8.2 sp3 Hybridization
A. Sp3 hybridization
VB theory proposes that one s and all three p-orbitals of the central atom mix and form four sp3 hybrid orbitals, which point toward the vertices of a tetrahedron. In methane CH4, the central C atom is sp3 hybridized. Its four valence electrons half-fill the four sp3 hybrids, which overlap the half-filled 1s orbitals of the four H atoms and form four C-H bonds.
Figure 8.6 Overlap of s-p atomic orbitals
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Figure 8.7 sp3 hybrid orbital shape
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Here is the electronic distribution
Figure 8.8 Electronic arrangement in sp3 hybrid orbital
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CH4 hybridization:
Figure 8.9 Electronic arrangement in Hybrid Orbitals of Methane
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Tetrahedral (sp3) hybridization
Let us now look at several tetravalent molecules, and see what kind of hybridization might be involved when four outer atoms are bonded to a central atom. Perhaps the commonest and most important example of this bond type is methane, CH4.
In the ground state of the free carbon atom, there are two unpaired electrons in separate 2p orbitals. In order to form four bonds (tetravalence), need four unpaired electrons in four separate but equivalent orbitals. We assume that the single 2s, and the three 2p orbitals of carbon mix into four sp3 hybrid orbitals which are chemically and geometrically identical; the latter condition implies that the four hybrid orbitals extend toward the corners of a tetrahedron centered on the carbon atom.
Figure 8.10 Formation of Hybrid Orbitals of Methane
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Methane is the simplest hydrocarbon; the molecule is approximately spherical, as is shown in the space-filling model:
Figure 8.11(a) Ball and Stick 3D model of Methane
Figure 8.11(b) Space filling 3D model of Methane
[source]
By replacing one or more of the hydrogen atoms in CH4 with another sp3 hybridized carbon fragments, hydrocarbon chains of any degree of complexity can be built up. The simplest of these is ethane:
Figure 8.12 Overlapping of sp3 hybrid orbitals in ethane
This shows how an sp3 orbital on each of two two carbon atoms join (overlap) to form a carbon-carbon bond, and then the remaining carbon sp3 orbital overlaps with six hydrogen 1s orbitals to form the ethane molecule. [source – Yale U. ]
4 Lone pair electrons in hybrid obitals
If lone pair electrons are present on the central atom, these can occupy one or more of the sp3 orbitals. This causes the molecular geometry to be different from the coordination geometry, which remains tetrahedral.
In the ammonia molecule, for example, the nitrogen atom normally has three unpaired p electrons, but by mixing the 2s and 2p orbitals, we can create four sp3-hybrid orbitals just as in carbon. Three of these can form shared-electron bonds with hydrogen, resulting in ammonia, NH3.
Figure 8.13 Lone pair in ammonia
The fourth of the sp3 hybrid orbitals contains the two remaining outer-shell electrons of nitrogen which form a non-bonding lone pair. In acidic solutions these can coordinate with a hydrogen ion, forming the ammonium ion NH4+.
Figure 8.14 sp3 Hybrid Orbitals in ammonia
Although no bonds are formed by the lone pair in NH3, these electrons do give rise to a charge cloud that takes up space just like any other orbital.
Figure 8.15 Lone pair in water
In the water molecule, the oxygen atom can form four sp3 orbitals. Two of these are occupied by the two lone pairs on the oxygen atom, while the other two are used for bonding. The observed H-O-H bond angle in water (104.5°) is less than the tetrahedral angle (109.5°); one explanation for this is that the non-bonding electrons tend to remain closer to the central atom and thus exert greater repulsion on the other orbitals, thus pushing the two bonding orbitals closer together.
Figure 8.16 Atomic Orbital mixing in water
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sp3 hybridized orbitals and sigma bonds | Structure and bonding | Organic chemistry | Khan Academy
The following video is about sp3d and sp3d2 hybridization model.