Group VII Elements Flashcards
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.
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
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.
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.
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.
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.
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.
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)
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
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)
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)
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
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
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)
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.
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)
̅− 2𝐹(𝑔)+4𝑂𝐻(𝑎𝑞)→4𝐹 (𝑎𝑞)+2𝐻𝑂(𝑙)+𝑂(𝑔)
