Character Tables & Selection Rules MCQ Quiz in తెలుగు - Objective Question with Answer for Character Tables & Selection Rules - ముఫ్త్ [PDF] డౌన్లోడ్ కరెన్
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Character Tables & Selection Rules Question 1:
The character table for the point group D3h is given below.
| D3h | E | 2C3 (z) | \(\rm 3C_{2}^{'}\) | σh(xy) | 2S3 | 3σv | ||
| \(\rm A_{1}^{'}\) | +1 | +1 | +1 | +1 | +1 | +1 | - | x2 + y2, z2 |
| \(\rm A_{2}^{'}\) | +1 | +1 | −1 | +1 | +1 | −1 | Rz | - |
| E' | +2 | −1 | 0 | +2 | −1 | 0 | (x, y) | (x2 − y2, xy) |
| \(\rm A_{1}^{''}\) | +1 | +1 | +1 | −1 | −1 | −1 | - | - |
| \(\rm A_{2}^{''}\) | +1 | +1 | −1 | −1 | −1 | +1 | z | - |
| E'' | +2 | −1 | 0 | −2 | +1 | 0 | (Rx, Ry) | (xz, yz) |
In the electronic ground state, BF3 has D3h symmetry. Therefore,
Answer (Detailed Solution Below)
Character Tables & Selection Rules Question 1 Detailed Solution
Concept:-
- The general form of a character table is
(a) Gives the Schonflies symbol for the point group.
(b) Lists the symmetry operations by class) for that group.
(c) Lists the characters, for all irreducible representations of each class of operation.
(d) Shows the irreducible representation for which six vectors x, y, z, and Rx, Ry, and Rz provide the basis.
(e) Shows how functions that are binary combinations of x,y, and z (xy or z2) provide bases for certain irreducible representations.
- A fundamental transition to a particular state is IR active if these three vectors x, y or z provide the basis.
- While, A fundamental transition to a particular state is Raman active if binary combinations of x,y, and z (xy or z2) provide bases for certain irreducible representations.
Explanation:-
- The character table for the point group D3h is given below.
| D3h | E | 2C3 (z) | \(\rm 3C_{2}^{'}\) | σh(xy) | 2S3 | 3σv | ||
| \(\rm A_{1}^{'}\) | +1 | +1 | +1 | +1 | +1 | +1 | - | x2 + y2, z2 |
| \(\rm A_{2}^{'}\) | +1 | +1 | −1 | +1 | +1 | −1 | Rz | - |
| E' | +2 | −1 | 0 | +2 | −1 | 0 | (x, y) | (x2 − y2, xy) |
| \(\rm A_{1}^{''}\) | +1 | +1 | +1 | −1 | −1 | −1 | - | - |
| \(\rm A_{2}^{''}\) | +1 | +1 | −1 | −1 | −1 | +1 | z | - |
| E'' | +2 | −1 | 0 | −2 | +1 | 0 | (Rx, Ry) | (xz, yz) |
- A fundamental transition to an \(\rm A_{1}^{'}\) state is not IR active as these three vectors x, y or z does not provide the basis.
- Thus statement A is not correct.
- A fundamental transition to the \(\rm A_{2}^{'}\) state is neither IR active nor Raman active as neither three vectors x, y or z nor the binary combinations of x,y, and z (xy or z2) provide bases.
- Thus statement B is correct.
- A fundamental transition to the \(\rm A_{2}^{''}\) state is not Raman active, the binary combinations of x,y, and z (xy or z2) does not provide bases.
- Thus statement A is not correct.
- A fundamental transition to the E'' state is not both IR and Raman active as only the binary combinations of x,y, and z (xy or z2) provide bases.
- Thus, statement B is also not correct.
Conclusion:-
Hence, a fundamental transition to the \(\rm A_{2}^{'}\) state is neither IR active nor Raman active.
