The Periodic Table consists of blocks of elements

s block

p blockd block

There is a clear link between the Periodic Table and the electronic configuration of an element

1s 1s

2s 2p

3s 3p

4s 3d 4p

ATOMIC RADIUS

Atomic radius decreases from Na to Ar◦ Nuclear charge (number of protons) increases

◦ Electrons are added to the same main energy level

◦ Shielding is constant

The electrons experience a greater electrostatic force of attraction to the nucleus

FIRST IONISATION ENERGY

First ionisation energy increases from Na to Ar◦ Nuclear charge (number of protons) increases

◦ Atomic radius decreases

◦ Shielding is constant

The outermost electron experiences a greater electrostatic force of attraction to the nucleus

ELECTRONEGATIVITY

Electronegativity increases from Na to Cl◦ Nuclear charge (number of protons) increases

◦ Atomic radius decreases

◦ Shielding is constant

It becomes easier to attract the electrons of a covalent bond.

MELTING POINT / K Period 2

0

1000

2000

3000

4000

5000

Giant lattice Simple

molecular

Atomic

MELTING POINT / K Period 3

0

500

1000

1500

2000

Giant lattice Simple

molecular

Atomic

MELTING POINT

Metals have giant lattice structures

Strong electrostatic forces of attraction between positive metal ions and their delocalised electrons

Melting point increases from Na to Al◦ Size of ionic charge increases

◦ Number of delocalised electrons increases

◦ Size of ion (ionic radius) decreases

MELTING POINT

The Group IV element has the highest melting point in each period

Giant lattice structure

Strong covalent bonds between atoms

MELTING POINT

Group V, VI, and VII elements have very low melting points

Simple molecular structure

Covalent bonding

Weak van der Waals’ forces of attraction between molecules

OXIDATION

Loss of electrons (OILRIG)

Increase in oxidation state

REDUCTION

Gain of electrons (OILRIG)

Decrease in oxidation state

OXIDISING AGENT

An electron acceptor

The oxidising agent is always reduced.

REDUCING AGENT

An electron donor

The reducing agent is always oxidised.

Oxidation states in a compound add up to zero

Oxidation states in an ion add up to the charge

The most electronegative element has a negative oxidation state and all the other elements have positive oxidation states

Element Oxidation state

Uncombined element 0

Hydrogen –1 or +1

Oxygen –2

Group I +1

Group II +2

Group VII –1 to +7

Used to represent oxidation or reduction alone

Note the reagent and product species given.

Balance the element oxidised or reduced.

Balance charge by adding electrons to one side. (OILRIG)

Make the number of electrons in each half equation the same.

Add the two half equations together.

Atomic radius increases from Be to Ba

Atomic number increases so there are more electrons

More main energy levels are needed to accommodate these

Each additional main energy level is further from the nucleus

For the elements of Group II, ionic radius is

always smaller than the atomic radius.

On ionising, the Group II atom loses both the electrons in its outermost energy level

It loses its outermost energy level

First ionisation energy decreases from Be to Ba

The outermost electron is easier to remove◦ Despite more protons in the nucleus

◦ Because atomic radius increases

◦ Because there are more occupied energy levels shielding it from the nucleus

It experiences a weaker electrostatic force of attraction

Electronegativity decreases from Be to Ba

Nuclear charge (number of protons) increases

Atomic radius increases

Shielding increases

Thus it becomes more difficult to attract the electrons of the bond

Melting point decreases from Be to Ba

Size of ionic charge stays the same

Number of delocalised electrons stays the same

Size of ion (ionic radius) increases

When the electrons are further from the centres of positive charge in the ions, the electrostatic forces of attraction are weaker

Atoms react by losing two electrons to form an M2+ ion

The oxidation state of the elements increases from 0 to +2

The elements are oxidised

Group II elements are reducing agents (electron donors)

Reactivity of the elements increases from beryllium to barium.

