Chapter 4 Answers to Self-test Questions

4.1 Use Eqn. 4.3 and the reduced masses to calculate the ratio of the harmonic vibrational wavenumbers,

 

If  is at 2991.0 cm⁻¹, use your ratios to predict the positions of and .

 

When carrying out isotopic substitution the key assumption is that the force constant k is the same in both isotopomers as it is related to the bonding which is determined by the protons and electrons, and changing the number of neutrons makes no difference. For diatomics the ratio of the vibrational frequencies or wavenumbers is the ratio of the masses as shown below. While using integer masses was sufficiently illustrative in Example 4.1, for this more quantitative analysis, more accurate masses of the individual isotopes are required. As the wavenumber value is given to five significant figures, it is good practice to use one more significant figure in the calculations, and to round at the end, rather than at each stage to avoid cumulative rounding errors. The accurate masses are: ¹H, 1.007825 g mol⁻¹; ²H/D, 2.014102 g mol⁻¹; ³⁵Cl, 34.968852 g mol⁻¹, ³⁷Cl, 36.965903 g mol⁻¹.

     Eqn. 4.3

 

Therefore, as expected from the answer to Example 4.1, there is only a small shift of 2.3 cm⁻¹ when ³⁵Cl is substituted by ³⁷Cl in HCl, but a much larger change of 845.8 cm⁻¹ when ¹H is replaced by ²D.

 

 

4.2 On isotopic substitution it is assumed that k remains constant. Making use of this, predict the position of the CO stretching mode in (a) ¹³C¹⁶O and (b) ¹²C¹⁸O.

 

This self test requires the value of k calculated in Example 4.2 (1855 N m⁻¹), and that the experimental anharmonic, vibrational wavenumber, , for ¹²C¹⁶O is 2143 cm⁻¹. As for the previous self-test, the accurate masses of the isotopes are required. To five significant figures these are: ¹²C, 12.000 g mol⁻¹; ¹³C, 13.003 g mol⁻¹, ¹⁶O, 15.995 g mol⁻¹; ¹⁸O, 17.999 g mol⁻¹.

 

As might be expected, the greater change in mass on going from ¹²C¹⁶O to ¹²C¹⁸O compared to ¹³C¹⁶O results in a larger shift of the vibrational wavenumber.

4.3 Determine the IR and Raman activity of the stretching modes of BF₃ shown in Figure 4.3ST.

 

In the symmetric stretching mode, all the B-F bonds are getting longer and shorter in phase, therefore the dipole moment is zero at all stages of the molecular vibration, so it is IR inactive. However, as the polarisability envelope around the molecule is getting larger and smaller, there is a change in the polarisability and so the stretching symmetric mode is Raman active (and as this is a totally symmetric mode it is polarised). In the asymmetric stretching mode there is a dipole change in the y direction in the top representation and a dipole in the x direction in the bottom one, so this degenerate vibrational mode is IR active. It is also Raman active as there is a change in the polarisability as well. What is clear is that whilst it is usually reasonably easy to spot IR activity, Raman activity is harder to visualise. It is for this reason that the group theoretical approach developed in later sections is more useful for determining IR and Raman activity.

 

 

4.4 What is the position for a 200 cm⁻¹ Stokes shift (in nm) when using 1064 nm excitation?

 

1064 nm is equivalent to 9398.5 cm⁻¹, so the Stokes peak at 200 cm⁻¹ is at 9198.5 cm⁻¹, which is 1087 nm. Therefore, a Stokes shift of 200 cm⁻¹ corresponds to 23 nm at 1064 nm excitation but only 5.7 nm at 532 nm excitation, indicating the non-linear relationship between cm⁻¹ and nm units.

 

4.5 Explain why trans-[Os(bipy)₂(O)₂] has one n_M=O IR band at 872 cm⁻¹, whilst cis-[Os(bipy)₂(O)₂] (bipy = 2,2′-bipyridine) has two n_M=O IR bands at 863 and 833 cm⁻¹.

The structures of the two isomers are given in Figure 4.5ST. In the trans isomer the O=Os=O unit is linear. There is a substantial change in the dipole moment for the antisymmetric n_M=O vibrational mode resulting in an IR active mode. The symmetric n_M=O mode results in no change in the dipole moment and hence this is IR inactive. (If the centre of symmetry is removed by the arrangement of the bipy ligands, then it might be very very weak). In the cis isomer, with a bent O=Os=O unit, there is a change in the dipole moment during both the symmetric and antisymmetric stretching modes, so both are IR active.

Figure 4.5ST Structures of trans-[Os(bipy)₂(O)₂] and cis-[Os(bipy)₂(O)₂] (bipy = 2,2′-bipyridine)

 

 

4.6 The solid state IR spectrum of [Co₂(CO)₈] displays n_CO bands at 2071, 2044, 2042, 1866 and 1857 cm⁻¹, but in a non-coordinating solvent there are only n_CO bands at 2069, 2055 and 2032 cm⁻¹. Account for these observations.

