Group VII Elements

Question 1

What is the trend of Non-metallic character of group VII members?
All group members are non – metallic with increasing metallic character down the group.

Question 2

What is the state of group VII elements at room temperature?
Fluorine and chlorine- gases (pale yellow and greenish yellow respectively)

Bromine- brown volatile liquid

Iodine- black shiny solid and Astatine (radioactive, short – lived) black solid

Question 3

How do group VII elements bond?
(i) gaining one electron to form ions with a single negative charge e.g. F-, Cl-, Br-, I-
(ii) sharing of electrons forming one covalent bond with themselves or other elements e.g. F2, Cl2, Br2, HCl, HBr e.t.c
(iii) Chlorine, bromine and iodine have easily accessible d – orbitals, they are able to form covalent compounds in which the octet of electrons is expanded e.g. iodine shows valences of 1, 3, 5 and 7 respectively in ICl, ICl3, IF5 and IF7.
(iv) Fluorine cannot expand its octet due to lack of empty d-orbitals and therefore restricted to a covalency of 1.

Question 4

What is the trend of melting and boiling points of group VII elements?

Trend: It increases down the group

Explanation:
- As atomic radius of the elements increases down the group, their molecular masses and the number of electrons they contain increases leading to increase in Van der Waal forces attraction down the group.
- As the Van der Waal forces increase, the melting and boiling points also increase because more heat energy will be required to break the increased Van der Waal forces before the individual molecules can separate for melting or boiling to take place.

Question 5

Why are fluorine and chlorine gases at room temperature while iodine and bromine are solids and liquids respectively?

• In fluorine and chlorine, the Van der Waal forces are too weak and can even break at room temperature while in bromine and iodine these forces are sufficiently strong to keep the elements as liquid and solid respectively at room temperature.

Question 6

What is the trend of electron affinity of group VII elements?

• It increases from fluorine to chlorine and decreases from chlorine to iodine.
• The electron affinity of fluorine is abnormally lower than that of chlorine due to its very small atomic radius which causes unexpectedly stronger repulsion of the incoming electron by its outer electrons that are very close to each other.

Question 7

What is the trend of bond energy of group VII elements?
Trend: It increases from fluorine, F2 to chlorine Cl2 and then decreases from chlorine, Cl2 to iodine, I2.

Explanation:
- Fluorine has abnormally low bond energy. Due to the small atomic radius of fluorine, the lone pairs of electrons on the two bonded fluorine atoms repel themselves strongly and this weakens the fluorine – fluorine bonding resulting into abnormally low bond energy.
- From Cl2 to I2, the bond energy decreases with increase in bond length.
- As atomic radius increases down the group, the halogen – halogen bonds become longer and weaker resulting into decreased bond energy.

Question 8

What is the trend of reactivity of group VII elements?

The elements are generally very reactive due to:
• the low bond dissociation energy of their molecules which allows them to easily
dissociate into atoms.
• High electron affinity of the individual atoms which allows them to easily form ions.

Pb2+(aq)+2Br−(aq) → PbBr2(s)

• A yellow precipitate is formed with iodide ions.
Pb2+(aq)+2I−(aq) → PbI2(s)

Question 9

Describe the preparation of fluorine
• Fluorine is prepared by electrolysis of an equimolar mixture of hydrogen fluoride and potassium fluoride at 250oC between a carbon anode and a steel cathode.
Fluorine is liberated at the anode: 2F− (l) − 2e− → F (g) 2

Question 10

Describe the preparation of chlorine
• Chlorine can be prepared by:
(i) oxidation of cold concentrated hydrochloric acid by potassium manganate (VII)
2𝑀𝑛𝑂4−(𝑠) + 16𝐻+(𝑎𝑞) + 10𝐶𝑙−(𝑎𝑞) → 2𝑀𝑛+(𝑎𝑞) + 5𝐶𝑙2(𝑔) + 8𝐻2𝑂(𝑙)

(ii) heating a mixture of manganese(IV) oxide with concentrated hydrochloric acid MnO2 (s) + 4HCl(aq) → MnCl2 (aq) + Cl2 (g) + H2O(l)

Question 11

How may bromine and iodine be prepared?
• Bromine and iodine can be prepared by heating potassium bromide or potassium iodide respectively with concentrated sulphuric acid in presence of enough manganese(IV) oxide.

