Chapter 7: Elements of Groups 16, 17 and 18

For most of the nineteenth century, one element defeated every chemist who went after it. Fluorine could be located easily enough in its compounds — in the mineral fluorspar, in the enamel of teeth — but no one could pull the free element out of them. The attempts were dangerous. Several chemists were badly poisoned by the hydrogen fluoride they were handling, and some died. Later writers called them the martyrs of fluorine.

The difficulty was chemical, not technical. The moment fluorine was set free, it attacked whatever was holding it — glass, metal, water, the apparatus itself.

In 1886, Henri Moissan, working in Paris, found a way through. Instead of relying on a chemical reagent he used electrolysis: potassium hydrogen fluoride dissolved in anhydrous hydrogen fluoride, chilled to about −50 °C, in a platinum vessel closed with stoppers cut from fluorspar. A pale yellow gas collected at one electrode, and it set alight almost everything brought near it. Two decades later, Moissan received the Nobel Prize for this work.

Fluorine belongs to group 17. One column further along, in group 18, sit elements that for more than sixty years afterwards refused to form even a single compound. Same block of the periodic table, opposite extremes of behaviour — and the whole difference lies in how many electrons short of a filled shell each atom is. Fig. 7.0 places the three families of this chapter side by side, from the most electron-hungry to the completely satisfied.

Diagram

Fig. 7.0 – Three vertical strips of the p-block placed side by side, labelled Group 16 (ns²np⁴, “two short”), Group 17 (ns²np⁵, “one short”) and Group 18 (ns²np⁶, “complete”), with a left-to-right arrow beneath reading “reactivity falls as the octet fills”; show O, S, Se, Te, Po / F, Cl, Br, I, At / He, Ne, Ar, Kr, Xe, Rn in the strips.

NoteCuriosity Corner

Q1. Fluorine is the most electronegative element in the periodic table, yet the F–F bond is weaker than the Cl–Cl bond. Why should the greediest atom hold on to its own partner so loosely? Q2. Oxygen and sulfur belong to the same family, but water is a drinkable liquid while hydrogen sulfide is a foul-smelling poisonous gas. Why does the first member of a group behave so unlike the rest? Q3. Sulfur combines with fluorine to give \(\mathrm{SF_6}\), but oxygen manages only \(\mathrm{OF_2}\). What stops oxygen from taking on six fluorine atoms? Q4. Noble gases were believed to be completely unreactive for over sixty years. Why did xenon, and not helium or neon, finally break that record? Q5. Oxygen and ozone are built from the same kind of atom, yet one sustains life at ground level and the other protects it from far above — while ozone near the ground is treated as a pollutant. How can one element behave in two such different ways?

Keep these questions in mind as you read; the chapter will give you everything you need to reason out each answer.

7.1 Introduction

How does the valence shell electronic configuration of the elements vary in the p-block of the periodic table? Name the first element of groups 16, 17 and 18.

In Std. XI you learnt that the elements of the periodic table are sorted into blocks according to which orbital receives the last electron added to the atom. This last-filling electron is called the differentiating electron, because it is what sets an element apart from the element immediately before it.

NoteDefinition

The differentiating electron is the last electron that enters the atom of an element as we move from one element to the next in the periodic table; the type of orbital it occupies decides the block to which the element belongs.

In the p-block elements, the differentiating electron enters a p-orbital of the outermost shell. Every shell from the second one onwards contains three p orbitals, and each orbital can hold at most two electrons. A p-subshell can therefore accommodate a maximum of six electrons.

That number six is exactly why the p-block is six columns wide. As the p-subshell is filled one electron at a time, it gives rise to six groups, numbered 13 to 18 in the modern periodic table.

Q. If the p-block is six columns wide, why is the d-block ten columns wide?

A. Because a d-subshell has five orbitals, and five orbitals holding two electrons each can take ten electrons in all. The width of a block is simply the number of electrons its subshell can hold.

In this chapter we shall study the properties of the elements of groups 16, 17 and 18 — the last three of these six columns. Their outermost p-subshells are, in order, two electrons short of full, one electron short of full, and exactly full. Keep that simple difference in mind, because it will explain almost everything that follows.

The electronic configuration of an element is \(\mathrm{[Ne]\,3s^2 3p^5}\). Identify the block and the group to which it belongs.

The last electron enters a 3p orbital, so the differentiating electron is in a p-subshell and the element belongs to the p-block.

The outermost shell is the third shell (\(n = 3\)), so the element lies in period 3.

