Transition metals are located in the center of our modern day periodic table. Aside from the ions that are
formed, and some of the more common ionic compounds involving species like copper and iron, we
have ignored these species all year. The transition elements are remarkably useful in their uncombined
form, and are also able to form a variety of unique and special compounds. Copper and iron are used
worldwide everyday. Copper is used in wires and as a coating on pennies (however some of your older
pennies are all copper!). Iron is used in industry as steel.
Many other transition metals are
indispensable as well: chromium is used in automobile parts, gold and silver are used in jewelry,
tungsten is used in lightbulb filaments, platinum in automobile catalytic converters, titanium is now
used in bicycle frames, and zinc in batteries might be some of the more “common” transition metals and
their uses. But there is more! Zirconium is used in nuclear reactor liners, vanadium is used in axels and
crankshafts, molybdenum is used in boiler plates, nickel is used in coinage, tantalum for organreplacement parts, and palladium in some telephone parts. The d-block contains the densest elements
(osmium d = 22.48 g/cm3 and iridium d=22.41 g/cm3), and metals with the highest and lowest melting
points (tungsten m.p. = 3410oC and mercury m.p. = -38.9oC). The list goes on and on. Transition metals
are everywhere! As ions, many transition metals also play vital roles in our bodies!
The “d-block” on the periodic table contains transition metals as well as the “f-block” which contain the
inner transition metals. We will concentrate mainly in the fourth period – from scandium to zinc.
Transition elements have
elements. In some ways,
and non-metals. But all
diverse properties. Most
tend to be highly colored.
very different physical and chemical properties compared to the main group
their properties are more uniform. Main group elements contain both metals
transition elements are metals. In other ways, transition metals have more
main group ionic compounds are colorless. But transition metal compounds
Vanadium compounds vary in color based on their oxidation state.
From left to right: V+5, V+4, V+3, V+2
Most main group species are diamagnetic while transition metals tend to be paramagnetic. If you
remember, diamagnetic means that the electrons are paired, and thus the species is not attracted to (and
slightly repelled by) a magnetic field. While paramagnetic species contain unpaired electrons and are
attracted to a magnetic field.
With the exception of mercury, transition metals exist in the solid phase and have high melting and
boiling points. They have a metallic sheen and conduct electricity and heat well. Gold, silver and
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platinum are resistant to oxidation, which makes them useful for jewelry and decorative items. While
the other transition metals react with oxidizing agents to form ionic compounds.
Certain d-block elements are very important biologically. Cobalt is important in vitamin B12,
hemoglobin and myoglobin contain iron (a heme group). As compounds that are highly colored, they
are used as pigments in paints and dyes. If anyone remembers Bob Ross, a painter on TV, he made
much use of paint that contains transition metals: Prussian blue, which contains iron, cadmium yellow,
which contains cadmium, and white, which contains titanium. If you have some spare time, flip to the
jewelry channel on cable and you will see transition metals in precious gems: iron (II) in citrine,
chromium (III) in ruby. Transition metals also give color to glass: cobalt (II) gives the brilliant blue
color, chromium can be used to turn glass green, and glass that has a purple color contains traces of
manganese.
As with any element, the properties of the transition elements and their compounds arise from the
electrons associated with that element – specifically – the electron configuration of the atom.
The
generic representation of the valence electrons is ns2(n-1)d. Transition metal ions form from the loss of
the HIGHEST principal quantum number electrons first. It is the electron configuration of the transition
metal atom that correlates and determines the physical properties of the element, such as density and
magnetic properties. It is the electron configuration of the ion that determines the properties of the ionic
compounds that are formed from that element.
Categories of electrons:
1.) inner (core) electrons: those electrons up to the previous noble gas and any completed transition
series. These make up the lower energy levels of the atom
2.) outer electrons: those in the highest energy level (highest n value), these electrons are the
farthest away from the nucleus.
3.) valence electrons: those electrons involved in chemical reactions. For the main group (A)
elements, they are the outer electrons, for the transition metals (B), sometimes inner d electrons
can be involved as well.
Ion formation
For transition elements, d block, remove ns electrons before (n-1)d electrons
For example, let’s examine the two ions formed by Tin (Sn)
Sn+2: 1s22s22p63s23p64s23d104p65s24d105p2
rewrite the electron configuration grouping the n values together.
Sn+2: 1s22s22p63s23p64s23d104p64d105s25p2
Sn+4: 1s22s22p63s23p64s23d104p64d105s25p2
1s22s22p63s23p64s23d104p64d105s2
1s22s22p63s23p64s23d104p64d10
2
3
Concept Test:
Write the electron configurations for:
Zr:
V+3:
Mo+3:
We have previously studied the atomic properties of the main group elements (1 A - 8 A) and in doing so,
we skipped over a large chunk of the periodic table, ignoring the transition metals. The atomic
properties of the transition metals differ in several ways from the main group elements.
Atomic Size Across a Period
Recall that as you travel across a row (period) for the main group elements, the atomic size decreased
rather nicely. We were adding electrons into the same principal quantum level (e.g. adding electrons to
the 3s and then the 3p) while increasing the number of protons. The electrons were not getting any
farther away from the nucleus as they were put into the outer 3p orbitals but the number of protons was
also increasing. Thus, they were able to more efficiently “pull” on those electrons and the size of the
atoms therefore decreased.
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For the transition metals, electrons are put into the 3d or 4d level. Recall the electron configurations of
the transition metals: the d orbitals lie at a lower energy state than the s shell which is filled before the 3d
or 4d. (e.g. 4s23d5). This means that all the electrons are going into the d level and shielding the s shell
from feeling the full nuclear charge. The electrons in this case are being added to a shell that is inside the
s shell. These inner d orbitals shield the outer electrons very well and as such the 4s electrons do not get
any farther away from the nucleus, meaning that the size of the transition metals remains fairly constant
as you go from left to right across a row.
Electronegativity (X) Across a Period
Recall that the general trend in electronegativity is that X values increase from left to right across a row
and from bottom to top up a column. Recall also, that fluorine is the most electronegative atom on the
periodic table and the elements increase their values as we work our way up and across towards F. The
reason for this is going to depend on those same factors that we used to explain the trends in atomic size,
ionization energy, and electron affinity. This trend works well for the main group elements, but again,
transition metals show very little variation in their X values. This is consistent with their relatively
constant atomic size. Remember X is a propensity to be attracted to electrons. More X atoms want
electrons more than less X atoms. Given that they are located in the middle of the table, the transition
metals have intermediate X values, but they all have similar (relatively) X values.
