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What is chemical hybridization?

The chemical hybridization is the “mix” of the atomic orbitals, whose concept was introduced by chemist Linus Pauling in 1931 to cover imperfections of the theory of the valence bond (TEV). What imperfections? These are: molecular geometries and equivalent bond lengths in molecules such as methane (CH 4 ).

According to the TEV, in methane the C atomic orbitals form four σ bonds with four H atoms. The 2p orbitals, with ∞ shapes (lower image) of C are perpendicular to each other, so the H should be about a few from others at a 90º angle.

Additionally, the 2s (spherical) orbital of C binds to the 1s orbital of H at an angle of 135º with respect to the other three H. However, experimentally it has been found that the angles in CH 4 are 109.5º and that Furthermore, the lengths of the C – H bonds are equivalent.

To explain this, a combination of the original atomic orbitals must be considered to form four degenerate hybrid orbitals (of equal energy). Here chemical hybridization comes into play. What are hybrid orbitals like? It depends on the atomic orbitals that generate them. They also exhibit a mixture of their electronic characteristics.

Sp 3 hybridization

For the case of CH 4 , the hybridization of C is sp 3 . From this approach, the molecular geometry is explained with four sp 3 orbitals separated at 109.5º and pointing towards the vertices of a tetrahedron.

In the upper image you can see how the sp 3 orbitals (green) establish a tetrahedral electronic environment around the atom (A, which is C for CH 4 ).

Why 109.5º and not other angles, in order to “draw” a different geometry? The reason is because this angle minimizes the electronic repulsions of the four atoms that bind to A.

Thus, the CH 4 molecule can be represented as a tetrahedron (tetrahedral molecular geometry).

If, instead of H, C formed bonds with other groups of atoms, what then would its hybridization be? As long as the carbon forms four σ bonds (C – A), their hybridization will be sp 3 .

It can consequently be assumed that in other organic compounds such as CH 3 OH, CCl 4 , C (CH 3 ) 4 , C 6 H 12 (cyclohexane), etc., the carbon has sp 3 hybridization .

Interpretation

What is the simplest interpretation for these hybrid orbitals without addressing the mathematical aspects (the wave functions)? The sp 3 orbitals imply that they were originated by four orbitals: one s and three p.

Because the combination of these atomic orbitals is assumed to be ideal, the resulting four sp 3 orbitals are identical and occupy different orientations in space (such as in the p x , p, and p z orbitals ).

The above is applicable for the rest of the possible hybridizations: the number of hybrid orbitals that is formed is the same as that of the atomic orbitals that are combined. For example, sp 3 d 2 hybrid orbitals  are formed from six atomic orbitals: one s, three p and two d.

Bond angle deviations

According to the Valencia Shell Electronic Pair Repulsion Theory (RPECV), a pair of free electrons occupies more volume than a bonded atom. This causes the links to move apart, reducing the electronic tension and deviating the angles from 109.5º:

For example, in the water molecule the H atoms are bonded to the sp 3 orbitals (in green), and likewise the unshared pairs of electrons “:” occupy these orbitals.

The repulsions of these pairs of electrons are usually represented as “two globes with eyes”, which, due to their volume, repel the two σ O – H bonds.

Thus, in water the bond angles are actually 105º, instead of the 109.5º expected for tetrahedral geometry.

What geometry then does H 2 O have? It has an angular geometry. Why? Because although the electronic geometry is tetrahedral, two pairs of unshared electrons distort it to an angular molecular geometry.

Sp 2 hybridization

When an atom combines two p and one s orbitals, it generates three sp 2 hybrid orbitals ; however, one p orbital remains unchanged (because there are three of them), which is represented as an orange bar in the upper image.

Here, the three sp 2 orbitals are colored green to highlight their difference from the orange bar: the “pure” p orbital.

An atom with sp 2 hybridization can be visualized as a flat trigonal floor (the triangle drawn with the sp 2 orbitals colored green), with its vertices separated by 120º angles and perpendicular to a bar.

And what role does the pure p orbital play? That of forming a double bond (=). The sp 2 orbitals allow the formation of three σ bonds, while the pure p orbital one π bond (a double or triple bond involves one or two π bonds).

For example, to draw the carbonyl group and the structure of the formaldehyde molecule (H 2 C = O), proceed as follows:

The sp 2 orbitals of both C and O form a σ bond, while their pure orbitals form a π bond (the orange rectangle).

It can be seen how the rest of the electronic groups (H atoms and the pairs of electrons not shared) are located in the other sp 2 orbitals , separated by 120º.

Sp hybridization

In the upper image an A atom with sp hybridization is illustrated. Here, one s orbital and one p orbital combine to give rise to two degenerate sp orbitals. However, now two pure p orbitals remain unchanged, which allow A to form two double bonds or one triple bond (≡).

In other words: if in a structure a C meets the above (= C = or C≡C), then its hybridization is sp. For other less illustrative atoms – such as transition metals – the description of electronic and molecular geometries is complicated because the d and through f orbitals are also considered.

The hybrid orbitals are 180 ° apart. For this reason the bonded atoms are arranged in a linear molecular geometry (BAB). Finally, the lower image shows the structure of the cyanide anion:

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