Valence Bond Theory

Chapter 09

Valence bond (VB) theory is a basic bonding theory in chemistry that attempts to rationalize chemical bonding as a phenomenon arising from the overlap of atomic orbitals. These electrons are localized in the bonds they form.

Consider the simple case of H2. A single bond forms due to the overlap of the 1s orbitals from each hydrogen atom, resulting in electron density localized between each hydrogen atom. An energetically stable molecule forms at the lowest energy on the energy curve.


H2 potential energy surface. Source: Openstax


The lowest energy possible between the two atoms give rise to the bond length in H2. If the atoms are too far apart, the orbitals are unable to overlap resulting in no bond.


Atomic orbital overlap

Bonds that involve a localization of electron density between the atomic nuclei are called sigma (σ) bonds. Sigma bonds can form due to the overlap of atomic orbitals such as

  1. two s orbitals
  2. an s and a p orbital
  3. two “head-on” p orbitals

as shown in the image below. The orbitals involved in the overlap must be in the same phase (constructive interference; distinguished with black and white colors). Orbitals that do not result in any significant overlap due to their orientations, or results in an overlap of lobes of different phase (destructive interference), a bond is not formed.



Some simple overlap/bonding examples are illustrated below.



See also interactive examples of s and p overlap at ChemTube3D.

Bonds involve electron density that is localized in an area that is not directly between the atomic nuclei (i.e. in an area that is “spread out over” two atomic nuclei) are called pi (π) bonds. The electrons are localized in an area that would be “above” or “below” (or to the sides), for example, the two atoms involved in the bond. These bonds can form from an overlap of two “side-by-side” p orbitals (see image below). σ bonds are stronger than their accompanying π bonds.


Source: Openstax


According to valence bond theory, the overlap of atomic orbitals that lead to bonds involve two electrons of opposite spin, typically one from each atom involved in the bond.


Atomic orbital hybridization

Simple VBT works well for simple bonding schemes. However, a more robust theory is needed to explain the bonding in more complex geometries. Orbitals in central atoms are “hybridized” to create hybrid orbitals. These hybrid orbitals are constructed by taking linear combinations of atomic orbitals and are degenerate (have the same energy). Hybridization of terminal atoms should be avoided or performed with caution.

sp Hybridization - 2 electron domains

For example, simple VBT is not able to properly describe the bonding in H2O, a bent molecule. Experimentally, the bond angle in water is about 104.5°. According to simple VBT, the bond angle would be 90° due to the overlap of s orbitals with two orthogonal p orbitals.


Water is a bent molecule with a bent geometry. This should result in a bond angle that is < 109.5°. Simple VBT would predict that the bond angle would be 90° due to the overlap of the 1s orbitals from each hydrogen atom and two 2p orbitals from oxygen which are orthogonal to each other.


A new theory was formulated to try and explain these more complex bonding schemes. Linus Pauling developed the theory of orbital hybridization, a concept that involves mixing atomic orbitals to form new hybrid orbitals that results in different shapes, energies, etc. A set of hybrid orbitals are degenerate (have the same energy).


Example: BeCl2

Experiment determined, and VSEPR theory predicted, the bond angle in BeCl2 to be 180° (linear).



According to VBT, for these two single bonds to form, electrons of opposite spin must be involved in the overlap of two atomic orbitals, one from beryllium and one from chlorine. The ground state configuration of beryllium contains two electrons that are already paired in the 2s orbital. These electrons would need to be unpaired in order to form covalent bonds. Perhaps beryllium could enter into an excited state to generate these two unpaired electrons.


The excitation of a 2s electron being promoted into a 2p orbital in Be


Now the two single electrons in Be could form single bonds with each Cl. This alone does not properly explain the geometry of BeCl2.

Consider the overlaps involved. A 2s orbital in Be would overlap with a 3p orbital in one chlorine to form one bond, and a higher energy 2p orbital would overlap with a 3p orbital in another electron, resulting in two different bonds. However, BeCl2 should have two identical bonds, both in length and bond energy.



The orbitals involved in the overlap must be the same to give rise to identical bonds. A set of new, hybrid orbitals was imagined by mixing the 2s and 2p orbitals to give a pair of sp orbitals. The mixing would result in the sp orbitals being slightly higher in energy than an s orbital and slightly lower in energy than a p orbital. The two sp hybrid orbitals are degenerate.


The mixing of one s and three 2p orbitals to form two sp hybrid orbitals in beryllium


Conservation of orbitals

Two atomic orbitals, a 2s and a 2p, mix to form two sp hybridized orbitals.

These sp orbitals are similar in shape to a p orbital and are “in line” (or 180°) with each other.



The unhybridized 2p orbitals are still present on beryllium and remain 90° to each other.

The sp hybrids, both containing a single, unpaired electron, can now overlap with a singly-occupied 3p orbital on chlorine to form a linear molecule.

Another depiction of this sp hybridization is given below.

  • (top left) an s and p orbital
  • (top right) combine to form two sp orbitals that are facing 180° away from each other
  • (bottom right) giving a linear arrangement

sp hybridization. Source: Openstax


In summary, BeCl2 forms when

  1. Be is excited to obtain two unpaired electrons
  2. one s and one p orbital in Be hybridize to form two sp hybrid orbitals
  3. two equivalent σ bonds form with contribution of one unpaired electron in a 3p orbital from each of the two terminal Cl atoms


Example: Acetylene, C2H2

Acetylene forms when

  1. C is excited to obtain four unpaired electrons
  2. one s and one p orbital in C hybridize to form two sp hybrid orbitals
  3. σ bonds form with contribution of one unpaired electron in a 1s orbital from each of the terminal H atoms
  4. two π bonds form between each carbon from the overlap of two parallel 2p orbitals
  5. 1 σ + 2 π = triple bond
  6. a linear arrangement of sp hybrid orbitals give 180° bond angles.




Other Examples

Molecules with a linear electron domain geometry have a central atom that is sp hybridized.


Molecules with central atoms that are sp hybridized.


Summary:

  • one s orbital and one p orbital mix to produce two sp hybrid orbitals
  • sp hybrids give a linear arrangement (180° orientation)
  • two atomic p orbitals remain unchanged


sp2 Hybridization - 3 electron domains

The mixing of one s and two p atomic orbitals gives three sp2 hybrid orbitals.


An example of one s and two 2p orbitals mixing to form three sp2 hybrid orbitals


These sp2 hybrids have a trigonal planar arrangement.

  • (top left) an s and two p orbitals
  • (top right) combine to form three sp2 orbitals that are facing 120° away from each other
  • (bottom right) giving a trigonal planar arrangement

sp2 hybridization. Source: Openstax


Below is an alternative view of the sp2 hybrid (bottom right in above image) showing a trigonal planar geometry.


sp2 hybridization alternative view. Source: Openstax


Example: Borane, BH3

Borane forms when

  1. B is excited to obtain three unpaired electrons
  2. one s and two p orbital in B hybridize to form three sp2 hybrid orbitals
  3. three equivalent σ bonds form with contribution of one unpaired electron in a 1s orbital from each of the three terminal H atoms
  4. the trigonal planar arrangement of sp2 hybrid orbitals give 120° bond angles.




Example: Boron dihydride, BH2

Borane forms when

  1. B is excited to obtain three unpaired electrons
  2. one s and two p orbital in B hybridize to form three sp2 hybrid orbitals
  3. two equivalent σ bonds form with contribution of one unpaired electron in a 1s orbital from each of the two terminal H atoms
  4. a single electron remains in an sp2 hybrid orbital (a non-bonding hybrid orbital)
  5. the trigonal planar arrangement of sp2 hybrid orbitals give a bond angle of < 120°




Example: Ethene, C2H4

Ethene forms when

  1. C is excited to obtain four unpaired electrons
  2. one s and two p orbital in C hybridize to form three sp2 hybrid orbitals
  3. two equivalent σ bonds form with contribution of one unpaired electron in a 1s orbital from each of the two terminal H atoms and a π bond forms between the two carbon atoms
  4. the trigonal planar arrangement of sp2 hybrid orbitals give a bond angle of 120°




The π bond between the two carbon atoms prevents the CH2 groups from freely rotating unlike that observed in ethane where the CH3 groups do freely rotate.

Visualize ethene, C2H4


Other Examples

Molecules with a trigonal planar electron domain geometry have a central atom that is sp2 hybridized.


Molecules with central atoms that are sp2 hybridized.


Summary:

  • one s orbital and two p orbitals mix to produce three sp2 hybrid orbitals
  • sp2 hybrids give a trigonal planar arrangement (120° orientation)
  • one atomic p orbital remains on the central atom


sp3 Hybridization - 4 electron domains

The mixing of one s and three p atomic orbitals gives four sp3 hybrid orbitals.


An example of one s and three 2p orbitals mixing to form three sp2 hybrid orbitals.


These sp3 hybrids have a tetrahedral arrangement.

  • (top left) an s and three p orbitals
  • (top right) combine to form four sp3 orbitals that are facing 109.5° away from each other
  • (bottom right) giving a tetrahedral arrangement

sp3 hybridization. Source: Openstax


Example: Methane, CH4

Methane forms when

  1. C is excited to obtain four unpaired electrons
  2. one s and three p orbital in C hybridize to form four sp3 hybrid orbitals
  3. four equivalent σ bonds form with contribution of one unpaired electron in a 1s orbital from each of the four terminal H atoms
  4. The tetrahedral arrangement of sp3 hybrid orbitals give a bond angle of 109.5°


Example: Ammonia, NH3

Ammonia has bond angles that are < 109.5°. Without orbital hybridization, the bond angles would be 90° due to the s and p orbital overlaps.

