Hybridization

Sections

Overview
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Overview

Hybridization

• Mathematical model that describes the pattern of bonds found in a molecule (geometry)
• Explains difference between the electron configuration of a free atom and the electron configuration when an atom is part of a molecule

Hybrid orbitals

• Combine orbitals within the same atom
• Equal the number of atoms and lone pairs bound to the atom (the steric number)
• Average the properties (shape and arrangement) of their component atomic orbitals
• Are degenerate (equivalent)
• Position themselves as far apart as possible

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• Here, we will learn about hybridization.

• Start a table.

• Denote that hybridization is a mathematical model that describes the pattern of bonds found in a molecule.

• Denote that the electron configuration of a free atom (its orbital) is different from its electron configuration when it is a part of a molecule; hybrid orbitals explain this difference.

• Denote that hybrid orbitals combine orbitals within the SAME atom.
  • The number of hybrid orbitals equals the number of atoms and lone pairs bound to the atom (the steric number).
  • Hybrid orbitals are an average of the free atomic orbitals and are degenerate (equivalent).
  • Hybrid orbitals have a different shape and arrangement than their component orbitals.
  • They position themselves as far apart as possible: as if we blew up several balloons and tied them together.

Thus, through hybridization we arrive at the observed geometry of bonds in molecules. Now, let's study each of these truisms in 4 steps.

Step I.

First, we'll learn that the number of hybrid orbitals equals the number of the atoms and the number of lone pairs bound to the atom. Let's use the molecule ethylene (C2H4) as our example.

• Draw the Lewis structure of ethylene.

• First let's select a single carbon, because hybrid orbitals combine orbitals within the SAME atom.
  • Write that each carbon is bound to 2 hydrogens and 1 carbon (total of 3 atoms).
  • And each has 0 lone pairs.
  • Thus, the # of hybrid orbitals is 3.

Step II.

Let's use an orbital energy diagram to show that hybrid orbitals are an average of the free atomic orbitals (they are degenerate orbitals).

• Draw the diagram for carbon in its excited state electron configuration of [1s^2 2s^1 2p^3].
  • In this configuration, one of the 2s electrons has moved up to the 2p orbitals.

• Circle the 2s, 2px, and 2py orbitals that will be hybridized (recall that only valence electrons are used in bonding).

• Now, show the energy orbital diagram for carbon in its hybridized state. It will have the following orbitals, in order of increasing energy:
  • One 1s orbital, with 2 paired electrons
  • Three degenerate sp2 orbitals, each with one unpaired electron of up spin
  • One 2pz orbital, with one unpaired electron of up spin

• Write that the designation of sp2 indicates that the hybrid orbitals have 33% s character and 66% p character.
  • Thus, hybrid orbitals are an average of the free atomic orbitals.

Step III.

Now, let's see how the sp2 hybrid orbitals have a different shape than their component orbitals.

• Draw the 2s atomic orbital of carbon.

• Next, draw the 2px orbital.
  • Shade its right lobe in the same color as the 2s orbital to indicate that the right lobe is the positive phase.

• Now, draw the 2py orbital on a third Cartesian coordinate system.
  • Shade its top lobe (just like we shaded the right lobe in the previous orbital).

• Now, draw an sp2 hybrid orbital, which has an enlarged, shaded, positive lobe along the x-y plane and a reduced negative lobe.
  • The lopsided shape of the hybrid orbital does not match the shapes of either the 2s, the 2px, or the 2py orbitals, but is, instead, a combination of all of them.

Step IV.

Now, let's see how hybrid orbitals position themselves as far apart as possible.

• Redraw the 2s atomic orbital of carbon.

• Then, draw 2px, 2py, and 2pz orbitals of carbon.
  • Using two representative lobes, show that these lobes have a 90 degree separation: we randomly selected the y-z planes, here, any two planes would do.

• Then, draw 3 sp2 hybrid orbitals and give them maximum separation: 120 degree angles in the x-y plane.
  • Being degenerate, hybrid orbitals position themselves as far apart as possible: they repulse each other for maximal spatial separation, in this instance, the orbitals point to the corners of a planar triangle.

• Finally, redraw the Lewis structure for ethylene.
  • Indicate that the H-C-H and the H-C-C bond angles are 120 degrees.
  • Thus, through hybridization we can account for the placement of electrons and predict molecular structures.