Chapter 4 Answers to Problem Questions

4.1 Describe how IR spectroscopy could be used to determine the following.

(a) Whether the water molecules in CrCl₃(H₂O)₆ are water of crystallisation or coordinated to the metal centre?

Coordinated water will give rise to sharper features at higher wavenumber (3500 – 3400 cm⁻¹) than water of crystallisation which will be identified by a broader peak at lower wavenumber (3300 – 3200 cm⁻¹) due to hydrogen bonding.

(b) Whether the hydrogen bond in solid β-CrOOH was stronger or weaker than that in a different phase, α-CrOOH?

Weaker hydrogen bonds would shift the ν_OH mode to higher wavenumber, stronger hydrogen bonding would shift the ν_OH mode to lower wavenumber.

(c) The mode of coordination of SeCN⁻ ligands in transition metal complexes.

SeCN⁻ is analogous to SCN⁻, and is ambidentate so that harder metals tend to bind via the N, and softer metals via the Se (see Section 4.3.10). The behaviour is very similar to NCS⁻ where the ν_CN mode for N-bonded complexes is below 2080 cm⁻¹ but is higher than this for Se-bonded complexes. The value for K[NCSe] is 2070 cm⁻¹. The ν_CSe mode is at 700 – 620 cm⁻¹ for N-bonded and 550 – 500 cm⁻¹ Se-bonding, compared to 558 for K[NCSe].

For [Co(NCSe)₄]²⁻ the ν_CN and ν_CSe modes are at 2053 and 672 cm⁻¹, respectively indicating N-bonded ligands. In contrast the ν_CN and ν_CSe modes are at 2130 and 519 cm⁻¹ in [Pt(SeCN)₆]²⁻ indicating Se-bonded ligands. In [HgCo(NCSe)₄] (which is often used as a calibrant for magnetic susceptibility measurements) the ν_CN mode is at 2146 cm⁻¹, but this is a result of the NCSe⁻ acting as bridging ligands.

(d) The rate of an isomerisation reaction of the [Ru(NH₃)₄(H₂O)(SO₂)]²⁺ cation, which involves a change in the coordination by the SO₂ ligand, from sulfur to oxygen, to the ruthenium centre.

This is analogous to nitrite complexes in Section 4.3.11, especially as SO₂ is isoelectronic to NO₂⁻. For NO₂⁻ a larger separation of the two ν_NO modes was associated with oxygen ligation, and a smaller separation with nitrogen ligation. For SO₂ complexes a variety of coordination modes are known. ν_SO modes at 1279 and 1126 cm⁻¹ for [Ru(NH₃)₄(H₂O)(SO₂)]²⁺ have been shown by single-crystal X-ray diffraction to be associated with an η¹-SO₂ ligand bonded via the sulfur atom. Photocrystallography of the short lived excited state of [Ru(NH₃)₄(H₂O)(SO₂)]²⁺ has shown that the ν_SO modes at 1181 and 946 cm⁻¹ are associated with an η²-SO₂ side bound ligand. Subsequent photocrystallography studies have identified a second excited state with an η¹-SO₂ ligand bonded via the oxygen atom. Therefore, the kinetics of the photochemical isomerisation of the ground state into the excited states, and their subsequent decay can be followed by IR spectroscopy. (D. A. Johnson and V. C. Dew, Inorg. Chem. 18 3273 (1979), A. Y. Kovalevsky, K. A. Bagley and P. Coppens, J. Am. Chem. Soc. 124 9241 (2002), A. Y. Kovalevsky, K. A. Bagley, J. M. Cole and P. Coppens, Inorg. Chem. 42 140 (2003), K. F. Bowes, J. M. Cole, S. L. G. Husheer, P. R. Raithby, T. L. Savarese, H. A. Sparkes, S. J. Teat and J. E. Warren, Chem. Commun. 2448 (2006).)

4.2 (a) Explain the ordering of the ν_CO modes in the following fac-isomers: [Mo(CO)₃(PF₃)₃], 2090, 2055 cm⁻¹; [Mo(CO)₃(PPh₃)₃], 1934, 1835 cm⁻¹; [Mo(CO)₃(py)₃], 1905, 1775 cm⁻¹.

