IB syllabus > bonding (hl) > 14.1 

These notes were written for the old IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

14.1 - Shapes of molecules and ions


14.1.1:- Predict the shape and bond angles for for species with 5- and 6-negative charge centres using the VSEPR theory.
Examples should include: PCl5, SF6, XeF4 and PF6-. AIM 7: Interactive simulations are available to illustrate this.


Repulsion of electrons

Electrons are negatively charged and, as such, repel each other. When electrons are in orbitals with the same energy (degenerate) these orbitals then orient themselves to be as far apart as possible so as to minimise repulsion. The orientations that are adopted will depend on the number of regions of electron density (orbitals) surrounding the nucleus.

If there are two regions of electron density attached to a nucleus then they will be on opposite sides of the nucleus i.e. they will be arranged 180 degrees to each other.

The following table summarises the shapes adopted by the regions of electron density

No of regions shape adopted angles between orbitals
2 linear 180º
3 trigonal planar 120º
4 tetrahedral 109,5º
5 trigonal bipyramid 120º and 90º
6 octahedral 90º and 180º

It must be remembered that we are talking about the regions of electron density here and that this may not be reflected in the molecular shape. If the electrons are not being used to bond then they cannot be seen and the molecule must be described as if they were not there.


Phosphorus pentachloride, PCl5

In this molecule the phosphorus is the central atom and provides 5 electrons.

Each of the bonded chlorines provide one electron making 5 + 5=10 in total around the central atom.

Each bond has two electrons and there are five bonds using up 5 x 2 =10 electrons.

Hence there are no electron pairs left over and the central atom arranges the five regions of electron density into the ideal orientation - a trigonal pyramidal shape.

Phosphorus pentachloride

Phosphorus(V) chloride


Xenon hexafluoride, XeF6

Xenon from group 0 has eight electrons in its outer shell and each fluorine provides one for the bond maing a total of 8 + 6=14 electrons.

It may be seen from the formula that the Xe-F bonds require a total of 6 x 2=12 electrons therefore there is also a lone pair to make up the 14 electrons.

Electronically the seven oritals orientate themselves to form a pentagonal bipyramidal shape with two different bond angles 90º and 72º. The lone pair will occupy the one of the orbitals at 90º to the plane of the central 5 orbitals but will distort them downwards according to the valence shell electron pair repulsion theory. The final result is a sort of umbrella shape.


Xenon tetrafluoride, XeF4

Xenon from group 0 has eight electrons in its outer shell and each fluorine provides one for the bond maing a total of 8 + 4 = 12 electrons.

It may be seen from the formula that the four Xe-F bonds require a total of 4 x 2 = 8 electrons therefore there are also two lone pairs to make up the 12 electrons.

Electronically the six oritals orientate themselves to form an octahedral shape with bond angles 90º . The lone pairs will occupy the orbitals as far apart as possible i.e. on oppositesides of the octahedral shape according to the valence shell electron pair repulsion theory.

The final result is a square planar arrangement of Xe-F bonds with lone pairs above and below.

Xenon tetrafluoride


Useful links

 


Resources

VSEPR flash chart

VSEPR flash