In the recent question about the molecular orbitals of NOXx; x∈{−1,0,+1} the follow-up title question arose:
Just a random side-question: What’s the dipole moment’s negative and positive sides in NOX+?
At the time I was unable to answer that question sufficiently, therefore I share the addendum Q&A style. The question further deconstruct the molecular orbital components to the dipole moment.
Answer
In non-neutral species the dipole moment in calculations is dependent on the choice of the co-ordinate system. For this reason the origin is chosen to be the centre of mass. With r=107.8 pm on RMP2/def2-QZVPP previously determined by a geometry optimisation. The program used for the computations is Gaussian 09 Rev. D.01 and used input files are attached at the bottom of the post.
I am loosely following the same procedure/ conclusions as in Gernot Frenking, Christoph Loschen, Andreas Krapp, Stefan Fau, and Steven H. Strauss, J. Comp. Chem., 2007, 28 (1), 117-126. (I believe it is available free of charge.) since NOX+ is isoelectronic with CO.
According to the definition of the dipole moment ⊖p⟶⊕ positive values in the following table indicate that the negative end is at the nitrogen atom.
OrbitalBond MomentMO 1 (σ)−4.8336MO 2 (σ)5.5232MO 3 (σ)−0.7891MO 4 (σ)−3.0061MO 5 (π)−1.1642MO 6 (π)−1.1642MO 7 (σ)4.7383MO 8 (π,virt.)(2.3831)MO 9 (π,virt.)(2.3831)Total ∑7iMO i−0.6956
While we see the same polarisation of the molecular orbitals as in carbon monoxide, the magnitude differs and the HOMO of NOX+ cannot compensate for the polarisation of the electron density towards the oxygen.
Appendix
Base calculation for NOX+ after previous optimisation, E(RHF)=-128.972090453
input file:
%chk=mp2qzvpp.chk
#p MP2/def2QZVPP
scf(verytight)
symmetry(loose)
gfinput gfoldprint iop(6/7=3)
base
+1 1
N 0.0 0.0 -0.575312
O 0.0 0.0 0.503398
Analysis calculation input file for MO 1
%oldchk=mp2qzvpp.chk
#p MP2/def2QZVPP
geom=check
guess=(read,only,alter)
symmetry(loose)
gfinput gfoldprint iop(6/7=3)
analysis
+13 1
(Both empty lines are necessary.)
Analysis calculation input file for MO 2, etc
%oldchk=mp2qzvpp.chk
#p MP2/def2QZVPP
geom=check
guess=(read,only,alter)
symmetry(loose)
analysis
+13 1
2,1
The last line needs to contain the complete reordering procedure. The MO to be analysed needs to be in first place. Therefore for MO 3 it needs to read 3,1,2
, for 4 4,1,2,3
, etc.
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