Question

Determine the oxidation number of chromium (cr) in each of the following compounds. (1) cro2 4 (2) cr2o72− 6 (3) cr2(so4)3 −3

Answer 1

The oxidation number of chromium (Cr) in CrO₂, Cr₂O₇²⁻ and Cr₂(SO₄)₃³⁻ is +4, +6 and +1.5 respectively.

What is oxidation number?

Oxidation number of any compound gives idea about the number of exchangeable electrons from the outer most shell of an atom.

• CrO₂

Let the oxidation state of chromium is x and charge on the whole molecule is zero, so it will be calculated as:

x + 2(-2) = 0

x - 4 = 0

x = +4

• Cr₂O₇²⁻

Let the oxidation state of chromium is x and charge on the whole molecule is -2, so it will be calculated as:

2x + 7(-2) = -2

2x - 14 = -2

2x = 12

x = +6

• Cr₂(SO₄)₃³⁻

Let the oxidation state of chromium is x and charge on the whole molecule is -3, so it will be calculated as:

2x + 3(-2) = -3

2x - 6 = -3

2x = 3

x = +1.5 (which is impossible)

Hence oxidation state of chromium atom in CrO₂, Cr₂O₇²⁻ and Cr₂(SO₄)₃³⁻ is +4, +6 and +1.5 respectively.

To know more about oxidation state, visit the below link:

brainly.com/question/8990767

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