VII.C.8. (XII.E.1.) |
Notes on converting between angles in tetravalent centers
C_{3V} molecules
In a C_{3v} molecule such as CH_{3}F, the angles FCH and HCH are
related by symmetry.
The angles a (FCH) and b (HCH)
are related using the following formulas:
Please note that the latter solution is not unique, and that solving for angle a may return 180° - a |
C_{2V} pentatomic molecules
In a C_{2V} molecule such as CH_{2}F_{2}
the angle HCF can be related to the two angles HCH and FCF by symmetry.
The angle e (HCF)
is related to the angles c (HCH)
and d (FCF) by:
Please note that the latter solution is not unique, and that solving for angle e may return 180° - e |
C_{s} tetravalent molecules
There are six possible angles, but only three are needed to define the tetravalent center. Given three angles what is the fourth?
We use fluorochloromethane as an example. We know:
angle c (H-C-H) (there is only one) angle d (F-C-Cl) (there is only one) angle e (H-C-Cl) (there are two equivilant) We want angle f (F-C-H) (there are two equivilant) |
Arbitrary tetravalent center
Given five of the possible angles what is the sixth?
In a arbitrary tetravalent center their are six angles.
We define these angles as:
a is angle H-C-Cl b is angle F-C-Br c is angle H-C-Br d is angle F-C-Cl e is angle Cl-C-Br f is angle H-C-F Assume we want angle f as a function of the other five angles. We can work with the cosines of the angles: ca = cos(a), cb = cos(b),etc. Define p and q: p = sqrt( (ca^{2} + cc^{2} + ce^{2} - 2*ca*cc*ce - 1) * ((ca^{2} + cb^{2} + cd^{2} - 2*ca*cb*cd - 1) ) q = ca*(cb*ce + cc*cd) - cb*cc - cd*ce The two solutions for cos(f) are: cf = ( p + q)/(ca^{2} - 1) cf = (-p + q)/(ca^{2} - 1) |
Dihedral from three angles
Given the three angles:
a is angle F-C-Cl b is angle H-C-Cl c is angle H-C-F We want the dihedral angle d (the angle between the two planes defined by F-C-Cl and H-C-Cl). |