 

In a non-coordinating solvent, the n_CO bands at 2069, 2055 and 2032 cm⁻¹ indicate that all the CO ligands are terminal (see Figure 4.13 in the main text), and these are consistent with a D_3d geometry shown in Figure 4.6ST (The D_2d geometry is observed after photolysis in low temperature matrices). Figure 4.13 indicates that in solid state [Co₂(CO)₈] all five of the n_CO bands at 2071, 2044, 2042, 1866 and 1857 cm⁻¹ could in principle be associated with terminal CO ligands. However, those at 1866 and 1857 cm⁻¹ fall in the region where they could also be assigned to bridging CO ligands. In this case it is more reasonable to explain the differences in the n_CO modes to a mixture of terminal and bridging modes, rather than to such large differences in the back-bonding to terminal CO ligands, especially with only cobalt and CO present. These data are consistent with the C_2v structure. While an 18-electron analysis would indicate the presence of a Co-Co bond in the carbonyl bridged C_2v structure, the consensus of theoretical studies is that there is no appreciable metal-metal bonding in this (or [Fe₂(CO)₉) and that the dimer is held together by 3c-2e bonds in the CO-Co-CO bridging units. (see J. C. Green, M. L. H Green, G. Parkin, Chem. Comm. 48, 11482 (2012))

 

Figure 4.6ST Structures of cobalt carbonyl complexes

4.7

(a) Account for the changes in v_CO in the following complexes: [Ni(CO)₃(PMe₃)], 2064.1 cm⁻¹; [Ni(CO)₃(PPh₃)], 2068.9 cm⁻¹; [Ni(CO)₃(PF₃)], 2110.8 cm⁻¹.

The position of the n_CO modes indicates the extent of back-bonding from the metal to the CO ligand. The data are for the A₁ modes and are taken from C A Tolman, J. Am. Chem. Soc. 92, 2953 (1970). Figure 4.15 (in the main text) shows that greater back-bonding increases the population of the π* orbital on the CO, which results in a weaker C≡O bond and hence lower n_CO mode (The M-C bond becomes stronger). The extent of back-bonding is determined by the extent of electron density on the metal, which in turn is dependent on the metal formal oxidation state, and the Lewis base/acid behaviour of the other, non-carbonyl ligands and the availability of suitable acceptor orbitals on the ligands. For CO the acceptor orbital is the π* orbital, and for phosphane ligands it is thought to be a σ* orbital rather than P 3d orbital that acts as the acceptor. (The IUPAC nomenclature for PR₃ ligands is phosphane, with phosphine restricted to PH₃, although phosphine is still in very widespread use for the alkyl and aryl containing ligands) In this case the nickel is in the same oxidation state for all three complexes, so any differences are due to the three phosphorus containing ligands. Therefore, in terms of n_CO modes the complexes can be ordered: [Ni(CO)₃(PMe₃)], 2064.1 cm⁻¹ < [Ni(CO)₃(PPh₃)], 2068.9 cm⁻¹ < [Ni(CO)₃(PF₃)], 2110.8 cm⁻¹. The lowest v_CO mode associated with PMe₃ indicates that there is a largest degree of backbonding in this complex compared to the PPh₃ complex, with lowest amount occurring in the PF₃ complex. As expected on electronegativity grounds PF₃ is a weak σ donor ligand, but a very effective π acceptor ligand, often thought of as similar to CO. Although PPh₃ and PR₃ ligands have similar σ base and π acceptor behaviour, in general the alkyl phosphanes are better σ bases and weaker π acids than the aryl phosphanes. Therefore, the v_CO mode is at lower wavenumber for[Ni(CO)₃(PMe₃)] (2064.1 cm⁻¹) than [Ni(CO)₃(PPh₃)] (2068.9 cm⁻¹) However, care needs to be exercised when looking at the experimental data as a change in solvent can result in shifts of the v _CO modes that are larger than the differences between the phosphane ligands

(b) Account for the changes in the IR active n_CO modes in the following tetrahedral complexes: [Cu(CO)₄]⁺, 2184 cm⁻¹; [Ni(CO)₄], 2058 cm⁻¹; [Co(CO)₄]⁻, v_CO 1883 cm⁻¹; [Fe(CO)₄]²⁻, v_CO 1729 cm⁻¹; [Mn(CO)₄]³⁻, 1670 cm⁻¹.

These examples show that as the electron density increases on the metal with a reduction in formal oxidation state the n_CO modes move to lower wavenumber indicating greater backbonding.

4.8 The ν_NO mode in [RhCl₂(NO)(PPh₃)₂] is at 1630 cm⁻¹, in [CoCl₂(NO)(PPh₃)₂] there are two ν_NO modes at 1725 and 1640 cm⁻¹. Predict whether the NO ligand is bent or linear in these complexes.

