Modifications to basic shapes: considering the effect of lone pairs
|
•
|
Consider the molecule [XeF5]-. Xe is in group 18 and possesses 8 electrons in its valence shell. There are 5 fluorine atoms, each contributing 1 electron, this brings our electron count to 13. Finally, we add one electron to account for the single negative charge on the Xe. This brings our total to 14. Again, we divide this number by 2 to give the total number of electron pairs (i.e. 7) The parent shape is therefore a pentagonal bipyramid.
|
|
•
|
However, once the basic shape of a molecule has been identified, adjustments must be made to account for the differences in electrostatic repulsion between bonding regions and lone pairs. Repulsions lie in the order: lone pair/lone pair > lone pair/bonding pair > bonding pair/bonding pair The greater repelling effect of a lone pair is explained by supposing that the lone pair is on average closer to the nucleus than a bonding pair and therefore repels other electrons more strongly.
|
|
•
|
Thus for [XeF5]- the parent shape is a pentagonal bipyramid with the lone pairs opposite to each other in order to minimise lone pair/lone pair repulsions. The [XeF5]- anion is therefore pentagonal planar.
|
|
•
|
Di-bromodimethylselenium (CSD refcode RIZMIW) has 5 electron pairs (4 bonding pairs, and 1 lone pair), the parent shape is therefore trigonal-bipyramidal. Notice that the lone pair occupies the equatorial site in the trigonal-bipyramidal array. In the equatorial position the lone pair is repelled by two bonding pairs at 90 degrees, whereas in the axial position it would be repelled by 3 bonding pairs at 90 degrees. RIZMIW therefore adopts a disphenodial, or “see-saw” shape.
|
|
•
|
Water (CSD refcode MUSIMO01) has 4 electron pairs (2 bonding pairs and 2 lone pairs). The parent shape is therefore tetrahedral and the molecule will adopt an angular, or “bent” shape. Notice that the HOH angle is decreased relative to that expected when all pairs are bonding. The two lone pairs repel each other more strongly and move apart thus forcing the HOH angle to be less than the ideal tetrahedral angle of 109.5 degrees.
|
|