The valence shell configuration is \(ns^2np^5\), that is, five electrons in the p-subshell. Counting across the p-block, \(np^1\) is group 13, so \(np^5\) is the fifth column of the block — group 17.

The element is therefore a p-block element of period 3, group 17 (chlorine). Identifying the block and group straight from a configuration like this is a standard one-mark question in board and entrance papers, so practise reading the valence shell before anything else.

The differentiating electron of an element enters the 4p subshell, and its valence shell configuration is \(4s^2 4p^4\). State the block, period and group of the element, and name it.

7.2 Occurrence

Group 16 — The Oxygen Family

The elements oxygen (\(_{8}\mathrm{O}\)), sulfur (\(_{16}\mathrm{S}\)), selenium (\(_{34}\mathrm{Se}\)), tellurium (\(_{52}\mathrm{Te}\)) and polonium (\(_{84}\mathrm{Po}\)) together constitute Group 16. The group is called the oxygen family, after its first member.

A large number of metal ores are oxides or sulfides, that is, compounds in which a metal is joined to oxygen or to sulfur. Because these elements are the ones that build ores, the group 16 elements are also called chalcogens, or ore-forming elements.

Oxygen is the most abundant of all the elements on earth. It forms 20.95 % by volume of air and about 46.6 % by mass of the earth’s crust, so it is plentiful both in the free state and in combination.

Sulfur, in contrast, forms only 0.034 % by mass of the earth’s crust. It occurs mainly in combined forms — as sulfates such as gypsum (\(\mathrm{CaSO_4{\cdot}2H_2O}\)), epsom salt (\(\mathrm{MgSO_4{\cdot}7H_2O}\)) and baryte (\(\mathrm{BaSO_4}\)).

Sulfur also occurs widely as sulfides, among them galena (\(\mathrm{PbS}\)), zinc blende (\(\mathrm{ZnS}\)) and copper pyrites (\(\mathrm{CuFeS_2}\)). These three are the principal ores from which lead, zinc and copper are obtained, which is a good illustration of why the family earned the name chalcogen.

Selenium and tellurium are found in the same company. They occur as metal selenides and tellurides present within sulfide ores, and are recovered as by-products when those ores are processed.

Polonium is radioactive and has no ore of its own. It is a decay product of thorium and uranium, and is found only in traces wherever those two elements occur.

Group 17 — The Halogens

Fluorine (\(_{9}\mathrm{F}\)), chlorine (\(_{17}\mathrm{Cl}\)), bromine (\(_{35}\mathrm{Br}\)), iodine (\(_{53}\mathrm{I}\)) and astatine (\(_{85}\mathrm{At}\)) constitute Group 17. They are known collectively as the halogens, that is, salt-producing elements.

The name halogen is built from two Greek roots — halo, meaning salt, and gene, meaning born. The elements were named this way because they combine with metals to give salts, the most familiar of them being common salt, sodium chloride.

Halogens have very high electronegativities, so each atom pulls electrons strongly towards itself and reacts readily with almost anything that will part with an electron. As a direct result, the halogens are not found in the free state in nature; they occur only in the form of compounds.

Fluorine occurs mainly as insoluble fluorides: fluorspar (\(\mathrm{CaF_2}\)), cryolite (\(\mathrm{Na_3AlF_6}\)) and fluorapatite (\(\mathrm{3Ca_3(PO_4)_2{\cdot}CaF_2}\)). Cryolite is industrially valuable because molten cryolite is used as the medium in which alumina is dissolved during the electrolytic extraction of aluminium.

Small quantities of fluorine are also present in soil, in fresh water plants, and in the bones and teeth of animals — the same fluoride that hardens tooth enamel.

Sea water contains chlorides, bromides and iodides of Na, K, Mg and Ca. However, it mainly contains sodium chloride, \(\mathrm{NaCl}\), at about 2.5 % by mass.

Where old sea beds have dried up, the deposits left behind contain sodium chloride together with carnallite, \(\mathrm{KCl{\cdot}MgCl_2{\cdot}6H_2O}\).

Marine life also carries iodine within its systems. Sea weed, for example, contains up to 0.5 % iodine, and chile saltpetre contains up to 0.2 % of sodium iodate.

The iodine that sea weed concentrates is the same iodine your body needs to make thyroid hormones. Soils far from the coast are often poor in it, and a long-term shortage causes the thyroid gland to swell — the condition called goitre. Adding a small amount of potassium iodate or potassium iodide to common salt corrects the deficiency cheaply, which is why iodised salt is sold across India.