First, the horizontal comparison. As you go across a period from left to right, the atoms of each element
all have the same number of energy levels and the same number of shielding electrons. Thus the factor
that predominates is the increased nuclear charge. When the nuclear charge increases, so will the
attraction that the atom has for electrons in its outermost energy level and that means the
electronegativity will increase. Transition metals are not getting smaller, so their effective nuclear
charge is not increasing! In fact, the nuclei of the transition metals do not seem to be as attracted to their
outmost electrons as the size is NOT decreasing. Thus, Zn and Mn have about the same attraction to
their respective outer electrons.
You might notice that there are some high values in the middle of the transition group. Those elements
have fairly high electronegativities for metals. The reason for that ties in with the arrangement of
electrons and the fact that the atoms are using d orbitals. The way that the d orbitals are shielded is
different than the way that s and p orbitals are shielded so there is some variation in the transition metals
that is not as easily explained.
5
Ionization Energy (IE) Across a Period
Recall that ionization energy refers to the ability of an atom to LOSE an electron (or two or three). We
know that metals have a much easier time losing their electrons than do non-metals (which prefer to gain
electrons so we referred to that situation as electron affinity). Since it is easier for metals to lose their
electrons, the energy required for that to happen is lower (in kJ/mole) than for nonmetals. For example,
the IE of K is 419 kJ/mole, but for Kr it is 1351 kJ/mole. MUCH easier to take that electron away from K
than from Kr! For transition metals, recall that most have the ns2 (n-1)dx electron configurations, and
since the d electrons are shielding the ns2 electrons from feeling the full nuclear charge, it is relatively
easy to strip those electrons away. Thus, for the first IE values (removing the first electron) the transition
metals all have very similar IE1 values.
Atomic Size Within a Group
Recall that atomic size increases as you travel down a group – the reason being that you are
adding electrons to new and larger quantum numbers/levels that are farther from the nucleus.
For example, as you travel down group 1 you add electrons into the n=1 level for H, then n=2
level for Li, then n=3 for Na. These principal quantum levels are farther from the nucleus (much
like Earth is farther from the sun than either Mercury or Venus). However, for the transition
metals, the lanthanides are sandwiched in-between the transition metals on the periodic table and
the electrons are added to the f orbitals – specifically the 4f (and then the 5f for the actinides).
The electrons are added to inner shells (just like the d orbitals are inside the s orbitals for the
transition metals), and in addition to adding inner electrons, the number of protons increases,
which results in an increased nuclear charge. What results is that the addition of inner electrons
and more protons keeps the elements smaller than you might initially predict. This is termed the
lanthanide contraction which is an effect that causes sixth period elements with filled 4f subshells
to be smaller than otherwise expected. The intervention of the lanthanides increases the effective
nuclear charge, which offsets the size increase expected from filling the n=6 valence shell. As a
consequence, sixth period transition metals are about the same size as their fifth period
counterparts.
6
Electronegativity (X) Down a Group
The vertical trend seen in most of the transition metal groups is opposite what is seen with elements in
the main groups. Here, X values increase in X as you go down a group from Period 4 to Period 5. The
heavier elements become quite X, more so than you might expect, with values that exceed most
metalloids and even some of the non-metals! Although atomic size increases slightly, X increases
because of the larger nuclear charge. The heavier members of the transition metal groups exhibit more
covalent character in their bonds and thus have a greater attraction for electrons than do the main group
elements. In fact, in certain situations elements like gold will actually gain electrons! (ex. The formation
of saltlike CsAu which contains the Au-1 ion).
Ionization Energy (IE) Down a Group
Recall again that the trend for main-group elements is that X values decrease going down a group. As
the atom’s size increases, the electrons are located further and further away from the nucleus, and as
such, they are easier to remove from the atom. Thus, IE values decrease down a group. For transition
metals, the size does not really change much traveling down a group and comparing atomic sizes from
Period 5 to 6 the atoms are almost the same size!! Thus, IE1 values increase traveling down a group.
This means that it is harder to remove the first electron from the transition metals as principal quantum
numbers increase.
Density
Atomic size and therefore volume, are inversely related to density (d=m/v). As the volume stays fairly
constant across a row but the atomic mass increases, densities increase traveling across a row. When you
compare densities down a group, again, volumes increase very little but the atomic mass increases
greatly, thus densities down a group increase dramatically. As a result, Period 6 contains some of the
7
densest elements known: tungsten, rhenium, osmium, iridium, platinum, and gold have densities about
20 times that of water and twice that of lead!
Like the atomic properties, the chemical properties of transition metals differ greatly from the maingroup elements.
Oxidation States
One of the most characteristic chemical properties of the transition metals is the ability of different
metals to lose different numbers of electrons, thus having a variety of different oxidation states. For
example, there is Cr+2, Cr+3, and also Cr+6! Remember that the valence electrons for transition metals are
the ns2 electron and also sometimes the (n-1)d electrons. The table below shows the Period 4 elements
and the oxidation states that are possible.
Metallic Behavior
Atomic size and oxidation number/state have a major effect on the nature of bonding in transition metal
compounds. When the transition metal is in a lower oxidation state, it behaves like a metal. That is,
ionic bonding dominates. When the transition metal exists in its higher oxidation state, it behaves more
like a covalent compound. For example, at room temperature, TiCl2 is an ionic solid, while TiCl4 is a
molecular liquid. In this higher oxidation state, Ti+4 is a small highly charged ion – so its charge density
is very large. This high charge density is better able to polarize the electron clouds of non-metal species
and the bonding becomes more covalent in nature.
TiCl2
MnO
m.p. = 1035oC
m.p. = 1785oC
TiCl4 and Mn2O7, on the other hand, are both liquids at room temperature, with melting points below
0oC and relatively low boiling points, as might be expected for covalent compounds.
