Question
The order of the oxidation state of the phosphorus atom in $${H_3}P{O_2},{H_3}P{O_4},{H_3}P{O_3}\,{\text{and}}\,{{\text{H}}_4}{P_2}{O_6}\,\,{\text{is}}$$
A.
$${H_3}P{O_3} > {H_3}P{O_2} > {H_3}P{O_4} > {H_4}{P_2}{O_6}$$
B.
$${H_3}P{O_4} > {H_3}P{O_2} > {H_3}P{O_3} > {H_4}{P_2}{O_6}$$
C.
$${H_3}P{O_4} > {H_4}{P_2}{O_6} > {H_3}P{O_3} > \,{H_3}P{O_2}$$
D.
$${H_3}P{O_2} > {H_3}P{O_3} > {H_4}{P_2}{O_6} > {H_3}P{O_4}$$
Answer :
$${H_3}P{O_4} > {H_4}{P_2}{O_6} > {H_3}P{O_3} > \,{H_3}P{O_2}$$
Solution :
Let oxidation states of phosphorus in $${H_3}P{O_2},{H_3}P{O_4},{H_3}P{O_3}\,{\text{and}}\,{H_4}{P_2}{O_6}\,\,{\text{be}}\,\,w,x,y,\,\,{\text{and}}\,z\,\,{\text{respectively}}$$
Thus, in $${H_3}P{O_2}$$ :
$$3\, \times \left( { + 1} \right) + w + 2 \times \left( { - 2} \right) = 0\,\,\,\,\therefore w = + 1$$
In $${H_3}P{O_4}:$$
$$3\, \times \left( { + 1} \right) + x + 4 \times \left( { - 2} \right) = 0\,\,\,\,\therefore x = + 5$$
In $${H_3}P{O_3}:$$
$$3\, \times \left( { + 1} \right) + y + 3 \times \left( { - 2} \right) = 0\,\,\,\,\therefore y = + 3$$
In $${H_4}{P_2}{O_6}:\,$$
$$4\, \times \left( { + 1} \right) + 2z + 6 \times \left( { - 2} \right) = 0\,\,\,\,\therefore z = + 4$$
Thus, the order of oxidation state is:
$${H_3}P{O_4} > {H_4}{P_2}{O_6} > {H_3}P{O_3} > \,{H_3}P{O_2}$$