Chemical Applications of Group Theory MCQ Quiz in বাংলা - Objective Question with Answer for Chemical Applications of Group Theory - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:-

• A point group is a term used in the field of chemistry to describe the symmetry of a molecule or crystal.
• It refers to the group of symmetry operations that leave the molecule or crystal unchanged. These operations include rotations, reflections, and inversions. The point group is determined by the shape of the molecule or crystal, as well as the positions of its atoms.

• Group theory is the study of the set of elements in a molecule to determine its symmetry. In group theory symmetry, symmetry elements, and point groups are studied.

Symmetry: It is the distribution pattern of masses or atoms in 3D space

Symmetry elements: The geometrical patterns which can be utilized to define the distribution pattern of atoms or molecules. There are 3 elements

• Axis of symmetry
• plane of symmetry
• center of symmetry

• Each element of symmetry has a specific operation (that is applied on the element to define its symmetry) which are:

  Elements of symmetry	Operations of symmetry elements
  axis of symmetry (Cn)	rotation
  plane of symmetry (σ )	reflection
  improper axis of symmetry (Sn)	roto-reflection
  centre of symmetry (i)	inversion
  identity (E)	nothing to do

• These are the different point group assignments in group theory.

  C-type or Cn-type	D-type or Dn-type	S-type	High symmetry
  Cn	Dn	Sn	Oh (octahedral)
  Cs	Dnd		Td (tetrahedral)
  Cnv	Dnh		Ih (icosahedron)
  Cnh

where,

Cnv = Cn + σv

Cnh = Cn + σh

Dn = Cn + nC2

Dnd = Cn + nC2 + nσv + σd

Dnh = Cn + nC2 + nσv + σh

Explanation:-

[PtCl4]2- :

• [PtCl4]2- has square planar geometry.

• [PtCl4]2- has the following symmetry elements-

  ]

• There are five mirror planes passing through Pt and all four chlorine atoms (perpendicular to the molecular plane).
• Through Pt and bisecting the angles made by Cl1-Pt-Cl2 and Cl3-Pt-Cl4 and perpendicular to the molecular plane.
• It has a C₄ principal axis standing perpendicular to the square plane of the molecule. There is an additional C2 axis where the C4 axis runs.
• So, the symmetry elements present in are

E, 2C4, C2, 2C2’, 2C2”, i, 2S4,

• The molecule [PtCl4]2- belongs to the D4h point group.

B2H6 :

• Diborane molecule has 3C2 rotation axis of symmetry and 2\(\sigma \)  vertical planes of symmetry and 1 \(\sigma\) horizontal planes of symmetry.

• Thus, B2H6 belongs to D2h point group.

Conclusion:-

• Hence, [PtCl4]2- and B2H6 belong to D4h, and D2h point groups.

Concept:-

• A point group is a term used in the field of chemistry to describe the symmetry of a molecule or crystal.
• It refers to the group of symmetry operations that leave the molecule or crystal unchanged. These operations include rotations, reflections, and inversions. The point group is determined by the shape of the molecule or crystal, as well as the positions of its atoms.

• Group theory is the study of the set of elements in a molecule to determine its symmetry. In group theory symmetry, symmetry elements, and point groups are studied.

Symmetry: It is the distribution pattern of masses or atoms in 3D space

Symmetry elements: The geometrical patterns which can be utilized to define the distribution pattern of atoms or molecules. There are 3 elements

• Axis of symmetry
• plane of symmetry
• center of symmetry

Each element of symmetry has a specific operation (that is applied on the element to define its symmetry) which are:

• These are the different point group assignments in group theory.

where,

Cnv = Cn + σv

Cnh = Cn + σh

Dn = Cn + nC2

Dnd = Cn + nC2 + nσv + σd

Dnh = Cn + nC2 + nσv + σh

Explanation:-

• The cubane shown below is a bi-substituted isotopomer of cubane, where two hydrogens are substituted by deuterium.

• In Triphenylene C3 principal axis is passing perpendicular to the center of the middle ring.
• While the plane of the paper represents the σd plane that bisects the C3 axis.
• Its elements of symmetry are as follows:
  • C3 principal axis,
  • Three C2 axis,
  • Three σv planes, and a
  • σd plane
• Thus, Triphenylene belongs to the D3d point group.
• D3d = C3 + 3C2 + 3σv + σd

Conclusion:-

Hence, D3d is correct.