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TiCl4
Mn2O7
m.p. = 24.1oC
m.p. = -20oC
b.p = 136.4oC
b.p = 25oC
The principal difference between these compounds is the oxidation state of the metal. As the oxidation
state of an atom becomes larger, so does its ability to draw electrons in a bond toward itself. In other
words, titanium atoms in a +4 oxidation state and manganese atoms in a +7 oxidation state are more
electronegative than titanium and manganese atoms in an oxidation state of +2.
As the oxidation state of the metal becomes larger, the difference between the electronegativities of the
metal and the nonmetal with which it combines decreases. The bonds in the compounds these elements
form therefore become less ionic (or more covalent).
For example: if the X of Ti+2 = 1.7 and X of Cl = 3.0 then X = 1.3
If X of Ti+4 = 2.2 and X of Cl = 3.0 then X = 0.8
The bond became more covalent in nature!
Color and Magnetism
Most main group ionic solids are colorless – or white (boring!!) because the metal has a filled outer level
(noble gas configuration). When electrons are promoted into higher energy levels, the higher energy
levels are so high, the ion does not absorb light in the visible range – which means it does not give off
light in the visible range. In contrast, electrons in a partially filled d orbital can absorb visible light
(wavelengths, frequencies or energies associated with light in the visible region) and move to slightly
higher d orbital energy levels. As a result, which we will discuss later in more detail, many transition
metal compounds have strikingly beautiful colors (not boring!). Exceptions are transition metal ions that
have either empty d sublevel or a filled d sublevel (e.g. Sc+3: 4s03d0, Ti+4: 4s03d0, Zn+2: 4s03d10).
Magnetic properties are also related to how filled (or unfilled) the d orbitals are. Recall that a
paramagnetic substance has unpaired electrons (as an atom or as an ion) while a diamagnetic substance
has only paired electrons. Most main-group ions are diamagnetic and thus are repelled by a magnetic
field. In contrast, many transition metal compound are paramagnetic because the transition metal ions
have unpaired d electrons. Transition metal ions with a d0 or a d10 are also colorless and diamagnetic.
9
The most distinctive aspect of transition metal chemistry is the ability of transition metal ions to form
coordination compounds (also called complexes). These substances contain at least one complex ion (a
central metal ion that is bonded to molecules and/or anions called ligands). In order to maintain charge
neutrality, coordination complexes are typically associated with other ions, called counter ions. A
positively charged coordination complex will have a negative counter ion, while a negatively charged
coordination complex will have positive counter ions.
A typical coordination compound, such as [Co(NH3)6]Cl3 gives us information about the complex ion
and the counter ions present. The complex ion is contained inside the brackets, thus [Co(NH3)6]+3 is the
complex ion and the 3 chloride ions are the counter ions. The ligands are the 6 NH 3 molecules that are
bound to the central Co+3 ion.
In water, the coordination compound behaves like an electrolyte with the complex ion and the counter
ion(s) separating from one another. The ligands and the central metal ion will remain together, thus the
complex ion behaves much like a polyatomic ion.
A complex ion is described by the metal ion and the number and type of ligands attached to it. Its
overall structure depends on three characteristics: coordination number, geometry, and number of
donor atoms per ligand:
Coordination Number: the number of ligand atoms/molecules that are bonded directly to the
central metal ion and the coordination number is specific for a given metal ion in a particular
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oxidation state and compound. The coordination number of the Co+3 in [Co(NH3)6]+3 is 6 because
there are 6 nitrogens (from the ammonia) that are bonded to the central cobalt ion. In general, the
most common coordination number is 6, but 2 and 4 are often seen.
Geometry: The geometry or shape of a complex ion depends on the coordination number and
nature of the metal ion. If there are 2 ligands attached to a central metal ion, then the geometry is
linear, if there are 4 ligands, then square planar or tetrahedral, and 6 will result in octahedral
geometry. The difference between the square planar and tetrahedral geometries depends on the
central metal ion. Most d8 ions are square planar while most d10 ions are tetrahedral.
Donor atoms per ligand: The ligands of complex ions are molecules and/or anions with one or
more donor atoms that donate a lone pair of electrons to the metal ion to form a covalent bond.
Remember that the metal ions are electron deficient (and as such as classified as Lewis acids) and
can accept electron pairs from other species (called Lewis bases). Because the species that are
donating electrons have at least one lone pair, they will be atoms from Group 5A, 6A, or 7A.
Ligands are classified in terms of the number of donor atoms or “teeth” that each uses to bond with the
central metal ion. Chloride and NH3 ligands use a lone pair of electrons from a single atom and are thus
terms monodentate ligands. Bidentate ligands have two donor atoms, polydentate ligands have more
than 2 donor atoms. Bidentate and polydentate ligands give rise to ring formation in the complex ion, as
the ligand loops around so that both atoms can bond to the central metal ion. Such ligands appear to
grab the central metal ion like a claw, as if they are pinching the metal ion between their fingers, and
thus are termed chelating ligands (from the Greek meaning “crab claw”)
11
12
There are three important rules for writing the formulas for coordination compounds. The first two
should look familiar as we applied these rules to formulas for ionic compounds:
1. The cation is written before the anion
2. The charge of the cation is balanced by the charge of the anion(s)
3. In the complex ion, neutral ligands are written before anion ligands, and the formula of the ion is
placed inside brackets.
Remember that the whole ion complex can be either positive or negatively charged depending on the
metal ion and the ligands that are present.
For example: the complex ion [Co(NH3)2Cl4] would exist with Co+2, NH3 are neutral molecules and Cl-1
is the anion ligand. Notice the cobalt is written first, followed by the neutral NH 3 ligands and then the
chloride ligands. Thus the charge of this complex ion would be -2 as the Co+2 only cancels out with 2 of
the chlorides leaving a net -2 charge behind. Thus the ion looks like this: [Co(NH3)2Cl4]-2. Since we
cannot have an ion floating around all by itself, the counter ions that would be present to make this a
neutral compound must be positively charged. As positive ions, they should be written first also!
Therefore, K2[Co(NH3)2Cl4] is the complete formula for this coordination compound.
What about this complex ion: [Co(NH3)2Cl2]+1 indicates a Co+3 ion, two neutral NH3 ligands and 2 Cl-1
ions which gives us the net +1 charge for the ion. Again, there must be a counter ion present, in this
case, we need an anion, and they are written last.
