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The oxidation number of Cr in [Cr(NH3)4C...

The oxidation number of `Cr` in `[Cr(NH_3)_4Cl_2]^(+)` is:

A

`+3`

B

`+2`

C

`+1`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
A
`a +(4 xx 0) +2 x-1=1 " " therefore a= +3`
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BRILLIANT PUBLICATION-REDOX REACTIONS-LEVEL -I
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