Atomic Structure

1.8 recall that atoms consist of a central nucleus, composed of protons and neutrons, surrounded by electrons, orbiting in shells

The centre of an atom is the nucleus, which is composed of protons and neutrons. Their masses are roughly equal and since the mass of the electron is pretty much negligible, most of the mass of an atom is in the nucleus. 

The electrons are found in a series of energy levels which you call shells at IGCSE. Each 'shell' can only hold a certain number of electrons, these shells can be thought of as getting progressively further from the nucleus. Electrons will always go into the lowest possible energy level, provided there is space. The first shell can only hold 2 electrons, then the shells after that can hold a maximum of 8. 

In a diagram, the electrons are shown on circles around the nucleus. Beware that these circles are just imaginary lines to help you understand that the electrons orbit around the nucleus, at IGCSE level you just need to accept that. 

For those moving on to higher levels of Chem, the 'truth' is different, it's actually not really possible to plot the path of an electron. For more information look at the following link, to be honest I don't really understand it myself, it's so confusing. 

http://www.chemguide.co.uk/atoms/properties/atomorbs.html#top

Later in the course you will come across  dot-and-cross diagrams. Dots or crosses are used to represent electrons, in the above, Carbon has 4 outer shell electrons. They are drawn far apart even though you could draw them close to each other like in the first shell, this is because electrons would repel each other as they have the same negative charge. (Remember like charges repel). So only if you have more than 4 OSE, then do you draw them in pairs. Remember, draw the 4 OSE like in the diagram of carbon above, then pair up any OSE left. The following diagram might help you understand:

This way of drawing electrons is clear and makes it easy to count too. When you learn about ions, dot-and-cross diagrams are useful and they help you see how the electrons are transferred. Like in the above diagram, the Chlorine atom gains one electron (the cross) from the sodium atom to become a chloride ion (Cl^-), whilst the sodium atom becomes a sodium ion (Na⁺). It is positive because it lost one electron, so it has one more proton than electron now. :)

1.9 recall the relative mass and relative charge of a proton, neutron and electron

	Relative mass	Relative charge
Proton	1	+1
Neutron	1	0
Electron	1/1840 (negligible)	-1

Extra (you don't need to know this): 

Protons and neutrons don't actually have exactly the same mass - neither of them has a mass of exactly 1 on the carbon-12 scale (the scale on which the relative masses of atoms are measured). On the carbon-12 scale, a proton has a mass of 1.0073, and a neutron a mass of 1.0087. 

1.10 understand the terms atomic number, mass number, isotopes and relative atomic mass (A_r)

Atomic number is the number of protons there are in the nucleus, it is sometimes called the proton number, though atomic number should be more accurate because atoms are electrically neutral, the number of protons and electrons are equal. (Protons have a charge of +1 whilst electrons are -1, so they cancel each other out.) So the atomic number tells you the number of protons and the number of electrons. 

Mass number is the number of protons and neutrons. It is sometimes known as the nucleon number, because protons/neutrons are nucleons. So if a question asks you for the number of neutrons in an atom, mass number - atomic number = no. of neutrons. 

Isotopes: these are atoms which have the same atomic number but different mass numbers, i.e. same number of protons but different number of neutrons.

The number of neutrons in an atom can vary a little. For instance, there are three kinds of carbon atoms ¹²C, ¹³C and ¹⁴C. Their number of neutrons varies but they have the same number of protons, because each element's atomic number is unique. If it has a different number of protons, it wouldn't be the same element anymore. So these atoms are isotopes of carbon. Bear in mind that the fact that they have varying numbers of neutrons makes no difference whatsoever to the chemical reactions of the carbon. Though their physical properties may vary. 

	Protons	Neutrons	Mass number
Carbon-12	6	6	12
Carbon-13	6	7	13
Carbon-14	6	8	14

Relative atomic mass (web definition): the ratio of the average mass per atom of the naturally occurring form of an element to one-twelfth the mass of an atom of carbon-12. Symbol A_r Abbreviation r.a.m. I find it easier to think of it as:
The number of times the mass of one atom of an element is heavier than 1/12 of the mass of a carbon-12 atom. Anyways, this number is always at the top of an element's symbol, it's always either mass number or RAM at the top, atomic number at the bottom - as shown below:

mass number is always on the top, atomic number at the bottom, don't mix them up. at least you know that the mass number is always greater than atomic number

1.11 calculate the relative atomic mass of an element from the relative abundances of its isotopes
You multiply the relative abundance of each isotope by its mass number, add these together, and divide by 100. It's easier to understand through an example, in this case I'll use chlorine, since it's pretty common.

   and  

Chlorine consists of 75% Chlorine-35 and 25% Chlorine-37. You can think of the data as 100 atoms, 75 having a mass of 35 and 25 with a mass of 37. So the calculation is:
[(75 x 35) + (25 x 37)] / 100 = 35.5
So the RAM of chlorine, or A_r(Cl) is 35. 
(There are tiny percentages of other chlorine isotopes but the two shown above are the most common, and so the rest are ignored at IGCSE level.)