 

Figure 4.16 (in the main text) shows that the position of the n_NO mode is diagnostic for linear or bent coordination of the nitrosyl ligand. The n_NO mode at 1630 cm⁻¹ in [RhCl₂(NO)(PPh₃)₂] is diagnostic of a bent metal nitrosyl unit, as confirmed by single crystal X-ray diffraction where the bent nitrosyl ligand is in the axial position of a square based pyramid where the PPh₃ and Cl⁻ ligands are trans to each other.

In [CoCl₂(NO)(PPh₃)₂] the _NO mode at 1725 cm⁻¹ is characteristic of a liner metal nitrosyl linkage, and that at 1640 cm⁻¹ of a bent nitrosyl unit. Therefore, there is a mixture of linear and bent nitrosyl ligands in [CoCl₂(NO)(PPh₃)₂] indicating an equilibrium between 18 electron (linear M-NºO) and 16 electron (bent M-N=O) complexes.

(S. Z. Goldberg, C. Kubiak, C. D. Meyer and R. Eisenberg, Inorg. Chem. 14 1650 (1975), C. P. Brock, J. P. Collman, G. Dolcetti, P. H. Farnham, J. A. Ibers, J. E. Lester and C. A. Reed, Inorg. Chem. 12 1304 (1973).)

4.9 Identify and sketch the C₃ and C₂ symmetry operations in tetrahedral SnCl₄.

(Hint: look for axes that are along bonds, bisect bonds, or are perpendicular to planes)

In Figure 4.9ST there is a C₃ rotation axis (symmetry element) down each Sn-Cl bond which gives rise to 120° (C₃¹) and 240° rotations (C₃²), which are the symmetry operations. Therefore, there are a total of eight C₃ symmetry operations in a tetrahedral molecule.

 

To identify the C₂ symmetry element and operations it is convenient to reorient the molecule, so that C₂ axis is vertical. In this case the C₂ rotation axis bisects the Sn-Cl bonds, and there are three C₂ symmetry elements. Therefore, there are three C₂ symmetry operations in a tetrahedral molecule.

 

 

Figure 4.9ST C₃ and C₂ rotations in SnCl₄.

 

4.10 Sketch the effect of the E, C₃, C₂, σ₄, and σ_d symmetry operations for tetrahedral molecules such as SnCl₄ and the E, C₃, C₂, σ₃, σ_h and σ_v symmetry operations for trigonal bipyramidal molecules such as PF₅.

As for all molecules, the identity, E, is the “do nothing” symmetry operation in SnCl₄, so all the atoms are in the same location before and after the symmetry operation as shown in Figure 4.10ST(a). The C₃ and C₂ symmetry operations are discussed in the previous self-test. The σ₄, improper rotation, symmetry operation involves a 90° rotation about an axis that bisects the bonds, followed by a reflection in a plane perpendicular to this. As there is also a 270° rotation directed along the same direction, followed by a reflection in a plane perpendicular, there are a total of six σ₄ symmetry operations in a tetrahedral molecule. The σ_d mirror plane are in the planes of the Cl-Sn-Cl units, so there are six in total.

 

 

Figure 4.10ST(a) Effect of E, C₃, C₂, σ₄, and σ_d symmetry operations on SnCl₄.

 

For PF₅ the identity symmetry operation, E, is as usual the do nothing one. As shown in Figure 5.1^ST(b), the C₃ rotation axis is aligned along the axial F-P-F direction, and there are two symmetry operations, C₃¹ (120°) and C₃² (240°). The C₂ symmetry operations are directed along the equatorial bonds, so there are a total of three of these. The σ₃ improper rotation involves either a 120° or 240° rotation along the axial bonds, followed by a reflection in the equatorial plane so all the F atoms move. In the character tables there are only two of these listed, which are the σ₃¹ and σ₃⁵ symmetry operations as σ₃² and σ₃⁴ are equivalent to C₃² and C₃¹, respectively. In the σ_h symmetry operation, the reflection plane contains the central P and the three equatorial F atoms, and there is just one of these. In contrast there are three σ_v symmetry operations, as the reflection plane now contains the central P, the two axial F and one of the equatorial F atoms.

 

Figure 4.10ST(b) The effect of E, C₃, C₂, σ₃, σ_h and σ_v symmetry operations on PF₅.

 

4.11 Assign the following linear molecules to their point groups: N₂O, NO, HCºCH, C₃O₂ (O=C=C=C=O).

As all of these molecules are linear they will only belong to either the D_∞h point group or the C_∞v point group. The centrosymmetric molecules will have a D_∞h point group, whereas those without a centre of symmetry will have a C_∞v point group.

