The chemical hybridization it is the "mixture" of atomic orbitals, the concept of which was introduced by chemist Linus Pauling in 1931 to cover the imperfections of the Valencia Bond Theory (TEV). What imperfections? These are: molecular geometries and equivalent bond lengths in molecules such as methane (CH4).
According to TEV, in methane the C atomic orbitals form four σ bonds with four H atoms. The 2p orbitals, with formas shapes (bottom 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 CH4 are 109.5º and that, in addition, 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 the electronic characteristics of these.
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In the case of CH4, hybridization of C is sp3. From this approach, molecular geometry is explained with four sp orbitals3 109.5º apart and pointing towards the vertices of a tetrahedron.
In the image above you can see how the sp orbitals3 (colored green) establish a tetrahedral electronic environment around the atom (A, which is C for CH4).
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 bond to A.
Thus, the CH molecule4 can be represented as a tetrahedron (tetrahedral molecular geometry).
If, instead of H, C formed bonds with other groups of atoms, what would then be their hybridization? As long as the carbon forms four σ (C-A) bonds, their hybridization will be sp3.
It can consequently be assumed that in other organic compounds such as CH3OH, CCl4, C (CH3)4, C6H12 (cyclohexane), etc., the carbon has sp hybridization3.
This is essential for sketching organic structures, where single bonded carbons represent points of divergence; that is, the structure does not remain in a single plane.
What is the simplest interpretation for these hybrid orbitals without addressing the mathematical aspects (the wave functions)? The sp orbitals3 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 four sp orbitals3 resulting are identical and occupy different orientations in space (such as in the p orbitalsx, pY And pz).
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 hybrid orbitals3dtwo are formed from six atomic orbitals: one s, three p, and two d.
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 voltage and deviating the angles from 109.5º:
For example, in the water molecule the H atoms are bonded to the sp orbitals3 (in green), and also 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 the HtwoOR? 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.
When an atom combines two p and one s orbitals, it generates three hybrid sp orbitalstwo; 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 orbitalstwo they are colored green to highlight their difference from the orange bar: the "pure" p orbital.
An atom with sp hybridizationtwo can be visualized as a flat trigonal floor (the triangle drawn with the sp orbitalstwo 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 orbitalstwo allow the formation of three σ bonds, while the pure p orbital one π bond (a double or triple bond implies one or two π bonds).
For example, to draw the carbonyl group and the structure of the formaldehyde molecule (HtwoC = O), it continues as follows:
The sp orbitalstwo 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 unshared pairs of electrons) are located in the other sp orbitals.two, separated by 120º.
In the upper image an A atom with sp hybridization is illustrated. Here, an s orbital and a 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 the electronic and molecular geometries is complicated because the d orbitals and even the f orbitals are also considered..
The hybrid orbitals are 180 ° apart. For this reason the bonded atoms are arranged in a linear molecular geometry (B-A-B). Finally, the lower image shows the structure of the cyanide anion:
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