1.Why does  the following reactions occur?

   What conclusion about the compound Na₄XeO₆ (of which Xe   is a part) can be drawn from the reaction.

       It is a redox reaction in which Xenon (ox. no. 8) from Xe   is reduced to XeO₃ [Ox. no. of  Xe = 6] and F is oxidized to F₂ [Ox. no. increases from ^-1 in F^- to O in F₂]

       Sodium Xenate (Na₄XeO₄] is an ionic compound in which (ox. no. 8) of sodium does not undergo a change and it is xenon whose ox. no. changes from +8 to +6.  It is a powerful oxidizing agent which oxidizes even F to F₂.

 2. Consider the reactions :

   2S₂O₃^2-(aq) + I₂(s) → S₄O₆^2-(aq) + 2I^-(aq)

   S₂O₃^2-(aq) + 2Br₂(I) + 5H₂O(I) → 2SO₄^2-(aq) + 4Br^-(aq) + 10H⁺(aq)

   Why does the same reductant, thiosulphate react differently with iodine and bromine?

       S₂O₃^2-(aq) +2Br₂(I)+5H₂O(I)→ 2SO₄^2-(aq)+4Br^-(aq)+10H⁺(aq) the reductant, i.e., S₂O₃^2-(aq) is to be same, but among the halogens Br₂ is a stronger oxidizing agent as compared to I₂ which is a much weaker oxidizing agent. The ox. state of S in SO₄^2- is higher than in S₄O₆^2-.

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

       (a) Let the Ox no. of P be x

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

                         +1 +2 + x - 8 = 0

                                        x = +5

       O.No. of P in NaH₂PO₄ = +5.

       (b) Let the oxidation no. of S be x

              +1 + 1 + x + 4(-2) = 0

                           + 2 + x - 8 = 0

                                           x = +6

       O. No. of S in NaHSO₄ = +6

       (c) Let the oxidation number of P be x.

             4(+1) + 2x + 7(-2) = 0

                        +4 + 2x - 14 = 0

                                         2x = 10

                                            x = +5

       O.No. of P in H₄P₂O₇ = +5.

       (d) Let the Ox. no. of Mn be x

              2(+1) + x + 4(-2) = 0

                         + 2 + x - 8 = 0

                                         x = +6

       O. No. of Mn is K₂MnO₄ = +6.

       (e) Let the oxidation number of oxygen in CaO₂ be x

                              +2 + 2x = 0                               [  Ox. No. of Ca = +2]

                                       2x = -2

                                         x = -1

       O. No. of O in CaO₂ = -1

       (f) Let the Ox No. of B in NaBH₄ be x

                +1 + x + 4(-1) = 0                [ Ox no. of H = -1]

                                x - 3 = 0

                                      x = +3

       O. No. of B in NaBH₄ = +3.

       (g) Let the Ox no. of S in H₂S₂O₇ be x

              2(+1)+2x + 7(-2) = 0

                        +2+2x - 14 = 0

                                       2x = 12

                                          x = +6

       O. No. of S in H₂S₂O₇ = +6

       (h) Let the Ox no. of S in KAl(SO₄)₂.24H₂O = x

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

              +4 + 2x - 16 = 0 [Ox. No. of K = 1, Ox. No. of Al = 3]

                                                2x = 12

                                                  x = +6

         O. No. of S is +6.

4. Identify the substance oxidized, reduced, oxidizing agent and reducing agent for each of the following reactions.

   (a) 2AgBr(s) + C₆H₆O₂(aq)→ 2Ag(s) + 2HBr(aq) + C₆H₄O₂(aq)

   (b) HCHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 3OH^-(aq) → 2Ag(s) + HCOO^-(aq) + 4NH₃(aq) + 2H₂O(l)

   (c) HCHO(l) + 2Cu²⁺(aq) + 5OH^-(aq) → Cu₂O(s) + HCOO^-(aq) +H₂O(l)

   (d) N₂H₄(l)+2H₂O₂(l)→ N₂(g)+4H₂O(l)

   (e) Pb(s) + PbO₂(s) + 2H₂SO₄(aq)→2PbSO₄(s)+2H₂O(l)

       Sol.

       (a) 2AgBr(s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr(aq) + C₆H₄O₂(aq)

             AgBr is reduced to Ag(s) [Removal of Electronegative element Br]

             C₆H₆O₂ is oxidised to C₆H₂O₂ [Removal of hydrogen]

             AgBr is an oxidizing agent & C₆H₆O₂ is a reducing agent.

       (b) HCHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 3OH^-(aq)→ 2Ag(s) + HCOO^-(aq) + 4NH₃(aq) + 2H₂O(l)

             Here HCHO is  oxidized to HCOO^-, whereas [Ag(NH₃)₂]⁺ is reduced to Ag(s)

             HCHO is a reducing agent and [Ag(NH₃)₂]⁺ is an oxidizing agent.

       (c) HCHO(l) + 2Cu²⁺(aq) + 5OH^-(aq) → Cu₂O(s) + HCOO^-(aq) + 3H₂O(l)

             Here HCHO has been oxidized to HCOO^-1

             whereas Cu²⁺(aq) has been reduced to Cu(I) state.

       (d) N₂H₄(I)+2H₂O₂(I)→ N₂(g)+4H₂O(I)

             N₂H₄ has been oxidized to N₂(g)

             H₂O₄ is reduced to H₂O

             H₂O₂ is an oxidizing agent, whereas N₂H₄ is a reducing agent.

       (e) Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(I)

             Pb(s) has been oxidized  to PbSO₄(aq)

             [Increase of ox. no. from O in Pb to +2 in PbSO₄] and PbO₂ has been reduced to PbSO_4.