Therefore [Co(NH3)2Cl2]Cl is the complete formula for this coordination compound. Again, remember
that all this is INSIDE the bracket is the complex ion – anything appearing on the outside is a counter ion
that is present to make a neutral compound.
Complex ions can be paired together to create a neutral coordination compound. For example, consider
the complex ions below
[Co(NH3)5Br]+2 and [Fe(CN)6]-4
Knowing the charges of the complex ions: Calculate the oxidation numbers of the metals and ligands:
Co =
NH3 =
Br =
Fe =
CN =
Pairing these two complex ions together means that we need 2 of the positive complex ions to charge
balance one of the negatively charged complex ions:
The formula becomes: [Co(NH3)5Br]2[Fe(CN)6]
13
Coordination compounds used to be named based on who discovered them or based on their color, and
some of these names are still used, however, there is a systematic naming system that is based on a set of
rules:
1. the cation is named before the anion: thus [Co(NH3)4Cl2]+1 would be named before the counter
ion Cl-1. The only space in the names appears between the cation and the anion.
2. Within the brackets which contain the components of the complex ion, the ligands are named
alphabetically, before the metal ion. Thus in [Co(NH3)4Cl2]+1, the four NH3 ligands and the 2 Cl-1
ligands would be named before the cobalt +3 ion.
3. Neutral ligands generally have the molecule name (or a derivation thereof) while anion ligands
do not end in ide (like the typical ionic naming system) but use the root name and end in o
instead. Thus Cl is not chloride but chloro. Thus in [Co(NH3)4Cl2]+1, there are four ammine (NH3)
ligands and two chloro ligands bonded to the cobalt (III) ion.
4. A numerical prefix indicates the number of ligands present. Mono is not used, but the others
should look familiar: di = 2, tri = 3, tetra =4, penta = 5, hexa =6. The numerical prefixes does
NOT affect the alphabetical order. Some ligand names already contain a numerical prefix (such
as ethylenediamine) so if two of these types of ligands are present, the numerical prefix bis is
used (or tris for 3, or tetrakis for 4). Thus a complex ion with two ethylenediamine ligand has
bis(ethylenediamine) in its name.
5. The oxidation state of the central metal ion is given by a Roman numeral: such as cobalt (III),
ONLY if the metal ion can have more than one oxidation state.
6. If the complex ion is an anion (has a net negative charge), the ending of the metal in the complex
ion is changed such that the suffix is –ate. Thus the name for K[Pt(NH3)Cl5] is:
Potassium amminepentachloroplatinate (IV)
We must use the Roman numeral because there is another oxidation state for Pt.
We know that it is in the +4 oxidation state based on the charges of the counter ion and the
chloride ligands. Check this!
For some metals, using their recent names sounds really funny – for example, ironate?!! So we go
back to their Latin names: Na4[FeBr6]
Sodium hexabromoferrate (II)
14
Concept Test:
What are the systematic names for:
Na3[AlF6]
[Co(en)2Cl2]NO3
Concept Test:
What are the formulas for:
Tetraamminebromochloroplatinum (IV) chloride?
hexamminecobalt (III) tetrachloroferrate (III)
Previously we have discussed isomerism as it related to organic structures. We identified structural
isomers, which were species that had the same molecular formula but different connectivity of atoms
(e.g. hexane and 2,2-dimethyl butane). We also identified geometric isomers. These isomers have the
same molecular formula but their spatial orientation (geometry) was different. In this case, the
connectivity is the same, but the location of the atoms or functional groups was different in space. (e.g.
cis-1,2-dibromoethene and trans-1,2-dibromoethene). While the molecular formulas for isomers are the
same, their properties are not. Isomers can have very different boiling or melting points (think IMFs!).
They can also have different chemical properties as well. Coordination compounds can also exhibit
isomerism.
CH3
H
H
CH3CH2CH2CH2CH2CH3
CH3CCH2CH3
CH3
C
Br
15
H
Br
C
C
Br
H
C
Br
Constitutional Isomers: same atoms, different connectivity (counterpart: structural isomer). These
isomers have the same molecular formula but the atoms are connected differently. There are two subgroups to the constitutional isomers but they should be fairly easy to spot:
Coordination Isomers: in the case of coordination isomers, the composition of the complex ion changes
but the compound stays the same. Remember that a complete formula for a coordination compound
includes the complex ion and the counter ions that are present in order for the overall compound to be a
neutral species. Knowing this, one type of isomer exists when the ligands of the complex ion switch
places with the counter ions. For example: [Pt(NH3)4Cl2](NO2)2 is an isomer of [Pt(NH3)3(NO2)2]Cl2. In
the first compound, the two chloride ions are the ligands attached to the platinum central metal ion, and
the NO2-1 are the counter ions. In the second compound, the NO2-1 ions are now the ligands attached to
the central platinum ion and the chloride ions are the counter ions.
Another type of isomerism exists when you have pairs of complex ions that exist as an overall neutral
compound and the ligands again switch places. For example: consider the following complex ion and
its isomer: [Cr(NH3)6][Co(CN)6] and [Cr(CN)6][Co(NH3)6]. In the first compound, NH3 is a ligand of Cr+3
cyanide ions are ligands of Co+3. In the isomer, the cyanide ions are ligands of the Cr+3 and the NH3 are
now ligands to Co+3.
Linkage Isomers: in the case of linkage isomers, the identity of the ligand stays the same, it is the
attachment that is different. For example, some ligands are able to bond with the central metal ion in
more than one way. The nitrite ion (NO2-1) can bind to the central metal ion either through the nitrogen,
or through one of the oxygens. It is still the nitrite ion, but the attachment is different. When the nitrite
ion binds through the nitrogen O2N: it is called a nitro group. When it binds through one of the oxygens
ONO: it is called a nitrite group.
Two linkage isomers are shown below. The first gives the orange compound, pentaamminenitrocobalt
(III) chloride [Co(NH3)5(NO2)]Cl2 and the second gives a red compound, pentaamminenitritocobalt (III)
chloride [Co(NH3)5(ONO)]Cl2. (remember that the nitrite ion has a charge of -1, the amines are neutral,
and so two chlorides are present as counter ions).
