Assign oxidation number to the underlined elements in each of the following species

(a) Let the oxidation no. of P in Na${{H}*{2}}$P${{O}*{4}}$ be x.

- 1 + 2(+ 1) + x + 4(- 2) =0
- 1 + 2 + x-8 =0

x = + 5

Oxidation number of P in Na${{H}*{2}}$P${{O}*{4}}$ = + 5.

(b) Let the oxidation no. of S in NaHS${{O}_{4}}$ be x. - 1 + 1 + x + 4(- 2) =0
- 2 + x - 8=0

x = + 6

Oxidation number of S in NaHS${{O}*{4}}$ = + 6.*{4}}$${{P}

© Let the oxidation number of P in ${{H}*{2}}$${{O}*{7}}$ be x.

4(+ 1) + 2x + 7(-2) =0 - 4 + 2x -14 =0

2x =10

x = + 5

Oxidation number of P in ${{H}*{4}}$${{P}*{2}}$${{O}*{7}}$ = + 5.*{2}}$Mn${{O}_{4}}$ be x.

(d) Let the oxidation no. of Mn in ${{K}

2(+ 1) + x + 4(- 2) =0 - 2 + x-8 =0

x = + 6

Oxidation number of Mn in ${{K}*{2}}$Mn${{O}*{4}}$ = + 6.

(e) Let the oxidation number of oxygen in Ca${{O}_{2}}$ be x. - 2 + 2x =0

[oxidation no. of Ca = + 2]

2x = -2

x =-1

Oxidation number of O in Ca${{O}*{2}}$ =-1.*{4}}$ be x.

(f) Let the oxidation no. of B in NaB${{H} - l+x + 4(-l) =0

[•.• oxidation no. of H = -1]

x-3 =0

x = + 3

Oxidation number of B in NaB${{H}*{4}}$ = + 3.*{2}}$${{S}

(g) Let the Oxidation no. of S in ${{H}*{2}}$${{O}*{7}}$ be x.

2(+1)+ 2x + 7(-2) =0 - 2 + 2x -14 =0

2x =12

x = + 6

Oxidation number of S in ${{H}*{2}}$${{S}*{2}}$${{O}*{7}}$ = + 6.*{4}})}

(h) Let the oxidation no. of S in KAl(S${{({{O}*{2}}$).24${{H}*{2}}$O = x - 1 + 3 + 2x + 8(- 2) + 24(+ 2) + 24(- 2) = 0
- 4 + 2x -16 = 0

[O. No. of K = 1,O. No. of Al = 3]

2x =12

x = +6

Oxidation number of S in KAl(S${{({{O}*{4}})}*{2}}$).24${{H}_{2}}$O=+6