Chemistry Form 6 Sem 2 06

Element

Chlorine, Cl2

Bromine, Br2

Iodine, I2

0.099

0.114

0.133

Physical state at room temperature Electronic configuration

Yellow greenish gas

Red brown liquid

3s2 3p5

4s2 4p5

5s2 5p5

Melting point (oC)

-102

-7.1

--

Boiling point (oC)

-34.5

58.6

184

242.5

191.5

150.0

-364

-340

-297

3.0

2.8

2.5

soluble

Sparingly soluble

Atomic radius (nm)

Bond energy (kJ/mol) Electron affinity (kJ/mol) Electronegativity Solubility in water

soluble

Black solid

1. Atomic radius
When goes down to group 17, atomic radius increase. This is due to, when going down to Group 17, nuclear charge increase, as the proton number increase. As number of proton increase, number of electrons also increased, which caused more shells are required to fill in electrons and gradually increase the screening effect. As a result, the effective nuclear charge decreased and caused the atomic radius increase 2. Melting point, boiling point, physical state at room temperature, colour intensity & density of Group 17
All Group 17 elements exist as diatomic molecules, where fluorine, chlorine, bromine, iodine and astatine exist as F2, Cl2,Br2,I2,and At2 respectively
The physical state of chlorine, bromine and iodine are different despite they are from the same classification of elements, nonmetal. Chlorine exist as yellow-greenish gas, bromine exist as brown liquid, while iodine exist as purplish-black solid. These observations were supported by the data of melting point and boiling point of Group 17 elements


All these were due to when going down to Group 17, relative

molecular mass increased from Cl2 to Br2 to I2 and caused the weak Van Der Waals' forces increase. As a result, the melting point and boiling point increase down the Group and each elements of Group 17, hence exist as different physical forms.
At the same time, as the weak intermolecular forces increased, molecules become more close packed, causing the number of molecules per unit volume increased, hence increase the density of elements, which consequently increased the colour intensity of halogen. This will cause the colour of halogens to become darker. This can also be used to explain the volatility of molecules, where going down Group 17, volatility decreased which were also due to stronger intermolecular forces between particles hence harder to vapourise.

3. Electronegativity, bonding energy and electron affinity
Electronegativities measure the ability to pull the bonding pair towards the direction of the atom.
Electronegativity of the group 17 decrease when goes down to group due to the increase of covalent radius of the elements. Fluorine has the highest electronegativity among all Group 17 elements. This will influence the bonding length of the Group 17 which directly causing the decrease in bonding energy.
Bonding energy measure the heat (energy) required to break the bond formed between 2 atoms. Example : Cl - Cl (g) → 2 Cl (g) ∆H = + 242.5 kJ mol-1
Bonding energy decreased, as the bond length increased from chlorine to iodine. As the bond length of the covalent molecules increased, it is getting easier to break the bond, thus lower the bonding energy.


Fluorine has a lower bonding energy compare to chlorine as the

fluorine atom is too small, which result the lone pair electrons between F-F repel each other, causing the bond length increased. (since lone pair - lone pair electrons repulsion is stronger than lone pair - bond pair electrons repulsion).
Electron affinity is the heat liberated when one mole of gaseous atom accept one mole of electron to form a gaseous anion under standard condition.
Equation : F (g) + e- → F- (g) ∆HEA = - 364 kJ mol-1
The electron affinity of Group 17 become less negative when goes down to group 17. This is due to, as atomic size increased, nucleus is further away from the outermost shell. As a result, electron received is further away and the electrostatic forces of attraction is weaker, hence less exothermic

4. Solubility of halogens in water
Solubility of halogen decrease when goes down to Group 17. Fluorine is the most soluble halogen among the Group 17 because enthalpy change of hydration of fluorine is very exothermic. Furthermore, fluorine can form hydrogen bond with water.
Both chlorine and bromine are considerably soluble in water and form an acid and acid halide via disproportionation reaction (oxidation and reduction occur simultaneously on the same substance), while iodine is sparingly soluble in water. Equation : Cl2 (g) + H2O (l) → HCl (aq) + HOCl (aq) Br2 (g) + H2O (l) → HBr (aq) + HOBr (aq)
Iodine aqueous solution can be prepared by dissolving iodine in potassium iodide (KI) aqueous solution and form a brown solution of I3-. Equation : I2 (s) + I- (aq) ↔ I3- (aq)

bleaching agent.
The hypohalous acid formed can be used as …………………
When hypohalous is exposed to sunlight, it will decomposed and

oxygen is evolved. Hypochlorous acid and hypobromous acid both give different reaction. 2 HOCl 2 HCl (aq) + O2 (g) Overall : 2 Cl2 + 2 H2O

4 HCl (aq) + O2 (g)


As extra notes, halogens also dissolve in organic solvent such as

tetrachloromethane, CCl4, or benzene solution.