Centrosymmetric, D_∞h

HCºCH, C₃O₂ (O=C=C=C=O)

Non-centrosymmetric, C_∞v

N₂O, NO

 

4.12 Determine the point groups of (a) NH₃, (b) SO₂Cl₂, (c) XeOF₄, and (d) cis-[PtCl₂(NH₃)₂].

All of these have C_nv point groups as they have a principal rotation axis (C_n), no C₂ rotation axes perpendicular to C_n, no horizontal mirror plane (σ_h), but they have vertical mirrors planes, σ_v. The point groups are shown in Figure 4.12ST.

While it is conventional to draw cis-[PtCl₂(NH₃)₂] in a square planar arrangement as in the first illustration, the principal rotation axis bisects the Cl-Pt-Cl and N-Pt-N bonds, so to get the z axis in the vertical position, the second representation is required.

 

Figure 4.12ST Point groups of NH₃, SO₂Cl₂, XeOF₄, and cis-[PtCl₂(NH₃)₂].

4.13 Determine the point groups of (a) [PtCl₄]²⁻, (b) trans-[PtCl₂(NH₃)₂], (c) XeF₄ and (d) [XeF₅]⁻.

These are all examples of D_nh point groups as there is a principal rotation axis, C_n, there is at least one C₂ rotation axis perpendicular to the principal C_n (this is the requirement for D_nh point groups), and there is a horizontal mirror plane, σ_h, perpendicular to the principal C_n rotation axis. The point groups are shown in Figure 4.13ST 

For trans-[PtCl₂(NH₃)₂] there are three C₂ axes that could all be taken as the principal rotation axis. These are perpendicular to the molecular plane, directed along the Cl-Pt-Cl bond or the N-Pt-N bond.

[XeF₅]⁻ is planar, with the C₅ axis perpendicular to the molecular plane.

 

Figure 4.13ST Point groups of [PtCl₄]²⁻, trans-[PtCl₂(NH₃)₂], XeF₄, [XeF₅]⁻.

 

4.14 Use the data in Table 4.14 and the mutual exclusion principle to identify whether the molecules are linear or bent.

 

Molecule	Vibrational mode	IR / cm−1	Raman / cm−1	Mutual Exclusion principle	Geometry
XeF2	nXe-F	555	497	Yes	Linear
O3	nO-O	1135, 1089	1135, 1089	No	Bent
BrOBr	nBr-O	587, 504	587, 504	No	Bent

Table 4.14

4.15 Sketch the IR and Raman spectra in the ν_N-H region for solid NH₃ (A₁, 3300 cm⁻¹, E 3375 cm⁻¹)

 

In general “symmetric” modes are most intense in the Raman spectra, and the “asymmetric” modes most intense in the IR spectra. The E mode is doubly degenerate, and therefore contains two bands at identical wavenumbers.

In this case the A₁ mode in NH₃ at 3300 cm⁻¹ is the symmetric N-H breathing mode where all the N-H bonds are getting longer and shorter in phase and is expected to be more intense than the asymmetric E mode at 3375 cm⁻¹ in the Raman spectrum. In the IR spectrum the intensities will be reversed.

(The observed spectra are more complex due to the presence of Fermi resonance between ν₁ and 2ν₄ (see Section 4.7.5) M. Buback and K. R. Schulz, J. Phys. Chem. 80 2478 (1976), W. H. Koehler, J. W. Lundeen, A. Moradi-Araghi, B. De Bettignies, L. D. Schultz and M. Schwartz, J. Phys. Chem. 83 3264 (1979))

 

 

4.16 Use the D_∞h character table in the Appendix to show that for XeF₂ Γ_Xe-F = Σ_g⁺ + Σ_u⁺ and assign the bands in Table 4.14 to their symmetry species.  

 

D∞h	E	2C∞F	∙∙∙	∞v	i	2σ∞F	∙∙∙	∞C2
Gstr	2	2		2	0	0		0
Assume 1 Σg+	1	1		1	1	1		1
residual	1	1		1	−1	−1		−1	ÞΣu+

 

Therefore, Γ_str =  Σ_g⁺ + Σ_u⁺. The Σ_g⁺ symmetric stretching mode is only Raman active and the Σ_u⁺ antisymmetric mode is only IR active. (This is an example of the mutual exclusion principle.) Therefore, the following assignments can be made.

 

	IR / cm−1	Raman / cm−1
XeF2	555 Σu+ antisymmetric stretching mode	497 Σg+ symmetric stretching mode

 

 

4.17

(a) The IR spectrum of ClF₃ has Cl-F stretching modes at 752, 702 and 529 cm⁻¹, with the 702 cm⁻¹ band being the most intense. The Raman spectrum also contained three bands at the same wavenumbers, of these those at 752 and 529 cm⁻¹ were the most intense, while the 702 cm⁻¹ band was barely visible. Use these data to identify whether ClF₃ is planar, pyramidal or “T-shaped”, and assign the vibrational modes to their symmetry species.

