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QAIPMT-2010-Ch-1992: NEET 2010 Chemistry - Class 12 AIPMT-2010-Ch-1992 Solution: Oxidation states of P in \(H_4P_2O_5\), \(H_4P_2O_6\),…
Oxidation states of P in , , are respectively
Multiple Choice Options
- A +3, +5, +4
- B +5, +3, +4
- C +5, +4, +3
- D +3, +4, +5
Step-by-Step Verified Solution
The oxidation state can be calculated as :
H 4 P 2 O 5 : +4 + 2x + 5(� 2) = 0 _ 2x � 6 = 0 _ x = +3
H 4 P 2 O 6 : +4 + 2x + 6(� 2) = 0 _ 2x � 8 = 0 _ x = +4
H 4 P 2 O 7 : +4 + 2x + 7(� 2) = 0 _ 2x � 10 = 0 _ 2x = 10 _ x = +5