Let the oxidation no. of underlined element in all the given compounds = x (a) $\overset{+1}{\mathrm{Na}}\overset{+1}{\mathrm{H}_2}\overset{x}{\mathrm{P}}\overset{-2}{\mathrm{O}_4}$ Since the sum of oxidation number of various atoms in $\mathrm{NaH_2PO_4}$ (neutral)is zero. 1(+1) + 2(+1) + (x) + 4(–2) = 0 or x = +5 Thus, the oxidation number of P in $\mathrm{NaH_2PO_4}$ = +5. (b) In $\overset{+1}{\mathrm{Na}}\overset{+1}{\mathrm{H}}\overset{-2}{\mathrm{SO}_4}$ : 1(+1) + 1(+1) +x + 4 (–2)= 0 or x = + 6 Thus, the oxidation number of S in $\mathrm{NaHSO_4}$ = +6. (c) In $\overset{+1}{\mathrm{H}_4}\overset{x}{\mathrm{P}_2}\overset{-2}{\mathrm{O}_7}$ : 4(+1) + 2(x) + 7(–2) = 0 or x = +5 Thus, the oxidation number of P in $\mathrm{H_4P_2O_7}$ = +5. (d) In $\overset{+1}{\mathrm{K}_2}\overset{x}{\mathrm{Mn}}\overset{-2}{\mathrm{O}_4}$ : 2(+1) +1(x) + 4(–2) = 0 or x = +6 Thus, the oxidation number of Mn in $\mathrm{K_2MnO_4}$ = +6. (e) In $\overset{+2}{\mathrm{Ca}}\overset{x}{\mathrm{O}_2}$ : 2 + 2x = 0 or x = –1 Thus, the oxidation number of oxygen in $\mathrm{CaO_2}$ = –1. (f) In $\mathrm{NaBH_4}$, hydrogen is present as hydride ion. Therefore, its oxidation number is –1. Thus, In $\overset{+1}{\mathrm{Na}}\overset{x}{\mathrm{B}}\overset{-1}{\mathrm{H}_4}$ : 1(+1) + x + 4(–1) = 0 or x = +3 Thus, the oxidation number of B in $\mathrm{NaBH_4}$ = +3 (g) In $\overset{+1}{\mathrm{H}_2}\overset{x}{\mathrm{S}_2}\overset{-2}{\mathrm{O}_7}$ : 2(+1)+2(x) + 7(–2) = 0 or x = +6 Thus, the oxidation number of S in $\mathrm{H_2S_2O_7}$ = +6. (h) In $\overset{+1}{\mathrm{K}}\overset{+3}{\mathrm{Al}}\left(\overset{x}{\mathrm{S}}\overset{-2}{\mathrm{O}_4}\right)_2 \cdot 12\mathrm{H_2O}$ : +1 + 3 + 2x + 8(–2) + 12 × 0 = 0 or x = +6 Thus, the oxidation number of S in $\mathrm{KAl(SO_4)_2} \cdot 12\mathrm{H_2O}$ = +6