Unit 2 – group 2 & 7 (final)

• Electronegativity the ability of an atom to
attract electrons in a covalent bond
• Trend in electronegativity  decreases as
you go down the group
• Explanation
1. As you down the group electron density
increases
2. So shielding increases
3. This results in the outer electrons being
further away from the nucleus so there is
a weaker electrostatic attraction
Topic 5&6 group 2
& 7 (halogens)
Group 7- trends in physical properties
 understand the trends in electronegativity and boiling point of the halogens
Halogen
Electronegativity
F 4.0
Cl 3.0
Br 2.8
I 2.5

• Trend in boiling (&melting)
point increases as you go down
the group
• Explanation
1. The molecule size gets larger
2. As the number of electrons
increase increasing shielding
3. Increasing the strength of van
der waals forces between
molecules requiring more
energy to overcome
Topic 5&6 group 2
& 7 (halogens)
Group 7- trends in physical properties
 understand the trends in electronegativity and boiling point of the halogens
Halogen Melting point
/oC
Boiling point
/oC
Fluorine -220 -188
Chlorine -101 -35
Bromine -7 59
Iodine 114 184

• F2(g) + 2e  2F-(aq)
• Cl2(g) + 2e  2Cl-(aq)
• Br2(l) + 2e  2Br-(aq)
• I2(aq) + 2e  2I-(aq)
• All the halogens are oxidising
agents, as they can accept
electrons and get reduced
Topic 5&6 group 2
& 7 (halogens)
Group 7- Trends in the oxidising abilities of the halogens
 understand that the ability of the halogens (from fluorine to iodine) to oxidise
decreases down the group (e.g. the displacement reactions with halide ions in
aqueous solution)
Oxidising
agent 
reduced
Oxidising
agent a
reagent that
oxidises
(removes
electrons
from another
species)

• Trend in oxidising ability of halogens increases as we go
up the group (decreases as you go down)
Fluorine is the most oxidising agent > chlorine> bromine>
iodine is the least oxidising agent
• Explanation
1. As you go up the group there are fewer numbers of
electrons
2. So there is less shielding
3. Distance from the outer electron and nucleus is less =
stronger electrostatic attraction
4. This therefore means it is easier to attract an electron
Topic 5&6 group 2
& 7 (halogens)
Group 7- Trends in the oxidising abilities of the halogens
 understand that the ability of the halogens (from fluorine to iodine) to oxidise
decreases down the group (e.g. the displacement reactions with halide ions in
aqueous solution)

Displacement reaction-
KCl KBr KI
Cl2 ------------------------- Orange solution
= Br 2
Brown solution =
I 2
Br2 Nothing -------------------------- Brown solution
I2 Nothing Nothing --------------------------
K – The potassium ion is a spectatar ion and does
nothing we are looking at the Cl- chlorine ion
(halide ions)
Cl2 + KBr  Br2 + 2KCl
Ionic equation take away spectatar ion
Cl2  Br2
Cl2 + 2Br-  Br2 + 2Cl-
Displacement = redox reaction
Cl2 2Cl- [0  -1] = reduction
 oxidising agent
2Br-  Br2 [-1  0] = oxidation
 Reducing agent

Formation of bleach
• 2NaOH + Cl2 NaClO + NaCl +H2O
• Sodium chlorate NaClO  bleach – used for cleaning
• Chlorine oxidation state [0 -1] (cl2 NaCl) and [0 +1] (cl2
NaClO)
• Chlorine is both oxidised and reduced= disproportionation reaction
• Cl2 + H20 < = = >HClO + HCl
• 0 +1 -1
• In UV light (sunlight) this reaction happens
• 2cl2 + 2H2o  4HCl +O2
• Chlorine Is used to treat water because the benefit outweighs the
risk of being toxic as it disinfects water killing bacteria

• Halides = reducing agents
• Trends in reducing ability increases as
we go down the group
• 2F-(aq)  F2(g) + 2e
• 2Cl-(aq)  Cl2(g) + 2e
• 2Br-(aq)  Br2(aq) + 2e
• 2I-(aq)  I2(aq) + 2e
Topic 5&6 group 2
& 7 (halogens)
Group 7- Trends in the reducing abilities of the halide ions
 understand the trend in reducing ability of the halide ions
 know the different products formed by reaction of NaX and H2SO4
Reducing
agent
oxidised
*ions

