chemistry Lecture Notes IV

8/8/2019 chemistry Lecture Notes IV

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NORTHERN CARIBBEAN UNIVERSITY

DEPARTMENT OF BIOLOGY AND CHEMISTRY

LECTURE NOTES

GENERAL CHEMISTRY I

Assistant Professor: Dr. Nicole White

5. Molecular Orbital (MO) Theory

This model was developed due to the insufficiencies of the valence bond theory, as it sometimes

lead to an incorrect electronic description. MO theory is a theory of the electronic structure of

molecules in terms of molecular orbitals, which may spread over several atoms or the entire

molecule; in VB theory, we were concerned with atomic orbitals. Here too, as in VB theory,

orbitals overlap; unlike VB theory MO theory involves bonding and antibonding orbitals. Again,

this model is derived from the quantum mechanical wave equation. The need to learn this

equation is not relevant for this course. However, the square of the wave function in said

equation gives the probability of finding an electron in a given region of space.

Orbital- a solution to the Schrdinger wave equation describes a region of space where an

electron is likely to be found.

Atomic orbital a wave function whose square gives the probability of finding the electron

within a given region of space in an atom.

Molecular orbital a wave function whose square gives the probability of finding the electron

within a given region in space in a molecule.

In MO theory orbitals can interact 2 ways: additive (electrons are concentrated in areas between

nuclei called bonding orbitals, ) and subtractive (electrons are concentrated in areas other than

between the two nuclei, *). The means that the MO has a cylindrical shape about the bond

and the star (*) tells us that the molecular orbital is antibonding. The orbitals involved in

overlapping must contain 1 electron each and when they overlap these electrons must be in

opposite spin. MOs have specific size, shape and energy level.


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The energy associated with the bonding orbital is always lower than the isolated orbitals; hence it

is more stable, while the anitbonding orbital has a higher energy. This higher energy makes the

antibonding orbital unstable as electrons cannot occupy this region and thus do not contribute to

bonding.

To better understand this model let us consider the H2 molecule its MO is shown below.

Electrons are added to molecular orbitals, one at a time, starting with the lowest energy

molecular orbital. Recall the following criteria:

The aufbau principle, lowest energy MOs fill first, or fill the orbitals from below, IF

there are large orbital energy differences.

The Pauli exclusion principle, a maximum of two electrons per orbitals and these must

be of opposite spin.


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The bonding orbital is denoted as 1s which tells us that there is a bonding orbital obtained from

the 1s atomic orbital. As is observed from the diagram the energy of the anitbonding orbital

(excited state) is indeed higher than that of the bonding orbital (ground state).

As by now realized, the number of molecular orbitals formed is the same as the number of

atomic orbitals combined.

For helium; both 1s and *1s each have 2 electrons and the stability (decrease in energy) gainedby the bonding MO is cancelled by the antibonding MO energy. This He2 would be an unstable

molecule for this reason helium exists as a monatomic gas.

Bond order

This is tells the number of bonds formed between two atoms.

electronsgantibondinofnumbernelectronsbondingofnumbern

nnorderbond

a

b

ab

=

2

)(

What is the bond order for H2?

What about He2?


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If you have done both calculations you will see that for hydrogen the bond order (BO) is 1. For

helium, the BO is zero (0).

MOs for other diatomic molecules.

The above examples concerned bonding between 1s orbitals which are not the only orbitals

involved in molecular bonding. They also represent homonuclear bonding (molecules composed

of two like nuclei); heteronuclear bonding also occurs between molecules composed of two

different nuclei. For orbitals to interact they must have approximately equal energies and they

will interact to give maximum overlapping in order to have strong bonds the orbitals must

overlap greatly.

Let us consider the p orbitals. There are 2 ways in which 2p orbitals can interact 1 set of 2p

orbitals can overlap along their axes, head to head, to give one bonding and one antibonding

orbital (2p and 2p*); the other 2 sets then overlap sideways to give two bonding and 2

antibonding orbitals (2p and 2p*). The p orbitals are denoted as px, py and pz and when they

overlap they match up for example, x bonding molecular orbital and a x* antibonding one.

The 2 remaining 2p orbitals interact sideways to give y and y* and x and x orbitals have the

same energy and are known as degenerate, oriented at 90o. Because they are degenerate filling of

these orbitals must follow Hunds rule - when there are equal energy or "degenerate"

orbitals, these fill one electron at a time before pairing begins. Or, if two orbitals are not very

different in energy, near-degenerate, or even exactly degenerate, then two electrons will fill both

orbitals with parallel spin.


