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Question

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

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

The correct option is D +5,+6 and +6
(I) xPO34x+4×(2)=3
x=3+8=+5
x=+5
Oxidation number of P=+5
(II) xSO24x+4×(2)=2
x=2+8
x=+6
Oxidation number of S=+6
(III) xCr2O272x+7×(2)=2
2x=2+14
2x=12
x=122=+6

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