Chemical Bonding (5) Bond Hybridisation Theory: Benzene

Benzene is a marvellous molecule.

C6H6 is a bonding and structural puzzle.

Try building a structural or display formula for C6H6.

You will have to produce a structure in which there are multiple carbon carbon bonds.

Your molecule will be unsaturated.

And therein lies our problem.

Because benzene's chemical properties do not suggest it is unsaturated.

Simply, it will not decolorize red bromine water or acidified purple potassium manganate (VII)!!

Then again take its molecular structure.

If it is a linear unsaturated molecule then some carbon-carbon bonds will be longer than others since
double bonds are shorter than single bonds.

Augustus Kekule struggled with the structure of benzene in the mid 19th century.

He could not see how the molecule could be saturated C6H6.

This is the quotation where Kekule describes how he came to see for the first time what benzene's structure could be:

I was sitting writing at my textbook but the work did not progress; my thoughts were elsewhere. I turned my chair to the fire and dozed. Again the atoms were gambolling before my eyes. This time the smaller groups kept modestly in the background. My mental eye, rendered more acute by the repeated visions of the kind, could now distinguish larger structures of manifold confirmation: long rows, sometimes more closely fitted together all twining and twisting in snake like motion. But look! What was that? One of the snakes had seized hold of its own tail, and the form whirled mockingly before my eyes. As if by a flash of lightning I awoke; and this time also I spent the rest of the night in working out the rest of the hypothesis. Let us learn to dream, gentlemen, then perhaps we shall find the truth... But let us beware of publishing our dreams till they have been tested by waking understanding.

What Kekule had realised in his dozing dream was that the molecule of C6H6 was not linear but circular.

"One of the snakes seized its own tail."

This is one of those cases of serendipity in scientific discovery.

This flash of brilliance did not fully resolve the problem of benzene's structure but it went a good deal of the way to doing so.

Now it was possible to suggest that the structure of benzene could be this:

without the snake you understand!!

But as you can see, this does not adequately solve the problem of benzene because we still have double and single bonds in an unsaturated structure for a molecule that does not behave as if it were unsaturated.

Kekule's suggestion is not much of an improvement, hailed at the time for its brilliance though it was.

Now, if we look at hydrogenation of benzene to cyclohexane and compare its value with that of a related molecule cyclohexene we see these results:

:










If we were to think of a molecule of benzene with three distinct carbon-carbon double bonds then what would the enthalpy of hydrogenation be?

We would expect a value of around three times that for cyclohexene: say -360 kJ/mol.

Why do we not see that value but measure the enthalpy of hydrogenation to be only -208kJ/mol?

If less energy is released then the structure of benzene requires more energy to break apart its carbon carbon bonding.

That difference is 152kJ/mol, the difference between the hydrogenation energy for a hypothetical Kekule structure and the hydrogenation energy of the actual benzene molecule.

Consider the diagram below: (there is a mistake in this diagram, the end product of hydrogenation is always cyclohexAne not cyclohexene)

with this diagram which includes the hypothetical Kekule structure:

You can see in the middle the hypothetical Kekule structure called here 1, 3, 5-cyclohexatriene.

(The hydrogenation energies are given in kJ/mol and in kcal/mol for our American friends!!)

So how do we now account for the fact that the bonding between carbon atoms in the benzene ring structure is much stronger than expected, requiring 152kJ/mol more energy to disrupt than expected?

We need to use the Molecular Orbital Theory for ethene as a template here.

Remember in ethene the model for the arrangement of three bonds around the carbon atoms a double and two single bonds is described as being sp2 hybridised.

Well a similar model is applied to the situation involving benzene.

Energy promotes an electron from the ground state in each of the six carbon atoms in benzene.

The resultant atomic orbitals in each carbon atom then hybridise to form three sp2 hybrid orbitals and leave one remaining p orbital

This model allows then for the formation of three sigma bonds between the six carbon atoms and the six hydrogen atoms.

But what about the p orbital?

Here is the genius of the model for benzene.

