Answer
Please see the work below.
Work Step by Step
The oxidation number of hydrogen in all cases except in hydrides is +1.
The oxidation number of oxygen in all cases except in peroxides and superoxides is -2.
Thus in below oxoacids of phosphorus oxidation number of ‘P’ is may not the same. We can arbitrarily take it as ‘x’.
a)
$HPO_{3}$
$1 + x +(3\times -2) = 0$
Then x = +5
Hence oxidation number of Phsphorus in $HPO_{3}$ = +5
b)
$H_{3}PO_{2}$
$(3\times 1) + x +(2\times-2) = 0$
Then x = +1
Hence oxidation number of Phsphorus in $H_{3}PO_{2}$ = +1
c)
$H_{3}PO_{3}$
$(3\times 1) + x +(3\times-2) = 0$
Then x = +3
Hence oxidation number of Phsphorus in $H_{3}PO_{3}$ = +3
d)
$H_{3}PO_{4}$
$(3\times 1) + x +(4\times-2) = 0$
Then x = +5
Hence oxidation number of Phsphorus in $H_{3}PO_{4}$ = +5
e)
$H_{4}P_{2}O_{7}$
$(4\times 1) +(2 \times x) +(7 \times-2) = 0$
Then x = +5
Hence oxidation number of Phsphorus in $H_{4}P_{2}O_{7}$ = +5
f)
$H_{5}P_{3}O_{10}$
$(5\times 1) +(3\times x) +(10\times-2) = 0$
Then x = +5
Hence oxidation number of Phsphorus in $H_{5}P_{3}O_{10}$ = +5
