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
Martin Breza Saeedreza Emamian Miguel Mompeán Sir, I have a doubt on NBO calculation of my system. A particular bond is absent in NBO but is showing bond when I visualize the system in avogadro and the bond should present. Is the NBO calculation wrong in my case? or NBO can not give all bonds that a molecule has?
Hallo, Sashi!
NBO analysis does not show any (prevailingly) ionic bonds because only (prevailingly) Lewis bonding is supposed. If you visualize the system, the bonds are ascribed on the basis of interatomic distances.
Best wishes
mb
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