The six p orbitals overlap side-on to form a ring molecular orbital called a π bond above and below the ring of six carbon atoms.

In this ring there are six electrons and they are free to move: they are said to be delocalised.

The π bond is sometime referred to as a delocalised system.  The diagram below helps to visualise this new and original situation.

This diagram just shows the π bond, the σ bonds have been rendered as lines for clarity.

Now the genius of the model is to suggest that in real benzene molecules something like this actually happens and a π bond of this kind requires that extra 152 kJ/mol to disrupt it before benzene will react with hydrogen.

The 152kJ/mol is called the delocalisation energy.

Here are the representations of the benzene structure:

The structure of benzene has since been confirmed to be a perfect hexagonal ring.

Note that all six Carbon Carbon bonds are of equal length as the model predicts: each is 140pm.

The Carbon Carbon Carbon internal bond angle is 120o as the model predicts.

The molecule is planar in shape and is represented by the circle in the hexagon although most courses and exam authorities accept the Kekule structures as well in equations and mechanisms.

Lastly, and beautifully, if the π bond has six delocalised moving electrons then these six electrons moving ought to set up a magnetic field.

The ring current should do this and this is actually what is observed and has real consequences for measurements of benzene type molecules in nuclear magnetic resonance studies.

We have Kekule to thank for starting the ball rolling on benzene and he is rightly celebrated as one more famous Belgian.

Here is how he has sometime been seen:

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Chemical Bonding (4) Bond Hybridisation Theory: Ethene

In a previous post I discussed bond hybridisation theory in the context of the Methane molecule.

Now we come to examine how the model is used to both describe and predict the properties we observe about the ethene molecule.

Here is a displayed formula of ethene showing the bond angles and bond lengths:

Ethene has a trigonal planar shape about each carbon atom i.e. it is a flat molecule.

Lets look closely at ethene.

What stands out is the double bond and the knowledge that one bond is weaker than the other.

We know this from studies of the reactions of ethene.

Addition reactions involve the addition of a molecule across the double bond in which the weaker one of the two bonds breaks.

I have blogged on this phenomenon here.

Ethene has three equivalent covalent bonds around each carbon atom and the additional weaker bond between the two carbon atoms.

It has been calculated and shown experimentally that a mole of the C-H bonds takes  about 416 kJ to break.

A mole of the C=C bonds takes around 614kJ and a C-C bond 347kJ leaving 267kJ to fit the other bond in the double bond.

How can we explain this data?

When we examine the electron arrangement of carbon we find this:

The lowest energy carbon atom has its 6 electrons in the following configuration:

Carbon   1s2    2s2   2p2

The four electrons in the second energy level are in two sub shells of slightly different energy.

Now there's the issue.

For these four electrons to remain in these two subshells would not make four covalent bonds with hydrogen electrons around the carbon atom.

To arrive at three equivalent bonds, plus the weaker one we observe in ethene, the hybridisation model suggests the promotion of one electron into the empty p orbital.

But then - and this is new in ethene - hybrid orbitals form from two p orbitals and an s orbital.

This model is illustrated below.

In this model an electron is promoted to the empty 2pz orbital and then the 2s and two 2p orbitals "hydridise" to form three equal energy atomic orbitals in the carbon outer shell.

These three atomic orbitals are designated sp2 because each one is formed from an s orbital and two p orbitals.

That leaves an electron in a 2p orbital - in the example above it is in the 2pz orbital.

When it comes to bonding with hydrogen, the model suggests that the two hydrogen atoms overlap their 1s atomic orbitals with two of the sp2 orbitals of carbon.

That gives us two C-H σ bonds.

The other sp2 orbital overlaps with that of the other carbon atom, again giving a σ bond, only this time joining two carbon atoms together.

Question is what does the model suggest gives rise to the other bond of the double bond?

Here we have a very new and original suggestion.

And it is a model that fits the reality of ethene.

Ethene has another property we've not yet mentioned.

This property is that the double bond cannot rotate freely, its rotation is severly restricted.

