# What is the IUPAC name for the compound "FeS"?

Jul 15, 2016

Iron(II) sulfide.

#### Explanation:

The first thing to recognize here is that you're dealing with an ionic compound made up of iron, $\text{Fe}$, a metal, and sulfur, $\text{S}$, a nonmetal.

Right from the start, you should know that the metal will form a positively charged ion, i.e. a cation, and the nonmetal will form a negatively charged ion, i.e. an anion.

Now, grab a periodic table and look for sulfur. You'll find it located in period 3, group $16$. For main-group elements, the group number gives you the number of valence electrons.

In this case, sulfur has $6$ valence electrons, which means that it needs $2$ more electrons to complete its octet. You can thus say that sulfur will take these two electrons from iron and form $\textcolor{b l u e}{2 -}$ anions.

Keep in mind that anions are always named using the -ide suffix. In this case, the sulfur ion is called the sulfide anion, ${\text{S}}^{2 -}$.

If you take $\textcolor{red}{x +}$ to be the charge on the iron cation, you can say that you have

"FeS" = ["Fe"^(color(red)(x+))]_ color(blue)(2)["S"^(color(blue)(2-))]_ color(red)(x)

Since the chemical formula contains $1$ cation and $1$ anion, you can say for a fact that the positive charge on the iron cations must balance out the negative charge on the sulfur anion.

Therefore, you have

$\textcolor{red}{x +} = 2 +$

This is important because iron is a transition metal, which means that it can form multiple cations. To distinguish between the possible charges on the iron cation, you must use Roman numerals.

In this case, the iron cation has a $2 +$ charge, which means that you'll use the (II) Roman numeral. The name of the compound will thus be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{FeS " -> " iron(II) sulfide}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

It's worth mentioning that this compound is commonly called ferrous sulfide.