Assign the oxidation number to the underlined element in each of the following species:
$1$. $K_2\underline{Mn}O_4$
$2$. $Ca\underline{O}_2$
$3$. $Na\underline{B}H_4$
$4$. $KAl(\underline{S}O_4)_2 \cdot 12H_2O$

(N/A) $1$. $K_2MnO_4$: $2(+1) + Mn + 4(-2) = 0 \implies Mn = +6$
$2$. $CaO_2$: $Ca$ is $+2$,so $2(O) = -2 \implies O = -1$ (Peroxide ion)
$3$. $NaBH_4$: $1(+1) + B + 4(-1) = 0 \implies B = +3$
$4$. $KAl(SO_4)_2 \cdot 12H_2O$: Since $H_2O$ is neutral,$1(+1) + 3 + 2(S) + 8(-2) = 0 \implies 4 + 2S - 16 = 0 \implies 2S = 12 \implies S = +6$