We know this from the  existence of E-Z or cis - trans isomers.

The hybridisation model suggests that another kind of bond forms between the two carbon atoms that stops rotation.

It is called a π bond and it is formed from the side ways overlap of the two remaining 2pz atomic orbitals.

So a spatial representation of the atoms in ethene looks like this:

Now the strange thing is that the π bond is both the blue and red parts shown in the diagram.

Both spaces have a probability of finding the electron of about 95%.

The two comprise the whole bonding molecular orbital (MO).

This diagram shows us a model of the ethene molecule that is the right shape with the right bond angles between the two C-H bonds at either end of the molecule .

In the model, molecular orbitals (MO) result from the overlap of atomic orbitals.

VSEPR theory is supported on this hybridisation model because there is a higher electron density between the carbon atoms due to the π bond.

This higher electron density means that the two C-H sigma bonds at each end of the molecule will be pushed closer together than in a normal trigonal planar molecule.

We ought to see the H-C-H angle drop below 120º and we do: it is 117.4º.

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Chemical Bonding (3) Bond Hybridisation Theory: Methane

What is Bond Hybridisation Theory and why does it matter?

In a previous post I discussed the bond angles in molecules like methane CH4.  

Bond hybridisation theory seeks to offer explanations for these bond angles and bond lengths in a way that tries not to violate the actual observed facts about molecules like methane and ethene.

So let's start with what we can observe about methane and ethene and then try and relate those facts to the atomic electron structures of carbon and hydrogen.

Here are in picture form some of those observed features of both methane and ethene molecules.

Methane

As we know a methane has a tetrahedral shape

Ethene

Ethene has a trigonal planar shape about each carbon atom i.e. it is a flat molecule.

Lets look closely now at Methane.

A couple of things stand out.

Methane has four equivalent covalent bonds.

Each is 108.70pm long and therefore each is of the same energy.

Dr Carr’s Rescue Box

This box occurs just when you need it to explain some unusual appearance on one of my blogs. In this case the bond length of the C-H bond is 108.7 pm.

What is a "pm"?  It is a pico metre. Divide a meter into a million million bits and each bit is a picometer wide or 10^-12 m.

Hope that helps. It is pretty small!!

Each is a C-H bond and it has been calculated and shown experimentally that a mole of these bonds takes 416 kJ to break.

How can we explain this data?

When we examine the electron arrangement of carbon.

The lowest energy carbon atom has its 6 electrons in the following configuration:

Carbon   1s2    2s2   2p2

The four electrons in the second energy level are in two sub shells of slightly different energy.


Now there's the issue.

These four electrons to remain in these two subshells would not make four covalent bonds with hydrogen electrons that were of equal energy.

The four electrons would have to be in the same energy level for that to happen - but they are not.

In hybridisation theory, hybrid atomic orbitals are assumed to form from the 2s and 2p atomic orbitals when the hydrogen atoms bond to the carbon atom and create the methane molecule.

This effectively puts the carbon atom in an energy level higher than its ground state ( or lowest energy level.)

Diagrammatically its often drawn like this:

In this model an electron is promoted to the empty 2p orbital and then the 2s and the three 2p orbitals "hydridise" to form four degenerate or equal energy atomic orbitals in the carbon outer shell.

These four atomic orbitals are designated sp3 because each one is formed from an s orbital and three p orbitals.  

Each of these sp3 orbitals can now accept one electron from each hydrogen atom and the molecule of methane is formed. 

Thus this model accounts for the formation of four equivalent C-H bonds around the carbon atom in methane.

The model is not the reality it is one of the best representations of the reality of the methane molecule.

The model's capacity to predict what we observe about the properties of methane is what makes it very useful.

The shapes of these sp3 orbitals have been calculated from studies of the quantum mechanics of the methane molecule.

So a spatial representation of the carbon atom in methane looks like this:

This diagram shows us a model of the methane molecule that is the right shape with the right bond angles between the four C-H bonds.