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The order of filling differs depending on the molecule being studied. For O2, F2, and Ne2 the

below diagram is obtained. This leads to less electron charge between the nuclei for the pi

bonding molecular orbital than for the sigma bonding molecular orbital. Less electron character

between the nuclei means less plus-minus attraction, less stabilization, and higher potential

energy for the pi bonding molecular orbital compared to the sigma bonding molecular orbital.

When the interaction is out-of-phase, less overlap leads to less shift of electron charge from

between the nuclei. This leads to more electron charge between the nuclei for the pi antibonding

molecular orbital than for the sigma antibonding molecular orbital. More electron charge

between the nuclei means more plus-minus attraction and lower potential energy for the pi

antibonding molecular orbital compared to the sigma antibonding molecular orbital. The

molecular configuration is written as: 1s 1s* 2s 2s* 2p x y x* y * 2p*.


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For , Li2, B2, C2, and N2 the following MO diagram is obtained.

The molecular configuration is written as: 1s 1s* 2s 2s* x y 2p x* y* 2p*. As should be

noted, there is a lower shell of electrons that is not represented in the orbital diagram, which is

because they are filled both their boding and antibonding orbitals and so they are not considered

because the stability gained by the bonding orbitals is cancelled by the instability caused by the

antibonding orbital. These inner shells are denoted as KK inner shells of 2 atoms.

What is the difference between nonbonding and antibonding?

A molecule can be determined as paramagnetic (substances with unpaired electrons and are

attracted by magnetic fields) or diamagnetic (substances whose electrons are all spin paired and

are weakly repelled or not attracted by magnetic fields) from their molecular orbital.


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Which of the following diatomic molecules are diamagnetic and which are paramagnetic:

O2, N2 and F2.

6. Chemical Reactions

In a chemical reaction there is a rearrangement of the atoms present in the reacting substances to

give new chemical combinations present in the substances formed by the reaction.

2HCl(aq) + Zn(s) H2(g) + ZnCl2(aq)

It is also important that you know the correct state symbols to place behind each species in your

reaction. The state symbols id basically an indication of the state of the substances in the

reaction: (s) solid, (l) liquid, (g) gas, (aq) aqueous. An aqueous solution means that the

substance was dissolved in water.

There are, in essence, 3 main types of chemical reactions and as students of chemistry it is

important that you are able to distinguish one type from another.

Precipitation/ exchange

Neutralization

Reduction-oxidation (redox)

Precipitation/Exchange reaction: a precipitate is an insoluble solid compound formed during a

chemical reaction in solution. In order to know whether or not a precipitate will occur, the

solubilities of the potential products must be known. In this type of reaction ions will exchange

partners as it were.

2AgNO3(aq) + Na2CO3(aq) Ag2CO3(s) + 2NaNO3(aq)


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Neutralization reaction: which occurs in acid-base reactions, the H+ ion in the acid reacts with

the OH- ion in the base causing the formation of water. Generally, the product of this reaction is

some ionic salt and water.

saltbaseacid

O(l)2H(aq)CaCl(aq)Ca(OH)2HCl(aq)222 ++

Reduction-oxidation reactions: this type is further divided into combination/synthesis,

decomposition, displacement and combustion reactions.

Combination/synthesis is a reaction in which two substances combine to form a more

complicated substance and has the general form - A + B AB

2Na(s) + Cl2(g) 2NaCl(s)

In the above, sodium is oxidized and chlorine was reduced.

Decomposition - A decomposition reaction is the opposite of a synthesis reaction; a

complex molecule breaks down to make simpler ones AB A + B.

)()(2)(2 2 gOlHgsHgO +

In the above, mercury is reduced and oxygen is oxidized.

Displacement a reaction in which an element reacts with a compound, displacing anelement from it.

Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s)

In the above, copper is oxidized and silver is reduced.

Combustion - A combustion reaction is when oxygen combines with another compound

to form water and carbon dioxide. These reactions are exothermic, meaning they produce

heat.

2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(g)

Balancing equations


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When balancing a chemical reaction it is important to remember the law of conservation of mass

no mass is lost or gained during a chemical reaction. In other words the reactants and products

must be in the same proportions. In balancing an equation it is good to take note of the

following:

Never touch subscripts when balancing equations since that will change the

composition and therefore the substance itself.

Make an element inventory. How are you going to know if the equation is balanced if

you don't actually make a list of how many of each atom you have? You won't. You have

to make an inventory of how many atoms of each element you have, and then you have to

keep it current throughout the whole problem.

Check to be sure that you have included all sources of a particular element that you are

balancing on a particular side since there may be two or more compounds that contain the

same element on a given side of an equation.

Example:

Let us balance the following equation.

Make an inventory of the elements noting how much of each you have on either side. The

numbers should be the same as shown below.

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chemistry Lecture Notes IV