Oxidation states of P in H 4 P 2 O 5- H 4 P 2 O 6- H 4 P 2 O 7 are respectively-A- -3-5-4B- -5-3-4C- -3-4-5D- -5-4-3

The correct option is C +3, +4, +5Let oxidation number of P be Z 1. In H_4P_2O_5, (4 × 1)+(2 × Z)+(5 × 2)=0, x=+3 2. In H_4P_2O_6, (4 × 1)+(2 × Z)+(6 × 2)=0 ...

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Standard XII

Chemistry

Stoichiometry

Oxidation sta...

Question

Oxidation states of P in $H_{4} P_{2} O_{5}, H_{4} P_{2} O_{6}, H_{4} P_{2} O_{7}$ are respectively:

+3, +5, +4

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+5, +3, + 4

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+5, +4, +3

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+3, +4, +5

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Solution

The correct option is D +3, +4, +5
Let oxidation number of P be Z
1. In $H_{4} P_{2} O_{5}, (4 \times 1) + (2 \times Z) + (5 \times - 2) = 0, x = + 3$
2. In $H_{4} P_{2} O_{6}, (4 \times 1) + (2 \times Z) + (6 \times - 2) = 0, x = + 4$
3. In $H_{4} P_{2} O_{7}, (4 \times 1) + (2 \times Z) + (7 \times - 2) = 0, x = + 5$

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Oxidation states of P in H4P2O5,H4P2O6,H4P2O7 are respectively:

The correct option is D +3, +4, +5Let oxidation number of P be Z 1. In H4P2O5,(4×1)+(2×Z)+(5×−2)=0, x=+3 2. In H4P2O6,(4×1)+(2×Z)+(6×−2)=0, x=+4 3. In H4P2O7,(4×1)+(2×Z)+(7×−2)=0, x=+5

Oxidation states of P in $H_{4} P_{2} O_{5}, H_{4} P_{2} O_{6}, H_{4} P_{2} O_{7}$ are respectively: