Group 7 elements (The halogens)

• Trends going down Group 7:

  • The colour of the elements becomes darker.
  • The atomic radius increases.
  • The reactivity decreases.
  • The boiling point of the elements increases.
  • The electronegativity decreases.
• Trends going down Group 7: Atomic radius

  More energy levels → Outer electrons are further away from the nucleus → Increases the size of the atom.
• Trends going down Group 7: Reactivity

  Outer electrons move further away from nuclease because of added energy levels → Increased shielding → Weaker attraction between the positive nucleus and outer negative electrons.
• Trends going down Group 7: Boiling point

  Size increases → Larger molecules because of more electrons → Greater induced dipole-dipole forces → Greater van der Waals forces → More energy needed to overcome forces.
• Trends going down Group 7: Electronegativity

  For smaller atoms, electrons in covalent bonds are closer to the nucleus → Less shielding → Strong attraction between positive nucleus and negatively charged electrons.
• Trends going down Group 7: Oxidising ability

  The oxidising power decreases as the atomic number increases. Therefore the oxidising ability decreases from fluorine to chlorine to bromine to iodine.

  1. A substance which gains electrons easily will be a good oxidising agent. The halogen atoms are gaining electrons into their outer energy levels to complete them.
  2. The electron which is gained by a fluorine atom completes an energy level closer to the nucleus than the electron gained by a chlorine atom. The electron in the fluorine atoms has stronger attraction and is less affected by shielding.
• Displacement reactions of halide ions in aqueous solution:

  Fluorine displaces all other halides from a solution of a halide compound.

  So…

  Fluorine will displace chlorine.

  Chlorine will displace bromine.

  Bromine will displace iodine.
• Trends going down Group 7: Reducing ability

  The reducing power of the halide ions increases as the atomic number increases, so the reducing ability increases going down the group.

  1. A substance which loses electrons easily is a good reducing agent, the halide ions are losing electrons from their outer energy level.
  2. The electron which is lost by an iodide ion comes from an energy level further from the nucleus than an electron lost from a bromide ion. The electron being lost by the iodide ion will have weaker attraction and greater shielding between itself and the nucleus.
• Reaction of $NaF$ with concentrated $H_2SO_4$:

  $NaF\ +$$\ H_2SO_4\ \rightarrow\ NaHSO_4\ +$$\ HF$

  This reaction produces sodium hydrogen sulfate and hydrogen fluroide and misty white fumes can be observed.
• Reaction of $NaCl$with concentrated $H_2SO_4$:

  $NaCl\ +$$\ H_2SO_4\ \rightarrow\ NaHSO_4\ +$$\ HCl$

  This reaction produces sodium hydrogen sulfate and hydrogen chloride (or hydrochloric acid) and misty white fumes can be observed.
• Reaction of $KBr$with concentrated $H_2SO_4$:

  $KBr\ +$$\ H_2SO_4\ \rightarrow\ KHSO_4\ +$$\ HBr$ NOT REDOX

  $2HBr\ +$$\ H_2SO_4\ \rightarrow\ Br_2\ +$$\ SO_2\ +$$\ 2H_2O$ REDOX

  Ionic equations of the redox equations:

  NOT REDOX = $Br^-+$$H_2SO_4\ \rightarrow\ HSO_4^-+$$HBr$

  REDOX = $2Br^-+$$SO_4^{2-}+$$4H^+$$\ \rightarrow\ Br_2+$$SO_2\ +$$2H_2O$

  The bromine ion reduces the sulfur in the sulfuric acid (+6 to +4).
• Reaction of $KBr$ with concentrated $H_2SO_4$:

  Overall equation:

  $2KBr+$$3H_2SO_4\ \rightarrow\ 2KHSO_4+$$Br_2+$$SO_2+$$2H_2O$

  Ionic equation of the overall equation:

  $2Br^-+$$3SO_4^{2-}+$$6H^+$$\rightarrow\ 2HSO_4^-+$$Br_2+$$SO_2+$$2H_2O$
• Reactions of $KI(s)$with concentrated $H_2SO_4$: First equation

  $KI+$$H_2SO_4\ \rightarrow\ KHSO_4+$$HI$

  This reaction produces sodium hydrogen sulfate and hydrogen iodide. Misty white fumes can be observed.

