It is a mathematical operation that involves the formation of mixed electron orbitals by a linear combination of appropriate wave functions. The work on that issue dates back to 1931, when Linus Pauling suggested the answer to the structure of methane – as a spatial arrangement of bonds. The bond in such molecules as diatomic hydrogen is rectilinear, but the geometry of organic compounds, containing tetravalent carbon atoms, is much more complex.
Methane: sp3 hybridisation
The simplest organic compound is the molecule of methane, which contains one atom of carbon. Its valence shell has four electrons, so it is capable of producing four bonds, and in a methane molecule – with four hydrogen atoms. Initially it was assumed that due to the use of two types of orbitals (2s and 2p) while forming the bonds, methane has two different types of C-H bonds. However, further research revealed a very high probability that, despite those assumptions, each C-H bond existing in methane is identical and spatially directed towards the corners of a regular tetrahedron. It was Linus Pauling who answered the question why it is so. He has proved mathematically how hybridisation, that is mixing an s-orbital with three p-orbitals, is possible. This causes the formation of four equivalent atomic orbitals whose spatial geometry has the shape of tetrahedron. This type of hybridisation is called sp3. The mere term “hybridisation” logically explains how exactly different orbitals mix with one another, but it does not answer the question why such transformations actually occur. However, this can also be explained. When an s-orbital hybridises, it mixes with three p-orbitals, and the resulting hybridised orbitals are not arranged symmetrically in relation to the nucleus. This is because the created sp3-orbital turns out to have one smaller and one bigger loop. The latter, being much larger, overlaps considerably better with the orbital of another atom while the bond is being formed. As a result, the orbitals or such an sp3 hybrid create bonds which are much stronger compared to the non-hybridised s- and p-orbitals.
The mechanism of sp3 hybridisation
The sp3-orbitals feature an asymmetry that relates to the wave equation defining the p-orbitals; in consequence, the two loops have opposite signs: plus and minus. Due to that characteristic and the overlapping of p- and s-orbitals, one of the p-orbital loops is additive, while the other is subtractive with the s-orbital. As a result, these loops are added to or subtracted from the s-orbital, which forms a hybridised orbital that is strongly oriented in one direction.
When identical orbitals of a carbon atom, with sp3 hybridisation, overlap with 1s-orbitals of four hydrogen atoms, the resulting C-H bonds are identical. In methane, their bonding energy is 438 kJ/mol and their length is up to 1.10Å. These are characteristic values that are fixed for a particular bond in that molecule. Another characteristic for the geometry of that molecule is the bond angle. It determines the angle formed between two consecutive H-C-H bonds and equals exactly 109.5o. We call it the tetrahedral angle.
Ethane: sp3 hybridisation
Another compound which can be considered in the same way is the ethane, which contains a bond between carbons (C-C). The carbon atoms existing in its structure link with each other as a result of overlapping σ-orbitals in the sp3 hybrid of each of them. The other three hybridised orbitals of each carbon atom overlap with the 1s-orbitals of hydrogen atoms. This creates six identical C-H bonds. Such bonds are characterised by an energy of 420 kJ/mol. The C-C bonds are 276 kJ/mol in energy and 1.54Å in length. The angles formed in such a configuration are tetrahedral (109.5o).
Ethylene: sp2 hybridisation
The most common electron state for carbon is the sp3 hybridisation, but there are also other variants. Research has revealed that in ethylene, for example, the carbon atoms show an appropriate quantity of bonds only if they link with one another to share four electrons. Then they form a double bond between themselves. Another fact is that ethylene has a flat structure, and the angles between its bonds are 120o. This is because in this case the 2s-orbital mixes with only two out of three existing 2p-orbitals. The result is the presence of three hybridised orbitals which are called sp2. There also exists one 2p-orbital, which is not hybridised. Then the geometric structure is as follows: three hybridised orbitals are located on the same plane, at 120o to one another, while the non-hybridised p-orbital is perpendicular to the sp2 plane.
