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By stoichiometry the composition of bleaching powder is Ca²⁺(OCl- )Cl^- . Here O.N. of Cl in OCl is +1 while that in Cl is -1 .So that the average of O.N. is zero

Mn has +7 O.S. which is maximum O.S. so MnO₄ ^- can undergo reduction only while MnO₄ ^2- has +6 O.S. of Mn so it can undergo oxidation as well as reduction.

P₄(s) + 3OH^– (aq) ----------> 3H₂O(l) PH₃(g) + 3H₂PO₂ ^- .

 (i) Valency of an element cannot not be zero but oxidation number can be. 

(ii) Valency is just a number but oxidation number can be negative or positive. 

(iii) Valency is the combining capacity whereas oxidation number refers to charge which an atom carry when present in the combined state.

In MnO₄^- , Mn is in the highest oxidation state i.e. +7. Therefore, it does not undergo disproportionation. MnO₄^2– undergoes disproportionation as follows :

3MnO₄^2– + 4H⁺ → 2MnO₄^– + MnO₂ + 2H₂O

Among the Oxoanions of chlorine listed above, ClO₄ ^– does not dis proportionate because in this Oxo anion chlorine is present in its highest oxidation state that is, +7. 

The disproportionation reactions for the other three Oxo anions of chlorine are as follows: 

3ClO^– 2Cl^– --------->ClO^– ₃ 

6 ClO2^– -----------> 4ClO3^– + 2Cl^– 

4ClO^– ₃ ---------> Cl^– + 3 ClO₄ ^-

a.  HPO₃^2- = (+1)+x+3(-2) =-2

= +1+x-6=-2 

X=+3

b. PO₄^3- = x+4(-2)=-3

= x-8=-3

X = +5

Cu₂O helps sulphur in Cu₂S to increase its oxidation number, therefore, Cu(I) is an oxidant; and sulphur of Cu₂S helps copper both in Cu₂S itself and Cu₂Oto decrease its oxidation number; therefore, sulphur of Cu₂S is reductant.

(a) Fe₂O₃ =  Fe₂(III)O₃ 

(b) CuO =  Cu(II) 

(c) MnO =  Mn(II)O 

(d) MnO2 =  Mn(IV)O₂

I reacts with Cl₂ to form I₂ which form blue coloured complex with starch.

 As per rule, the oxidation state of Fe comes out to be 8/3 Which is not possible. So,here the oxidation state of Fe is +2 & +3 since it is mixed oxide of FeO and Fe₂O₃

(a) Nickel(II) Sulphate = Ni(II)SO₄ 

(b) Tin(IV)Oxide = Sn(IV)O₂ 

(c) Thallium(I)Sulphate = Tl₂(I)SO₄ 

(d) Iron(III)Sulphate = Fe₂(III)(SO₄)₃.

Cu gets dissolved indil HNO₃ due to oxidation of Cu by NO₃ - ions.

A chemical reaction in which both oxidation and reduction takes place simultaneously is called redox reaction.

Zn_(s) + Cu⁺² (aq) → Zn⁺² _(aq)  + Cu_(s) 

+2  is the ON of oxygen in OF₂

-1  is the ON of hydrogen in hydrides

-1   is the ON of oxygen in peroxides.

During Oxidation ON of an element increases. During reduction ON of an element decreases.

Consider the half reaction, 

2H⁺(aq) + 2e^– → H₂(g); E° = 0.0V

Au³⁺(aq) + 3e^– → Au(s); E° = 1.50V

Since E° (1.50V) for (Au³⁺/Au) is higher than that H⁺/-\(\frac{1}{2}\)H₂ (0.0V), therefore, Au³⁺ can be more easily reduced than H⁺ ions Au³⁺ can be reduced to Au metal by H₂ but H⁺ cannot oxidize metallic gold to Au³⁺ ions. In other words, metallic gold does not dissolve in 1M HCl instead H₂ gas can reduce gold salt to metallic gold.

The cell which converts chemical energy into electrical energy . It is called as galvanic cell and Voltaic cell after the name of scientists who first tried to convert chemical energy into electrical energy .

MnO₂ and BrO₃ ^- are formed.

Ox. state of Ni in Ni(CO)₄ is 

x + 4(0) = 0

x = 0

Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belongs entirely to more electronegative element.

Fe³⁺ is strongest oxidising agent because it has standard reducing potential. (i.e., +ve value) value.

(iii) A is true but R is false.

(c)A is true but R is False

(a) Potassium is undergoing oxidation

(d) None of the species are undergoing oxidation or reduction

Tellurium (Te), semimetallic chemical element in the oxygen group, closely allied with the element selenium in chemical and physical properties. 

Possible maximum and minimum oxidation states of tellurium are −2, +2, +4, +6

The correct option is (ii) NH₄NO₃

(i) Bromine is a stronger oxidant than iodine.

(i) Oxidation number of hydrogen is -1 in metal hydrides like NaH.

(a)Both A and R are true and R is the correct explanation of the A.

(a) Both A and R are true and R is the correct explanation of the A.

(ii) Both A and R are true but R is not the correct explanation of A.

2I^–(aq) + Cl₂(g) → 2Cl^–(aq) + I₂(s) 

Oxidant is Cl₂(g) reductant is I^–(aq)

Zn is reducing agent (reductant) whereas H⁺ is oxidizing agent (oxidant).

SnCl₂ is a reductant and HgCl₂ is an oxidant.

Strongest reducing agent is Zn and weakest reducing agent is Ag.

Au, Ag, Cu, Fe, Zn, Ca 

This is based on standard reduction potential. From Ca to Au value of E⁰ cell goes on increasing. Greater the E⁰ value lesser is the reactivity of a metal.