Astatine, the last member of the halogen family, is radioactive and has a half life of 8.1 hours. A half life that short means half of any sample decays away in a little over eight hours, so astatine can never be collected and stored in weighable amounts.

Group 18 — The Noble Gases

The elements helium (\(_{2}\mathrm{He}\)), neon (\(_{10}\mathrm{Ne}\)), argon (\(_{18}\mathrm{Ar}\)), krypton (\(_{36}\mathrm{Kr}\)), xenon (\(_{54}\mathrm{Xe}\)) and radon (\(_{86}\mathrm{Rn}\)) constitute Group 18.

All the noble gases except radon occur in the atmosphere. Their abundance in dry air is about 1 % by volume, with argon as the major constituent of that one per cent.

The main commercial source of helium is not air but natural gas, from which it is separated. Helium and neon are also found in minerals of radioactive origin, for example pitchblende, monazite and cleveite, where they have accumulated as the products of radioactive decay.

Xenon and radon are the rarest elements of the group. Radon is a decay product of \(^{226}\mathrm{Ra}\).

Q. Why are the noble gases found free in the atmosphere while the halogens are found only as compounds?

A. A noble gas atom already has a completely filled outermost p-subshell, so it gains nothing by bonding and stays as separate atoms in the air. A halogen atom is one electron short of that arrangement, so it reacts at the first opportunity and ends up locked into a compound.

Helium was identified in the sun before it was found on the earth. During a solar eclipse in 1868, astronomers observed a yellow line in the sun’s spectrum that matched no known element, and the new element was named after helios, the Greek word for sun. About twenty-five years later, Lord Rayleigh noticed that nitrogen separated from air was consistently a little denser than nitrogen prepared from chemical compounds. Working with William Ramsay, he traced the difference to an unreactive gas mixed in with atmospheric nitrogen, and named it argon, from the Greek for idle. Ramsay went on to obtain helium on earth from the mineral cleveite, and within a few years the whole of group 18 had been added to the periodic table.

Occurrence of group 16 vs group 17 vs group 18

Feature Group 16 (chalcogens) Group 17 (halogens) Group 18 (noble gases)
Found free in nature? Oxygen yes (20.95 % by volume of air); others chiefly combined No member occurs free Yes, as separate atoms in air
Reason Oxygen is abundant enough to remain in air; heavier members bind to metals Very high electronegativity makes them react at once Completely filled valence shell, so no tendency to combine
Chief natural sources Air and crust (O); sulfates and sulfides (S); sulfide ores (Se, Te); decay of Th and U (Po) Fluorides (F); sea water and dried sea beds (Cl, Br, I); marine life (I) Atmosphere (~1 % by volume of dry air, mainly Ar); natural gas (He); radioactive minerals (He, Ne)
Radioactive member Polonium Astatine (\(t_{1/2}\) = 8.1 hours) Radon (decay product of \(^{226}\mathrm{Ra}\))

Gypsum has the formula \(\mathrm{CaSO_4{\cdot}2H_2O}\). Calculate the percentage of water of crystallisation in it. (Molar masses in \(\text{g mol}^{-1}\): Ca = 40.08, S = 32.06, O = 16.00, H = 1.008)

First find the molar mass of the anhydrous part, \(\mathrm{CaSO_4}\):

\[40.08 + 32.06 + (4 \times 16.00) = 136.14\ \text{g mol}^{-1}\]

Next, the mass of the two water molecules. One water molecule has molar mass \((2 \times 1.008) + 16.00 = 18.02\ \text{g mol}^{-1}\), so

\[2 \times 18.02 = 36.04\ \text{g mol}^{-1}\]

The molar mass of gypsum is therefore

\[136.14 + 36.04 = 172.18\ \text{g mol}^{-1}\]

The percentage of water of crystallisation is the mass of water divided by the total mass, expressed as a percentage:

\[\%\ \mathrm{H_2O} = \frac{36.04}{172.18} \times 100 = 20.9\ \%\]

So roughly one-fifth of the mass of gypsum is water held inside the crystal. This is why gypsum loses mass on heating — the water of crystallisation is driven off, leaving plaster of Paris behind.

Sea water contains 2.5 % sodium chloride by mass. Calculate (i) the mass of \(\mathrm{NaCl}\) present in 1.00 kg of sea water and (ii) the mass of chlorine in that quantity of \(\mathrm{NaCl}\). (Molar masses in \(\text{g mol}^{-1}\): Na = 23.0, Cl = 35.5)

7.3 Electronic Configuration of Elements of Group 16, 17 and 18

Within each of these three groups every element carries the same kind of outermost arrangement, and only the shell number changes as we move down. The general electronic configuration of the group 16 elements is \(\mathrm{ns^2np^4}\), that of the group 17 elements is \(\mathrm{ns^2np^5}\), and the group 18 elements are shown by the \(\mathrm{ns^2np^6}\) configuration, where \(n\) is the number of the outermost shell.

