Simulating Exchange Coupled Biradicals

[jchen3]

Hi All,

I've asked several questions about a room temperature epr simulation on the forum, while many suggestions have been offered and tried with no luck. I decide to post it one more time with all information, and to seek helpful advice on that. The raw room temperature epr spectrum is shown below

SQVD.jpg (23.26 KiB) Viewed 6818 times

The system is expected to be exchange coupled at ground state, and the computation agrees with that. Compared to single radical, it is lacking certain degree of central symmetry. There are 4 nitrogens, and maybe 2 hydrogens responsible to the hyperfine.
I started with building a system with exchange coupling information using pepper, and only to find it was to large to compute. Given the fact it is exchange coupled, I removed exchange coupling information but still with pepper, and the spin system is shown as below:

Code: Select all

Sys.S=1;
Sys.g=2.0040;
Sys.Nucs='14N,14N,14N,14N,1H,1H'
Sys.A=[14.88;14.88;18.23;18.23;3.79;3.79];
Sys.lw=0.138;
Exp.mwFreq=9.3953
Exp.Range=[331,339]

This gives me everything consistent with the experimental except for the symmetry. In other words, this looks exactly like a single radical spectrum. And the hyperfine values are almost identical to those of single radical, while what I have learnt all the time is the measured hyperfine constant should be halved compared to the intrinsic ones.
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Simulation.jpg (27.59 KiB) Viewed 6818 times

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I did collect half field transition at 77K to assure it is a biradical.
Initially, based on previous work on Nitronyl Nitroxide radicals which does not have that much hyperfine features, I tried to increase the linewidth to distort the spectrum. It did work on the distortion part, but certainly I lost smaller hyperfines as well.
After posting the problem on the forum, I have tried settings from Stefan as Rotational correlation time. In order to include this feature, I switched to "chili", and since it was a isotropic spectrum, the simulation returned as a flat line.
Although it didn't look reasonable, I tried to include zero-field splitting. This gave an asymmetrical looking while I was not able to simulate the hyperfine features.
So all in all, I have lost my mind about this problem, and I'd appreciate all suggestions on this! Thank you.

Best,
Ju

[Stefan Stoll]

The deviation of the simulation form the experiments has two aspects: (a) the exp has a broad underlying line that the sim doesn't capture; (b) there is a slight asymmetry in the experiment that the sim doesn't capture.

As for (a), you can attempt to subtract a broad Lorentzian line from the experimental data (or add it to the sim). You will need to figure out where it comes from.

As for (b), the asymmetries are really small, and your simulation looks quite good. If you are only after the isotropic hyperfine constants, then I wouldn't worry about the residual asymmetry. Extracting anything more than that is likely not possible. If you include a rotational correlation time, for simulation with chili, you will need to include the full anisotropic hyperfine tensors.

[jchen3]

The deviation of the simulation form the experiments has two aspects: (a) the exp has a broad underlying line that the sim doesn't capture; (b) there is a slight asymmetry in the experiment that the sim doesn't capture.

As for (a), you can attempt to subtract a broad Lorentzian line from the experimental data (or add it to the sim). You will need to figure out where it comes from.

As for (b), the asymmetries are really small, and your simulation looks quite good. If you are only after the isotropic hyperfine constants, then I wouldn't worry about the residual asymmetry. Extracting anything more than that is likely not possible. If you include a rotational correlation time, for simulation with chili, you will need to include the full anisotropic hyperfine tensors.

Dear Stefan,

Thanks for your reply! I have 2 more questions accordingly.
1. For substracting Lorentzian line to the data for example, are you talking about adding it onto the original data instead of the1st derivative? Since simply adding it to the data will make the left side more positive, and the right side less negative, while what I am looking for is to make left side more positive, and more negative on the right side.
2. What is your opinion that my simulation gives me "observed" A values of biradical almost identical to those of the single radical? What I have learnt is that for an exchange coupled system, the observed A values is roughly half of the intrinsic A values. I have not performed susceptibility measurement yet, but computation indicates it is strongly exchange coupled at ground state(triplet 100 wavenumber above)?
Thank you!

[nwili]

What I have learnt is that for an exchange coupled system, the observed A values is roughly half of the intrinsic A values.

This depends on the formalism you use. The hf coupling is only halved if you use the coupled representation with S=1. If you use two S=1/2 with isotropic exchange, it is expected that the hf couplings roughly stay the same.