Ammonia forms when

  1. one s and three p orbital in N hybridize to form four sp3 hybrid orbitals
  2. three equivalent σ bonds form with contribution of one unpaired electron in a 1s orbital from each of the three terminal H atoms
  3. the tetrahedral arrangement of sp3 hybrid orbitals give a < 109.5° bond angle

Notice that nitrogen does not need to be excited first before hybridization since there already exists three unpaired electrons available for bonding.



Example: Water, H2O

Water has bond angles that are << 109.5°. Without orbital hybridization, the bond angles would be 90° due to the s and p orbital overlaps.

Water forms when

  1. one s and three p orbital in N hybridize to form four sp3 hybrid orbitals
  2. two equivalent σ bonds form with contribution of one unpaired electron in a 1s orbital from each of the two terminal H atoms
  3. the tetrahedral arrangement of sp3 hybrid orbitals give a << 109.5° bond angle

Notice that oxygen does not need to be excited first before hybridization since there already exists two unpaired electrons available for bonding.





The oxygen in water is not actually sp3 hybridized.

Oxygen (in water) does not have “bunny ears”.

VSEPR and Valence Bond Theory is a convenient tool to easily visualize and explain the bonding behavior within molecules to obtain the proper geometry by considering atomic and hybrid orbitals. However, this is simply a tool and not necessarily reflective of reality.

According to VBT, the lone electron pairs on oxygen in water are degenerate in two of the sp3 hybrid orbitals (giving the “bunny ears”); however, photoelectron spectroscopy demonstrates that they are not degenerate. One lone pair is in a 2s orbital and the other lone pair is in a 2p orbital (giving different energies). The oxygen is actually not sp3 hybridized.

The tetrahedral arrangement of atoms in water is not properly described as Pauli repulsions between lone electron pairs in localized domains. It is more appropriate to describe the σ bonds being formed as an overlap of the 1s orbital on hydrogen with a 2p orbital on oxygen. The 104.5° bond angle occurs due to the van der Waals repulsion of the two hydrogen atoms.

Source: “No rabbit ears on water. The structure of the water molecule: What should we tell the students?” (J. Chem. Educ. 1987, 64 (2), 124.) Link


Example: Ethane, C2H6

Ethane is a multi-centered molecule that has bond angles that are 109.5°.

Ethane forms when

  1. one s and three p orbital in each C hybridize to form four sp3 hybrid orbitals
  2. seven equivalent σ bonds form between the carbon atoms and the carbon-hydrogen atoms
  3. the tetrahedral arrangement of sp3 hybrid orbitals give a 109.5° bond angle

Notice how carbon does not have any unhybridized p orbitals.

The image below shows how one of the carbon atoms (in black) in ethane obtains its hybrid sp3 orbitals. The second carbon (in gray) undergoes the same treatment.





It is known experimentally that the methyl (–CH3) groups in ethane can rotate freely and rapidly. The σ bond between the two carbon atoms allows this rotation to happen.


Methyl groups freely rotate about a carbon-carbon σ bond.


Visualize staggered ethane, C2H6


Other Examples

Molecules with a tetrahedral electron domain geometry have a central atom that is sp3 hybridized.


Molecules with central atoms that are sp3 hybridized.


Summary:

  • one s orbital and three p orbitals mix to produce three sp3 hybrid orbitals
  • sp3 hybrids give a tetrahedral arrangement (109.5° orientation)


sp3d Hybridization - 5 electron domains

d orbital hybridization is wrong!

The oversimplified way of teaching VBT and hybridization is to state that larger, main group elements have an expanded octet due to the ability for d orbitals to be involved in bonding.

This is a myth and should be avoided.

The d orbitals are too low in energy to participate in bonding (source), and therefore, absolutely do not mix with s and p valence orbitals as simple VBT would suggest. However, due to the entrenched nature of general chemistry education, you will be taught this myth (“to keep things simple”) and will be expected to apply it. Do so as an academic exercise but know that this has no basis in reality.

The mixing of one s, three p, and one d atomic orbitals gives five sp3d hybrid orbitals. These sp3d hybrids have a trigonal bipyramidal arrangement.


sp3d hybridization. Source: Openstax


Any atom with five electron domains is sp3d hybridized.

