Answer
d.
Work Step by Step
(a) In iron (II) oxide, we see that this particular atom of iron (Fe) has an oxidation of $2+$ by looking at the Roman numeral next to it. We know that oxygen (O) has an oxidation number of $2-$. In this compound, iron and oxygen combine in a one-to-one ratio, which means we have one iron atom for every one oxygen atom. The formula for this compound should be $FeO$; therefore, the formula given is correct.
(b) For titanium (IV) sulfate, we see that this particular atom of titanium (Ti) has an oxidation number of $4+$, and sulfate (SO_4) has an oxidation number of $2-$. This means that we need to have two titanium atoms for every four sulfates; we reduce this ratio to the lowest whole number ratio, which means we need one titanium atom for every two sulfate ions. Therefore, the formula for this compound should be $Ti(SO_4)_2$; therefore, the formula given is correct.
(c) In cobalt (II) chloride, this particular atom of cobalt (Co) has an oxidation number of $2+$, and chlorine (Cl) has an oxidation number of $1-$. Cobalt and chlorine combine in a one-to-two ratio, which means we will need one atom of cobalt for every two atoms of chlorine. Therefore, the formula for this compound should be $CoCl_2$; therefore, the formula given is correct.
(d) For vanadium (IV) oxide, we see that this particular atom of vanadium (V) has an oxidation number of $4+$, and oxygen (O) has an oxidation number of $2-$. This particular atom of vanadium combines with oxygen in a two-to-four ratio; we reduce this ratio to its lowest whole number equivalent, meaning we need one vanadium atom for every two oxygen atoms. Therefore, the formula for this compound should be $VO_2$; therefore, the formula given is incorrect.
