Multiple Bonds and Molecular Shapes

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 12:40

In a double bondAttraction between two atoms (nuclei and core electrons) that results from sharing two pairs of electrons between the atoms; a bond with bond order = 2., two electronA negatively charged, sub-atomic particle with charge of 1.602 x 10-19 coulombs and mass of9.109 x 1023 kilograms; electrons have both wave and particle properties; electrons occupy most of the volume of an atom but represent only a tiny fraction of an atom's mass. pairs are shared between a pair of atomic nuclei. Despite the fact that the two electron pairs repel each other, they must remain between the nuclei, and so they cannot avoid each other. Therefore, for purposes of predicting molecular geometry, the two electron pairs in a double bond behave as one. They will, however, be somewhat “fatter” than a single electron-pair bond. For the same reason the three electron pairs in a triple bondAttraction between two atoms (nuclei and core electrons) that results from sharing of three pairs of electrons between the atoms; a bond with bond order = 3. behave as an “extra-fat” bond.

As an example of the multiple-bond rules, consider hydrogen cyanide, HCN. The Lewis structure is

Image:HCN.jpg

Treating the triple bond as if it were a single “fat” electron pair, we predict a linear moleculeA set of atoms joined by covalent bonds and having no net charge. with an H―C―H angle of 180°. This is confirmed experimentally. Another example is formaldehyde, CH2O, whose Lewis structure is

Image:Formaldehyde.jpg

Since no lone pairs are present on C, the two H’s and the O should be arranged trigonally, with all four atomsThe smallest particle of an element that can be involved in chemical combination with another element; an atom consists of protons and neutrons in a tiny, very dense nucleus, surrounded by electrons, which occupy most of its volume. in the same plane. Also, because of the “fatness” of the double bond, squeezing the C—H bond pairs together, we expect the H―C―H angle to be slightly less than 120°. Experimentally it is found to have the value of 117°.



EXAMPLE 1 Predict the shape of the two molecules (a) nitrosyl chloride, NOCl, and (b) carbon dioxide, CO2.


SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture.


a) We must first construct a skeleton structure and then a Lewis diagram. Since N has a valence of 3, O a valence of 2, and Cl is monovalent, a probable structure for NOCl is

Image:ONCl bonding.jpg

Completing the Lewis diagram, we find

Image:ONCl.jpg

Since N has two bonds and one lone pair, the molecule must be angular. The O—N—Cl angle should be about 120°. Since the “fat” lone pair would act to reduce this angle while the “fat” double bond would tend to increase it, it is impossible to predict at this level of argument whether the angle will be slightly larger or smaller than 120°.

b) The Lewis structure of CO2 was considered in the previous chapter and found to be

Image:Carbon dioxide.jpg

Since C has no lone pairs in its valence shell and each double bond acts as a fat bond pair, we conclude that the two O atoms are separated by 180° and that the molecule is linear.