JEE Chemistry The P Block Elements Notes Set 21

Group 17: Occurrence and Preparation

The halogens are fluorine, chlorine, bromine, iodine and radioactive astatine. They are too reactive to occur free. Fluorine is found in fluorspar \( \ce{CaF2} \), cryolite \( \ce{Na3AlF6} \) and fluorapatite; chlorine in rock salt, carnallite and sea water; bromine in sea water and brine wells; iodine in sea weed and in Chile saltpetre as sodium iodate.
1. Fluorine cannot be freed by any chemical oxidant, so it is obtained by electrolysis of \( \ce{KHF2} \) dissolved in anhydrous HF (anode: \( \ce{2F- -> F2 + 2e-} \); hydrogen collects at the cathode).
2. Chlorine: electrolysis of brine (chlor-alkali), Deacon process \( \ce{4HCl + O2 ->[CuCl2][700 K] 2Cl2 + 2H2O} \), or in the laboratory \( \ce{MnO2 + 4HCl -> MnCl2 + Cl2 + 2H2O} \) and \( \ce{2KMnO4 + 16HCl -> 2KCl + 2MnCl2 + 8H2O + 5Cl2} \).
3. Bromine: chlorine is passed into bromide-rich brine, \( \ce{2Br- + Cl2 -> 2Cl- + Br2} \), and the bromine is blown out by air.
4. Iodine: from iodate with sodium hydrogensulphite, \( \ce{2IO3- + 5HSO3- -> 3HSO4- + 2SO4^2- + H2O + I2} \), or by oxidising iodide with chlorine.

Periodic Trends of the Halogens

Fluorine shows only the -1 state because it has no d-orbitals and is the most electronegative element. The others show +1, +3, +5 and +7 in oxides, oxoacids and interhalogens, using their d-orbitals to expand the octet. Iodine is violet in vapour and in \( \ce{CS2} \), brown in water and in KI solution (as \( \ce{I3-} \)).

Chemical Reactions of the Halogens

1. With water: \( \ce{2F2 + 2H2O -> 4HF + O2} \); chlorine and bromine give \( \ce{X2 + H2O <=> HX + HOX} \) (disproportionation); the equilibrium for iodine lies to the left.
2. With alkali: cold dilute NaOH gives \( \ce{Cl2 + 2NaOH -> NaCl + NaOCl + H2O} \); hot concentrated NaOH gives \( \ce{3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O} \). Fluorine gives \( \ce{2F2 + 2NaOH -> 2NaF + OF2 + H2O} \) (dilute) or oxygen (concentrated).
3. With ammonia: excess ammonia gives \( \ce{8NH3 + 3Cl2 -> 6NH4Cl + N2} \); excess chlorine gives explosive \( \ce{NCl3} \) (\( \ce{NH3 + 3Cl2 -> NCl3 + 3HCl} \)).
4. Bleaching: moist chlorine gives HOCl, which releases nascent oxygen, an oxidising and permanent bleach. Sulphur dioxide bleaches by reduction and the colour returns on standing in air.
5. Displacement: a more electronegative halogen displaces a heavier halide: \( \ce{Cl2 + 2KI -> 2KCl + I2} \). Fluorine, however, reacts with water instead.
6. Iodine and thiosulphate: \( \ce{I2 + 2S2O3^2- -> 2I- + S4O6^2-} \), the basis of iodometric titration.
7. With hydrogen sulphide, hydrocarbons: \( \ce{H2S + Cl2 -> 2HCl + S} \), and \( \ce{C10H16 + 8Cl2 -> 10C + 16HCl} \) for turpentine oil.

Hydrogen Halides

Preparation: \( \ce{NaCl + H2SO4 -> NaHSO4 + HCl} \) (heated further to give \( \ce{Na2SO4} \)), and \( \ce{CaF2 + H2SO4 -> CaSO4 + 2HF} \). HBr and HI cannot be made with concentrated sulphuric acid, since they reduce it: \( \ce{2HBr + H2SO4 -> Br2 + SO2 + 2H2O} \). Use phosphoric acid, or \( \ce{PBr3} \) and \( \ce{PI3} \) made in situ.
Trends from HF to HI: bond length increases, bond enthalpy and thermal stability decrease, acid strength and reducing power increase. HF has the highest boiling point because of hydrogen bonding (and exists as chains, with the ion \( \ce{HF2-} \) in KHF2). HF attacks glass: \( \ce{SiO2 + 4HF -> SiF4 + 2H2O} \), so it is stored in wax or plastic bottles. AgF is soluble while AgCl, AgBr and AgI are precipitated; AgCl dissolves in ammonia, AgBr partly, AgI not.
Aqua regia is a 3:1 mixture of concentrated HCl and HNO3; it dissolves gold and platinum: \( \ce{Au + 4H+ + NO3- + 4Cl- -> [AuCl4]- + NO + 2H2O} \).

Oxides and Oxoacids of Chlorine

Acid strength of the oxoacids: \( \ce{HOCl < HClO2 < HClO3 < HClO4} \) (more terminal oxygen atoms delocalise the negative charge of the anion). Thermal and oxidising behaviour: hypochlorite and chlorate disproportionate on heating; chlorate \( \ce{4KClO3 -> 3KClO4 + KCl} \) (without catalyst) and \( \ce{2KClO3 ->[MnO2] 2KCl + 3O2} \). Fluorine forms only the unstable \( \ce{HOF} \). For the same oxidation state, acid strength falls from chlorine to iodine (\( \ce{HClO4 > HBrO4 > HIO4} \)); hypohalous acid strength also falls from chlorine to iodine: \( \ce{HOCl > HOBr > HOI} \).