Read the three together and the pattern is plain: moving from group 16 to group 17 to group 18 adds one p-electron at each step, until the p-subshell is completely filled.

Compare these with the nearest noble gas and a second pattern appears. The elements of groups 16 and 17 respectively have two and one electrons less than the stable electronic configuration of the nearest noble gas, while the group 18 elements already possess it.

That shortfall is the most useful single fact about these families. It decides how many electrons an atom must gain or share before it is satisfied, and therefore what charge its ion carries and how many covalent bonds it forms.

Reading a Condensed Configuration

Writing out a full configuration for a heavy element such as polonium would take a long line of symbols. In the condensed form, the symbol of the preceding noble gas is written in square brackets to stand for all the inner electrons, and only the electrons added after it are written out.

Sulfur, for instance, has 16 electrons; the first ten of them match neon exactly, so sulfur is written as \(\mathrm{[Ne]\,3s^2 3p^4}\) instead of \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^4}\). The part inside the brackets is the inert core; the part outside it is the valence shell that does the chemistry. Table 7.1 shows the condensed electronic configuration of the elements of group 16, 17 and 18.

Table 7.1 : Condensed electronic configuration of elements of group 16, 17 and 18

Group 16 (Oxygen family) Condensed Electronic Configuration Group 17 (Halogen family) Condensed Electronic Configuration Group 18 (Noble gases) Condensed Electronic Configuration
— — — — \(_{2}\mathrm{He}\) \(\mathrm{1s^2}\)
\(_{8}\mathrm{O}\) \(\mathrm{[He]\,2s^2 2p^4}\) \(_{9}\mathrm{F}\) \(\mathrm{[He]\,2s^2 2p^5}\) \(_{10}\mathrm{Ne}\) \(\mathrm{[He]\,2s^2 2p^6}\)
\(_{16}\mathrm{S}\) \(\mathrm{[Ne]\,3s^2 3p^4}\) \(_{17}\mathrm{Cl}\) \(\mathrm{[Ne]\,3s^2 3p^5}\) \(_{18}\mathrm{Ar}\) \(\mathrm{[Ne]\,3s^2 3p^6}\)
\(_{34}\mathrm{Se}\) \(\mathrm{[Ar]\,3d^{10}4s^2 4p^4}\) \(_{35}\mathrm{Br}\) \(\mathrm{[Ar]\,3d^{10}4s^2 4p^5}\) \(_{36}\mathrm{Kr}\) \(\mathrm{[Ar]\,3d^{10}4s^2 4p^6}\)
\(_{52}\mathrm{Te}\) \(\mathrm{[Kr]\,4d^{10}5s^2 5p^4}\) \(_{53}\mathrm{I}\) \(\mathrm{[Kr]\,4d^{10}5s^2 5p^5}\) \(_{54}\mathrm{Xe}\) \(\mathrm{[Kr]\,4d^{10}5s^2 5p^6}\)
\(_{84}\mathrm{Po}\) \(\mathrm{[Xe]\,4f^{14}5d^{10}6s^2 6p^4}\) \(_{85}\mathrm{At}\) \(\mathrm{[Xe]\,4f^{14}5d^{10}6s^2 6p^5}\) \(_{86}\mathrm{Rn}\) \(\mathrm{[Xe]\,4f^{14}5d^{10}6s^2 6p^6}\)

Notice in the table that from selenium onwards a completely filled \(d\) subshell — and, in the last row, a filled \(f\) subshell as well — sits inside the valence shell. These inner subshells are full and take no part in ordinary bonding, but they shield the outer electrons from the nucleus rather poorly, and that will matter when we come to atomic size and ionisation enthalpy in the next section.

Q. Helium has the configuration \(\mathrm{1s^2}\) and no p electrons at all. Why is it placed in group 18 with the other \(\mathrm{ns^2np^6}\) elements?

A. Because what matters is a completely filled outermost shell, not the presence of p electrons. The first shell has only one orbital, the 1s, and two electrons fill it completely. Helium is therefore just as satisfied as neon or argon, and behaves the same way chemically.