 

The point groups for the three geometries are shown in Figure 4.17ST(a) given below.

Figure 4.17ST(a) Possible shapes and point groups of ClF₃.

 

As there are only three Cl-F stretches to consider it is easier to do it by eye, certainly for D_3h and C_3v, but the tabular method will also be used for the C_2v example.

 

D3h	E	2C3	3C2	σh	2σ3	3σv
GCl-F	3	0	1	3	0	1
assume A1'	1	1	1	1	1	1
residual	2	−1	0	2	−1	0	→ E'

 

∴ Γ_Cl-F = A₁' + E'

 

A₁′ is Raman active (and polarised (p)). E′ is Raman active (depolarised (dp)) and IR active. Therefore, there will be one IR active Cl-F stretching mode and two Raman active stretching modes (one p, one dp) for D_3h ClF₃.

 

 

C3v	E	2C3	3σv
ΓCl-F	3	0	1
assume A1	1	1	1
residual	2	−1	0	→ E

∴ Γ_Cl-F = A₁ + E

 

A₁ is Raman active (p) and IR active. E' is Raman active (dp) and IR active. Therefore, there will be two IR active Cl-F stretching modes and two Raman active stretching modes (one p, one dp) for C_3v ClF₃.

For the C_2v, ‘T’-shaped geometry the atoms lie in the yz plane. In this case both the “by-eye” and tabular methods have been used.

“By-eye” method

C2v	E	C2	σv(xz)	σv′(yz)
ΓCl-F	3	1	1	3
assume 1 A1	1	1	1	1
residual	2	0	0	2	this does not correspond to a single irreducible representation
assume 2nd A1	1	1	1	1
residual	1	-1	-1	1	→B2

 

∴ Γ_Cl-F = 2A₁ + B₂

Tabular method

C2v	E	C2	σv(xz)	σv′(yz)		h = 4
ΓCl-F	3	1	1	3	S	S/h
n(A1)	3	1	1	3	8	2
n(A2)	3	1	−1	−3	0	0
n(B1)	3	−1	1	−3	0	0
n(B2)	3	−1	−1	3	4	1

 

∴Γ_Cl-F = 2A₁ + B₂

 

A₁ is Raman active (p) and IR active. B₂ is Raman active (dp) and IR active. Therefore, there will be three IR active Cl-F stretching modes and three Raman active stretching modes for C_2v ClF₃.

 

Therefore, the IR and Raman data are consistent with ClF₃ having C_2v point group. (This is consistent with VSEPR as there are three bond pairs and two lone pairs, and the lone pairs go in the equatorial positions.) The B₂ mode is likely to be the most intense in the IR spectrum, and weakest in the Raman spectrum, so the 702 cm⁻¹ bands can be assigned to B₂, with the 752 and 529 cm⁻¹ bands assigned to the A₁ modes.

(H. Selig, H. H. Claassen and J. H. Holloway, J. Chem. Phys. 52 3517 (1970))

(b) The Raman spectrum of [XeF₅]⁺ has Xe-F stretching modes at 671 (29) cm⁻¹, 664 (5) cm⁻¹, 629 (100) cm⁻¹ and 623 (5) cm⁻¹ (the relative intensities are given in parentheses). Use these data to determine whether [XeF₅]⁺ has trigonal bipyramidal (D_3h) or square based pyramidal (C_4v) geometry. Explain how IR data could be used to assign the vibrational modes to their symmetry species.

The possible shapes and point groups for [XeF₅]⁺ are shown in Figure 4.17ST(b)

 

Figure 4.17ST(b) Possible shapes and point groups of [XeF₅]⁺.

 

As there are five stretching modes it is much more efficient to use the tabular method.

 

D3h	E	2C3	3C2	σh	2σ3	3σv		h = 12
GXe-F	5	2	1	3	0	3	S	S/h
n(A1')	5	4	3	3	0	9	24	2
n(A2')	5	4	−3	3	0	−9	0	0
n(E')	10	−4	0	6	0	0	12	1
n(A1'')	5	4	3	-3	0	−9	0	0
n(A2'')	5	4	−3	−3	0	9	12	1
n(E'')	10	−4	0	−6	0	0	0	0

 

Therefore, Γ_Xe-F = 2A₁' + E' + A₂''.

Of these the 2A₁' + E' modes are Raman active, hence expect 3 Raman active Xe-F stretching modes for D_3h.

 

C4v	E	2C4	C2	2σv	2σd		h = 8
GXe-F	5	1	1	3	1	Σ	Σ/h
n(A1)	5	2	1	6	2	16	2
n(A2)	5	2	1	−6	−2	0	0
n(B1)	5	−2	1	6	−2	8	1
n(B2)	5	−2	1	−6	2	0	0
n(E)	10	0	−2	0	0	8	1

 

Therefore, Γ_Xe-F = 2A₁ + B₁ + E.

