Nickel atom can lose two electrons to form Ni | vedclass

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(B) The electronic configuration of neutral nickel $(_{28}Ni)$ is $1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 3d^{8}, 4s^{2}$.When forming a cation,elect

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$4s$ orbital

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(B) The electronic configuration of neutral nickel $(_{28}Ni)$ is $1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 3d^{8}, 4s^{2}$.When forming a cation,electrons are removed from the orbital with the highest principal quantum number $(n)$ first.Since the $4s$ orbital has a higher principal quantum number $(n=4)$ compared to the $3d$ orbital $(n=3)$,the two electrons are lost from the $4s$ orbital.Therefore,the configuration of $Ni^{2+}$ is $1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 3d^{8}, 4s^{0}$.

Nickel atom can lose two electrons to form $Ni^{2+}$ ion. The atomic number of nickel is $28$. From which orbital will nickel lose two electrons?

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