             [ decrease of ox. no. of Pb from +4 in PbO₂ to +2 in PbSO₄]

               PbO₂ is an oxidizing agent, whereas Pb is a reducing agent.

5. Consider the reactions :

   (a) H₃PO₂(aq) + 4AgNO₃(aq) + 2H₂O(l) → H₃PO₄(aq) + 4Ag(s) + 4HNO₃(aq)

   (b) H₃PO₂(aq) + 2CuSO₄(aq)+2H₂O(l)→ H₃PO₄(aq) + 2Cu(s) + H₂SO₄(aq)

   (c) C₆H₅CHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 3OH^-(aq) → C₆H₅COO^-(aq) + 2Ag(s) + 4NH₃(aq) + 2H₂O(l)

   (d) C₆H₅CHO(l) + 2Cu²⁺(aq) + 5OH^-(aq)→ No change observed.

   What inference do you draw about the behavior of Ag⁺ and Cu²⁺ from the reactions?

       (a) H₃PO₂(aq) +4AgNO₃(aq) +2H₂O(l) → H₃PO₄(aq) + 4Ag(s) + 4HNO₃(aq)

             In this reaction Ag⁺ ions have been reduced to Ag(s).  It is oxidizing H₃PO₂ to H₃PO₄.

       (b) H₃PO₂(aq) + 2CuSO₄(aq)+2H₂O(I)→ H₃PO₄(aq)+2Cu(s) +H₂SO₄(aq)

             In this reaction Cu⁺⁺ ions are reduced to Cu(s)

             Thus it acts as an oxidizing agent and oxidizes H₃PO₂ to H₃PO₄.

       (c) C₆H₅CHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 3OH^-(aq) → C₆H₅COO^-(aq) + 2Ag(s) + 4NH₃(aq) +2H₂O(l)

        In this reactions Ag⁺ present in the complex [Ag(NH₃)₂]⁺ is reduced to Ag(s).  It oxidizes C₆H₅CHO to C₆H₅COO^-1

       (d) C₆H₅CHO(l) + 2Cu²⁺(aq) + 5OH^-(aq)→ No change observed.

             Here Cu²⁺ ions are not in a position to oxidize C₆H₅CHO.

6. How do you count for the following observation?

   (a) Though alkaline potassium permanganate and acidic potassium permanganate are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant.  Why?  Write a balanced redox equation for the reaction.

   (b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCI, but if the mixture contains bromide, then we get red vapours of bromine, Why?

       (a) Toluene, being a non-polar, liquid will not dissolve in aqueous alkaline or acidic KMnO4.  Therefore alcoholic KMnO4 is preferred in the manufacture of benzoic acid from toluene.

       (b) When con.H₂SO₄ is added to an inorganic mixture containing chloride, it gives colourless pungent smelling HCI gas.

             MCI + H₂SO₄ → MHSO₄ + HCI

           But when con.H2SO₄ is added to inorganic mixture containing bromide, we get red vapours of bromine.

                MBr + H₂SO₄  → MHSO₄ + HBr

              2HBr + O → H₂O + Br₂

                        (from air )           (red vapour of bromine)

7. The compound AgF₂ is unstable compound.  However, if formed, the compound acts as a very strong oxidizing agent. Why?

       AgF₂ → Ag + F₂

       Due to the release of F₂(g), AgF₂ will act as a very powerful oxidizing agent.

 8. Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

       Among the halogens best oxidant is F₂ and oxidizing character decreases in the order F₂> CI₂ > Br₂ > I₂

        Therefore F₂(g) + 2e^- → 2F^-(aq)   = +2.87  is maximum among the halogens

   so that F₂ can displace CI^-, Br^-, I^- from their salt solutions to CI₂, Br₂, I₂, respectively.

                             F₂ + 2NaCl(aq) → 2NaF(aq) + Cl₂

                             F₂ + 2NaBr(aq) → 2NaF(aq) + Br₂

                             F₂ + 2NaI(aq) → 2NaF(aq) + I₂

                             CI₂ + 2NaF → No reaction

       i.e., F₂ can be reduced most readily [Therefore  its reduction potential is maximum that is +2.87V]

       Hence it is the strongest oxidizing agent.

   But among the hydrohalic acids HI is the strongest reducing agent.

             The reducing power of halogen acids depends upon the case with which they decompose to give H₂ and X₂.

       This, is turn, depends upon the bond dissociation energy.  Since the bond dissociation energy of the halogen acids increases in the order.

                                        HI < HBr < HCI < HF

        Therefore reducing power of these acids increases in the reverse direction.

                                        HF < HCI < HBr < HI

                                          2HI → H₂ + I₂

9. Whenever a reaction between an oxidizing agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess. Justify this statement giving three illustrations.

       Whenever a reaction between an oxidizing agent and reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidizing agent is in excess. It is clear from the following illustrations:

       (i) (a) Fe₃O₄(s) + 4H₂(g)    → 3Fe(s) + 4H₂O(g)

                 (Iron (II). Iron (III)      (excess)                    Iron

                                oxide)

            (b) CuO + H₂  → Cu + H₂O

              Copper(II) oxide (excess)                ox.no. = 0

            (c) SnCl₂(aq) + HgCl₂(aq)   → Hg₂Cl₂(s) + SnCl₄(aq)

                                                  excess mercury (II) chloride           Mercury (I) Chloride

       (ii) (a) FeCl₂  +  Cl₂       →           FeCl₃

                  Iron (II)Chloride       (excess)                              from(II) Chloride                                 

            (b) 4NH₃(g) + 5O₂(g)   →   4NO(g)  + 6H₂O

                                  excess

            (c) H₂S  +  Br₂             →  2HBr  + S

                            (excess)