In Niobium (Z = 41),the number of electrons w | vedclass

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(B) The electronic configuration of Niobium $(Z = 41)$ is: $[Kr] 4d^4 5s^1$ or $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 4d^4, 5s^1$.For $m

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$8$ or $9$

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(B) The electronic configuration of Niobium $(Z = 41)$ is: $[Kr] 4d^4 5s^1$ or $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 4d^4, 5s^1$.For $m = -1$,we consider the orbitals where the magnetic quantum number $m$ can be $-1$:$2p$ subshell $(2p_y)$: $2$ electrons$3p$ subshell $(3p_y)$: $2$ electrons$3d$ subshell ($3d_{xy}$ or $3d_{yz}$): $2$ electrons$4p$ subshell $(4p_y)$: $2$ electrons$4d$ subshell ($4d_{xy}$ or $4d_{yz}$): $0$ to $1$ electron depending on the specific arrangement of the $4$ electrons in $4d$.Summing these,we get $2 + 2 + 2 + 2 + (0 	ext{ to } 1) = 8$ or $9$ electrons.

In Niobium $(Z = 41)$,the number of electrons with $m = -1$ will be:

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