Notice too that combining the sp3 atomic orbitals of the carbon atom and the four 1s hydrogen atomic orbitals leads to the structure of the methane molecule.

In the model molecular orbitals (MO) result from the overlap of atomic orbitals.  

As the ends of these atomic orbitals overlap the resulting molecular orbital is called a sigma σ
molecular orbital.

As the diagram notes the molecular orbital has σ symmetry i.e. it is symmetrical about the central axis that connects both hydrogen and carbon atoms. 

Chemical Bonding (2) VSEPR and non-bonding electron pairs

In a previous post I asked this question: 

How come the angles between the bonds in methane are all 109.5°?

When the molecule of methane is examine closely it has a particular shape.

We can see that shape on this picture of a model of the molecule:

The shape is called a tetrahedron or a triangular based pyramid. 

Imagine the carbon atom is in the centre and the hydrogen atoms are at the points of the tetrahedron: 

Now how come this configuration fits the molecule? 

Remember each of the bonds is formed from a pair of electrons that are negatively charged.

Like all negative charges these electrons will repel each other to the point where they cannot repel any further.

Therefore, the molecule adopts the tetrahedral shape that results from the maximum repulsion of the bond pairs of electrons.  

In the case of methane the hydrogens end up at the points of a tetrahedron.

The bond angle can be calculated from geometrical theory to be 109.5°.

But what if the molecule contains both bonds and non bonding electrons, what happens then?

Let's look at the example of ammonia NH3

What is the shape of the ammonia molecule and what is its bond angle?

Here is the dot and cross diagram of ammonia:

You can see there are three bond pairs but also one pair of electrons is not bonded to hydrogen.  

This is a non bonding pair of electrons.

Non-bonding pairs of electrons exist closer to the nucleus than bonding pairs of electrons or you can say they take up more "bonding space".

Therefore, non-bonding pairs of electrons exert a greater repulsive force on the bond pairs than the bond pairs do on each other.  

How does that affect the shape of the ammonia molecule and its bond angle?

As you can see from the diagram below the 4 pairs of electrons are like a tetrahedron but the non-bonding pair push the bond pairs closer to each other.

So the angle between the N-H bonds is not 109.5° but they exist a little closer to each other at 107.3° apart.   

The resultant shape is called pyramidal or trigonal pyramidal.

Non bonding pairs of electrons repel bond pairs more than bond pairs repel bond pairs.

And if a molecule has two bond pairs and two non-bonding pairs then the angle between the bond pairs will be reduced even further.

Here is the example of water H2O:

As you can see the effect of the two non-bonding pairs is to reduce the H-O-H bond angle even further than in ammonia to 104.4°.

Water is described as a bent or angular molecule.

The theory that gives rise to these shapes is called Valence Shell Electron Pairs Repulsion theory or VSEPR for short.

Here is a formal statement of the VSEPR theory:

The VSEPR theory assumes that each atom in a molecule will achieve a geometry that minimizes the repulsion between electrons in the valence or outer shell of that atom. Non- bonding electron pairs exert greater repulsion than bond pairs and double bonds (2 electron pairs) exert greater repulsion than do single bonds.  

So how can we work out the shape of a simple molecule or ion?

1) Work out the number of electrons in the outer shell of its central atom.

2) Work out the number of atoms that each add one electron to fill the outer shell

3) Add together answers to 1 and 2 then divide by two to find the number of bond pairs

4) Use the table below to read off the shape of the simple molecule

Electron pairs Molecule geometry      Example

2                      linear                           BCl₂

3                      trigonal planar             BF₃

4                      tetrahedral                   CH₄

5                      trigonal bipyramidal     PF₅

6                      octahedral                   SF₆

The theory has to be modified to account for non-bonding pairs and multiple bonds.

a) Non bonding electrons exert a greater force on bond pairs and have the effect of distorting the basic geometries of molecules.

Calculating the number of electrons in the outer shell of the central atom that are not bonded will usually determine the effect of these non bonding electrons on the molecule shape.

e.g.  H2S in which there are two non bonding pairs and two bonding pairs giving rise to a shape similar to that of water.  

