A covalent bond is a chemical bond formed by the sharing of a
pair of electrons between two atoms.
The Lewis structure of a covalent compound or polyatomic ion shows how the valence electrons are arranged among the atoms in the molecule to show the connectivity of the atoms.
Instead of using two dots to indicate the two electrons that comprise the covalent bond, a line is substituted for the two dots that represent the two electrons.
Below is shown the Lewis structure for water. Two hydrogens (H) are separately covalently bonded to the central oxygen (O) atom. The bonding electrons are indicated by the dashes between the oxygen (O) and each hydrogen (H) and the other two pairs of electrons that constitute oxygens octet, are called non-bonding electrons as they are not involved in a covalent bond.
Rules for getting Lewis Structures
1. Determine whether the compound is covalent or ionic. If covalent, treat the entire molecule. If ionic, treat each ion separately. Compounds of low electronegativity metals with high electronegativity nonmetals (DEN 1.6) are ionic as are compounds of metals with polyatomic anions. For a monoatomic ion, the electronic configuration of the ion represents the correct Lewis structure. For compounds containing complex ions, you must learn to recognize the formulas of cations and anions.
2. Determine the total number of valence electrons available to the molecule or ion by:
(a) summing the valence electrons of all the atoms in the unit and
(b) adding one electron for each net negative charge or subtracting one electron for each net positive charge. Then divide the total number of available electrons by 2 to obtain the number of electron pairs (E.P.) available.
3. Organize the atoms so there is a central atom (usually the least electronegative) surrounded by ligand (outer) atoms. Hydrogen is never the central atom.
4. Determine a provisional electron distribution by arranging the electron pairs (E.P.) in the following manner until all available pairs have been distributed:
a) One pair between the central atom and each ligand atom.
b) Three more pairs on each outer atom (except hydrogen, which has no additional pairs), yielding 4 E.P. (i.e., an octet) around each ligand atom when the bonding pair is included in the count.
c) Remaining electron pairs (if any) on the central atom.
Types of Electrons Pairs
Bond pair: electron pair shared between two atoms.
Lone pair: electron pair found on a single atom.
Molecules obeying the octet rule.
In many molecules, each atom (except hydrogen) is surrounded by eight bonding or lone-pair electrons. There is a special stability associated with this configuration. Examples are water, ammonia and methane.
The ground state (g.s.) configuration of N has three unpaired electrons.
Each hydrogen atom has one. No rearrangement is necessary to make the three
N-H bonds. Be sure to mark the lone pair on the Lewis diagram.
The ground state of carbon has only two unpaired electrons, but it is
necessary to make four bonds to the hydrogens. The solution, in this case,
is to promote a 2s electron to the empty p orbital. Then four bonds can
SiH4 has the same Lewis Structure as CH4 since Si and C are in the same group.
Lewis structure of CCl4
i) Valence electrons: 4 (from one carbon)+ 28 (7 from each chlorine) = 32 = 16 electron pairs
ii) Central atom
iii) Octet on C is already complete iv). Count electrons
4 x 3 lone pairs = 12 pairs
4 x 4 bond pairs = 4 pairs
16 electron pairs
Lewis Structure of CCl4. CF4 and CBr4 have similar structures.
Lewis Structure of PCl3
i) Valence electrons: 5 + 3 x 7 = 26 (13 pairs)
ii) Central atom is P
iii) Connect to terminal atoms to central atom
vi) Give octet to P and give octets to Cl
v) Count electron pairs:
3 bond pairs = 3 pairs
1 + 3 x 3 = 10 lone pairs = 10 pairs
NF3 and NCl3 have similar Lewis Structures.
Molecules with multiple bonds:
Examples: Consider carbon dioxide CO2
|The central atom is still electron deficient, so share another pair.|
The bond length is the distance between those two atoms. The greater the number of electrons between two atoms, the closer the atoms can be brought towards one another, and the shorter the bond.
The BO is an indication of the bond length, the greater the bond order, the shorter the bond.
Formal charge is an accounting procedure. It allows chemists to determine the location of charge in a molecule as well as compare how good a Lewis structure might be. The formula for calculating formal charge is shown below:
Consider the molecule H2CO2. There are two possible
Lewis structures for this molecule. Each has the same number of bonds.
We can determine which is better by determining which has the least formal
charge. It takes energy to get a separation of charge in the molecule (as
indicated by the formal charge) so the structure with the least formal
charge should be lower in energy and thereby be the better Lewis structure.
There are two possible Lewis structures for this molecule. Each has the
same number of bonds. We can determine which is better by determining which
has the least formal charge. It takes energy to get a separation of charge
in the molecule (as indicated by the formal charge) so the structure with
the least formal charge should be lower in energy and thereby be the better
Lewis structure. The two possible Lewis structures are shown below. They
are connected by a double headed arrow and placed in brackets. The non-zero
formal charge on any atoms in the molecule have been written near the atom.
Electron deficient molecules: atoms have less than
octet of electrons
Do not have enough electrons to satisfy the the octet rule. Simple examples are beryllium hydride and boron trichloride:
Lewis Structure of BeCl2
Beryllium, an element of the second column of the periodic classification, has for electronical
configuration : [He] 2s2. It has two single bonds with two atoms in BeCl2.
Berillium do not have a octet, therfore, BeCl2 is an electron deficient compound. Be has only four electorns. BeCl2 violate octet rule.
Lewis Structure of BCl3
The boron must be in a suitable valence state to bind to the three chlorines. In the molecule the boron is associated with only six electrons. Much of the chemistry of this molecule and ones like it is connected with the resulting strong elecrophilic nature. Other examples include the boron hydrides such as diborane and alkyl-lithium, beryllium and aluminum compounds, which will be described later.
Lewis Structure of BF3
Lewis Structure of AlCl3
Similarly AlCl3 has the Lewis Structure:
Total of 17 valence electrons are present.
24 electrons would be needed to satisfy the valencies independently.
24-17 = 7. The odd number of electrons is a signal that there will be a place where we would normally expect to find another electron. When you have some experience with these situations, you probably will have no problem in assigning the Lewis structure directly. If this is a problem for you, the easiest way to treat these is to assume that we have one extra valence electron to begin with and remove that electron at the end. In this case, it means drawing the Lewis diagram for NO2-. For NO2- the answer to question 1 is 18 valence electrons and the answer to 3 become 6 electrons, or 3 bonds.
We draw one N-O single bond and one N=O double bond in order to achieve the correct number of bonds.
For NO2- there will be 12 electrons left to make into nonbonding pairs that can be assigned as shown in the picture below. This will give octets at each atom. To derive the Lewis structure for NO2, the neutral molecule, we must then remove one electron from NO2-. The choice preferred is to remove the electron from the lone pair that resides on the least electronegative atom. Since N is less electronegative than O, we will take the electron from the N lone pair.
AsCl5 have similar structure. PCl5 and AsCl5 violates octet rule having 10 electrons on P and As.
There are a number of compounds and polyatomic ions that cannot be written using one single
structure. This was known even back to the early beginnings of structural chemistry in the
mid-1850s. These substances must be described in terms of "intermediate" structures, possessing
non-integral bonds such as one and one-half bonds or one and one-third bonds.