16
Another couple of examples are the cyanate ion and the thiocyanate ion:
O
C
N
-1
S
C
N
-1
Stereoisomers: compounds that have the same molecular formula but different spatial arrangements of
atoms. The atoms will all be connected in the same manner, so ligands and counter ions are not
exchanging places, but their location in space will be different. Again there are two sub-groups to the
stereoisomer classification.
Geometric Isomers: (called cis-trans isomers). This type of isomer occurs when the groups are arranged
in differently in space relative to the central metal ion. For example, think of the square planar geometry
and how atoms in space might be arranged. Perhaps the ligands are located next to one another or they
could be located across from one another. This gives rise to the cis geometry (same side) compared to
the trans geometry (on opposite sides)
Naming the species means that you must include the cis or trans designation as these species are NOT
the same!
Cis-diamminedichloroplatinum (II) is used as an anti-tumor drug while transdiamminedichloroplatinum (II) has no antitumor effect at all.
Octahedral complexes also exhibit cis trans isomerism. It might initially be difficult to see, but remember
the octahedral complex is very similar to the square planar in geometry. The reason that it might
initially be difficult to identify the cis trans isomerism is because there are several planes in the
17
octahedral complex to consider. But examining the species even after rotating them the isomerism is
there and should be apparent. For the cis isomer, rotation keeps the ligands 90o apart no matter how
many times you rotate the species around. While for the trans, the ligands are always 180o apart.
Examine the cis and trans figures of the [Co(NH3)4Cl2]+1 complex ions. An interesting fact about these
species is that the color changes due to the location of the ligands. The cis isomer is violet while the trans
is green!
Optical Isomers: Occurs when a molecule and its mirror image cannot be superimposed on one another.
Unlike the other types of isomers, which have distinct and unique physical properties, optical isomers
are physically identical in all ways but one: the direction in which they rotate the plane of polarized
light.
When a beam of light is passed through a certain type of filter, all of the waves except those in one plane
are removed. The figure below shows this plane-polarized light impinging upon and being rotated by
two optical isomers. One of the optical isomers rotates the light in one direction, the other rotates the
light in the opposite direction but by the same amount. In every other way, such as boiling point,
density, refractive index, viscosity, etc., the two optical isomers are identical.
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Optical isomers are vitally important in organic chemistry. Optical isomers can be pharmaceutically
important as well. Often only one of the enantiomers exhibits the desired activity but both contribute to
the side-effects. Thalidomide was a drug given to pregnant women in the 1960's to alleviate morning
sickness. It did reduce the symptoms of morning sickness, but resulted in severe birth defects. It was
later determined that one of the isomers may have been reducing symptoms, while the other was
producing the birth defects.
In coordination compounds, when examining a structure to determine if it is an optical isomer, you need
to make sure that the mirror images are NOT superimposable on one another. If they are
superimposable, then the structures are the same structure, and therefore are NOT isomers. When
indicating an optical isomer, the designation d or l is used to indicate whether the compound rotates
light to the right, or to the left (d for dextrorotatory(right) and l for levorotatory(left)).
Previously we have discussed hybridization of orbitals using the ideas that orbitals come together and
mix, creating a hybrid of equal energy. Let’s review briefly:
Valence bond theory, based on quantum mechanics, is used to describe the formation of covalent bonds.
Previously, we have discussed the formation of ionic bonds by showing the transfer of electrons from
one species (the metal) to another (the non-metal). Now we will examine the sharing of electrons
between two or more species. In the formation of a complex ion, the filled ligand orbitals overlap with
the empty metal ion orbitals. Remember the ligand acts like a Lewis base, donating its electrons to the
metal ion which acts like a Lewis acid and accepts the electrons to form a bond.
The valence bond theory assumes that the each bond is localized, that is, each bond connects two atoms
together, just like we draw them in the Lewis structure. The valence bond theory assumes that the
atomic orbitals do not change during the bond formation – instead, they simply overlap as the atoms
approach one another. Electron density is higher in the region of overlap, and the negative charge
between the electrons in the overlapping orbitals and the positive nuclei attracts and holds the two
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atoms together. We talked about covalent bond formation and the distance between the two atoms in a
bond must be one that balances out repulsion of the nuclei, repulsion of the electron clouds, and the
attraction of the nuclei with the electrons. In general, the more extensive the overlap between the two
orbitals, the stronger the covalent bond.
As per the Pauli Exclusion Principle:
A covalent bond is formed by the pairing of a maximum of two electrons with opposite spins in the
overlap region of the atomic orbitals from the two atoms.
Some important points to remember about valence bond theory:
Most of the electrons in a molecule remain in the same orbital location that they occupied in the
separated atoms
Bonding electrons are localized in the region where the atomic orbitals overlap; the greatest
probability of finding the bonding electrons is in this region of overlap
For orbitals with directional lobes (the p and d and f orbitals), the maximum overlap occurs when
the atomic orbitals overlap end to end, along the axes of the orbitals. Recall that the s orbital is a
sphere and has no directional lobes.
The molecular geometry depends on the geometric relationships among the orbitals of the central
atom that participate in bonding. Thus, the valence bond theory combined with VSEPR theory
will account for the true geometric shape of the molecule.
The number of hybrid orbitals that results must equal the number of atomic orbitals that are
mixed together
The type of hybrid orbitals obtained varies depending on the types of atomic orbitals that are
mixed (do we get sp, sp2, sp3 etc. . . ) and now we will investigate hybridizations that are d2sp3,
dsp3, and sp3 in nature
What is hybridization?
In order to account for bonding, the valence bond theory is based on the hybridization of the bonding
orbitals. Hybridization is an imaginary mixing process in which the orbitals of an atom rearrange to
form new atomic orbitals called hybrid orbitals. In forming the hybrid orbitals, if possible, shared pairs
of electrons will be promoted so that all electrons are unpaired. The promotion will occur between the s,
p, d, and f orbitals in order to maximize the number of unpaired electrons.
Electron promotion accounts for the bonding between certain atoms to be identical in energy. Why, for
instance, are all bonds in the CH4 molecule identical? The ground state electron configuration for C is
[He]2s22p2. It is easy to see where two bonds come from – by interaction with the 2p electrons with the
1s1 electrons from H. But why are there 4 bonds and why are they all equal? The 2s2 electrons are paired
and at a different energy level than the 2p electrons.
