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By the end of this section, you will be able to:
  • Outline the basic premise of crystal field theory (CFT)
  • Identify molecular geometries associated with various d-orbital splitting patterns
  • Predict electron configurations of split d orbitals for selected transition metal atoms or ions
  • Explain spectral and magnetic properties in terms of CFT concepts

The behavior of coordination compounds cannot be adequately explained by the same theories used for main group element chemistry. The observed geometries of coordination complexes are not consistent with hybridized orbitals on the central metal overlapping with ligand orbitals, as would be predicted by valence bond theory. The observed colors indicate that the d orbitals often occur at different energy levels rather than all being degenerate, that is, of equal energy, as are the three p orbitals. To explain the stabilities, structures, colors, and magnetic properties of transition metal complexes, a different bonding model has been developed. Just as valence bond theory explains many aspects of bonding in main group chemistry, crystal field theory is useful in understanding and predicting the behavior of transition metal complexes.

Crystal field theory

To explain the observed behavior of transition metal complexes (such as how colors arise), a model involving electrostatic interactions between the electrons from the ligands and the electrons in the unhybridized d orbitals of the central metal atom has been developed. This electrostatic model is crystal field theory    (CFT). It allows us to understand, interpret, and predict the colors, magnetic behavior, and some structures of coordination compounds of transition metals.

CFT focuses on the nonbonding electrons on the central metal ion in coordination complexes not on the metal-ligand bonds. Like valence bond theory, CFT tells only part of the story of the behavior of complexes. However, it tells the part that valence bond theory does not. In its pure form, CFT ignores any covalent bonding between ligands and metal ions. Both the ligand and the metal are treated as infinitesimally small point charges.

All electrons are negative, so the electrons donated from the ligands will repel the electrons of the central metal. Let us consider the behavior of the electrons in the unhybridized d orbitals in an octahedral complex. The five d orbitals consist of lobe-shaped regions and are arranged in space, as shown in [link] . In an octahedral complex, the six ligands coordinate along the axes.

This figure includes diagrams of five d orbitals. Each diagram includes three axes. The z-axis is vertical and is denoted with an upward pointing arrow. It is labeled “z” in the first diagram. Arrows similarly identify the x-axis with an arrow pointing from the rear left to the right front, diagonally across the figure and the y-axis with an arrow pointing from the left front diagonally across the figure to the right rear of the diagram. These axes are similarly labeled as “x” and “y.” In this first diagram, four orange balloon-like shapes extend from a point at the origin out along the x- and y- axes in positive and negative directions covering just over half the length of the positive and negative x- and y- axes. Beneath the diagram is the label, “d subscript ( x superscript 2 minus y superscript 2 ).” The second diagram just right of the first is similar except the x, y, and z labels have been replaced in each instance with the letter L. Only a pair of the orange balloon-like shapes are present and extend from the origin above and below along the vertical axis. An orange toroidal or donut shape is positioned around the origin, oriented through the x- and y- axes. This shape extends out to about a third of the length of the positive and negative regions of the x- and y- axes. This diagram is labeled, “d subscript ( z superscript 2 ).” The third through fifth diagrams, similar to the first, show four orange balloon-like shapes. These diagrams differ however in the orientation of the shapes along the axes and the x-, y-, and z-axis labels have each been replaced with the letter L. Planes are added to the figures to help show the orientation differences with these diagrams. In the third diagram, a green plane is oriented vertically through the length of the x-axis and a blue plane is oriented horizontally through the length of the y-axis. The balloon shapes extend from the origin to the spaces between the positive z- and negative y- axes, positive z- and positive y- axes, negative z- and negative y- axes, and negative z- and positive y- axes. This diagram is labeled, “d subscript ( y z ).” In the fourth diagram, a green plane is oriented vertically through the x- and y- axes and a blue plane is oriented horizontally through the length of the x-axis. The balloon shapes extend from the origin to the spaces between the positive z- and negative x- axes, positive z- and positive x- axes, negative z- and negative x- axes, and negative z- and positive x- axes. This diagram is labeled “d subscript ( x z ).” In the fifth diagram, a pink plane is oriented vertically through the length of the y-axis and a green plane is oriented vertically through the length of the x-axis. The balloon shapes extend from the origin to the spaces between the positive x- and negative y- axes, positive x- and positive y- axes, negative x- and negative y- axes, and negative x- and positive y- axes. This diagram is labeled, “d subscript ( x y ).”
The directional characteristics of the five d orbitals are shown here. The shaded portions indicate the phase of the orbitals. The ligands (L) coordinate along the axes. For clarity, the ligands have been omitted from the d x 2 y 2 orbital so that the axis labels could be shown.

In an uncomplexed metal ion in the gas phase, the electrons are distributed among the five d orbitals in accord with Hund's rule because the orbitals all have the same energy. However, when ligands coordinate to a metal ion, the energies of the d orbitals are no longer the same.