For example, certain molecules it is necessary to draw several Lewis structures (resonance structures) to adequately describe the structures of the molecules: Examples are CO32-, NO3-, NO2-
Resonance Lewis Structures of CO32-:
Refinements to molecular geometry. The details of the predicted geometry can be further refined by considering what the objects are more carefully. Is another atom bound by a single bond or higher bond order? Is the bond polar? What about the difference between a lone pair and a bond pair? The table below shows how these questions and others can be answered.
|Largest Repulsion||Lone Pairs
Multiple or shorter bonds
or bonds polarized towards
the centre atom
|Medium Repulsion||Normal single bonds|
|Smallest Repulsion||Longer single bonds and
bonds polarized away from
the central atom
Polarity of Molecules
The degree of polarity of a molecule is described by its dipole moment, m = Q * r
Dipole moments are generally reported in Debye units
1 debye = 3.33 x 10-30 coulomb meters (C m)
Example: H-Cl a covalent polar compound
The actual dipole of H-Cl is 1.08 debye. The reason for this
is that the compound is covalent and not ionic, thus the charges of the
dipole are less that +1, and -1 (values expected for a fully ionic compound)
Although the bond length is increasing, the dipole is decreasing as you move down the halogen group. The electronegativity decreases as we move down the group. Thus, the greater influence is the electronegativity of the two atoms (which influences the charge at the ends of the dipole).
The Polarity of Polyatomic Molecules
ABn molecules and non-polar geometries
For ABn molecules, where the central atom A is surrounded by identical atoms for B, there a certain molecular geometries which result in no effective dipole, regardless of how polar the individual bonds may be. These geometries are:
Valence Bond Theory: Assumes that covalent bonds are formed when atomic orbitals (with electrons) on different atoms overlap and the electrons are shared. The atomic orbitals can be the original atomic orbitals of the atoms, but often the geometry of these orbitals is such that effective overlap cannot occur in the known geometry of the molecule. Under these circumstances, the atomic orbitals on an atom can reconfigure themselves into a different configuration, and the reconfigured orbitals are said to be hybridized. This theory is similar to molecular orbital theory which is concerned with the formation of covalent bonds. The difffrence being molecular orbitals are formed by empty atomic orbitals and later electrons are filled ino them.
The Valence Electrons and Valence Shell OrbitalsThe valence shell orbitals of an atom are taken to be the ns, np and nd orbitals where n is the number of the period of the element. The valence electrons are the electrons contained in these orbitals.
Linear sp hybrids. These are composed of the valence
shell s-orbital and one of the three p-orbitals. The other two p-orbitals
remain unhybridized and may hold lone pairs or participate in p-bonding.
The two equivalent sp hybrid orbitals pointat 180o to each other
and their formation is depicted graphically and mathematically below:
The diagram below shows the 2s and 2px orbitals superimposed in the same space before hybridization. The colours represent the relative signs of y. The signs/colors for the 2p orbital would be reversed by the negative sign in first equation which follows the diagram:
y1 = 1/Ö3ys + 1/Ö6ypx + 1/Ö2ypy
y2 = 1/Ö3ys + 1/Ö6ypx - 1/Ö2ypy
y3 = 1/Ö3ys - 2/Ö6ypx
Below, the hybrids corresponding to the three equations above are shown separately (in order from top to bottom):
In the diagrams below, the trio of hybrids are superimposed in the same
space as they should be. Two diagrams are shown: On the left the antinodes
are shown transparent. The small knob of each hybrid is buried deeply.
The right hand diagram shows the equivalent hybrid enclosure surfaces and
hides the internal structure. Note the "three-fold" symmetry:
sp2 hybridization was postulated for the possibility of an s orbital hybridizing with two p orbitals:
y1 = 1/Ö4ys
y2 = 1/Ö4ys - 1/Ö4ypx - 1/Ö4ypy + 1/Ö4ypz
y3 = 1/Ö4ys + 1/Ö4ypx - 1/Ö4ypy - 1/Ö4ypz
y4 = 1/Ö4ys - 1/Ö4ypx + 1/Ö4ypy -1/Ö4ypz
Below, the hybrids corresponding to the four equations above are shown
separately (in order, left to right and top to bottom):
In the diagrams below, the quartet of hybrids are superimposed in the
same space as they should be. Two diagrams are shown: On the left the antinodes
are shown transparent. Again, the small knob of each hybrid is buried deeply.
The right hand diagram shows the equivalent hybrid enclosure surfaces and
hides the internal structure. Note the tetrahedral symmetry of the group:
Notes on the equations: Each component atomic orbital must be "shared out" completely between the hybrids which leads to the condition that the sum of the squares of the coefficients for a particular atomic orbital add to one. In additon, individual hybrid orbitals must be "normalized", that is, the probability of an electron occupying one of them somewhere must be 1. This leads to the condition that the sum of the squares of the coefficients in one of the formulae must also add to 1.
In order to produce angles which differ from the ideal values,
the equations are modified. Increasing the ratio of p to s character in
any subset will decrease the appropriate inter-hybrid angles.
These sp3-hybridized orbitals were centered about the nucleus, pointing toward the corners of a regular tetrahedron.
Trigonal bipyramidal sp3d hybrids. These are
composed of the valence s-orbital, all the p-orbitals and one of the d-orbitals
(normally, the dz2 or one of the two in the xy plane
i.e. dxy or dx2-y2).
The choice depends on the atoms involved so it is not possible to write
down a unique set of equations like those above.) The remaining d-orbitals
could be involved in p-bonding.
sp3d hybrids in PF5
Octahedral sp3d2 hybrids. These are composed of the valence s-orbital, the three p-orbitals and two d-orbitals (normally the dz2 and one of the two in the xy plane.) The remaining d-orbitals can participate in p-bonding.
sp3d2 hybrids in SF6
Linear Molecular Geometry
Trigonal Planar Molecular Geometry
Central atom sp2 hybridized
Bent Molecular Geometry
Tetrahedral Molecular Geometry
single and double bonds
Trigonal Bipyramidal Molecular Geometry
single bonds only
Single and double bonds
Seesaw Molecular Geometry
One Lone Pair in an equatorial hybrid
Central atom sp3d hybridized
One Lone Pair in an equatorial hybrid
Central atom sp3 hybridized
Doen met eers hibridisasie en dan die mol geometrie
|Linear||AX2 sp, or AX2E3 sp3d|
3.5 Homonuclear diatomic molecules
3.4 The hydrogen molecule
A bond between two atoms is formed when a pair of electrons is shared by two overlapping
orbitals, according to the VB theory. For example, in a hydrogen molecule, the two 1s orbitals from
each H atoms overlap and share electrons.
3.6 Polyatomic molecules
In all of the examples given above used as examples for Lewis structures, it was implied that the bonds were formed from two of the basic atomic orbitals. For example, the bonding in ammonia was shown as involving three (mutually perpendicular) 2p orbitals on nitrogen and the 1s orbitals of each hydrogen. (For any one of the bonds yA could be a 2p orbital and yB a 1s orbital in the valence bond theory notes mentioned above.) This simple treatment is clearly not adequate because it predicts a H­N­H bond angles of 90o: the observed angle is 107o.
The method of handling this in valence bond theory is via the mixing of the basic atomic orbitals mathematically to form hybrid orbitals pointing in the required directions. This mixing is achieved mathematically by taking linear combinations of the atomic orbitals as indicated below: The examples given above are set out again using one more step to include geometry and thus the hybridization of the central atom. Where appropriate the mechanism for p-bonding is indicated. Comments on distortions from ideal angles are added.
