The number of electrons with magnetic quantum | vedclass

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(A) For $Z=24$ (Chromium): The electronic configuration is $[Ar] 3d^5 4s^1$. Electrons with $m_l=0$: $1s^2$ ($2$ electrons),$2s^2$ ($2$ electrons),$2p

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$12, 13$

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(A) For $Z=24$ (Chromium): The electronic configuration is $[Ar] 3d^5 4s^1$. Electrons with $m_l=0$: $1s^2$ ($2$ electrons),$2s^2$ ($2$ electrons),$2p^6$ ($2$ electrons in $2p_z$),$3s^2$ ($2$ electrons),$3p^6$ ($2$ electrons in $3p_z$),$3d^5$ ($1$ electron in $3d_{z^2}$),$4s^1$ ($1$ electron). Total = $2+2+2+2+2+1+1 = 12$. For $Z=29$ (Copper): The electronic configuration is $[Ar] 3d^{10} 4s^1$. Electrons with $m_l=0$: $1s^2$ $(2)$,$2s^2$ $(2)$,$2p^6$ $(2)$,$3s^2$ $(2)$,$3p^6$ $(2)$,$3d^{10}$ ($2$ electrons in $3d_{z^2}$),$4s^1$ $(1)$. Total = $2+2+2+2+2+2+1 = 13$. Thus,the values are $12$ and $13$.

The number of electrons with magnetic quantum number,$m_l=0$ in the elements with atomic numbers $Z=24$ and $Z=29$ are respectively

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