Linus Pauling answered the question by stating that the overlapping atomic orbitals are actually
different in character than the atomic orbitals in the individual atoms. Quantum mechanical calculations
show that the mixing of atomic orbitals gives rise to new atomic orbitals. The spatial organization of the
overlapping orbitals leads to a more stable situation and leads to consistency in assigning shape.
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Write the electron configuration for C, and identify the valence electrons for C
C: 1s22s22p2
valence electrons are 2s22p2 electrons
Draw the orbital box diagram for the valence electrons
2s
2p
However, when hybridizing the orbitals, we will take the paired electrons, in this case an electron from
the 2s and promote it to the 2p. All electrons in the hybridized orbitals are considered to be the same
energy.
2s
2p
Thus, these three orbitals, the 2s and the 2p orbitals come together to make 4 orbitals, all of the same
energy, known as sp3 (notice there is an electron in the s and one electron in each of the 3 p orbitals)
sp3 Hybridization
Tetrahedral geometry, indicating an AX4 geometry, (or an AX3E, or AX2E2). Notice there are 3 p
orbitals and 1 s orbital coming together from the same n value (principal quantum number).
Notice 4 orbitals are available for bonding, and when 4 things – or 4 electron domains surround
a central atom we know that we have a tetrahedral VSEPR geometry.
sp3
Explain, using valence bond theory, why N wants to form 3 bonds and has a lone pair of electrons.
N: 1s22s22p3
2s
2p
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In this case, no electrons need to be promoted, all orbital boxes are filled with at least one electron.
Combining the orbital boxes as they are shows us the lone pair electrons and the 3 electrons that will be
used for bonding
sp3
Keep in mind that the hybridization explains the type of bonding that we see with these atoms as the
central atom. It can generally be used to explain the other atoms in covalent bonds, but it is most often
used to describe the central atom. It is also used to explain the type of bonding that the central atom is
participating in that particular molecule. Keep in mind that expanding the valence shells will mean
expanding the hybridization, thus, this method is used to describe that after the fact, not really in
predicting that expansion will occur.
sp2 Hybridization
Occurs when an s orbital and 2 p orbitals are of the same energy. The third p orbital is left unhybridized
(left alone). Examine Boron and its valence electrons.
B: 1s22s22p1
2s
2p
Hybridization occurs with the promotion of one of the paired electrons to an empty p orbital
sp2
2p unhybridized
3 bonding electrons means AX3 geometry which means trigonal planar VSEPR geometry.
sp Hybridization
Two valence electrons, one s and one p come together having the same energy to participate in bonding.
In this case, examine Beryllium and its valence electrons.
Be: 1s22s2
2s
2p empty
In this case, we have a location to promote an electron to, as all the p orbital boxes are empty.
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sp
2p unhybridized
2 bonding electrons means AX2 geometry which means linear VSEPR geometry
Any hybridization scheme that only allows for s and p orbitals means that we are confined by the octet
rule. However, we know that we do not always follow the octet rule when drawing Lewis structures.
Initially, you were just told that the d orbitals were available to participate in bonding. Using valence
bond theory you can determine exactly where the electrons are going to go in the d subshell. It is also
difficult to predict that this would happen unless you knew that your molecule violated the octet rule.
For example, consider Phosphorus.
Writing out the electron configuration for P: 1s22s22p63s23p3 we would predict phosphorus to behave
very similarly to Nitrogen – after all, they are in the same group, they each have 5 valence electrons, but
P can have an expanded octet while N cannot. Thus, it is important to know the molecule that you are
dealing with – you cannot just know P and not what is attached, as PH3 will have a different
hybridization than PCl5.
Write out the hybridization for PCl5 knowing that P will have no lone pairs and have 5 electrons
available for bonding.
3s
3p
All the orbital boxes are filled. There are 3 electrons available for bonding, but we need 5 electrons
available for bonding in order to be able to make PCl5. If we just promote a 3s electron up to the 3p then
we will just move our pair from the s to the p, we need to unpair the electrons as we did for CH4.
Knowing that P will be surrounded by 5 bonds, which means 10 electrons, which indicates we will be
violating the octet rule. In order to do that, we will use the d orbitals, and here is how:
3s
3p
3d
Creating a hybridization scheme like this:
sp3d
d unhybridized orbitals
Notice again, that the electrons are all contained in the same n value (principal quantum number). This
sp3d hybridization indicates one s, 3 p’s, and 1d orbital are used for bonding. That is a total of 5 orbitals
for bonding indicating the AX5 geometry or the trigonal bipyramidal VSEPR geometry.
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Finally, what about Sulfur. It is directly underneath oxygen, and will behave in the tetrahedral manner
(e.g. H2S) but what about when sulfur expands its octet making SF6?
Write out the hybridization for Sulfur knowing that it will form 6 bonds.
S: 1s22s22p63s23p4
3s
3p
In this case, S appears to only want to form 2 bonds (thus explaining H2S). But we need to move the
electrons around so that S will form 6 bonds.
3s
3p
3d
Hybridization indicates:
sp3d2
d unhybridized
An sp3d2 indicates 6 bonding orbitals for 6 bonding electrons which indicates AX6 which indicates
octahedral VSEPR geometry.
The same method will be applied to coordination compounds to explain their bonding and geometry.
Octahedral complexes: the hexaamminechromium (III) ion illustrates the application of the valence
bond theory to octahedral complexes. The six lowest, available empty orbitals of the Cr+3 ion can be
determined by writing the electrons configuration for the ion:
Cr+3 = 1s22s22p63s23p64s13d5 which becomes 1s22s22p63s23p64s03d3 when the ion is formed. Remember,
once you get a level you get all of the available orbitals, even if they are empty.
As such, the lowest energy orbitals are the 3d, 4s and 4p orbitals in chromium (the 4 d are also there but
are not going to be used in the hybridization as they are not needed!)
NOTE: As electrons fill the d orbitals, they actually drop in energy below the s orbitals making them the
lowest in energy.
When the NH3 ligands come in and act like a Lewis base, they will be bringing 2 electrons per :NH 3
group, and therefore will be donating BOTH electrons to the central chromium ion. This means that we
must find a “home” for 12 electrons that will be coming from :NH 3. In order to accommodate all 12, let’s
examine what we have in terms of orbital boxes from the chromium ion.