Example: Hypothetical phosphorus pentahydride, PH5

A hypothetical phosphorus pentahydride molecule forms when

  1. P is excited to obtain five unpaired electrons
  2. one s, three p, and one d orbital in P hybridize to form five sp3d hybrid orbitals
  3. five equivalent σ bonds form with contribution of one unpaired electron in a 1s orbital from each of the five terminal H atoms
  4. the square bipyramidal arrangement of sp3d hybrid orbitals give bond angles of 90°, 120°, and 180 °


Examples


Molecules containing central atoms that are sp3d hybridized.


sp3d 2 Hybridization - 6 electron domains

d orbital hybridization is wrong!

The oversimplified way of teaching VBT and hybridization is to state that larger, main group elements have an expanded octet due to the ability for d orbitals to be involved in bonding.

This is a myth and should be avoided.

The d orbitals are too low in energy to participate in bonding (source), and therefore, absolutely do not mix with s and p valence orbitals as simple VBT would suggest. However, due to the entrenched nature of general chemistry education, you will be taught this myth (“to keep things simple”) and will be expected to apply it. Do so as an academic exercise but know that this has no basis in reality.

The mixing of one s, three p, and two d atomic orbitals gives five sp3d 2 hybrid orbitals giving an octahedral arrangement.

Any atom with six electron domains is sp3d 2 hybridized.

Example: Hypothetical sulfur hexahydride, SH6

A hypothetical sulfur hexahydride molecule forms when

  1. S is excited to obtain five unpaired electrons
  2. one s, three p, and two d orbital in S hybridize to form six sp3d 2 hybrid orbitals
  3. six equivalent σ bonds form with contribution of one unpaired electron in a 1s orbital from each of the six terminal H atoms
  4. the octahedral arrangement of sp3d 2 hybrid orbitals give a bond angles of 90° and 180°


Examples


Molecules containing central atoms that are sp3d 2 hybridized.


Hybridization of terminal atoms

Hybridization of terminal atoms should be used with caution. Various schemes can be applied which all seem rationale and it is difficult to pinpoint which one is more accurate. One should use quantum mechanical computations to reliably determine the hybrid nature of terminal atoms and/or experimental data.

Example: CO2

Consider the case of carbon dioxide. The central atom, carbon, in CO2 is sp hybridized and two double bonds (each consisting of a σ and π bond) form between carbon and each terminal oxygen. The hybridization of the terminal atoms in CO2 could be described in three different ways.

Oxygen as unhybridized



The σ bond results from the overlap of an sp hybrid orbital from C and p atomic orbital on the terminal oxygens and lies between the atoms involved in the bond. The π bond results from the overlap of an unhybridized p atomic orbital on C and left O (in the image below) and lies above and below the σ bond axis while another π bond forms from the overlap of an unhybridized p atomic orbital on C and right O (in the image below) that lies “on each side” of the molecule.



Oxygen as sp hybridized



The σ bond results from the overlap of an sp hybrid orbital from C and an sp hybridized orbital on the terminal oxygens and lies between the atoms involved in the bond. The π bond results from the overlap of an unhybridized p atomic orbital on C and left O (in the image below) and lies above and below the σ bond axis while another π bond forms from the overlap of an unhybridized p atomic orbital on C and right O (in the image below) that lies “on each side” of the molecule.



Oxygen as sp2 hybridized



The σ bond results from the overlap of an sp hybrid orbital from C and an sp2 hybridized orbital on the terminal oxygen atoms and lies between the atoms involved in the bond. The π bond results from the overlap of an unhybridized p atomic orbital on C and O and lies above and below the σ bond axis.



Which is correct?

Quantum mechanical calculations indicate that oxygen is unhybridized or, if hybridization is applied, has some sp character but certainly not sp2 character.


Aromaticity

Aromaticity is a chemical property that results in an increase in stabilization of a molecule. Aromaticity can occur in molecules that exhibit a delocalization of π electrons such as the case of benzene.

Benzene is a cyclic, planar molecule. Each carbon atom is sp2 hybridized resulting in a trigonal planar geometry about the carbon backbone.


Representations of benzene. Source:Wikipedia


The unhybridized 2p orbitals on each carbon are parallel to each other, resulting in a delocalized π system. These p orbitals all overlap giving a cyclic π bond above and below the carbon ring. Two Lewis structures are needed to describe this resonance in benzene. Benzene is said to be aromatic due to the π electrons in the ring.


Hybridization summary

The hybridization of a central atom in a molecule can be determined by counting the number of electron domains around the atom.

Summary of hybridization of central atom based on the number of surrounding electron domains
Electron domains Hybridization Geometry

2

sp

linear

3

sp2

trigonal planar

4

sp3

tetrahedral

5

sp3d

trigonal bipyramidal

6

sp3d 2

octahedral

Bond types:

  • single bond → σ bond (freely rotates)
  • double bond → σ bond + π bond (no free rotation)
  • triple bond → σ bond + 2π bonds (no free rotation)