NIOS Class 12 Chemistry Worksheet 2 Solutions: Atomic Structure

This post explains clear and step by step solutions to NIOS Class 12 Chemistry Worksheet 2 from the chapter Atomic Structure. The questions in this worksheet focus on key ideas such as the dual nature of light, Bohr’s model, quantum numbers, electronic configuration. Each answer is explained in a simple exam-oriented way so you can understand the concepts as well as the method.

 

Main topics Covered in the Chapter Atomic Structure

    1. Discovery of electron,proton, neutron

    2. Atomic number,mass number, isotopes and isobars

    3. Thomson model and Rutherford α scattering

    experiment.

    4. Nature and parameters of electromagnetic radiation

    5. Hydrogen line spectrum and Rydberg equation

    6. Bohr’s model and energy levels

    7. De Broglie wave particle duality

    8. Heisenberg uncertainty principle

    9. Schrodinger wave mechanical model and orbitals

   10.  Quantum numbers and shape of s, p, d orbitals

   11.  Electronic configuration

   12.Stability of half-filled and fully filled subshells

 

 

Q1. What experimental evidences shows the dual nature of light?

(a)       Calculate the energy of the FM radio signal transmitted at a frequency of 100 MHz.

(b)      What is the energy of the red coloured wave with 670mm wavelength?

Answer

 

(a)       Energy of FM radio signal at 100 MHz

 Energy of a photon is given by

 

    h= 6.626 ×10^-34 Js

   v= 100 MHz = 1.0 10⁸ s^-1

    E= 6.626 ×10^-34   ×1.0 10⁸

     E= 6.626 ×10^-26   J

(b)      Energy of red light of wave length 670 nm

  

  

      

      

 

Q2. How is the Bohr model superior to the Rutherford model?

Answer

The Bohr model removed the major drawbacks of Rutherford’s model in the following way:

1. Stability of atom: Rutherford’s model could not explain why the revolving electron does not lose energy and fall into the nucleus. Bohr proposed that electrons revolve only in fixed circular orbits (energy levels) without radiating energy thus explain atomic stability.

2. Quantization of energy: Rutherford gave no idea about fixed energy of electrons. Bohar stated that each orbit has a definite quantized energy.

3. Explanation of hydrogen spectrum: Rutherford’s model failed to explain the line spectrum of hydrogen. Bohr successfully explained it by electron transmission between energy levels.

4. Concept of stationary orbits: Bohr introduced the idea of stationary states where electrons do not emit energy.

5. Calculation of energy and radius: Bohr’s model provided mathematical expression to calculate the radius and energy of electron orbits.

Hence, Bohr’s model was superior because it explained atomic stability and hydrogen line spectrum, which Rutherford’s model could not.

 

Q3. Wavelength to green light is 335 nm. Calculate the energy of green photons.

Answer

We use the following formula

 

Where

H= 6.626 ×10^-34Js

C =3.0 ×10⁸ ms^-1

The wavelength of green light is clearly intended as 355nm.

 

 

 

 

Energy of one green photon =

 

 

 

Q4. How did the wave mechanics model of Atom develop?

Answer

The wave mechanics model developed step by step from the ideas that showed matter and radiation both have wave and particle character.

1. Dual nature of radiation was stabilised by experiment like interference /diffraction (wave nature) and the photoelectric effect (particle nature).

2. Louis de Broglie proposed that electrons(matter) also possess wave nature and have an associated wavelength  .

3. The wave behaviour of electrons was experimentally confirmed by C.J. Davisson and G.P. Thomson through electron diffraction.

4. Warner Heisenberg stated that the exact position and momentum of an electron cannot be determined simultaneously, rejecting the idea of fixed circular orbits.

5. Finally, Erwin Schrodinger developed a mathematics equation (wave equation) to describe the electrons as a wave its solutions gave orbitals- regions of high probability of finding an electron.

So, the wave mechanics model evolved from de Broglie’s matter wave, Heisenberg’s uncertainty principle and Schrodinger’s wave equation, replacing Bohr’s fixed orbits with probability-based orbitals.

 

Q5. Calculate the wavelength corresponding to the balmar line n=3.

Answer

For the Balmar series, the electron falls to n₁=2

Given line:

We use the following formula

Where

R= 1.097 × 10⁷ m^-1

Wavelength of the Balmar line (n=3) = 656 nm

 

Q6. If a 380 gram cricket ball is thrown at a speed of 140 kilometres per hour, calculate the de Broglie wavelength.

