Thursday 28 January 2016

Periodicity (2) Melting and boiling points of the elements of Period 3


Here we go again with the concept of periodicity.

We’ve already said what periodicity is:

Periodicity is the regular recurrence of similar properties of the elements across the periodic table.

I think one of the ways we can get this circularity or regular recurrence of similar properties is to view the periodic table in a 3D circular form as here in this diagram:



As you can see from both views the first three rows are connected; there is no break in the flow of atomic numbers.  

Now as the elements with similar properties appear above each other we can see the regular recurrence of their properties. 

So this graph I made of the melting and boiling points ought to be circular–yes!!

























What we see first about these values is that there is a pattern but unlike ionisation energy and electronegativity the trends in melting and boiling point across a period are not straightforward.

Let’s define boiling and melting first and then get to an explanation of what is going on here.

Then you can construct the chart for the Period 2 values and see how they compare with Period 3. 

What is meant by boiling point?

Boiling point is the temperature at which the vapour pressure of the element is equal to the current atmospheric pressure.  The energy being supplied to the element is used to break interatomic or intermolecular bonds in the liquid element to allow its particles to exist much further apart and move at great velocity in the gaseous state.  The stronger these forces of attraction are the higher the boiling point. 

What is meant by melting point?

Melting point is the temperature at which the particles of the element transition from the solid into the liquid state. To transition, energy is supplied to break bonds between particles in the solid state instead of raising the temperature of the element. The particles remain about as close as before but they now move randomly rather than vibrating on the spot. 

What these two definitions tell you is that the explanation for the variation in melting and boiling point is going to be to do with the bonding and structure of these elements. 

Explanation

Let’s now look at and try to explain the periodic pattern.

Sodium, Magnesium, Aluminium

The first thing we can see is that there are three metallic elements: sodium, magnesium and aluminium, with relatively low melting and boiling points.

Metallic bonds hold the atoms together in these three metals. 

Here is a pictorial description of metallic bonding for a group 1 and a group 2 metal.



So notice this as we go from sodium to magnesium to aluminium the charge on the metal ion increases from +1 to +3 that also means that the number of delocalised electrons per atom increases and that has the effect of increasing the strength of the metallic bonds. 


Silicon

So what happens with silicon?

Well, silicon is not a metal it is a metalloid i.e. it has some metallic characteristics (it looks like a metal grey and shiny) but crucially its structure and bonding is not metallic. 

The atoms of silicon are held together by strong directional covalent bonds in a huge atomic network. 

This is much like the structure of diamond that you might already be familiar with.

So each silicon atoms as we can see is covalently bonded to four other atoms in a tetrahedral arrangement. 

Each of these four covalent bonds is strong.

This tetrahedral arrangement extends to edge of a silicon crystal and is sometimes called a giant structure for obvious reasons. 

To merely melt this stuff, energy will need to be given to it to break each covalent bond if the atoms are to be set free to move randomly around each other. 

As you can see that energy value and hence the temperature at which silicon melts will be very high. 

Phosphorus, Sulfur, Chlorine and Argon

Again these are not metal but non–metals. 

They have molecular structures. 

They exist in small groups of atoms: P4, S8, Cl2 and Ar.

















There are strong forces of attraction between their atoms within each molecule; what are called intra–molecular bonds. 

But what matters for our explanation are the inter–molecular bonds.

These inter molecular bonds are very weak.

These are called van der Waals forces.

The strength of these van der Waals forces depends n the number of electrons in each molecule and so the melting and boiling points follow this pattern: S8 > P4 > Cl2 > Ar.


The bigger the molecule, more electrons, the stronger the van der Waals force, the higher the melting and boiling point.

Tuesday 26 January 2016

Periodicity (1) Ionisation energy and electronegativity of the elements


The Periodic Table is a wonderful thing.


It’s not just that there are these elements in groups that reflect their physical properties like you look at the Alkali Metals (Group 1) and see that they get more reactive with water as you go down the group but the properties are periodic.

The periodic nature of the physical properties is called periodicity.

Here’s a definition of periodicity:  Periodicity is the name given to the regular recurrence of similar physical and chemical properties of the elements and their compounds in the periodic table. 

It’s like this: your school timetable is periodic because the same chemistry lesson with the same teacher occurs at the same time each week. 



Your timetable is a periodic table of sorts!! (See, you just can’t get away from Chemistry its everywhere!!)

In life periodic things take place year on year: we go on holiday in the summer, we give presents to those we love at Christmas, we hope that people remember our birthday when it comes round every year. 

So how do the properties of the elements reveal their periodicity?

Let’s look at some of the obvious properties of the elements and see if we can see periodic patterns in them.

