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VII.C.8. (XII.E.1.)

Notes on converting between angles in tetravalent centers

C3V molecules

CH3F angles In a C3v molecule such as CH3F, the angles FCH and HCH are related by symmetry. The angles a (FCH) and b (HCH) are related using the following formulas:
cos(b)=1-3/2sin^2(a)
sin(a)=SQRT(2/3(1-cos(b))
Please note that the latter solution is not unique, and that solving for angle a may return 180 - a
enter an angle greater than 0 and less than 180
a= Click to solve for b

enter an angle greater than 0 and less than 120
b= Click to solve for a
Click to clear

C2V pentatomic molecules

CH2F2 angles In a C2V molecule such as CH2F2 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:
cos(e)=cos(c/2)cos(d/2)
Please note that the latter solution is not unique, and that solving for angle e may return 180 - e
enter two angles between 0 and 180 degrees
c=
d=
e=
Solve for angle c
Solve for angle d
Solve for angle e
Clear

Cs tetravalent molecules

There are six possible angles, but only three are needed to define the tetravalent center. Given three angles what is the fourth?
four angles in tetravalent center 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)
cos(f) = cos(d)*cos(e)-sin(d)*sqrt((1+cos(c))/2-cos(e)cos(e))
ugly equation

Arbitrary tetravalent center

Given five of the possible angles what is the sixth?
six angles in tetravalent center 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( (ca2 + cc2 + ce2 - 2*ca*cc*ce - 1) * ((ca2 + cb2 + cd2 - 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)/(ca2 - 1)
cf = (-p + q)/(ca2 - 1)

Dihedral from three angles

dihedral as a function of 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).
cos(d)=(cos(c)-cos(a)cos(b))/(sin(a)sin(b))