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Cr
3d
4s
4p
MIX:
Electron pairs donated from
NH3
3d2 + 4s + 4p3 = d2sp3 hybridization
The three unpaired electrons belonging to the chromium ion mean that the complex should be
paramagnetic, which it is – so our hybridization scheme explains the formation of a complex with a
central metal ion and 6 ligands. The d2sp3 hybridization scheme corresponds to the octahedral shape.
Square Planar: metal ions with a d8 electron configuration usually form square planar complexes. In the
[Ni(CN)6]-2 ion, for example, the model proposes that one 3d, one 4s, and two 4p orbitals mix and form
the hybrid orbitals dsp2 which point to the 4 corners of a square. Again, we must make room for the
donated electron pair from the cyanide ligands.
Examining the electron configuration for the Ni+2 ion might make one initially ask how this
hybridization can occur when there are electrons in all of the 3d orbitals. Remember, electrons can and
do move on occasion, and the nickel complex is diamagnetic – which tells us that all the electrons are
paired. A way to explain this, is to have the unpaired electrons of the Ni+2 ion to pair, leaving behind an
empty d orbital which can then be used in the hybridization.
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Tetrahedral: metal ions that have filled d sublevels often form tetrahedral complexes. (called d10 metal
ions). Examining [Zn(OH)4]-2 shows us that when the d orbitals of the metal ion are filled, the next
lowest available (empty) orbitals are the s and p orbitals of the next quantum number – in this case, the
4s and 4 p orbitals. If 4 ligands are going to come in, all three empty orbitals will be hybridized together
to form sp3 hybridized orbitals.
The valence bond depiction of electrons is easy to picture and it rationalizes or explains he bonding and
shape that results in coordination compounds, but it really just treats the orbitals as empty homes to put
electrons pairs into as we need them. As such, it does not explain one of the key features of coordination
compounds – their beautiful color. Also, unless you know whether a compound is para or diamagnetic
in nature, valence bond theory has a difficult time predicting (if we didn’t know that the nickel complex
above was diamagnetic, how would we have known to pair up the electrons??!!??)
Crystal field theory does a much better job explaining color and magnetism. In order to do this, it
highlights the effects on the d-orbitals energies of the metal ion as the ligands come in and bond.
What is C – O – L – O – R
?
Remember that light contains all forms of electromagnetic radiation. As we (humans) tend to focus on
what we see, our depiction of light limits us to the visible region of the EM spectrum – which we call
color. Objects appear to have color because they absorb certain wavelengths and reflect or transmit
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others: an opaque object reflects like while a clear one transmits light. The reflected or transmitted light
enters our eye, is processed by our brain and is then perceived by us to be a certain color. A substance
that absorbs all wavelengths of visible light appears black, a substance that reflects all wavelengths of
visible light appears white.
Each color has a complimentary color that appears opposite on an artist’s color wheel. Thus, the
opposite or complimentary color of red is green. The opposite or complimentary color of
yellow is violet. The opposite or complimentary color of orange is blue. And vice versa.
A object has a particular color for one of two reasons:
It reflects light of that color: thus it absorbs light of all different colors but reflects one,
and that color is sent back to our eye. Thus, if a substance reflects violet but absorbs red,
orange, yellow, green, and blue, the violet is sent into our eye and that is what we see!
It absorbs light of the complimentary color. Thus, if an object absorbs only red, the
complement is green. The remaining mixture of reflected light enters our eyes and is
interpreted as green also.
A familiar example of this is seen in the color of leaves on trees. During spring and summer, the
leaves on deciduous trees appear green. The leaves contain chlorophyll in high concentration
and other compounds called xanthophylls. Chlorophyll strongly absorbs some colors of light
and reflects mostly green back at us – so the leaves appear green. In the fall, the concentration
of chlorophyll decreases and the xanthophylls are now present. The xanthophylls absorb some
colors of light but most strongly reflect back at us the yellow and reds that we see in fall.
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The crystal field theory is used to explain the properties of complexes which result from the
splitting of the d orbitals into d orbitals that lie at a lower energy state than other d orbitals.
Recall there are 5 d orbitals, the dxy, dxz, dyz, dz2, and the dx2-y2. This splitting results when the
ligands approach and bond with the central metal ion. The electrostatic attraction between the
approaching ligands can be either a partial negative charge (as in the neutral NH3 ligand) or
full, as in an anionic ligand like Cl-1. The ligands approach the metal ion along the x, y, and z
axes, which will minimize the overall energy of the system, meaning it would not be good
energetically for all the ligands to approach the metal ion in an assembly-line type process!
In an octahedral complex, the six ligands will move towards the metal from a –y, +y, -x, +x, -z,
and +z type direction (as seen below Figure 23.17). Initially, in the metal ion, all of the d orbitals
are of equal energy. As the ligands approach, they bring with them electrons, which are going
to cause the d electrons to be repelled unequally due to the orientation of the individual d
orbitals (remember the d orbitals are named based on their orientations!). Because the ligands
approach along the x, y, and z axes, their electrons will repel the d orbitals that lie ON those
axes, which are the dx2-y2 and the dz2 orbitals. These orbitals will experience stronger repulsions
and thus will be at a higher energy state than the dxy, dyz, and dxz orbitals which lie in-between
the axes. How much splitting occurs depends on the ligand. Some ligands will cause the
splitting to be small, other ligands will cause a much larger splitting to occur.
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Some terminology: All of the orbitals in the ligand-metal ion complex lie at a higher energy
state than when the metal ion is alone – due to electron electron repulsions. The orbitals that are
sent to the higher energy state are termed the eg orbitals (dz2 and dx2-y2). The remaining 3
orbitals are termed the t2g orbitals (dxz, dyz, dxy). The difference between the higher energy
orbitals and the lower energy orbitals is called the crystal field splitting energy (). Ligands that
cause a large split are termed strong field ligands. Ligands that cause a smaller split are termed
weak field ligands.
Why is splitting important? The splitting, meaning , is directly related to the color and
magnetic properties of the complexes.
Remember that color is just visible light. It is the form of light that we, as humans, are able to
see. Remember that light, be it visible or IR, or UV comes from the transitions of electrons.
