 	ext{ or } -1/2$.

The set of quantum numbers for the $19^{th}$ electron in chromium $(Cr)$ is:

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