Question

What are the symmetry species of vibration of BF3 that are both IR and Raman active?

Answer 1

Ans- 1) Determine the point group. BF3 is in the D3h point group.

2) Degrees of freedom. BF3 is a non-linear molecule, with 4 atoms. Using the equation 3N, we see that BF3 has 12 degrees of freedom. Using the equation 3N – 6, we see that BF3 has (12 – 6 =) 6 vibrational degrees of freedom. 3) Determine irreducible representations of Γtot. Three axes put on each atom. Γ4 atoms calculated by seeing the effect on the axes by all the symmetry operations

A1’: 1/12 [ (12x1x1) + (0x1x2) + (-2x1x3) + (4x1x1) + (-2x1x2) + (2x1x3) ] = 1/12 [ 12 – 6 + 4 – 4 + 6 ] = 1

A2’: 1/12 [ (12x1x1) + (0x1x2) + (-2x-1x3) + (4x1x1) + (-2x1x2) + (2x-1x3) ] = 1/12 [ 12 + 6 + 4 - 4 - 6 ] = 1

E’: 1/12 [ (12x2x1) + (0x-1x2) + (-2x0x3) + (4x2x1) + (-2x-1x2) + (2x0x3) ] = 1/12 [ 24 + 8 +4 ] = 3

A1”: 1/12 [ (12x1x1) + (0x1x2) + (-2x1x3) + (4x-1x1) + (-2x-1x2) + (2x-1x3) ] = 1/12 [ 12 – 6 –4 +4 –6 ] = 0

A2”: 1/12 [ (12x1x1) + (0x1x2) + (-2x-1x3) + (4x-1x1) + (-2x-1x2) + (2x1x3) ] = 1/12 [ 12 + 6 –4 +4 +6 ] = 2

E”: 1/12 [ (12x2x1) + (0x-1x2) + (-2x0x3) + (4x-2x1) + (-2x1x2) + (2x0x3) ] = 1/12 [ 24 – 8 – 4 ] = 1

Therefore Γtot = A1’ + A2’ + 3E’ + 2A2” + E” Γtot has twelve degrees of freedom. This agrees with our earlier answer.

3) Determine Γvib. We know that Γtot = Γtrans + Γrot + Γvib From character table we see that… Γtrans = E’ + A2” Γrot = A2’ + E” Therefore using Γvib = Γtot – Γtrans - Γrot Γvib = A1’ + 2E’ + A2” Γvib has six degrees of freedom. This agrees with our earlier answer.

4) Split into stretches and bends. BF3 has three bonds, so therefore has 3 stretches and 3 bends.

5) Determine irreducible representations of Γstretch. One axis put on each bond. Bond calculated by seeing the effect on the axes by all the symmetry operations.

A1’: 1/12 [ (3x1x1) + (0x1x2) + (1x1x3) + (3x1x1) + (0x1x2) + (1x1x3) ] = 1/12 [ 3 + 3 + 3 +3 ] = 1

A2’: 1/12 [ (3x1x1) + (0x1x2) + (1x-1x3) + (3x1x1) + (0x1x2) + (1x-1x3) ] = 1/12 [ 3 – 3 +3 - 3 ] = 0

E’: 1/12 [ (3x2x1) + (0x-1x2) + (1x0x3) + (3x2x1) + (0x-1x2) + (1x0x3) ] = 1/12 [ 6 + 6 ] = 1

A1”: 1/12 [ (3x1x1) + (0x1x2) + (1x1x3) + (3x-1x1) + (0x-1x2) + (1x-1x3) ] = 1/12 [ 3 + 3 – 3 –3 ] = 0

A2”: 1/12 [ (3x1x1) + (0x1x2) + (1x-1x3) + (3x-1x1) + (0x-1x2) + (1x1x3) ] = 1/12 [ 3 – 3 – 3 + 3 ] = 0

E”: 1/12 [ (3x2x1) + (0x-1x2) + (1x0x3) + (3x-2x1) + (0x1x2) + (1x0x3) ] = 1/12 [ 6 - 6 ] = 0

Therefore Γstretch = A1’ + E’ 6) Determine Γbend. We know that Γbend = Γvib - Γstretch Therefore Γbend = E’ + A2” 7) Assign irreducible representations to spectra. From character tables we see that only E’ will be visible in both IR and Raman spectra.

E’ stretch will be at higher energy (1505 cm-1)

E’ bend will be at lower energy (482cm-1)

From character tables we see that

A1’ will not be visible in the IR. A1’ stretch will be at 888 cm-1 From character tables we see that

A2” will not be visible in Raman.

A2” bend will be at 718cm-1 1505 (R, IR)

E’ stretches 888 (R) A1’ stretch 718 (IR)

A2” bend 482 (R, IR)E’ bends 8) Sketch these vibrations (although not specifically asked for, this is a common question).