Questions & Answers

what is a balanced equation 4 trioxonitrate (V)acid and sodium hydroxide?
Marcel Reply
proved ur Worth: If A is a of trioxonitrate(V)acid,HNO3' of unknown concentration .B is a standard solution of sodium hydroxide containing 4.00g per dm cube of solution.25cm cube portions solution B required an average of 24.00cm cube of solution A for neutralization,using 2 drops of methyl orange.
Marcel
calculate the concentration of solution B in moles per dm cube
Marcel
calculate the concentration of solution A and B in moles per DM cube
Marcel
finally calculate the concentration in g/dm cube of HNO3 in solution A (H=1,N=14,O=16,Na=23)
Marcel
calculate the standard enthalpy of formation for propane(C3H8) from the following data; 1), C3H8+5O2->3CO2+4H2O; -222.0kJ/mol 2), C+O2->CO2;-395.5kJ/mol 3),H2+O->H2O; 285.8kJ/mol
Josephine
let eventually of formation of propane = X X + (-222)=3×(-395.5)+4×(-286) rearrange to find X
Paul
wat is electrolysis?
Mgbachi Reply
it is the chemical decomposition of a substance when electric current is passed through it either in molten form or aqueous solution
Nuru
list the side effect of chemical industries
Chelsea Reply
how do you ionise an atom
Rabeka Reply
many ways ,but one of them is when the atom becomes heated to a certain temperature the surface electron becomes too energetic and leaves the atom because the attraction between the nucleus and the electron becomes overpowered by the energetic eletron
sunday
also hitting of two atoms can cause transfer of surface electrons
sunday
and when this transfers occur the atom becomes ionised
sunday
who is doing Cape chemistry tomorrow?
caramel Reply
What is hybridization
edmondnti Reply
the mix between different breeds of species in one
Jared
it is the blending of orbitals.
stanley
the mixing of orbital
caramel
are covalent bonds influenced by factors such as temperature and pressure?
patrick Reply
what is catalyst used for mirror test
Sanjay Reply
when an atom looses electron, what does it become?
Abdullahi Reply
it's oxidized and called an ion
Anora
thanks
Abdullahi
Now, I get it
Abdullahi
cation
Anora
can you give an example please, if you don't mind
Abdullahi
a positive ion,become positively charged/a cation.
Janis
sodium plus one is simple cation is exmpl
ajmal
Taking Sodium as example..... it carries a positive charge which means it is positively charged.....when it gains an electron, it is reduced cuz an electron is negatively charged.....also when an atom looses an electron, it becomes positively charged and when it gains, it becomes negatively charged.
Nuru
typically, ionization is the process where an atom looses or gains electron(s) to form ion(s) either a positively or negatively
Nuru
what is copper
Bryan Reply
just an element
Power
Cu
daniel
Why is water a single covalent bond?
Mohamed Reply
nitrogen is a gas whereas phosphorus is solid .Explain.
Jacky Reply
can you explain what you are needing it now better than maybe I'm just not interpreting it what you're needing to know
Alex
cool nitrogen down to around negative 270 °F and it will be solid. now they are both solid
daniel
they are different elements and dats how they are pal.....check the periodic table
Nuru
Nitrogen is a diatomic molecule with relatively weak van de waals forces between the molecules. These forces are overcome when the solid melts or liquid evaporates. Phosphorus forms larger molecules consisting of four phosphorus atoms in a tetradedral shape. The intermolecular forces are stronger
Paul
whats a base
Daksalma Reply
A base is a substance which will neutralize an acid to yield salt and water only
Zainab
base is a substance that produces OH(aq) ions in aqueous solution. Strong soluable bases are in water and are completely dislocated. Therefore weak base ionize slightly...
Roy
a base is a substance that neutralise and acid to form salt and water
Daksalma
write electrolysis of bright solution using either carbon or platinum and write the reaction at the anode or at the cathode
Abdullah Reply
what is the H3O of a solution with the pH of 2.5
Sgt.Elliott_98 Reply
pH<7, therefore there are only H3O+HX3OX+particles in the solution. [H3O+]=10−pH=10−6.99=1.02⋅10−7[HX3OX+]=10−pH=10−6.99=1.02⋅10−7 When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking  and I'll give you another one if this is ki
Alex
if I'm answering and interpreting what you're asking correctly
Alex
When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking 
Alex
sorry I don't know why that sent again
Alex
We have [H3O+]=10−pH=10−6.99=1.02⋅10−7[HX3OX+]=10−pH=10−6.99=1.02⋅10−7 and [OH−]=10−pOH=10−7.01=9.77⋅10−8[OHX−]=10−pOH=10−7.01=9.77⋅10−8.  Because of H3O++OH−⟶2H2OHX3OX++OHX−⟶2HX2O we are left with [H3O+]=1.02⋅10−7−9.77⋅10−8=4.6⋅10−9
Alex

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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