How many d-electrons are in the outer shell of the metals in following compounds:


Do I just use Zn^2+ and as I know that in Ions the s orbitals are removed first I count 10 d electrons?

However ZnS makes a metal lattice. Can I use Zn^2+ then?

Help would be really appreciated

  • $\begingroup$ Don't ask about the same stuff twice! Don't you know you can edit it? $\endgroup$ – Mithoron Mar 7 '16 at 21:35

In zinc sulfide, two s-electrons have been removed from $\ce{Zn(0)}$ to make $\ce{Zn^{2+}}$. Therefore, yes, it does have 10 d-electrons.

In its lattice, zinc is bound to four sulfur atoms which fulfills the 18-electron rule (10 from $\ce{Zn^{2+}}$ and 2 each from four separate $\ce{S^{2-}}$; the 18-electron rule for transition metals is the equivalent of the octet rule for main-block elements). In this case, however, the metal (zinc) still only has 10 d-electrons. The additional electrons are in hybrid, non-d orbitals.

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