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(a) F. Fluorine being the most electronegative element shows only a -ve oxidation state of -1.

(b) Cs. Alkali metals because of the presence of a single electron in the valence shell, exhibit an oxidation state of +1.

(c) I. Because of the presence of seven electrons in the valence shell, I shows an oxidation state of -1 (in compounds of I with more electropositive elements such as H, Na, K, Ca, etc.) or an oxidation state of +1 compounds of I with more electronegative elements, i.e., O, F, etc.) and because of the presence of d-orbitals it also exhibits +ve oxidation states of +3, +5 and +7.

(d) Ne. It is an inert gas (with high ionization enthalpy and high positive electron gain enthalpy) and hence it neither exhibits -ve nor +ve oxidation states.

The unbalanced chemical reaction is:
Mn³⁺ (aq) → Mn²⁺ (aq) + MnO₂? (s) + H⁺ (aq)
The oxidation half reaction is,  
Mn³⁺ (aq) → MnO₂? (s).
To balance oxidation number, one electron is added on R.H.S.
Mn³⁺ (aq) → MnO₂? (s) + e⁻
4 protons are added to balance the charge.
Mn³⁺ (aq) → MnO₂? (s) + 4H⁺ (aq) + e⁻
2 water molecules are added to balance O atoms.
The reduction half reaction is Mn³⁺ (aq) → Mn²⁺ (aq).
An electron is added to balance oxidation number.
Mn³⁺ (aq) + e⁻ → Mn²⁺ (aq)
Two half-cell reactions are added to obtain balanced chemical equation.
2Mn³⁺ (aq) + 2H₂? O (l) → Mn2+ (aq) + MnO₂? (s) + 4H⁺ (aq)

Let x be the O.N. of C.
O.N. of C in cyanogen, (CN)₂ = 2 (x – 3) = 0 or x = +3

O.N. of C in cyanide ion, CN^- = x – 3 = -1 or x = +2

O.N. of C in cyanate ion, CNO =x-3-2 = -l or x: = +4

The four information about the reaction are:
(i) The reaction involves decomposition of cyanogen, (CN)₂ in the alkaline medium to cyanide ion, CN and cyanate ion, CNO^–.

(ii) The O.N. of C decreases from +3 in (CN)₂ to +2 in CN^–ion and increases from +3 in (CN)₂ to +4 in CNO^– ion. Thus, cyanogen is simultaneously reduced to cyanide ion and oxidised to cyanate ion.

(iii) It is an example of a redox reaction in general and a disproportionation reaction in particular.

(iv) Cyanogen is a pseudohalogen (behaves like halogens) while cyanide ion is a pseudohalide ion (behaves like halide ion).

(a) Oxidation number method:
The oxidation number of P decreases from 0 to -3 and increases from 0 to +2. Hence,  P₄?  is oxidizing as well as reducing agent.
During reduction, the total decrease in the oxidation number for 4 P atoms is 12.
During oxidation, total increase in the oxidation number for 4 P atoms is 4.

The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying H_2? PO₂⁻? with 3. 

P₄? (s) + OH⁻ (aq) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

To balance O atoms, multiply OH⁻ ions by 6.

P₄? (s) + 6OH⁻ (aq) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

To balance H atoms, 3 water molecules are added to L.H.S and 3 hydroxide ions on R.H.S.

P₄? (s) + 6OH⁻ (aq) + 3H₂? O (l) → PH₃? (g) + 3H₂? PO₂⁻? (aq) + 3OH⁻ (aq)
Subtract 3 hydroxide ions from both sides.

P_4? (s) + 3OH⁻ (aq) + 3H₂? O (l) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

Ion electron method:
The oxidation half reaction is 

P₄? (s) → H₂? PO₂⁻? (aq).
The P atom is balanced.

P₄? (s) → 4H₂? PO₂⁻? (aq)

The oxidation number is balanced by adding 4 electrons on RHS.

P₄? (s) → 4H₂? PO₂⁻? (aq) + 4e⁻

The charge is balanced by adding 8 hydroxide ions on LHS.

P₄? (s)+8OH⁻ (aq)→4H₂? PO₂⁻? (aq)
The O and H atoms are balanced.

The reduction half reaction is P₄? (s)→PH₃? (g).
The oxidation number is balanced by adding 12 electrons on LHS.

P_4? (s)+12e⁻→PH₃? (g)
The charge is balanced by adding 12 hydroxide ions on RHS.

