Solids with the exception of mercury
Most have high melting and boiling points
Conduct electricity and heat
Many are colored or form colored compounds (d orbitals)
Many are paramagnetic (contain unpaired electrons)
Most transition metals can have more than one oxidation state. Some, such as Mn, have
MP ˚C BP ˚C Oxidation States
K alkali metal 64 759 +1
P nonmetal 44 280 -3
Cr transition 1863 2679 2,3,6
Fe transition 1538 2862 2,3
Cu transition 1085 2563 1,2
Mn transition 1246 2062 2,3,4,6,7
Radii : in general size decreases going from left to right, increase going down
lanthanide contraction : the 4d and 5d metals are similar in size as a result of the
Between La and Hf the electrons being added are going into the 4f orbitals which
are closer to the nucleus than the 5d orbitals so the extra electrons aren’t adding to
The nuclear charge is increasing as the number of protons increases. The
increased charge pulls the electrons closer to the nucleus causing the radii to
Place the following in order from smallest to largest : Sc, Y, Zn
Zn (133 pm) Sc (161pm) Y (178 pm)
Ionization Energy : increases going left to right, and bottom to top (→↑)
As the charge in the nucleus increases it becomes harder and harder to remove an
Reducing agents : ability to act as a reducing agent decreases, generally, from left to right
and bottom to top.
Reducing agents are oxidized, they lose electrons. The ability to lose electrons
decreases as we go right for the same reason ionization energy increases.
Oxidation will remove the outermost s electrons first, followed by d electrons.
Some metals are not very prone to oxidation, including Ag, Au, Pt which are
often used for jewelry
Radii Ionization Reduction
(pm) Potential Potential
Sc 161 6.56 -2.007 Sc 3+ + 3e- → Sc
Ti 145 6.83 -1.37 Ti 3+ + 3e- → Ti
Cr 125 6.77 -0.744 Cr 3+ + 3e- → Cr
Mn 137 7.43 1.679 MnO4- + 4H+ + 3e- → MnO2 + 2 H2O
Co 135 7.88 1.92 Co 3+ + 3e- → Co
Place the following in order from lowest to highest ionization energy :Pd, Zr, Mo, Ac
Ac < Zr < Mo < Pd
Place the following in order from best to worst reducing agent: Cu ,Mo, Ru
Mo is better than Ru is better than Cu
Which of the following would be the best oxidizing agent, Zn or Nb?
Zn is harder to oxidize than Nb, and easier to reduce than Nb. Things that are easily
reduced are good oxidizing agents. It would be the best oxidizing agent.
Unexpected filling patterns occur with Cr, Cu, Ag, Au, and Pt.
(Also with Nb, Mo,Ru,Rh,)
In these metals an s electron is moved into a d orbital. The energy level for the s orbital
is close to that of the d orbital. The difference in these energies is less than the energy
caused by electron repulsion. So rather than have 2 electrons in one orbital we move an s
electron into a d orbital.
When making ions we remove the electrons from the highest energy orbital, largest n
Write the shorthand electron configurations for the following:
Zn , Zn+2 , Au , Au+1
Zn : [Ar] 3d104s2
Zn+2 : [Ar] 3d10
Au : [Xe] 4f145d106s1
Au+1 : [Xe] 4f145d10
Made of a complex ion and a counter ion
complex ions are made of a metal and ligands
Complex Ion : A metal and ligands which associate to form a cation or anion
Ligand : a molecule or ion that has a lone pair of electrons that can be used to form a
bond to a metal atom
Monodentate : can form 1 bond
Bidentate : can form 2 bonds
Polydentate : can form multiple bonds
Chelator / Chelating agents : bidentate or polydentate
An example of a common chelating agent found in many products is EDTA
N CO2 -
EDTA (ethylenediaminetetraacetic acid) can use both nitrogens and all four
carboxylic acid groups to chelate metal ions.