Concept:

A cube is a three-dimensional shape with six identical square faces.   It possesses following symmetry axis:

1. C₁ axis: The identity element, which does not change the cube's orientation.

2. C₂ axis: A two-fold rotation axis that passes through the center of opposite faces of the cube. This axis generates a 180⁰ rotation of the cube about the axis.

3. C₃ axis: There are four C₃ axes, each passing through the centers of two opposite edges of the cube. These axes generate a 120⁰ rotation of the cube about the axis.

4. C₄ axis: A four-fold rotation axis that passes through the centers of opposite faces of the cube. This axis generates a 90⁰ rotation of the cube about the axis.

5. C₆ axis: passing vertically through the center of the molecule.

Explanation: 

→  The cube contains three different types of symmetry axes: three 4-fold axes, each of which passes through the centers of two opposite faces, four 3-fold axes, each of which passes through two opposite vertices, and. six 2-fold axes, each of which passes through the midpoints of two opposite edges i.e.C2, C₃, C₄ axis are are present but C₆ axis is absent_.

Conclusion: 
The correct answer is option 4.

Additional Information
 

Concept:

• The "Great Orthogonality Theorem" States that the rows of characters are orthogonal vectors.
• The matrices of the different Irreducible Representations (IR) possess certain well-defined interrelationships and properties. The "Great Orthogonality Theorem" is concerned with the elements of the matrices which constitute the IR of a group. Kronecker delta can have values 0 and 1.​
• If one irreducible representation Γ is orthogonal to A it ​must satisfy,

 ꭍ AΓdτ = 0

Explanation:

The character of the irreducible representation A1 in the C3v point group is given below

We have to Identify one irreducible representation orthogonal to A1 among the following.

• Now for the integral ꭍ A1Γ2 dτ,   

= A1\( \times \)Γ1 = 1×1×2 + 2×1×(-1) + 3×1×0

 = 2 - 2 + 0

= 0

• Thus, the integral ꭍ A1Γ2 dτ is zero (0).

​​Conclusion:

• Hence, the irreducible representation orthogonal to A1 is Γ2

Concept:-

• An action that leaves an object looking the same after it has been carried out is called a symmetry operation.
• The symmetry operations include rotation, reflection, the alternating axis of symmetry, and inversions.
• A molecule possessing an alternative axis of symmetry (S_n) of n fold represents the rotation of the molecule about the axis \({{{{360}^ \circ }} \over n}\) followed by reflection through a plane perpendicular to this axis produces an indistinguishable structure.
• S_n It can be represented by,

\({S_n} = {C_n} \times \sigma \) 

• The trace of a matrix A, designated by tr(A), is the sum of the elements on the main diagonal.
• If matrix A is represented by the matrix,

\(A = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 2&0&{ - 1} \end{array}}\right] \)

then the trace of matrix A is given by,

tr(A)= 1+1-1

=1

Explanation:-

• The alternative axis of symmetry (Sn) is represented by,

\({S_n} = {C_n} \times \sigma \)​

• Now, \(S^2_3\) is given by,

​\(S_n^2 = C_n^2 \times {\sigma ^2}\)

\( = C_n^2 \times E\left( {As,{\sigma ^2} = E} \right)\)

\( = C_n^2\)

• Thus, trace of the matrix \(S^2_3\) is given by,
   

\(\rm S_3^2=C_3^2\sigma_n^2=C_3^2\)

\(\rm C_3 = \left[ {\begin{array}{*{20}{c}} { - 1/2}\\ { \sqrt 3 /2}\\ 0 \end{array}\begin{array}{*{20}{c}} { - \sqrt 3 /2}\\ { - 1/2}\\ 0 \end{array}\begin{array}{*{20}{c}} 0\\ 0\\ 1 \end{array}} \right] \)

\(\rm C_3^2 = \left[ {\begin{array}{*{20}{c}} { - 1/2}\\ { \sqrt 3 /2}\\ 0 \end{array}\begin{array}{*{20}{c}} { - \sqrt 3 /2}\\ { - 1/2}\\ 0 \end{array}\begin{array}{*{20}{c}} 0\\ 0\\ 1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 1/2}\\ { \sqrt 3 /2}\\ 0 \end{array}\begin{array}{*{20}{c}} { - \sqrt 3 /2}\\ { - 1/2}\\ 0 \end{array}\begin{array}{*{20}{c}} 0\\ 0\\ 1 \end{array}} \right] \)

