(a) NaH${}_2$PO4

Let the oxidation number of P be x.
We know that,
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = –2

$\Rightarrow$$\stackrel{+1}{\mathrm{Na}}{\stackrel{+1}{\mathrm{H}}}_2\stackrel{\mathrm{x}}{\mathrm{P}}{\stackrel{-2}{\mathrm{O}}}_4$

Then, we have

1(+1)+2(+1)+1(x)+4(-2)=0

$\Rightarrow$1+2+x-8=0

$\Rightarrow$x=+5

Hence, the oxidation number of P is +5.

(b) $\mathrm{NaH}\underline{\mathrm{S}}{\mathrm{O}}_4$

$\stackrel{+1}{\mathrm{Na}}\stackrel{+1}{\mathrm{H}}\stackrel{\mathrm{x}}{\mathrm{S}}\stackrel{-2}{O_4}$

Then, we have

1(+1)+1(+1)+1(x)+4(-2)=0

$\Rightarrow$1+1+x-8=0

$\Rightarrow$x=+6

Hence, the oxidation number of S is + 6.

(c) ${\mathrm{H}}_4{\underline{\mathrm{P}}}_2{\mathrm{O}}_7$

${\stackrel{+1}{\mathrm{H}}}_4{\stackrel{x}{P}}_2{\stackrel{-2}{O}}_7$

Then, we have

4(+1)+2(x)+7(-2)=0

$\Rightarrow$4+2 x-14=0

$\Rightarrow$2x=+10

$\Rightarrow$x=+5

Hence, the oxidation number of P is + 5.

(d) ${\mathrm{K}}_2\underline{\mathrm{M}}{\mathrm{nO}}_4$

${\stackrel{+1}{\mathrm{K}}}_2\stackrel{\mathrm{x}}{\mathrm{M}}\mathrm{n}{\stackrel{-2}{\mathrm{O}}}_4$

Then, we have

2(+1)+x+4(-2)=0

$\Rightarrow$2+x-8=0

$\Rightarrow$x=+6

Hence, the oxidation number of Mn is + 6.

(e) $\mathrm{Ca}{\underline{\mathrm{O}}}_2$

$\stackrel{+2}{\mathrm{Ca}}{\stackrel{\mathrm{x}}{\mathrm{O}}}_2$

Then, we have

(+2)+2(x)=0

$\Rightarrow$2+2 x=0

$\Rightarrow$x=-1

Hence, the oxidation number of O is – 1.

(f) $\mathrm{Na}\underline{\mathrm{B}}{\mathrm{H}}_4$

$\stackrel{+1}{\mathrm{Na}}\stackrel{\mathrm{x}}{\mathrm{B}}{\stackrel{-1}{\mathrm{H}}}_4$

Then, we have

1(+1)+1(x)+4(-1)=0

$\Rightarrow$1+x-4=0

$\Rightarrow$x=+3

Hence, the oxidation number of B is + 3.

(g) ${\mathrm{H}}_2{\underline{\mathrm{S}}}_2{\mathrm{O}}_7$

${\stackrel{+1}{\mathrm{H}}}_2{\stackrel{\mathrm{x}}{\mathrm{S}}}_2{\stackrel{-2}{\mathrm{O}}}_7$

Then, we have

2(+1)+2(x)+7(-2)=0

$\Rightarrow$2+2x-14=0

$\Rightarrow$2x=12

$\Rightarrow$x=+6

Hence, the oxidation number of S is + 6.

(h) $\mathrm{KAl}{\left(\underline{\mathrm{S}}{\mathrm{O}}_4\right)}_2·12{\mathrm{H}}_2\mathrm{O}$

$\stackrel{+1}{\mathrm{K}}\stackrel{3+}{ \mathrm{Al}}{\left(\stackrel{\mathrm{x}}{\mathrm{S}}{\stackrel{2-}{\mathrm{O}}}_4\right)}_2·12{\stackrel{+1}{\mathrm{H}}}_2\stackrel{-2}{\mathrm{O}}$

Then, we have

1(+1)+1(+3)+2(x)+8(-2)+24(+1)+12(-2)=0

$\Rightarrow$1+3+2 x-16+24-24=0

$\Rightarrow$2x=12

$\Rightarrow$x=+6

Or,
We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the water molecule, we have

1(+1)+1(+3)+2(x)+8(-2)=0

$\Rightarrow$1+3+2 x-16=0

$\Rightarrow$2x=12

$\Rightarrow$x=+6

Hence, the oxidation number of S is + 6.