22.6: Valence Bond Theory of Complexes

Valence Bond Theory of Complexes:

Remember from Chapter 10 that a covalent bond is usually formed by the overlap of the a bonding orbital from each atom, each containing one electron. The subsequent bond then has two electrons.

Remember also that the bonds in complexes are coordinate covalent bonds, where both electrons are donated from the ligand to the metal ion. With the coordinate covalent bond in a complex, a ligand orbital with two electrons overlaps an empty orbital on the metal.



This bonding requires hybrid orbitals.

Example:

What are the hybridized orbitals on Co in [Co(H₂O)₆]³⁺?

First we need to find the configuration of the metal ion. Start by writing the configuration of the cobalt atom. Here's a helpful table to remember how to do it:



Co: [Ar]4s²3d⁷

If our ion has a charge of +3, and H₂O is a neutral molecule, we know that the charge on the cobalt ion is +3. To reflect this in the configuration, remove 3 electrons. Remember the outer s electrons are removed first:

Co³⁺: [Ar] 3d⁶

Now draw the orbital diagram of the ion. Remember to follow Hund's Rule, where each electron is placed in a separate orbital with the same spin before pairing it with an electron of opposite spin:



In the complex ion [Co(H₂O)₆]³⁺ each water molecule wants to donate one pair of electrons to the metal. In order to do this, it needs an empty orbital. Since there are six H₂O, it needs six empty orbitals. Starting with the first empty orbital, work forward until you get six, and these are the hybridized orbitals:



It is called sp³d² because one orbital comes from s, three come from p, and two come from d.

Example:

What are the hybridized orbitals on Mn in [Mn(NH₃)₆]⁴⁺?

First we need to find the configuration of the metal ion. Start by writing the configuration of the manganese atom.

Mn: [Ar] 4s²3d⁵

If our ion has a charge of +4, and NH₃ is a neutral molecule, we know that the charge on the manganese ion is +4. To reflect this in the configuration, remove 4 electrons. Remember the outer s electrons are removed first:

Mn⁴⁺: [Ar] 3d³

Now draw the orbital diagram of the ion, following Hund's Rule:



In the complex ion [Mn(NH₃)₆]⁴⁺ each NH₃ molecule wants to donate one pair of electrons to the metal. In order to do this, it needs an empty orbital. Since there are six, it needs six empty orbitals:



It is called d²sp³ because two orbitals come from d, one comes from s, and three come from p.