Oxidation number = primary valence, used to form ionic bonds
Coordination number = secondary valence, used to form bonds with ligands (a coordinate
For each of the complexes determine the oxidation number and the coordination number
for the metal.
Our complex ion is Pt(NH3)2Cl4 with no net charge. NH3 has no charge and each
chloride has a charge of -1. The charge on Pt must therefore be +4
There are 6 ligands around the Pt (2 NH3 + 4 Cl- = 6 ligands)
Oxidation number = +4 CN = 6
Our complex ion is [Co(NH3)5Cl]+2. The SO42- is a counter ion, the complex ion
will be the structure given in the brackets. The net charge is +2, NH3 is zero
charge, and the chloride is -1. Therefore the Co must be +3.
Co charge + 5(0) + 1(-1) = +2
There are 6 ligands around the Co.
Oxidation number = +3 CN = 6
Nomenclature of Coordination Compounds
There are 2 possibilities. Your complex ion can be acting as the cation (listed first in the
formula) or as the anion (listed second in the formula).
Case 1: Complex ion is the cation
Just like any salt, you name the cation first and then the anion.
Cation = ligands (in alphabetical order) followed by the metal ion (with Roman Numeral
1. When a ligand is used more than once, indicate the number of ligands using di,
tri, tetra, penta, or hexa as a prefix to the ligand
2. IF the ligand has di or tri or tetra etc in it’s name, use the terms bis, tris,
tetrakis, pentakis etc. to avoid confusion.
OR if the ligand is polydentate you will use bis, tris etc.
3. structural information prefixes ligands, and can be for each ligand if necessary
(cis-; trans-; fac-; mer-)
Small ligands Ligands with di, tri etc in name
2 di bis
3 tri tris
4 tetra tetrakis
5 penta pentakis
6 hexa hexakis
7 hepta heptakis
Anionic Ligands Names
Neutral Ligands Names
Here is an example.....
[Co(NH3)6]Cl *notice the brackets used to indicate the complex ion
This breaks apart into a cation Co(NH3)6 + and an anion Cl-
We name the cation (the complex ion) first.
name the ligands and tell how many there are : hexaammine
now name the metal and give it’s charge : cobalt (III)
put it together, our cation is : hexaamminecobalt (III)
Now name the anion.
Cl- is chloride
All together we have hexaaminecobalt(III) chloride
[Ni(H2O)6]SO4 hexaaquanickel(II) sulfate
[Cu(NH3)4]SO4 tetraamminecopper(II) sulfate
[Cr(en)2(CN)2]Cl bis(ethylenediamine)dicyanochromium (I) chloride
[Co(en)2(H2O)Cl]Cl2 aquachlorobis(ethylenediamine)cobalt(II) chloride
[Co(NH3)5Br]SO4 pentaamminebromocobalt (III) sulfate
Case 2 : Complex ion is the anion
Name the cation first and then the anion. Follow the same rules for naming the complex
ion but now the metal will end in –ate.
For some metals the latin name is used.
Metal Ion Latin name
The cation is Na+ ; sodium. Which means the anion has an overall charge of -2. We can
use this to calculate the charge on the iron
Fe(CN)5NO charge Fe + 5(charge cyanide) + charge NO = -2
charge Fe + (-5) + (0) = -2
charge Fe is +3
The anion is written with the ligands first (in alphabetical order) followed by the metal.
five cyanos and 1 nitrosyl would be pentacyanonitrosyl
and the metal is ferrate (III) giving us pentacyanonitrosyl ferrate (III) as the name of the
The complete name would therefore be sodium pentacyanonitrosyl ferrate (III).