Group 16 vs Group 17 vs Group 18 — the valence shell picture

Feature Group 16 Group 17 Group 18
General valence configuration \(\mathrm{ns^2np^4}\) \(\mathrm{ns^2np^5}\) \(\mathrm{ns^2np^6}\)
Valence electrons 6 7 8 (2 in helium)
Electrons short of the nearest noble gas 2 1 none
Simple negative ion normally formed \(\mathrm{E^{2-}}\) \(\mathrm{E^{-}}\) none
Covalent bonds normally formed 2 1 normally none

Write the condensed electronic configurations of the sulfide ion, \(\mathrm{S^{2-}}\), and the chloride ion, \(\mathrm{Cl^{-}}\). Which noble gas is each of them isoelectronic with?

A negative ion is formed by adding electrons to the neutral atom, and the added electrons enter the outermost p subshell.

Sulfur is \(\mathrm{[Ne]\,3s^2 3p^4}\). Adding two electrons gives

\[\mathrm{S^{2-}} : \mathrm{[Ne]\,3s^2 3p^6}\]

Chlorine is \(\mathrm{[Ne]\,3s^2 3p^5}\). Adding one electron gives

\[\mathrm{Cl^{-}} : \mathrm{[Ne]\,3s^2 3p^6}\]

Argon is \(\mathrm{[Ne]\,3s^2 3p^6}\), so both ions have exactly the electron arrangement of argon. Two species with the same number of electrons arranged in the same way are said to be isoelectronic, and both \(\mathrm{S^{2-}}\) and \(\mathrm{Cl^{-}}\) are isoelectronic with argon.

This is simply the shortfall of Table 7.1 being made good — sulfur needed two electrons and took two, chlorine needed one and took one.

Write the condensed electronic configuration of the telluride ion, \(\mathrm{Te^{2-}}\), and name the noble gas with which it is isoelectronic. State also how many electrons an iodine atom must gain to reach the same arrangement.

7.4 Atomic and Physical Properties of Elements of Group 16, 17 and 18

7.4.1 Atomic Properties of Group 16, 17 and 18 Elements

These properties are given in Tables 7.2, 7.3 and 7.4. Before reading the trends, look at the numbers in the tables themselves — every statement that follows is a summary of what those columns actually show.

i. Atomic and Ionic Radii

In groups 16, 17 and 18, atomic and ionic radii increase down the group. The reason is that each successive element has one more quantum shell than the element above it, so the outermost electrons sit further from the nucleus.

The nuclear charge also rises as we go down a group, and that pulls electrons inwards. But the addition of a whole new shell, together with the screening provided by all the inner electrons, outweighs it. The net result is a steady increase in size.

Across a period, atomic or ionic radii decrease with increasing atomic number. Here no new shell is added; electrons keep entering the same shell while protons keep being added to the nucleus. The effective nuclear charge (\(Z_{\text{eff}}\)) — the net pull actually felt by an outer electron after screening by the inner electrons — therefore increases, and the electron cloud is drawn in tighter.

Group 17 elements (halogens) have the smallest atomic radii in their respective periods, since they experience the highest effective nuclear charge among the elements that still belong to the same shell.

Q. If nuclear charge increases both across a period and down a group, why does size fall in one case and rise in the other?

A. Across a period the added electrons go into the same shell, so screening barely changes and the stronger nuclear pull wins. Down a group a new shell is added and the inner electrons screen the nucleus much more effectively, so the increase in shell number wins.

atomic and ionic radius (across a period →: decreases, because electrons enter the same shell while \(Z_{\text{eff}}\) rises and pulls the cloud inward; down a group ↓: increases, because a new quantum shell is added at every step and inner-shell screening offsets the higher nuclear charge)

ii. Ionisation Enthalpy

The group 16, 17 and 18 elements have high ionisation enthalpy. Ionisation enthalpy is the energy needed to remove the most loosely held electron from an isolated gaseous atom, and these elements hold their electrons tightly because their nuclear pull on a small valence shell is strong.

The ionisation enthalpy decreases down the group due to increase in the atomic size. As the valence electron moves further from the nucleus, less energy is needed to pull it away.

Across a period ionisation enthalpy increases with increase of atomic number. This is due to addition of electrons in the same shell, which leaves the electron more and more firmly held as the nuclear charge builds up.

However, the elements of group 16 have lower ionisation enthalpy values compared to those of group 15 in the corresponding periods, owing to the extra stable half filled electronic configuration of p-orbitals in elements of group 15.

first ionisation enthalpy (across a period →: increases, because \(Z_{\text{eff}}\) rises while the shell stays the same; down a group ↓: decreases, because the valence electron lies in a larger, better-screened shell. Local dip: group 16 falls below group 15 in the same period because group 15 has the extra-stable half-filled \(ns^2np^3\) arrangement)

Elements of group 16 generally show lower values of first ionisation enthalpy compared to the elements of the corresponding period of group 15. Why?