All of these modes are Raman active, hence expect 4 Raman active Xe-F stretching modes for C_4v.

 

Therefore, the Raman data predicts that [XeF₅]⁺ is C_4v.

 

On the basis of the Raman intensities, the 671 and 629 cm⁻¹ modes can be assigned to A₁ (and will both polarised) and the weaker bands at 664 and 623 cm⁻¹ can be assigned to the E and B₁ modes, respectively. IR data can be used to confirm this assignment as the B₁ mode is IR inactive, and the E mode is expected to be very intense in the IR spectrum.

(S. S. Nabiev, Spectrochim. Acta A 56 1589 (2000).)

(c) The Raman spectrum of (n-Bu₄N)₂[PtCl₄]²⁻ in chloroform has n_Pt-Cl modes at 328 cm⁻¹ (polarised) and 305 cm⁻¹ (depolarised), whereas the IR spectrum has one band at 313 cm⁻¹. Use these data to confirm a square planar geometry for [PtCl₄]²⁻.

 

Square planar [PtCl₄]²⁻ has D_4h point group as shown in Figure 4.17ST(c)

 

Figure 4.17ST(c) Shape and point group of [PtCl₄]²⁻.

 

D4h	E	2C4	C2 (C42)	2C2'	2C2''	i	2σ4	h	2v	2d		h = 16
GPt-Cl	4	0	0	2	0	0	0	4	2	0	σ	σ/h
n(A1g)	4	0	0	4	0	0	0	4	4	0	16	1
n(A2g)	4	0	0	−4	0	0	0	4	−4	0	0	0
n(B1g)	4	0	0	4	0	0	0	4	4	0	16	1
n(B2g)	4	0	0	−4	0	0	0	4	−4	0	0	0
n(Eg)	8	0	0	0	0	0	0	−8	0	0	0	0
n(A1u)	4	0	0	4	0	0	0	−4	−4	0	0	0
n(A2u)	4	0	0	−4	0	0	0	−4	4	0	0	0
n(B1u)	4	0	0	4	0	0	0	−4	−4	0	0	0
n(B2u)	4	0	0	−4	0	0	0	−4	4	0	0	0
n(Eu)	8	0	0	0	0	0	0	8	0	0	16	1

(The C₂ rotation axis is the same as the C₄ axis. The C₂' rotation axis is directed along the Cl-Pt-Cl bonds, which also define σ_v mirror planes. The C₂'' rotation axis bisects the Cl-Pt-Cl bonds, which also define the σ_d mirror planes.)

∴ Γ_Pt-Cl = A_1g + B_1g + E_u

A_1g and B_1g are Raman active only, and E_u is only IR active (as an inversion centre, i, is present the mutual exclusion principle applies). The Raman active A_1g mode will be polarised, while the B_1g mode will be depolarised.

Therefore, the IR active mode at 313 cm⁻¹ can be assigned to the E_u mode. The polarised Raman band at 328 cm⁻¹ to the A_1g mode and the Raman depolarised band at 305 cm⁻¹ to the B_1g mode.

Using the character table the Raman activity of the A_1g mode is indicated by x² + y², indicating this is a breathing mode where all the Pt-Cl bonds are increasing and decreasing together in phase. The B_1g Raman active mode is indicated by x² − y² indicating that in this vibrational mode the Pt-Cl bonds in the x direction are increasing while those in the y direction are reducing. The degenerate IR active E mode has one vibrational mode with a dipole change in the x direction and one with a dipole change in the y direction. Schematic diagrams are given in Figure 4.17ST(d).

 

Figure 4.17ST(d) Representations of the stretching modes in [PtCl₄]²⁻.

 

(P. L. Goggin and J. Mink, J. Chem. Soc., Dalton Trans. 1479 (1974))

 

4.18 Calculate Γ_vib, Γ_str and Γ_def and determine the number of IR and Raman active modes in: (a) BF₃, (b) GeF₄, (c) XeF₄, (d) SeF₄.

 

While calculating Γ_str is relatively straight forward by considering the number of unshifted stretching modes (which is usually the same as the number of bonds), using the same methodology to calculate Γ_vib would involve determining the number of unshifted vibrational modes, including the deformation modes which are usually hard to visualise. Therefore, an alternative approach is used of considering the number of unshifted Cartesian vectors on the atoms to give Γ_3N and the subtraction of Γ_rot and Γ_trans from this to give Γ_vib. While it is possible to calculate the number of unshifted Cartesian vectors, it is much easier to determine the number of unshifted atoms (N_us), and then multiply this by the contribution or character per unshifted atom (χ_us) to get Γ_3N. Γ_def is calculated by subtracting Γ_str from Γ_3N.

