Question:
The oxidation states of ' $\mathrm{P}^{\prime}$ in $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}, \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{5}$ and $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}$, respectively, are :
Correct Option: , 3
Solution:
Oxidation state of $\mathrm{P}$ in $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}, \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{5}$ and $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}$ is $5,3 \& 4$ respectively
$2 x+4(+1)+7(-2)=0$
$x=+5$
$\mathrm{H}_{4} \underline{\mathrm{P}_{2}} \mathrm{O}_{5}$
$2 x+4(+1)+5(-2)=0$
$x=+3$
$\mathrm{H}_{4} \underline{\mathrm{P}_{2}} \mathrm{O}_{6}$
$2 x+4(+1)+6(-2)=0$
$x=+4$