Group 7, Halogens

Question 1

what are the colours of the group 7 elements

Answer 1

fluorine = pale yellow gas
chlorine = pale green gas
bromine = brownish/redish liquid
iodine = black solid, purple vapour

Question 2

what are the properties of the halogens

Answer 2

DOWN A GROUP
* atomic radius increases
* boiling point increases - due a higher nuclear charge, stronger VDWs to overcome
* electronegativity decreases
* bond enthalpy decreases (except for F)
* reactivity increases

Question 3

what is the oxidising ability of the halogens

Answer 3

• as you go down the group, oxidising ability DECREASES
• fluorine has the most powerful oxidising ability and so reactions cannot be observed

Question 4

how is Br extracted from seawater

Answer 4

• involves oxidation of halide ions by a more reactive halide
• only Cl can be used as F is too reactive and so will reactive with the seawater

Question 5

what is the reducing ability of the halogens

Answer 5

• reducing power INCREASES down the group
• due to ionic radius increasing and therefore electrons are lost more easily as they are less strongly attracted to the nucleus

Question 6

what is the equation for the reaction of Sodium Chloride with Sulfuric Acid

Answer 6

NaCl (s) + H2SO4 (l) = NaHSO4 (l) + HCl (g)

Question 7

what is the equation for the reaction of Sodium Bromide with Sulfuric Acid

Answer 7

NaBr (s) + H2SO4 (l) = NaHSO4 (l) + HBr (g)
in this case the acid base reaction occurs again, but the hydrogen bromide formed can react with sulfuric acid

half equations -
* 2Br = Br2 + 2e-
* H2SO4 + 2H+ + 2e- = SO2 + 2H2O
combined -
* 2HBR (g) + H2SO4 (l) = SO2 (g) + Br2 (l) + 2H2O (l)

Question 8

what are the 3 products of Sodium Bromide and Sulfuric Acid

Answer 8

sodium bromide is a powerful enough reducing agent to reduce sulfuric acid to sulfuric dioxide
PRODUCTS:
* misty fumes ( HBr )
* brown fumes (Br2)
* colourless, choking gas (SO2)

Question 9

what is the equation for the reaction of Sodium Iodide with Sulfuric Acid

Answer 9

NaI (s) + H2SO4 (l) = NaHSO4 (l) + HI (g)
the first reaction is the acid base reaction to produce the hydrogen iodide

the iodide ion in HI is capable of reducing the H2SO4, further than the bromide can

combined : 3H2SO4 + 16HI = SO2 + H2S + S +8H2 + 10H2O

Question 10

what are the 5 products of Sodium Iodide and Sulfuric Acid

Answer 10

iodide ions are a powerful reducing agent, the products are:
* misty fumes ( HI )
* purple fumes and black solid ( I2)
* colourless, choking gas ( SO2 )
* yellow solid ( S )
* colourless gas with rotten egg smell ( H2S )