The valence bond (VB) theory

The overlapping AOs can be of different types, for example, a sigma bond may be formed by the overlapping the following AOs.

However, the atomic orbitals for bonding may not be "pure" atomic orbitals directly from the solution of the Schrodinger Equation. Often, the bonding atomic orbitals have a character of several possible types of orbitals. The methods to get an AO with the proper character for the bonding is calledhybridization. The resulting atomic orbitals are calledhybridized atomic orbitalsor simplyhybrid orbitals.

We shall look at the shapes of some hybrid orbitals first, because these shapes determine the shapes of the molecules.

Hybridization of atomic orbitals

The solution to the Schrodinger Equation provides the wavefunctions for the following atomic orbitals:

Quantum mechanical approaches by combining the wave functions to give new wavefunctions are calledhybridizationof atomic orbitals. Hybridization has a sound mathematical fundation, but it is a little too complicated to show the details here. Leaving out the jargons, we can say that an imaginary mixing process converts a set of atomic orbitals to a new set ofhybrid atomic orbitalsorhybrid orbitals.

At this level, we consider the following hybrid orbitals:

Thesphybrid atomic orbitals

Thesphybrid atomic orbitalsare possible states of electron in an atom, especially when it is bonded to others. These electron states have half 2sand half 2pcharacters. From a mathematical view point, there are two ways to combine the 2sand 2patomic orbitals:sp₁= 2s+ 2p
sp₂= 2s- 2p
These energy states (sp₁andsp₂) have a region of high electron probability each, and the two atomic orbitals are located opposite to each other, centered on the atom. The sp hybrid orbitals are represented by this photograph.

H-Be-H	1s 1s Hsp1Besp2H 1s 1s

For example, the molecule H-Be-H is formed due to the overlapping of two 1sorbitals of 2 H atoms and the twosphybridized orbitals of Be. Thus, the H-Be-H molecule is linear. The diagram here shows the overlapping of AOs in the molecule H-Be-H.

The ground state electronic configuration of Be is 1s²2s², and one may think of the electronic configuration "before" bonding as 1s²sp². The two electrons in thesphybrid orbitals have the same energy.

Linear molecules
ClBeCl HCCH HCN O=C=O

You may say that the concept of hybridizing AOs for the bonding is just a story made up to explain the molecular shape of Cl-Be-Cl. You are right! The story is lovely and interesting, though.

In general, when two and only two atoms bond to a third atom and the third atom makes use of thesphybridized orbitals, the three atoms are on a straight line. For example,sphybrid orbitals are used in the central atoms in the molecules shown on the right.

Thesp²hybrid orbitals

The energy states of the valence electrons in atoms of the second period are in the 2sand 2porbitals. If we mix two of the 2porbitals with a 2sorbital, we end up withthreesp²hybridized orbitals. These three orbitals lie on a plane, and they point to the vertices of a equilateral triangle as shown here.

When the central atom makes use ofsp²hybridized orbitals, the compound so formed has a trigonal shape. BF₃is such a molecule:

Molecules withsp2Hybrid orbitals
F | B / \ F F	. . -2 :O: | C / \\ :O:: O	. N // \\ O O	. . O // \\ O O	. . S // \\ O O

Not all threesp²hybridized orbitals have to be used in bonding. One of the orbitals may be occupied by a pair or a single electron. If we do not count the unshared electrons, these molecules are bent, rather than linear. The three molecules shown together with the BF₃molecule are such molecules.

Carbon atoms also makes use of thesp²hybrid orbitals in the compound H₂C=CH₂. In this molecule, the remainingporbital from each of the carbon overlap to form the additional pi,p, bond.

Planar molecules withsp2Hybrid orbitals
H H \ / C = C / \ H H	O 2- \ C = O / O	O 1- \ N = O / O

Other ions such as CO₃^2-, and NO₃^-, can also be explained in the same way.

Thesp³hybrid orbitals

Mixing onesand all threepatomic orbitals produces a set of four equivalentsp³hybrid atomic orbitals. The foursp³hybrid orbitals points towards the vertices of a tetrahedron, as shown here in this photograph.

Whensp³hybrid orbitals are used for the central atom in the formation of molecule, the molecule is said to have the shape of a tetrahedron.

The typical molecule is CH₄, in which the 1sorbital of a H atom overlap with one of thesp³hybrid orbitals to form a C-H bond. Four H atoms form four such bonds, and they are all equivalent. The CH₄molecule is the most cited molecule to have a tetrahedral shape. Other molecules and ions having tetrahedral shapes are SiO₄^4-, SO₄^2-,

As are the cases withsp², hybrid orbitals, one or two of thesp³hybrid orbitals may be occupied by non-bonding electrons. Water and ammonia are such molecules.

Tetrahedral arrangements of CH4, NH3E and OH2E2
H H \ / C /\ H H	H H \ / N /\ : H	H H \ / O /\ : :

The C, N and O atoms in CH₄, NH₃, OH₂(or H₂O) molecules use thesp³hybrid orbitals, however, a lone pair occupy one of the orbitals in NH₃, and two lone pairs occupy two of thesp³hybrid orbitals in OH₂. The lone pairs must be considered in the VSEPR model, and we can represent a lone pair by E, and two lone pairs by E₂. Thus, we have NH₃E and OH₂E₂respectively.

TheVSEPR numberis equal to the number of bonds plus the number of lone pair electrons. Does not matter what is the order of the bond, any bonded pair is considered on bond. Thus, theVSEPR numberis 4 for all of CH₄, :NH₃, ::OH₂.

According the the VSEPR theory, the lone electron pairs require more space, and the H-O-H angle is 105 deegrees, less than the ideal tetrahedral angle of 109.5 degrees.

Thedsp³hybrid orbitals

The fivedsp³hybrid orbitals resulted when one 3d, one 3s, and three 3patomic orbitals are mixed. When an atom makes use of ficedsp³hybrid orbitals to bond to five other atoms, the geometry of the molecule is often atrigonalbipyramidal. For example,The molecule PClF₄displayed here forms such a structure. In this diagram, the Cl atom takes up an axial position of the trigonalbipyramid. There are structures in which the Cl atom may take up the equatorial position. The change in arrangement is accomplished by simply change the bond angles. This link discusses this type of configuration changes of this molecule.

Some of thedsp³hybrid orbitals may be occupied by electron pairs. The shapes of these molecules are interesting. In TeCl₄, only one of the hybriddsp³orbitals is occupied by a lone pair. This structure may be represented by TeCl₄E, where E represents a lone pair of electrons. Two lone pairs occupy two such orbitals in the molecule BrF₃, or BrF₃E₂. These structures are given ina VSEPR table of 5 and 6 coordinations.

The compound SF₄is another AX₄E type, and many interhalogen compounds ClF₃and IF₃are AX₃E₂type. The ion I₃^-is of the type AX₂E₃.

Thed²sp³hybrid orbitals

The sixd²sp³hybrid orbitals resulted when two 3d, one 3s, and three 3patomic orbitals are mixed. When an atom makes use of sixd²sp³hybrid orbitals to bond to six other atoms, the molecule takes the shape of an octahedron, in terms ofmolecular geometry. The gas compound SF₆is a typical such structure. This link provides other shapes as well.

There are also cases that some of thed²sp³hybrid orbitals are occupied by lone pair electrons leading to the structures of the following types:

AX₆, AX₅E, AX₄E₂ AX₃E₃ and AX₂E₄
IOF₅, IF₅E, XeF₄E₂

No known compounds of AX₃E₃and AX₂E₄are known or recognized, because they are predicted to have a T shape and linear shape respectively when the lone pairs of electrons are ignored.

Molecular shapes of compounds

A summary in the form of a table is given here to account for the concepts ofhybrid orbitals, valence bond theory, VSEPR, resonance structures, andoctet rule. In this table, the geometric shapes of the molecules are described bylinear, trigonal planar, tetrahedral, trigonal bypyramidal, and octahedral. The hybrid orbitals use aresp, sp², sp³, dsp³, andd²sp³.

TheVSEPR numberis the same for all molecules of each group. Instead of using NH₃E, and OH₂E₂, we use :NH₃, ::OH₂to emphasize the unshared (or lone) electron pairs.

Only Be and C atoms are involved in linear molecules. In gas phase, BeH₂and BeF₂are stable, and these molecules do not satisfy the octet rule. The element C makes use ofsphybridized orbitals and it has the ability to form double and triple bonds in these linear molecules.

Carbon compounds are present in trigonal planar and tetrahedral molecules, using different hybrid orbitals. The extra electron in nitrogen for its compounds in these groups appear as lone unpaired electron or lone electron pairs. More electrons in O and S lead to compounds with lone electron pairs. The five-atom anions are tetrahedral, and many resonance structures can be written for them.

Trigonal bipyramidal and octahedral molecules have 5 and 6 VSEPR pairs. When the central atoms contain more than 5 or 6 electrons, the extra electrons form lone pairs. The number of lone pairs can easily be derived using Lewisdot structuresfor the valence electrons.

In describing the shapes of these molecules, we often ignore the lone pairs. Thus, •NO₂, N₃^-, :OO₂(O₃), and :SO₂arebent moleculeswhereas :NH₃, :PF₃, and :SOF₂are pyramidal. You already know that ::OH₂(water) and ::SF₂are bent molecules.

The lone electron pair takes up the equatorial location in :SF₄, which has the same structure as :TeF₄described earlier. If you lay a model of this molecule on the side, it looks like abutterfly. By the same reason, ::ClF₃and ::BrF₃have aTshape, and :::XeF₂, :::I₃^-, and :::ICl₂^-are linear.

Similarly, :BrF₅and :IF₅are square pyramidal whereas ::XeF₄is square planar.

The Center Atom

Which atom in the formula is usually the center atom?

Usually, the atom in the center is more electropositive than the terminal atoms. However, the H and halogen atoms are usually at the terminal positions because they form only one bond.

Take a look at the chemical formulas in the table, and see if the above statement is true.

However, the application of VSEPR theory can be expanded to complicated molecules such as

    H H         H   O
    | |         |  //
  H-C-C=C=C-C=C-C-C
    |           |  \
    H           N   O-H
               / \
              H   H

• H-C-C bond angle = 109^o
• H-C=C bond angle = 120^o, geometry around C trigonal planar
• C=C=C bond angle = 180^o, in other words linear
• H-N-C bond angle = 109^o, tetrahedral around N
• C-O-H bond angle = 105 or 109^o, 2 lone electron pairs around O
• Thanks & Regards
• Rinkoo Gupta
• AskIITians Faculty