Interhalogens, Polyhalides and Pseudohalogens

Interhalogens have the formulae XX', \( \ce{XX'3} \), \( \ce{XX'5} \) and \( \ce{XX'7} \), where X is the larger, less electronegative halogen. They are more reactive than the halogens (except fluorine) because the X-X' bond is weaker than the X-X bond, and they hydrolyse to halide and oxohalide: \( \ce{ICl + H2O -> HCl + HOI} \) and \( \ce{IF5 + 3H2O -> HIO3 + 5HF} \).
Shapes: \( \ce{ClF3} \), \( \ce{BrF3} \) T-shaped (sp3d, two lone pairs); \( \ce{IF5} \) and \( \ce{BrF5} \) square pyramidal (sp3d2, one lone pair); \( \ce{IF7} \) pentagonal bipyramidal (sp3d3). ClF3 is a fluorinating agent; liquid BrF3 conducts through self-ionisation \( \ce{2BrF3 <=> BrF2+ + BrF4-} \).
Polyhalide ions such as \( \ce{I3-} \), \( \ce{ICl2-} \) and \( \ce{IBr2-} \) are formed with large cations; this is why iodine dissolves in KI solution.
Pseudohalogens such as \( \ce{(CN)2} \), \( \ce{(SCN)2} \) and pseudohalide ions \( \ce{CN-} \), \( \ce{SCN-} \), \( \ce{N3-} \), \( \ce{OCN-} \) mimic halogens: they form insoluble silver and lead salts, dimerise, and disproportionate in alkali: \( \ce{(CN)2 + 2OH- -> CN- + OCN- + H2O} \).

Bleaching Powder

Chlorine passed over dry slaked lime gives \( \ce{Ca(OH)2 + Cl2 -> Ca(OCl)Cl + H2O} \). It is a mixed salt of hypochlorite and chloride; with dilute acid it liberates chlorine ('available chlorine'), \( \ce{Ca(OCl)Cl + H2SO4 -> CaSO4 + Cl2 + H2O} \). It is used for bleaching and disinfecting.

Group 18: Noble Gases

The noble gases are helium, neon, argon, krypton, xenon and radioactive radon, with closed-shell configurations (ns2np6, and 1s2 for helium). They are monoatomic gases with very low boiling points, which rise down the group because dispersion forces grow with size. Argon makes up about 0.93 percent of dry air, and the others occur in traces. Helium is obtained from natural gas, and is used in cryogenics, balloons (non-flammable, but heavier than hydrogen) and in deep-sea breathing mixtures.
Ionisation enthalpy falls down the group, while the solubility in water, atomic radius, polarisability and density increase. Helium does not solidify at ordinary pressure and shows superfluidity (helium II) below 2.17 K.
Only Kr and Xe (and Rn) form chemical compounds, with fluorine and oxygen. Neon and helium have no known stable compounds and, being very small, do not form clathrates. Argon, krypton and xenon form clathrates by being trapped in cages of quinol or ice. The first compound, \( \ce{Xe+[PtF6]-} \), was made by Bartlett, since the ionisation energies of \( \ce{O2} \) and Xe are almost the same.

Compounds of Xenon

1. Hydrolysis: \( \ce{2XeF2 + 2H2O -> 2Xe + 4HF + O2} \); \( \ce{6XeF4 + 12H2O -> 4Xe + 2XeO3 + 24HF + 3O2} \); \( \ce{XeF6 + 3H2O -> XeO3 + 6HF} \).
2. In alkali, xenate disproportionates: \( \ce{2HXeO4- + 2OH- -> XeO6^4- + Xe + O2 + 2H2O} \) (perxenate, Xe in +8).
3. Fluoride acceptors and donors: \( \ce{XeF2 + SbF5 -> [XeF]+[SbF6]-} \), \( \ce{XeF6 + RbF -> Rb[XeF7]} \).
4. Reduction: \( \ce{XeF6 + 3H2 -> Xe + 6HF} \), and \( \ce{XeF4 + 2H2 -> Xe + 4HF} \). All the fluorides are strong oxidising and fluorinating agents; \( \ce{XeF2 + S8} \) gives \( \ce{SF6} \).

Worked Examples

Example 1. Excess ammonia is treated with chlorine. Name the products and state the bond order in the gaseous one.
Answer:
Excess \( \ce{NH3} \) gives \( \ce{NH4Cl} \) and \( \ce{N2} \): \( \ce{8NH3 + 3Cl2 -> 6NH4Cl + N2} \). In N2 the bond order is 3 (one sigma and two pi bonds).

Example 2. Compound A hydrolyses to give HF and HIO3. Identify A and count the lone pairs on all atoms in A.
Answer:
A is \( \ce{IF5} \) (\( \ce{IF5 + 3H2O -> HIO3 + 5HF} \)). Iodine has 7 valence electrons, five used in bonds, leaving one lone pair; each fluorine has 3 lone pairs, so 1 + 15 = 16 lone pairs in all.

Example 3. Find the number of lone pairs on the central atom in \( \ce{XeF2} \), \( \ce{XeF4} \) and \( \ce{XeF6} \).
Answer:
Xenon has 8 valence electrons and each Xe-F bond uses one of them. So the lone pairs are (8 - 2)/2 = 3, (8 - 4)/2 = 2 and (8 - 6)/2 = 1 respectively.