Question

Oxidation states of $P$ in $H_4 \, P_2 \, O_5, H_4, P_2, O_6 , H_4 P_2 O_7$ respectively are

• A. #ERROR!
• B. - 5, + 3 and + 4
• C. #ERROR!
• D. #ERROR!

Answer 1

Answer: (D)

#ERROR!

Answer 2

Key Idea Oxidation state of H is + 1 and that of O is - 2. Let the oxidation state of P in the given compounds isx. In $H_4 P_2 O_5$ $( + 1) \times 4 + 2 \times x + ( - 2) \times 5 = 0$ $4 + 2x - 10 = 0$ $2x = 8$ $\therefore$ $x = + 4$ In $H_4 P_2 O_7$ $( + 1 ) \times 4 + 2 \times x + ( - 2) \times 7 = 0$ $4 + 2x - 14 = 0$ $2x = 10$ $\therefore$ $x = + 5$ Thus, the oxidation states of $P$ in $H_4 P_2 O_6, \, H_4 P_2 O_6$ and $H_4 P_2 O_7$ are $+3, + 4$ and $+5$ respectively