The decrease in the A₁ and E ν_CO modes indicates that there is an increase in back-bonding to the CO π* orbital. PF₃ is a poor σ base and almost as good a π acid as CO. PPh₃ and pyridine are both better σ bases, and pyridine is a worse π acid so there is more electron density available on the molybdenum for back-donation to the CO ligands.

(b) Explain the position of the ν_CO and ν_NO modes in [RuCl(CO)(NO)(P^tBu₂Me)₂] (ν_CO 1914 cm⁻¹, ν_NO 1570 cm⁻¹) and [Ru(CO)(NO)(P^tBu₂Me)₂]⁺ (ν_CO 1966 cm⁻¹, ν_NO 1709 cm⁻¹).

The value of ν_NO 1570 cm⁻¹ for [RuCl(CO)(NO)(P^tBu₂Me)₂] indicates that the Ru-NO group is bent, and can be considered as NO⁻ and a two electron donor. Therefore, the Ru is formally Ru(II) and has 4d⁶ electronic configuration, resulting in a 16 electron complex.

In contrast, ν_NO of 1709 cm⁻¹ for [Ru(CO)(NO)(P^tBu₂Me)₂]⁺ implies linear NO⁺ and a two electron donor. Therefore, the Ru is formally Ru(0) with a 4d⁸ electronic configuration, and overall it is also a 16 electron complex.

Although one would normally expect a Ru(0) complex to have lower a ν_CO mode than a Ru(II) complex, in this case the difference in the ν_CO modes reflects the different back-bonding characteristics of linear and bent NO ligands, where linear NO⁺ is a better π acid than bent NO⁻.

Alternatively one could treat NO as neutral. In [RuCl(CO)(NO)(P^tBu₂Me)₂] with bent M-NO acting as a one electron donor, and in [Ru(CO)(NO)(P^tBu₂Me)₂]⁺ with linear M-NO acting as a three electron donor, the ruthenium is formally present as Ru(I) and both are 16 electron compounds. The differences in the ν_CO modes are due to the different σ and pi bonding effects in linear and bent NO.

(M. Ogasawara, D. J. Huang, W. E. Streib, J. C. Huffman, N. GallegoPlanas, F. Maseras, O. Eisenstein and K. G. Caulton, J. Am. Chem. Soc. 119 8642 (1997).)

4.3 Use the IR spectra of isotopically enriched [Co(N₂)] and [Ni(N₂)] in Figure 4.3P to identify whether the N₂ is side-on or end-on bonded in these compounds. The isotopic enrichment was ¹⁴N₂:¹⁴N¹⁵N:¹⁵N₂ 3:2:1.

Figure 4.3P(a) IR spectra of isotopically enriched [Co(N₂)] and [Ni(N₂)] in solid argon. (Adapted from G. A. Ozin and A. van der Voet, Can. J. Chem., 51, 637 (1973))

Figure 4.3P(b) shows that if the N₂ ligand is side-on bonded three bands are expected when N₂ isotopically enriched with ¹⁵N is used as although there are two possible arrangements for M(¹⁴N¹⁵N) and M(¹⁵N¹⁴N) these occur at the same energy. In contrast for end-on bonded N₂ there will be four bands in the spectrum because M(¹⁴N¹⁵N) and M(¹⁵N¹⁴N) are no longer equivalent and will vibrate at different frequencies. The vibrational frequency for M(¹⁵N¹⁴N) will be higher than that for M(¹⁴N¹⁵N). Therefore, in [Co(N₂)] the N₂ ligand is side-on bonded as there are three bands, but in [Ni(N₂)] the N₂ must be end-on bonded as there are four bands in the  ν_NN region.

Figure 4.3P(b). Different bonding modes for dinitrogen complexes.

4.4 The carbonylation of RuCl₃ using a mixture of HCl and HCOOH followed by precipitation with CsCl gave complexes Cs₂[Ru(CO)₂Cl₄] A after four hours and Cs[Ru(CO)₃Cl₃] B after twenty hours. The IR spectrum of complex A contained two ν_CO modes at 2074 (vs) and 2006 (vs) cm⁻¹, and the IR spectrum of complex for B had ν_CO modes at 2141 (s) and 2077 (vs) cm⁻¹. Use these data to identify the isomers formed and comment on the position of the ν_CO modes.

The shapes and point groups of the isomers of A and B are shown in Figure 4.4P.

