Question
Oxidation states of $$P$$ in $${H_4}{P_2}{O_5},{H_4}{P_2}{O_6}$$ and $$\,{H_4}{P_2}{O_7},$$ respectively are
A.
$$+3,+ 5$$ and $$+ 4$$
B.
$$+5,+ 3$$ and $$+4$$
C.
$$+5, + 4$$ and $$+3$$
D.
$$+3, + 4$$ and $$+5$$
Answer :
$$+3, + 4$$ and $$+5$$
Solution :
Oxidation state of $$H$$ is + 1 and that of $$O$$ is - 2.
Let the oxidation state of $$P$$ in the given compounds is $$x.$$
$$\eqalign{
& {\text{In}}\,\,{H_4}{P_2}{O_5} \cr
& \left( { + 1} \right) \times 4 + 2 \times x + \left( { - 2} \right) \times 5 = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 + 2x - 10 = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x = 6 \cr
& \therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = + 3 \cr
& {\text{In}}\,\,{H_4}{P_2}{O_6} \cr
& \left( { + 1} \right) \times 4 + 2 \times x + \left( { - 2} \right) \times 6 = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 + 2x - 12 = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x = 8 \cr
& \therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = + 4 \cr
& {\text{In}}\,\,{H_4}{P_2}{O_7} \cr
& \left( { + 1} \right) \times 4 + 2 \times x + \left( { - 2} \right) \times 7 = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 + 2x - 14 = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x = 10 \cr
& \therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = + 5 \cr} $$
Thus, the oxidation states of $$P$$ in $${H_4}{P_2}{O_5},{H_4}{P_2}{O_6}$$ and $${H_4}{P_2}{O_7}$$ are + 3, + 4 and + 5 respectively.