P₄? (s)+12e⁻→PH_3? (g)+12OH⁻
The oxidation half reaction is multiplied by 3 and the reduction half reaction is multiplied by 2.
The half reactions are then added to obtain balanced chemical equation.

P_4? (s)+3OH⁻ (aq)+3H₂? O (l)→PH₃? (g)+3H₂? PO₂⁻? (aq)

(b) The oxidation number of N increases from – 2 in N₂H₄ to + 2 in NO and the oxidation number of Cl decreases from + 5 in CIO⁻₃ to – 1 in Cl^–. Hence, in this reaction,  N₂H₄ is the reducing agent and CIO⁻₃ is the oxidizing agent.
Ion - electron method:
The oxidation half equation is:

−2               +2

N₂H₄ (l)? NO (g)
The N atoms are balanced as:

N₂H₄ (l)?2NO (g)+8e⁻
The charge is balanced by adding 8 OH^– ions as:

N₂H₄ (I)+8OH⁻ (aq)?2NO (g)+8e⁻
The O atoms are balanced by adding 6H2O as:

N₂H₄ (I)+8OH⁻ (aq)?2NO (g)+6H₂O (I)+8e⁻. (i)
The reduction half equation is:

+5            

(a) The balanced half reaction equations are:
Oxidation half equation:

I⁻ (aq) → I₂ (s)       - (i)

Reduction half reaction equation:

MnO₄⁻ (aq) → MnO₂ (aq) - (ii)

Balance I atoms and charges in the oxidation half reaction.

2I⁻ (aq) → I₂ (s) + 2e⁻

In the reduction half reaction, the oxidation number of Mn changes from +7 to +4. Hence, add 3 electrons to reactant side of the reaction.

MnO₄⁻ (aq) + 3e⁻→ MnO₂ (aq)

Balance charge in the reduction half reaction by adding 4 hydroxide ions to product side.

MnO4⁻ (aq) + 3e⁻ → MnO₂ (aq)+4OH⁻

To balance O atoms, add 2 water molecules to reactant side.

MnO4⁻ (aq)+ 3e⁻ +2H₂O→MnO₂ (aq) + 4OH⁻

To equalize the number of electrons, multiply the oxidation half reaction by 3 and multiply the reduction half reaction by 2.

6I⁻ (aq) → 3I₂ (s)+6e⁻

2MnO₄⁻ (aq) + 6e⁻+4H₂O → 2MnO₂ (aq)+8OH⁻

Add two half-cell reactions to obtain the balanced equation.

2MnO₄⁻? (aq) + 6I⁻ (aq) + 4H₂? O₂? (l) → 2MnO₂? (s) + 3I₂? (s) + 8OH⁻

(b) The balanced half reaction equations are:
Oxidation half equation:

SO₂ (g) + 2H₂O (l) → HS0₄^–  (aq) + 3H⁺ (aq) +2e^–                        … (i)
Reduction half equation:

MnO₄^– (aq) + 8H⁺ (aq) + 5e^– → Mn²⁺ (aq) + 4H₂O (l) ………. (ii)
Multiply Eq. (i) by 3 and Eq. (ii) by 2 and add, we have,

2MnO₄^– (aq) + 5S0₂ (g) + 2H₂0 (l) + H⁺ (aq) → 2Mn²⁺ (aq) + 5HSO₄^– (aq)

(c) Oxidation half equation: Fe²⁺ (aq) → Fe³⁺ (aq) + e^– … (i)
Reduction half equation: H₂O₂ (aq) + 2H⁺ (aq) + 2e^– → 2H₂O (l) … (ii)
Multiply Eq. (i) by 2 and add it to Eq. (ii), we have,

H₂O₂ (aq) +2Fe²⁺ (aq) +2H⁺ (aq) → 2Fe³⁺ (aq) + 2H₂O (l)

(d)Oxidation half equation:

SO₂ (g) + 2H₂O (l) → SO₄^2- (aq) + 4H⁺ (aq) + 2 e^–  … (i)
Reduction half equation:

Cr₂O₇^2– (aq) + 14H⁺ (aq) + 6e^– → 2Cr³⁺ (aq) + 7H₂O (l) … (ii)
Multiply Eq. (i) by 3 and add it to Eq. (ii), we have,

Cr₂O₇^2– (aq) + 3SO₂ (q) + 2H⁺ (aq) → 2Cr³⁺ (aq) + 3SO₄^2- (aq) + H₂O (l)