I’ve been asked to study using DFT (wb97xd/lanl2dz). Prior to introducing Cu(2+), my structure seems reasonable enough yet when Cu(2+) is introduced at a particular site, the C-C bond prior to the carboxylate (we knocked off the hydrogen from the COOH group due to its absence from our mass spec data) group yields a distance of 1.88 angstrom. The structures have been tightly optimised, and I have tried this out on a different control with the same result.
My question is how do I go about understanding whether there is a bond, or no bond at all using NBO. I have looked at my NBO output and found the BD and BD* entries:
1. (0.97584) BD ( 1) C 1 – C 2
( 47.03%) 0.6858* C 1 s( 25.63%)p 2.90( 74.37%)
0.0001 -0.4963 0.0999 0.5583 0.0288
0.6491 0.0140 0.0981 0.0049
( 52.97%) 0.7278* C 2 s( 11.14%)p 7.98( 88.86%)
-0.0001 -0.3314 0.0399 -0.6464 -0.0426
-0.6385 -0.0367 -0.2441 -0.0184
I am not a pure chemist to any degree, but I assume the C-C occupancy of 0.97584 implies that no bond is formed? i.e., a single electron on the bonding valence suborbital. However, when I investigate further, it appears as though no occupancies for BD appear to be above 1, which seems odd as I would have expected a lot of values about 1.9. Below are a few more NBOs straight out of the log file:
(Occupancy) Bond orbital/ Coefficients/ Hybrids
---------------------------------------------------------------------------------
1. (0.97584) BD ( 1) C 1 - C 2
( 47.03%) 0.6858* C 1 s( 25.63%)p 2.90( 74.37%)
0.0001 -0.4963 0.0999 0.5583 0.0288
0.6491 0.0140 0.0981 0.0049
( 52.97%) 0.7278* C 2 s( 11.14%)p 7.98( 88.86%)
-0.0001 -0.3314 0.0399 -0.6464 -0.0426
-0.6385 -0.0367 -0.2441 -0.0184
2. (0.99644) BD ( 1) C 1 - O 13
( 31.38%) 0.5602* C 1 s( 37.17%)p 1.69( 62.83%)
0.0001 0.6079 0.0465 -0.2135 -0.0086
0.6909 -0.0369 -0.3219 0.0202
( 68.62%) 0.8284* O 13 s( 37.72%)p 1.65( 62.28%)
0.0000 0.6136 0.0266 0.3119 -0.0100
-0.5957 0.0211 0.4122 -0.0156
3. (0.99695) BD ( 1) C 1 - O 14
( 29.66%) 0.5446* C 1 s( 14.77%)p 5.77( 85.23%)
0.0000 0.3830 0.0321 0.0542 -0.0281
0.1042 0.0036 0.9151 -0.0150
( 70.34%) 0.8387* O 14 s( 14.01%)p 6.14( 85.99%)
0.0000 0.3737 0.0205 -0.7304 0.0181
0.3934 -0.0082 0.4138 0.0086
4. (0.99576) BD ( 2) C 1 - O 14
( 30.93%) 0.5561* C 1 s( 23.68%)p 3.22( 76.32%)
-0.0002 0.4838 0.0527 0.7945 -0.0296
-0.2866 -0.0135 -0.2199 -0.0185
( 69.07%) 0.8311* O 14 s( 22.60%)p 3.42( 77.40%)
0.0000 0.4747 0.0261 -0.2006 0.0201
0.0712 -0.0080 -0.8532 0.0161
Many thanks
Anthony
I have looked into this further. If I remove copper from the system (an earlier calculation I had performed as a control), I always find that my BD values are greater an than 1.
Hallo, Anthony!
I suppose you are using an unrestricted formalism. In such case you are working with spinorbitals (containing 1 electron - alpha or beta) and not with orbitals (containing 2 electrons - alpha + beta in closed shell systems). It means that your NBO output contains two parts - for alpha and beta electrons. It means you must make a summation of both these parts to btain complete results.
I propose to start with the NBO calculations on more simple systems (such as CH3COO(-) in singlet spin state and CH3COO-Cu(+) in doublet spin state).and only afterwards to continue with more complex systems
Best wishes
mb
Hi Anthony,
I agree with Martin, you are dealing with an open-shell system and that is why you see those occupancy numbers. Regarding your first question in the former post, your output indeed tells that there is a C1-C2 bond, and you can see what the overlap between the hybrids is (sp^2.90 & sp^7.98, respectively).
As Martin pointed, you should look for both the alpha and beta spin orbitals. You can achieve this by searching for these two entries in your output file:
"NATURAL BOND ORBITAL ANALYSIS, alpha spin orbitals"
"NATURAL BOND ORBITAL ANALYSIS, beta spin orbitals"
(In the case of closed-shell systems you only have " NATURAL BOND ORBITAL ANALYSIS")
Cheers,
Miguel
I totally agree with Martin whose comments are quite straightforward and reliable.
Best,
Saeed
- Badges
- Science topic