Figure 4.4P Shapes and point groups of the isomers of [Ru(CO)₂Cl₄]²⁻ and [Ru(CO)₃Cl₃]⁻.

In general for carbonyl complexes it is possible to deal with the carbonyl ligands in isolation (i.e. Γ_CO) as their stretching modes are well separated from the other vibrational modes. As there are only two or three carbonyl ligands to consider, the “by-eye” method is appropriate in this case.

Complex A, cis-[Ru(CO)₂Cl₄]²⁻ (assume that the CO ligands lie in the yz plane.

C2v	E	C2	σv(xz)	σv′(yz)
ΓCO	2	0	0	2
assume 1A1	1	1	1	1
residual	1	−1	−1	1	→B2

 

Therefore, Γ_CO = A₁ + B₂. Both of these are IR active, therefore expect two IR active CO stretching modes for the cis- complex.

Complex A, trans-[Ru(CO)₂Cl₄]²⁻

D4h	E	2C4	C2 (C42)	2C2'	2C2''	i	2S4	σh	2σv	2σd
ΓCO	2	2	2	0	0	0	0	0	2	2
assume 1A1g	1	1	1	1	1	1	1	1	1	1
residual	1	1	1	−1	−1	−1	−1	−1	1	1	→A2u

 

Therefore, Γ_CO = A_1g + A_2u. Of these only the A_2u mode is IR active, therefore expect one IR active CO stretching mode for the trans- complex

Therefore, IR data indicates that complex A is cis-[Ru(CO)₄Cl₂]²⁻.

Complex B, fac-[Ru(CO)₃Cl₃]⁻

C3v	E	2C3	3σv
ΓCO	3	0	1
assume 1A1	1	1	1
residual	2	−1	0	→E

Therefore Γ_CO = A₁ + E. These are both IR active, so expect two IR active CO stretching modes for the fac- complex.

Complex B, mer-[Ru(CO)₃Cl₃]⁻ (assume that the CO ligands are in the yz plane)

C2v	E	C2	σv(xz)	σv′(yz)
ΓCO	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

 

Therefore Γ_CO = 2A₁ + B₂. These are all IR active, so expect three IR active CO stretching modes for the mer- complex.

Therefore, complex B is fac-[Ru(CO)₃Cl₃]^−.

The ν_CO modes are at lower values for the dicarbonyl complex, than the tricarbonyl complex. This is to be expected as there will be more back-bonding per CO, for the smaller number of CO ligands. In addition, there are more Cl⁻ ligands in the dicarbonyl complex, and as these act as both σ and π bases, there is more electron density on the metal for donation than in the tricarbonyl complex.

4.5 The IR spectrum from [Rh(CO)₃] is shown in Figure 4,5P. Determine whether [Rh(CO)₃] is trigonal planar, pyramidal or T-shaped.

Figure 4.5P. IR spectrum of [Rh(CO)₃] in an argon matrix at 10 K.

As there are only three CO stretching modes, the “by-eye” method is satisfactory in this case.

Trigonal planar, D_3h [Rh(CO)₃]

D3h	E	2C3	3C2	σh	2S3	3σv
ΓCO	3	0	1	3	0	1
Assume 1 A1'	1	1	1	1	1	1
residual	2	−1	0	2	−1	0	→ E'

Therefore, Γ_CO = A₁' + E'.

The A₁' mode is only Raman active (and polarised), whereas the E' mode is both IR and Raman active.

Therefore expect one ν_CO band in the IR spectrum, (and two ν_CO bands in the Raman spectrum) for a trigonal planar geometry.

Pyramidal, C_3v [Rh(CO)₃]

C3v	E	2C3	3σv
ΓCO	3	0	1
assume 1A1	1	1	1
residual	2	−1	0	→ E

Therefore Γ_CO = A₁ + E. The A₁ mode is both IR and Raman active, as is the E mode. Therefore, expect two IR bands (and two Raman bands) in the CO stretching region (ν_CO).

‘T’-shaped, C_2v [Rh(CO)₃]

Assume the molecule lies in the yz plane.

C2v	E	C2	σv(xz)	σv'(yz)
ΓCO	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, therefore try one more A1
assume 2nd A1	1	1	1	1
residual	1	-1	-1	1	→ B2

 

Therefore, Γ_CO = 2A₁ + B₂ (If the molecule is in the xz plane, then get 2A₁ + B₁ which is just as correct). As all are both IR and Raman active, expect 3 CO stretching modes in the IR and 3 in the Raman spectra.