Answer

We use formula

where

mass (m) = 380 g= 0.380kg

speed (v) = 140 km h^-1

 

De Broglie wavelength =

 

Q7. Describe the Hunds rule of maximum multiplicity with five examples.

Answer

Hund’s rule: In a group of orbitals having the same energy (degenerate orbitals), electrons fill each orbital singly first with parallel spins and only then pairing takes place.

This happens because this arrangement has maximum number of unpaired electrons and hence greater stability.

 

Examples

Carbon (Z=6)

Electronic configuration: 1s²2s²2p²

2p orbitals  

↑

↑

Nitrogen (Z=7)

Electronic configuration: 1s²2s²2p³

2p orbitals  

↑

↑

↑

Oxygen (Z=8)

Electronic configuration: 1s²2s²2p⁴

2p orbitals  

↑↓

↑

↑

Phosphorus (Z=15)

Electronic configuration: [Ne]3s²3p³

3p orbitals  

↑

↑

↑

Sulphur (Z=16)

Electronic configuration: [Ne]3s²3p⁴

3p orbitals  

↑↓

↑

↑

In all these, electrons first occupy different orbitals singly begore pairing which proves Hund’s rule.

 

 

Q8. Which oxidation states more stable and why?

(a)       Fe²⁺ or Fe³⁺

(b)      Mn²⁺ or Mn³⁺

Answer

Stability of oxidation states can be explained using electronic configuration and the extra stability of half filled and fully filled subshells.

(a)       Fe²⁺ or Fe³⁺

Iron (Z=26)

Electronic configuration: [Ar]3d⁶4s²

Fe²⁺ loses 2 electrons: [Ar]3d⁶

Fe³⁺ loses 3 electrons: [Ar]3d⁵

The 3d⁵ configuration is half filled and there more stable.

Hence, Fe³⁺ is more stable than Fe²⁺

(b)      Mn²⁺ or Mn³⁺

Manganese (Z=25)

Electronic configuration: [Ar]3d⁵ 4s²

Mn²⁺ loses 2 electrons: [Ar]3d⁵

Mn³⁺ loses 3 electrons: [Ar]3d⁴

The 3d⁵ configuration is half filled and there more stable.

Hence, Mn²⁺ is more stable than Mn³⁺

 

Q9. Which of the following class has the first storage and why?

(a)       2p or 3s

(b)      3d or 4p

(c)       4s or 3d

Answer

The order of filling of subshells follows the Aufbau principle and the  rule:

The subshell with lower t  value fills first. If   is same, the subshell with lower ‘n’ fills first.

(a)       2p or 3s

2p :

3s :

Since :   is same , the subshell with lower ‘n’ fills first, so  2p fills first.

(b)      3d or 4p

3d :

4p :

3s :

Since :   is same , the subshell with lower ‘n’ fills first, so  3d  fills first.

(c)       4s or 3d

4s :

3d :

3s :

the subshell with lower    fills first, so 4s fills first.

 

Q10. What is the significance of the azimuthal magnetic and spinning quantum numbers?

(a)    Write the four quantum numbers for 3p³(3^rd electron),4d⁵ (4^th electron),6s²(2^nd electron).

(b)      How many electrons are s= +1/2 and ml=0 for n=4

Answer

Azimuthal quantum number  tells the subshell (s,p,d) and shape of the orbital.

Magnetic quantum number  tells the orientation of the orbital in space.

Spin quantum number   tell the direction of spin of the electron (

(a)       Four quantum numbers

(i)       3p³ (3^rd electron)

For p subshell:

Electrons fill singly with parallel spins (Hund’s rule)

3^rd electron goes to

(ii)         4d⁵ (4^th electron)

 For d subshell:

Electrons fill singly

4^th  electron goes to

(iii)     6s² (2^nd electron)

  

For s  subshell :

Second electron pairs with opposite spin

 

(b)   Number of electrons with

For n=4 , subshells are : 4s,4p ,4d ,4f

(a)   Orbitals with  in each

  

4s=1 orbital , 4p= 1 orbital , 4d= 1 orbital , 4f= 1 orbital

Each such orbital can have one electron with

Total electrons =4