2KBr(s) + 2H2SO4 (aq) + MnO2 (s) → MnSO4 (aq) + K2SO4 (aq) + Br2 (g) + 2H2O(l) 2KI(s) + 2H2SO4 (aq) + MnO2 (s) → MnSO4 (aq) + K2SO4 (aq) + I2 (g) + 2H2O(l)

Question 12

Why does fluorine behave differently from the rest of halogens?

(i) Fluorine has the lowest bond dissociation energy.

(ii) Fluorine has the smallest small atomic radius

(iii) Fluorine has the highest electronegativity.

(iv) Fluorine has the highest positive electrode potential

Question 13

In what ways does fluorine behave differently from other halogens

• Fluorine has only one oxidation state
• Fluorine does not form oxo-acids
• Calcium fluoride is insoluble in water while the other calcium halides are soluble in water
• Hydrofluoric acid is the weakest acid compared to other halogen acids which are strong acids
• Fluorine combines directly with carbon at room temperature to form carbon tetra fluoride
• Silver fluoride is soluble in water while other halides of silver are insoluble in water
• Fluorides are more ionic than other halides because of the high electronegativity of fluorine resulting into large electronegative difference between fluorine and other elements.
• Hydrogen fluoride forms very strong hydrogen bonds because of its high electronegativity while the rest hydrogen halides form weak Van der Waal forces.
• Fluorine is very reactive compared to the rest of halogens due to its very low F – F bond dissociation energy.
• Fluorine forms very stable compounds compared to other group VII elements.
• Fluorine tends to bring out the maximum valency in other elements e.g. SF6, IF7

Question 14

How does fluorine react with water?
• Fluorine reacts with water vigorously forming hydrogen fluoride and oxygen gas.

2𝐹 (𝑔) + 2𝐻2𝑂(𝑙) ⇌ 4𝐻𝐹(𝑎𝑞) + 𝑂2 (𝑔)

Question 15

How do chlorine, bromine and iodine react with water?
• Chlorine, bromine and iodine slowly disproportionate in water forming hydronium ions, halide ions and hypohalous acid.

𝐶𝑙2(𝑔) + 2𝐻2𝑂(𝑙) ⇌ 𝐻3𝑂+(𝑎𝑞) + 𝐶𝑙−(𝑎𝑞) + 𝐻𝑂𝐶𝑙(𝑎𝑞) or
𝐶𝑙2(𝑔) + 𝐻2𝑂(𝑙) ⇌ 𝐻𝐶𝑙(𝑎𝑞) + 𝐻𝑂𝐶𝑙(𝑎𝑞)
𝐵𝑟 (𝑔) + 2𝐻 𝑂(𝑙) ⇌ 𝐻 𝑂+(𝑎𝑞) + 𝐵𝑟−(𝑎𝑞) + 𝐻𝑂𝐵𝑟(𝑎𝑞) 223
𝐼2(𝑔) + 2𝐻2𝑂(𝑙) ⇌ 𝐻3𝑂+(𝑎𝑞) + 𝐼−(𝑎𝑞) + 𝐻𝑂𝐼(𝑎𝑞)

The extent of disproportionation decreases from chlorine to iodine with iodine being only slightly soluble in water.

Question 16

How does fluorine react with sodium hydroxide (alkalis)?
• Fluorine reacts with cold dilute sodium hydroxide forming a sodium fluoride, oxygen difluoride and water.
2𝐹 (𝑔) + 2𝑁𝑎𝑂𝐻(𝑎𝑞) → 2𝑁𝑎𝐹(𝑎𝑞) + 𝐻 𝑂(𝑙) + 𝑂𝐹 (𝑔)
̅− 2𝐹(𝑔)+2𝑂𝐻(𝑎𝑞)→2𝐹 (𝑎𝑞)+𝐻𝑂(𝑙)+𝑂𝐹(𝑔)