K2[CoCl4] potassium tetrachlorocobalt (II)
Na[Pt(NH3)Cl3] sodium amminetrichloro platinate (II)
Common Coordination Geometries
The common coordination numbers are 2,4, and 6
-All ML2 are linear
examples : [Ag(NH3)2] + or [CuCl2]-
- ML4 are often tetrahedral
examples : TiCl4 or [CoCl4]2- or [Zn(NH3)4]2+
-ML4 can also be square planar, most often Pt or Ni
examples : Pt(NH3)2 Cl2 or [Ni(CN)4]2-
-ML6 will be octahedral
examples : [Ni(H2O)6]SO4 or Pt(NH3)2Cl4
1. Structural Isomers : same atoms but different bonds
a. Coordination Isomers : pieces swapped between the complex ion and
pentaamminesulfatochromium(III) bromide and
[Cr(NH3)5SO4]Br and [Cr(NH3)5Br]SO4
b. Linkage Isomers : a different atom of the ligand is used to form the bond
sodium pentacyanonitroferrate(III) and sodium pentacyanonitritoferrate (III)
We could use the N or the O as a ligand forming linkage isomers.
2. Stereoisomers : same atoms, bonds between the same atoms, different 3D shape
a. Geometric Isomers : (cis/trans)
When two identical groups are on the same side they are cis. If they are on
opposite sides they are trans.
b. Optical Isomers : (chiral)
When two isomers are non-superimposable mirror images they are optical
These two molecules are mirror images. There is no way that we can rotate the second
molecule to make it look identical to the first, so they are not superimposable.
What type of isomers are the following pairs?
a. [Cr(NCS)(NH3)5]Cl2 and [Cr(SCN)(NH3)5]Cl2 linkage isomers
b. [CrCl(H2O)5]Cl2·H2O and [CrCl2(H2O)4]Cl·2H2O coordination isomers
c. geometric isomers
Are the following pairs identical molecules or optical isomers?
a. optical isomers
b. identical molecules
We will limit our discussion of stereochemistry, with regards to naming, to square planar
complex ions with 2 identical substituents, octahedral complex ions with 2 identical
substituents, and octahedral complexes with 3 identical substituents.
Compound type No. of isomers
Ma2bc 2 (cis and trans)
Ma4b2 2 (cis and trans)
Ma3b3 2 (fac and mer)
here a, and b, represent monodentate ligands
MAA2b2 3 (2*cis and 1 trans)
here a, and b, represent monodentate ligands and AA is a bidentate ligand.
In square planar geometries we can have cis and trans molecules when 2 of the ligands
are the same (Ma2bc or Ma2b2). If the 2 identical ligands are on the same side the
molecule is cis, if on opposite sides it is trans.
In octahedral geometries a cis molecule will have two identical ligands in adjacent
positions (they will define the ends of an edge of the octahedron). The trans molecule
will have the two identical ligands on opposite sides of the metal.
When there are 3 identical ligands we can describe them as being fac or mer. Fac is used
if the 3 identical ligands define a face of the octahedron. Mer is used when the 3
identical ligands define a meridian of the octahedron.
Crystal Field Model
Crystal Field Theory, also called the Ligand Field Model, can be used to explain the color
of complex ions as well as their magnetic properties.
We are going to think of our ligands as negative point charges.
The ligands closest to a d orbital will have the highest energy interactions. The negative
charge of the ligands is repelled by the negative charge of the electron(s) in the orbital.
These are called the eg orbitals
The ligands furthest from the d orbitals will have a lower energy.
These are called the t2g orbitals
Without ligands all of our d orbitals have the same energy.
With ligands our d orbitals now have different energies.
The difference in energy is is called the d orbital splitting (Δ).
When the size of Δ is large we refer to it as the strong field case. The pattern of electron
filling will be to pair up electrons in the lower energy level first.
When the size of Δ is small we refer to it as the weak field case. Electrons will fill up all
of the orbitals before pairing up. (tetrahedral?)
For octahedral complexes (CN = 6) d orbital splitting looks like:
If a weak field ligand is used we see a small amount of splitting, if the ligand is stronger
we see a larger Δ.
