Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Write correctly balanced equation for the following reaction using half reactions:

$bold 1 bold. bold space bold Bi bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold NO subscript bold 3 superscript bold minus bold space bold plus bold space bold H to the power of bold plus bold space bold space bold space bold rightwards arrow bold space bold space bold space bold NO subscript bold 2 bold space bold plus bold space bold Bi to the power of bold 3 bold plus end exponent bold space bold plus bold space bold H subscript bold 2 bold O bold 2 bold. bold space bold I to the power of bold minus bold O subscript bold 2 bold left parenthesis bold g bold right parenthesis bold space bold plus bold space bold H subscript bold 2 bold O bold space bold space bold space bold rightwards arrow bold space bold space bold space bold I subscript bold 2 bold space bold plus bold space bold OH to the power of bold minus bold 3 bold. bold space bold Cu bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold Au to the power of bold plus bold space bold space bold rightwards arrow bold space bold space bold space bold Au bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold Cu to the power of bold 2 bold plus end exponent$

119 Views

72. Define oxidation number. What are the rules for assigning oxidation number to an element?

Oxidation number: The term oxidation number represents the positive or negative character of an atom in a compound. It may be defined as the charge which an atom has or appears to have when present in the combined state with another atom in the formula of a compound or an ion. Oxidation number can be zero, positive, negative or fraction.

Rules for assigning oxidation numbers:
(i) The oxidation number of an element in the free atomic state (Ma, HCl, Fe, Ag) or in its poly-atomic state (P₄, Sg, graphite, H₂, Cl₂, etc.) is always zero.

(ii) The oxidation number of a monatomic ion is the same as the charge on it e.g. oxidation numbers of Na⁺, Mg²⁺ and Al³⁺ are +1, +2 and +3 respectively; oxidation numbers of Cl^-, S^2- and N^3-ions are –1, –2 and –3 respectively.

(iii) In a binary compound, the more electronegative element has negative oxidation number whereas less electronegative element has positive oxidation number

+1 -1 +3 -1
Cl F Br Cl₃

(iv) The oxidation number of hydrogen in its compounds is always +1 except in metallic hydrides (e.g. LiH, NaH, MgH₂) where it is -1.

(v) The oxidation number of oxygen in most compounds is -2. However in peroxides like H₂O₂Na₂O₂, BaO_z etc., the oxidation number of oxygen is -1. In OF₂ the oxidation number is +2 because F is more electronegative than O.

(vi) The oxidation number of F is always -1 in all its compounds.

(vii) The oxidation number of alkali metals i.e. Li, Na, K etc. is always +1 in their compounds and that of alkaline earth metals i.e. Be, Ca, Sr and Ba are always +2 in their compounds.

(viii) The algebraic sum of the oxidation numbers of all atoms in a neutral molecule is zero.

(xi) The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge on the ion.

144 Views

73. Define the term oxidation, reduction, oxidation agent and reducing agent on the basis of oxidation number.

142 Views

Short Answer Type

74.

Consider the elements:
Cs, Ne, I and F.

(a) Identify the element that exhibits only negative oxidation state.

(b) Identify the element that exhibits only positive oxidation state.

(c) Identify the element that exhibits both positive and negative oxidation states.

(d) Identify the element which exhibits neither negative nor positive oxidation state.

142 Views

75. Fluorine reacts with ice and results in the change:
H₂O(s) + F₂(g) → HF(g) + HOF (g)
Justify that this reaction is a redox reaction.

820 Views

76. Justify that the reaction

bold space bold space bold 2 bold Na bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold H subscript bold 2 bold left parenthesis bold g bold right parenthesis bold space bold space bold space bold rightwards arrow bold space bold space bold space bold 2 bold NaH bold space bold left parenthesis bold s bold right parenthesis bold space bold is bold space bold a bold space bold redox bold space bold change bold.

131 Views

77. Justify that the reaction

bold 2 bold Cu subscript bold 2 bold O bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold Cu subscript bold 2 bold S bold left parenthesis bold s bold right parenthesis bold space bold space bold rightwards arrow bold space bold space bold 6 bold Cu bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold SO subscript bold 2 bold left parenthesis bold g bold right parenthesis

is a redox reaction. Identify the species oxidised/ reduced, which acts as an oxidant and which acts as a reductant.

143 Views

78. Identify the oxidising agent, reducing agent and the substance undergoing oxidation and reduction in the following reactions:

bold space bold space bold left parenthesis bold a bold right parenthesis bold space bold H subscript bold 2 bold SO subscript bold 4 bold space bold plus bold space bold 2 bold HBr bold space bold space bold rightwards arrow bold space bold space bold space bold space bold SO subscript bold 2 bold space bold plus bold space bold Br subscript bold 2 bold space bold plus bold space bold 2 bold H subscript bold 2 bold O bold space bold space bold left parenthesis bold b bold right parenthesis bold space bold 3 bold CuO bold space bold plus bold space bold 2 bold NH subscript bold 3 bold space bold space bold rightwards arrow bold space bold space bold space bold space bold 3 bold Cu bold space bold plus bold space bold N subscript bold 2 bold space bold plus bold space bold 2 bold H subscript bold 2 bold O

207 Views

Long Answer Type

79. Identify the substance oxidised, reduced, oxidising 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) + 2 Cu²⁺ (aq) + 5 OH^–(aq) → Cu₂O(s) + HCOO^-(aq) + 3H₂O(l)

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

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

986 Views

80. Identify the oxidising agent, reducing agent and the substance undergoing oxidation and reduction in the following reactions:

bold left parenthesis bold a bold right parenthesis bold space bold space bold space bold 2 bold H subscript bold 2 bold S bold space bold plus bold space bold SO subscript bold 2 bold space bold space bold rightwards arrow bold space bold space bold space bold 3 bold S bold plus bold space bold 2 bold H subscript bold 2 bold O bold left parenthesis bold b bold right parenthesis bold space bold 3 bold MnO subscript bold 2 bold space bold plus bold space bold 4 bold Al bold space bold rightwards arrow bold space bold space bold space bold 3 bold Mn bold space bold plus bold space bold 2 bold Al subscript bold 2 bold O subscript bold 3 bold left parenthesis bold c bold right parenthesis bold space bold 2 bold MnO subscript bold 4 superscript bold minus bold space bold plus bold space bold 10 bold Cl to the power of bold minus bold space bold plus bold 16 bold H to the power of bold plus bold space bold space bold rightwards arrow bold space bold space bold space bold space bold 2 bold Mn to the power of bold 2 bold plus end exponent bold space bold plus bold space bold 5 bold Cl subscript bold 2 bold space bold plus bold space bold 8 bold H subscript bold 2 bold O

134 Views