Subject

Chemistry

Class

CBSE Class 12

Pre Boards

Practice to excel and get familiar with the paper pattern and the type of questions. Check you answers with answer keys provided.

Sample Papers

Download the PDF Sample Papers Free for off line practice and view the Solutions online.
Advertisement

 Multiple Choice QuestionsShort Answer Type

Advertisement

11.

Give reasons for the following:
(i) N2 is less reactive at room temperature.

(ii) H2Te is the strongest reducing agent amongst all the hydrides of Group 16 elements.

(iii) Helium is used in diving apparatus as a diluent for oxygen.


Two nitrogen atoms are joined by triple bonds. The nitrogen atom is very small, therefore the bond length is also quite small (109.8pm) and as the result, the bond dissociation energy is quite high (946Kj/mol) Therefore, N2 is less reactive at room temperature.

H2Te is the strongest reducing agent among the hydrides of group 16. The size of Te is very large due to which the bonding between hydrogen and Te is not strong. On the other hand, the electronegativity of Te is very less. So it will easily loose hydrogen. As the size of the elements increases in the order O < S < Se < Te, thus bond strength decreases from H2O to H2Te and therefore, the bond dissociation enthalpy decreases.  Hence, due to the increase in the tendency to release proton, the element's reducing tendency also increases.


(iii) Helium is used in diving apparatus as a diluent for oxygen because it is chemically inert and does not participate in the chemical reaction. Helium has low solubility in water than many other gases, such as nitrogen. Due to low solubility means it does not enter the bloodstream, even under pressure commonly experienced by deep sea divers.

2642 Views

Advertisement
12.

(a) Write the hybridization and shape of the following complexes:

(i) [CoF6]3–

(ii) [Ni (CN)4]2–

(Atomic number: Co = 27, Ni = 28)


(b) Out of NH3 and CO, which ligand forms a more stable complex with a transition metal and why?

5420 Views

13.

Calculate the freezing point of the solution when 31 g of ethylene glycol (C2H6O2) is dissolved in 500 g of water. (Kf for water = 1.86 K kg mol–1)

1111 Views

14.

Define the following terms:

(i) Primitive unit cells

(ii) Schottky defect

(iii) Ferromagnetism

1223 Views

Advertisement
15.

The rate constant of a first order reaction increases from 2 — 10-2 to 4 — 10-2when the temperature changes from 300 K to 310 K. Calculate the energy of activation (Ea).

(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)

1347 Views

16.

Define the following terms:
(i) Brownian movement
(ii) Peptization
(iii) Multimolecular colloids

1219 Views

17.

Calculate E.M.F. and G for the following cell:
Mg left parenthesis straight s right parenthesis space vertical line Mg to the power of 2 plus end exponent left parenthesis 0.001 space straight M right parenthesis space vertical line vertical line space Cu to the power of 2 plus end exponent left parenthesis 0.0001 space straight M right parenthesis space vertical line Cu space left parenthesis straight s right parenthesis space

Given space colon space straight E subscript open parentheses bevelled Mg to the power of 2 plus end exponent over Mg close parentheses end subscript superscript 0 space equals negative 2.37

straight E subscript open parentheses bevelled Cu to the power of 2 plus end exponent over Cu close parentheses end subscript superscript 0 space equals plus 0.34 space straight V.

1285 Views

18.

The conductivity of 0.20 mol L-1 solution of KCl is 2.48 x 10-2 S cm-1. Calculate its molar conductivity and degree of dissociation  (K+) = 73.56 S cm2 mol-1 and  (Cl-)= 76.5 S

(b) What type of battery is mercury cell? Why is it more advantageous than dry cell?

1372 Views

Advertisement
19.

Which would undergo SN1 reaction faster in the following pair?

1908 Views

20. Write the IUPAC name of the given compound:


1002 Views

Advertisement