diatomic so have weak
V der W's IM forces or
attraction
these increase in
strength as you move
down the group
Halogen bonds - where the
electronegative halogen becomes
delta -ve and it is attracted to a
delta +ve on a nearby molecule.
Ionic in halide salts
Covalent in
between 2
halogen atoms
to form
diatomic
molecules and
also covalent
in hydrogen
halides
Electronic configuration
Fluorine: [He] 2s2 2p5
Bromine: [Ar] 4s2 3d10 4p5
Chlorine: [Ne] 3s2 3p5
Iodine: [Kr] 4d10 5s2 5p5
Astatine: [Xe] 4f14 5d10 6s2 6p5
Disproportionation Reactions
Chlorine and water
Cl2 + H2O --> HCl + HClO
Chlorine w/ cold dilute NaOH
2NaOH + Cl2 --> NaCl + NaClO + H2O
Chlorine w/ hot conc. NaOH
6NaOH + 3Cl2 --> 5NaCl + NaClO3 + 3H2O
Colours
At RTP: Fl = pale yellow, Cl =
pale green, Br = dark red, I =
Grey/violet, At = black
Colour of organic solvent
solution: Fl x (reacts w/ solvents),
Cl pale green, Br orange, I purple,
At darker than I
in water: Fl = x (reacts w/ water), Cl =
pale green, Br = orange, I = insoluble but
forms brown sol. if excess KI is present
As a vapour (if not
already gasseous at RTP)
Br = brown, I = violet, At =
darker purple / black
Half cell reactions /
electrode reactions
2Br-(aq) --> Br2(g) + 2e-
2Cl-(aq) --> Cl2(g) + 2e-
2F-(aq) --> F2(g) + 2e-
I-(aq) --> I(s) + e-
At-(aq) --> At(s) + e-
Tests
Ag+ ion forms insol. ppts w/ Cl-,Br- and I- ions (silver
fluoride is soluable so you don't get a ppt).Add HNO3 to
aqueous sol. of halide ion followed by silver nitrate sol.
Ag+(aq) + Cl-(aq) --> AgCl(s)
which is a white ppt.
Ag+(aq) +Br-(aq) --> AgBr(s)
which is a cream ppt
Ag+(aq) +I-(aq) --> AgI(s)
which is a yellow ppt.
Nitric acid is added to ensure
any carbonate or hydroxide
ions are removed as CO2 or
H20
Add dilute ammonia sol. to te ppt formed
in the AgNO3(aq) test. The Cl ppt will
dissolve, the Br ppt is partially soluable,
and the I ppt will not dissovle.
NH3 + AgCl --> [Ag(NH3)2]+ + Cl-
Add conc.
ammonia sol.
to the ppt
formed in the
AgNO3 test.
Cl and Br
ppts will
dissovle bu
the I ppt will
not.
NH3 + AgBr --> [Ag(NH3)2]+ + Br-
Add conc. sulphuric acid to a solid sample
of the halide. The concentrated sulphuric
acid gives a hydrogen ion to the halide ion
to produce a hydrogen halide. Because
this is a gas, it immediately escapes from
the system. If the hydrogen halide is
exposed to moist air, you see it as steamy
fumes.
Steamy acidic fumes of HF
if F- ions were present
Steamy white acidic
fumes of HCl if Cl-
was present. Turn
blue litmus red.
steamy white acidic fumes (of
HBr) contaminated with
brown bromine vapour.
This is because Br = a
strong enough reducing
agent to reduce the H2SO4
which in the process
oxidises the Br- ions
creating Br gas
2Br- --> Br2(g) + 2e-
Some steamy fumes
(of HI), but lots of
purple iodine vapour.
I = an even more
powerful reducing
agent than Br so they
reduce the H2SO4
first to SO2, then to S,
then to H2S (which
has a rotten egg
smell)
H2SO4 + 8H+ + 8I- --> 4I2 + H2S + 4H2O
States at RTP
Fl + Cl = gas Br = liquid I + At = solid
Displacement reactions
If a more reactive halogen is placed into a solution containing a
less reactive halide a displacement reaction is seen. The more
reactive halogens - i.e. the strongest oxidising agents - will
displace the more reactive halides - i.e. the strongest reducing
agents - from sols. of their ions.
Fluorine = most reactive, Astatine = least reactive
Reaction with metals
With Na
All react w/ Na
to produce Na
halides. In
each case you
get an orange
flame and a
white solid.
With Fe(II)
With the exception of iodine, iron burns in
the halogen vapour to give iron(III) halides.
Iodine is less reactive, and only produces
iron(II) iodide.
Fluorine: Cold iron wool burns in cold fluorine to give
iron(III) fluoride. Anhydrous iron(III) fluoride is variously
described as white or pale green. This reaction is very rapid.
2Fe + 3F2 --> 2FeF3
Chlorine: If you pass chlorine gas over hot iron, the iron burns
to form iron(III) chloride. Iron(III) chloride forms black crystals if
it is anhydrous. Any trace of water present in the apparatus, or
in the chlorine, reacts with this to give reddish-brown colours.
2Fe + 3Cl2 --> 2FeCl3
Bromine: If you pass bromine vapour over hot
iron, a similar but slightly less vigorous reaction
happens, this time producing iron(III) bromide.
Anhydrous iron(III) bromide is usually produced as
a reddish-brown solid.
2Fe + 3Br2 --> 2FeBr3
Iodine: The reaction
between hot iron and
iodine vapour only
produces iron(II) iodide,
and is much less vigorous.
The iron(II) iodide is said to
be grey. If you do this
reaction, the product is
always contaminated with
iodine, so it is difficult to be
sure.
Fe + 2I2 --> FeI2
W/ Fe(III)
Chlorine and bromine are strong enough oxidising agents to
oxidise iron(II) ions to iron(III) ions. In the process, the
chlorine is reduced to chloride ions; the bromine to bromide
ions: 2Fe2+ + Cl2 --> 2Fe3+ + 2Cl-. This is the same
reaction with bromine. The very pale green solution
containing the iron(II) ions will tur into a yellow or orange
solution containing iron(III) ions.Iodine isn't a strong enough
oxidising agent to oxidise iron(II) ions, so there is no
reaction. In fact the reverse reaction happens. Iron(III) ions
are strong enough oxidising agents to oxidise iodide ions to
iodine: 2Fe3+ + 2I- --> 2Fe2+ + I2
BP's and MP's
BP's: Fl=-188, Cl=-35, Br=59, I=184, At=337
MP's: Fl=-220,
Cl=-101, Br=-7,
I=114, At=302
Polarisability
More and more easily polarised
anions as you move down group 7