Now if you have a Data Book of elements’ properties you can draw up charts and tables to reveal this periodicity.

Here’s one of them:

1. First Ionisation Energy

Here is a plot of first ionisation energy (Em1) against atomic number (Z).

You can see the regular repetition of similar (not identical) properties.

Every eighth element has the highest Em1 starting with Helium. 

We notice the break for the first transition series at atomic number 21 Scandium.

But the pattern picks up again at atomic number 30 and peaks at Kr Krypton.

The alkali metals are also labelled because they have the lowest Em1 values.

Their values repeat at regular intervals: every eighth element. 

We can see from this 3D image the trends in the first ionisation values:

Em1 decreases down a group, e.g. lithium (Li) to caesium (Cs), because the outer shell electrons get further from the positive electrostatic pull of the nucleus.

Em1 values generally increase across a period, e.g. Lithium (Li) to Neon (Ne), because the size of the positive nucleus increases, increasing the positive pull of the nucleus on the outer electron shell.


2. Electronegativity

What is electronegativity?

This is the power of an atom of an element within a covalent bond to attract the bonding pair of electrons to itself. 

So here is a water molecule which we can represent like this to show how the oxygen atom tends to pull electrons in the two covalent bonds towards itself ( the arrows) and lead to an internal dipole in the water molecule.
The two hydrogen atoms end up slightly positively charged relative to the oxygen atom. 

This ability to attract the bonding electrons to itself is called an element’s electronegativity.

Electronegativity is measured on a scale from 0-4 first developed by one of the greatest ever chemists Linus Pauling. 

Here is how electronegativity varies across the periodic table and as you can see it is a periodic function of the elements.  


The values and trends are easier to see on the next diagram.


The value increases across the periodic table from left to right and it decreases down a group of the periodic table. 

It is highest at Fluorine (F).  (Why are there no values for the Noble Gases?)

It is at its lowest at Caesium and Francium. 

Metals tend to have low values and non–metals have high values, hence the colour densities in the diagram above. 

The reason why the values tend in these directions is due to the dependence of electronegativity on the atomic radii values: the higher the electronegativity the smaller the atomic radius. 

And the atomic radius tends to depend on the size of the nucleus, the number of electron shells and the tendency for the inner shells to shield the outer shell electrons from the influence of the nucleus.

So take the Fluorine atom that has the highest electronegativity, it has two shells with electron arrangement 1s2, 2s2, 2p5.  There is little chance of shielding of the outer shell electrons so it pulls bonding electrons easily to itself. 

But look at Caesium with a very low electronegativity.  It has 6 electron shells.  It is in Period 6.  Its outer shell electron is well shielded from the positive nucleus even though the nucleus is huge (Z=55) with 55 protons. So there is a great tendency of the outer electron to be lost and caesium atoms are easily oxidised.


Things to do:  

So why do the Noble gases not have an electronegativity value? 

Can you construct a chart to reveal the periodicity of the melting points and boiling points of the elements?


Can you calculate the atomic volumes (the volume of one mole of atoms of the element, hint: use the molar mass and the element’s density.) of the elements and show that this is also a periodic property. 

Thursday 14 January 2016

Volumetric Analysis (6) Estimating the % of copper in brass–a redox titration.


In this extensive blog about this titration I’m going to discuss the theory behind what happens then describe a simple procedure and work you though the calculation that you will need to do.

Theory

Brass is an alloy of copper and zinc.

I love this titration because it involves some superb colours and a lovely toxic and corrosive concentrated acid–nitric acid.

The reaction involves several redox reactions:

First, the reaction between copper and concentrated nitric acid

Cu + 4H+ + 2NO3  =   Cu2+ + 2NO2 + 2H2O

Or more simply

Cu (s)     =    Cu 2+ (aq)  +   2e

Second, the reaction between copper ions and iodide ions to release iodine

2Cu 2+ (aq)    +    4I (aq)   =   2CuI(s)   +   I2 (aq)

Third, the reaction between iodine and sodium thiosulphate solution

2Na2S2O3 (aq)    +   I2 (aq)   =    2NaI (aq)      +    Na2S4O6 (aq)

The basic procedure goes like this:

We measure out a given mass of brass and dissolve it in an excess of concentrated nitric acid. 

You can of course break British law and use a copper coin instead of brass and so find out the % of copper in coinage.

You could work on coins from different times to see if the Government has gone cheap in its production of coinage.

I wouldn’t be surprised to see that the % of copper has fallen over the years. 

You will need to wait longer than you think until all the copper dissolves to form a green solution; it is a reaction best done in a fume cupboard since the brown fumes evolved are toxic nitrogen dioxide (NO2).

