Read Ebsworth et al pp115-128

Problems, Ebsworth p 136                  3.3, 3.6, 3.8, 3.10, 3.11, 3.14

Solutions to 3.3 and 3.11 are given on p 483

3.6
Spectrum consists of an octet of 1:2:1 triplets. Since V is ~100 ⁵¹V (I = 7/2), the electron interacts with a V nucleus to give 8 hyperfine lines, and with two equivalent P atoms. This suggests that two phosphine ligands are coordinated to the V in the product.

(a) If the solution were frozen, the spectrum would become more complicated with 8 parallel and 8 perpendicular sets of lines, each set with different hyperfine coupling constants. All lines would also be split by the P atoms.

(b) AsEt₃ instead of PEt₃. Since As is 100% ⁷⁵As (I = 3/2) each of the 8 V hyperfine lines would be split into a heptet (2nI+1) with intensities 1:2:3:4:3:2:1. In practice some of these lines might overlap

(c) Nb instead of V. Since Nb is 100% ⁹³Nb (I=9/2), the spectrum would consist of ten 1:2:1 triplets.

3.8
The spectrum shows a large signal (at g-perpendicular) without hyperfine coupling, and a smaller g-parallel signal with 8 hyperfine lines. The complex (Rh^II, d⁷ low spin) therefore has axial symmetry and is presumably a trans isomer.
Possible magnetic nuclei are four ¹⁴N (~100%, I=1), ¹⁰³Rh (100%, I=1/2), and two ^35,37Cl (100%, I=3/2)

I can think of two possible interpretations, assuming a trans isomer.

(a) Two equivalent Cl nuclei should give 7 lines of intensity 1:2:3:4:3:2:1. The Rh splits each of these into a doublet to give 1:1:2:2:3:3:4:4:3:3:2:2:1:1. If the coupling constants are the same for Cl and Rh, then adjacent lines will overlap and the intensities become 1:(1+2):(2+3):(3+4):(4+3):(3+2):(2+1):1 which agrees with the observed spectrum.

Why is there no splitting from the N's? Perhaps because the unpaired electron is in a d(z²) orbital.

(b) Four equivalent N nuclei give 9 lines of intensity 1:4:10:16:19:16:10:4:1. The Rh splits each into a doublet. 1:1:4:4:10:10 etc. Again, if the coupling constants are equal, adjacent lines overlap to give 1:(1+4):(4+10):(10+16):(16+19):(19+16) etc

i.e.    1:5:14:26:35:35:26:14:5:1  (10 lines)

If the two weakest lines cannot observed, the ratio of the remaining lines is  1:~3:~5:7:7:~5:~3:1.

Why no splitting by the Cl's? Perhaps because the unpaired electron is in a d(x²-y²) orbital.

Which is correct? Difficult to tell. In practice you could prepare the compound with ¹⁵N-labeled pyridine. If spectrum is unchanged, then (a) is correct. What would the hyperfine pattern look like if (b) were correct? Remember that ¹⁵N has I = 1/2.

3.10
(a) Co₄ cluster. ⁵⁹Co, 100%, I=7/2. 2nI+1 = 29 hyperfine lines

(b) Rh₄ cluster. ¹⁰³Rh, 100%, I=1/2. 2nI+1 = 5 hyperfine lines

(c) Co₂Rh₂ cluster. (2n_CoI_Co+1)(2n_RhI_Rh+1) = 45 hyperfine lines

3.14
(a) Fe^III high spin has S=5/2. Zero-field splitting gives ±½, ±3/2, ±5/2 states. Five possible transitions: (-5/2 <> -3/2) (-3/2<>-1/2) (-½<> +1/2) (+½<> +3/2)(+3/2<> +5/2).

(b) Cu^II has S = ½. One transition (may also be four hyperfine lines), (-½<> +1/2)

(c) V^II has S = 3/2. Three transitions (each may show eight hyperfine lines) as a result of zero-field splitting: (-3/2<>-½) (-½<> +½) (+½ <>+3/2)

Problems, Drago pp 597-603  1, 8, 11

1. a. g is taken as the midpoint of the two central lines (each measured halfway between the peaks)

My estimate of the four line positions (in G, 100 G = 13 mm) is:

2908, 2977, 3054, 3123,         So midpoint is 3015G

g = hn/bH = (6.63x10^-27[erg s] x 9.4x10⁹ [s^-1])/(9.274x10^-21 [erg G^-1] x 3015 [G]    = 2.23

A = (3123-2908)/3 = 72G

b. A is an energy of interaction between S and I, and should be exspressed in hertz to make it independent of the value of g.

Conversion to hertz

A(MHz) = A(G) x [2.80247(2.0023/g)] = 181 MHz

8(a) Co(H₂O)₆²⁺ has S = 3/2 and a T₁ ground state. As a result of zero-field splitting (see answer to 3.14c above) there should be three transitions. However the degeneracy of the ground state indicates that the electron spin relaxation time will be very short and no spectrum is likely to be observed except at very low temperature.

8(b) Cr(H₂O)₆³⁺ also has S=3/2, but with an A₂ ground state. Three transitions should be observable unless zero-field splitting is very large, when only the -½ +½ will be seen.

11a Four parameters are needed to decribe the spectrum: g-parallel , g-perpendicular , A-parallel and A-perpendicular. The features are the two sets of four lines and the two different fields at which they appear

b The high-field region of the spectrum is more intense because this is the perpendicular region which includes both x and y components of g-perpendicular and A-perpendicular.

c The value of g-parallel is determined as in question 1. The midoint of the four parallel lines is estimated to be 2810G

So g = hn/bH = (6.67x10^-27 x 9.12x10⁹)/(9.27x10^-21 x 2810) = 2.34

FINAL QUESTION

The ESR spectrum of a solution of [Fe(CN)₅NO]^3- consists of three equally-spaced hyperfine lines, while that of [Mn(CN)₅NO]^2- has six hyperfine lines. Interpret these spectra in terms of the electronic structures of the complexes.

The ESR spectrum of [Fe(CN)₅NO]^3- has three hyperfine lines. Shows unpaired electron interacting with N  (¹⁴N, I=1) of NO ligand. (Fe is 98% I=0, and ~2% ⁵⁷Fe I=1/2)

The spectrum of [Mn(CN)₅NO]^2- has six lines, showing electron interacting with ⁵⁵Mn nucleus (100%, I=5/2), but not with NO ligand.

Why are they different?

Count the number of valence electrons.

You can consider NO as neutral (with one "extra" electron) or as NO⁺ (isoelectronic with CN^-)

So [Fe(CN)₅NO]^3- is either an Fe^II (d⁶) complex with NO or an Fe^I (d⁷) complex with NO⁺. Either way, there are 7 valence electrons. Six will occupy the t_2g orbitals and the seventh will be in an e_g orbital directed towards the ligands (i.e. towards the N in NO).

[Mn(CN)₅NO]^2- is either a Mn^III (d⁴) complex with NO, or a Mn^II (d⁵) complex with NO⁺.
Either way, there are 5 valence electrons. These will occupy the t_2g orbitals, and the unpaired electron will be therefore be associated with the metal center.