6.5.1: Buckminsterfulerene

C₆₀ Has Icosahedral Symmetry

Buckminsterfulerene has four IR active vibrational modes (528, 577, 1180, 1430 cm^-1) and ten Raman active modes (273, 436, 496, 710, 773, 110, 1250, 1435, 1470, 1570 cm^-1). Demonstrate that the assumption of icosahedral symmetry for C₆₀ is consistent with this data.

\[ \begin{matrix} \begin{array} E & & E & C_5 & & C_5^2 & C_3 & & C_2 & i & & S_{10} & S_{10}^3 & & S_6 & \sigma \end{array} & ~ \\ \text{CIh} = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 3 & \frac{1 + \sqrt{5}}{2} & \frac{1- \sqrt{5}}{2} & 0 & -1 & 3 & \frac{1- \sqrt{5}}{2} & \frac{1+ \sqrt{5}}{2} & 0 & -1 \\ 3 & \frac{1 - \sqrt{5}}{2} & \frac{1+ \sqrt{5}}{2} & 0 & -1 & 3 & \frac{1+ \sqrt{5}}{2} & \frac{1- \sqrt{5}}{2} & 0 & -1 \\ 4 & -1 & -1 & 1 & 0 & 4 & -1 & -1 & 1 & 0 \\ 5 & 0 & 0 & -1 & 1 & 5 & 0 & 0 & -1 & 1 \\ 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 \\ 3 & \frac{1 + \sqrt{5}}{2} & \frac{1 - \sqrt{5}}{2} & 0 & -1 & -3 & - \frac{1 - \sqrt{5}}{2} & - \frac{1 + \sqrt{5}}{2} & 0 & 1 \\ 3 & \frac{1 - \sqrt{5}}{2} & \frac{1 + \sqrt{5}}{2} & 0 & -1 & -3 & - \frac{1 + \sqrt{5}}{2} & - \frac{1 - \sqrt{5}}{2} & 0 & 1 \\ 4 & -1 & -1 & 1 & 0 & -4 & 1 & 1 & -1 & 0 \\ 5 & 0 & 0 & -1 & 1 & -5 & 0 & 0 & 1 & -1 \end{bmatrix} & \begin{array} \text{Ag: }x^2 + y^2 + z^2 \\ \text{Eg: } 2z^2-x^2-y^2,x^2-y^2 \\ \text{T1g: Rx, Ry, Rz} \\ \text{T2g} \\ \text{Gg} \\ \text{Hg: }2z^2 -x^2-y^2,~x^2-y^2, ~xy,~yz,~xz \\ \text{Au} \\ \text{T1u: x, y, z} \\ \text{T2u} \\ \text{Gu} \\ \text{Hu} \end{array} \end{matrix} \nonumber \]

\[ \begin{matrix} \text{Ih:} \begin{pmatrix} 1 & 12 & 12 & 20 & 15 & 1 & 12 & 12 & 20 & 15 \end{pmatrix} & \text{Ih = Ih}^T & \Gamma_{uma} = \begin{pmatrix} 60 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 4 \end{pmatrix} & \Gamma_{uma} = \Gamma_{uma}^T \\ \Gamma_{bonds} = \begin{pmatrix} 90 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 8 \end{pmatrix} & \Gamma_{bonds} = \Gamma_{bonds}^T & \Gamma_{stretch} = \Gamma_{bonds} \end{matrix} \nonumber \]

\[ \begin{matrix} \text{Ag} = ( \text{CIh}^T )^{<1>} & \text{T}_{1g} = ( \text{CIh}^T )^{<2>} & \text{T}_{2g} = ( \text{CIh}^T )^{<3>} & \text{G}_{g} = ( \text{CIh}^T )^{<4>} & \text{H}_{g} = (\text{CIh}^T)^{<5>} \\ \text{A}_{u} = ( \text{CIh}^T )^{<6>} & \text{T}_{1u} = ( \text{CIh}^T )^{<7>} & \text{A}_{u} = ( \text{CIh}^T )^{<8>} & \text{G}_{u} = ( \text{CIh}^T )^{<9>} & \text{H}_{u} = (\text{CIh}^T)^{<10>} \end{matrix} \nonumber \]

\[ \begin{matrix} h = \sum \text{Ih} & h = 120 & \Gamma_{tot} = \overrightarrow{( \Gamma_{uma} T1u)} & \Gamma_{vib} = \Gamma_{tot} - T1g - T1u & \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} & i = 1 .. 10 \end{matrix} \nonumber \]

\[ \begin{matrix} \text{Vib}_i = \frac{\sum \overrightarrow{[\text{Ih} ( \text{CIh}^T)^{<i>} \Gamma_{vib}]}}{h} & \text{Vib} = \begin{bmatrix} 2 \\ 3 \\ 4 \\ 6 \\ 8 \\ 1 \\ 4 \\ 5 \\ 6 \\ 7 \end{bmatrix} \begin{array} \text{Ag: }x^2 + y^2 + z^2 \\ \text{Eg: } 2z^2-x^2-y^2,x^2-y^2 \\ \text{T1g: Rx, Ry, Rz} \\ \text{T2g} \\ \text{Gg} \\ \text{Hg: }2z^2 -x^2-y^2,~x^2-y^2, ~xy,~yz,~xz \\ \text{Au} \\ \text{T1u: x, y, z} \\ \text{T2u} \\ \text{Gu} \\ \text{Hu} \end{array} \end{matrix} \nonumber \]

The 4 T_1u modes are IR active and the 2 A_g and 8 H_g modes are Raman active. Also there are no coincidences. Thus the assumption of icosahedral symmetry is consistent with the spectroscopic data.