Introduction

Based on Valence Bond Theory, carbon would only be able to form two covalent bonds, making CH₂. However, as you will find out, we know that this is not true and that in reality, it makes CH₄. The hybridization of orbitals is also greatly favored because hybridized orbitals are lower in energy compared to their separated, unhybridized counterparts. This results in more stable compounds when hybridization occurs. Also, major parts of the hybridized orbitals, or the frontal lobes, overlap better than the lobes of unhybridized orbitals. This leads to better bonding.

Carbon is a perfect example showing the need for hybrid orbitals. As you know, Carbon's ground state configuration is:

According to the valence bond theory, carbon should form two covalent bonds, resulting in a CH₂. However, tests show that CH₂ is highly reactive and cannot exist outside of a reaction. Therefore, this does not explain how CH₄ can exist. However, you can excite a 2s electron and bump it into one of the 2p orbitals. This would give you the following configuration:

While this would allow us to have four covalent bonds, resulting in CH₄, it also implies that the C-H covalent bonds would have different energies due to the different levels of orbital overlap. However, with testing, it has been proven that in CH4, any hydrogen can be removed with the same amount of energy. This means that every C-H covalent bond should have equal energies. Once again, this means that the valence bond theory fails to explain the existence of CH₄. The only way it can be explained is if when we had the exited state above, the 2s and the 3 2p orbitals fused together to make four, equal energy sp³ hybrid orbitals. That would give us the following configuration:

This explains how a carbon can have four equal energy bonds. The next section will explain the various types of hybridization and how each type helps explain the structure of certain molecules.

sp Hybridization

sp Hybridization can explain the linear structure in molecules. In it, the 2s orbital and one of the 2p orbitals hybridize to form two sp orbitals, each consisting of 50% s and 50% p character. The front lobes face away from each other and form a straight line leaving a 180° angle between the two orbitals. This formation minimizes electron repulsion. Because only one p orbital was used, we are left with two unaltered 2p orbitals that the atom can use. These p orbitals are at right angles to one another and to the line formed by the two sp orbitals.

Energy changes occurring in hybridization

_{Notice how the energy of the electrons lowers when hybridized.}

These p orbitals come into play in compounds such as ethyne where they form two addition ? bonds, resulting in in a triple bond. This only happens when two atoms, such as two carbons, both have two p orbitals that each contain an electron. An sp hybrid orbital results when an s orbital is combined with p orbital (Figure 2). We will get two sp hybrid orbitals since we started with two orbitals (s and p). sp hybridization results in a pair of directional sp hybrid orbitals pointed in opposite directions. These hybridized orbitals result in higher electron density in the bonding region for a sigma bond toward the left of the atom and for another sigma bond toward the right. In addition, sp hybridization provides linear geometry with a bond angle of 180^o.

sp² hybridization

sp² hybridization can explain the trigonal planar structure of molecules. In it, the 2s orbitals and two of the 2p orbitals hybridize to form three sp orbitals, each consisting of 67% p and 33% s character. The frontal lobes align themselves in the trigonal planar structure, pointing to the corners of a triangle in order to minimize electron repulsion and to improve overlap. The remaining p orbital remains unchanged and is perpendicular to the plane of the three sp² orbitals.

Energy changes occurring in hybridization

Hybridization of an s orbital with two p orbitals (p_x and p_y) results in three sp² hybrid orbitals that are oriented at 120^o angle to each other (Figure 3). Sp² hybridization results in trigonal geometry.

sp³ hybridization

sp³ hybridization can explain the tetrahedral structure of molecules. In it, the 2s orbitals and all three of the 2p orbitals hybridize to form four sp orbitals, each consisting of 75% p character and 25% s character. The frontal lobes align themselves in the manner shown below. In this structure, electron repulsion is minimized.

Energy changes occurring in hybridization

Hybridization of an s orbital with all three p orbitals (p_x , p_y, and p_z) results in four sp³ hybrid orbitals. sp³ hybrid orbitals are oriented at bond angle of 109.5^o from each other. This 109.5^o arrangement gives tetrahedral geometry (Figure 4).

References

1. John Olmsted, Gregory M. Williams Chemistry: The Molecular Science Jones & Bartlett Publishers 1996. 366-371
2. Francis A. Carey Advanced Organic Chemistry Springer 2001. 4-6
3. L. G. Wade, Jr. Whitman College Organic Chemistry Fifth Edition 2003

Practice Problems

Using the Lewis Structures, try to figure out the hybridization (sp, sp², sp³) of the indicated atom and indicate the atom's shape.

1. The carbon.

2. The oxygen.

3. The carbon on the right.

Answers

1. sp²- Trigonal Planar

The carbon has no lone pairs and is bonded to three hydrogens so we just need three hybrid orbitals, aka sp².

2. sp³ - Tetrahedral

Don't forget to take into account all the lone pairs. Every lone pair needs it own hybrid orbital. That makes three hybrid orbitals for lone pairs and the oxygen is bonded to one hydrogen which requires another sp³ orbital. That makes 4 orbitals, aka sp³.

3. sp - Linear

The carbon is bonded to two other atoms, that means it needs two hybrid orbitals, aka sp.

An easy way to figure out what hybridization an atom has is to just count the number of atoms bonded to it and the number of lone pairs. Double and triple bonds still count as being only bonded to one atom. Use this method to go over the above problems again and make sure you understand it. It's a lot easier to figure out the hybridization this way.

Contributors

• Harpreet Chima (UCD), Farah Yasmeen

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