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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.


    This page titled 6.8: CH₄ is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Frank Rioux via source content that was edited to the style and standards of the LibreTexts platform.

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