            For D_3h expect 1 IR and 2 Raman active CO bands.

            For C_3v expect 2 IR and 2 Raman active CO bands

            For C_2v expect 3 IR and 3 Raman active CO bands

Therefore [Rh(CO)₃] is ‘T’-shaped as there are three bands in the IR spectrum.

(The spectrum in Figure 4.5P is based on a combination of experimental and calculated data (M. F. Zhou and L. Andrews, J. Phys. Chem. A 103 7773 (1999) and comparison with the experimental and calculated data for the isoelectronic [Rh(N₂)₃] molecule (X. Wang and L. Andrews, J. Phys. Chem. A 106 2457 (2002))

4.6 The IR spectrum of VF₅ has bands in the V-F stretching region at 810 and 784 cm⁻¹, and at 719 and 608 cm⁻¹ in the Raman spectrum, together with a very weak peak at 810 cm^-1. Use these data to determine the shape of VF₅ and assign the vibrational modes to their symmetry species.

The possible shapes and point groups for VF₅ are shown in Figure 4.6P

Figure 4.6P Possible shapes and point groups for VF₅

As there are five stretching modes to consider, it is better to use the tabular method, as otherwise a lot of time will be spent checking permutations. If the axial and equatorial V-F stretching modes are dealt with separately, this can simplify matters (slightly).

D3h	E	2C3	3C2	sh	2S3	3sv		h = 12
GV-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_V-F = 2A₁' + E' + A₂''

If the axial V-F are treated separately

D3h	E	2C3	3C2	sh	2S3	3sv		h = 12
GV-F(ax)	2	2	0	0	0	2	S	S/h
n(A1')	2	4	0	0	0	6	12	1
n(A2')	2	4	0	0	0	−6	0	0
n(E')	4	−4	0	0	0	0	0	0
n(A1'')	2	4	0	0	0	−6	0	0
n(A2'')	2	4	0	0	0	6	12	1
n(E'')	4	−4	0	0	0	0	0	0

 

Therefore Γ_V-F(ax) = A₁' + A₂''

If the equatorial V-F are treated separately

D3h	E	2C3	3C2	sh	2S3	3sv		h = 12
GV-F(eq)	3	0	1	3	0	1	S	S/h
n(A1')	3	0	3	3	0	3	12	1
n(A2')	3	0	−3	3	0	−3	0	0
n(E')	6	0	0	6	0	0	12	1
n(A1'')	3	0	3	−3	0	−3	0	0
n(A2'')	3	0	−3	−3	0	3	0	0
n(E'')	6	0	0	−6	0	0	0	0

Therefore Γ_V-F(eq) = A₁' + E'

	2A1'	A2''	E'
IR active	no	yes	yes	2 IR active V-F stretching modes, A2'' and E'
Raman active	yes	no	yes	3 Raman active V-F stretching modes 2A1' + E'
Raman depolarisation ratio	polarised r < 3/4	n/a	depolarised r = 3/4	2 polarised (p) 1 depolarised (dp)

 

C4v	E	2C4	C2	2sv	2sd		h = 8
GV-F	5	1	1	3	1	S	S/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_V-F = 2A₁ + B₁ + E

If the axial V-F is considered separately, this transforms (Γ_V-F(ax)) as A₁, therefore Γ_V-F(eq) = A₁ + B₁ + E

	2A1	B1	E
IR active	yes	no	yes	3 IR active V-F stretching modes
Raman active	yes	yes	yes	4 Raman active V-F stretching modes
Raman depolarisation ratio	polarised r < 3/4	depolarised r = 3/4	depolarised r = 3/4	2 polarised (p) 2 depolarised (dp)

 

The IR spectrum has bands in the V-F stretching region at 810 and 784 cm⁻¹, and the Raman spectrum has bands at 719 and 608 cm⁻¹ in the Raman spectrum, together with a very weak peak at 810 cm^-1. The two IR bands and three Raman bands indicates that VF₅ has D_3h symmetry, also 2 polarised Raman bands would confirm this.

The 784 cm^-1 band is only present in the IR spectrum and can therefore be assigned to the A₂'' mode, likewise the 719 and 608 cm^-1 bands are only observed in the Raman spectrum (and would be polarised), therefore these are the A₁' modes. Therefore, the 810 cm^-1 band which is intense in the IR spectrum but very weak in the Raman spectrum is the E' mode. (In practice it is hard to discern this above the noise in the Raman spectrum.