Ammonia - NH3
The nitrogen is bound to three hydrogen atoms. In addition, there is a lone-pair on N, therefore 4-coordination i.e. tetrahedral geometry (with sp3 hybrids) is predicted. (The molecule is sometimes refered to as pyramidal if the lone-pair is not considered part of the geometry.)
The lone-pair demands more space than the bond pairs so that the H-N-H angles will be less than the ideal 109.5o. The actual experimental value is 107.3o.
The appropriate hybridization is sp3 (but with the lone pair in a hybrid orbital having somewhat greater s character than the other three which are used for the bonds, and which, correspondingly, have slightly higher p character.)
Some related molecules are NF3 with a bond angle of
102.5o and PH3 with a bond angle of 93.5o.
This latter molecule uses nearly pure 3p orbitals to bind the hydrogens
and a nearly pure 3s orbital to house the lone pair. This optimizes the
overlap with the H 1s orbitals. Other isoelectronic/isostructural molecule/ions
are H3O+ and CH3-.
Methane - CH4
With four hydrogens bound to carbon and no lone pairs, the structure is predicted to be tetrahedral, and the appropriate hybridization would be sp3.
Some isostructural molecule/ions are BF4-
Nitrosyl Chloride - NOCl
The nitrogen has three "things" attached: the lone pair, the chlorine and the oxygen . N.B. The oxygen only counts as one "thing" despite the double bond.) Therefore, the shape is based on trigonal, and disregarding the lone pair, might be called bent or angular. Like COCl2, the appropriate hybridization, sp2 leaves one 2p orbital on nitrogen to form the p bond. This is depicted in the lower diagram on the right.
The hybrid orbitals would all be inequivalent in such a way as
to reduce the Cl-N=O angle to make room for the lone-pair and to allow
for the differences between bonding orbital sizes on Cl and O.
Nitrate - NO3-
The predicted geometry is trigonal because there are three oxygens bound
to nitrogen, and no lone-pairs. The appropriate hybridization is sp2.
The single Lewis structure drawn shows the unhybridized p-orbital of nitrogen
interacting with a p-orbital on oxygen to form the p-bond.
Molecular orbital theory, see below, treats this situation differently
to avoid the need to draw three canonical structures to account for the
observed geometry where all angles are 120o and the bond lengths
Boron trichoride - BCl3
The approprite hybridization is sp2 since the boron is attached to three "things". In certain cases (for example BF3), p-bonding using the empty boron p-orbital can be invoked. To do this, an electron must be moved from the more electronegative terminal atom to boron. This runs contrary to what one expects from the electronegativities so the exact extent to which this occurs is the subject of some controversy.
Chlorine Trifluoride - ClF3
Because the chlorine carries the three fluorines and two lone pairs, the geometry will be trigonal bipyramidal and the hybridization will be sp3d. Three isomers are possible.
It is highly unfavorable for the lone pairs to be 90o apart which rules one of the three. In order to rationalize the obervation that the isomer with both lone pairs in equatorial positions is the observed form, it is necessary to count the number of 90o bond pair - lone pair interactions. The observed isomer (in the box) has four such angles and the isomer with the lone pairs in axial positions has six which is (presumably) less favorable. The molecule might be called T-shaped based on the atomic positions only. The appropriate hybridization would be sp3d where the d orbital will have a greater contribution to the axial hybrids than the equatorial ones.
A few more examples are:
Iodine pentafluoride - IF5
The five unpaired electrons needed on iodine to bind the fluorines can only be obtained (in the valence bond approximation) by invoking the use of the 5d orbitals. After the bonding with fluorine, the iodine is left with one lone pair, so there are six "things" in all. The geometry will be based on an octahedron, and might be referred to as (square) pyramidal if the lone pair is not counted. The F-I-F angles (basal to apical) will be less than 90o due to the steric influence of the lone pair.
The appropriate hybridization would be sp3d2. By using a molecular orbital approach involving 3-centre 4-electron bonds the use of the d-orbital in understanding the bonding can be avoided, but the predictive value of the valence bond VSEPR approach would be lost.
Some isolectronic/isostructural species are: TeCl5- and XeF5+.
Carbonyl Chloride (Phosgene) - COCl2
The two chlorines and the oxygen require four unpaired electrons for bonding on the carbon, ie the usual tetravalent excited state. Since there are three "things" on the carbon the predicted shape is trigonal, and the appropriate hybridization is sp2. This leaves one unhybridized 2p-orbital on the carbon, and another on the oxygen, to form the p component of the double bond.
It is difficult to predict distortions from the ideal angles because on the one hand the chlorines are larger than the oxygen which would tend to widen the Cl-C-Cl angle, but on the other hamd there is a double bond involving the oxygen which would tend to widen the Cl-C=O angles.
Thionyl Chloride - SOC l2
Unless a canonical structure with charge separation (S+ and O-) is constructed, the sulphur must form four bonds, one to each of the chlorines and the double bond to oxygen. It is necessary to invoke a valence state using one of the 3d orbitals to do this. Sulphur is attached to four "things": the lone pair, two chlorines and the oxygen, so the shape is based on the tetrahedral geometry, and the structure might be called pyramidal if the lone pair is disregarded.
The hybridization on sulphur for the sigma bonds is sp3, which leaves the d electron available to form the p bond by overlap with one of the oxygen 2p orbitals. This is depicted in the lower diagram on the right. (For clarity, the lone pair is not shown in this diagram.)
Sulphate - SO42-
If the negative charge on the ion is distributed on two of the oxygen atoms there are six unpaired electrons needed on the sulphur. Promotion of two electrons to the 3d orbitals achieves this. The four oxygens attached to sulphur cause it to adopt a tetrahedral geometry. The appropriate hybridization to account for the s bonds is sp3, leaving the two d orbitals for use forming the p bonds. In a full molecular orbital calculation, if the d orbitals really played a part, they would also be hybridized to point in the appropriate directions. The situation is too complicated to attempt to draw meaningful diagrams.
The canonical structures shown at the left correspond to the bonding model described above. It is possible to conceive of other canonical forms where three or four of the oxygen atoms carry a negative charge and the sulphur one or two positive charges. The second of these canonical forms would not require d-orbital involvement.
Transition Metal Complexes
It uses "outer orbital" and "inner orbital" hybridization to explain the formation of compounds.
Both fragments are 3 electrons short of making maximum use of their 4 valence orbitals (s and 3´ p) by achieving an octet. The clusters C4H4 or (CH)4 (known as tetrahedrane) and P4 have the same number of cluster electrons and therefore adopt the same shape:
The organometallic fragment Co(CO)3 has 15 electrons:
Co has 9 valence electrons and each CO donates 2e to the metal
The fragment is 3 electrons short of the stable electron count for an organometallic fragment of eighteen Eighteen electrons corresponds to making maximum use of the 9 valence orbitals of the transition metal (s + 3´ p and 5´ d). It can achieve 18 electrons by sharing 3 electrons to make 3 covalent bonds. The Co(CO)3 fragment thus has the same requirements as the C-H unit and the P atom. The cluster [Co(CO)3]4 thus also has the same tetrahedral shape as that shown by C4H4 and P4. It is also possible to swap the three fragments around:
B and C are more conventionally thought of as (m3-P)[CoCO)3]3 and (h3-C3H3)Co(CO)3 respectively but this hides the structural and electronic links between all these clusters.