Halogen Colour in aqueous solution Colour in organic solvent (CCl4)

Chlorine, Cl2 Bromine, Br2

Iodine, I2

Pale yellow

Brown

Brown

Colourless

Brown

Purple

6.2 Chemical properties of Group 17 elements
Halogens are strong oxidising agent. This is supported with the value of standard reduction potential where Eo = + 2.87 V ½ F2 (aq) + e-
F– (aq) ½ Cl2 (aq) + e-
Cl– (aq) Eo = + 1.36 V Stronger oxidising ½ Br2 (aq) + e-
Br– (aq) Eo = + 1.07 V agent – o ½ I2 (aq) + e-
I (aq) E = + 0.54 V
A displacement reaction can take place between halogens. Using this method, the reactivity of halogen can also be deduced. The following mixture of halide and halogen are added and observation is made

Observation

Equation

Chlorine in Tetrachloromethane Colourless solution is added to aqueous in CCl4 turned potassium bromide brown (KBr).

Reactants : Cl2 + BrSo the half equation for reaction: Cl2 + 2e-
2 Cl- E0 = + 1.36 V 2 Br-
Br2 + 2e- E0 = - 1.07 V -------------------------------------------Cl2 + 2 Br-
2 Cl- + Br2 Ecell = + 0.29 V Since the Ecell is positive, the reaction is spontaneous.

Bromine in tetrachloromethane is added to aqueous potassium iodide (KI)

Reactants : Br2 + ISo the half equation for reaction: Br2 + 2e-
2 Br- E0 = +1.07 V 2 I-
I2 + 2eE0 = - 0.54 V -------------------------------------------Br2 + 2 I-
2 Br- + I2 Ecell = + 0.53 V Since the Ecell is positive, the reaction is spontaneous.

Brown colour in CCl4 turn purple colour.

Iodine is tetrachloromethane Purple colour in is added to aqueous CCl4 remain potassium chloride unchanged (KCl)

Reactants : I2 + ClSo the half equation for reaction: I2 + 2e-
2 IE0 = + 0.54 V 2 Cl-
Cl2 + 2e- E0 = - 1.36 V -------------------------------------------I2 + 2 Cl-
2 I- + Cl2 Ecell = – 0.82 V Since the Ecell is negative, the reaction is non-spontaneous.

5.2.1 Changing of iron (II) ion to iron (III) ion and vice versa
Given the half equation of iron (II) ion and iron (III) ion and its standard reduction potential.
Fe3+ (aq) + e-
Fe2+ (aq) E∅ = + 0.77 V
If the halogen aqueous solution is added to iron (II) ion, Fe2+, solution respectively
Chlorine in iron (II) ion aqueous solution. Reactants : Cl2 + Fe2+ ; So the half equation for reaction: Cl2 + 2e-
2 ClE0 = + 1.36 V Fe2+
Fe3+ + eE0 = – 0.77 V Cl2 + 2 Fe2+
2 Cl- + 2 Fe3+ Ecell = + 0.59 V Since the Ecell is positive, the reaction is spontaneous. Obs : pale yellow solution dissolve in green solution to form yellow solution
Bromine in iron (II) ion aqueous solution. Reactants : Br2 + Fe2+ ; So the half equation for reaction: Br2 + 2e-
2 BrE0 = + 1.07 V Fe2+
Fe3+ + eE0 = – 0.77 V Br2 + 2 Fe2+
2 Br- + 2 Fe3+ Ecell = + 0.30 V Since the Ecell is positive, the reaction is spontaneous. Obs : brown solution first dissolve in green solution and form yellow solution




Iodine in iron (II) ion aqueous solution.

Reactants : I2 + Fe2+ ; So the half equation for reaction: E0 = + 0.54 V I2 + 2e-
2 IE0 = – 0.77 V Fe2+
Fe3+ + eI2 + 2 Fe2+
2 I- + 2 Fe3+ Ecell = – 0.23 V Since the Ecell is negative, the reaction is non-spontaneous. Obs : no changes occur *Conversely when iron (III) ion is added to iodide ion, 2 I- + 2 Fe3+
I2 + 2 Fe2+ Ecell = + 0.23 V iodide ion (which act as reducing agent) are able to reduce iron (III) ion to form iron (II) ion and iodine is formed.