The A₂'' mode has a dipole moment change in the z direction, and as the earlier analysis indicated this belonged to the axial stretching modes this can be considered as the antisymmetric ν_VF(ax) out-of-plane stretching mode. The E' mode has dipole moment changes in the x and y directions, and this belongs to the degenerate in-plane antisymmetric stretching modes. While it might be attractive to assign one of the A₁' modes to the axial out-of-plane symmetric stretching mode and the other A₁' mode to the in-plane symmetric stretching mode, in practice they will be reasonably strongly coupled as they are quite close in energy, so that one of A₁' modes is the totally symmetric mode where all five V-F bonds increase and decrease in phase, and the second A₁' mode is where the axial V-F increase and the equatorial V-F decrease.

(H. H. Claassen and H. Selig, J. Chem. Phys. 44 4039 (1966), E. G. Hope, J. Chem. Soc., Dalton Trans. 723 (1990))

4.7 The planar molecule N₂F₂ can be isolated as both cis (C_2v) and trans (C_2h) isomers. One of these isomers yielded the following IR and Raman data.

 

IR:                    1525, 950, 890, 728, 341 cm⁻¹

Raman:           1525(p), 950, 890(p), 728, 550, 341(p) cm⁻¹

Use the mutual exclusion principle to identify which isomer has a structure consistent with these data. Calculate Γ_vib, Γ_str and Γ_def for both isomers and use this to confirm the assignment. Assign all the bands to their symmetry species.

The shapes and point groups of cis-N₂F₂ and trans-N₂F₂ are shown in Figure 4.7P.

As there are bands that are present at the same wavenumber in both the IR and the Raman spectra, the mutual exclusion principle rules out the presence of an inversion centre, and hence a C_2h point group so the data belongs to the C_2v cis isomer.

 

Figure 4.7P Shapes and point groups of cis-N₂F₂ and trans-N₂F₂

 

cis-N₂F₂, C_2v (assume molecule is in yz plane)

Γ_str first

 

C2v	E	C2	σv(xz)	σv′(yz)		h = 4
Γstr	3	1	1	3	Σ	Σ/h
nA1	3	1	1	3	8	2
nA2	3	1	−1	−3	0	0
nB1	3	−1	1	−3	0	0
nB2	3	−1	−1	3	4	1

 

Therefore, Γ_str = 2A₁ + B₂.

Although the N-N stretching mode appears to be rotated by the C₂ symmetry operation and reflected by the σ_v(xz) symmetry operation, it is indistinguishable before and after as stretching modes are not vectors, therefore it still contributes as a non-shifted stretching mode. As the reducible representation for this is 1 under each symmetry operation, the ν_NN mode transforms as A₁. Therefore, the two ν_NF modes transform as A₁ + B₂, which can be thought of as a symmetric (A₁) and antisymmetric (B₂) pair. Although there will be some interaction between the two A₁ modes, in this the separation is of the order of 600 cm⁻¹ so it is reasonable to factor them into ν_NN and ν_NF modes.

To get Γ_vib, it is necessary to calculate Γ_3N first, and then subtract Γ_rot and Γ_trans.

 

C2v	E	C2	σv(xz)	σv′(yz)		h = 4
Nus	4	0	0	4
χus	3	−1	1	1
Γ3N	12	0	0	4	Σ	Σ/h
n(A1)	12	0	0	4	16	4
n(A2)	12	0	0	−4	8	2
n(B1)	12	0	0	−4	8	2
n(B2)	12	0	0	4	16	4

 

 

 

 

 

 

Therefore, Γ_3N = 4A₁ + 2A₂ + 2B₁ + 4B₂.

Γ_rot = A₂ + B₁ + B₂

Γ_trans = A₁ + B₁ + B₂

∴Γ_vib = 3A₁ + A₂ + 2B₂

as Γ_str = 2A₁ + B₂, Γ_def = A₁ + A₂ + B₂

	A1	A2	B1	B2
IR active	yes	no	yes	no
Raman active	yes	yes	yes	yes

Therefore, expect three IR and Raman active stretching modes, and 2 IR active and 3 Raman active deformation modes for cis-N₂F₂.

trans-N₂F₂, C_2h

Γ_str first

C2h	E	C2	i	σh		h = 4
Γstr	3	1	1	3	Σ	Σ/h
n(Ag)	3	1	1	3	8	2
n(Bg)	3	−1	1	−3	0	0
n(Au)	3	1	−1	−3	0	0
n(Bu)	3	−1	−1	3	4	1

 

 

 

 

 

Therefore, Γ_str = 2A_g + B_u.