The similarity between these fragments and between other groups of fragments has been investigated in detail by Roald Hoffmann. Fragments are deemed to be isolobal if
"the number, symmetry properties, approximate energy and shape of the
frontier orbitals and the number of electrons in them are similar"
(R. Hoffmann, Angew. Chem. Int. Ed. Engl., 21, 711, 1982)
The isolobal relationship is symbolized by a double-headed arrow with a tear-drop,
The isolobal analogy relates the orbitals and bonding in inorganic, organometallic and cluster chemistry to that in organic and main group chemistry. The utility of the isolobal analogy is that one should be able to replace replace a (transition metal) MLn fragment in a molecule with the isolobal (main group) AHn fragment, and vice versa, to produce new molecules with very similar bonding.
The transition metal fragments are generated in an analogous way. For example, from the starting point of CrL6, 2, where L is a two electron donor such as CO, (or any molecule obeying the eighteen electron rule such as 3), the fragments 2a, 2b and 2c are generated by successive homolytic cleavage of M–L bonds on one octahedral face. As L is a two-electron donor, homolytic cleavage of CrL6 gives CrL5– and L+. To remove the charge, the metal is then replaced by Mn (the element one to the right in the 3d series):
As long as the electron count is maintained or consistent changes are
made, the metal or main group element can be substituted. Thus,
It should be noted that the main use of the isolobal analogy is in generating alternative fragments in molecules. The geometry of the fragment in a molecule not as an isolated species is important. Thus, CH3 and BH3 are considered as pyramidal (not planar) species and Cr(CO)5 is considered as a square-based pyramid (not trigonal bipyramid).
The isolobal relationship of CH2, with Fe(CO)4 generates the compounds illustrated below. The transition metal molecule has been drawn as a metallocyclopropane. A more common description is to consider it as an ethene complexes, (h2–C2H4)Fe(CO)4. The isolobal relationship shows that cyclopropane itself can be consider as (h2–C2H4)CH2.
Other known compounds generated using this relationship include:
Here are some common isolobal fragments
Molecular orbital theory
3.7 An introduction to the theory of Linear Combination of Atomic
A theory which treats bonding as an over lapping of ligand orbitals with those of the central atom.
By summing the original wavefunctions for the bonding orbitals in constituent species, "hybrid" molecular orbitals of the compound can be generated. These new orbitals have an intermediate character between the original ,, and orbitals (if available) in the outer energy level, and produce additional bond sites. The hybridization is named on the basis of the orbitals involved, and the hybrid wavefunction is the (renormalized) sum of the individual wavefunctions, where each addition may be with an arbitrary sign. The composite wavefunctions with differing signs are orthogonal, since
where the wavefunctions on the right are the solutions to Schrödinger's equation, and the normalization constants are needed so that the hybrid wavefunction is normalized. has the electron density is greatest between the two nuclei. It will therefore bind the nuclei together, and is called a bonding molecular orbital. has the electron density greatest on the sides of the nuclei. It will therefore pull the nuclei apart, and is called an antibonding molecular orbital. In some instances, a nonbonding molecular orbital may be generated for which the electron density is uniformly distributed between and on the sides of the nuclei. A measure of the stability of a compound based on the occupancy of its molecular orbitals is given by the body order.
More complicated bonding interactions will involve ,, and orbitals. For a homonuclear diatomic compound with hybrid orbitals constructed from ,, and p orbitals, the molecular orbital have the following form.
There are 12 electrons in the valence shell, so the levels are filled through the nonbonding orbitals. The compound is therefore stable, with a bond order of 4. For an even more complicated example, consider benzene. For certain compounds, electrons are delocalized. Such compounds have an extremely large number of molecular orbitals. The result, as the number of levels goes to infinity, is a band of bonding orbitals, and band of antibonding orbitals (known as the conduction band, since free electrons will exist here), possibly overlapping or possibly separated by a gap. In metals, the levels overlap, and the bonding orbitals are completely filled. In semiconductors, the levels are separated by a small "forbidden zone." The addition of a small amount of energy will therefore remove an electron from the filled bonding orbital, through the forbidden zone, and into the conduction band.
y1 - y4 represent solutions of increasing energy. In three dimensions, the equation determines the energy and defines the spatial distribution of each electron. Solutions of the wave equations in three-dimensions allows calculation of the "shape" of each orbital. The first five solutions of the wave equation for an electron associated with a proton can be shown in the figure below:
In the hydrogen atom, the 1s atomic orbital has the lowest energy,
while the remainder (2s, 2px, 2py and 2pz)
are of equal energy (ie.degenerate), but for all other atoms,
the 2s atomic orbital is of lower enegry than the 2px,
2py and 2pz orbitals, which are degenerate.
In atoms, electrons occupy atomic orbitals, but in molecules they occupy similar molecular orbitals which surround the molecule. The simplest molecule is hydrogen, which can be considered to be made up of two seperate protons and electrons. There are two molecular orbitals for hydrogen, the lower energy orbital has its greater electron density between the two nuclei. This is the bonding molecular orbital - and is of lower energy than the two 1s atomic orbitals of hydrogen atoms making this orbital more stable than two seperated atomic hydrogen orbitals. The upper molecular orbital has a node in the electronic wave function and the electron density is low between the two positively charged nuclei. The energy of the upper orbital is greater than that of the 1s atomic orbital, and such an orbital is called an antibonding molecular orbital.
Normally, the two electrons in hydrogen occupy the bonding molecular orbital, with anti-parallel spins. If molecular hydrogen is irradiated by ultra-violet (UV) light, the molecule may absorb the energy, and promote one electron into its antibonding orbital (s*), and the atoms will seperate. The energy levels in a hydrogen molecule can be represented in a diagram - showing how the two 1s atomic orbitals combine to form two molecular orbitals, one bonding (s) and one antibonding (s*). This is shown below - by clicking upon either the s or s* molecular orbital in the diagram - it will show graphically in a window to the right:
This molecule has twelve electrons, two more than nitrogen - and these extra two are placed in a pair of degenerate pg orbitals. The atomic orbitals combine to produce the following molecular orbital diagram:
3.9 Heteronuclear diatomic molecules
If the molecule is heteronuclear, the parent atomic orbitals will have different energy levels. The more easily ionized (less electronegative) atom will have the atomic orbital level closer to E = 0 in the arrangement depicted below:
The bonding molecular orbital has an energy and a wave function which
approximates the more electronegative atom. The antibonding molecular orbital
will have an energy and wavefunction which resembles that of the less electronegative
atom. As an example, consider the molecule hydrogen chloride. The hydrogen
1s orbital (one electron) would be f1
and the chlorine 3p orbital that bonds with it (one electron) will be f2.
The molecular orbital y will look very much
like the chlorine 3p orbital and will end up holding both electrons,
while the y* orbital will look like the original
H 1s orbital and will end up empty. Thus, the molecular orbital theory
correctly represents Hd+Cld-.
(Note that the above is a rather treatment because it ignores the possible
involvement of the 3s orbital of Cl.)