The RAM of an element will be closer to the mass number of the more abundant isotope. For example, the RAM of chlorine is 35.5, which is closer to chlorine-35, because it is the more abundant isotope. Obviously 75% > 25%!.


1.12 understand that the Periodic Table is an arrangement of elements in order of atomic number
The number of protons in the element's atom increases across the Periodic Table as you've probably noticed in yours.


1.13 deduce the electronic configurations of the first twenty elements from their positions in the Periodic Table
To work out the electronic arrangement of an atom:

• Look up the atomic number in the Periodic Table - making sure that you choose the right number if two numbers are given. The atomic number will always be the smaller one and tends to be below the symbol.
• This tells you the number of protons, and hence the number of electrons.
• Arrange the electrons in levels, always filling up an inner level before you go to an outer one. 

e.g. to find the electronic arrangement in oxygen

• the Periodic Table gives you the atomic number of 8.
• Therefore there are 8 protons and 8 electrons.
• The arrangement of the electrons will be 2,6. (First shell only holds 2 electrons, then there's 6 left which occupy the second shell.)

1.14 deduce the number of outer electrons in a main group element from its position in the Periodic Table
If you look at the patterns in the table:

• The number of electrons in the outer level is the same as the group number. (Except with helium which has only 2 electrons. The noble gases are also usually called group 0 - not group 8.) This pattern extends throughout the Periodic Table for the main groups (i.e. not including the transition elements). 
• So if you know that barium is in group 2, it has 2 outer shell electrons (btw, outer shell electrons which I will abbreviate to OSE are also known as valence electrons); iodine is in group 7, so it has 7 OSE, lead is in group 4, so surprise surprise, it has 4 OSE. 
• Noble gases have full outer shells. Thus they are unreactive. More on them in this post:  http://askmichellechemistry.blogspot.com/2012/03/periodic-table.html

More on Electrolysis

Note: This post is mainly for Single Science although it could be good background information for Double Award anyway. :)

SS 1.53 describe simple experiments for the electrolysis, using inert electrodes, of aqueous solutions of sodium chloride, copper (II) sulfate and dilute sulfuric acid and predict the products

So, electrolysis can be used to decompose molten compounds as described in an earlier post on Electrolysis: 

http://askmichellechemistry.blogspot.com/2012/04/electrolysis.html

However, for Single Science, you also need to know about electrolysis of compounds in aqueous solutions. Predicting the reactions and working out the products for aqueous solutions are less straightforward than for molten compounds. 

An aqueous solution of a compound is a mixture of two electrolytes, it's a compound dissolved in water really (so water is the solvent). For example, an aqueous solution of copper (II) sulphate contains two electrolytes: copper (II) sulphate and water. It therefore contains copper (II) sulphate ions (Cu²⁺) and sulphate ions (SO₄^2-), and also small amounts of hydrogen ions (H+) and hydroxide ions (OH^-) from the dissociation of water molecules.

H₂O (l) à H⁺(aq) + OH^-(aq)

These ions compete with the ions from copper (II) sulphate for discharge at the electrodes. 

In general, when an aqueous solution of an ionic compound is electrolysed, a metal or hydrogen is produced at the cathode. At the anode, a non-metal, for example oxygen or a halogen, is given off. 

Let's see if this is the case in the electrolysis of dilute sodium chloride solution. 

Electrolysis of Dilute Sodium Chloride Solution

Note: There is a difference in the products between the electrolysis of dilute sodium chloride solution and concentrated sodium chloride solution. I will elaborate later.

An aqueous solution of sodium chloride contains four different types of ions. They are

• Ions from sodium chloride – Na⁺ (aq) and Cl^- (aq)
• Ions from water – H⁺ (aq) and OH^- (aq)

When dilute sodium chloride solution is electrolysed using inert electrodes, the Na⁺ and H⁺ ions are attracted to the cathode. The Cl^- and OH^- ions are attracted to the anode. 

