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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Latest Character Tables & Selection Rules MCQ Objective Questions
Top Character Tables & Selection Rules MCQ Objective Questions
Character Tables & Selection Rules Question 1:
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 1 Detailed Solution
The given vibrational is not symmetric w.r.t. σh
Correct option is (d)
Character Tables & Selection Rules Question 2:
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 2 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 3:
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 3 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 4:
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 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 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.