• Fluorine reacts with hot concentrated sodium hydroxide forming a sodium fluoride, oxygen gas and water.
2𝐹 (𝑔) + 4𝑁𝑎𝑂𝐻(𝑎𝑞) → 4𝑁𝑎𝐹(𝑎𝑞) + 𝐻 𝑂(𝑙) + 𝑂 (𝑔) 222
̅− 2𝐹(𝑔)+4𝑂𝐻(𝑎𝑞)→4𝐹 (𝑎𝑞)+2𝐻𝑂(𝑙)+𝑂(𝑔)
222

How do chlorine, bromine and iodine react with sodium hydroxide (alkalis)?
• Chlorine, bromine and iodine disproportionate in cold dilute sodium hydroxide forming a halide, a halate (I) and water.
𝐶𝑙2(𝑔) + 2𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝑁𝑎𝐶𝑙(𝑎𝑞) + 𝑁𝑎𝑂𝐶𝑙(𝑎𝑞) + 𝐻2𝑂(𝑙)
𝐵𝑟 (𝑙) + 2𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝑁𝑎𝐵𝑟(𝑎𝑞) + 𝑁𝑎𝑂𝐵𝑟(𝑎𝑞) + 𝐻 𝑂(𝑙) 22
𝐼2(𝑠) + 2𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝑁𝑎𝐼(𝑎𝑞) + 𝑁𝑎𝑂𝐼(𝑎𝑞) + 𝐻2𝑂(𝑙)
OR
̅−̅
𝐶𝑙2(𝑔) + 2𝑂𝐻(𝑎𝑞) → 𝐶𝑙 (𝑎𝑞) + 𝑂𝐶𝑙(𝑎𝑞) + 𝐻2𝑂(𝑙)
̅−̅
𝐵𝑟 (𝑙) + 2𝑂𝐻(𝑎𝑞) → 𝐵𝑟 (𝑎𝑞) + 𝑂𝐵𝑟(𝑎𝑞) + 𝐻 𝑂(𝑙)
22
̅−̅
𝐼2(𝑠) + 2𝑂𝐻(𝑎𝑞) → 𝐼 (𝑎𝑞) + 𝑂𝐼(𝑎𝑞) + 𝐻2𝑂(𝑙)

• Chlorine, bromine and iodine disproportionate in hot concentrated sodium hydroxide forming a halide, a halate(V) and water.

3𝐶𝑙2(𝑔) + 6𝑁𝑎𝑂𝐻(𝑎𝑞) → 5𝑁𝑎𝐶𝑙(𝑎𝑞) + 𝑁𝑎𝐶𝑙𝑂3(𝑎𝑞) + 3𝐻2𝑂(𝑙)
3𝐵𝑟 (𝑙) + 6𝑁𝑎𝑂𝐻(𝑎𝑞) → 5𝑁𝑎𝐵𝑟(𝑎𝑞) + 𝑁𝑎𝐵𝑟𝑂 (𝑎𝑞) + 3𝐻 𝑂(𝑙) 232
3𝐼2(𝑠) + 6𝑁𝑎𝑂𝐻(𝑎𝑞) → 5𝑁𝑎𝐼(𝑎𝑞) + 𝑁𝑎𝐼𝑂3(𝑎𝑞) + 3𝐻2𝑂(𝑙)
OR
̅−−
3𝐶𝑙2(𝑔) + 6𝑂𝐻(𝑎𝑞) → 5𝐶𝑙 (𝑎𝑞) + 𝐶𝑙𝑂3 (𝑎𝑞) + 3𝐻2𝑂(𝑙)
−−
3𝐼2(𝑠) + 6𝑂𝐻(𝑎𝑞) → 5𝐼 (𝑎𝑞) + 𝐼𝑂3 (𝑎𝑞) + 3𝐻2𝑂(𝑙)

How do fluorine and chlorine react with potassium iodide?
• Both fluorine and chlorine displace iodine from potassium iodide solution forming iodine and this changes the solution from colourless to brown.

