The number of electrons in a nickel (Ni) atom | vedclass

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(A) The atomic number of nickel $(Ni)$ is $28$. The electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8$.For $m = 0$:- $1s^2$: $2$ electro

Vedclass · Accepted Answer

$14$

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(A) The atomic number of nickel $(Ni)$ is $28$. The electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8$.For $m = 0$:- $1s^2$: $2$ electrons $(l=0, m=0)$- $2s^2$: $2$ electrons $(l=0, m=0)$- $2p^6$: $2$ electrons ($m=0$ for $2p_z$)- $3s^2$: $2$ electrons $(l=0, m=0)$- $3p^6$: $2$ electrons ($m=0$ for $3p_z$)- $4s^2$: $2$ electrons $(l=0, m=0)$- $3d^8$: In $d$-orbitals,$m$ values are $-2, -1, 0, +1, +2$. For $d^8$,filling $8$ electrons: $d_{xy}^2, d_{yz}^2, d_{zx}^2, d_{x^2-y^2}^1, d_{z^2}^1$ (or similar distribution). The $d_{z^2}$ orbital corresponds to $m=0$. With $8$ electrons,$d_{z^2}$ will have $2$ electrons.Total electrons with $m=0$ is $2+2+2+2+2+2+2 = 14$.

The number of electrons in a nickel $(Ni)$ atom having $m = 0$ is:

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