Q69 / 200 · NEET 2010 (Screening) · Chemistry
Oxidation states of $\mathrm{P}$ in $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_5, \mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6$, $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7$, are respectively
- A$+3,+5,+4$
- B$+5,+3,+4$
- C$+5,+4,+3$
- Correct answerD$+3,+4,+5$
Solution
Key Idea Oxidation state of $\mathrm{H}$ is +1 and that of $\mathrm{O}$ is -2.
Let the oxidation state of $\mathrm{P}$ in the given compounds is $\mathrm{x}$.
In $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_5$,
$\begin{aligned}
(+1) \times 4+2 \times x+(-2) \times 5 & =0 \\
4+2 x-10 & =0 \\
2 x & =6 \\
\therefore x & =+3
\end{aligned}$
$\begin{array}{l}
\text {In } \mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6, \\
(+1) \times 4+2 \times \mathrm{x}+(-2) \times 6 =0 \\
4+2 \mathrm{x}-12 =0 \\
2 \mathrm{x} =8 \\
\therefore \mathrm{x} =+4
\end{array}$
In $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7$
$\begin{aligned}
(+1) \times 4+2 \times x+(-2) \times 7 & =0 \\
4+2 x-14 & =0 \\
2 x & =10 \\
\mathrm{x} & =+5
\end{aligned}$
Thus, the oxidation states of $\mathrm{P}$ in $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_5$, $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6$ and $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7$ are $+3,+4$ and +5 respectively.
Hence, the correct answer is option: 'D'.