A simple diatomic molecule is Hydrogen fluoride. There are eight valence electrons which occupy four molecular orbitals. The two highest energy MO's are degenerate, are p-type and have no electron density associated with the hydrogen atom, ie. they are Non-Bonding Orbitals (NBO) and in Lewis Theory are represented as two "Lone Pairs". Another important difference between Hydrogen Fluoride and previous molecules is that the electron density is not equally distributed about the molecule. There is a much greater electron density around the fluorine atom. This is because fluorine is an exremely electronegative element, and in each bonding molecular orbital, fluorine will take a greater share of the electron density.
For the energy diagram and pictorial view of the orbitals - please see below:
3.10 Bond properties in the molecular orbital formalism
Bond Lengths and Covalent Radii There are several points to be made here:
Bond lengths are derived from the sums of covalent radii. The covalent radi can be obtained in the first instance by taking half the length of a homonuclear bond:
The single-bond covalent radius of an atom is influenced by the type of hybrids that it is using. The more p-character in the hybrid, the larger will be the radius:
In the water molecule the highest occupied orbital, (1b1) is non-bonding and highly localized on the oxygen atom, similar to the non-bonding orbitals of hydrogen fluoride. The next lowest orbital (2a1) can be thought of as a non-bonding orbital, as it has a lobe pointing away from the two hydrogens. From the lower energy bonding orbitals, it is possible to see that oxygen also takes more than its "fair share" of the total electron density.
The atomic energy levels are shown under Be and 2H in the figure above. Note the hydrogen orbitals are shown to have a lower energy than the beryllium orbitals because the non-metal, hydrogen, is more electronegative (harder to ionize than the metal, Be.
There are two linear combinations of the two hydrogen 1s orbitals:
y1 = 1/Ö2(y1sHa + y1sHb) and y2 = 1/Ö2(y1sHa - y1sHb)
The beryllium 2s orbital is of the correct symmetry to form combinations
with y1 and the Be 2p orbital directed
along the internuclear axis (say z) is of the correct symetry to form combinations
s1 = 1/Ö2(y1 + y2sBe) and s*3 = 1/Ö2(y1 - y2sBe)
s2 = 1/Ö2(y2 + y2pzBe) and s*4 = 1/Ö2(y2 - y2pzBe)
The beryllium 2px and 2py orbitals do not have matching symmetry ligand combinations.
An Electron Deficient Molecule witn Bridging Hydrogen - Diborane,
Diborane has the structure shown on the right. At first sight, it seems to have 8 bonds, but between the two boron atoms and the 6 hydrogen atoms, there are only 12 electrons - enough to make only 6 "conventional" 2-centre - 2-electron bonds.
The figure below shows how this situation is handled using molecular orbital theory:
The hydrogen 1s orbital can be combined with y1 yielding a bonding and an antibonding combination. The other ligand combination, y2 is not used.
s = 1/Ö2(y1
and s* = 1/Ö2(y1
y1 = 1/Ö2(y2pFa + y2pFb) and y2 = 1/Ö2(y2pFa - y2pFb)
The diagram below shows the combinations graphically.
The diagram below shows the energy levels in the axial orbital system
only. There are two electrons in the bonding orbital (s)
which leads to a bond order of ½ in each phosphorus - fluorine
"connection", since the two non-bonding electrons (in y2)
do not contribute. Therefore, the axial bonds are expected to be weaker
than the equatorial bonds. This is supported by the experimental evidence.
3.13 Molecular shape in terms of molecular orbitals
The Delocalized Approach to Bonding: Molecular Orbital Theory
In molecular orbital theory, molecular orbital wavefunctions are constructed by taking linear combinations of atomic orbitals. If there are only two atoms involved, this means the sum and the difference of the atomic orbital wavefunctions. If there are more than two atoms involved, the combinations are formed in a more complicated way, and usually the symmetry properties of the molecular or molecular ion are used to simplify the problem. (Of course, this does not simplify much if the mathematical theory which covers symmetry (group theory) has not been covered!) The first part deals with diatomic species. This is followed by several more complicated sample systems.
Molecules and Ions with double bonds
In molecules where the number of bonding electron pairs exceeds the number of unions between atoms, the extra electrons occupy higher energy molecular orbitals than the orbitals found in molecules where the number of bonding electron pairs equals the number of unions between atoms. These are double bonds, and the orbitals have a nodal plane containig the atoms sharing these p-type orbitals.
The simplest alkene is ethene. Its chemistry is dominated by two "frontier orbitals", that is the Highest Occupied Molecular Orbital (HOMO) and the Lowest Unoccupied Molecular Orbital (LUMO). For the ethene orbital energy diagram these are shown as pCC for the HOMO, and p*CC for the LUMO.
An important property of the ethene molecule, and alkenes in general is the existence of a high barrier to rotation about the C=C which tends to hold the molecule flat.
For the energy diagram and pictorial view of the orbitals - please see below:
A Trigonal-planar molecule/ion - CO32-
(or NO3- or BF3 which are isoelectronic)
This is a case where the s-bonding is usually handled with valence bond theory. If the three-fold axis of the molecule/ion is considered the z axis, the s-bond framework involves hybridizing the carbon 2s 2px and 2py orbitals (sp2) and using them to attach the oxygens by a 2p orbital lying in the molcular plane. This accounts for 6 of the valence electrons.
The p-bonding molecular orbitals are formed from the carbon and three oxygen 2pz orbitals and there will be six more electrons to accomodate.
The resulting molecular orbitals are shown under CO32-
on the energy level diagram below.
The 6 electrons occupy p1, y2 and y3 as shown in the energy level diagram. The only bonding p-orbital, p1 contains 2 electrons spread over 4 centres involved in 3 C-O connections. The non-bonding y2 and y3 orbitals have no contribution. Therefore, the bond order for each carbon - oxygen connection is 1/3. Do not forget the s-bonds which were not included in this scheme.
The energies of the p-molecular orbitals of conjugated molecules like butadiene, (see below) - occur in pairs, with their energies equal to (a±xb), where a and b are constants. For each bonding orbital of and energy a-xb there is a corresponding antibonding orbital of energy a+xb. The p-molecular orbitals are extended over the whole molecule.
For butadiene, the p manifold contains four electrons, leading to an electronic configuration of p12p22.
For the energy diagram and pictorial view of the p-molecular orbitals - please see below:
The radicals allyl:
have the same arrangement of p-orbitals,
(ie. the occur in pairs of energy a±xb),
but because there is an odd number of carbon atoms in the conjugate chain,
there must be a non-bonding orbital with energy x=0. Also, because
of the pairing properties of the p-molecular
orbitals of conjugated chains, there will be a node at every alternate
carbon atom in the non-bonding orbital. This is important for the unpaired
electron of allyl, which will occupy this non-bonding orbital. If an electron
is added to the allyl radical to form the anion, the negative charge will
appear at the terminal carbon atoms. If the unpaired electron is removed
forming the cation, the resulting positive charge is also spread over the
termial carbon atoms.
