Download Unit 2.7: Periodic Table Group1 Group2 Li Be Na Mg K Ca Rb Sr Cs

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Group 2 note
Unit 2.7: Periodic Table
Group 2
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Group 1 and 2 elements are known as alkali metals and alkaline earth metals
respectively.
All are Solids at room temperature
Elements of group 1 and 2 together known as S block elements
All are metallic elements.
Group 2 elements form cations with charge 2+ and are conductors of electricity
Physical properties
Group1
Li
Na
K
Rb
Cs
Fr
Group2
Be
Mg
Ca
Sr
Ba
Ra
Group 2 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 decrease down the group because the
number of shells and distance between the nucleus and the outer electrons increase down the group, as a result
the electrons in the inner shells shield outer 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 less
energy is required to remove them
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 increases down ward
and become easier to remove the two outer electrons,
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
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Group 2 note
Group
Species
Flame colour
Li+
Crimson red (magenta)
Na+
Yellow
K+
Lilac
1
2
Ca2+
Brick 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 of visible 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
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Group 2 note
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 .
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Group 2 note
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
Thermal stability of group 1 and group 2 Nitrates and Carbonates
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Cations with small size and high charge 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 most stable
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
Be(NO3)2
Mg(NO3)2
Ca(NO3)2
Sr(NO3)2
Ba(NO3)2
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Decomposition
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Group 2 note
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 carbonates decompose 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
BeCO3
MgCO3
CaCO3
SrCO3
BaCO3
Decomposition
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.
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Group 2 note
Indicator
Methyl Orange
Bromophenol blue
Methyl red
Bromothymol blue
Phenolphthalein
Colour in acid
Red
Yellow
Red
Yellow
colourless
Colour in alkali
Yellow
Blue
Yellow
Blue
Purple
End point colour
Orange
Green
Orange
Green
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 NaOH solution.
Calculate concentration of NaOH solution.
2. 25.0 cm3of 0.0504 mol/dm3 solution 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.
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Group 2 note
5.Calculate the volume of 0.100mol/dm3 HCl solution required to neutralize 25.0cm3of 0.0567mol/dm3 Na2CO3
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 25.0cm 3 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
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Group 2 note
 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 a 0.0485 mol/dm3 solution of Phosphoric
(III) acid. Work out equation for this acid base reaction.
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