Group 15 elements have extra stable, half filled p-orbitals with electronic configuration \(ns^2np^3\). Therefore a greater amount of energy is required to remove an electron from them, compared to the partially filled orbitals \(ns^2np^4\) of the group 16 elements of the corresponding period.

In short, breaking into a half-filled set costs extra energy; removing the one paired-up electron of an \(np^4\) arrangement actually relieves some electron–electron repulsion, so it comes off more easily.

Table 7.2 : Atomic and physical properties of group 16 elements

Element | Atomic number | Atomic mass g/mol | Atomic radius (pm) | Ionic radius E\(^{2-}\) (pm) | Ionization enthalpy (\(\Delta_i H_1\)) kJ/mol | Electronegativity | Electron gain enthalpy kJ/mol | Density g/cm\(^3\) | M.P. (K) | B.P. (K) |

|—|—|—|—|—|—|—|—|—|—|—|

O | 8 | 16.00 | 66 | 140 | 1314 | 3.50 | -141 | 1.32 | 55 | 90 |
S | 16 | 32.06 | 104 | 184 | 1000 | 2.44 | -200 | 2.06 | 393 | 718 |
Se | 34 | 78.96 | 117 | 198 | 941 | 2.48 | -195 | 4.19 | 490 | 958 |
Te | 52 | 127.60 | 137 | 221 | 869 | 2.01 | -190 | 6.25 | 725 | 1260 |
Po | 84 | 210.00 | 146 | 230 | 813 | 1.76 | -174 | - | 520 | 1235 |

Table 7.3 : Atomic and physical properties of group 17 elements

Element | Atomic number | Atomic mass g/mol | Atomic radius (pm) | Ionic radius E\(^{-}\) (pm) | Ionization enthalpy (\(\Delta_i H_1\)) kJ/mol | Electronegativity | Electron gain enthalpy kJ/mol | Density g/cm\(^3\) | M.P. (K) | B.P. (K) |

|—|—|—|—|—|—|—|—|—|—|—|

F | 9 | 19.00 | 64 | 133 | 1680 | 4.0 | -333 | 1.5 | 54.4 | 84.9 |
Cl | 17 | 35.45 | 99 | 184 | 1256 | 3.2 | -349 | 1.66 | 172.0 | 239.0 |
Br | 35 | 79.90 | 114 | 196 | 1142 | 3.0 | -325 | 3.19 | 265.8 | 332.5 |
I | 53 | 126.90 | 133 | 220 | 1008 | 2.7 | -296 | 4.94 | 386.6 | 458.2 |
At | 85 | 210 | - | - | - | 2.2 | - | - | - | - |

Table 7.4 : Atomic and physical properties of group 18 elements

Element | Atomic number | Atomic mass g/mol | Atomic radius (pm) | Ionization enthalpy (\(\Delta_i H_1\)) kJ/mol | Electron gain enthalpy kJ/mol | Density g/cm\(^3\) | M.P. (K) | B.P. (K) | Atmospheric content (% by volume) |

|—|—|—|—|—|—|—|—|—|—|

He | 2 | 4.00 | 120 | 2372 | 48 | 1.8 × 10\(^{-4}\) | - | 4.2 | 5.24 × 10\(^{-4}\) |
Ne | 10 | 20.18 | 160 | 2080 | 116 | 9.0 × 10\(^{-4}\) | 24.6 | 27.1 | 1.82 × 10\(^{-3}\) |
Ar | 18 | 39.95 | 190 | 1520 | 96 | 1.8 × 10\(^{-3}\) | 83.8 | 87.2 | 0.934 |
Kr | 36 | 83.80 | 200 | 1351 | 96 | 3.7 × 10\(^{-3}\) | 115.8 | 119.7 | 1.14 × 10\(^{-4}\) |
Xe | 54 | 131.30 | 220 | 1170 | 77 | 5.9 × 10\(^{-3}\) | 161.3 | 165.0 | 8.7 × 10\(^{-6}\) |
Rn | 86 | 222.00 | - | 1037 | 68 | 9.7 × 10\(^{-3}\) | 202 | 211 | - |

Compare the ionisation enthalpy columns of the three tables and one point stands out: for any period, the noble gas has the highest value of all. Helium, at 2372 kJ/mol, is the hardest atom in the periodic table to ionise.

