Group 7 note / 1

Unit 2.7: Periodic Table

Group1 / Group2
Li / Be
Na / Mg
K / Ca
Rb / Sr
Cs / Ba
Fr / Ra

Group 2

  • Group 1 and 2 elements are known as alkalimetals andalkalineearthmetals respectively.
  • All are Solids at room temperature
  • Elements of group 1 and 2 together known asSblock elements
  • All are metallic elements.
  • Group 2 elements form cations with charge 2+ and are conductorsof electricity

Physical properties

Group2 metals have higher melting temperature than group1 metals in the same period.This is because each atom loses two electrons to form the metallic bond, which is therefore stronger than metallic bond in group 1 metal and also metallic radius of group2 elements is smaller than group1 elements in the same period. The delocalized electrons in the metal are mobile. Therefore they can move across an electric potential and hence group 2 metals act as good conductors of electricity.

The first and second ionization energies of elements in group2 decreasedown the group because the number of shells and distance between the nucleusand the outer electrons increase down the group, as a result the electrons in the inner shellsshieldouter electrons from the nuclear charge. The effect of the increase in nuclear charge is outweighed by the increase in shielding so that the outer electrons are held less firmly and lessenergy is required to remove them

Electronegativities of group 2 elementsdecreases down the group from 1.5 for Be to 0.9 forRa. Polarising power of group 2 metal ions decreases down the group

Elements / Ionic radius / Lattice energy of chlorides KJ/mol / Difference
Born-Haber value / Calculated value
Mg / 0.065 / -2526 / -2326 / 200
Ca / 0.099 / -2258 / -2223 / 35
Sr / 0.11 / -2156 / -2127 / 29
Ba / 0.14 / -2056 / -2033 / 23

Greater the difference in lattice energy value more covalent the molecule.

REACTIONS OF GROUP 2 METAL

All the metals of group 2 show 2+ oxidation state in their compounds.

Reactivity increases down the group, this is because the size of the atom increasesdown ward and become easierto remove the twoouterelectrons,

Reaction with Oxygen

All the group 2 metals burn in air and form ionic oxides of formula MO.

Eg:

Flame test

Flame colours of compounds of group 1 and 2 elements

Group / Species / Flame colour
1 / Li+ / Crimson red (magenta)
Na+ / Yellow
K+ / Lilac
2 / Ca2+ / Yellow red
Sr2+ / Red
Ba2+ / Pale green (Apple Green)

Explanation of Flame colour

These colours are caused because heat causes the compound to vapourise and produce some atoms of metals with electron in a higher orbital than the ground state.

When the electrons falls back to its ground state energy is released in the form ofvisible light

The light that is emitted is of a characteristic frequency and hence colour dependent on the energy level difference between the two shells.

Electronic transition on heating in a flame

Reaction with water

Beryllium does not react with water

Mg reacts slowly with cold water to produce an alkaline suspension of Mg(OH)2 and H2

Equation:

When heated in steam, Mg burns producing MgO and H2

Equation:

Ca,Sr and Ba react rapidly with cold water to produce alkaline solutions of the metal─ hydroxide and bubbles of hydrogen gas.

Equation:

Reaction with Chlorine

All the group 2 metals react when heated in Chlorine.

Equation:

Be forms a covalent anhydrous chloride. The other group 2 metals form ionic chlorides.

The chlorides have the general formula MCl2

Preparation of chlorides

Direct synthesis: The metal and chlorine heated together.

Equation:

Neutralization: Neutralization of the oxide or carbonate by dilute HCl produces metal chlorides.

Equation:

Chlorides other than BeCl2 dissolves in water producing solutions that containing hydrated cation of formula [M(H2O)6]+ 2 .

Group 2 oxides and hydroxides

Group 2 oxides and hydroxides are bases. The general formulae are MO and M(OH)2.

Oxides and hydroxides of group 2 metals form salts and water with acids

Equation:1

Equation: 2

Beryllium oxide does not react with water.

MgO reacts slowly and incompletely to form a slightly alkaline suspension of Mg(OH)2

Equation:

CaO is called quick lime, reacts very exothermically with water to form alkaline suspension of Ca(OH)2 .

Equation:

Solid Ca(OH)2 is called slaked lime and solutions of it are called lime water

Sr and Ba oxides react with water to form alkaline solution of the hydroxide.

Equation: 1.

2.

Reaction of lime water with CO2

CO2 is an acidic gas reacts with the base Ca(OH)2 to form salt, calcium carbonate and water. calcium carbonate is insoluble and appears as a milky ppt. This reaction is the test for CO2

Equation:

Solubility of group 2 hydroxides

Solubility / PH
Be(OH)2 / insoluble / 7
Mg(OH)2 / 0.0002 / 10.6
Ca(OH)2 / 0.015 / 12.5
SrOH)2 / 0.083 / 13.2
Ba(OH)2 / 0.15 / 13.5

Solubility increases down the group

Be(OH)2 is insoluble

Mg(OH)2 isvery slightly soluble

Ca(OH)2 is slightly soluble

Sr(OH)2 and Ba(OH)2 more s

Solubility of group 2 sulfates

The solubility of group 2 sulfates decreases down the group

Be and Mg sulfates are very soluble

CaSO4 slightly soluble

SrSO4 isvery slightly soluble

BaSO4 is insoluble

Thermal stability of group 1 and group 2 Nitrates and Carbonates

  • Cations with smallionic radius and highcharge has high polarizing power
  • They can polarize the anion effectively, which weakens the bonds in the anion.