At the cathode: 

The H⁺ and Na⁺ ions are attracted to the platinum cathode. H⁺ ions gains electrons from the cathode to form hydrogen gas. (The hydrogen ions accept electrons more readily than the sodium ions. As a result, H⁺ ions are discharged as hydrogen gas, which bubbles off. I will explain why H⁺ ions are preferentially discharged later.)

2H⁺(aq) + 2e^- à H₂(g)

Na⁺ ions remain in solution. 

At the anode:

OH^- and Cl^- are attracted to the platinum anode. OH^- ions give up electrons to the anode to form water and oxygen gas. 

4OH^-(aq) à 2H₂O(l) + O₂(g) + 4e^-

Cl^- ions remain in solution. 

Summary: 

The overall reaction is: 

  2H₂O(l)   à 2H₂(g) + O₂(g)

Since water is being removed (by decomposition into hydrogen and oxygen), the concentration of sodium chloride solution increases gradually. The overall reaction shows that the electrolysis of dilute sodium chloride solution is equivalent to the electrolysis of water.

Another important thing to note is that twice as much hydrogen is produced as oxygen. This is because for every 4 electrons that flows around the circuit, you would get one molecule of oxygen. But four electrons would produce 2 molecules of hydrogen. Hence in a diagram, you would see the volume of hydrogen produced is twice that of oxygen. Refer to the equations above and note the number of electrons involved to help you understand. 

This diagram is just to illustrate how twice as much hydrogen gas is produced. 

Electrolysis of Concentrated Sodium Chloride Solution

The only difference is that at the anode, Cl^-  ions are more numerous than OH^- ions. Consequently, Cl^- ions are discharged as chlorine gas, which bubbles off. 

2Cl^- (aq) à Cl₂(g) + 2e^-

The OH^- ions remain in solution.   

One volume of hydrogen gas is given off at the cathode and one volume of chlorine gas is produced at the anode. The resulting solution becomes alkaline because there are more OH^- than H⁺ ions left in the solution.    

Comparison:

Compare the electrolysis of molten sodium chloride, dilute sodium chloride solution and concentrated sodium chloride solution:

Molten sodium chloride:

• Cathode: Na⁺ ions discharged
• Anode: Cl^- discharged

Dilute NaCl solution: 

• Cathode: H⁺ ions discharged
• Anode: OH^- discharged

Concentrated NaCl solution:

• Cathode: H⁺ discharged
• Anode: Cl^- discharged

So you can see that Na⁺ and Cl^- ions are not always discharged even though in all 3 of the above, the electrolytes contained these ions. For example in the electrolysis of dilute NaCl solution, H⁺ are discharged in preference to Na⁺ ions. OH^- ions are discharged in preference to Cl^- ions. Before I talk about the electrolysis of copper(II) sulfate and dilute sulfuric acid, I will discuss why one type of cation (or anion) in the electrolyte is more readily discharged than another type. If you already know this, just scroll right down. :)

Note: Most of the following is taken from a G.C.E. 'O' Level textbook, but I find it useful. :) 

Reactivity Series and Selective Discharge of Ions

In electrolysis, when more than one type of cation or anion is present in a solution, only one cation and one anion are preferentially discharged. This is known as the selective discharge of ions. 

How do you predict which ions are discharged in the electrolysis of a compound in aqueous solution?

If inert electrodes are used during electrolysis, the ions discharged and hence the products formed depend on three factors:

1. The position of the metal (producing the cation) in the reactivity series. 
2. The relative ease of discharge of an anion. 
3. The concentration of the anion in the electrolyte. 

The ease of discharge of cations and anions during electrolysis is shown below.

Cations	NB: Ease of discharge increases as you go down the table	Anions
Potassium ion, K+		Chloride ion, Cl-
Sodium ion, Na+		Bromide ion, Br-
Calcium ion, Ca2+		Iodide ion, I-
Magnesium ion, Mg2+		Hydroxide ion, OH-
Zinc ion, Zn2+		Note: sulphate ions (SO42-) and nitrate ions (NO3-) will not be discharged during electrolysis.
Iron ion, Fe2+
Lead ion, Pb2+
Hydrogen ion, H+
Copper ion, Cu2+
Silver ion, Ag+

Selective discharge of cations during electrolysis

The cations of an element lower in the reactivity series are discharged at the cathode in preference to cations above it in the solution. This is because cations of a less reactive element accept electrons more readily. For example, if a solution containing Na⁺ and H⁺ ions is electrolysed,  H⁺ ions are discharged in preference to Na⁺ ions. The more reactive the metal, the more stable its compound. They have lost a lot of energy and have lost electrons to form stable cations, so cations lower down the reactivity series are more readily discharged.

