Created by Adam R. Johnson (adam_johnson@hmc.edu) and posted on VIPEr in July 2010. Copyright
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There are 3 ways that inorganic chemists often classify complexes: metal
oxidation state, dn count, and total valence electron count. While a useful formalism, you
should be aware that an isolated Cr+6 ion is really not present in a complex ion. For this
reason, metal ions are usually written with roman numerals like Cr(VI).
The oxidation state is the formal charge on the central metal. The total valence
electron count is the total number of electrons around the central metal (think about the
octet rule for p-block elements). The dn count is the number of electrons in the metal's
valence shell (remember that metals lose their s electrons first; if there are any s-electrons
on the metal, (Cu(I) for example) then these count for the purposes of dn count, since they
are in the valence shell). The hardest to count is the total valence electrons, and 2
methods for doing this are outlined on this handout.
X vs. L type ligands. An X-type ligand is a ligand that would be anionic by itself
if removed from the metal. Examples include Cl-, H-, CH3-, OH-. Some ligands are X2
ligands, like ox2-. An L-type ligand is a ligand that is by itself a stable molecule, not ion.
Examples include CO, PF3, PMe3, NH3. Some ligands are L2 ligands, such as bipy.
Some ligands are both X and L, such as Cp = XL2. A bridging Cl counts as an X for one
metal and an L for the other.
Methods of counting total valence electrons. There are two common methods
for counting total valence electrons: the covalent and the ionic method. In the covalent
method, the metal is said to have as many electrons as its group number (Ti = 4, Cu = 9).
The X-ligands donate 1 electron each, and the L ligands donate 2 electrons each.
Examples include: H· [1], ·CH3 [1], Cl· [1], :NH3 [2], :CO [2]; Cp [XL2 = 5]). If there is
a charge on the complex, the number of electrons must be reduced for each + charge and
increased for each - charge.
In the ionic method, the metal is said to have its group number minus its oxidation
state (for example, Ti(II) has 4-2 = 2 electrons). X ligands are treated as anionic and thus
donate 2 electrons each. L ligands also donate 2 electrons each. Cp would donate 6,
since it is XL2.
Here is a worked example of each method. Note, there is only one correct
answer, the two methods are equivalent. Don’t mix-n-match methods, however!
covalent
M(X)m(L)n+q
M e: group # (gn)
X e: 1*m
L e: 2*n
total e: gn+m+2*n-q
ionic
M(X)m(L)m+q
M e: (gn-OS) (note, OS = q+m)
X e: 2*m
L e: 2*n
total e: (gn-q-m)+2*m+2*n
= gn-q+m+2*n
so, TiCl3(OH2)3+
4+3+2*3-1
valence electron count =12
4-1+3+2*3
=12