Consider the ground state of Cr (Z = 24). The | vedclass

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(B) The electronic configuration of $Cr$ $(Z = 24)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1$.For $l = 1$ (which corresponds to $p$-orbitals),the

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$12$ and $5$

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

Consider the ground state of $Cr$ $(Z = 24)$. The numbers of electrons with the azimuthal quantum numbers,$l = 1$ and $l = 2$ are,respectively:

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