When electrons are excited, they will absorb some exact amount of energy (+E), termed
quantized E which will cause the electron to be promoted to a higher energy level. It used to be
that all the d orbitals were of the same energy, but now, due to splitting, we have an energy
discrepancy! The electrons in the lower (t2g) orbitals can be excited (by absorbing E) up to the
higher energy (eg) d orbitals. After some time, the electrons fall back down to their lower
energy state and they release (-E) the energy in the form of light. What kind of light will
depend on the ! If the E is of the right value, frequency, or wavelength, it will be seen as
visible light and we will see color. If the species absorbs light in the visible range, when the
electron falls down, it will then emit light in the visible range and we will see a colored species.
Remember that the color we see will depend on the color(s) of light absorbed by the species.
For example, if a species absorbs green and yellow light –what is left behind? Red, blue and
violet, therefore red and blue make purple so what we will see will be a purple species.
What color(s) of light are absorbed by a species depends on two factors: the oxidation state of
the metal ion and also the type of ligand. As both of these factors will ultimately affect , which
is the splitting of the d orbitals.
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Color is a difficult thing to predict, but we can, using our knowledge of strong field ligands and
weak field ligands, attempt to predict what type of light will be absorbed by a species.
Remember that a strong field ligand causes the d orbitals to split farther apart than the weak
field ligands. Species that are father apart have a larger E between them.
Recall: Ephoton = hυ = hc/
As the strong field ligands have a larger E then, since E and  are inversely related, that would
correspond to a shorter wavelength absorbed/emitted– so light in the blue to indigo to violet
range perhaps. Given that the weak field ligands have a smaller E, and since E and  are
inversely related, that would correspond to a larger wavelength absorbed/emitted – so perhaps
light in the red, orange or yellow range. Of course, the middle range, yellow, green, blue and
how large or small the wavelength absorbed or emitted is what is difficult to predict.
When the d orbitals split, the magnetic properties will be affected as well. Recall that whether
or not a species is magnetic or not has to do with whether or not the electrons are all paired, or if
there are unpaired electrons present. If the electrons are all paired, the species is not attracted to
a magnetic field (in fact it is slightly repulsed by one). If there are unpaired electrons, the
species will be attracted to a magnetic field. The splitting is going to affect the number of
unpaired electrons. Recall Hund’s Rule: electrons fill empty orbitals singly before they pair up.
However, when we were studying Hund’s rule, all of the d orbitals were of equal energy.
When the splitting occurs, the orbitals are of different energies. The difference in energy will
help us decide how we will fill the d orbitals. There are two factor to consider, the E due to the
crystal field splitting () and the energy that must be overcome in order to pair the electrons.
For strong field ligands, the E between the low energy orbitals and the higher energy orbitals
is too great. Just like we completely fill up the 2 p and pair up electrons before we moved on to
the higher energy 3s orbital we will do the same now that the d orbitals have split in energy –
we will completely fill the lower energy t2g orbitals before we start putting electrons into the
higher eg orbitals.
Think about the statements above. Yes, electrons like to exist singly (as opposed to paired, since
pairing electrons involves overcoming electron- electron repulsions. BUT – when examining
electron configurations, we do not say 1s1 2s1 2p3 3s1 for a species with 6 electrons on order to
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make sure that we do not pair up any electrons. Why? We fill the lower energy orbitals
completely because pairing (electron electron repulsions) are still a lower energy state than
filling higher energy orbitals. So the configuration for a species with 6 electrons would be
1s22s22p2. We pair up the electrons in the 2s orbital – which a favored meaning lower energy
state than just keeping the electrons as singles.
We will follow the same mantra now that the d orbitals have split due to a strong field ligand.
We will fill the t2g (the lower energy orbitals) before we send electrons up to the eg orbitals as
pairing is the lower energy state. Epairing < . This means, that compared to the free ion (as the d
orbitals are of the same E), the complex has fewer unpaired electrons, and they may lose their
magnetic properties completely if all the electrons end up paired. This is called a low-spin
complex (remember ms is called the spin quantum number and thus, we have fewer electrons
with spin that can be affected by a magnetic field)
Conversely, weak field ligands do not cause a large energy discrepancy between the different d
orbitals. This means that Epairing > . The complex finds the lower energy state to be promoting
the electrons to the eg orbitals before pairing. Pairing is thus, unfavored until absolutely
necessary. In this case, we follow Hund’s rule to the letter (same as before) so the number of
unpaired electrons in the complex will not differ from the free metal ion. This is called a highspin complex.
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Crystal Field Splitting in Tetrahedral and Square Planar Complexes
For tetrahedral complexes, the ligands will not approach in-line with the dz2 and dx2-y2 orbitals.
They also will not directly approach in-line with the dxz, dyz, or dxy orbitals. There is no direct
overlap, but there is still repulsion. The lower energy orbitals will actually be the dx2-y2 and dz2
orbitals. Examining the figure below should show you that the ligands will have a bit more (not
much more, but a little more) interaction/repulsion with the orbitals in the planes of the axes –
and thus the dxy, dyz, dxz orbitals will be the higher energy orbitals. This is OPPOSITE the
octahedral situation. However, the  (crystal field splitting) is low. In fact, only high spin
complexes are seen for tetrahedral complexes. This means that the complex will keep the
electrons unpaired as long as possible (just like the free metal ion).
For square planar complexes, the ligands approach on the y and x axes and there is NO
repulsion along the z axes. The dxz and dyz orbitals are going to be the lowest in energy as there
is no overlap. Remember that the dz2 has a little donut, so while it mostly lies on the z axes, it
too is low in energy, but not as low as the d orbitals that have no interaction with the incoming
ligands. Since the ligands are approaching along the x and y axes, the orbitals that involve the x
and y axes, the dxy and the dx2-y2 are going to experience the most repulsion and be the highest
in energy. However, remember that the dxy lies in the plane and thus there is not direct overlap.
It is the dx2-y2 that lie on the x and y axes that will experience the most repulsion and therefore
be the higher in energy. The  in the square planar complexes is greater than Epairing. Thus, the
electrons will pair all up the split orbitals. The only unpairing will occur at the first level, in the
dyz and dxz orbitals. Electrons 1 and 2 will go in unpaired, after that, they will pair as they go.
This means that square planar complexes are low spin – pairing electrons is favored due to the
large  values.
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