3.2.3 - the halogens Flashcards
Diatomic halogen molecules are held together by …
Weak vdw
Fluroine exists as
Pale yellow gas
Chlorine exists as
Green gas
Bromine exists as
Red brown liquid
Iodine exists as
Grey solid
Trend in boiling points down group 7
Boiling points increase down group 7
Why do bp increase down group 7
Increasing strength of vdw
Due increasing size and increased number of electrons more chance of dipoles inducing dipoles in the neighbouring molecules
Chlorine has a low bp bc forces bet mol are weak
Explain how these forces arise between molecules of chlorine (3)
Imbalance of of e- density leads to temporary dipole in the mol which induces a dipole im a neighbouring mol
These temporary dipoles attract
Trend in bond enthalpies down group 7
Decrease down down the group
So i-i bonds easier to break than cl-cl bodnds
Why do bond enthalpies decrease down group 7
This is due to the increasing atomic radius, increased shielding and
reduced attraction between the nucleus and bonding pair of electrons
Exception to trend in bond enthalpies down group 7
The exception to this is fluorine.
● Fluorine has such a small atomic radius that bonding and non-bonding
electrons repel each other, reducing the strength or bond enthalpy.
Define electronegativity
Power of an atom to attract a bonding pair of electrons in a covalent bond
Trend in electronegativity going down group 7
● Electronegativity of the halogens decreases going
down the group.
Explain trend of decreasing electronegativity down group 7
As the atomic radius increases the shared pair of
electrons in the covalent bond are further from the
nucleus, reducing the strength of the attraction.
● Increasing atomic radius also means increasing
shielding effect, again reducing the electronegativity.
● These factors have a greater effect than the increased
nuclear charge
trend in boiling points down group 7
increase down the group
explain trend of increasing boiling points down group 7
B.p.s increase down the group because…
* Size of diatomic molecules increase down the group.
* Larger molecules have more electrons, leading to greater induced
dipole-dipole forces.
* Therefore greater van der Waals’s forces between molecules.
* Therefore more energy required to overcome the greater van der Waals’s
forces as you go down the group.
define oxidation agent
electron acceptor
trend in oxidising ability down group 7
decreases down the group
explain trend in decreasing oxidising ability down the group
Oxidising power decreases down the group because…
* Size of ions increase down the group.
* Therefore outer electrons are more shielded and further away from the
nucleus.
* Therefore electrostatic force of attraction by nucleus on the
additional electron becomes weaker down the group.
* Therefore harder to gain an electron.
reactivity down group 7
decreases
explain decreasing reactivity down group 7
increased atomic rad
outershel has increased dist from nuclues
less electrostatic attraction harder to attract e-
whats is displacment
more reactive halogen will displace a less reactive one
whats observed when chlorine water reacts w
kcl
kbr
ki
kcl - no reac
kbr - orange br2
ki - brown i2
whats observed when bromine water reacts w
kcl
kbr
ki
kcl - no reac
kbr - no reac
ki - brown i2 formed
whats observed when iodine water reacts w
kcl
kbr
ki
no reaction for all 3
what is a reducing agnent
electron donor
it itslef is oxidised
trend in reducing ability down grouo 7
increases down thr grouo
explain trend in reducing ability of halogens down the group
Reducing power increases down the group because:
* Size of the ions increase down the group.
* Therefore outer electrons are more shielded and further away from the
nucleus.
* Therefore electrostatic force of attraction by nucleus on outer
electrons becomes weaker down the group.
* Therefore easier to lose an electron.
equation for sodium fluroide + conc sulfuric acid
NaF + H2SO4 —> NaHSO4 + HF
ionic equation for when chlorine diplaces bromine and iodine
cl2 (aq) + 2br-(aq) –> 2cl-(aq) + br2 (aq)
cl2(aq) + 2i- (aq) –> 2cl- + i2
ionic equation when bromine displaces iodine
br2 + 2i- –> 2br- + i2
observation fro soild sodium fluroide w conc sulfuric acid
misty white fumes (HF) are evolved.
equation for reaction w solild nacl and conc h2so4
NaCl + H2SO4 → NaHSO4 + HCl
observations w solild nacl and conc h2so4
misty white fumes (HCl) are evolved.
2 equations for reac of nabr w conc h2so4
1 - NaBr + H2SO4 → NaHSO4 + HBr
2 - 2HBr + H2SO4 → Br2 + SO2 + 2H2O
obs for reac 1 NaBR + h2so4 —-> naso4 + HBR
misty white fumes (HBr) and red-brown vapour (Br2) are evolved.
Change in ox number for 2Hbr + H2SO4 —> Br2 + So2 + 2H2O
Br is oxidised from -1 (in HBr) to 0 (in Br2) therefore bromine =
oxidation product. - S is reduced from +6 (in H2SO4) to +4 (in SO2) therefore sulphur
dioxide = reduction product.