First and second ionisation energies decrease from beryllium to barium◦ There is an increase in atomic radius ◦ There is an increase in the number of occupied

energy levels shielding the nucleus◦ It becomes easier to lose two electrons◦ They experience a weaker electrostatic force of

attraction

Reactions of elements (Redox reactions)

With oxygen2Ca(s) + O2(g) 2CaO(s) With waterCa(s) + 2H2O(l) Ca(OH)2(aq) + H2(g)

alkaline With acidCa(s) + 2HCl(aq) CaCl2(aq) + H2(g)

Reactions of oxides

With water

CaO(s) + H2O(l) Ca(OH)2(aq)

alkaline

With acid

CaO(s) + 2HCl(aq) CaCl2(aq) + H2O(l)

Reactions of hydroxides

With water

Ca(OH)2(s) + (aq) Ca(OH)2(aq)

alkaline

With acid

Ca(OH)2(s) + 2HCl(aq) CaCl2(aq) + 2H2O(l)

Reactions of carbonates

With heat

CaCO3(s) CaO(s) + CO2(g)

With acid

CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

Testing for carbon dioxide

Ca(OH)2(aq) + CO2(g) CaCO3(s) + H2O(l)

Atomic radius increases from fluorine to

iodine.

Atomic number increases so there are more electrons

More main energy levels are needed to accommodate these

Each additional main energy level is further from the nucleus

For the elements of Group VII, ionic radius is

always larger than the atomic radius.

A halide ion always has one more electron than the atom from which it was formed. This electron enters the outermost energy level of the atom and there is an increase in the mutual electrostatic forces of repulsion between the negatively charged electrons.

First Ionisation Energy decreases from fluorine

to iodine.

The outermost electron is easier to remove◦ Despite more protons in the nucleus

◦ Because atomic radius increases

◦ Because there are more occupied energy levels shielding it from the nucleus

It experiences a weaker electrostatic force of attraction

Electronegativity decreases from fluorine to

iodine.

Nuclear charge (number of protons) increases

Atomic radius increases

Shielding increases

Thus it becomes more difficult to attract the electrons of the bond

Boiling point increases from fluorine to iodine.

Size of molecules increases

Number of electrons increases

More temporary induced dipoles occur

Strength of van der Waals’ forces increases

More heat energy is needed to overcome these forces.

Volatility decreases from fluorine to iodine.

Size of molecules increases

Number of electrons increases

More temporary induced dipoles occur

Strength of van der Waals’ forces increases

More heat energy is needed to overcome these forces.

Oxidising agents are electron acceptors

Fluorine, chlorine, bromine and iodine are all powerful oxidising agents

The ability to behave as oxidising agents decreases from fluorine to iodine◦ There is an increase in atomic radius

◦ There is an increase in the number of occupied energy levels of electrons shielding the nucleus

◦ It becomes more difficult to attract an electron.

Reducing agents are electron donors

Fluoride, chloride, bromide and iodide ions are all powerful reducing agents

The ability to behave as reducing agents increases from fluoride to iodide ions◦ There is an increase in ionic radius ◦ There is an increase in the number of occupied

energy levels of electrons shielding the nucleus◦ It becomes easier to donate an electron.

Names of the Halogen elements end in INE

Names of the Halide ions end in IDE

IDE

Cl-(aq) Br-(aq) I-(aq)

INE

Cl2(aq) x √ √

Br2(aq) x x √

I2(aq) x x x

Redox reactions (displacement)

Cl2(aq) + 2Br-(aq) 2Cl-(aq) + Br2(aq)

Cl2(aq) + 2I-(aq) 2Cl-(aq) + I2(aq)

Br2(aq) + 2I-(aq) 2Br-(aq) + I2(aq)

Orders of reactivity:

Chlorine > Bromine > Iodine

Iodide > Bromide > Chloride

Redox reactions (disproportionation)

Cl2(aq) + H2O(l) HCl(aq) + HClO(aq)

Cl2(aq) + 2NaOH(aq) NaCl(aq) + NaClO(aq) + H2O(l)

Cl2(aq) + 2OH-(aq) Cl-(aq) + ClO-

(aq) + H2O(l)

In a disproportionation reaction, one species is

both oxidised and reduced

Testing for halide ions in aqueous solution

Ag+(aq) + Cl-(aq) AgCl(s) White precipitate soluble in dil. NH3(aq)

Ag+(aq) + Br-(aq) AgBr(s) Cream precipitate soluble in conc. NH3(aq)

Ag+(aq) + I-(aq) AgI(s) Pale yellow precipitate insoluble in NH3(aq)