The p-Block Elements
Halogens contain fluorine, chlorine, bromine, iodine (and highly unstable astatine).
Characteristic properties of the halogens
All halogens have
i) high electronegativities.
ii) high electron affinities.
These show that halogens have a high tendency to attract electrons.
Usually halogens have the oxidation state of –1 in their ionic or covalent compounds.
e.g. HF , HCl, NaCl
Fluorine being the most electronegative element should have the oxidation state of –1 in its
compound. However, Cl , Br anf I may have the oxidation state other –1.
e.g. Cl2 O , ClO 3 , BrO - , IO 3 - ,..
Variation in properties of the halogens and their compounds
i) Melting point and boiling
The melting /boiling point increases down the group because the molecular mass increases down
the group and hence the intermolecular (van der Waals’) force increases.
The electronegativity decreases down the group because the outer electrons become progressively
Better shielded from the nucleus as the atomic size and number of inner electron shells increase.
Hence the electrons in a covalent bond are attracted less to the halogens.
iii) Electron affinity
The electron affinity is the enthalpy cha nge when 1 mole of a gaseous atoms acquire an electron to
give gaseous ions.
X(g) + e- ---à X-(g)
The electron affinity increases from fluorine to chlorine and then decreases from chlorine to
a) the increase in atomic size and number of electron shells down the group lead to the decrease in
effective nuclear charge.
b) The atomic size is very small, the addition of an electron produces an important
electron-electron repulsion. Hence its electron affinity is lower than expected.
iv) Bond enthalpy
The bond enthalpy is the energy required to break 1 mole of covalent bonds in gaseous state.
X2 (g) ---à 2 X(g)
The bond enthalpy of fluorine is lower than expected. Since the size of fluorine atom is very small,
the unusually short F-F bond length leads to very high repulsion between the non-bonded
electrons of each fluorine atom. Hence F-F bond is weaker than expected.
The bond enthalpy for other halogens decreases in the order of: Cl2 > Br2 > I2 . It is because the
increase in atomic size down the group leads to the increase in bond length. Hence the strength
Relative oxidizing power of halogens
Relative oxidizing power: F2 > Cl2 > Br2 > I2
i) Reaction with sodium
2 Na(s) + F2 ---à 2 NaF(s) explosive
2 Na(s) + Cl2 ---à 2 NaCl(s) violent
2 Na(s) + Br2 ---à 2 NaBr(s) moderate
2 Na(s) + I2 ---à 2 NaI(s) moderate
ii) Reaction with iron(II) ion
Cl2 (aq) + 2 e- ---à 2 Cl-(aq) E = +1.36V
Br2 (aq) + 2 e- ---à 2 Br-(aq) E = +1.07V
I2 (aq) + 2 e- ---à 2 I-(aq) E = +0.54V
Fe3+(aq) + e- ---à Fe2+(aq) E = +0.77V
iii) Reaction with phosphorus
2 P(s) + 5 F2 (g) ---à 2 PF5 (s)
2 P(s) + 5 Cl2 (g) ---à 2 PCl5 (s)
2 P(s) + 3 Cl2 (g) ---à 2 PCl3 (s)
2 P(s) + 3 Br2 (g) ---à 2 PBr3 (s)
2 P(s) + 3 I2 (g) ---à 2 PI3 (s)
Disproportionation of the halogens in alkalis
Disproportionation is a chemical change in which one particular species (molecule, atom or ion) is
simultaneously oxidized and reduce.
Cl2 (aq) + H2 O(l) ---à HCl(aq) + HOCl(aq)
For F2 ,
2 F2 (g) + 2 NaOH(aq) ---à 2 NaF(aq) + OF2 (g) + H2 O(l) cold, dilute
2 F2 (g) + 4 NaOH(aq) ---à 4 NaF(aq) + O2 (g) + 2 H2 O(l) hot, conc entrated
For Cl2 ,
Cl2 (aq) + 2 NaOH(aq) ---à NaCl(aq) + NaOCl(aq) + H2 O(l) cold, dilute
3 Cl2 (aq) 6 NaOH(aq) ---à 5 NaCl(aq) + NaClO 3 (aq) + 3 H2 O(l) hot, concentrated
For Br2 ,
3 Br2 (aq) 6 NaOH(aq) ---à 5 NaBr(aq) + NaBrO 3 (aq) + 3 H2 O(l)
For I2 ,
3 I2 (aq) 6 NaOH(aq) ---à 5 NaI(aq) + NaIO 3 (aq) + 3 H2 O(l)
Comparative study of the reactions of halide ions
i) with halogens
The stronger oxidizing halogen can displace (oxidize) other halide ions.