There are three p-molecular orbitals for allyl, the p1 is bonding, the p2 orbital is non-bonding and the p3 is anti-bonding. In the neutral allyl species - there are a total of seventeen valence electrons - of which three fill the p-orbital manifold. A pictorial representation of the energy diagram for the neutral, cationic and anionic allyl species are shown below - (orbitals are shown only for the cationic species):
These ions are represented in resonance theory as two or three canonical
In all cyclic polyenes (CnHn), the p-molecular
orbitals occur in degenerate pairs, except for the lowest p-orbital,
and for the cyclic polyenes with even numbers of carbon atoms, the highest
Tropylium ion: C3H3+
From the cyclic polyene diagram - the square molecule cyclobutadiene (C4H4) has four p-orbitals, a bonding orbital (p1), two degenerate non-bonding orbitals (p2 and p3) and an anti-bonding orbital (p4). Four electrons are placed into these four orbitals; twon into the bonding orbital, and one each with parallel spins into the degenerate non-bonding orbitals (Hund's rule) - see below:
This molecule is a relatively acidic hydrocarbon, and the anion
is formed by the treatment of cyclopentadiene with a strong base. From
the cyclic polyene diagram it can be seen that cyclopentadiene has three
bonding orbitals which are delocalised over the five carbon atoms. The
uppermost p bonding orbitals are a degenerate
pair, and are the highest occupied molecular orbitals (HOMO's).
These orbitals are much higher in energy than those in neutral aromatic species such as benzene, indicating that this anion is far more susceptible t attack by electrophiles. The anion is far more capable of coordinating to transition metals with available empty d-orbitals. The anion has six p-electrons, making the system aromatic. The six electrons are arranged as in the diagram below:
Benzene is the archetypal aromatic compound. It has a symmetrical
system and so is not over reactive on any one site. From the cyclic polyene
diagram it can be seen that benzene has six p-molecular
orbitals, (which contain the six p-electrons),
three bonding and three anti-bonding. The upper bonding degenerate pair
of orbitals are the HOMO's of benzene. The p
orbital manifold, is shown below - but also of interst is that the pattern
of the p orbitals is repeated within the s
s functions - like the p
orbitals are delocalized throughout the carbon skeleton.
The six p electrons are arranged as in the diagram below:
Bonding in benzene
From the above diagram it can be seen that the lowest lying orbital, p1, the orbital coefficients are such that the bonding charachter between each pair of adjacent carbon atoms is equal. In p2 bonding only occurs between atoms C2 and C3 and between C5 and C6 since the coefficients on C1 and C4 are zero. In p3, C1, which is bonded to C2 and C6 and C4 is bonded to C3 and C5, there are anti-bonding interactions between C2 and C3 and between C5 and C6. Therefore if we consider the pair of orbitals p2 and p3 the contribution to the C-C p bonding is equal for each bond. Since there are three occupied bonding orbitals and six CC linkages - the p bond order is 1/2. This description is in accord with the two resonating mesomeric forms (or Kekulé structures in a) below in which single and double bond characters alternate around the ring. Conventionally, the diagram in b) is used to show that the six electrons are delocalized around the ring:
C7H7+ (tropyllium) and C8H82+
Do them yourself!
The molecular orbital theory of solids
The third major type of chemical bond is the bond between two metal atoms. Metals lose electrons and cannot normally accept them. This means that, in a metallic bond, there are no atoms to accept the electrons. Instead, the electrons are given up to a "sea" of electrons that surrounds the metal atoms. In a way, this is similar to ionic bonding, except that the "ions" are electrons. The attraction between the electrons and the metal ions keeps the metal together.
The metals are the most numerous of the elements. About 80 of the 100 or so elements are metals. You know from your own experience something about how metallic atoms bond together. You know that metals have substance and are not easily torn apart. They are ductile and malleable. That means they can be drawn into shapes, like the wire for this paper clip, and their shape can be changed. They conduct heat and electricity. They can be mixed to form alloys.How is it that metallic bonding allows metals to do all these things?
The nature of metals and metallic atoms is that they have loosely held electrons that can be taken away fairly easily. Let's use this idea to create a model of metallic bonding to help us explain these properties. I will use potassium as an example. Its valence electron can be represented by a dot. When packed in a cluster it would look like this (also in example 31 in your workbook). The valence electron is only loosely held and can move to the next atom fairly easily. Each atom has a valence electron nearby but who knows which one belongs to which atom. It doesn't matter as long as there is one nearby.
Metals are held together by delocalized bonds formed from the atomic orbitals of all the atoms in the lattice. The orbitals spread over many atoms and blend into a band of molecular orbitals. The range of energies of these orbitals are closely spaced. The band is composed of as many levels as there are contributing atomic orbitals and each level can hold electrons of opposite spin. The idea that the molecular orbitals of the band of energy levels are spread or delocalized over the atoms of the piece of metal accounts for bonding in metallic solids. This theory of metallic bonding is called the band theory. The band is split into two regions, the upper portion being the empty levels or the antibonding and the lower portion is the filled levels or the bonding orbitals. In a metal the band of energy levels is only partly filled. The highest filled level right before going to the empty level is called the Fermi level. The trend of melting points of the transition metals is based on the electrons in the metal.
Bonding Models for Metals
3.14 Molecular orbital bands
Band Theory of Bonding in Solids
Bonding in solids such as metals, insulators and semiconductors may be understood most effectively by an expansion of simple MO theory to assemblidges of scores of atoms. If we recall, in simple MO theory we assumed that atomic orbitals on two atoms could come together to form bonding and antibonding orbitals.
If we bring three atoms together we can create a string of atoms with bonding that connects all three. Here we have a bonding orbital, an antibonding orbital, and a curious critter called a nonbonding orbital. Essentially a nonbonding orbital is an orbital that neither increases nor decreases the net bonding in the molecule. The important feature here is that three atomic orbitals in must give three molecular orbitals out. The total number of orbitals must remain constant.
Now let's expand these idea by considering combinations of four and ten atoms. As shown below, four atoms (four atomic orbitals) will give four molecular orbitals, two bonding and two antibonding. Notice that the two bonding (and two antibonding) orbitals are not exactly the same energy. The lower bonding orbital is slightly more bonding than the other (and one antibonding orbital is slightly more antibinding than the other). For the ten atoms we'd get a set of five bonding and five antibonding orbitals, each slightly different in energy.
If we now jump to a huge number of atoms, n, where n is perhaps as big as Avogadro's number, we can see that we're going to have a nhge number of bonding and antibonding orbitals. These orbitals will be so close together in energy that they begin to blur creating bands of bonding and bands of antibonding orbitals. It is the existance of these bands of orbitals that underlie our understanding of the properties of solids.
Notice, by the way, that there is some point at which the properties of an assemblidge of atoms makes a transition from separate discrete orbitals (and hence quantum properties like atoms), to bands of orbitals. The technology built upon tiny clusters called nanodots in which groups of atoms with perhaps twenty atoms acts like a quantum object is built upon this concept. By the time one has even a tiny cluster of atoms such as the submicron objects in a computer chip, the properties are best described by band theory.
MO Theory of Solids. Consider a linear chain of n identical atoms, each bringing in a valence s orbital for MO formation. If n = 2, 2 MOs are formed, one bonding and one antibonding. If n = 3, we obtain 3 MOs, bonding, nonbonding, and antibonding. If we run n right up to Avogadro's number, we expect to obtain No MOs, ranging from fully bonding (+++++...) to fully antibonding (+-+-+-...), with a whole bunch of other MOs between these extremes. This is shown in Figure MO-20. The energy spacing between lowest and highest MOs is determined primarily by the overlap between neighboring atoms, so will stay finite even though the number of atoms in the
chain reaches toward the infinite! We thus have a huge number of MOs crammed into a finite energy interval. They will be so close together in energy that they will form, for all practical purposes, a continuous band of energy levels. For this reason the MO theory of solids is often called Band Theory. A very important result of this treatment is that each of the MOs in the band is delocalized over all of the atoms in the chain.
