Are $Z n^{2 +}$ $Zn^(2+)$ and $N i$ $Ni$ isoelectronic?

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Are $Z n^{2 +}$ $Zn^(2+)$ and $N i$ $Ni$ isoelectronic?

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Answer 1 · 2016-07-27T22:07:07

Isoelectronic means the same number of electrons, or the same electronic structure.

The electron configuration of $Ni$ $"Ni"$ is $[A r] 3 d^{8} 4 s^{2}$ $[Ar] 3d^8 4s^2$ .
The electron configuration of $Zn$ $"Zn"$ is $[A r] 3 d^{10} 4 s^{0}$ $[Ar] 3d^10 4s^0$ .

It may seem tempting to remove electrons from the $3 d$ $3d$ orbitals, but the $4 s$ $4s$ orbitals are higher in energy for these first-row transition metals. Hence, it makes sense to remove electrons from them upon ionization.

Since the electron configurations are not the same, they do not have the same electronic structure. They do, however, have the same number of electrons in general.

So, sure, they're isoelectronic, but not because of having the same electronic structure; instead, it's purely due to how many electrons they have, not where they are.

Another example of isoelectronic species is $^{(-)} : C \equiv O :^{(+)}$ $""^((-)):"C"-="O":^((+))$ vs. $^{(-)} : C \equiv N :$ $""^((-)):"C"-="N":$ . They have the same number of valence electrons, but they aren't identical compounds.

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Are $Z n^{2 +}$ $Zn^(2+)$ and $N i$ $Ni$ isoelectronic?

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