The equivalent symmetry operations for S36 and S63 are, respectively,

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CSIR-UGC (NET) Chemical Science: Held on (16 Feb 2022)
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  1. C3 and C2
  2. σh and i
  3. σh and E
  4. i and E

Answer (Detailed Solution Below)

Option 4 : i and E
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Detailed Solution

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Concept:

  • Sn represents the improper axis of symmetry which is includes application of rotation symmetry followed by the reflection or vice-versa.
  •  Therefore, we can write, \(S_n=C_n\times\sigma \). Cn here represents the axis of rotation symmetry (of order n). rotation by \(\frac{360^0}{n}\), gives same arrangement (identity)\(\sigma\) is the reflection plane.

 

Explanation :

Equivalent symmetry of \(S_{6}^{3}\)

\(S_{6}^{3}\) can be written as:

 \(S_{6}^{3} = C_{6}^{3}\times \sigma^{3} \)

Even times application of plane symmetry gives identity and odd times application gives \(\sigma\) itself.

 \(\sigma^{even}=E\;\; and\; \;\sigma^{odd} =\sigma \)

So, \(S_{6}^{3}\) can be rewritten as: 

\(S_{6}^{3} = C_{6}^{3}\times \sigma \)

 \(S_{6}^{3} = C_{2}^{1}\times \sigma \)

\( C_{2}^{1}\times \sigma \) represents application of \(C_2\) rotation symmetry followed by reflection using plane symmetry, which is nothing but the improper axis of symmetry of order 2.

 \(i.e.,\; C_{2}^{1}\times \sigma =S_2 \)

Also, \(S_2 \) results in inversion of the molecule (i.e., \(S_2 =i\) )

 \(\therefore C_{2}^{1}\times \sigma =S_2 =i \)

Equivalent symmetry of \(S_{3}^{6}\)

 \(S_{3}^{6}\) can be written as: 

\(S_{3}^{6} = C_{3}^{6}\times \sigma^{6} \)

application of Cn symmetry, same number of times as multiple of its order results in identity(E). so we can write:

\(S_{3}^{6}=C_{3}^{6}\times E\;\;\;\;(\because\sigma^{even}=E) \)

\(S_{3}^{6}=C_{1}^{1}\times E \) 

(here, \(C_{1}^{1}\) is just another way of writing identity (E) as rotation by \(\frac{360^0}{1}\) gives the same orientation of molecule.)

\(S_{3}^{6}=E\times E =E\)

Conclusion:

The equivalent symmetry of \(S_{6}^{3}\) and \(S_{3}^{6}\) is i and E respectively.

 

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