As Γ_3N will have an order of twelve in BF₃ and fifteen in GeF₄, XeF₄ and SeF₄, it is only viable to use the tabular method, but the “by-eye” route can be used for Γ_str.

The shapes and point groups of BF₃, GeF₄, XeF₄ and SeF₄ are shown in Figure 4.18ST.

Figure 4.18ST Shapes and point groups of BF₃, GeF₄, XeF₄ and SeF₄

(a) BF₃ has D_3h symmetry.

Step 1 Calculate G_3N by determining the number of unshifted atoms, N_usand multiplying this by the contribution per unshifted atom, χ_us.

Step 2 Calculate the irreducible representation using the tabular method

 

D3h	E	2C3	3C2	σh	2σ3	3σv		h = 12
Nus	4	1	2	4	1	2
χus	3	0	−1	1	−2	1
Γ3N	12	0	−2	4	−2	2	S	S/h
n(A1')	12	0	−6	4	−4	6	12	1
n(A2')	12	0	6	4	−4	−6	12	1
n(E')	24	0	0	8	4	0	36	3
n(A1'')	12	0	−6	−4	4	−6	0	0
n(A2'')	12	0	6	−4	4	6	24	2
n(E'')	24	0	0	−8	−4	0	12	1

 

∴ Γ_3N = A₁' + A₂' + 3E' + 2A₂'' + E''

There are 12 Cartesian vectors in BF₃,and Γ_3N should span 12 representations, which it does as both E' and E'' are doubly degenerate.

Step 3 Identify Γ_trans (x, y, z) as E' and A₂''and G_rot (R_x, R_y, R_z) as A₂' and E'' from the character table and subtract these from Γ_3N to give Γ_vib.

 

Γ3N	A1'	A2'	3E'	2A2''	E''
−Γtrans			E'	A2''
−Γrot		A2'			E''
Γvib.	A1'		2E'	A2''

 

∴ Γ_vib = A₁' + 2E' + A₂''

Step 4 Calculate Γ_str for BF₃ using the “by-eye” method, so that this can be subtracted from Γ_vib to give Γ_def.

 

D3h	E	2C3	3C2	σh	2σ3	3σv
ΓB-F, Γstr	3	0	1	3	0	1
assume A1'	1	1	1	1	1	1
residual	2	−1	0	2	−1	0	ÞE'

 

∴ Γ_str = A₁' + E'

 

∴Γ_def = Γ_vib −Γ_str = (A₁' + 2E' + A₂'') – ( A₁' + E') = E' + A₂''

 

A₁'  is Raman active, E' is both IR and Raman active, A₂'' is IR active.

Therefore, there is one IR active stretching mode (E') and two IR active deformation modes (E' + A₂'')

Therefore, there are two Raman active stretching modes (A₁' + E'), and one Raman active deformation mode (E')

(b) GeF₄ has T_d point group

Td	E	8C3	3C2	6σ4	6σd		h = 24
Nus	5	2	1	1	3
cus	3	0	−1	−1	1
Γ3N	15	0	−1	−1	3	Σ	Σ/h
n(A1)	15	0	−3	−6	18	24	1
n(A2)	15	0	−3	6	−18	0	0
n(E)	30	0	−6	0	0	24	1
n(T1)	45	0	3	−6	−18	24	1
n(T2)	45	0	3	6	18	72	3

 

∴Γ_3N = A₁ + E + T₁+ 3T₂

 

Γ_trans = T₂, Γ_rot = T₁

 

∴Γ_vib = A₁ + E + 2T₂

 

Td	E	8C3	3C2	6σ4	6σd
Gstr	4	1	0	0	2
assume 1 A1	1	1	1	1	1
residual	3	0	−1	−1	1	Þ T2

 

∴Γ_str = A₁ + T₂

∴Γ_def = E + T₂

A₁ is Raman Active, E is Raman active, T₂ is both Raman and IR active.

Therefore, expect one IR active stretching mode and one IR active deformation mode, two Raman active stretching modes and two Raman active deformation modes.

(c) XeF₄ has D_4h symmetry.

