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Oxidation states of P in $H_4P_2O_5$, $H_4P_2O_6$, $H_4P_2O_7$ are respectively
A
+3, +4, +5
B
+3, +5, +4
C
+5, +3, +4
D
+5, +4, +3
Detailed Solution
Take H = +1, O = −2 and let the oxidation state of P be x.
$H_4P_2O_5$: $4(+1) + 2x + 5(-2) = 0$, so $2x = 6$, x = +3
$H_4P_2O_6$: $4(+1) + 2x + 6(-2) = 0$, so $2x = 8$, x = +4
$H_4P_2O_7$: $4(+1) + 2x + 7(-2) = 0$, so $2x = 10$, x = +5
The oxidation states are +3, +4, +5 respectively.
$H_4P_2O_5$: $4(+1) + 2x + 5(-2) = 0$, so $2x = 6$, x = +3
$H_4P_2O_6$: $4(+1) + 2x + 6(-2) = 0$, so $2x = 8$, x = +4
$H_4P_2O_7$: $4(+1) + 2x + 7(-2) = 0$, so $2x = 10$, x = +5
The oxidation states are +3, +4, +5 respectively.