Character Tables

Question 1

a. E

b. C₂(x)

c. C₃

d. σ_h

e. C₄

f. i

Question 2

		1	0	0
E =		0	1	0
		0	0	1

		-1/2	-√3/2	0
C3 =		√3/2	-1/2	0
		0	0	1

		-1	0	0
C2 =		0	1	0
		0	0	-1

		1	0	0
σh =		0	1	0
		0	0	-1

		-1/2	-√3/2	0
S3 =		√3/2	-1/2	0
		0	0	-1

		1	0	0
σv(xz) =		0	-1	0
		0	0	1

Now block diagonalise the matricies. Since C₃ and S₃ have off-diagonal elements, each matrix must be divided into 2×2 and 1×1 matrices:

			1	0			0
E =			0	1			0
			0	0			1

			-1/2	-√3/2			0
C3 =			√3/2	-1/2			0
			0	0			1

			-1	0			0
C2 =			0	1			0
			0	0			-1

			1	0			0
σh =			0	1			0
			0	0			-1

			-1/2	-√3/2			0
S3 =			√3/2	-1/2			0
			0	0			-1

			1	0			0
σv(xz) =			0	-1			0
			0	0			1

You now have the following after summing the 2x2 and 1x1 matrices:

Four rows are missing, since there needs to be the same number of rows as there are columns of operations, including the all symmetric row (Property 7):

Three rows are still missing. The sum of the squares of the characters under E must equal the order of the group (Property 4), so:
2² + 1² + 1² + x² + y² + z² = 12
x² + y² + z² = 6
x = 2, y = 1, z =1

Irreducible representations are orthogonal to each other (Property 6). Starting with the x = 2 row, which we will start by assuming is simialr to the (x,y) row above, we could have either of the following two possibilities (the character under E is always positive):

Both options are orthognal with the (x,y) row and the all symmetric row, but only the second combination is orthognal with the z row.

Two rows are still missing, each with a character of 1 under E. So they will all have characters of +/- 1. Using trial-and-error and some educated guessing (along with Property 6), you can solve for the missing rows. Aim for +6 and -6 when you add them up:

Now add the symmetry labels (A, B, E, T) and any modifiers (₁, ₂, ', ", g, u)

Finally, by considering the symmetry of the d orbitals and rotations, add them to the character table:

Question 3

		-1/2	-√3/2	0				-1	0	0						1/2	-√3/2	0
Missing Operation = C3(C2) =		√3/2	-1/2	0				0	-1	0			=			√3/2	1/2	0			= C6
		0	0	1				0	0	1						0	0	1

Vibrational Spectroscopy

Question 1

a. SiCl₄

3N = 15
3N-6 = 9

Reduces to: A₁ + E + T₁ + 3T₂

IR = 2T₂
Raman = A₁ + E + 2T₂

b. trans-SF₂Cl₄

3N = 21
3N-6 = 15

Reduces to: 2A_1g + A_2g + B_1g + B_2g + 2E_g + 3A_2u + B_2u + 4E_u

IR = 2A_2u + 3E_u
Raman = 2A_1g + B_1g + B_2g + E_g

c. SbCl₅

3N = 18
3N-6 = 12

Reduces to: 2A₁' + A₂' + 4E' + 3A₂" + 2E"

IR = 3E' + 2A₂"
Raman = 2A₁' + 3E' + E"

d. CH₂Cl₂

3N = 15
3N-6 = 9

Reduces to: 5A₁ + 2A₂ + 4B₁ + 4B₂

IR = 4A₁ + 2B₁ + 2B₂
Raman = 4A₁ + A₂ + 2B₁ + 2B₂

Resonance Spectroscopy

Question 1

a. PF₆^-
19F NMR:

Peak Height Ratio: 1 : 1

Explaination: All F atoms are equivalent (i.e. exchangable by symmetry operations) and couple with one P atom (³¹P: I = ½, 100% abundant).

b. cis-¹⁹⁵PtF₂Cl₂^2-
19F NMR:

Peak Height Ratio: 1 : 1

Explaination: Both F atoms are equivalent (i.e. exchangable by symmetry operations) and couple with one Pt atom that is only the Pt-195 isotope (¹⁹⁵Pt: I = ½).

c. fac-MoF₃(PMe₃)₃
¹⁹F NMR:

Peak Height Ratio: 1 : 3 : 3 : 1

Explaination: The three F atoms are equivalent (i.e. exchangable by symmetry operations) and couple with the three equivalent P atoms (³¹P: I = ½, 100% abundant).

d. DF
¹⁹F NMR:

Peak Height Ratio: 1 : 1 : 1

Explaination: The F atom couples with the D atom (D = ²H: I = 1)

e. SiFCl₃
¹⁹F NMR:

Peak Height Ratio: 2.35 : 95.3 : 2.35

Explaination: The F atom couples with ²⁹Si when present (resulting in satellites 4.7% of the time), the other isotopes of silicon are not NMR active (29Si: I = ½, 4.7% abundant)

f. SeF₄
¹⁹F NMR:

Peak Height Ratio: 3.8 : 7.6 : 3.8 : 92.4 : 184.8 : 92.4 : 3.8 : 7.6 : 3.8

Explaination: The are two unique sets of F atoms. Each set will couple with the Se 7.2% of the time and couple with the other set of two F atoms always (⁷⁷Se: I = ½, 7.6% abundant; ¹⁹F: I = ½, 100% abundant).

g. mer-PF₃Cl₃^-
19F NMR:

Peak Height Ratio: 1 : 2 : 1 : 1 : 2 : 1 : 4 : 4 : 4 : 4

Explaination: Two of the three F atoms are equivalent (i.e. exchangable by symmetry operations), while the third atom is unique. The F atoms will couple with the P atoms and the non-equivalent F atoms (³¹P: I = ½, 100% abundant; ¹⁹F: I = ½, 100% abundant). The signal for the unique F atom is one the left and the signal for the two equivalent F atoms is on the right.

h. ⁵⁷FeF₄^-
19F NMR:

Peak Height Ratio: 1 : 1

Explaination: All F atoms are equivalent (i.e. exchangable by symmetry operations) and couple with one Fe atom that is only the Fe-57 isotope (⁵⁷Fe: I = ½).

i. AgFPMe₃
¹⁹F NMR:

Peak Height Ratio: 24.1 : 24.1 : 25.9 : 25.9 : 25.9 : 25.9 : 24.1 : 24.1

Explaination: The F atom couples to the Ag atom, which has two NMR active isotopes (¹⁰⁷Ag: I = ½, 51.8% abundant; ¹⁰⁹Ag: I = ½, 48.2% abundant) and then couples with the P atom (³¹P: I = ½, 100% abundant).

j. AlF₄^-
19F NMR:

Peak Height Ratio: 1 : 1 : 1 : 1 : 1 : 1

Explaination: All F atoms are equivalent (i.e. exchangable by symmetry operations) and couple with one Al atom (²⁷Al: I = 5/2, 100% abundant).