HALOGENS
answers
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b)
c)
d)
2.
a)
i)
The relative ability of an atom to attract electrons towards
itself
In a covalent bond
ii)
Size increases down a group
So outer electrons are less closely held
So electronegativity decreases from fluorine to iodine
The size of the molecule increases, and the number of electrons
increases
So the chance of a temporary dipole increases
So there will be more Van der Waal’s forces
Nothing happens when bromine is added to an aqueous solution of
sodium chloride
A dark brown colour will be observed, and the red-brown colour will
disappear
Br2 + 2I-  2Br- + I2
Cl- is the weakest reducing agent, and I- is the strongest reducing
agent
Reducing power increases down the group
I- is the largest ion so gives up its electrons most easily
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i)
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ii)
iii)
b)
i)
ii)
iii)
Iodide ions have a larger ionic radius (0.219 nm) than bromide
ions (0.196 nm)
and so their outer electrons are less closely held.
Therefore iodide ions lose electrons more easily than bromide
ions are hence more easily oxidised
Atomic radius increases from fluorine (0.064 nm) to chlorine
(0.099 nm)
So outer electrons are less closely held
So electronegativity decreases from fluorine to chlorine
The number of electrons in the molecule decreases from Br2
(70) to Cl2 (34)
So the chance of a temporary dipole decreases
So there will be less Van der Waal’s forces
Reducing power increases down group VII,
so At will be a good reducing agent.
So hydrogen sulphide will be evolved.
Astatine is less reactive than chlorine
so chlorine will displace astatine from a solution of its ions.
Chlorine is smaller so removes the electrons from astatide ions
So astatine will be precipitated
Solubility of silver halides decreases down the group,
so silver astatide will be the least soluble
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So it will not dissolve in aqueous ammonia
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3.
a)
b)
c)
d)
4.
a)
b)
c)
5.
a)
b)
c)
d)
e)
6.
a)
b)
i)
ii)
iii)
S, SO2, H2S
I2
Iodine - a purple vapour
Hydrogen sulphide – a gas smelling of rotten eggs
iv)
Sulphuric acid behaves as an acid in the formation of hydrogen
iodide
v)
H2SO4 + 8H+ + 8e  H2S + 4H2O
Bromide ion, Bri)
A white precipitate
ii)
A cream precipitate
The white precipitate will dissolve in dilute aqueous ammonia
The cream precipitate will not dissolve in dilute aqueous ammonia
but will dissolve in concentrated aqueous ammonia
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An electron acceptor
Chloride ion, ClH2SO4 + Cl-  HSO4- + HCl
Iodide ion, I2I-  I2 + 2e
H2SO4 + 8H+ + 8e  H2S + 4H2O
H2SO4 + 8H+ + 8I-  H2S + 4H2O + 4I2
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Cl2(g) + H2O(l) == HCl(aq) + HClO(aq)
The green colour is the dissolved chlorine, Cl2(aq)
The Cl2 disproportionates
Cl2(aq) + 2OH-(aq)  Cl-(aq) + ClO-(aq) + H2O(l)
Hence the green colour disappears
Cl-(aq) + ClO-(aq) + 2H+(aq)  Cl2(aq) + H2O(l)
ClO-(aq), usually in the form of NaClO
Oxidation number of Cl = +1
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A brown colour will be seen in the solution
Cl2 + 2I-  2Cl- + I2
Chlorine is an oxidizing agent
i)
Mistake in question – should say sodium hydroxide
Cl2(aq) + 2OH-(aq)  Cl-(aq) + ClO-(aq) + H2O(l)
ii)
Cl2(g) + 2e  2Cl-(aq)
Cl2(g) + 6H2O(l)  2ClO3-(aq) + 12H+(aq) + 10e
6Cl2(g) + 6H2O(l)  2ClO3-(aq) + 12H+(aq) + 10Cl-(aq)
3Cl2(g) + 3H2O(l)  ClO3-(aq) + 6H+(aq) + 5Cl-(aq)
in alkaline solution
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3Cl2(g) + 3H2O(l) + 6OH-(aq)  ClO3-(aq) + 6H+(aq) + 6OH-(aq) + 5Cl-(aq)
3Cl2(g) + 3H2O(l) + 6OH-(aq)  ClO3-(aq) + 6H2O(l) + 5Cl-(aq)
3Cl2(g) + 6OH-(aq)  ClO3-(aq) + 6H2O(l) + 5Cl-(aq)
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7.
From experiment 1:
Bromine and iodine are produced
So mixture contains iodide and bromide ions
Cl2(g) + 2I-(aq)  2Cl-(aq) + I2(s) black precipitate
Cl2(g) + 2Br-(aq)  2Cl-(aq) + Br2(g) orange-brown fumes
From experiment 2:
AgBr and AgI are formed.
Ag+(aq) + Br-(aq)  AgBr(s)
Ag+(aq) + I-(aq)  AgI(s)
Total mass of AgBr and AgI = 0.902 g
AgBr is removed on addition of concentrated aqueous ammonia
AgBr(s) + 2NH3(aq)  [Ag(NH3)2]+(aq) + Br-(aq)
Mass of AgI only = 0.546 g
So mass of AgBr = 0.902 – 0.546 = 0.356 g
Mass of I = 127/235 x 0.546 = 0.295 g
Mass of Br = 80/188 x 0.356 = 0.151 g
% of I = 0.295/0.545 x 100 = 54.1%
% of Br = 0.151/0.545 x 100 = 27.8%
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