The acidic green solution consists of copper complexes with nitrogen oxides in the solution.

The acidic solution contains a high concentration of nitrate ions (NO3 ) and under these conditions these ions are oxidising.

If nitrate ions and iodide ions are in the same solution then a redox reaction occurs because iodide ions are reducing agents like so:

2NO3- + 4H+ + 2I-   =  I2 + 2NO2 + 2H2O.

If left the nitrate ions will affect the end point of the titration as they will remove iodide ions from the solution. 

So we need to neutralise the solution with sodium carbonate, a non oxidising compound.

You will see a precipitate for when you add the sodium carbonate this precipitate is copper hydroxide. 

Adding a weak acid like a dilute solution of ethanoic acid will redissolve this precipitate.

Now if you add potassium iodide to the reaction mixture the iodide ions will only react with copper(II) ions and not the nitrate ions present. 

Iodine is liberated together with copper(I) iodide which is a white precipitate.

Your solution should go a thick creamy colour looking like toffee yogurt (ugh!).

As you titrate this stuff with sodium thiosulphate the mixture goes pale then you can add the starch indicator to give the usual purple colour. 

The end point occurs when this purple colour disappears with the addition of one drop of sodium thiosulphate. 

Here you will find a youtube video of the titration procedure

Here are some instructions you could follow to carry out this titrimetric determination.

Procedure

First, weigh accurately a piece of brass (or your copper coin) it should be about 2.5g.

Second, add the brass to about 20ml of concentrated nitric acid in a 250ml beaker., do this in the fume cupboard.

Third, let the beaker and contents stand in a fume cupboard until the copper has all dissolved. (this take ages!!)

Fourth, very carefully transfer all this mixture to a 250ml volumetric flask with all the washings—the solution should look green, but do not make up to the mark. 

Fifth, now add 2M sodium carbonate solution drop by drop until a faint precipitate appears; follow the sodium carbonate with 2M ethanoic acid drop by drop until the precipitate just dissolves. 

Sixth, now you can make up to the mark with distilled water!!

Seventh, pipette 25ml of your green copper(II) solution into a 250ml conical flask, add about 10ml of 1M potassium iodide solution (measuring cylinder will do) to liberate the iodine and precipitate copper(I) iodide and then titrate with standardised 0.100M sodium thiosulphate solution.      

Eighth, when the iodine colour in the flask is pale brown (remember this looks like toffee yoghurt!) add 1ml starch solution to turn it purple.

Ninth, continue titrating with 0.100Msodium thiosulphate solution until one drop removes the purple starch/iodine colour and leaves a creamy-white colour in the flask.

You will concordant results i.e. three titres to with 0.1ml.

Here is a typical set of results

Mass of brass sample 2.67g

Titration results:

Pipette solution:    copper(II)                                       x  mol/dm3
Burette solution:    Sodium thiosulphate                    0.100mol/dm3
Indicator:                 starch solution

Burette readings:               Rangefinder            1                     2                     3
Final reading (ml):           25.30                        25.10              25.20           25.20
First reading (ml):              0.00                         0.00               0.00               0.10
Volume NaOH (ml):           25.30                        25.10             25.20           25.10
Average titre (ml):                        25.1(3)

1.    Work out amount of sodium thiosulphate used.

Amount thiosulphate   =    25.1  ×  0.100   =   2.51mmol.

2.    This amount of thiosulphate reacts with a tenth of the mixture in the flask so amount thiosulphate equivalent to all the liberated iodine is 25.1mmol.

3.    Use the reaction equations to determine the amount of copper in the sample of brass

2Na2S2O3 (aq)    +   I2 (aq)   =    2NaI (aq)      +    Na2S4O6 (aq)
25.1mmol          12.55mmol

2Cu 2+ (aq)    +    4I (aq)   =   2CuI(s)   +             I2 (aq)
25.1mmol                                                          12.55mmol

So the amount copper in the brass sample was 25.1mmol.

4.    Work out the mass of copper in this amount.

Mr (Cu)  =  63.5 g/mol  therefore mass of copper is

63.4  ×  25.1/1000  =   1.59g  copper.

5.    Work out percentage of copper in the brass sample as

Mass of copper ×  100/mass of sample   =   % copper

1.59  ×   100/2.67    =   59.6%


Extension questions:

1.    You could analyse the redox equations and work out the oxidation number changes
2.    You could estimate the percentage error in the final result and work out where the greatest error is likely to be. 
3.    You could work out the kind of complex ions copper makes with nitrogen compounds not just nitric acid.
4.    You could look up the way copper reacts with nitric acid under different conditions.




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