inorganic 1 - group 7 Flashcards
(25 cards)
What is the trend in atomic radius as you go down group 7 and why?
Atomic radius increases.
There are more electron shells so bigger atoms.
What is the trend in electronegativity as you go down group 7 and why?
Electronegativity decreases.
There are bigger atoms (they have more shells) and so have a weaker attraction between the nucleus and the 2 electrons in a covalent bond.
What is the trend in 1st ionisation energy as you go down group 7 and why?
1st ionisation energy decreases.
Atoms get bigger and there is more electron shielding. So there is weaker attraction between the positive nucleus and the outermost electron.
What is the trend in mp/bps as you go down group 7 and why?
The mp/bps increase.
There are bigger molecules (more electrons) and so more VDWs forces between molecules.
Write an equation for the oxidisation of Cl2 using Br-(aq) and state what you would observe.
Cl2 + 2Br- –> 2Cl- + Br2
yellow solution forms.
Write an equation for the oxidisation of Cl2 using I-(aq) and state what you would observe.
Cl2 + 2I- –> 2Cl- + I2
brown solution forms.
Write an equation for the oxidisation of Br2 using Cl-(aq) and state what you would observe.
no reaction.
Write an equation for the oxidisation of Br2 using I-(aq) and state what you would observe.
Br2 + 2I- –> 2Br- + I2
brown solution forms.
Write an equation for the oxidisation of I2 using Cl-(aq) and state what you would observe.
no reaction
Write an equation for the oxidisation of I2 using Br-(aq) and state what you would observe.
no reaction
What is the trend in oxidisation power as you go down group 7 any why?
Cl2 > Br2 > I2
Halogen atoms gain an electron when they oxidise the halide ion.
The smaller the halogen atom, the easier it is to gain an electron as it is smaller and has less shielding.
What are the half equations you need to know to help write equations to show the reducing power of halide ions?
Cl- —> Cl2
2Cl- —> Cl2 + 2e-
Br- —> Br2
Br- —> Br2 + 2e-
I- —> I2
I- —> I2 + 2e-
H2SO4 —> SO2
H2SO4 + 2H+ + 2e- —> SO2 + 2H2O
H2SO4 —> S
H2SO4 + 6H+ + 6e- —> S + 4H2O
H2SO4 —> H2S
H2SO4 + 8H+ + 8e- —> H2S + 4H2O
Reducing power of Cl-
halide - KCl
product - HCl
observation - white fumes
equation - H2SO4 + KCl —> HCl + KHSO4
reaction type - acid-base
reducing power of Br-
halide - KBr
product - HBr
observation - white fumes
equation - H2SO4 + KBr —> HBr + KHSO4
reaction type - acid-base
halide - KBr
product - Br2
observation - orange fumes
equation - H2SO4 + 2Br- + 2H+ —> Br2 + SO2 + 2H2O
reaction type - redox
halide - KBr
product - SO2
observation - colourless gas
equation - H2SO4 + 2Br- +2H+ —> Br2 + SO2 + 2H2O
reaction type - redox
reducing power of I-
halide - KI
product - HI
observation - white fumes
equation - H2SO4 + KI —> HI + KHSO4
reaction type - acid-base
halide - KI
product - I2
observation - purple fumes + black solid
equation -
H2SO4 + 2I- + 2H+ —> I2 + SO2 + 2H2O
H2SO4 + 6I- + 6H+ —> 3I2 + S + 4H2O
H2SO4 + 8I- + 8H+ —> 4I2 + H2S + 4H2O
reaction type - redox
halide - KI
product - SO2
observation - colourless gas
equation - H2SO4 + 2I- + 2H+ —> I2 + SO2 + 2H2O
reaction type - redox
halide - KI
product - S
observation - yellow solid
equation - H2SO4 + 6I- + 6H+ —> 3I2 + S + 4H2O
reaction type - redox
halide - KI
product - H2S
observation - bad egg smell
equation - H2SO4 + 8I- + 8H+ —> 4I2 + H2S + 4H2O
reaction type - redox
how far can Cl- reduce S+6 in H2SO4?
Cl- does not reduce H2SO4
how far can Br- reduce S+6 in H2SO4?
Br- reduces H2SO4 from S+6 to S+4 in SO2
how far can I- reduce S+6 in H2SO4?
I- reduces H2SO4 FROM S+6 to S-2 in H2S
What is the trend in reducing power as you go down group 7 and why?
I- > Br- > Cl-
halide ions lose an electron when they reduce H2SO4. the bigger the halide ion, the easier it is to lose an electron as it is bigger and has more electron shielding.
how do you test for halide ions?
- add nitric acid (aq) to the solution being tested. this removes any other ions (e.g. carbonate ions) that could give a ppt with silver nitrate solution.
- add silver nitrate (aq) to the solution being tested. This produces a ppt for Cl-, Br- and I- ions.
- add ammonia (aq) (dilute and concentrated) to the ppt. This used to see if the ppts ‘redissolve’ to help confirm their identity as the colours of the ppt can be hard to distinguish.
Write an equation and the observation you would see when you add AgNO3 (aq) to each halide ion.
F- (aq)
no reaction
Cl- (aq)
white ppt
Ag+ (aq) + Cl- (aq) —> AgCl (s)
Br- (aq)
cream ppt
Ag+ (aq) + Br- (aq) —> AgBr (s)
I- (aq)
yellow ppt
Ag+ (aq) + I- (aq) —> AgI (s)
Write an equation and the observation you would see when you add dilute NH3 (aq) to each halide ion.
Cl-
colourless solution
AgCl (s) + 2NH3 (aq) —> Ag(NH3)2+(aq) + Cl- (aq)
Br-
no reaction
ppt remains
I-
no reaction
ppt remains
Write an equation and the observation you would see when you add conc NH3 (aq) to each halide ion.
Br-
colourless solution
AgBr (s) + 2NH3 (aq) —> Ag(NH3)2+(aq) + Br- (aq)
I-
no reaction
ppt remains
reactions of chlorine with water
equation for reaction-
Cl2 + H2O —> HCl + HOCl
in bright light-
HOCl —> HCl + 0.5O2
so overall in bright light -
Cl2 + H2O —> 2HCl + 0.5O2
use of reaction -
HOCl acts as a bleach and kills bacteria.
Very small amounts of chlorine are added to water to kill microbes. Although chlorine is toxic, it is added in low enough amounts to kill microbes but not harm humans.
redox nature of reaction -
disproportionation reaction as Cl is both oxidised and reduced.
other notes -
Chlorine (aq) appears pale green due to the presence of some dissolved Cl2 in the equilibrium mixture.
If an indicator is added to chlorine (aq), it will go the acid colour first because of the presence of the acids HOCl and HCl, but then it will be bleached by HOCl.