This assists decomposition and makes the compound less stable to heat. This means that, in both groups 1 and 2, the ease of decomposition decreases down the groups.

  • Therefore if the cation has a greater polarizing power, its compound is thermally less stable. Thus in group 1, lithium compound would be the least stable while Cs compounds would the most stable
  • In group 2 Be compounds would be the least stable while Ba compounds would the moststable

Decomposition of Nitrate

Group 1 nitrates, except LiNO3 , decomposes to form metal nitrites and oxygen gas.LiNO3 decomposes in a similar manner to group2 nitrites

Nitrates / Decomposition
LiNO3
NaNO3
KNO3
RbNO3
CsNO3

Group 2 nitrates decomposes to form metal oxide, nitrogen dioxide and oxygen gas

Nitrates / Decomposition
Be(NO3)2
Mg(NO3)2
Ca(NO3)2
Sr(NO3)2
Ba(NO3)2

Decomposition of Carbonates of Group 1 and 2

Group 1 carbonates are stable to heat except lithium carbonate, which decomposes in a similar manner to group 2 carbonates. Lithium carbonatesdecompose when heated this is because Li+ cation is very small and polarizes the O-C bond in the CO32- ion sufficiently for it to break and form an O2- ion

Equation:

Group 2 carbonates except barium carbonates decomposes to give metal oxide and carbon dioxide

Carbonates / Decomposition
BeCO3
MgCO3
CaCO3
SrCO3
BaCO3

Experiments to study Thermal decomposition

The relative ease of decomposition of group 2 carbonates and nitrates can be studied by the following experiment.

Place the same amount of each of the salts in a series of hard glass test tubes.

Fix a delivery tube to the test tube and clamp in a stand

Light the burner and measure the time taken for the gas evolved to reach the mark on the test tube in the water bath

Repeat the experiments by changing the salts.

Titrations

If the accurate concentration of the one solution is known, a titration can be used to determine the concentration of another solution. The method involves having one solution in a burette and pipetting a known volume of the other solution into a conical flask. An indicator is then added to the solution in the conical flask and is titrated from the burette until the indicator changes colour.

Indicators

The common acid base indicators are methyl orange and phenolphthalein. Some indicators and their colours are given below.

Indicator / Colour in acid / Colour in alkali / End point colour
Methyl Orange / Red / Yellow / Orange
Bromophenol blue / Yellow / Blue / Green
Methyl red / Red / Yellow / Orange
Bromothymol blue / Yellow / Blue / Green
Phenolphthalein / colourless / Purple / Pale pink

Calculations in Titration

The concentration of one solution is known is called standard solution, the reacting volumes are determined by titration

Calculate the amount of the reagent of known con. Amount=Con.× Volume

Calculate the amount of the other reagent, using stoichiometry of the equation

Calculate the concentration of the other solution. Con=Moles/Volume

  1. In a titration, neutralisation of 25.0 cm3 of 0.100 mol/dm3 HCl solution required 26.8 cm3of NaOHsolution. Calculate concentration of NaOH solution.

2. 25.0 cm3of 0.0504 mol/dm3solution of sulfuric acid was titrated with the solution of NaOH. The mean titre was 27.3 cm3calculate concentration of NaOH solution.

3.Calculate the concentration in g/dm3 of a 0.152 mol/dm3solution of NaOH.

4.What is the concentration in ppm of a solution containing 45ng(nano grams) of a toxic chemical in 100cm3 of water.

5.Calculate the volume of 0.100mol/dm3 HCl solution required to neutralize 25.0cm3of 0.0567mol/dm3Na2CO3 solution.

Back Titration

A back titration is used is used when the substance being investigated is either insoluble or for some other reasons, cannot be titrated directly. The method is as follows.

Weigh a sample of the substance being investigated

Add excess of a standard solution of an acid or a base

Either titrate the excess or make up the solution to 250.cm3 and titrate a portion of the diluted solution

1. 1.41g of a sample of chalk, which is mostly CaCO3 with some inert impurities, was placed in a beaker and 50.0cm3 1.00mol/dm3 HCl solution was slowly added. The equation for the reaction is

When the fizzing had ceased, the contents of the beaker were washed into a standard 250cm3 volumetric flask, made up to the mark and thoroughly shaken. A pipette was used to transfer a 25.0cm3 sample into a conical flask . The sample was titrated against 0.100mol/dm3 NaOH solution .The equation for the reaction is

The titration was repeated. The mean titre was23.6cm3

Calculate the amount of NaOH in the mean titre

Calculate the total amount HCl in excess

Calculate amount of HCl originally taken

Calculate the amount of HCl that reacted with chalk

Calculate the amount of CaCO3 in the chalk sample

Calculate the mass of CaCO3 in the sample and hence the percentage purity

To find the stoichiometry of a reaction

Calculate the amount of one reagent

Calculate the amount of other reagent

Find the whole number ratio of these amounts. This is the reactant ratio which allows the equation to be written

1. 25.0cm3 of a 0.100mol/dm3 solution of NaOH required 25.8cm3 of a0.0485mol/dm3 solution of Phosphoric (III) acid. Work out equation for this acid base reaction.