[Stefan Stoll]

Be careful to cleanly distinguish between the hf coupling (A in the spin Hamiltonian) and the hf splitting (observed in the spectrum).

Here is an example that shows the effect of strong exchange coupling on the hf splitting:

Code: Select all

clear, clc

g = 2;
A = 80;
Nuc = '1H';
lw = 1;

% Individual spin
Sys1.S = 1/2;
Sys1.g = g;
Sys1.Nucs = Nuc;
Sys1.A = A;
Sys1.lw = lw;

% Two strongly exchange-coupled spins
Sys2.S = [1/2 1/2];
Sys2.g = [g g];
Sys2.J = 1e5; % MHz
Sys2.Nucs = [Nuc ',' Nuc];
Sys2.A = [A 0; 0 A]; % MHz
Sys2.lw = lw;

% Simulate frequency-domain spectra
Exp.Field = 350; % mT
nu0 = mt2mhz(Exp.Field,mean(Sys1.g));
Exp.mwRange = nu0/1e3 + [-1 1]*0.1; % GHz
[nu,spc1] = pepper(Sys1,Exp);
[nu,spc2] = pepper(Sys2,Exp);

% Plotting
dnu = 1e3*nu-nu0;
plot(dnu,spc1,dnu,spc2);
xlabel('frequency offset (MHz)');
legend('individual','coupled');

[jchen3]

What I have learnt is that for an exchange coupled system, the observed A values is roughly half of the intrinsic A values.

This depends on the formalism you use. The hf coupling is only halved if you use the coupled representation with S=1. If you use two S=1/2 with isotropic exchange, it is expected that the hf couplings roughly stay the same.

Yes, I am with S=1, and the hf splitting of the biradical is almost the same as the hf splitting of a single radical. I have no idea what is going on.

[jchen3]

Be careful to cleanly distinguish between the hf coupling (A in the spin Hamiltonian) and the hf splitting (observed in the spectrum).

Here is an example that shows the effect of strong exchange coupling on the hf splitting:

Code: Select all

clear, clc

g = 2;
A = 80;
Nuc = '1H';
lw = 1;

% Individual spin
Sys1.S = 1/2;
Sys1.g = g;
Sys1.Nucs = Nuc;
Sys1.A = A;
Sys1.lw = lw;

% Two strongly exchange-coupled spins
Sys2.S = [1/2 1/2];
Sys2.g = [g g];
Sys2.J = 1e5; % MHz
Sys2.Nucs = [Nuc ',' Nuc];
Sys2.A = [A 0; 0 A]; % MHz
Sys2.lw = lw;

% Simulate frequency-domain spectra
Exp.Field = 350; % mT
nu0 = mt2mhz(Exp.Field,mean(Sys1.g));
Exp.mwRange = nu0/1e3 + [-1 1]*0.1; % GHz
[nu,spc1] = pepper(Sys1,Exp);
[nu,spc2] = pepper(Sys2,Exp);

% Plotting
dnu = 1e3*nu-nu0;
plot(dnu,spc1,dnu,spc2);
xlabel('frequency offset (MHz)');
legend('individual','coupled');

Dear Stefan,

Thanks for the example. I quickly ran it and that is showing exactly what we expect for a strongly exchange coupled system. Hyperfine splitting is halved for the biradical compared to the single one.

Exchange coupled.jpg (23.06 KiB) Viewed 6754 times

In my case, the hf splitting for the biradical is almost the same as those of single radical, and they can essentially overlap with each other. Maybe I really have to do a susceptibility measurement to see what is going on.

[Stefan Stoll]

Typically, susceptibility measurements are much less informative than EPR spectra, esp.if the exchange coupling is really small. If you see no difference between the monoradical and the biradical, this is very strong evidence that the coupling is rather insignificant.

[jchen3]

Typically, susceptibility measurements are much less informative than EPR spectra, esp.if the exchange coupling is really small. If you see no difference between the monoradical and the biradical, this is very strong evidence that the coupling is rather insignificant.

Dear Stefan,

Indeed, I really need to perform a variable temperature susceptibility measurement to address this issue while the instrument is not ready yet. And that's what have been bugging me. Computation was done with good model and inferring strong exchange coupling (over 100 wavenumber).