# Question #57194

Feb 12, 2016

Chromium(II) oxide.

#### Explanation:

The first thing to recognize here is that you're dealing with an ionic compound, since you hae a transition metal, chromium, $\text{Cr}$, bonded to a nonmetal, oxygen, $\text{O}$.

The second important thing to recognize here is that chromium, being a transition metal, can have multiple oxidation states, which is another way of saying that it can form multiple cations.

This means that you're going to have to use a Roman numeral to describe its oxidation state in the name of the compound.

So, start with what you know. Oxygen, being located in group 16 of the periodic table, forms $2 -$ anions aclled oxide anions, ${\text{O}}^{2 -}$.

Now, ionic compounds are always neutral. This means that the positive charge on the chromium cation must balance out the negative charge on the oxide anion.

Since the formula unit for this compound, $\text{CrO}$, contains one chromium anion and one oxygen anion, you can say that the charge on the chromium cation must be equal to $2 +$.

When naming ionic compounds, the cation is always named first, followed by the Roman numeral, when needed, and the anion.

In this case, the Roman numeral that must be used is $\text{(II)}$, which means that the name of the compound will be

chromium(II) oxide $\to$ $\text{CrO}$