Orbital hybridisation

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Four sp³ orbitals.

Three sp² orbitals.

In chemistry, hybridisation (or hybridization) is the concept of mixing atomic orbitals within an atom to form new hybrid orbitals. It is an integral part of the valence bond theory.

Hybridisation of orbitals will change the shape of the atom. Hybridised orbitals are also useful in the explanation of the shape of molecular orbitals of molecules.

Contents [hide] 1 Historical development 2 Hybridisation of 2s and 2p orbitals 3 sp3 hybrids 3.1 Example methane 4 sp2 hybrids 4.1 Example ethene 5 sp hybrids 5.1 Example ethyne 6 Hybridisation and molecule shape 6.1 Controversy regarding d-orbital participation 7 Hybridisation theory vs. MO theory 8 See also 9 External links 10 References

[edit]Historical development

The hybridisation theory was promoted by chemist Linus Pauling^[1] in order to explain the structure of molecules such as methane (CH₄). This concept was developed for such simple chemical systems, but the approach was later applied more widely, and today it is considered an effective heuristic for rationalizing the structures of organic compounds.

Hybridisation theory is not as practical for quantitative calculations as Molecular Orbital Theory. Problems with hybridisation are especially notable when the d orbitals are involved in bonding, as in coordination compounds. Although hybridisation schemes in transition metal chemistry can be used, they are not generally as accurate.

It is important to note that orbitals are a model representation of the behaviour of electrons within molecules. In the case of simple hybridisation, this approximation is based on the atomic orbitals of hydrogen. Hybridised orbitals are assumed to be mixtures of these atomic orbitals, superimposed on each other in various proportions. Hydrogen orbitals are used as a basis for simple schemes of hybridisation because it is one of the few examples of orbitals for which an exact analytic solution to its Schrödinger equation is known. These orbitals are then assumed to be slightly, but not significantly, distorted in heavier atoms, like carbon, nitrogen, and oxygen. Under these assumptions is the theory of hybridisation most applicable. It must be noted that one does not need hybridisation to describe molecules, but for molecules made up fromcarbon, nitrogen and oxygen (and to a lesser extent, sulfur and phosphorus) the hybridisation theory/model makes the description much easier.

The hybridisation theory finds its use mainly in organic chemistry, and mostly concerns C, N and O (and to a lesser extent P and S). Its explanation starts with the way bonding is organized in methane.

[edit]Hybridisation of 2s and 2p orbitals

When within a carbon atom the 2s orbital is mixed with one or more 2p orbitals there are 3 possibilities:

2s orbital + 1 2p orbital → 2 hybridised sp orbitals + 2 remaining 2p orbitals

2s orbital + 2 2p orbitals → 3 hybridised sp² orbitals + 1 remaining 2p orbital

2s orbital + 3 2p orbitals → 4 hybridised sp³ orbitals

The hybridised orbitals will be used for σ bonds with other atoms, whereas the remaining p orbitals will be used for π bonds. sp orbitals will be used for bonding with two other atoms or groups, sp² orbitals for bonding with three and sp³ orbitals give four σ bonds.

_{Note: 2 in 2s and 2p refers to electrons in the second electron shell with the corresponding energy level.}

[edit]sp³ hybrids

[edit]Example methane

Methane

Hybridisation describes the bonding atoms from an atom's point of view. That is, for a tetrahedrally coordinated carbon like methane (CH₄), the carbon should have 4 orbitals with the correct symmetry to bond to the 4 hydrogen atoms.
Carbon's ground-state configuration is 1s² 2s² 2p_x¹ 2p_y¹ or more easily read:

The valence bond theory would predict, based on the existence of two half-filled p-type orbitals (the designationsp_x p_y or p_z are meaningless at this point, as they do not fill in any particular order), that C forms two covalent bonds, i.e., CH₂ (methylene). However, methylene is a very reactive molecule (see also: carbene) and cannot exist outside of a molecular system. Therefore, this theory alone cannot explain the existence of CH₄.

Furthermore, ground state orbitals cannot be used for bonding in CH₄. While exciting a 2s electron into a 2p orbital would, in theory, allow for four bonds according to the valence bond theory, (which has been proved experimentally correct for systems like O₂) this would imply that the various bonds of CH₄ would have differing energies due to differing levels of orbital overlap. Once again, this has been experimentally disproved: Any hydrogen can be removed from a carbon with equal ease.

To summarise, to explain the existence of CH₄ (and many other molecules) a method by which as many as 12 bonds (for transition metals) of equal strength (and therefore equal length) was required.

The first step in hybridisation is the excitation of one (or more) electrons (we consider the carbon atom in methane, for simplicity of the discussion):

The proton that forms the nucleus of a hydrogen atom attracts one of the lower-energy valence electrons on carbon. This causes an excitation, moving a 2s electron into a 2p orbital. This, however, increases the influence of the carbon nucleus on the valence electrons by increasing the effective core potential (the amount of charge the nucleus exerts on a given electron = Charge of Core − Charge of all electrons closer to the nucleus).

The combination of these forces creates new mathematical functions known as hybridised orbitals. In the case of carbon attempting to bond with four hydrogens, four orbitals are required. Therefore, the 2s orbital (core orbitals are almost never involved in bonding) "mixes" with the three 2p orbitals to form four sp³ hybrids (read as s-p-three). See graphical summary below.

becomes

In CH₄, four sp³ hybridised orbitals are overlapped by hydrogen's 1s orbital, yielding four σ (sigma) bonds (that is, four single covalent bonds). The four bonds are of the same length and strength. This theory fits our requirements.

translates into

An alternative view is: View the carbon as the C⁴⁻ anion. In this case all the orbitals on the carbon are filled:

If we now recombine these orbitals with the empty s-orbitals of 4 hydrogens (4 protons, H⁺) and allow maximum separation between the 4 hydrogens (i.e., tetrahedral surrounding of the carbon), we see that at any orientation of the p-orbitals, a single hydrogen has an overlap of 25% with the s-orbital of the C, and a total of 75% of overlap with the 3 p-orbitals (see that the relative percentages are the same as the character of the respective orbital in an sp³-hybridisation model, 25% s- and 75% p-character).

According to the orbital hybridisation theory, the valence electrons in methane should be equal in energy but its photoelectron spectrum ^[2]shows two bands, one at 12.7 eV (one electron pair) and one at 23 eV (three electron pairs). This apparent inconsistency can be explained when one considers additional orbital mixing taking place when the sp³ orbitals mix with the 4 hydrogen orbitals.

[edit]sp² hybrids

[edit]Example ethene

Ethene

In sp² hybridisation the 2s orbital is mixed with only two of the three available 2p orbitals:

forming a total of 3 sp² orbitals with one p-orbital remaining.

Other carbon based compounds and other molecules may be explained in a similar way as methane. Take, for example, ethene (C₂H₄). Ethene has a double bond between the carbons. The Kekule structure looks like this:

Carbon will sp² hybridise, because hybrid orbitals will form only σ bonds and one π (pi) bond is required for thedouble bond between the carbons. The hydrogen-carbon bonds are all of equal strength and length, which agrees with experimental data. In ethylene (ethene) the two carbon atoms form a σ bond by overlapping two sp² orbitals and each carbon atom forms two covalent bonds with hydrogen by s–sp² overlap all with 120° angles. The π bond between the carbon atoms perpendicular to the molecular plane is formed by 2p–2p overlap (however, the π bond may or may not occur).

The amount of p-character is not restricted to integer values; i.e., hybridisations like sp^2.5 are also readily described. In this case the geometries are somewhat distorted from the ideally hybridised picture. For example, as stated in Bent's rule, a bond tends to have higher p-character when directed toward a more electronegative substituent.

[edit]sp hybrids

[edit]Example ethyne

Ethyne

The chemical bonding in compounds such as alkynes with triple bonds is explained by sp hybridization.

In this model, the 2s orbital mixes with only one of the three p-orbitals resulting in two sp orbitals and two remaining unchanged p orbitals.

The chemical bonding in acetylene (ethyne) (C₂H₂) consists of sp–sp overlap between the two carbon atoms forming a σ bond and two additional π bonds formed by p–p overlap. Each carbon also bonds to hydrogen in a sigma s–sp overlap at 180° angles.

[edit]Hybridisation and molecule shape

Hybridisation, along with the VSEPR theory, helps to explain molecule shape:

• AX₁ (e.g., LiH): no hybridisation; trivially linear shape
• AX₂ (e.g., BeCl₂): sp hybridisation; linear or diagonal shape; bond angles are cos⁻¹(−1) = 180°
  • AX₂E (e.g., GeF₂): bent/V shape, <>
• AX₃ (e.g., BCl₃): sp² hybridisation; trigonal planar shape; bond angles are cos⁻¹(−1/2) = 120°
  • AX₃E (e.g.,NH₃): trigonal pyramidal, 107°
• AX₄ (e.g., CCl₄): sp³ hybridisation; tetrahedral shape; bond angles are cos⁻¹(−1/3) ≈ 109.5°
• AX₅ (e.g., PCl₅): sp³d hybridisation; trigonal bipyramidal shape
• AX₆ (e.g., SF₆): sp³d² hybridisation; octahedral (or square bipyramidal) shape

This holds if there are no lone electron pairs on the central atom. If there are, they should be counted in the X_i number, but bond angles become smaller due to increased repulsion. For example, in water (H₂O), the oxygen atom has two bonds with H and two lone electron pairs (as can be seen with the valence bond theory as well from the electronic configuration of oxygen), which means there are four such 'elements' on O. The model molecule is, then, AX₄: sp³ hybridisation is utilized, and the electron arrangement of H₂O is tetrahedral. This agrees with the experimentally-determined shape for water, a non-linear, bent structure, with a bond angle of 104.5 degrees (the two lone-pairs are not visible).

In general, for an atom with s and p orbitals forming hybrids h_i and h_j with included angle θ, the following holds: 1 + λ_iλ_j cos(θ) = 0. The p-to-s ratio for hybrid i is λ_i², and for hybrid j it is λ_j². In the special case of equivalent hybrids on the same atom, again with included angle θ, the equation reduces to just 1 + λ² cos(θ) = 0. For example, BH₃ has a trigonal planar geometry, three 120^o bond angles, three equivalent hybrids about the boron atom, and thus 1 + λ² cos(θ) = 0 becomes 1 + λ² cos(120^o) = 0, giving λ² = 2 for the p-to-s ratio. In other words, sp² hybrids, just as expected from the list above.

Note: Although sometimes taught together with the valence shell electron-pair repulsion (VSEPR) theory, valence bond and hybridization are in fact not related to the VSEPR model.^[3]