F2 (g) + 2I − (aq) → 2F − (aq) + I2 (aq) Cl2 (g) + 2I − (aq) → 2Cl − (aq) + I2 (aq)

True or false
Iodine is readily soluble in aqueous potassium iodide

True
Due to formation of a stable soluble complex of potassium tri-iodide. During the reaction the solution turns from colourless to brown.

𝐼2(𝑠) + 𝐾𝐼(𝑎𝑞) ⇌ 𝐾𝐼3(𝑎𝑞)
𝐼2(𝑠) + 𝐼−(𝑎𝑞) ⇌ 𝐼3−(𝑎𝑞)

True or false
Bromine does not react with potassium iodide solution.

True

How does fluorine react with hydrogen gas?
• Fluorine combines explosively with hydrogen even in the cold and dark to give hydrogen fluoride gas.
F2 (g) + H2 (g) → 2HF(g)

How does chlorine react with hydrogen gas?
• Chlorine and hydrogen explode if exposed to sunlight or a flame forming hydrogen chloride gas.
Cl2 (g) + H2 (g) → 2HCl(g)

How does bromine vapor react with hydrogen gas?
• Bromine vapour and hydrogen combine with a mild explosion if ignited forming hydrogen bromide gas at 200oC and in presence of platinum catalyst.
Br2 (g) + H2 (g) → 2HBr(g)

How does iodine react with hydrogen gas?
• Iodine and hydrogen when heated to about 400oC only combine partially forming hydrogen iodide gas.
𝐼2(𝑠) + 𝐻2(𝑎𝑞) ⇌ 2𝐻𝐼(𝑔)

What is the trend of bond energy of group VII hydrogen halides?

Trend: It decreases from hydrogen fluoride through hydrogen chloride and hydrogen bromide to hydrogen iodide (i.e. in order HF > HCl > HBr > HI)

Explanation:
- As the electronegativity of the halogen decreases from fluorine to iodine, the percentage of ionic character of the hydrogen – halogen bond decreases and this weakens the bond.
- Also the bonds become longer and weaker as atomic radius of the halogens increases from fluorine to iodine.
- Hence the energy required to break the H – X bond decreases down the group.

What is the trend of boiling points of hydrogen halides?
Trend: It decreases from hydrogen fluoride to hydrogen chloride and then increases from hydrogen chloride through hydrogen bromide to hydrogen iodide with hydrogen fluoride having abnormally higher boiling point than expected.
(It is in order HF&raquo_space;> HCl < HBr < HI)

Explanation:
- Hydrogen fluoride has abnormally high boiling point due to formation of very strong intermolecular hydrogen bonds owing to the high electronegativity of the fluorine atom. A lot of heat energy is required to break the strong hydrogen bonds in order to create enough vapour pressure equal to the atmospheric pressure for boiling to take place. Hence the very boiling point for hydrogen fluoride.
- From hydrogen chloride through hydrogen bromide to hydrogen iodide, the boiling point increases due to increase in the Van der Waal forces of attraction as the molecular mass of the compounds increases.

What is the trend of acid strength of hydrogen halides?

Trend: It increases from hydrogen fluoride through hydrogen chloride and hydrogen bromide to hydrogen iodide with hydrogen fluoride being far less acidic than expected. (It increases in order HF «< HCl < HBr < HI)

Explanation:
- The acid strength depends on the strength of the hydrogen – halogen bond.
- Due to the high electronegativity of fluorine, it forms a very strong H – F bonds; HF is therefore only slightly ionized in water and this makes it a weak acid.
- From hydrogen chloride to hydrogen bromide to hydrogen iodide the acid strength increases.
- This is because hydrogen - halogen bonds become longer and weaker as the atomic radius of the halogens increases.
- The hydrogen atom in hydrogen-halogen bond can easily be extracted by solvent water molecules forming hydronium ions.

Describe the ionization of hydrogen halides in water.
• Hydrogen chloride, hydrogen bromide and hydrogen iodide are all strong acids and therefore completely ionized in water forming hydronium and halide ions
𝐻𝑋(𝑎𝑞) + 𝐻2𝑂(𝑙) → 𝐻3𝑂+(𝑎𝑞) + 𝑋−(𝑎𝑞)

• Hydrogen fluoride is a weak acid and only partially ionized in water. Its degree of ionization depends on the concentration.