The same ideas apply to a 3-dimensional, close-packed aggregate of atoms. A band of MOs will be formed from each type of valence AO on the atoms. Thus we obtain an s band, a p band, a d band, and so on, as shown in Figure MO-20. All of the MOs are delocalized over all of the atoms in the aggregate, so electrons in them can be considered to be everywhere at once! The highest filled band of MOs is called the valence band; the lowest unfilled band is called the conduction band; and the energy separation between the top of the valence band and the bottom of the conduction band is called the band gap. A partially filled band is simultaneously the valence
band and the conduction band, so in this case the band gap is essentially zero.
Now let's apply this picture to understand the electronic nature of the various classes of materials given above.
A conductor (which is usually a metal) is a solid with a partially full band, as shown in the Figure. An electron in the highest occupied MO is easily promoted to the next higher empty delocalized MO, where it is then free to roam over the whole solid lattice under the influence of an applied electric field; i.e., the solid conducts electricity due to this facile electron movement. The high reflectivity of metals is also due to the availability of a proliferation of empty MOs above the HOMO. Electrons in the filled MOs of the partially-filled band can absorb and then re-emit light of many wavelenths in making transitions to empty MOs in the band. This gives the metal surface a shiny reflective appearance. An example of a conductor is Na metal. It has an s band consisting of N MOs, where N is the number of Na atoms in the crystal. The band contains N electrons
(one from each Na atom) arrayed in N/2 pairs. These N/2 pairs go in the N/2 bonding MOs, which leaves N/2 antibonding MOs empty but readily accessible. Thus Na exhibits the characteristic properties of a metal, and is a conductor.
An insulator is a solid with a full band and a large band gap,
as shown in Figure MO-21. The MOs in the conduction band are so high in
energy that they are not thermally populated by the Boltzmann distribution,
and there is no conductivity at ordinary temperatures. An example of an
insulator is solid carbon in the diamond modification. Diamond consists
of a covalently bonded network of carbon atoms (a fcc array of C atoms
with more C atoms in half the tetrahedral holes), constructed from sp3
hybrid orbitals. N carbon atoms contribute 4N sp3 hybrids, which overlap
strongly to give 2N bonding MOs and 2N antibonding MOs which are separated
in energy by 5.47 eV from the bonding MOs. The 4N electrons exactly fill
the band of bonding MOs. The antibonding band is not thermally accessible,
so diamond does not conduct.
A semiconductor is a solid with a full band and a small band gap, as shown in the Figure. There is a small thermal population of the conduction band at normal temperature, hence a small conductivity. For example, silicon has a diamond modification similar to that of carbon, but a band gap of only 1.12 eV, due to poorer overlap of the sp3 hybrids of the larger Si atoms. Since the antibonding band will be occupied to a small extent via the Boltzmann distribution, Si exhibits a small conductivity at room T.
The group 4A elements, which have a number of valence electrons equal to twice the number of MOs in the bonding band, are uniquely structured to show semiconductivity. The elements C through Sn all exhibit a diamondlike crystal form, but with a band gap which decreases in magnitude for the larger atoms as orbital overlap becomes weaker. The trend in band gap down family 14 is shown below:
Element Band Gap
Thus Si and Ge are semiconductors at room T, and Sn is a conductor.
Pure compounds which are electronically analogous to the group 14 elements
are also semiconductors. These include the compounds boron nitride, BN,
and gallium arsenide, GaAs. Note that these compounds contain one element
from group 13 and one element from group 15, in a 1:1 stoichiometric ratio.
They thus have exactly the same number of valence electrons as a group
14 element, and will arrange these electrons in a group-14 type band structure.
They are often called 3-5 compounds, to indicate that they consist of elements
taken from groups 13 and 15. Similarly, 2-6 compounds such as ZnS and CdS
(both of which have the zincblende structure, which is analogous to the
diamond structure) function as semiconductors. Generally, band gaps vary
with position in the periodic table, but tend to decrease with increasing
MW of the semiconductor.
The temperature dependence of conductivity is readily understood within the framework of band theory. For a conductor, promotion of electrons is facile within a band at any T. However, As T increases, vibrational motions of the metal atoms in the lattice increases and interferes with the motion of the conducting electrons. The result is a decrease in conductivity as T increases. For a semiconductor, an increase in T causes an exponential increase in the population of the conduction band, because of the Boltzmann distribution. Therefore the conductivity of semiconductors increases dramatically with T. Because an insulator is actually a semiconductor with a large band gap, the conductivity of an insulator should also increase markedly if the temperature is made
Because of the very large number of atoms that interact in a solid material, the energy levels are so closely spaced that they form bands. The highest energy filled band, which is analogous to the highest occupied molecular orbital in a molecule (HOMO), is called the valence band. The next higher band, which is analogous to the lowest unoccupied molecular orbital (LUMO) in a molecule, is called the conduction band. The energy separation between these bands is called the energy gap, Eg.
Idealized representation of energy bands and gaps
The filling of these bands and the size of the energy gap determine
if a material is a conductor (a metal), a semiconductor, or an insulator.
In metals there is no energy gap between filled and unfilled energy levels.
A significant number of electrons are thermally excited into empty levels,
creating holes in the filled band. The electrons in a conduction band and
the holes in a valence band can move throughout the material, allowing
it to easily conduct electricity. In semiconductors Eg is small,
but large enough so that a fairly small number of electrons are in the
conduction band due to thermal energy, and these materials conduct poorly.
In insulators Eg is large so that electrons are not promoted
to the conduction band due to thermal energy, and these materials do not
Three categories of materials may be easily understood by the energy gap between the bonding bands and the antibonding bands. If there is a large gap in energy, the material is called an insulator. If the gap is finite, but small, then the material is a semiconductor, and if there is effectively no gap between the bands, the materials are called conductors. These properties arise because electrons that enter the antibonding band are free to move about the crystal. Such behavior is associated with electrical conduction. To give you an idea of the energies involved, the band gap in diamond is 502 kJ/mol, while that in Si is 100 kJ/mol and that in Ge is 67 kJ/mol. Diamond is an insulator while the other two materials are semiconductors.
|Left: When Onnes cooled mercury to 4.15K, the resistivity suddenly dropped to zero|
In 1913, it was discovered that lead went superconducting at 7.2K. It
was then 17 years until niobium was found to superconduct at a higher temperature
Onnes also observed that normal conduction characteristics could be restored in the presence of a strong magnetic field.
|Above: The Meissner effect - a superconducting sphere in a constant applied magnetic field excludes the magnetic flux|
The limit of external magnetic field strength at which a superconductor
can exclude the field is known as the critical field strength, Bc.
Type II superconductors have two critical field strengths; Bc1, above which the field penetrates into the superconductor, and Bc2, above which superconductivity is destroyed, as per Bc for Type I superconductors.
Ginzburg-Landau theory has been largely superseded by BCS theory, which
deals with superconduction in a more microscopic manner.
BCS theory was proposed by J. Bardeen, L. Cooper and J. R. Schrieffer in 1957 - it is dealt with in the Theory section. BCS suggests the formation of so-called 'Cooper pairs', and correlates Ginzburg-Landau and London predictions well.
Cooper pair formation - electron-phonon interaction: the electron is attracted to the positive charge density (red glow) created by the first electron distorting the lattice around itself.