  Ionic equation: $I^-+$$H_2SO_4\ \rightarrow\ HSO_4^-+$$HI$
• Reactions of $KI(s)$with concentrated $H_2SO_4$: Second equation

  $2HI+$$H_2SO_4\ \rightarrow\ I_2+$$SO_2+$$2H_2O$

  The I is oxidised from -1 to 0 and the S is reduced from +6 to +4 (by the iodide ion).

  Ionic equation: $2I^-+$$SO_4^{2-}+$$4H^+$$\ \rightarrow\ I_2+$$SO_2+$$2H_2O$
• Reactions of $KI(s)$with concentrated $H_2SO_4$: Third equation

  $6HI+$$H_2SO_4\ \rightarrow\ 3I_2+$$S+$$4H_2O$

  The I is oxidised from -1 to 0 and the S is reduced from +6 to 0 (by the iodide ion).

  Ionic equation: $6I^-+$$8H^+$$+$$SO_4^{2-}\rightarrow\ 3I_2+$$S+$$4H_2O$
• Reactions of $KI(s)$with concentrated $H_2SO_4$: Fourth equation

  $8HI+$$H_2SO_4\ \rightarrow\ 4I_2+$$H_2S+$$4H_2O$

  The I is oxidised from -1 to 0 and the S is reduced from +6 to -2 (by the iodide ion).

  Ionic equation: $8I^-+$$10H^+$$+$$SO_4^{2-}\rightarrow\ 4I_2+$$H_2S+$$4H_2O$
• Reactions of $KI(s)$with concentrated $H_2SO_4$: Explanation

  The redox reactions between the solid sodium halides with concentrated sulfuric acid depend on the relative reducing power of the halides in the hydrogen halides.
• • Iodide ions are the best reducing agent, this is shown by it reducing S into sulfuric acid $\left(SO_2\right)$.
  • Bromine ions are a good reducing agent, this is shown by it reducing some of the sulfuric acid into $SO_2$ and $H_2O$.
  • Chlorine ions are not powerful for reducing agents, $HCl$ cannot reduce sulfuric acid, the chloride ions are simple displaced by the acid.
• Using concentration sulfuric acid to test for halide ions:

  • A solid bromide compound produces misty white fumes (of HBr) with brown gas (Bromine).
  • A solid iodide compound produces misty white fumes (of HI), a black solid (Solid iodine), purple fumes (Iodine gas) and a rotten egg smell (Hydrogen sulfide gas).

  This is not normally used as a test for halide ions as concentrated sulfuric acid is corrosive and toxic gasses are formed.
• Why is nitric acid added to the halide ions before adding silver nitrate solution?

  The dilute nitric acid removes other ions which would react with the silver ions in the silver nitrate solution, for example carbonate or hydroxide ions. This would interfere with the test and create inaccurate results.
• Silver nitrate solution does not form a precipitate with fluoride ions in solution as silver fluoride is soluble in water, therefore silver nitrate solution cannot be used to test for fluoride ions.
• The silver halides can be further identified by adding a dilute or concentrated ammonia solution, which shows the silver halides difference in solubility.

  • Silver chloride will dissolve in dilute and concentrated ammonia solution, forming a colourless solution.
  • Silver bromide will not dissolve in dilute ammonia but will dissolve in a concentrated ammonia solution, forming a colourless solution.
  • Silver iodide will not dissolve in dilute or concentrated ammonia solution.
• The other three silver halides are insoluble in water and the precipitates differ in colour.

  Chloride ions = White precipitate formed.

  $Ag^+$$(aq)+$$Br^-(aq)\ \rightarrow\ AgBr(s)$

  Bromide ions = Cream precipitate formed. 

  $Ag^+$$(aq)+$$Br^-(aq)\ \rightarrow\ AgBr(s)$

  Iodide ions = Yellow precipitate formed. 

  $Ag^+$$(aq)+$$I^-(aq)\ \rightarrow\ AgI(s)$
• Which halide ion is this based on this description?

  White precipitate formed which dissolves in dilute ammonia solution to form a colourless solution.

  Chloride ion
• Which halide ion is this based on this description?

  Cream precipitate formed which does not dissolve in dilute ammonia solution but dissolves in concentrated ammonia solution to form a colourless solution.

  Bromide ion
• Which halide ion is this based on this description?

  Yellow precipitate formed which does not dissolve in dilute or concentrated ammonia solution.

  Iodide ion
• Which halide ion is this based on this description?

  No precipitate formed

  No halide ions present