Character Tables & Selection Rules Question 2:
Given below is a specific vibrational mode of BCl3 with ⊕ and Θ denoting movements of the respective atoms above and below the plane of the molecule respectively. The irreducible represen-tation of the vibrational mode and its IR / Raman activity are
| D3h | E | 2C3 | 3C2 | σb | 2S3 | 3σv | ||
| A1 | 1 | 1 | 1 | 1 | 1 | 1 | x2 + y2, z2 | |
| A2 | 1 | 1 | -1 | 1 | 1 | -1 | Rz | |
| E' | 2 | -1 | 0 | 2 | -1 | 0 | (x, y) | (x2 - y2, xy) |
| \(\rm A^"_1\) | 1 | 1 | 1 | -1 | -1 | -1 | ||
| \(\rm A^"_2\) | 1 | 1 | -1 | -1 | -1 | 1 | z | |
| E" | 2 | -1 | 0 | 1 | 1 | 0 | (Rx, Ry) | (xz, yz) |
Answer (Detailed Solution Below)
Character Tables & Selection Rules Question 2 Detailed Solution
The given vibrational is not symmetric w.r.t. σh
Correct option is (d)
Character Tables & Selection Rules Question 3:
For a point group, an incomplete character table is given below with one irreducible representation missing
| E | 2C3 | 3σv | |
| A1 | 1 | 1 | 1 |
| - | - | - | - |
| E | 2 | -1 | 0 |
The Mulliken symbol and characters of the missing representation are
Answer (Detailed Solution Below)
Character Tables & Selection Rules Question 3 Detailed Solution
Concept:
- The "Great Orthogonality Theorem" (GOT) States that the rows of characters are orthogonal vectors.
- The matrices of the different Irreducible Representations (IR) possess certain well-defined interrelationships and properties.
- According to GOT theorem, any two IR must be orthogonal to each other.
- According to the "Great Orthogonality Theorem," the Γ4 must satisfy,
Γn × Γ2=0 (where n =1 or 3)
Explanation:
- If the three characters of Γ2 are respectively 1 1 -1, then Γ1 × Γ2 will be
= 1× 1 × 1 + 2 × 1 × 1 + 3 × 1 × (-1)
= 6 - 6
= 0
- Also, If the four characters of Γ2 are respectively 1 1 -1, then Γ3 × Γ2 will be
Γ3 × Γ2
= 1× 1 × 2 + 2 × 1 × (-1) + 3 × 1 × 0
= 2 - 2
= 0
The character table of the C3V points group and the complete table is
| E | 2C3 | 3σv | |
| A1 | 1 | 1 | 1 |
| A2 | 1 | 1 | -1 |
| A3 | 2 | -1 | 0 |
Conclusion:-
- Hence, the four characters of Γ2 are respectively 1 1 -1
- The correct option is (c)
Character Tables & Selection Rules Question 4:
The character table of the C3v point group is given below, along with an additional reducible representation, Γ.
|
C3v E 2C3 3σv |
|||
|
A1 |
1 |
1 |
1 |
|
A2 |
1. |
1 |
-1 |
|
E |
2 |
-1 |
0 |
|
Γ 12 0 2 |
|||
Γ is given by
Answer (Detailed Solution Below)
Character Tables & Selection Rules Question 4 Detailed Solution
Concept:
- The sum of the squares of the dimensions of all the irreducible representations is equal to the order of the group.
∑di2 = h, where h = order of the group, and d = dimension
Explanation:-
- The irreducible representations can be obtained from reducible representations from the formula,
\(\eta _{IR}=\frac{1}{h}\sum _i X_iY_iZ_i\)
where h is the order of the group,
Xi is the character of the reducible representation,
Yi is the character of the irreducible representation,
Zi is the coefficient of the symmetry element.