However, BCS theory does not account well for high temperature superconduction, which is still not fully understood.
|3.1 - 3.5||These are important.|
|3.6||Important. Ignore the bit about the vapour phase.|
|3.7 - 3.11||Aklthough not explicitly covered in the course, you should be able to tackle these.|
|3.12, 3.13||These are important.|
|3.14||Important. Assume S2 and Cl2 are like their equivalents in the first row.|
|3.16 - 3.28||These are beyond the scope of this course.|
|3.1||You should be able to answer this.|
|3.2||You would be able to do this, but in is not explicitly covered in the course.|
|3.3||You should be able to do this by adding on top of to the 1s - 1s interaction, the diagram for the homonuclear diatomic molecules of the first row (which use 2s and 2p interactions).|
|3.4 - 3.15||These are beyond the scope of this course.|
Y1 = N.(y1sA + y1sB) = (1/Ö2).(y1sA + y1sB)...(1)
Since y1sA and y1sB are positive everywhere, their sum must be too. Just as for the atomic orbitals, the value of Y2 gives the probability of finding the electron in a small region, or the electron density at a point. This particular orbital is referred to as a bonding molecular orbital for reasons that will shortly be explained.
There is another linear combination that we should have considered
according to LCAO theory:
Y2 = (1/Ö2).(y1sA-y1sB) ..................(2)
Use the applet
to have a look at what this combination looks like. Figure 3 shows a crude
representation of what you should see. This combination is referred to
as antibonding. Whenever we combine two atomic orbitals in a way which
produces a change in the sign of Y between the
two component atomic orbitals, anti-bonding results.
We can better understand the difference between Y1
and Y2 by examining the electron
density or probability. For the bonding combination this is given by:
Y12 = ½(y1sA + y1sB)2 = ½(y1sA2 + y1sB2 + 2.y1sAy1sB) ..(3)
If we compare this to the electron density contribution from two individual
hydrogen atoms, ½(y1sA2
+ y1sB2), it is obvious
that the electron density has been increased by a amount y1sAy1sB.
This effect, which accounts for the bonding because the negatively charged
electrons hold the positively charged nuclei together, is illustrated in
Question 1. The probability plot for Y22
is also shown in Figure 4. What would be its equation? (The equivalent
of equation 3) What do you notice about the electron density between the
nuclei as compared to two individual hydrogen atoms simply placed side
Further insight into the bonding of HA and HB can be obtained by considering the energies of the electrons in Y1 and Y2 compared to their energies in the non-interacting atoms. This can be done by plugging the LCAO wave functions for the molecule back into the appropriate Scrödinger equation (just as it can be done for the individual atoms using the atomic wave functions). The results are shown in Figure 5, where Y1 and Y2 are sometimes renamed s(1s) and s*(1s), respectively, to indicate the type of molecular orbital and their parentage.
These molecular orbitals are useful for any molecule, or molecule-ion, using only 1s orbitals for bonding. Several possibilities are: H2+, H2, H2-, H22-, He2+ and He2. The positive molecule-ions are unstable, but have been detected in the gas phase under high energy conditions. The negative molecule-ions and He2 have not been observed.
Question 2. Using Figure 5, comment on the observations about the stability of the diatomic species listed above. (Would you expect all the negative molecule-ions to be unstable? Are there other species not listed which might be observed? Your answer should make reference to the electronic configuration and bond orders in these species
Second Row : Homonuclear Diatomics
Second row atoms have 2s and 2p orbitals available for use in bonding. As in the case of two atoms with 1s orbitals interacting, two atoms with 2s orbitals interacting lead to two molecular orbitals called s(2s) and s*(2s).
The symbol s is used when the molecular orbital has no nodal plane which contains both nuclei. For the bonding combination, there is only one region of high electon density between the two nuclei. If there is a single nodal plane containing both nuclei, the orbital is of type p. In this case, the bonding combination will have two regions of high electron density separated by the node. There are rare cases in certain transition metal compounds, where two nodes per molecular orbital contain both nuclei. These are designated d orbitals.
In valence bond theory terms, a single bond would have only a s symmetry bond. A double bond consists of a s and a p bond, a triple bond would have a s and two p bonds, and the esoteric quadruple bond has one s, two p and one d combination. In a multiple bond, the various orbitals co-exist in the same region of space between the nuclei. Do not mistake the two regions of overlap of a p-bonding orbital for a double bond!
Use the applet to display the molecular orbitals derived from the 2s atomic orbitals on two atoms.
Question 3. Prepare sketches of the s(2s)
s*(2s) orbitals similar to Figures 2 and
3. Do not attempt to copy all shading; just show all the nodes and the
phase (sign) of Y. Do they differ at all from
figures 2 and 3?
Next use the computer program to display the overlap of two 2px orbitals. (The x axis is taken as the internuclear axis by the computer program.)
Question 4. Prepare crude sketches of the resulting orbitals.
Which combination, the sum or the difference, corresponds to the bonding
combination this time? What would be the names for these orbitals?
Now use the applet to examine the molecular orbitals that result from the linear combinations of the 2py orbitals. (Note that the results would be the same for the combinations of 2pz orbitals, which, if you have time you can check by varying the value of z.)
Question 5. Once again prepare simplified sketches of what
you see. Give the proper labels for the two combinations, including the
* to indicate which is the antibonding orbital.
At this point you should have seen all the all the molecular orbitals formed by pairs of 2s and 2p orbitals. We can construct an energy level diagram to illustrate the relative energies of all these molecular orbitals and the atomic orbitals from which they are derived (Figure 6).
Question 6. What is the bond order in Li2 and O2?
What homonuclear molecule-cation and molecule-anion species should have the same bond order as O2? Consider cases between X22- and X22+ where X = Li to Ne.
Predict which two second-row elements are unlikely to give a diatomic molecule. Does simple Lewis bonding theory agree with your predictions?
Figure 6 does not quite tell the whole story. It can be refined a little by considering some other types of combination. Use the computer program to examine the linear combination of a 2s orbital on one atom with a 2px orbital on the other.
Question 7. Again sketch the results. There is no special
way to label these combinations, but say if they are of type s
or p, and whether they are bonding or antibonding.
Use the computer to form a combination of a 2s orbital on one atom with a 2py orbital on the other.
Question 8. Sketch the resulting orbital. Can you
classify this orbital as s or p
and bonding or antibonding? Explain. Go back to the beginning
of this section if you are not sure.
Your answers to questions 7 and 8 should have convinced you that an LCAO of the type illustrated in question 7 is a reasonable combination, but that the other combination, described in question 8, is not useful. (Such a combination is called nonbonding.)
The results of including the additional overlap of 2s with 2p orbitals is shown in Figure 7. The main changes are shown in red.
In actual fact the extent of the 2s - 2p interaction changes from Li2
to F2, becoming gradually less important, so that we should
use the "improved" diagram (Figure 7) for Li2 to N2
and Figure 6 for O2 F2 (and cations of Ne2).
Figure 8 below shows what happens to the molecular orbitals as we cross
Question 9. What is the difference for B2?
Go back and briefly answer, in the context of the homonuclear diatomic molecules, the four questions which were posed at the beginning of this lab.
Feel free to play with the applet
to examine some of the other combinations of atomic orbitals that are possible.
Because the projection direction is down the z-axis certain combinations
cannot be displayed.