F2 (aq) + 2 Cl-(aq) ---à 2 F-(aq) + Cl2 (aq)
Cl2 (aq) + 2 Br-(aq) ---à 2 Cl-(aq) + Br2 (aq)
Cl2 (aq) + 2 I-(aq) ---à 2 Cl-(aq) + I2 (aq)
Br2 (aq) + 2 I-(aq) ---à 2 Br-(aq) + I2 (aq)
ii) with concentrated sulphuric(VI) acid
For F- and Cl- ions,
NaF(s) + H2 SO4 (l) ---à NaHSO4 (s) + HF(g)
NaCl(s) + H2 SO4 (l) ---à NaHSO4 (s) + HCl(g)
For Br- ion,
NaBr(s) + H2 SO4 (l) ---à NaHSO4 (s) + HBr(g)
2 HBr(g) + H2 SO4 (l) ---à Br2 (g) + SO2 (g) + 2 H2 O(l)
2 NaBr(s) + 3 H2 SO4 (l) ---à 2 NaHSO4 (s) + SO2 (g) + Br2 (g) + 2 H2 O(l)
For I- ion,
NaI(s) + H2 SO4 (l) ---à NaHSO4 (s) + HI(g)
8 HI(g) + H2 SO4 (l) ---à 4 I2 (g) + H2 S(g) + 4 H2 O(l)
8 NaI(s) + 9 H2 SO4 (l) ---à 8 NaHSO4 (s) + 4 I2 (g) + H2 S(g) + 4 H2 O(l)
The difference between reactions of halide ions with concentrated H2 SO4 is due to HF and HCl
cannot be oxidized by concentrated H2 SO4 . However HBr is fairly easily oxidized to bromine and
HI can be easily oxidized to iodine.
iii) with phosphoric(V) acid
Halide ions react with phosphoric(V) acid to form the hydrogen halides. Since phosphoric(V) acid
is not a oxidizing agent (concentrated H2 SO4 is a strong oxidizing agent), the hydrogen halides
will not be oxidized to halogens.
3 NaX(s) + H3 PO4 (l) ---à Na3 PO4 (s) + 3 HX(g)
e.g. 3 NaCl(s) + H3 PO4 (l) ---à Na3 PO4 (s) + 3 HCl(g)
3 NaBr(s) + H3 PO4 (l) ---à Na3 PO4 (s) + 3 HBr(g)
3 NaI(s) + H3 PO4 (l) ---à Na3 PO4 (s) + 3 HI(g)
iv) with silver ion
The presence of Cl- , Br- and I- can be tested by treatment of acidified silver nitrate solution
(acidified with dilute nitric acid). The addition of dilute nitric acid is to remove the interfering
ions like SO3 2- , CO32- and OH- ions. This is because if these ions are present, they will react with
silver ion to form precipitates.
2 Ag+(aq) + SO3 2-(aq) ---à Ag2 SO3 (s)
2 Ag+(aq) + CO32-(aq) ---à Ag2 CO3 (s)
2 Ag+(aq) + 2 OH-(aq) ---à Ag2 O(s) + H2 O(l)
If dilute nitric acid is added before, these ions will be removed.
2 H+(aq) + SO3 2-(aq) ---à SO2 (g) + H2 O(l)
2 H+(aq) + CO3 2-(aq) ---à CO2 (g) + H2 O(l)
H+(aq) + OH-(aq) ---à H2 O(l)
The different silver halides can be identified by their colours.
