(a) Let the oxidation number of \(P\) be \(x\). Writing the oxidation number of each atom above its symbol, we have,
\(\stackrel{+1}{\mathrm{Na}}\stackrel{+1}{\mathrm{H}_2} \stackrel{x}{\mathrm{P}}\stackrel{-2}{\mathrm{O}_4}\)
Sum of oxidation numbers of various atom in \(\mathrm{NaH}_2 \mathrm{PO}_4=1(+1)+2(+1)+1(\mathrm{x})+4(-2)=\mathrm{x}-5\) But the sum of oxidation number of various atoms in \(\mathrm{NaH}_2 \mathrm{PO}_4\) (neutral) is zero
\(\therefore \quad x-5=0\) or \(x=+5\) Thus, the oxidation number of \(\mathrm{P}\) in \(\mathrm{NaH}_2 \mathrm{PO}_4=+5\).
(b) \(\stackrel{+1}{\mathrm{Na}}\stackrel{+1}{\mathrm{H}}\stackrel{x}{\mathrm{S}}\stackrel{-2}{\mathrm{O}_4}\)
\(1(+1)+1(+1)+x+4(-2)=0\)
or \(x=+6\)
Thus, the oxidation number of \(\mathrm{S}\) in \(\mathrm{NaHSO}_4=+6\).
(c) \(\begin{aligned}
&\stackrel{+1}{\mathrm{H}_4} \stackrel{x}{\mathrm{P}_2} \stackrel{-2}{\mathrm{O}_7} \\
&4(+1)+2(\mathrm{x})+7(-2)=0 \\
&\text { or } \mathrm{x}=+5
\end{aligned}\)
Thus, the oxidation number of \(\mathrm{P}\) in \(\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7=+5\).
(d) \(\stackrel{+1}{\mathrm{K}_2} \stackrel{x}{\mathrm{Mn}}\stackrel{-1}{\mathrm{O}_4}\)
\(2(+1)+1(x)+4(-2)=0\)
or \(x=+6\)
Thus, the oxidation number of \(\mathrm{Mn}\) in \(\mathrm{K}_2 \mathrm{MnO}_4=+6\).
(e) Let the oxidation number of oxygen be \(x\).
Thus, \(\mathrm{+2}{\mathrm{Ca}}\stackrel{x}{\mathrm{O}_2}\)
\(\therefore \quad 2+2 \mathrm{x}=0, \quad \mathrm{x}=-1\)
Thus, oxidation number of oxygen in \(\mathrm{CaO}_2=-1\).
(f) In \(\mathrm{NaBH}_4, \mathrm{H}\) is present as hydride ion. Therefore, its oxidation number is \(-1\). Thus,
\(\stackrel{+1}{\mathrm{Na}}\stackrel{x}{\mathrm{B}}\stackrel{-1}{\mathrm{H}_4}\)
\(1+1(x)+4(-1)=0\) or \(x=+3\)
Thus, the oxidation number of \(\mathrm{B}\) in \(\mathrm{NaBH}_4=+3\).
(g) \(\stackrel{-1}{\mathrm{H}_2} \stackrel{x}{\mathrm{~S}_2} \stackrel{-2}{\mathrm{O}_7}\)
\(\begin{gathered}
2(+1)+2(x)+7(-2)=0 \\
\text { or } x=+6
\end{gathered}\)
Thus, the oxidation number of \(\mathrm{S}\) in \(\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7=+6\).
(h) \(\stackrel{+1}{\mathrm{K}}\stackrel{+3}{\mathrm{Al}}\stackrel{x-2}{\left(\mathrm{SO}_4\right)_2}\stackrel{+1}{ 12\left(\mathrm{H}_2 \mathrm{O}\right)^{-2}}\)
or \(+1+3+2 x-16=2 x-12\)
or \(x=+6\)
Alternatively, since \(\mathrm{H}_2 \mathrm{O}\) is a neutral molecule, therefore, sum of oxidation numbers of all the atoms in \(\mathrm{H}_2 \mathrm{O}\) may be taken as zero. As such water molecules may be ignored while computing the oxidation number of S.
\(\therefore+1+3+2 \mathrm{x}-16=0\) or \(\mathrm{x}=+6\)
Thus, the oxidation number of \(\mathrm{S}\) in \(\mathrm{KAl}\left(\mathrm{SO}_4\right)_2, 12 \mathrm{H}_2 \mathrm{O}=+6\)