The spectrochemical series is a list of ligands organized by how much orbital splitting
Cl-, Br-, I- < C2O42- < H2O < NH3 , en < phen < CN-
small Δ large Δ
weak field ligands strong field ligands
When a strong field ligand is used the splitting will be large. When a weak field lligand
is used the splitting will be small. If the splitting is small enough the electrons will fill
the orbitals as if they were all the same energy level before they start to pair up. This is
because the pair repulsion energy of placing two electrons in the same orbital is larger
than the difference in the energy levels.
Compare the magnetic properties of [Fe(CN)6]4- and [Fe(H2O)6]2+.
Both of these complexes are made with Fe2+ which has 6 d electrons.
Fe : [Ar] 4s23d6 so Fe2+ : [Ar] 3d6
Both have a coordination number of 6 and are therefore octahedral.
The only difference is the ligands. CN- is a stronger field ligand than H2O so it will
create a larger splitting.
The complexes [FeF6]3- and [Co(ox)3]4- have high spin configurations. How many
unpaired electrons are in each?
Fe3+ has 5 d electrons
To make the high spin case we will fill all of the orbitals singly before making pairs.
there are 5 unpaired electrons.
Co2+ has 7 d electrons
there are 3 unpaired electrons
The color of transition metal complexes is due to transitions between the d orbitals.
When light is absorbed by the complex an electron is excited into a higher orbital. The
energy of this transition is related to the wavelength absorbed.
Δo = E = hν = hc / λ
h = 6.626 x 10-34
The color we see is a result of an absence of the color absorbed. For example, if a
solution absorbs light that is red the solution will appear to be green. The color wheel
below can be used to approximate the colors absorbed and observed.
[Cr(H2O)6]2+ absorbs light at 700 nm. What is the value in kJ/mole for the ligand field
splitting? What color is the solution?
700 nm is red light, so the solution would appear green.
E = hc/λ = (6.626 x 10-34)(2.998 x 108) / (700 x 10-9)
E = 2.84 x 10-19 J or 2.84 x 10-22 kJ per electron transition
which multiplied by 6.022 x 1023 gives us 171 kJ / mole.
[Ti(H2O)6]3+ has a ligand field splitting of 2.35 x 105 J/mol. What color would a solution
of this complex appear?
2.35 x 105 J / mole divided by 6.022 x 1023 is 3.90 x 10-19 J.
3.90 x 10-19 = (6.626 x 10-34)(2.998 x 108) / λ
λ = 510 nm The solution would absorb green light and appear green.
[Cr(H2O)6]Cl3 and [Cr(NH3)6]Cl3
One of these is yellow and one is violet. Which one is yellow?
NH3 is a stronger field ligand than H2O is. The complex with NH3 will have a larger
splitting energy and will absorb light of a shorter wavelength. It will be the one that
absorbs violet and appears yellow.
For complexes with a CN of 4 we can have either a tetrahedral or square planar
geometry. The d orbital splitting patterns look like:
The Crystal Field Model is useful for explaining the magnetic properties of complex ions.
High Spin = maximum number of unpaired electrons
Low Spin = minimum number of unpaired electrons
definition : process of isolating a metal from an ore
Pyrometallurgy : uses a furnace, example would be isolating iron from iron ore
Iron ore is usually
Wavelength in nm Crystal Field
of Peak in Spectrum Splitting Energy
Ligand of Co(NH3)4X* (nm) for Co(NH3)5X (kJ/mol)
CN- 440 272
-NO2- 458 261
NH3 475 252
-NCS- 496 241
OH2 490 244
-ONO- 491 244
OH- 503 238
-ONO2- 500 239
F- 510 235
Cl- 533 224
Br- 550 217
I- 580 206
*This wavelength is that at which there is a maximum absorption of light in the spectrum
of the Cobalt(III) complex ion.
The complex ion Fe(CN)63- is paramagnetic with one unpaired electron. The complex
ion Fe(SCN)63- has five unpaired electrons. Where does SCN- lie in the spectrochemical
series relative to CN-?