Ag+(aq) + Cl-(aq) ---à AgCl(s) white precipitate
Ag+(aq) + Br-(aq) ---à AgBr(s) pale yellow precipitate
Ag+(aq) + I-(aq) ---à AgI(s) yellow precipitate
Acidic properties of hydrogen halides and the anomalous behaviour of hydrogen fluoride
All hydrogen halides are acidic because they all ionized when dissolved in water.
HX(g) + H2 O(l) H3 O+(aq) + X-(aq)
Acid strength: HI > HBr > HCl >> HF
Explanation: The ionization process involves the breakage of –X bonds. Since the bond strength is in
order: H-F > H-Cl > HBr > H-I . Hence this gives the above trend of acid strength.
Anomalous behaviour of HF
i) HF has a abnormally high boiling point. It is because it has intermolecular hydrogen bond.
ii) HF is a very weak acid in dilute aqueous solutio n but is a strong acid in concentrated solution
(5 to 15M).
The weak acidic property exhibited by HF in dilute aqueous solution is due to the high bond
energy of H-F bond and the exceptionally small electron affinity of fluorine.
In dilute solution: HF(aq) H+(aq) + F-(aq) (1)
In more concentrated solution: F-(aq) + HF(aq) [HF 2 ]-(aq) (2)
When in concentrated solution, F-(aq) ions form the stable complex ions with HF(aq). Hence this
favour the dissociation of the HF acid by shifting the equilibrium (1) to the right.
Uses of halogens and halogen containing compounds
1. Fluoride in fluoridation of water.
This decreases the incidence of tooth decay.
2. Chlorine in the manufacture of poly(chloroethene) PVC, bleach and disinfectant.
Bleach production: Cl2 (aq) + 2 NaOH(aq) ---à NaCl(aq) + NaOCl(aq) + H2 O(l)
3. Silver bromide used in photographic films.
Nitrogen and its compounds
Nitrogen occurs as free N2 molecules. Nitrogen is obtained industrially by fractional distillation of
Nitrogen is very unreactive because it has a very high bond enthalpy.
N¡ÝN +946 kJmol-1
Direct combination of nitrogen and oxygen:
This reaction requires high activation energy. (electric discharge in lightning and high temperature in
N2 (g) + O2 (g) ---à 2 NO(g)
2 NO(g) + O2 (g) ---à 2 NO 2 (g)
Manufacture of ammonia by Haber process and its underlying physicochemical principles:
Nitrogen and hydrogen can combine directly in the presence of a finely divided iron catalyst and 1t
550o C and 200 – 1000 atm.
N2 (g) + 3 H2 (g) 2 NH3 (g) £GH = -92.3 kJmol-1
The physicochemical principles and economic considerations in choice of reaction conditions
The following graphs show the formation of ammonia at different conditions
Choice of pressure: By Le Chatelier ’ principle, increase in pressure would shift the equilibrium to
right and increase the % yield of ammonia formed.
However, from the economic point of view, the cost of maintaining high pressure
is also very high, so that too high pressure are economically not justified.
Choice: 200 – 1000 atmospheres
Choice of temperature: By Le Chatelier ’ principle, lower the temperature would shift the
equilibrium to the right as the reaction is exothermic.
However, the reaction rate is very low at a low temperature and this would
increase the time to reach equilibrium.
Choice: 450 – 550 o C
The manufacturing plant for the synthesis of ammonia is shown below:
Ammonia as a reducing agent and a base
Ammonia can act as a reducing agent.
e.g. 4 NH3 (g) + 3 O2 (g) ---à
4 NH3 (g) + 5 O2 (g) ---à
2 NH3 (g) + 3 CuO(s) ---à
2 NH3 (g) + 3 Cl2 (g) ---à
Ammonia can act as a (weak) base.
NH3 (aq) + H2 O(l) NH4 +(aq) + OH-(aq)
Catalytic oxidation of ammonia in the manufacture of nitric(V) acid
Nitric(V) acid can be prepared by ammonia.
