2 Marks Questions

1.What is the mass (m) of an electron?

Ans: mass of an electron (m) =

So, the mass of an electron is  or  th of the mass of a hydrogen atom.

2.Which experiment led to the discovery of electrons and how?  

Ans:The cathode ray discharge tube experiment performed by J.J. Thomson led to the discovery of negatively charged particles called electron.

A cathode ray tube consists of two thin pieces of metals called electrodes sealed inside a glass tube with sealed ends. The glass tube is attached to a vacuum pump and the pressure inside the tube is reduced to 0.01mm. When fairly high voltage (10, 000V) is applied across the electrodes, invisible rays are emitted from the cathode called cathode rays. Analysis of this rays led to the discovery electrons.

3.Give the main properties of canal ray experiment.

Ans:The canal ray experiment led to the discovery of –

(i)The anode rays, travel in straight line

(ii)They are positively charged as they get deflected towards the –ve end when subjected to an electric and magnetic field.

(iii)They depend upon the nature of gas present in the cathode tube.

(iv)The charge to mass ration (e/m) of the particle is found to depend on the gas from which they originate.

(v)They are also material particles

The analysis of these proportions led to the discovery of positively charged proton.

4.Find out atomic number, mass number, number of electron and neutron in an element?

Ans: The mass no. of

The atomic no. of 

No. of proton is = Z – A = 40 – 20 = 20

No. of electron its (A) = 20

No. of proton is (A) = 20

5.Give the main features of Thomson’s Model for an atom.

Ans: J.J. Thomson proposed that an atom consists of a spherical sphere (radius of about 10^-10m)in which the positive charges are uniformly distributed the electrons are embedded into it in such a manner so as to give stable electrostatic arrangement.

This model is also called raisin pudding model.

6.What did Rutherford conclude from the observations of   scattering experiment?

Ans: Rutherford proposed the nuclear model of an atom as

(i) The positive charge and most of the mass of an atom was concentrated in an extremely small region. He called it nucleus.

(ii) The nucleus is surrounded by electrons that move around the nucleus with a very high speed in orbits.

(iii) Electron and nucleus are held together by electrostatic forces of attraction.

7.What is the relation between kinetic energy and frequency of the  photoelectrons?

Ans: Kinetic energy of the ejected electron is proportional to the frequency of the electromagnetic radiation.

8.What transition in the hydrogen spectrum would have the same  wavelength as the Balmer transition, n = 4 to n = 2 of He⁺ spectrum?

Ans: For the Balmer transition, n = 4, to n = 2 in a He⁺ ion, we can write.

For a hydrogen atom

Equating equation (ii) and (i), we get

This equation gives n₁ = 1 and n = 2. Thus the transition n = 2 to n = 1 in hydrogen atom will have same wavelength as transition, n = 4 to n = 2 in He⁺

9.Spectral lines are regarded as the finger prints of the elements. Why?

Ans: Spectral lines are regarded as the finger prints of the elements because the elements can be identified from these lines. Just like finger prints, the spectral lines of no two elements resemble each other.

10.Why cannot the motion of an electron around the nucleus be determined

accurately?

Ans: Because there is an uncertainty in the velocity of moving electron around the nucleus (Heisenberg’s Uncertainty Principle).

11.Calculate the uncertainty in the momentum of an electron if it is confined to a linear region of length

Ans: According to uncertainty Principle

12.Give the mathematical expression of uncertainty principle.    

Ans:Mathematically, it can be given as 

Where is the uncertainty in position and  is the uncertainty in momentum (or velocity) of the particle.

13.Which quantum number determines

(i)  energy of electron      

(ii) Orientation of orbitals.

Ans. (i) Principal quantum number (n), and

(ii) Magnetic quantum number (m).

14.Arrange the electrons represented by the following sets of quantem number in decreasing order of energy. 

1.   n = 4,  l = 0, m = 0, s = +1/2

2.   n = 3,  l = 1, m = 1,  s = -1/2

3.   n = 3,  l = 2, m = 0, s = +1/2

Ans.(i)Represents 4s orbital

(ii) Represents 3p orbital

(iii)Represents 3d orbital

(iv)Represents 3s orbital

The decreasing order of energy   3d > 4s > 3p > 3s

n = 3,  l = 0, m = 0, s = -1/2

3 Marks Questions

1.What designations are given to the orbitals having

(i)    n = 2, l = 1     (ii)   n = 2, l = 0     (iii) n = 4, l = 3

(iv)   n = 4, l = 2      (v)    n = 4, l = 1? 

Ans.    (i)           Here, n = 2, and l = 1

Since l = 1 it means a p-orbital, hence the given orbital is designated as 2p.

(ii) Here, n = 2 and l = 0

Since l = 0 means s – orbital, hence the given orbital is 2s.

(iii) Here, n = 4 and  l = 3

Since, l = 3 represents f – orbital, hence the given orbital is a 4f orbital.

(iv) Here, n = 4 and l = 2

Since, l = 2 represents d – orbital, hence the given orbital is a 4d – orbital.

(v) n = 4  and l = 1

 since, l = 1 means it is a p – orbital, hence the given orbital can be designated as – 4p orbital.

2.Write the electronic configuration of (i) Mn⁴⁺, (ii) Fe³⁺ (iii)  Cr²⁺ and Zn²⁺ Mention the number of unpaired electrons in each case.

Ans.(i)       Mn  (z = 25), Mn⁴⁺  (z = 21)

The electronic configuration of Mn⁴⁺ to Given by

1s² 2s² 2p⁶ 3s² 3p⁶ 3d³

As the outermost shell 3d has 3 electrons, thus the number of unpaired

electrons is 3.

 (ii) Fe   (z = 26), Fe³⁺ (z = 23)

The electronic configuration of Fe³⁺ is given lay

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

The number of unpaired electron is 5.

(iii) Cr (z = 24),  Cr²⁺ (z = 22)

The electronic configuration of Cr²⁺ is

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴

The number of unpaired electron is 4.

(iv) Zn (z = 30), Zn²⁺  (z = 28)

The electronic configuration of Zn²⁺ is

1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰

The number of unpaired electron is 0.