Assign oxidation number to the underlined elements in each of the following species:
$(a)$ $NaH_2\underline{P}O_4$
$(b)$ $NaH\underline{S}O_4$
$(c)$ $H_4\underline{P_2}O_7$
$(d)$ $K_2\underline{Mn}O_4$
$(e)$ $Ca\underline{O_2}$
$(f)$ $Na\underline{B}H_4$
$(g)$ $H_2\underline{S_2}O_7$
$(h)$ $KAl(\underline{S}O_4)_2 \cdot 12H_2O$

(N/A) $NaH_2\underline{P}O_4$: Let oxidation number of $P$ be $x$. $1(+1) + 2(+1) + x + 4(-2) = 0$ $\Rightarrow 3 + x - 8 = 0$ $\Rightarrow x = +5$.
$(b)$ $NaH\underline{S}O_4$: Let oxidation number of $S$ be $x$. $1(+1) + 1(+1) + x + 4(-2) = 0$ $\Rightarrow 2 + x - 8 = 0$ $\Rightarrow x = +6$.
$(c)$ $H_4\underline{P_2}O_7$: Let oxidation number of $P$ be $x$. $4(+1) + 2x + 7(-2) = 0$ $\Rightarrow 4 + 2x - 14 = 0$ $\Rightarrow 2x = 10$ $\Rightarrow x = +5$.
$(d)$ $K_2\underline{Mn}O_4$: Let oxidation number of $Mn$ be $x$. $2(+1) + x + 4(-2) = 0$ $\Rightarrow 2 + x - 8 = 0$ $\Rightarrow x = +6$.
$(e)$ $Ca\underline{O_2}$: Let oxidation number of $O$ be $x$. $1(+2) + 2x = 0$ $\Rightarrow 2 + 2x = 0$ $\Rightarrow x = -1$.
$(f)$ $Na\underline{B}H_4$: Let oxidation number of $B$ be $x$. $1(+1) + x + 4(-1) = 0$ $\Rightarrow 1 + x - 4 = 0$ $\Rightarrow x = +3$.
$(g)$ $H_2\underline{S_2}O_7$: Let oxidation number of $S$ be $x$. $2(+1) + 2x + 7(-2) = 0$ $\Rightarrow 2 + 2x - 14 = 0$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$.
$(h)$ $KAl(\underline{S}O_4)_2 \cdot 12H_2O$: Let oxidation number of $S$ be $x$. Ignoring neutral $12H_2O$,$1(+1) + 1(+3) + 2(x + 4(-2)) = 0$ $\Rightarrow 4 + 2x - 16 = 0$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$.