(i) In dilute solution it ionizes less forming hydronium and fluoride ions
𝐻𝐹(𝑔) + 𝐻2𝑂(𝑙) ⇌ 𝐻3𝑂+(𝑎𝑞) + 𝐹−(𝑎𝑞)

(ii) In concentrated solution the fluoride ions react with more hydrogen fluoride molecules forming hydrogen difluoride ions.
𝐻𝐹(𝑔) + 𝐹−(𝑎𝑞) ⇌ 𝐻𝐹 −(𝑎𝑞)

This causes more hydrogen fluoride to ionize in accordance to Le’ Chatelier’s
principle producing more hydronium ions and making it a stronger acid in
concentrated form than it is in dilute form.
Overall equation: 2𝐻𝐹(𝑔) + 𝐻 𝑂(𝑙) ⇌ 𝐻 𝑂+(𝑎𝑞) + 𝐻𝐹 −(𝑎𝑞)

Describe the preparation of hydrogen halides by direct synthesis
- All hydrogen halides can be prepared by direct combination between hydrogen and the halogens

• Fluorine explodes in dark forming hydrogen fluoride
H (g)+F (g)→ 2HF(l)

• Chlorine directly reacts with hydrogen in presence of sunlight/ultra-violet light sunlight
H2 (g) + Cl2 (g) ⎯⎯⎯⎯→ 2HCl(g)

• Bromine and iodine vapour directly react with hydrogen at 300oC in presence of
platinum catalyst to form the corresponding hydrogen halides
𝐵𝑟(𝑔)+𝐻(𝑔) ⇌ 2𝐻𝐵𝑟(𝑔)
𝐼2(𝑔) + 𝐻2(𝑔) ⇌ 2𝐻𝐼(𝑔)

Describe the preparation of hydrogen halides by displacement of a more volatile acid with a less volatile one

• Hydrogen chloride can be prepared by reacting solid sodium chloride with concentrated sulphuric acid
NaCl(s) + H2SO4 (l) → NaHSO4 (s) + HCl(g)

• Hydrogen bromide and hydrogen iodide cannot be prepared by the same method because both are strong reducing agents and readily oxidized by excess concentrated sulphuric acid to bromine and iodine respectively.
NaBr(s) + H2SO4 (l) → NaHSO4 (s) + HBr(g) NaI(s) + H2SO4 (l) → NaHSO4 (s) + HI(g)
8𝐻𝐼(𝑔) + 𝐻2𝑆𝑂4(𝑙) → 𝐻2𝑆(𝑔) + 4𝐻2𝑂(𝑙) + 4𝐼2(𝑔)

Describe the preparation of hydrogen halides by hydrolysis of phosphorus(III) halides
• Hydrogen chloride can be made by hydrolysis of phosphorus(III) chloride forming hydrogen chloride gas and orthophosphorous acid.
𝑃𝐶𝑙3(𝑙) + 3𝐻2𝑂(𝑙) → 𝐻3𝑃𝑂3(𝑎𝑞) + 3𝐻𝐶𝑙(𝑔)

(However, the most convenient method of making hydrogen chloride in the laboratory is by reacting concentrated sulphuric acid and sodium chloride).

• Hydrogen bromide can be conveniently prepared by hydrolysis of phosphorus(III) bromide.
• Method: A drop of bromine is added onto a mixture of red phosphorus and a little water. Bromine reacts phosphorus forming phosphorus(III) bromide which is then hydrolysed by water forming orthophosphorous acid and hydrogen bromide gas. 2𝑃(𝑠) + 3𝐵𝑟 (𝑙) → 2𝑃𝐵𝑟 (𝑙)
𝑃𝐵𝑟 (𝑙)+3𝐻 𝑂(𝑙)→𝐻 𝑃𝑂 (𝑎𝑞)+3𝐻𝐵𝑟(𝑔) 3233
2𝐻𝐵𝑟(𝑔)+𝐻 𝑆𝑂 (𝑙)→𝑆𝑂 (𝑔)+2𝐻 𝑂(𝑙)+𝐵𝑟 (𝑔)

