Considering the ground state of Cr atom (Z = | vedclass

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(B) The electronic configuration of $Cr$ $(Z = 24)$ is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$.For azimuthal quantum number $l = 1$,the electrons are in

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$12$ and $5$

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(B) The electronic configuration of $Cr$ $(Z = 24)$ is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$.For azimuthal quantum number $l = 1$,the electrons are in $p$-orbitals ($2p^6$ and $3p^6$). Total electrons = $6 + 6 = 12$.For azimuthal quantum number $l = 2$,the electrons are in $d$-orbitals $(3d^5)$. Total electrons = $5$.Therefore,the number of electrons for $l = 1$ and $l = 2$ are $12$ and $5$ respectively.

Considering the ground state of $Cr$ atom $(Z = 24)$,the number of electrons associated with azimuthal quantum numbers $l = 1$ and $l = 2$ are,respectively:

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