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Saturday, 9 April 2016
Sunday, 17 January 2016
Trend in reducing power of Hydrogen halide
**HCl doesn’t have enough reducing power
to reduce conc. H2SO4 Acid.
** HBr & HI have
enough reducing power to reduce concentrated H2SO4 further .
** HI is a stronger reducing agent than
HBr since HI reduce the sulphur in H2SO4 from +6 to -2
whereas HBr reduces sulphur from +6 to +4 .
Reaction of Concentrated sulfuric acid with metal iodide:
Again, Sulphuric acid
reacts with Metal iodide to form HI :
e.g. KI (s) +
H2SO4 (l) →
HI (g)
+ KHSO4 (s)
This is the primary
reaction that would occur in every reactions of Concentrated sulfuric acid with
Metal Halide.
If you have remembered
from early reaction with metal chloride, HCl can’t reduces H2SO4
further whereas HBr can. HBr can reduces H2SO4 once.
HI can ALSO reduces H2SO4 3 times.
In First reaction , HI further reduces H2SO4 to
SO2 .
(-1) (+6) ( 0 ) (+4)
2HI (g) + H2SO4 (l) → I2
(g) +
SO2 (g) + 2H2O
(l)
Here HI reduces Sulphur in sulphuric acid from +6
to +4 in SO2
In Second reaction , HI
further reduces H2SO4 to
S.
(-1) (+6) ( 0 ) ( 0 )
6HI (g) + H2SO4 (l) → 3I2
(g) + S (s)
+
4H2O (l)
Here HI reduces Sulphur in sulphuric acid from +6
to 0 in S .
In Third reaction, HI further reduces H2SO4 to
H2S.
(-1) (+6) ( 0 ) ( -2 )
8HI (g) + H2SO4 (l) → 4I2
(g) + H2S
(s) + 4H2O (l)
Here HI reduces Sulphur in sulphuric acid from +6
to -2 in H2S.
Here three reactions
are occurring.
Observations :
1) Steamy / Misty fumes due to HI gas.
2) Purple vapour due to I2 gas. Iodine
i.e. formed is in gaseous state since the reaction is exothermic.
3) Black solid due to formation of solid iodine.
4) Yellow solid due to
the formation of sulphur.
5) Rotten Egg smell / Smell
of bad eggs due to formation of H2S
Reaction of Concentrated sulfuric acid with metal bromide :
Again, Sulphuric acid
reacts with Metal bromide to form HBr :
e.g. KBr (s)
+ H2SO4
(l) → HBr (g) + KHSO4 (s)
This is the primary
reaction that would occur in every reactions of Concentrated sulfuric acid with
Metal Halide.
If you have remembered
from early reaction with metal chloride, HCl can’t reduces H2SO4
further whereas HBr can. HBr can reduces H2SO4 once.
This HBr then further
reduces H2SO4 to SO2
(-1) (+6) ( 0 ) (+4)
HBr (g) + H2SO4 (l) → Br2 (g) + SO2 (g) + 2H2O (l)
Here HBr reduces Sulphur in sulphuric acid from +6
to +4 in SO2
Here two reactions are occurring.
Observations :
1) Steamy / Misty fumes due to HBr gas.
2) Red brown vapour due to Br2 gas.
** Bromine i.e. formed here is in the gaseous state since the reaction is exothermic.
Reaction of Metal Chloride With Concentrated Sulphuric Acid:
Any metal chlorides
react with concentrated sulphuric acid to give Hydrochloric acid and Sodium
hydrogen sulphate.
e.g. NaCl (s) +
H2SO4 (l) → HCl (g) + NaHSO4 (s)
(concentrated)
HCl that is produced
cannot reduce H2SO4 Acid any further:
HCl (g) +
H2SO4 (l) → No reaction
Observation : Misty/
Steamy Fumes Due to the formation of HCl gas because HCl gas is extremely soluble in water vapour in the air.
Saturday, 16 January 2016
Chemical Test For Iodine
Chemical Test For Iodine 1:
Bleaching action: Iodine doesn’t
bleach damp litmus paper.
Chemical Test For Iodine 2:
Displacement reaction: Iodine cannot
displace bromine or chlorine from a solution of bromide or
chloride ions respectively.
chloride ions respectively.
Chemical Test For Iodine 3:
With Starch-iodide paper : Iodine turns starch- iodide paper deep blue / blue-black.
With Starch-iodide solution: Iodine makes a deep blue / blue-black colour with starch solution.
Chemical Test For Bromine 3 :
Action with starch-iodide paper :
Bromine gas also turns damp starch iodide
paper deep blue / blue black coloured complex.
Just like chlorine, Bromine first
reacts with KI (aq) in the starch iodide paper :
Br2 (l) + 2KI
(aq) → 2KBr
(aq) +
I2 (aq)
Here I3- Ion formed
which is actually soluble in water not I- Ion
Then iodine that is formed will
react with starch in the starch-iodine Paper :
I2 (aq) +
Starch
→ blue black / deep blue coloured complex
Chemical Test For Bromine 2:
2)
Displacement reactions : Bromine can also displace iodide ions from a metal iodide compounds.
