A-LEVEL CHEMISTRY: 3.2.3: Halogens (ChemRevise)

Question 1

describe:
-fluorine (F2)
-chlorine (Cl2)
-bromine (Br2)
-iodine (I2)

Answer 1

-fluorine (F2): pale yellow gas, very reactive

-chlorine (Cl2): green gas, reactive, poisonous in high concentrations

-bromine (Br2): red liquid, poisonous fumes

-iodine (I2): shiny grey solid, sublimes to purple gas

Question 2

explain the trend in melting & boiling point down group 7: (4)

Answer 2

increases

-number of shells increases. atomic radius increases

-atomic number increases. number of electrons increases

-stronger Van der Waals forces between the molecules, more energy needed. higher MP & BP

Question 3

explain the trend in electronegativity down group 7: (4)

Answer 3

decreases

-number of shells increases. atomic radius increases. shielding increases

-weaker electrostatic force of attraction between nucleus & external electrons

-less able to attract the bonding pair of electrons

Question 4

observation for reaction between chlorine (aq) and a chloride (aq):
why?

Answer 4

pale green solution.
no reaction (Cl does not displace Cl)

Question 4

observation for reaction between chlorine (aq) and a bromide (aq):
why?

Answer 4

yellow Solution
Cl displaces Br

Question 5

observation for reaction between chlorine (aq) and an iodide (aq):
why?

Answer 5

brown solution
Cl displaces I

Question 6

observation for reaction between bromine (aq) and a chloride (aq):
why?

Answer 6

yellow solution
no reaction (Br does not displace Cl)

Question 7

observation for reaction between bromine (aq) and a bromide (aq):
why?

Answer 7

yellow solution
no reaction (Br does not displace Br)

Question 8

observation for reaction between bromine (aq) and an iodide (aq):
why?

Answer 8

brown solution
Br displaces I

Question 9

observation for reaction between iodine (aq) and a chloride (aq):
why?

Answer 9

brown solution
no reaction (I does not displace Cl)

Question 10

observation for reaction between iodine (aq) and a bromide (aq):
why?

Answer 10

brown solution
no reaction (I does not displace Br)

Question 11

observation for reaction between iodine (aq) and an iodide (aq):
why?

Answer 11

brown solution
no reaction (I does not displace I)

Question 12

half equations for displacement reaction between chlorine (aq) and a bromide (aq):

Answer 12

Cl2 (aq) + 2e- —–> 2Cl- (aq)

2Br- (aq) —– Br2 (aq) + 2e-

Question 13

half equations for displacement reaction between chlorine (aq) and an iodide (aq):

Answer 13

Cl2 (aq) + 2e- —–> 2Cl- (aq)

2I- (aq) —–> I2 (aq) + 2e-

Question 14

half equations for displacement reaction between bromine (aq) and an iodide (aq):

Answer 14

Br2 (aq) + 2e- —–> 2Br- (aq)

2I (aq) —–> I2 (aq) + 2e-

Question 15

explain the reaction of halide ions with silver nitrate (AgNO3): (3)

Answer 15

-used as a test to identify which halide ion is present

-solution is made acidic with nitric acid, then silver nitrate solution is added dropwise

-used to differentiate between silver halide precipitates if they look too similar

Question 16

what is the role of nitric acid in the AgNO3 halide ions test?

Answer 16

to react with any carbonates present to prevent the formation of the precipitate Ag2CO3
this would otherwise mask the desired observations.

Question 17

silver chloride dissolves in dilute ammonia to form a complex Ion. reaction:

Answer 17

AgCl (s) + 2NH3 (aq) —–> Ag(NH3)2 + (aq) + Cl- (aq)

Question 18

silver bromide dissolved in dilute ammonia to form a complex ion. reaction:

Answer 18

AgBr (s) + 2NH3 (aq) —–> Ag(NH3)2 + (aq) + Br- (aq)

Question 19

silver iodide ___ ___ react with ammonia. why?

Answer 19

does not
because it is too insoluble

Question 20

what is a ‘reducing agent’?

Answer 20

a reducing agent donates electronse

(RIG)

Question 21

explain the trend in ‘reducing power’ as you go down group 7: (3)

Answer 21

-increases
-increased tendency to donate electrons

-because ions get bigger,
-increased shielding
-weaker electrostatic forces of attraction between nucleus & outermost electrons

-easier to loose them
-∴ become greater reducing agents

Question 22

F- and Cl- Ions are not strong enough Reducing Agents to Reduce the S in H2SO4.
No ___ Reactions occur. Only ___ ___ Reactions occur.

Answer 22

Redox, Acid-Base

Question 23

Acid-Base Reaction between Fluoride Ion & Sulfuric Acid:
(use Sodium Fluoride as the Fluoride)
Observations:

Answer 23

NaF (s) + H2SO4 (l) —–> NaHSO4 (s) + HF(g)

White Steamy Fumes of HF

Question 24

Acid-Base Reaction between Chloride Ion & Sulfuric Acid: (use Sodium Chloride as the Chloride) Observations:

Answer 24

NaCl (s) + H2SO4 (l) -----> NaHSO4 (s) + HCl (g) White Steamy Fumes of HCl

Question 25

Order these Halides in terms of Reducing Power (lowest to highest): Why?

Answer 25

-Fluoride -Chloride -Bromide Reducing Power Increases down the Group. (they become greater Reducing Agents)

Question 26

___ Ions are Stronger Reducing Agents than ___ & ___ Ions.

Answer 26

Br-, Cl- & F-

Question 27

Since Br- is such a strong Reducing Agent, it has an ___ ___ Reaction & then a ___ Reaction with Sulfuric Acid.

Answer 27

Acid-Base, Redox

Question 28

Br- Ion Reaction with Sulfuric Acid: (3)

Answer 28

Acid-Base step: NaBr (s) + H2SO4 (l) -----> NaHSO4 (s) +HBr (g) Redox step: 2H+ +2Br- + H2SO4 -----> Br2 (g) + SO2 (g) + 2H2O (l) Overall Equation: 2NaBr + 3H2SO4 -----> 2NaHSO4 + SO2 + Br2 + 2H2O

Question 29

Give the Oxidation & Reduction Half Equations for the Reaction between Br- and Sulfuric Acid: (2)

Answer 29

Ox: 2Br- -----> Br2 + 2e- Re: H2SO4 + 2H+ + 2e- -----> SO2 + 2H2O

Question 30

I- Ions are the ___ Halide Reducing Agents.

Answer 30

Strongest

Question 31

What is 'Disproportionation'?

Answer 31

Disproportionation is a type of Reaction where an Element simultaneously Oxidises & Reduces.

Question 32

eg of Disproportionation Reaction:

Answer 32

Chlorine with Water: Cl2 (g) + H2O (l) <-----> HClO (aq) + HCl (aq) Chlorine is simultaneously Oxidised & Reduced

Question 33

Reaction of Chlorine with Cold Dilute NaOH Solution:

Answer 33

Cl2 + 2NaOH -----> NaCl + NaClO + H2O