Question

Why does zinc only form a $2+$ ion?

Answer 1

We know that the valency is the combining capacity of an element with other elements to form a compound. The valency of zinc is two and also the valency of oxygen is two. Using these two hints we have to form the formula of the compound zinc oxide.

Complete answer:
The valency determines the combining capacity of an element to form a compound. If any two elements have the same valency, then one atom of each element will combine forming the compound. For the formation of an ionic compound, the two elements which will combine to form the ionic compound will either lose or gain the same number of electrons as their valency. The electropositive element will lose electrons while the electronegative element will gain electrons. Zinc has a valency of $2.$ Zinc is a metal and it is electropositive. That means it will readily lose electrons. Since the valency of zinc is $2,$ it will lose $2$ electrons. On the other hand, oxygen has a valency of $2$ but it is non-metal and it is highly electronegative. So, it will gain $2$ electrons due to its valency. Now, we know that zinc and oxygen combine to form zinc oxide. So, zinc will lose two electrons and the same electron will be accepted by oxygen to form zinc oxide.
Therefore, one atom of zinc and one atom of oxygen combines to form zinc oxide. Zinc can only form only $+2$ positive charged cation ion mainly because it has ability to lose/shed two electrons in its $4s$ subshell of the outermost shell which by yielding bare zinc ion with electronic configuration $\left[ Ar \right]3{{d}^{10}}.$ Here, ion will have a filled/closed three dimension subshell giving it stables the electronic configuration. In any case, the chemistry of zinc is mostly dominated by the $+2$ oxidation state.

Note:
Remember that the transition metal is one that forms one or more stable ions which have incompletely filled d orbitals. On the basis of this definition, scandium and zinc do not count as transition metals - even though they are members of the d block.