Assign the oxidation number to the underlined element in each of the following species:
$1$. $NaH_{2}\underline{P}O_{4}$
$2$. $NaH\underline{S}O_{4}$
$3$. $H_{4}\underline{P_{2}}O_{7}$

For $NaH_{2}PO_{4}$:
$1(+1) + 2(+1) + P + 4(-2) = 0$
$1 + 2 + P - 8 = 0$
$P = +5$
For $NaHSO_{4}$:
$1(+1) + 1(+1) + S + 4(-2) = 0$
$2 + S - 8 = 0$
$S = +6$
For $H_{4}P_{2}O_{7}$:
$4(+1) + 2(P) + 7(-2) = 0$
$4 + 2P - 14 = 0$
$2P = 10$
$P = +5$