The same numbers are easier to judge as a picture than as a column, so the first ionisation enthalpies of the halogens are plotted in Fig. 7.2.

Diagram

Fig. 7.2 – Line graph of first ionisation enthalpy (kJ/mol, y-axis 900–1800) against atomic number for F(9,1680), Cl(17,1256), Br(35,1142), I(53,1008); label each point with the element symbol and mark the y-axis “ΔiH₁ / kJ mol⁻¹”.

first ionisation enthalpy of the halogens, Fig. 7.2 (i. Does the curve fall steadily, or is the drop from F to Cl much larger than the later drops? ii. What does the size of that first drop tell you about the difference in atomic radius between F and Cl compared with the difference between Br and I? iii. Predict roughly where astatine would lie on this graph, and give your reasoning.)

Observe Table no. 7.3 and explain the trend in the following atomic properties of group 17 elements: i. atomic size, ii. ionisation enthalpy, iii. electronegativity, iv. electron gain enthalpy. Also: oxygen has less negative electron gain enthalpy than sulfur. Why?

The values of first ionisation enthalpy of S and Cl are 1000 and 1256 kJ mol\(^{-1}\), respectively. Explain the observed trend.

The elements S and Cl belong to the same period of the periodic table.

Across a period, effective nuclear charge increases and atomic size decreases with increase in atomic number.

Therefore the energy required for the removal of an electron from the valence shell (I.E.) increases in the order S < Cl.

iii. Electronegativity

In a group (16, 17 and 18) the electronegativity decreases down the group. Electronegativity measures how strongly an atom attracts a shared pair of electrons towards itself, and a larger atom, with its bonding pair further from the nucleus, attracts it less.

Oxygen has the highest electronegativity next to fluorine amongst all the elements.

Halogens have very high electronegativity. Fluorine is the most electronegative element in the periodic table.

Two members of group 17 were named for what the senses pick up first. Iodine was named from the Greek word for violet, after the colour of its vapour, and bromine from bromos, the Greek word for stench, after its sharp and unpleasant smell. The names are a reminder that these physical properties were the earliest handle chemists had on the family.

iv. Electron Gain Enthalpy

In groups 16 and 17, electron gain enthalpy becomes less negative down the group. Electron gain enthalpy is the enthalpy change when a gaseous atom accepts an electron; the more negative it is, the more readily the atom takes that electron.

electron gain enthalpy, \(\Delta_{eg}H\) (SI unit: kJ mol\(^{-1}\). Sign convention — a negative value means energy is released when the atom accepts an electron, so a more negative value indicates a stronger tendency to gain an electron. A positive value means energy must be supplied, that is, the atom resists gaining an electron. Note that “less negative” is not the same as “positive”: \(-296\) is less negative than \(-333\), but both still describe energy released.)

However, in group 16, oxygen has less negative electron gain enthalpy than sulfur due to its small atomic size. The incoming electron is forced into a compact 2p subshell that is already crowded, and the repulsion it meets there cancels part of the energy that would otherwise be released.

In group 17, fluorine has less negative electron gain enthalpy than that of chlorine. This is due to the small size of the fluorine atom, for exactly the same reason.

Group 18 elements (noble gases) have no tendency to accept electrons because of their stable electronic configuration \(ns^2np^6\), and thus have large positive electron gain enthalpy. An extra electron would have to start a new shell altogether, which costs energy rather than releasing it.

electronegativity and electron gain enthalpy (down a group ↓: electronegativity decreases because the bonding pair sits further from the nucleus; electron gain enthalpy becomes less negative for the same reason. Exceptions at the top: O is less negative than S, and F is less negative than Cl, because the small 2p subshell repels the incoming electron. Group 18 sits outside the pattern with large positive values, since its shell is already full)

The first ionisation enthalpies of nitrogen and oxygen are 1402 and 1314 kJ mol\(^{-1}\) respectively. Nitrogen comes before oxygen in the same period, yet its value is higher. Explain this observation, and name one other pair of neighbouring elements in the periodic table that behaves the same way.

7.4.2 Physical Properties of Group 16, 17 and 18 Elements

a. Group 16 Elements (Oxygen Family or Chalcogens)

Oxygen is a gas while the other elements are solids at room temperature.

Oxygen and sulfur are nonmetals, selenium and tellurium are metalloids, while polonium is a metal. Metallic character therefore increases down the group, which is the usual pattern in the p-block: the larger, less electronegative atoms lower down hold their electrons loosely enough to behave like metals. Polonium is radioactive, with a half life of 13.8 days.

