# What is the oxidation number of Cr in Cr2(SO4)3?

Oct 25, 2015

Should be $C {r}^{\text{+3}}$.

#### Explanation:

With finding oxidation numbers, you have to know the common oxidation states of an element. There might be some little algebra involved.

(1) Fast answer: the sulfate ion ($S {O}_{4}^{\text{-2}}$) is quite well-known to those who studied general chemistry. Since ionic bonds between ions are mostly "exchanging superscripts for subscripts", we can already glean that the oxidation state for chromium is +3 .

$C {r}_{2} {\left(S {O}_{4}\right)}_{3}$ = $C {r}^{\text{+3}}$ $S {O}_{4}^{\text{-2}}$

(2) Long answer: Most often than not, the oxidation state of oxygen is -2 (mainly due to it being a member of Group 6 in the periodic table and it being electronegative). Since sulfur and oxygen atoms are enclosed in a parenthesis, this means that they are in covalent bond with each other and are therefore "sharing" electrons. So if the oxygen is negatively charged, the sulfur must be positively charged.

Substituting the values of oxidation states in the formula,

$C {r}_{2} {\left(S {O}_{4}\right)}_{3}$ where O = -2, S = +6 and Cr = x then

$2 x + \left[\left(+ 6\right) \cdot \left(3\right) + \left(- 2\right) \cdot \left(4\right) \cdot \left(3\right)\right] = 0$

$2 x + \left(+ 18\right) + \left(- 24\right) = 0$

$2 x + \left(- 6\right) = 0$

$2 x = + 6 \implies x = + 3$

Therefore, correct answer is $C {r}^{\text{+3}}$

Addendum The element chromium, can only have two oxidation states: $C {r}^{\text{+3}}$ and $C {r}^{\text{+6}}$.