Selective discharge of anions during electrolysis
Sulphate (SO₄^2-) and nitrate (NO₃^-) ions remain in the solution and are not discharged during electrolysis. If a solution containing SO₄^2-, NO₃^- and hydroxide (OH^-) ions is electrolysed, the OH^- ions will be discharged in preference to SO₄^2- and NO₃^- ions. The OH^- ions give up electrons most readily during electrolysis to form water and oxygen.

4OH^- (aq) à 2H₂O (l) + O₂ (g) + 4e^-  

Effect of concentration on selective discharge of anions

An increase in the concentration of an anion tends to promote its discharge. For example, in the electrolysis of concentrated sodium chloride solution, two types of ions are attracted to the anode: Cl^- and OH^- ions. According to their relative ease of discharge, OH^- ions should be discharged preferentially. However, in concentrated sodium chloride solution,  Cl^- ions are far more numerous than OH^- ions and so are discharged at the anode instead. 

2Cl^- (aq) à Cl₂ (g) + 2e^-

What are the general rules for predicting selective discharge?

The following rules can be applied when predicting the products of electrolysis of any aqueous solution (using inert electrodes):

Rule 1	Identify the cations and anions in the electrolysis. Remember that an aqueous solution also contains H+ and OH- ions from the dissociation of water molecules.
Rule 2	At the anode, the product of electrolysis is always oxygen unless the electrolyte contains a high concentration of the anions, Cl-, Br­- or I- ions.
Rule 3	At the cathode, reactive metals such as sodium and potassium are never produced during electrolysis of the aqueous solution. If the cations come from a metal above hydrogen in the reactivity series, then hydrogen will be liberated (liberate=release). If the cations come from a metal below hydrogen, then the metal itself will be deposited.
Rule 4	Identify the cations and anions that remain in the solution after electrolysis. They form the product remaining in solution. Summarise the reactions. For example, in the electrolysis of dilute sodium chloride solution, Na+ and Cl- ions remain in solution after H+ and OH- ions have been discharged. Hence the solution of sodium chloride becomes more concentrated after electrolysis.

Electrolysis of Copper (II) sulphate solution
Copper (II) sulphate solution can be electrolysed using inert platinum electrodes. (Sometimes inert carbon electrodes in the form of graphite are used.)

During electrolysis, the cathode is coated with a layer of reddish-brown solid copper. The blue colour of the solution fades gradually as more copper is deposited. The resulting electrolyte also becomes increasingly acidic.

An aqueous solution of copper (II) sulphate contains four types of ions:

• Ions from copper (II) sulphate: Cu²⁺ and SO₄^2-
• Ions from water: H⁺ and OH^-

At the anode:
OH^- ions and SO₄^2- ions are attracted to the anode. OH^- ions give up electrons more readily than SO₄^2- ions. Consequently, OH^- ions are preferentially discharged to give oxygen gas.

4OH^- (aq) à 2H₂O (l) + O₂ (g) + 4e^-

The SO₄² ions remain in solution. 

At the cathode:

H⁺ ions and Cu²⁺ ions are attracted to the cathode. Copper is lower than hydrogen in the reactivity series. Cu²⁺ ions accept electrons more readily than H⁺ ions. As a result, Cu²⁺ ions are preferentially discharged as copper metal (atoms). 

Cu²⁺ (aq) + 2e^- à Cu (s)

The H⁺ ions remain in solution. 

Summary: 

When aqueous copper (II) sulphate is electrolysed using platinum electrodes, copper metal is deposited at the cathode and oxygen gas is given off at the anode. The overall reaction is:

2CuSO₄ (aq) + 2H₂O (l) à 2Cu (s) + O₂ (g) + 2H₂SO₄ (aq)

Electrolysis of dilute sulfuric acid solution
Inert carbon or platinum electrodes are used.
At the cathode:
In this case, the only positive ions arrive at the cathode are the hydrogen ions from the acid and the water. (Adding acid to water forces it to split up/hydrolyse.) These are discharged to give hydrogen gas.

2H⁺ (aq) + 2e^- à H₂ (g)

At the anode:

At the anode, SO₄^2- ions  and OH^- ions (from the water) accumulate. OH^- ions are discharged to give O₂ gas. 

4OH^- (aq) à 2H₂O (l) + O₂ (g) + 4e^-

The amount of hydrogen produced is twice that of oxygen. Just like in the electrolysis of dilute sodium chloride solution.  For every 4 e^- that flows around the circuit, you would get one molecule of O₂ . But four electrons would produce 2 molecules of H₂.