• Hydrogen iodide can be conveniently prepared by hydrolysis of phosphorus(III) iodide. Method: Water is dropped onto a mixture of red phosphorus and solid iodine. Iodine reacts phosphorus forming phosphorus(III) iodide. The phosphorus(III) iodide is hydrolyzed by water forming orthophosphorous acid and hydrogen iodide gas.
2𝑃(𝑠) + 3𝐼2(𝑠) → 2𝑃𝐼3(𝑠)
𝑃𝐼3(𝑙) + 3𝐻2𝑂(𝑙) → 𝐻3𝑃𝑂3(𝑎𝑞) + 3𝐻𝐼(𝑔)

How do hydrogen halides react with concentrated sulphuric acid?

• Hydrogen chloride is a weak reducing agent and therefore does not react with concentrated sulphuric acid.

• Hydrogen bromide is a stronger reducing agent than hydrogen chloride and therefore oxidized by concentrated sulphuric acid to bromine and the acid reduced to Sulphur dioxide gas.
2𝐻𝐵𝑟(𝑔)+𝐻 𝑆𝑂 (𝑙)→𝑆𝑂 (𝑔)+2𝐻 𝑂(𝑙)+𝐵𝑟 (𝑔)

• Hydrogen iodide is oxidized by concentrated sulphuric acid to iodine and the acid further reduced to hydrogen sulphide (because hydrogen iodide is a stronger reducing agent than hydrogen bromide).
8𝐻𝐼(𝑔) + 𝐻2𝑆𝑂4(𝑙) → 𝐻2𝑆(𝑔) + 4𝐻2𝑂(𝑙) + 4𝐼2(𝑔)

Increase in oxygen content of the oxoacids of chlorine leads to?

• increase in thermal stability i.e. HOCl < HClO2 < HClO3 < HClO4

• increase in acid strength i.e. HOCl < HClO2 < HClO3 < HClO4

• decrease in oxidizing power i.e. HOCl > HClO2 > HClO3 > HClO4

Describe the trend of acid strength of the oxoacids of chlorine

Trend: It increases in order HOCl < HClO2 < HClO3 < HClO4.

Explanation:
- Oxygen being more electronegative than chlorine it draws the bonding electrons away from chlorine and the more the oxygen atoms bonded to chlorine, the greater the bonding electrons are withdrawn away from chlorine and the weaker the oxygen-hydrogen bond (O – H) bond.
- As a result the ease with which the oxoacids release hydrogen ions increases as the number of oxygen atoms bonded to chlorine increases.

What is the result of reacting solid halides with concentrated sulphuric acid?

• White fumes (of hydrogen chloride) are observed with solid chlorides

• Reddish-brown fumes (of bromine) are observed with solid bromides

• Purple vapour (of iodine) is observed with solid iodides

Treating an aqueous solution of Cl-, Br- or I- with silver nitrate results in what?

• A white precipitate of silver chloride is formed with chloride ions. Ag+ (aq) + Cl− (aq) → AgCl(s)
The white precipitate is soluble in dilute aqueous ammonia but insoluble in dilute nitric acid.
AgCl(s)+2NH3(aq) → [Ag(NH3)2]+(aq)+Cl−(aq)

• A pale – yellow precipitate of silver bromide is formed with bromide ions.
Ag+ (aq) + Br− (aq) → AgBr(s)
The pale-yellow precipitate is insoluble in both dilute nitric acid and in dilute aqueous ammonia (but soluble in concentrated ammonia).

• A yellow precipitate of silver iodide is formed with iodide ions.
𝐴𝑔+(𝑎𝑞) + 𝐼−(𝑎𝑞) → 𝐴𝑔𝐼(𝑠)
The yellow precipitate is insoluble in both dilute nitric acid and concentrated ammonia solution.

What is the result of reacting aqueous lead(II) nitrate or aqueous lead(II) ethanoate with chloride, bromide and iodide ions?

• A white precipitate is observed with chloride and bromide ions. The precipitate is soluble in hot water.
Pb2+(aq)+2Cl−(aq) → PbCl2(s)