 

D4h	E	2C4	C2 (C42)	2C2'	2C2''	i	2σ4	Γh	2Γv	2Γd		h = 16
Nus	5	1	1	3	1	1	1	5	3	1
χus	3	1	−1	−1	−1	−3	−1	1	1	1
Γ3N	15	1	−1	−3	−1	−3	−1	5	3	1	S	Σ/h
n(A1g)	15	2	−1	−6	−2	−3	−2	5	6	2	16	1
n(A2g)	15	2	−1	6	2	−3	−2	5	−6	−2	16	1
n(B1g)	15	−2	−1	−6	2	−3	2	5	6	−2	16	1
n(B2g)	15	−2	−1	6	−2	−3	2	5	−6	2	16	1
n(Eg)	30	0	2	0	0	−6	0	−10	0	0	16	1
n(A1u)	15	2	−1	−6	−2	3	2	−5	−6	−2	0	0
n(A2u)	15	2	−1	6	2	3	2	−5	6	2	32	0
n(B1u)	15	−2	−1	−6	2	3	−2	−5	−6	2	0	0
n(B2u)	15	−2	−1	6	−2	3	−2	−5	6	−2	16	1
n(Eu)	30	0	2	0	0	6	0	10	0	0	48	3

 

Γ_3N = A_1g + A_2g + B_1g +B_2g + E_g + 2A_2u + B_2u+ 3E_u

(order = 15, therefore OK)

Γ_trans = A_2u + E_u

Γ_rot = A_2g + E_g

Therefore

Γ_vib = A_1g + B_1g +B_2g + A_2u + B_2u+ 2E_u

Γ_str = A_1g + B_1g + E_u (from Self test 4.17)

Γ_def = B_2g + A_2u + B_2u+ E_u

A_1g is Raman active, B_1g is Raman active, B_2g is Raman active, A_2u is IR active, B_2u is neither IR or Raman active, E_u is IR active. (Note the application of the mutual exclusion principle.)

Therefore expect one IR active stretching mode (E_u) and 2 IR active deformation modes (one in plane (E_u) the other out of plane (A_2u)). Two Raman active stretching modes (A_1g + B_1g) of which A_1g will be polarised are predicted, together with one Raman active deformation mode. The B_2u mode is neither IR or Raman active, but an estimate of its value can be obtained if the overtone is observed in the Raman spectra.

(d) SeF₄

SeF₄ has C_2v symmetry, with the z, C₂ axis along the direction of the lone pair in the VSEPR formalism.

C2v	E	C2	σv(xz)	σv'(yz)		h = 4
Nus	5	1	3	3
cus	3	−1	1	1
G3N	15	−1	3	3	S	S/h
n(A1)	15	−1	3	3	20	5
n(A2)	15	−1	−3	−3	8	2
n(B1)	15	1	3	−3	16	4
n(B2)	15	1	−3	3	16	4

 

∴ Γ_3N = 5A₁ + 2A₂ + 4B₁ + 4B₂

 

Γ_trans = A₁ + B₁ + B₂, Γ_rot = A₂ + B₁ + B₂

∴ Γ_vib = 4A₁ + A₂ + 2B₁ + 2B₂

 

C2v	E	C2	σv(xz)	σv'(yz)		h = 4
Γstr	4	0	2	2	S	S/h
n(A1)	4	0	2	2	8	2
n(A2)	4	0	−2	−2	0	0
n(B1)	4	0	2	−2	4	1
n(B2)	4	0	−2	2	4	1

 

∴Γ_str = 2A₁ + B₁ + B₂

 

∴ Γ_def = Γ_vib −Γ_str = (4A₁ + A₂ + 2B₁ + 2B₂) – (2A₁ + B₁ + B₂) = 2A₁ + A₂ + B₁ + B₂

 

A₁, B₁ and B₂ are IR active and A₁, A₂, B₁ and B₂ are Raman active.

 

Therefore, expect four IR active stretching modes and four IR active deformation modes.

Therefore, expect four Raman active stretching modes and five Raman active deformation modes.

 

4.19 The solid phase Raman spectrum of GeH₄ has bands at 2114, 2106, 931 and 819 cm⁻¹, with the 2106 band being the most intense and also polarised. The solid phase IR spectrum has bands at 2114 and 819 cm⁻¹ (Figure 4.9 in main text). Assign these to the correct symmetry species and label them in terms of v₁, v₂ etc.

Using the analysis for GeH₄ in Self Test 4.18 we can determine that

 

Γ_vib = A₁ + E + 2T₂

Γ_str = A₁ + T₂

Γ_def = E + T₂

Of these the T₂ modes are both IR and Raman active, and the A₁ and E are only Raman active. It should also be remembered that the stretching modes are usually at higher wavenumber than the deformation modes. Therefore, the bands at 2114 and 819 cm⁻¹ can be assigned to the T₂ modes as they are present in both the IR and Raman spectra. As the 2106 cm⁻¹ band is both the most intense and polarised in the Raman spectra this can be assigned to the A₁ mode, with the remaining band at 931 cm⁻¹ being assigned to the E mode.

The modes are assigned in terms of v₁, v₂ etc. by ranking the vibrational modes by the order they occur in the character table, and if there are more than one of each irreducible representation, they are ranked by wavenumber order within each irreducible representation.

 

symmetry label	vibrational label	observed wavenumber / cm−1	description of vibrational mode
A1	v1	2106	symmetric stretch
E	v2	931	deformation
T2	v3	2114	asymmetric stretch
T2	v4	819	deformation