mno4 calculate oxidation number
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Let oxidation number of Mn be x.
Net charge on molecule = 0.
Thus x - (2*4) = 0.
Thus x=8, but valenve electrons of Mn are 7, thus maximum ON of Mn can be 7, thus I think the compound is MnO4- with ON of Mn=7
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Question :
Calculate the oxidation number of Mn in MnO₄⁻ .
Answer:
+ 7
Explanation:
Let O.N. of Mn be y in MnO₄⁻ .
As we know :
O.N. of O = - 2
Here net change on compound is - 1 .
= > y + ( - 2 × 4 ) = - 1
= > y - 8 = - 1
= > y = - 1 + 8
= > y = + 7
Therefore , O.N. of Mn in MnO₄⁻ is + 7 .
# In d-block element Mn as highest number of oxidation state i.e. + 7 .
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