Chromium(iii) sulfate is a transition metal compound containing the metal chromium and the polyatomic ion sulfate.
the oxidation state of chromium in this compound is
, and the chemical formula of the compound is

The oxidation state is +3.

chemical formula of compound is Cr₂(SO₄)₃.

Explanation:

Calculation of oxidation state:

Cr₂(SO₄)₃

Cr= ?

The oxidation state of SO₄ = -2

2x + 3( -2) = 0

2x - 6 = 0

2x = 6

x = 6/2

x= 3

The oxidation state of Cr is +3.

The Cr has valency of 3 so three SO₄ will require while the valency of SO₄ is 2 so two Cr will require.

Cr×2 and SO₄×3

Cr₂(SO₄)₃

we know that oxidation state of Cr is +3 so (3×2) = 6

and the oxidation state of SO₄ is -2 so (-2× 3) = -6

6-6= 0

The overall charge is 0, the compound will be neutral.

a) oxidation number of chromium is +3

b) formula of the compound is Cr2(SO4)3

Explanation:

From the name of the compound, we can see that the oxidation state of chromium is +3

We know the sulphate ion to to be SO4^2-

According to the rules of writing the formula of compounds from valencies, the valency of chromium is 3 while that is the sulphate ion is 2.

The both species can now exchange valency

Cr2(SO4)3 This is the correct formula of the compound