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Question

Oxidation numbers of P in PO34 of S in SO24 and that of Cr in Cr2O27 are respectively:

A
+5, +5, +6
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B
+3, +6, +6
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C
+5, +6, +6
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D
+3, +6, +5
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Solution

The correct option is C +5, +6, +6
(I) PO34
x+4×(2)=3
x=3+8=+5
x=+5
Oxidation number of P = +5
(II) SO24
x+4×(2)=2
x=2+8
x=+6
Oxidation number of S = +6
(III) Cr2O27
2x+7×(2)=2
2x=2+14
2x=12
x==+6
Oxidation number of Cr = +6
Hence, option (d) is correct answer.

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