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The oxidation number of molybdenum in its oxo complex species [Mo2O4(C2H4)2(H2O2)]^2- is Explain???

The oxidation number of molybdenum in its oxo complex species [Mo2O4(C2H4)2(H2O2)]^2- is Explain???

Grade:11

4 Answers

Vikas TU
14149 Points
7 years ago
The maxm. CN can be 6.
Here it is 5. Hence oxidation state for Mo would be:
2x +2*-1 + 0 = – 2
x = 0 
Hence it is in 0 state.
RAHUL GURJAR
42 Points
6 years ago
answer of this question
 
The maxm. CN can be 6.
Here it is 5. Hence oxidation state for Mo would be:
2x +2*-1 + 0 = – 2
x = 0 
Hence it is in 0 state.
Swarnim Gupta
11 Points
6 years ago
Since C2H4 and H2O2 are neutal molecules so their oxudation state is 0. Therefore [2x + 4(-2) +0+0 = -2]x = 3.
Mahit james
14 Points
5 years ago
The answer is +3 ignore all the other answers h2o and c2h4 is neutral molecule hence by simple algebra or guessing u can see that the answer comes out to be 3

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