CBSE Class 12

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsLong Answer Type



(a) Complete the following chemical equation

(i) Cu + HNO3 (dilute) ---> 

(ii) XeF4 + O2F2 -->

(b) Explain the following observation:

(i) Phosphorus has a greater tendency for catenation than nitrogen.

(ii) Oxygen is a gas but sulphur a solid.

(iii) The halogens are coloured. Why?

(i) 3Cu + 8HNO3(dilute) ---->  3Cu (No3)2 + 2NO + 4H2O

(ii) XeF4 + O2F2 ---> XeF6 + O2


(i)Catenation is the property of atoms of an element to link together and in case of nitrogen the size of it being small and due to electron repulsion it can exist stably in diatomic form

Catenation is much more common in phosphorous compounds than in nitrogen compounds. This is because of the relative weakness of the N-N single bond as compared to the P-P single bond. Since nitrogen atom is smaller, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the N-N single bond.

(ii) Oxygen forms O2 which is a gas and sulphur forms S8 which is solid this can be explained as:

Due to the small size of oxygen, it has less tendency for catenation and the high tendency of pp-pp multiple bonds, hence forms stable O2 molecules whereas sulphur because of its higher tendency for catenation and lesser tendency to form pp-pp multiple bonds forms S8 molecules having 8-membered puckered ring. Held together by strong covalent bonds and exist as polyatomic molecule, so it exists solid

 (iii) All the halogens possess a valence shell electronic configuration s2, p5. This means that they contain unpaired electrons in their outermost p- orbital. These electrons absorb light and get promoted to higher orbitals. When they return to their ground state, they emit radiation which falls in the visible region of electromagnetic spectrum. Hence appear.



a) Define the following terms:

(i) Mole fraction

(ii) Ideal solution

(b) 15.0 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution freezes at - 0.34°C. What is the molar mass of the material? (Kf for water = 1.86 K kg mol-1)



(a) Explain the following:

(i) Henry’s law about the dissolution of a gas in a liquid.

(ii) Boiling point elevation constant for a solvent.

(b) A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C. What mass if glycerol was dissolved to make this solution? (Kb for water = 0.512 K kg mol-1)



(a) Write a suitable chemical equation to complete each of the following transformations:

(i) Butan-1-ol to butanoic acid

(ii) 4-methylacetophenone to benzene-1, 4-dicarboxylic acid

(b) An organic compound with molecular formula C9H10O forms 2, 4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the compound.



(a) Give chemical tests to distinguish between

(i) Propanol and propanone

(ii) Benzaldehyde and acetophenone

(b) Arrange the following compounds in an increasing order of their property as indicated:

(i) Acetaldeyde, Acetone, Methyl tert-butyl ketone (reactivity towards HCN)

(ii) Benzoic acid, 3, 4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength).

(iii) CH3CH2CH (Br) COOH, CH3CH (Br) CH2COOH, (CH3)2CHCOOH (acid strength)