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Assign Oxidation Numbers to the Underlined Elements in Each of the Following Species H4p2o7 - Chemistry

Assign oxidation numbers to the underlined elements in each of the following species

H4P2O7

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Solution 1

H4P2O7

Then, we have

4(+1) + 2(x) + 7(-2) = 0

=> 4 + 2x - 14 = 0

=> 2x = +10

=> x= +5

Hence, the oxidation number of P is + 5.

Solution 2

P in `H_4 P_2O_7`

 +1 x -2

`H_4 P_2 O_7`

4(+1) + 2x + 7(-2) = 0

2x - 10 = 10

x = +5

Concept: Oxidation Number - Introduction
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APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 8 Redox Reactions
Q 1.3 | Page 272
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