\(\rm =\begin{bmatrix}\frac{1}{4}-\frac{3}{4}+0&\frac{\sqrt3}{4}+\frac{\sqrt3}{4}+0&0+0+0\\\ -\frac{\sqrt3}{4}-\frac{\sqrt3}{4}+0&-\frac{3}{4}+\frac{1}{4}+0&0+0+0\\\ 0+0+0&0+0+0&0+0+0\end{bmatrix} \)

\(\rm =\begin{bmatrix}-\frac{1}{2}&\frac{\sqrt3}{2}&0\\\ -\frac{\sqrt3}{2}&-\frac{1}{2}&0\\\ 0&0&1\end{bmatrix} \)

Trace of matrix \(S^2_3\) is,

\( - {1 \over 2} - {1 \over 2} + 1\)

\( = 0\)

Conclusion:-

Hence, trace of the matrix representing for \(S^2_3\) is 0

Concept:

• The "Great Orthogonality Theorem" States that the rows of characters are orthogonal vectors.
• The matrices of the different Irreducible Representations (IR) possess certain well-defined interrelationships and properties. The "Great Orthogonality Theorem" is concerned with the elements of the matrices which constitute the IR of a group. Kronecker delta can have values 0 and 1.

​

Explanation:

The following is the character table for the C2v point group,

• The two functions f1 and f2 are A2 and B1 respectively.
• The product of two functions A2 and B1 are 

  A2	1	1	-1	-1
  B1	1	-1	1	-1
  	1	-1	-1	1

  =B2.
• Thus, from the character table, we can conclude that the product belongs to the B2 representation.
• Now for the integral ꭍ f1f2 dτ,   

           f₁\( \times \)f₂ = A₂\( \times \)B₁ = 1\( \times \)1 + 1 \( \times \)(-1) + (-1) \( \times \) 1 + (-1) \( \times \) (-1) = 1 - 1 - 1 + 1 = 0.

• Thus, the integral ꭍ f1f2 dτ is zero (0).
• So,  the product belongs to the B2 representation and the integral is zero.

​Conclusion:

Hence, the correct option for the product of two functions f1 and f2 and the integral ꭍ f1f2 dτ is the product belongs to B2 representation and the integral is zero. 

​Concept:

• Cn represents the axis of rotation symmetry of order n, and molecule doesn't change after rotation by \(\frac{360^0}{n}\)
• \(\sigma \) is the plane of symmetry. Any molecule possessing \(\sigma\)symmetry, either shows reflection or bisection when plane symmetry operation is applied. 
• The number of unshifted atoms gives the character of that symmetry which is being applied, in the irreducible representation.
• Character is calculated by adding the diagonal terms of matrix formed for that symmetry operation. 

• Cn represents the axis of rotation symmetry of order n, and molecule doesn't change after rotation by \(\frac{360^0}{n}\)
• \(\sigma \) is the plane of symmetry. Any molecule possessing \(\sigma\)symmetry, either shows reflection or bisection when plane symmetry operation is applied.

Explanation:

Character for \(C_4\) : 

Matrix for character of \(C_n\) operation is given by;

 \(\begin{vmatrix} cos\theta \;\;\;sin\theta \;\;0\\-sin\theta \;cos\theta \;0\\0\;\;\;\;\;\;\;0\;\;\;\;\;\;\;1 \end{vmatrix}\) 

This gives, \(\chi_{C_n}=2cos\theta +1 \)

for \(C_4\) symmetry, \(\theta \)= 90°

 \(\chi _{C_4}=2cos90 ^0+1 \)

\(\chi _{C_4}=1\)

Character for \(\sigma \): 

For \(\sigma \) symmetry, character = 1

Conclusion:

Therefore, the character for each unshifted atom in C₄ and \(\sigma \) in irreducible representation is 1 and 1 respectively.

CONCEPT:

Character Table and IR Activity in the C_2v Point Group

• The character table for the C_2v point group shows the symmetry elements and irreducible representations (A₁, A₂, B₁, B₂) along with their corresponding basis functions and transformations.
• In a molecule with C_2v symmetry, IR-active vibrational modes correspond to irreducible representations that transform like the Cartesian coordinates (x, y, z) or their combinations (e.g., xz, yz), as these represent the directions in which the dipole moment can change.
• In this question, vibrational modes belonging to the A₂ irreducible representation are given as IR inactive, so we only consider the modes associated with A₁, B₁, and B₂ representations for IR activity.

