Valence Bond Theory and Hybrid Atomic Orbitals
- Page ID
- 32843
H H H O | | | // H-C-C=C=C-C=C-C-C | | \ H N O-H / \ H H |
By applying the VSEPR theory, one deduces the following results:
- \(\ce{H-C-C}\) bond angle = 109o
- \(\ce{H-C=C}\) bond angle = 120o, geometry around \(\ce{C}\) trigonal planar
- \(\ce{C=C=C}\) bond angle = 180o, in other words linear
- \(\ce{H-N-C}\) bond angle = 109o, tetrahedral around \(\ce{N}\)
- \(\ce{C-O-H}\) bond angle = 105 or 109o, 2 lone electron pairs around \(\ce{O}\)
Confidence Building Questions
- In terms of valence bond theory, how is a chemical bond formed?
Discussion -
Molecular orbital theory considers the energy states of the molecule. - When one s and two p atomic orbitals are used to generate hybrid orbitals, how many hybrid orbitals will be generated?
Discussion -
Number of orbitals does not change in hybridization of atomic orbitals. - In the structures of \(\ce{SO2}\) and \(\ce{NO2}\), what are the values of the bond angles?
Discussion -
Since the lone electron pair in \(\ce{:SO2}\) and lone electron in \(\ce{.NO2}\) take up more space, we expect the structure to distort leaving a smaller angle than 120 between the bonds. - What is the geometrical shape of the molecule \(\ce{CH4}\), methane?
Discussion -
The 4 \(\ce{H}\) atoms form a tetrahedron, and methane has a tetrahedral shape. - What do you expect the bond angles to be in the \(\ce{NH4+}\) ion?
Discussion -
The structure of this ion is very similar to that of \(\ce{CH4}\). - What hybrid orbitals does the \(\ce{C}\) atom use in the compound \(\ce{H-C\equiv C-H}\), in which the molecule is linear?
Discussion -
Sigma (s) bonds are due to sp hybrid orbitals, and 2 p orbitals are used for pi (p) bonds. The two sigma bonds for each \(\ce{C}\) are due to overlap of sp hybrid orbitals of each \(\ce{C}\) atom. - What hybrid orbitals does \(\ce{C}\) use in the molecule:
H O=C< H
This is a trigonal planar molecule. It is called formaldehyde, a solvent for preserving biological samples. The compound has an unpleasant smell.Skill -
The \(\ce{C}\) atom has 3 sigma (s) bonds by using three sp2 hybrid orbitals and a pi (p) bond, due to one 2p orbital. - What is the shape of the molecule \(\ce{SF6}\)?
Discussion -
Since the \(\ce{S}\) atom uses d2sp3 hybrid orbitals, you expect the shape to be octahedral. The \(\ce{F}\) atoms form an octahedron around the sulfur. - Phosphorus often forms a five coordinated compound \(\ce{PX5}\). What hybrid orbitals does \(\ce{P}\) use in these compounds?
Discussion -
A total of 5 atomic orbitals are used in the hybridization: one 3d, one 3s and three 3p orbitals. The dsp3 hybrid orbitals of \(\ce{P}\) give rise to a trigonal bipyramidal coordination around the \(\ce{P}\) atom.The energy of d orbitals in \(\ce{N}\) is not compatible with 2s and 2p orbitals for hybridization. Thus, you seldom encounter a compound with formula \(\ce{NX5}\) with \(\ce{N}\) as the central atom.
Contributors and Attributions
Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo)