How does the position of a metal in the periodic table relate to whether it occurs primarily as an oxide or as a sulfide?

Metals are elements and their oxides that produce bases in water rather than acids. The periodic chart, where metals are usually readily defined by colour coding, is a straightforward way to determine if an element is metal or not.

To determine which metals are more likely to produce oxides and which are more likely to form sulfides, we must first examine the ions. Both oxygen and sulfur are members of the$16\mathrm{th}$group, although oxygen belongs to the second period and sulphur to the third. Despite having the same charge $(2-)$, ${\mathrm{S}}^{2-}$has a larger radius than ${\mathrm{O}}^{2-}$and is hence more polarizable.

Let's have a look at the metals now.

Metals on the left side of the periodic table are larger, have lower ionisation energies, and lower electronegativities, whereas metals on the right have greater ionisation energies, higher electronegativities, and smaller metal radius.

Because ${\mathrm{O}}^{2-}$is a smaller ion, it can easily approach massive metal cations on the periodic table's left side. Formed oxides have a high lattice energy and are stable.

The sulfide anion, ${\mathrm{S}}^{2-}$on the other hand, is more polarizable and so likes metals on the left side of the periodic table, which tend to form more covalent bonds.

Therefore, left side metals are oxides and right-side metals are sulfides.