To get Γ_vib, it is necessary to calculate Γ_3N first, and then subtract Γ_rot and Γ_trans.

 

C2h	E	C2	i	σh		h = 4
Nus	4	0	0	4
χus	3	−1	−3	1
Γ3N	12	0	0	4	Σ	Σ/h
n(Ag)	12	0	0	4	16	4
n(Bg)	12	0	0	−4	8	2
n(Au)	12	0	0	−4	8	2
n(Bu)	12	0	0	4	16	4

 

 

 

 

 

 

Therefore, Γ_3N = 4A_g + 2B_g + 2A_u + 4B_u.

Γ_rot = A_g + 2B_g

Γ_trans = A_u + 2B_u

∴Γ_vib = 3A_g + A_u + 2B_u

as Γ_str = 2A_g + B_u, Γ_def = A_g + A_u + B_u

	Ag	Bg	Au	Bu
IR active	no	no	yes	yes
Raman active	yes	yes	no	no

 

Therefore, expect 1 IR active and 2 Raman active stretching modes, 2 IR active and 1 Raman active deformation modes. As there is a centre of symmetry, the mutual exclusion principle operates, and no band is both IR and Raman active.

The experimental data is

IR:                    1525, 950, 890, 728, 341 cm⁻¹

Raman:            1525(p), 950, 890(p), 728, 550, 341(p) cm⁻¹

The 5 IR active modes and 6 Raman active modes confirm that the data belongs to cis-N₂F₂.

The three polarised bands at 1525, 890 and 341 cm⁻¹ in the Raman spectrum can be assigned to the A₁ modes. The 550 cm⁻¹ band is only present in the Raman spectrum, so this can be assigned to the A₂ mode. The two remaining bands at 950 and 728 cm⁻¹ are assigned to the B₂ modes. Stretching modes usually occur at higher wavenumber than bending or deformation modes, so the full assignment is as below.

	Symmetry species	cm−1	Type of vibrational mode
n1	A1	1525	nNN stretching mode
n2	A1	890	nNF symmetric stretching mode
n3	A1	341	symmetric in-plane bending mode
n4	A2	550	out-of-plane deformation mode
n5	B2	950	nNF antisymmetric stretching mode
n6	B2	728	antisymmetric in-plane bending mode

 

4.8 The Raman spectrum of white phosphorus (P₄) has three bands at 614, 467 and 372 cm^-1, the first of which is polarised. In the IR spectrum there is a very intense peak at 467 cm^-1, together with very weak peaks at 614 and 372 cm⁻¹. Determine Γ_str and Γ_vib for white phosphorus and assign the observed vibrational modes to their symmetry species. Comment on the apparent breakdown of separating the stretching and deformation modes in clusters such as P₄.

The structure of P₄ is based on a tetrahedron which as shown in Figure 4.8P has T_d point group.

 

Figure 4.8P Structure and point group of white phosphorus, P₄.

Γ_str first

(Remember that if the stretching mode is in the same place, but reversed, it still counts as being indistinguishable and therefore counts as one unshifted stretching mode)

Therefore, Γ_str = A₁ + E + T₂

Γ_vib

(As before it is necessary to determine Γ_3N first)

Therefore, Γ_3N = A₁ + E + T₁ + 2T₂.

Γ_rot = T₁, Γ_trans = T₂

∴Γ_vib = A₁ + E + T₂

In this case Γ_vib = Γ_str as in clusters such as P₄ it is not possible to treat the stretching modes in isolation, because for the P-P bonds to stretch or compress, some of the angles must change as well.

A₁ is Raman active and polarised, E is only Raman active, and T₂ is both IR and Raman active. Therefore the 614 cm⁻¹ band can be assigned to A₁. As the 467 cm⁻¹ band is present in both the IR and Raman spectra, this must be the T₂ mode. Therefore, the E mode is at 372 cm⁻¹.