• Trends of reducing ability of halogens increases as
you go down the group
• Explanation
1. The number of electrons increase in the ion
2. So shielding increases
3. The outer electrons and nucleus are further away
 so there is a weaker electrostatic attraction
4. This therefore means electrons are more easily
lost
I->Br->Cl-
Topic 5&6 group 2
& 7 (halogens)

Recations of halides with sulphuric acid
H2SO4-
• NaCl/ NaF (smallest = weak reducing agent)
• Form Sodium hydrogen sulphate
• NOT REDOX NO CHANGE IN OXIDATION
STATE
• NaCl +H2SO4  NaHSO4 +HCl
• NaF+ H2SO4  NaHSO4 + HF
• NaBr + H2SO4  NaHSO4 + HBr
Topic 5&6 group 2
& 7 (halogens)
Misty white fumes of
hydrogen chloride/
fluoride / Bromide

Recations of halides with sulphuric acid
H2SO4-
NaBr + H2SO4  NaHSO4 + HBr
Because hydrogen bromide is a stronger
reducing agent it can react further
brown fumes
2HBr + H2SO4  Br2 + SO2 + 2H20 (REDOX)
-1 +6 0 +4
Reduced sulphur
Topic 5&6 group 2
& 7 (halogens)

Recations of halides with sulphuric acid H2SO4-
NaI + H2SO4  NaHSO4 + HI (all )
Because hydrogen Iodide is a strongest reducing agent
it can react further even more
Black solid
2HI + H2SO4  I2 + SO2 + 2H20 (REDOX) (br- I-)
-1 +6 0 +4
Reduced sulphur
6HI + SO2  H2S + 3I2 + 2H20 (I-)
-1 +4 -2(toxic rotten egg smell)
Topic 5&6 group 2
& 7 (halogens)

• Testing for halides 
1. Halide containing solution + HNO3
2. Nitric acid removes unwanted ions
normally carbonate ions co3 2-
3. Then add silver nitrate AgNO3
• Cl- = white precipitate formed
• Br-= cream precipitate
• I-= yellow precipitate
• Ag+ + X-  AgX
Topic 5&6 group 2
& 7 (halogens)
Group 7-identification of halide ions using silver nitrate
 understand why acidified silver nitrate solution is used as a reagent to identify and
distinguish between F_, Cl_, Br_ and I_

• Atomic radius
• Trends atomic radius increases as you go
down the group
• Explanation
1. The number of electrons increase =
2. so shielding increases
3. Therefore the outer electrons are further
away from nucleus= weaker electrostatic
attraction
Topic 5&6 group 2
& 7 (halogens)
Trends in physical properties
 understand the trends in atomic radius, first ionisation energy and melting point of
the elements Mg – Ba

• First ionisation energies
• Trends generally decreases as you go down the group
• Explanation 
1. The number of electrons as well as protons increase
2. So the number of shells increases
3. The other electrons and nucleus distance increases =
resulting in a weaker electrostatic attraction
4. This means less energy is needed to remove an
electron
Topic 5&6 group 2
& 7 (halogens)

• Melting point
• Trends decreases as you go down the group
• Explanation
1. As you go down the group the size of the
cations increases and the charge stays the
same
2. The charge density therefore decreases
3. the electrostatic attraction between the
cations and delocalised sea of electrons
decreases
Topic 5&6 group 2
& 7 (halogens)

• Reaction with water
• M  M 2+ + 2e- [o+2] = oxidation
• M + H20  M(OH)2 +H2
• Mg +H20Mg (OH)2 + H2 (LIQUID
STATE)
• IN GASEOUS STATE
• Mg+ H2O(g)  MgO + H2
Topic 5&6 group 2
& 7 (halogens)
Trends in chemical properties
 know the reactions of the elements Mg – Ba with water and recognise the trend
 know the relative solubilities of the hydroxides of the elements Mg – Ba and that
Mg(OH)2 is sparingly soluble

• Reactivity increases as you go down the
group
1. Because atom gets bigger
2. More electrons so there’s more shielding
3. Outer electrons are further away from the
nucleus = weaker electrostatic attraction
• More rapid as you go down
Topic 5&6 group 2
& 7 (halogens)
Trends in chemical properties
 know the reactions of the elements Mg – Ba with water and recognise the trend
 know the relative solubilities of the hydroxides of the elements Mg – Ba and that
Mg(OH)2 is sparingly soluble

Unit 2 – group 2 & 7 (final)

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Unit 2 – group 2 & 7 (final)