CALCULATION:

The detailed solution is as follows:

The character table of the C_2v point group is given below. In the cis-butadiene molecule, the vibrational modes belonging to the A₂ irreducible representation are IR inactive. The remaining IR active modes are calculated as follows:

Calculation of irreducible representations based on contributions:

• n_A1 = 1/4 [30 + 0 + 10 + 0] = 10
• n_A2 = 1/4 [30 - 10] = 5
• n_B1 = 1/4 [30 + 0 + 10 + 20] = 15
• n_B2 = 1/4 [30 + 10] = 10

The reducible representation is therefore: R.R = 10A₁ + 5A₂ + 15B₁ + 10B₂

Subtracting the IR-inactive modes of A₂ yields the IR-active modes:

• Vibrational = Total - (Rotational + Translational)
• Vibrational modes: 9A₁ + 3B₁ + 8B₂

CONCLUSION:

The correct answer is 9A₁ + 3B₁ + 8B₂.

Concepts:

• Improper Rotation (S_n): An improper rotation operation, denoted as S_n, consists of a rotation by 360°/n followed by a reflection through a plane perpendicular to the axis of rotation. For example, S₄ involves a 90° rotation followed by reflection.
• Equivalent Symmetry Operations: Certain improper rotation operations are equivalent to other symmetry operations. 

Explanation:

1. For S₄², applying the improper rotation twice (90° rotation + reflection, repeated twice) results in a 180° rotation without a reflection. This operation is equivalent to a C₂ rotation.
2. For S₆⁴, applying the operation four times involves rotation and reflection combinations that lead to overall rotation of 240^o or twice C₃ rotation.  Thus, S₆⁴ is equivalent to \(C_3^2\).

Conclusion:

The correct answer is Option 1: C₂ and \(C_3^2\), as these are the equivalent operations for S₄² and S₆⁴, respectively.

Concepts:

• Vibrational Modes: The number of vibrational modes for a molecule is calculated by ( 3N - 6 ) for non-linear molecules, where ( N ) is the number of atoms. For NH₃, there are 4 atoms, so the number of vibrational modes is ( 3× 4 - 6 = 6 ).
• Character Table: The character table for the ( C_3v ) point group includes symmetry operations and corresponding characters for each irreducible representation (A₁, A₂, and E).
• Reduction Formula: To determine the coefficients for each irreducible representation in the reducible representation (Γ), we use the formula:
• \(n_i = \frac{1}{h} \sum (χ_{\text{reducible}} × χ_{\text{irreducible}} × \text{number of operations}) \) where ( h ) is the order of the group, (χ_reducible ) are the characters of the reducible representation, and (χ_irreducible ) are the characters of each irreducible representation.

Explanation:

Given the characters for the reducible representation (Γ =12, 0, 2) for NH₃, applying the reduction formula to find the coefficients for each irreducible representation (A₁, A₂, and E) as follows:

• \( n_{A_1} = \frac{1}{6} \left[ (1)(12)(1) + (2)(0)(1) + (3)(2)(1) \right] = \frac{1}{6} \left[ 12 + 0 + 6 \right] = \frac{18}{6} = 3 \)

• \( n_{A_2} = \frac{1}{6} \left[ (1)(12)(1) + (2)(0)(-1) + (3)(2)(-1) \right] = \frac{1}{6} \left[ 12 + 0 - 6 \right] = \frac{6}{6} = 1 \)

• \(n_{E} = \frac{1}{6} \left[ (2)(12)(1) + (2)(0)(-1) + (3)(2)(0) \right] = \frac{1}{6} \left[ 24 + 0 + 0 \right] = \frac{24}{6} = 4 \)

Thus, the reducible representation ( Γ(NH₃) ) is expressed as:

• \( \Gamma(NH_3) = 3A_1 + 1A_2 + 4E \)

Next, we determine the irreducible representations for the translational and rotational modes of NH₃:

• \( \Gamma(T) = 1A_1 + 1E \)

• \( \Gamma(R) = 1A_2 + 1E \)

• \(\Gamma(T + R) = 1A_1 + 1A_2 + 2E \)

Finally, to find the irreducible representation for the vibrational modes, we subtract the translational and rotational representations from the total reducible representation:

• \(\Gamma_{\text{vib}}(NH_3) = \Gamma(NH_3) - \Gamma(T + R) \)
• \((3A_1 + 1A_2 + 4E) - (1A_1 + 1A_2 + 2E) = 2A_1 + 2E\)

Conclusion:

The correct answer is Option 2: 2A₁ + 2E, as this is the reduced representation for the vibrational modes of NH₃.