Most Expected Questions for Chemical Science Students - Group Theory!

Group Theory - Most Important Questions! (Download PDF)

1. Which of the following is not the symmetry element that is not present in CH4?

1. S₄
2. C₂
3. σ_d
4. C₄

2. What is the point group of H₂O₂ ?

1. C₂V
2. C3
3. C₃h
4. C₂

3. Identify the Mulliken notation of the following irreducible representation.

1. A1
2. A2
3. B2’
4. A1’

4. No. of C2 axis in SiCl4 are ........

1. 2
2. 0
3. 3
4. 6

5. Point group of trans N2F2 is:

1. C₂h
2. C₂v
3. D₂h
4. None of above

6. The character of S6 is:

1. 0
2. 2
3. 1
4. –1

7. The pair of non-polar point group is:

1. C₂h, Ci
2. C₃,  D₄d
3. C₁, C₃v
4. S₄, D₄

8. What will be the final operation of C2(x).C2(y).C2(z). σxz ?

1. C2(z)
2. σxz
3. C4(z)
4. I

9. For H₂O, the electronic transition from the ground state to B₂ symmetry is :

1. Not allowed 
2. Allowed with x polarisation 
3. Allowed with y polarisation 
4. Allowed with z polarisation

10. Phosphorus pentachloride, PCl5 is a trigonal bipyramidal molecule. To what point group does it belong?

1. D₃v
2. D₃h
3. C₃h
4. D₅h

Solutions

Solution 1:

The point group of CH4 is tetrahedral. 

The symmetry elements present in this are: E, 4C₃ , 3C₂ , 6σ_d , 3S₄.

According to the above representation, C₄ is not present in methane.

Solution 2:

In this question, the form in which H2O2 exists is not given. In general, we take its gauche or open book form.

It has only C₂ axis and E symmetry operation. According to this, its point group is C₂. 

Solution 3:

For Mulliken notation, we need to see the character below the symmetry operation of a given point group. In the given table, it is symmetric w.r.t E, so, A. Since the subsidiary axis is absent, we will check the molecular plane (xz), it is antisymmetric to that, so, it will be A2. Further there is no horizontal plane present due to which ‘or ‘’ will not be applied.

Solution 4:

SiCl₄ is a tetrahedral molecule, just like CH₄.

As shown in the above figure, there are 3 C₂ axes present in this which is passing through all the pair of opposite faces of SiCl₄.

Solution 5:

The structure of trans – N₂F₂ is given below as:

It has one C₂ axis perpendicular to the plane of the paper. This is the only axis present in the molecule. There is no other C₂ axis present in the molecule. Also, it has a molecular plane which is also its horizontal plane. Based on symmetry operation, the molecule has a C₂h point group.

Solution 6:

Solution 7:

Polar molecules are those which do not have a plane of symmetry in molecules. Ci has a centre of inversion so, the molecule is nonpolar. C_2h molecule has a perpendicular plane of symmetry so it is also nonpolar. In general, all the molecules which have a C_n point group except C_i, C_nh are polar.

Solution 8:

For the application of the product of operations on Cartesian coordinates, we start applying from right side operations one by one.

σ_xz(x,y,z) = (x,–y,z)

C₂(z). (x,–y,z) = (–x,y ,z)

C₂(y). (–x,y,z) = (x,y,–z)

C₂(x). (x,y,–z) =(x,–y,z) 

The final result of the product of symmetry operation on (x,y,z) will be (x,–y,z) which can be directly obtained by σ_xz.

Solution 9:

Ground state is the most symmetric state so A₁ is the ground state.     

This direct product resembles the B₂, so, this transition is Y polarised.

Solution 10:

PCl5 has trigonal bipyramidal molecular geometry and it contains a C₃ main rotation axis and 3 perpendicular C₂ axes. There are 3 σv planes and a σh plane. Hence PCl₅ belongs to the D₃h point group.

Answer Key:

1. D
2. D
3. B
4. C
5. A
6. A
7. C
8. B
9. C
10. B

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