- The characters of a reducible representation ΓR under the C3v point group are presented below:
|
C3v E 2C3 3σv |
|||
|
A1 |
1 |
1 |
1 |
|
A2 |
1. |
1 |
-1 |
|
E |
2 |
-1 |
0 |
|
Γ 12 0 2 |
|||
- The order of the group is,
= 1×12 + 2×12 + 3×12
= 6
- The correct coefficients of the irreducible representations A1 will be,
\(\eta _{IR}=\frac{1}{6}\left [1 \times1 \times 12+ 1\times 2\times 0 + 1 \times3\times2 \right ]\)
= 3
- The correct coefficients of the irreducible representations A2 will be,
\(\eta _{IR}=\frac{1}{6}\left [1 \times1 \times 12+ 1\times 2\times 0 + (-1) \times3\times2 \right ]\)
= 1
- The correct coefficients of the irreducible representations E will be,
\(\eta _{IR}=\frac{1}{6}\left [2 \times1 \times 12+ (-1)\times 2\times 0 + 0 \times3\times2 \right ]\)
= 4
- Thus, the irreducible representations of the C3v point group are
3A1 + A2 + 4E
Conclusion:-
- Hence, option 1 is correct.
Character Tables & Selection Rules Question 5:
The reducible representation, Γ, in the table is equal to the following superposition of the irreducible representations of C2v point group.
|
C2v |
E |
C2 |
σv |
\(\rm\sigma_{v}^{\prime}\) |
|
A1 |
1 |
1 |
1 |
1 |
|
A2 |
1 |
1 |
−1 |
−1 |
|
B1 |
1 |
−1 |
1 |
−1 |
|
B2 |
1 |
−1 |
−1 |
1 |
|
|
|
|
|
|
|
Γ |
8 |
−2 |
−6 |
4 |
Answer (Detailed Solution Below)
Character Tables & Selection Rules Question 5 Detailed Solution
Concept:
- The sum of the squares of the dimensions of all the irreducible representations is equal to the order of the group.
∑di2 = h, where h = order of the group, and d = dimension
Explanation:-
- The irreducible representations can be obtained from reducible representations from the formula,
\(\eta _{IR}=\frac{1}{h}\sum _i X_iY_iZ_i\)
where h is the order of the group,
Xi is the character of the reducible representation,
Yi is the character of the irreducible representation,
Zi is the coefficient of the symmetry element.
- The characters of a reducible representation ΓR under the C2v point group are presented below:
|
C2v |
E |
C2 |
σv |
\(\rm\sigma_{v}^{\prime}\) |
|
A1 |
1 |
1 |
1 |
1 |
|
A2 |
1 |
1 |
−1 |
−1 |
|
B1 |
1 |
−1 |
1 |
−1 |
|
B2 |
1 |
−1 |
−1 |
1 |
|
|
|
|
|
|
|
Γ |
8 |
−2 |
−6 |
4 |
- The order of the group is,
= 12 + 12 + 12 + 12
= 4
- The correct coefficients of the irreducible representations A1 will be,
\(\eta _{IR}=\frac{1}{4}\left [1 \times1 \times 8+ 1\times 1\times -2 + 1 \times1\times-6 + 1 \times 1 \times 4 \right ]\)
= 1
- The correct coefficients of the irreducible representations A2 will be,
\(\eta _{IR}=\frac{1}{4}\left [1 \times1 \times 8+ 1\times 1\times -2 + 1 \times-1\times-6 + 1 \times -1 \times 4 \right ]\)
= 2
- The correct coefficients of the irreducible representations B1 will be,
\(\eta _{IR}=\frac{1}{4}\left [1 \times1 \times 8+ 1\times -1\times -2 + 1 \times1\times-6 + 1 \times -1 \times 4 \right ]\)
= 0
- The correct coefficients of the irreducible representations B2 will be,
\(\eta _{IR}=\frac{1}{4}\left [1 \times1 \times 8+ 1\times -1\times -2 + 1 \times-1\times-6 + 1 \times 1 \times 4 \right ]\)
= 2
Thus, the irreducible representations of the C2v point group is
A1 ⊕ 2A2 ⊕ 0B1 ⊕ 5B2
= A1 ⊕ 2A2 ⊕ 5B2
Conclusion:-
Hence, A1 ⊕ 2A2 ⊕ 5B2 is correct.