At the reactor a: 4 NH3 (g) + 5 O2 (g) 4 NO(g) + 6 H2 O(g)
At the reactor b: 2 NO(g) + O2 (g) ---à 2 NO2 (g)
At the absorber: 4 NO2 (g) + 2 H2 O(l) + O2 (g) ---à 4 HNO3 (aq)
Nitric(V) acid as an oxidizing agent
Reaction with copper:
Concentrated nitric(V) acid
Cu(s) 4 H+ (aq) + 2 NO3 -(aq) ---à Cu2+(aq) + 2 NO2 (g) + 2 H2 O(l)
Dilute nitric(V) acid
3 Cu(s) + 8 H+ (aq) + 2 NO3 -(aq) ---à 3 Cu2+(aq) + 2 NO(g) + 4 H2 O(l)
Reaction with iron(II) ion:
3 Fe2+(aq) + 4 H+(aq) + 2 NO3 -(aq) ---à 3 Fe3+(aq) + NO(g) + 2 H2 O(l)
Reaction with sulphur:
S(s) + 6 HNO3 (aq) ---à H2 SO4 (aq) + 6 NO2 (g) + 2 H2 O(l)
Action of heat on nitrate(V)
- For potassium nitrate and sodium nitrate, they decompose to form nitrate(III) and oxygen.
- For calcium nitrate to copper(II) nitrate, they decompose to form metal oxides, nitrogen dioxide
- For mercury(II) nitrate and silver nitrate, they decompose to from metals, nitrogen dioxide and
Brown ring test for nitrate(V) ions
The brown ring test is used to detect the presence of nitrate(V) ion NO3 - in aqueous solution.
Procedure: A freshly prepared iron(II) sulphate(VI) solution FeSO4 (aq) is mixed with a solution
suspected to contain the nitrate(V) ions in a test tube. While the test tube is tilted, concentrated
sulphuric acid is carefully added to the bottom of the test tube along the side. If a brown ring appears
at the interface of the upper aqueous solution and the lower concentrated sulphuric acid, it shows the
presence of nitrate(V) ions in the solution.
Sulphur and its compounds
Burning of sulphur:
If sulphur is burnt in oxygen, sulphur dioxide is formed.
S(s) + O2 (g) ---à SO2 (g)
If sulphur is heated first gently and then strongly, the following changes will be observed.
Oxidizing and reducing properties of sulphur dio xide
Sulphur dioxide can act as both an oxidizing agent and reducing agent.
Manufacture of sulphuric(VI) acid by contact process
The contact process involves the following steps:
1) production of sulphur dioxide
S(s) + O2 (g) ---à SO2 (g)
2) conversion of sulphur dioxide to sulphur trioxide
2 SO2 (g) + O2 (g) 2 SO3 (g)
conditions: 450 – 500 o C , vanadium(V) oxide V2 O5 catalyst , 1 atm.
3) conversion of sulphur trioxide to sulphuric(V) acid
sulphur trioxide is dissolved in concentrated sulphuric(VI) acid because SO3 (g) react violently
with water to form a mist of sulphuric(VI) acid.
Physicochemical and economic consideration of contact process
The most important process for the manufacture of sulphuric(VI) acid is the following equilibrium.
2 SO2 (g) + O2 (g) 2 SO3 (g)
By Le Chatelier’s principle, high pressure and low temperature would shift the equilibrium to the
right and favour the formation of sulphur trioxide.
However, too low temperature would increase the time to reach equilibrium and temperature of
about 450 o C is chosen. At this temperature, conversion is already high (>95%). Hence a low
pressure of 1 atm is sufficient.
Sulphuric(VI) acid as an oxidizing agent and a dehydrating agent
SO42-(aq) + 4 H+(aq) + 2 e- ---à SO2 (g) + 2 H2 O(l)
Test for sulphate(VI) ions
Procedure: To the aqueous solution to be tested, barium chloride solution is added. Followed by
excess dilute hydrochloric acid. If sulphate(VI) ions is present, a white precipitate can be observed.
Ba2+(aq) + SO4 2-(aq) ---à BaSO4 (s) white
Note that sulphate(IV) ios SO3 2- and carbonate ions CO3 2- also give white precipitate with barium
Hydrochloric acid is added to remove these interfering ions.
Uses of sulphuric(VI) acid
1. manufacture of fertilizers
2. manufacture of detergents.
3. manufacture of paints, pigments and dyestuffs.