C 3v Symmetry - BCl 3 The infrared spectrum of BCl 3 shows vibrational bands at 995, 480, and 244 cm -1 , while Raman bands appear at 995, 471, and 244 cm -1 . Is the geometry of the molecule trigonal pyramid (C 3v ) or trigonal planar (D 3h )? Is your answer to this question consistent with chemical bonding principles (VSEPR)? Assign symmetry labels to the vibrational bands and identify the stretches and bends. \[ \begin{matrix} ~ & \begin{array} E & C_3 & \sigma_v \end{array} \\ C_{C3v} = & \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 2 & -1 & 0 \end{pmatrix} & \begin{array} A_1:~ z,~x^2 + y^2,~z^2 \\ A_2:~R_z \\ E:~(x,~y)(R_x,~R_y),~(x^2 + y^2,~xy)(xz,~yz) \end{array} & C3v = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} & \Gamma_{uma} = \begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix} & \Gamma_{bonds} = \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} \end{matrix} \nonumber \] \[ \begin{matrix} A_1 = (C_{C3v}^T)^{<1>} & A_2 = (C_{C3v}^T)^{<2>} & E = (C_{C3v}^T)^{<3>} & h = \sum C3v \end{matrix} \nonumber \] \[ \begin{matrix} \Gamma_{tot} = \overrightarrow{[\Gamma_{uma} (A_1 + E)]} & \Gamma_{tot}^T = \begin{pmatrix} 12 & 0 & 2 \end{pmatrix} & \Gamma_{vib} = \Gamma_{tot} - A_1 - A_2 - 2E \end{matrix} \nonumber \] \[ \begin{matrix} i = 1 .. 3 & \text{Vib}_i = \frac{ \sum \overrightarrow{ [C3v (C_{C3v}^T)^{<i>} \Gamma_{vib} ]}}{h} & \text{Vib} = \begin{pmatrix} 2 \\ 0 \\ 2 \end{pmatrix} \begin{array} A_1:~ z,~x^2 + y^2,~z^2 \\ A_2:~R_z \\ E:~(x,~y)(R_x,~R_y),~(x^2 + y^2,~xy)(xz,~yz) \end{array} \\ \Gamma_{stretch} = \Gamma_{bonds} & \text{Stretch}_i = \frac{ \sum \overrightarrow{ [C3v (C_{C3v}^T)^{<i>} \Gamma_{stretch} ]}}{h} & \text{Stretch} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \begin{array} A_1:~ z,~x^2 + y^2,~z^2 \\ A_2:~R_z \\ E:~(x,~y)(R_x,~R_y),~(x^2 + y^2,~xy)(xz,~yz) \end{array} \\ \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} & \text{Bend}_i = \frac{ \sum \overrightarrow{ [C3v (C_{C3v}^T)^{<i>} \Gamma_{bend} ]}}{h} & \text{Bend} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \begin{array} A_1:~ z,~x^2 + y^2,~z^2 \\ A_2:~R_z \\ E:~(x,~y)(R_x,~R_y),~(x^2 + y^2,~xy)(xz,~yz) \end{array} \end{matrix} \nonumber \] This analysis predicts that there should be 4 IR and 4 Raman active vibrations. It also predicts that the IR and Raman vibrations should be coincident. This is not in agreement with the experimental data, so BCl3 does not have trigonal pyramidal geometry. This analysis also predicts that the stretches should have A1 and E symmetry, but since it has already been concluded that the molecule does not have C 3v symmetry this result will not be discussed further. D 3h Symmetry - BCl 3 \[ \begin{matrix} ~ & \begin{array} E & C_3 & C_2 & \sigma_h & S_3 & \sigma_v \end{array} \\ C_{C3v} = & \begin{pmatrix} 1 & 1 & 1 1 & 1 & 1 \\ 1 & 1 & -1 & 1 & 1 & -1 \\ 2 & -1 & 0 & 2 & -1 & 0 \\ 1 & 1 & 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & 1 \\ 2 & -1 & 0 & -2 & 1 & 0 \end{pmatrix} & \begin{array} A_1':~ x^2 + y^2,~z^2 \\ A_2':~Rz \\ E':~(x,~y), ~(x^2 + y^2,~xy) \\ A1": \\ A2":~z \\ \text{E": (Rx, Ry), (xz, yz)} \end{array} & D3h = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 1 \\ 2 \\ 3 \end{bmatrix} & \Gamma_{uma} = \begin{bmatrix} 4 \\ 1 \\ 2 \\ 4 \\ 1 \\ 2 \end{bmatrix} & \Gamma_{bonds} = \begin{bmatrix} 3 \\ 0 \\ 1 \\ 3 \\ 0 \\ 1 \end{bmatrix} \end{matrix} \nonumber \] \[ \begin{matrix} A_1 = (C_{D3h}^T)^{<1>} & A_2 = (C_{D3h}^T)^{<2>} & E = (C_{D3h}^T)^{<3>} & A_{11} = (C_{D3h}^T)^{<4>} \\ A_{21} = (C_{D3h}^T )^{<5>} & E_1 = (C_{D3h}^T )^{<6>} & h = \sum D3h & \Gamma_{tot} = \overrightarrow{[ \Gamma_{uma} (A_{21} + E)]} \end{matrix} \nonumber \] \[ \begin{matrix} \Gamma_{tot}^T = \begin{pmatrix} 12 & 0 & -2 & 4 & -2 & 2 \end{pmatrix} & \Gamma_{vib} = \Gamma_{tot} - A_2 - E - A_{21} - E_1 & i = 1 .. 6 \\ \Gamma_{stretch} = \Gamma_{bonds} & \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} \end{matrix} \nonumber \] \[ \begin{matrix} \text{Vib}_i = \frac{ \sum \overrightarrow{ [D3h (C_{D3h}^T)^{<i>} \Gamma_{vib} ]}}{h} & \text{Stretch}_i = \frac{ \sum \overrightarrow{ [D3h (C_{D3h}^T)^{<i>} \Gamma_{stretch} ]}}{h} & \text{Bend}_i = \frac{ \sum \overrightarrow{ [D3h (C_{D3h}^T)^{<i>} \Gamma_{bend} ]}}{h} \\ \text{Vib} = \begin{bmatrix} 1 \\ 0 \\ 2 \\ 0 \\ 1 \\ 0 \end{bmatrix} \begin{array} A_1':~ x^2 + y^2,~z^2 \\ A_2':~Rz \\ E':~(x,~y), ~(x^2 + y^2,~xy) \\ A1": \\ A2":~z \\ \text{E": (Rx, Ry), (xz, yz)} \end{array} & \text{Stretch} = \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \begin{array} A_1':~ x^2 + y^2,~z^2 \\ A_2':~Rz \\ E':~(x,~y), ~(x^2 + y^2,~xy) \\ A1": \\ A2":~z \\ \text{E": (Rx, Ry), (xz, yz)} \end{array} \\ \text{Bend} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} \begin{array} A_1':~ x^2 + y^2,~z^2 \\ A_2':~Rz \\ E':~(x,~y), ~(x^2 + y^2,~xy) \\ A1": \\ A2":~z \\ \text{E": (Rx, Ry), (xz, yz)} \end{array} \end{matrix} \nonumber \] This analysis shows that there are 3 IR active modes (2E', A2") and 3 Raman active modes (A1', E'). It also show that there are two coincidences (2E'). Thus, D 3h symmetry for BCl 3 is consistent with the spectroscopic data. Thus the specific assignments are: A1' = 471; A2" = 480; E' = 995; E' = 244. From the analysis above it can be seen that the stretches occur at 995 and 471, and the bends are at 480 and 244 cm -1 . VSEPR theory would predict a trigonal planar arrangement of chlorine atoms around the electron deficient boron, so the symmetry analysis and the bonding theory are in agreement with each other and the spectroscopic measurements.