6.8: CH₄

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Tetrahedral Symmetry for Methane

The infrared spectrum of methane shows two absorptions: a bend at 1306 cm^-1 and a stretch at 3019 cm^-1. Demonstrate that a symmetry analysis assuming tetrahedral symmetry for methane is consistent with this spectroscopic data. Also predict how many Raman active modes methane should have.

\[ \begin{matrix} ~ & \begin{array} E E & C_3 & C_2 & S_4 & \sigma \end{array} \\ C_{Td} = & \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & -1 & -1 \\ 2 & -1 & 2 & 0 & 0 \\ 3 & 0 & -1 & 1 & -1 \\ 3 & 0 & -1 & -1 & 1 \end{bmatrix} & \begin{array} A A_1:~x^2 + y^2 + z^2 \\ A_2 \\ E:~ 2z^2 - x^2 - y^2,~x^2-y^2 \\ T_1:~ (R_x,~R_y,~R_z) \\ T_2:~ (x,~y,~z),~(xy,~xz,~yz) \end{array} & Td = \begin{bmatrix} 1 \\ 8 \\ 3 \\ 6 \\ 6 \end{bmatrix} & \Gamma_{uma} = \begin{bmatrix} 5 \\ 2 \\ 1 \\ 1 \\ 3 \end{bmatrix} & \Gamma_{bonds} = \begin{bmatrix} 4 \\ 1 \\ 0 \\ 0 \\ 2 \end{bmatrix} \end{matrix} \nonumber \]

\[ \begin{matrix} A_1 = (C_{Td}^T)^{<1>} & A_2 = (C_{Td}^T )^{<2>} & E = (C_{Td}^T )^{<3>} & T_1 = ( C_{Td}^T )^{<4>} \\ T_2 = (C_{Td}^T )^{<5>} & \Gamma_{tot} = \overrightarrow{( \Gamma_{uma} T_2)} & h = \sum Td & \Gamma_{tot}^T = \begin{pmatrix} 15 & 0 & -1 & -1 & 3 \end{pmatrix} & i = 1 .. 5 \end{matrix} \nonumber \]

\[ \begin{matrix} \Gamma_{vib} = \Gamma_{tot} - T_1 - T-2 & \text{Vib}_i = \frac{ \sum \overrightarrow{ \left[ Td (C_{Td}^T )^{<i>} \Gamma_{vib} \right] }}{h} & \text{Vib} = \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 2 \end{bmatrix} \begin{array} A A_1:~x^2 + y^2 + z^2 \\ A_2 \\ E:~ 2z^2 - x^2 - y^2,~x^2-y^2 \\ T_1:~ (R_x,~R_y,~R_z) \\ T_2:~ (x,~y,~z),~(xy,~xz,~yz) \end{array} \\ \Gamma_{stretch} = \Gamma_{bonds} & \text{Stretch}_i = \frac{ \sum \overrightarrow{ \left[ Td (C_{Td}^T )^{<i>} \Gamma_{stretch} \right] }}{h} & \text{Stretch} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \begin{array} A A_1:~x^2 + y^2 + z^2 \\ A_2 \\ E:~ 2z^2 - x^2 - y^2,~x^2-y^2 \\ T_1:~ (R_x,~R_y,~R_z) \\ T_2:~ (x,~y,~z),~(xy,~xz,~yz) \end{array} \\ \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} & \text{Bend}_i = \frac{ \sum \overrightarrow{ \left[ Td (C_{Td}^T )^{<i>} \Gamma_{bend} \right] }}{h} & \text{Bend} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 1 \end{bmatrix} \begin{array} A A_1:~x^2 + y^2 + z^2 \\ A_2 \\ E:~ 2z^2 - x^2 - y^2,~x^2-y^2 \\ T_1:~ (R_x,~R_y,~R_z) \\ T_2:~ (x,~y,~z),~(xy,~xz,~yz) \end{array} \end{matrix} \nonumber \]

Thus the vibrational modes have A₁, E, and T₂ symmetry. Only the two T₂ modes are infrared active which is consistent with the experimental data quoted above. One of the T₂ modes is a stretch (3019 cm^-1) and the other is a bend (1306 cm^-1).

This symmetry analysis predicts that all of vibrational modes are Raman active - one singly degenerate mode (A₁), one doubly degenerate mode (E), and two triply degenerate modes (T₂). Indeed four Raman active modes are found at 3019, 2917, 1534, and 1306 cm^-1. Note, as expected from the symmetry analysis, there are two coincidences between the IR and Raman spectra.