The mechanism of sp2 hybridisation
As a result of the overlapping of sp2-sp2 orbitals, two carbon atoms with sp2 hybridisation form a σ-bond. The non-hybridised p-orbitals of atoms overlap laterally with one another, which causes the formation of a π-bond. In such a bond, the areas of electron density are present on both sides of the line between the nuclei, though not directly between them. Such a configuration, which contains a σ-bond of the sp2 hybrid and a π-bond of non-hybridised atoms, leads to four electrons being shared by two carbon atoms, so to forming a double bond C=C.
Therefore, the structure of ethylene contains four atoms of hydrogen, which form a σ-bond with four sp2-orbitals that remain once the double bond has been formed. The molecule has a flat geometry, and the bond angles are approximately 120o. The values specific to the C-H bond are the length of 1.076Å and the energy of 444 kJ/mol. Since only two electrons are shared, not four (unlike in, for instance, the structure of ethane), the double C=C bond is shorter and stronger than a single C-C bond. In ethylene, it is 1.33Å in length and 611 kJ/mol in energy. With the theory of molecular orbitals, we can also observe that the combination of two atomic p-orbitals forms bonding and antibonding molecular π-orbitals. A bonding orbital does not have a node between the nuclei due to the additive combination of the p-loop with the same algebraical sign. On the contrary, an antibonding orbital does have a node between the nuclei due to the subtractive effect of the loop with different algebraic signs. In consequence, only the less energetic, bonding molecular orbital gets filled.
Acetylene: sp hybridisation
Another possibility of linking between carbon atoms is by forming a triple bond with six electrons being shared. For this purpose, we need to introduce another orbital hybridisation, called sp hybridisation. In this configuration, the 2s-orbital of the carbon atom mixes only with a single p-orbital. This leads to the formation of two orbitals with sp hybridisation and two p-orbitals. The sp-orbitals form a linear structure, and the angle between them is 180o along the x axis. The other p-orbitals are perpendicular to the other axes (y and z). When two carbon atoms with sp hybridisation overlap, this causes frontal overlapping that leads to the formation of a strong σ-bond (sp-sp type). In addition, there occurs lateral overlapping of both the py– and the pz-orbitals, which forms π-bonds (py-py type) and π-bonds (pz-pz type) in that particular order. As a result, six electrons are shared, which form a triple C≡C bond. The other orbitals of the sp hybrids form σ-bonds with the hydrogen atoms.
Due to its sp hybridisation, ethine is a linear molecule with H-C-C bond angles of 180o. The C-H bond in acetylene has the length of 1.06Å and the energy of 552 kJ/mol. The length of that bond is smaller and its energy is higher compared to single and double bonds. These values are 1.20Å and 835 kJ/mol, respectively. It is the shortest and strongest bond that may exist between carbon atoms.
Hybridisation of other atoms
The concepts of three types of hybridisation (sp, sp2 and sp3) are applicable not only in structures that contain carbon atoms. Other elements can also be described in molecules with the use of hybridised orbitals.
- The ammonia molecule (NH3) – the nitrogen atom has five valence electrons and produces three atomic bonds, approaching an octet. The H-N-H bond angle has been measured experimentally; it equals 107.3o, so it is close to the tetrahedral angle. This suggests that ammonia should be considered in the context of sp3 The nitrogen atom hybridises with the formation of four sp3-orbitals; one of them has two non-bonding electrons, while every other orbital contains one bonding electron. The overlap of hybridised orbitals with 1s-orbitals produces a σ-bond with 1.008Å in length and 449 kJ/mol in energy.
- The molecule of water (H2O) – the oxygen atom also shows sp3 However, it contains six valence electrons, so it forms two atomic bonds, leaving two free electron pairs. The H-O-H bond angle in the molecule is 104.5o, so it is also similar to the tetrahedral angle, which suggests hybridisation. Any value of that angle that is lower is probably the result of as many as two free electron pairs, which push each other away. The O-H bond is 0.958Å in length and 498 kJ/mol in energy.
- The molecule of boron fluoride (BF3) – the boron atom includes three valence electrons, so it can produce only three bonds without reaching an octet. However, the existing atoms of fluorine form B-F bonds with it, spatially arranged as far as possible from each other. The result is a trigonal structure of the molecule, and sp2 Each of the three fluorine atoms bonds with a hybridised orbital of boron, leaving its p-orbital not filled.