Bromine liquid
is added to the solution of Iodide ions:
|
Notes
|
|
·
Aqueous
iodine i.e formed here is insoluble
in water but soluble in
KI
Solution forming I3- Ions.
·
To
Identify the spectator ions, we
need
to consider some facts:
1)
State Symbols
2)
Oxidation no. of components in
a redox reaction.
|
Br2
(l) +
2KI (aq) → 2KBr
(aq) + I2 (aq)
Ionically,
Br2 (l) + I - (aq) → 2Br- (aq)
+ I2 (aq)
At
the end of the reaction , if the product mixture is shaken with CCl4 /
Hexane/ Cyclohexane /, Two coloured layers are observed.
Iodine is
less soluble in water but more soluble in CCl4 or hexane or
cyclohexane since iodine is non-polar .
Chemical Test For Bromine 1 :
The chemical test of Bromine is similar to Chlorine but differs by rate.
1)
Bleaching action : Bromine also turns damp
/ moist blue litmus paper red and then bleaches
but much slower than that of Chlorine gas . This is because of acid is produced.
but much slower than that of Chlorine gas . This is because of acid is produced.
At
first….
Br2 + H20 → HBr + HOBr
Bromine gas also reacts
with Water in moist litmus paper to
form Hydrobromic acid and unstable
bromic (I) acid.
Then ……..
HOBr → HCl + [O]
Again,
Bromic (I) acid decompose to form Hydrobromic acid and Oxygen Free Radical
Friday, 15 January 2016
Chemical Test For Chlorine 3
Action with starch-iodide paper :
Chlorine
Gas turns damp starch iodide paper deep blue / blue black coloured complex.
First
chlorine reacts with KI (aq) in the starch iodide paper :
Cl2
(g) + 2KI (aq) →
KCl (aq) + I2 (aq)
Here I3- Ion formed which is actually
soluble in water
Then iodine
that is formed will react with starch in the starch-iodine Paper :
I2
(aq) + Starch
→ blue black / deep blue coloured
complex
Chemical Test For Chlorine 2
2) Displacement
reactions : Chlorine can displace bromide and iodide ions from a metal halide compounds.
Chlorine
gas is passed through the solution of bromide ions:
Cl2
(g) +
2KBr (aq) →
2KCl(aq) + Br2 (aq)
Notes
|
·
Bromine
gas i.e formed here is soluble in water.
·
To
Identify the spectator ions, we need
to consider some facts:
1)
State Symbols
2)
Oxidation no. of components in a redox reaction.
|
Ionically,
Cl2
(g) + 2Br-
(aq) → 2Cl- (aq)
+ Br2 (aq)
At the end of the
reaction , if the product mixture is shaken with CCl4 / Hexane/
Cyclohexane /, Two coloured layers are observed.
Chlorine
gas is passed through the solution of iodide ions:
Cl2
(g) +
2KI (aq) → 2KCl(aq) + I2
(aq)
Notes
|
·
To
Identify the spectator ions, we
need
to consider some facts:
1)
State Symbols
2)
Oxidation no. of components in a redox reaction.
|
Ionically,
Cl2
(g) + 2I-
(aq) → 2Cl- (aq)
+ I2 (aq)
* Iodine i.e formed here is actually soluble in KI Solution but insoluble
in water.
I2 (s) + KI (aq)
→
KI3 (aq)
Potassium triiodide
Ionically,
I2
(s) + I- (aq) → I3- (aq)
Triiodide
Ion
At the end of the
reaction , if the product mixture is shaken with CCl4 / Hexane/
Cyclohexane /, Two coloured layers are observed.
** It is the I3- (aq) triiodide ion that gives the yellow colour not the iodine itself.
Chemical Test For Chlorine 1
Chemical Test
For Chlorine:
1) Bleaching
action : Chlorine turns damp / moist
blue litmus paper red and then bleaches. This is because of acid is produced.
At first….
Cl2
+ H20 → HCl + HOCl
Chlorine gas reacts with Water in moist litmus paper to form Hydrochloric acid and unstable
Chloric (I) acid.
Then ……..
HOCl → HCl + [O]
Chloric (I) acid decompose
to form Hydrochloric acid and
Oxygen Free Radical
Group 7 Halogens
Periodic table: Group 7 element members
Elements
|
Formula
|
Physical
states at rtp
|
Colour
|
Fluorine
|
F2
|
Gas
|
Yellow
|
Chlorine
|
Cl2
|
Gas
|
Green
|
Bromine
|
Br2
|
Liquid
|
Red-brown/Red
|
Iodine
|
I2
|
Solid
|
Black
|
Chlorine
water / Aqueous solution of Chlorine , Cl2 (aq): colourless
Bromine water / aqueous solution of Bromine , Br2 (aq): yellow
Bromine Vapour : red-brown
Aqueous solution of Iodine, I2 (aq) :
Brown
Iodine Vapour : Purple vapour
Tuesday, 12 January 2016
Saturday, 9 January 2016
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