Why the angle is smaller than that of water I'm going to leave to an upcoming blog on bond hybridisation theory.

Here you can use water as an analogue to that of hydrogen sulphide.

b) Multiple bonds have greater electron density and also exert a greater force on single bond pairs.

Let's look at the bond angles around one of the carbon atoms in ethene, C2H4.

There are 4 electrons around the carbon atom

There are two electrons from the two hydrogen atoms attached to this carbon

There are two single bonds then and a double bond

The molecule has 3 bonds attached to each carbon so it is basically trigonal in shape

But the double bond has a greater density of electrical charge so pushes the two bond pairs together slightly reducing the 120° angle. Like this:

You can calculate the H-C-H bond angle to be 117.4° slightly less than the standard trigonal planar angle of 120°.

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Chemical Bonding (1) Covalent Bonds in Alkanes like Methane

Chemical Bonding: covalent bonds in alkanes like methane.

Looking at some of those formulae we've used in the previous oil fractionation posts should leave us puzzled.

What do the lines joining the element symbols stand for?

Here's a displayed formula for Butene that shows all the atoms and all the bonds in the molecule:

So what do the lines stand for?

Our explanation goes right back to the Periodic Table.

Let's look at a Periodic Table of just the first 20 elements.

Here we have names and symbols of the first twenty elements each with a couple of numbers:

It's the numbers we need to be looking at first especially the top one called the atomic number (symbol Z) 

This is the number of protons (positively charged)  in each atom's nucleus.

And since all atoms are electrically neutral it is also the number of each atom's electrons (negatively charged).

Your bog standard helium atom (He), you guessed it, has two protons and two electrons.

If you want to know about the other number (called Relative Atomic Mass or RAM for short) you'll need to go here.

Why is the number of electrons important to us?

Because that's what allows atoms to combine or stick or bond to each other.

There are several different ways in which the bonds can form but we are just thinking in this blog about the way atoms like hydrogen and carbon bond together.

And they bond like this:

The atoms in the Periodic Table hold their electrons in specific groups called shells or energy levels.

These shells or energy levels obey fixed rules.

The first shell can only hold two electrons, the second shell eight electrons and the third shell 18 electrons.

Let's see how this works out for the first twenty elements:

Here you can see the numbers in red are the numbers of electrons in each shell of that atom.

For example carbon 2)4 has six electrons, 2 in the first shell and 4 in the outer shell.

No prizes for guessing which element is missing form this illustration (Thanks BBC Bitesize).  Yes, you guessed it, Hydrogen.

Hydrogen has just one electron orbiting the nucleus.

So how is it that carbon and hydrogen can bond together to make methane?

methane formulas and models

If the outer shell electrons are shared between hydrogen and carbon, both elements gain full outer shells.

That's eight for carbon and two for hydrogen.

A full outer shell is considerably more energetically stable because the shells are at lower energies so that's why they form.

We construct what we call dot and cross diagrams to show this new situation, though the Americans call the diagrams Lewis Diagrams after GN Lewis who first used them.

Here's the one for methane:

So now you should be able to see that the line in the displayed formula for methane stands for a pair of electrons shared between hydrogen and carbon.

You can see them shown as red and green dots in the diagram on the left.

Pairs of shared electrons like this are called covalent bonds.

Sometimes you'll see them as dot and cross diagrams where the electron from one atom is shown by a cross and the electron from the other atom is shown by a dot.
Like here:

You could try to construct Dot and Cross diagrams for other simple molecules, why not try water (H2O), ammonia (NH3) and chlorine (Cl2)?

Students working on more advanced courses could also build Lewis diagrams for ethane (C2H6), ethene C2H4) with its double carbon carbon bond, and chloromethane (CH3Cl).

In the next blog, Chemical Bonding (2), I will talk about the shapes of molecules composed of covalent bonds, asking how it is that the angle between the covalent bonds in methane is exactly 109.5°?

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