4 eq for reaction of NaI w conc h2so4
NaI + H2SO4 → NaHSO4 + HI NOT REDOX
2HI + H2SO4 → I2 + SO2 + 2H2O REDOX
6HI + H2SO4 → 3I2 + S + 4H2O REDOX
8HI + H2SO4 → 4I2 + H2S + 4H2O REDOX
obs w NaI w conc h2so4
(all these reactions happen in succession - not in
isolation): misty white fumes evolved (HI), purple vapour evolved (I2),
yellow solid formed (S), rotten egg smell (H2S) + black solid formed
(I2).
explain the redox eq w soloid sodium iodide
➜ For the 1st redox reaction.
- the I is oxidised from -1 (in HI) to 0 (in I2) - the S is reduced from +6 (in H2SO4) to +4 (in SO2)
➜ For the 2nd redox reaction,
- the I is oxidised from -1 (in HI) to 0 (in I2) - the S is reduced from +6 (in H2SO4) to 0 (in S)
➜ For the 3rd redox reaction,
- the I is oxidised from -1 (in HI) to 0 (in I2) - the S is reduced from +6 (in H2SO4) to -2 (in H2S)
Therefore oxidation product = iodine; reduction products = sulphur
dioxide, sulphur and hydrogen sulphide.
general equation for tests for halide ions where X is eith
Ag+ (aq) + X- (aq) –> AgX(s)
test for halide ions
Add dilute nitric acid (HNO3), followed by silver(I) nitrate (AgNO3)
solution.
colour of solid halide ppt
Ag+(aq) + Cl-(aq) → AgCl(s) = white ppt. formed
Ag+(aq) + Br-(aq) → AgBr(s) = cream ppt. formed
Ag+(aq) + I-(aq) → AgI(s) = yellow ppt. formed
reason for adding dilute nitric acid
- The dilute nitric acid removes other ions which would react with the
silver nitrate solution (e.g. carbonates/sulphates/hydroxides)
what is further test in halide ion test
A further test can be carried out with ammonia solution to confirm the
identity of the halide ion:
results of further test
AgCl: will redissolve in dilute and concentrated ammonia solution to form
a colourless solution.
AgBr: does redissolve
in conc. ammonia solution forming a colourless solution.
AgI does not redissolve in dilute or conc. ammonia solution.
what kind of reaction occurs when you mic chlorine with water
disproportionation (This
means Chlorine, Cl2 is both oxidised and reduced).
give equation for reaction between chlorine and water and explain whats oxidised and reduced
Cl2 + H2O ⇌ HClO + HCl
* The Cl is oxidised from 0 (Cl2) to +1 (HOCl)
* The Cl is reduced from 0 (Cl2) to -1 (HCl)
give equation for the reaction of chlorine with water to form chloride ions
and oxygen.
2Cl2 + 2H2O ⇌ 4HCl + O2
adv and disad of formation of chlorate (i) ions when reacting chlorine with water
Chlorate(I) ions kill bacteria, this is why chlorine is used in water
treatment to kill bacteria. It’s been used to treat drinking water and
the water in swimming pools. - The benefits to health (including, the irradiation of bacterial
diseases like cholera) from using chlorine OUTWEIGH the fact it’s toxic
/ carcinogenic to humans.
When you mix chlorine gas with ____, _____ sodium hydroxide solution at
room temperature it undergoes ___________
When you mix chlorine gas with cold, dilute, sodium hydroxide solution at
room temperature it undergoes disproportionation.
equation for reaction of chlorine with cold dilute naoh and whats oxidised and reducee
Cl2 + 2NaOH → NaCl + NaClO + H2O
* The Cl is oxidised from 0 (Cl2) to +1 (NaClO)
* The Cl is reduced from 0 (Cl2) to -1 (NaCl)
observation when reacting chlorine w cold dilute naoh
Observation: green gas forms a colourless solution.
what is sodium chlorate
One of the products formed is sodium chlorate(I), NaClO. It’s in solution
in this case and that is bleach (which kills bacteria)
explain why
-silver nitrate solution is used to identify halide ions
- Silver ions combine with halide ions to produce silver halide precipitates with different colours.
explain why the silver nitrate solution is acidified
to react with any impurities (mainly carbonate ions) that may be present in the solution. Carbonate ions, for example, would form a white precipitate with the silver ions, making the identification of silver halide salts difficult.
explain why
ammonia solution is added.
The silver halide precipitates have different solubilities in ammonia (NH3), which can be used to help further identify the halides.
Explain why chlorine is used to kill bacteria in swimming pools even though chlorine is toxic (2)
Only used in small amounts
Health benefits outweigh risks
Write an equation for reaction of chlorine and water in absence of sunlight and give oxidation states of chlorine in each chlorine containing compound
Cl2 + H2O —> HCL + HOCl
0 -1 +1
State why silver nitrate solution is acidified when testing for iodide ions
Remove ions that would interfere with test
Giving a false positive
Explain why dilute HCL isn’t used to acidify silver nitrate solution in this test for iodide ions
HCL would form white ppt
Give the formula of 2 different chlorine containing compounds that are formed when chorine reacts with cold dilute aqueous sodium hydroxide
NaCl
NaClO