It can be conclude that

Chlorine and bromine can act as oxidising agent while iodine is not ………………………………………………………………………………………………………..…… a good oxidising agent. Conversely, iodide ion can act as a good reducing agent.


The oxidation of thiosulphate ions, S2O32-, by halogen is quantitative.

Using volumetric analysis, the exact concentration of the reagent (S2O32-) used can be found.
Since chlorine and bromine are strong oxidising agent, when chlorine and bromine react with thiosulphate solution, thiosulphate ion is sulphate ion, SO42oxidise to its highest oxidation state, which is ………………………………… S2O32- + 5 H2O
2 SO42- + 8 e– + 10 H+ Cl2 + 2 e–
2 Cl– -------------------------------------------------------------------S2O32- + 5 H2O + 4 Cl2
2 SO42- + 8 Cl– + 10 H+
But when it react with a weaker oxidising halogen like iodine, it only slightly oxidised. Thiosulphate ion will be oxidised to become tetrathionate ion, S4O62-, which can be used in the volumetric analysis 2 S2O32-
S4O62- + 2 e– I2 + 2 e –
2 I– --------------------------------------2 S2O32- + l2
S4O62- + 2 l–


Example : The carbon monoxide in a sample of polluted air can readily be

determined by passing it over solid iodine(V) oxide, I2O5, to give carbon dioxide and iodine. (a) Write a balanced equation reaction between carbon monoxide and iodine(V) oxide. I2O5 + 5 CO
5 CO2 + I2 ............................................................................................................................................ (b) The iodine produced is removed and titrated with aqueous sodium thiosulphate: 2 Na2S2O3 (aq) + I2 (s) → Na2S4O6 (aq) + 2 NaI (aq) A 1.0 dm3 sample of air produced iodine that required 20.0 cm3 of 0.10 mol dm-3 sodium thiosulphate to discharge the iodine colour. Calculate the mass of carbon monoxide in this sample of polluted air. [4] Mol of Na2S2O3 = MV / 1000 ; mol = (0.10)(20.0) / 1000 Mol of Na2S2O3 = 0.002 mol From eq (b) since 2 mol of Na2S2O3 = 1 mol of I2 Mol of I2 = 0.001 mol In (a) since 1 mol of I2 = 5 mol of CO Mol of CO = 0.005 mol Mass of CO = mol x RMM @ mass = 0.005 x (12.0 + 16.0) Mass of CO = 0.14 g

6.3

Hydrogen halide and hydrides.

Halides Boiling point (oC) Bonding enthalpy (kJ/mol)

HF 23 +560

HCl - 85 +433

HBr - 66 +366

HI - 35 +298


When chlorine is mixed with hydrogen gas under the presence of

sunlight, an explosive reaction occur and yellow-greenish (chlorine) gas turned to white fumes (hydrogen chloride). Equation : Cl2 (g) + H2 (g) → 2 HCl (g)
The reactivity decreased when going down to Group 17 (as suggested by their respective E0 value) where bromine react with hydrogen upon heating to 2000C to form hydrogen bromide
While iodine react partially with hydrogen until reaction established an

equilibrium between hydrogen, iodine and hydrogen iodide, at 450oC under the presence of platinum as catalyst


All hydrogen halide exist as gas under room temperature and pressure,

except hydrogen fluoride, which exist as a volatile liquid at room temperature. Hydrogen fluoride exist as volatile liquid because H-F has strong hydrogen bond between them, which are stronger than the weak Van Der Waals forces that exist between HCl, HBr and HI. The boiling points eventually increased from HCl < HBr < HI, since the relative molecular mass increased, which increase the weak intermolecular forces.
The thermal stability of hydrogen halide decreased down Group 17. This can be explained by the fact of the increasing bond length of H X. When going down Group 17, the atomic radius increased. As a result, the bond length between hydrogen atom and halogen atom increased, which eventually caused lesser heats were required to break the covalent bond between H - X. When heated at high temperature, all hydrogen halide will decomposed to form back their respective elements. Some hydrogen iodide can even decomposed to form hydrogen and iodine under room temperature pressure until equilibrium is established. Equation : 2 HX (g) → H2 (g) + X2 (g)


The factor of bond length can also be used to explain the acidity of













hydrogen halide. When dissolved in water, acidity increase from HF