Column I   Column II (A) He (i) High electron af

Subject

Chemistry

Class

JEE Class 12

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.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

1.

The percentage of an element M is 53 in its oxide of molecular formula M2O3. Its atomic mass is about

  • 45

  • 9

  • 27

  • 36


2.

The electronic configuration of the element with maximum electron affinity is

  • 1s2, 2s2,2p3

  • 1s2, 2s2,2p5

  • 1s2, 2s2,2p6, 3s2, 3p5

  • 1s2, 2s2,2p6, 3s2, 3p3


3.

The incorrect statement/s among the following is/are

I. NCl5, does not exist while PCl5, does.

II. Lead prefers to form tetravalent compounds.

III. The three C-O bonds are not equal in the carbonate ion.

IV. Both O+and NO are paramagentic

  • I, III and IV

  • I and IV

  • II and III

  • I and III


4.

The first ionisation energy of oxygen is less than that of nitrogen. Which of the following is the correct reason for this observation?

  • Lesser effective nuclear charge of oxygen than nitrogen.

  • Lesser atomic size of oxygen than nitrogen

  • Greater interelectron repulsion between two electrons in the same p-orbital counter balances the increase in effective nuclear charge on moving from nitrogen to oxygen.

  • Greater effective nuclear charge of oxygen than nitrogen.


Advertisement
Advertisement

5.

The correct match of contents in Column I with those in Column II is

  Column I   Column II
(A) He (i) High electron affinity
(B) Cl (ii) Most electropositive element
(C) Ca (iii) Strongest reducing agent
(D) Li (iv) Highest ionisation energy
  • A-iv, B-i, C-ii, D-iii

  • A-iii, B-i, C-ii, D-iv

  • A-iv, B-iii, C-ii, D-i

  • A-ii, B-iv, C-i, D-iii


A.

A-iv, B-i, C-ii, D-iii

(i) For noble gases (eg, He), ionisation energy is highest due to their completely filled electronic configuration.

(ii) Generally electron affinity increases in a period (from IA to VII A group) and decreases in a group but electron affinity is highest for chlorine (Cl) (due to smaller size and high electron density of fluorine).

(iii) The ionisation energy is lowest for Li, so it can lose electrons very easily, thus it behaves as a strongest reducing agent.

(iv) Electropositive character generally decreases in a period (from left to right) and increases in a group (on moving down), thus Ca is the most electropositive element among the given.

Hence, on the basis of above facts, the correct matches are

(A)-iv (B)-i (C)-ii (D)-iii


Advertisement
6.

Identify the nuclear reaction that differs from the rest.

  • γ- decay

  • κ- capture

  • Positron emission

  • β-decay


7.

Using the following thermochemical equations

(i) S(rh) + 3/2O2(g) → SO3 (g)       ΔH = -2x kJ mol-1

(ii) SO2(g) + 1/2O2 (g)  → SO3 (g)  ΔH = -y kJ mol-1

Find out the heat of formation of SO2 (g) in kJ mol-1.

  • (2x + y)

  • (x + y)

  • (2x/y)

  • (y- 2x)


8.

The lattice enthalpy and hydration enthalpy of four compounds are given below
Compound Lattice enthalpy (in kJ mol-1) Hydration
enthalpy
(in kJ mol-1)
P +780 -920
Q +1012 -812
R +828 -878
S +632 -600

The pair of compound which is soluble in water is

  • P and R

  • Q and R

  • P and Q

  • Rand S


Advertisement
9.

1.6 moles of PCl5(g) is placed in 4 dm closed vessel. When the temperature is raised to 500 K, it decomposes and at equilibrium 1.2 moles of PCl5(g) remains. What is the K, value for the decomposition of PCl5(g) to PCl3(g) and Cl5(g) at 500 K?

  • 0.013

  • 0.050

  • 0.033

  • 0.067


10.

For a concentrated solution of a weak electrolyte AxBy of concentration 'C', the degree of dissociation 'a' is given as

  • α = √ Keq/ C ( x + y)

  • α = √ Keq/ C ( xy)

  • α =( Keq/ Cxy)

  • α =( Keq/ Cxy)


Advertisement