The first ionisation energy of oxygen is less than that of nitrog

Subject

Chemistry

Class

JEE Class 12

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 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


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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.


C.

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.

The electronic configuration of nitrogen is

                  7N = 1s2, 2s2, 2p2

                    

Due to presence of half filled p-orbital, (more stable) a large amount of energy is required to remove an electron from nitrogen. Hence, first ionisation energy of nitrogen is greater than that of oxygen.

The electronic configuration of oxygen is

            8O =1s2, 2s2, 2p4

              

The other reason for the greater IP of nitrogen is that in oxygen, there is a greater interelectronic repulsion between the electrons present in the same p-orbital which counter- balance the  increase in effective nuclear charge from nitrogen to oxygen.         


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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


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


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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)


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