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Sagot :
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:
https://brainly.com/question/8990767
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