Melting and boiling points increase with increasing atomic number.

All the elements of group 16 exhibit allotropy, a property taken up in detail in section 7.7.

Why is there a large difference between the melting and boiling points of oxygen and sulfur?

Oxygen exists as a diatomic molecule (\(\mathrm{O_2}\)), whereas sulfur exists as a polyatomic molecule (\(\mathrm{S_8}\)).

The van der Waals forces of attraction between \(\mathrm{O_2}\) molecules are relatively weak, owing to its much smaller size.

The large van der Waals attractive forces in the \(\mathrm{S_8}\) molecules can be noticed because of the large molecular size. Therefore oxygen has low m.p. and b.p. as compared to sulfur.

Melting and boiling are about separating molecules from one another, not about breaking bonds inside them — which is why molecular size, and the strength of the forces between molecules, decides these values.

b. Group 17 Elements (Halogen Family)

Fluorine and chlorine are gases, bromine is a liquid and iodine is a solid at room temperature. \(\mathrm{F_2}\) is yellow, \(\mathrm{Cl_2}\) greenish yellow, \(\mathrm{Br_2}\) red and \(\mathrm{I_2}\) is violet, in colour.

Fluorine and chlorine react with water. Bromine and iodine are only sparingly soluble in water and are soluble in various organic solvents such as chloroform, carbon disulfide, carbon tetrachloride and hydrocarbons, which give coloured solutions.

Bond dissociation enthalpies of halogen molecules follow the order:

\[\mathrm{Cl-Cl} > \mathrm{Br-Br} > \mathrm{F-F} > \mathrm{I-I}\]

Read that order carefully. It is not a smooth trend down the group — fluorine, the smallest and most electronegative halogen, sits out of place near the bottom. The reason is worked out in Solved Example 7.8.

Bromine is the only nonmetallic element that is a liquid at room temperature, and it is one of the more hazardous reagents in a school laboratory. Its dense red-brown vapour is corrosive to the eyes and the respiratory tract, and liquid bromine causes deep, slow-healing burns on skin. It is therefore handled only in a fume cupboard, with gloves, and stored in tightly stoppered dark bottles. Iodine is gentler but still stains skin and sublimes readily, so it too is kept closed — the violet vapour rising from a warm iodine crystal is a sign that the sample is escaping.

Fluorine has less negative electron gain affinity than chlorine. Why?

The size of the fluorine atom is smaller than that of the chlorine atom. As a result, there are strong inter-electronic repulsions in the relatively small 2p orbitals of fluorine, and therefore the incoming electron does not experience much attraction. Thus fluorine has less negative electron gain affinity than chlorine.

Bond dissociation enthalpy of \(\mathrm{F_2}\) (158.8 kJ mol\(^{-1}\)) is lower than that of \(\mathrm{Cl_2}\) (242.6 kJ mol\(^{-1}\)). Why?

Fluorine has a smaller atomic size than chlorine. The lone pairs on each F atom in the \(\mathrm{F_2}\) molecule are so close together that they strongly repel each other, and make the F–F bond weak, as shown in Fig. 7.1.

Diagram

Fig. 7.1 – Two fluorine atoms bonded as F–F, each carrying three lone pairs drawn as shaded lobes; curved double-headed arrows between the facing lone pairs of the two atoms, labelled “lone pair – lone pair repulsion”.

Thus it requires a smaller amount of energy to break the F–F bond.

In the \(\mathrm{Cl_2}\) molecule the lone pairs on each Cl atom are at a larger distance and the repulsion is negligible. Thus the Cl–Cl bond is comparatively stronger. Therefore the bond dissociation enthalpy of \(\mathrm{F_2}\) is lower than that of \(\mathrm{Cl_2}\).

Q1. Fluorine is the most electronegative element in the periodic table, yet the F–F bond is weaker than the Cl–Cl bond. Why should the greediest atom hold on to its own partner so loosely?

A. Electronegativity describes how strongly an atom pulls on electrons it shares with a different atom. In \(\mathrm{F_2}\) both atoms are identical, so that pull cancels out and what remains is the geometry. Each fluorine atom carries three lone pairs, and because the atom is so small, those lone pairs are crowded very close to the lone pairs on the other atom. Their mutual repulsion works against the bond and weakens it. Chlorine is larger, its lone pairs are further apart, and the repulsion is negligible — so Cl–Cl survives with a bond dissociation enthalpy of 242.6 kJ mol\(^{-1}\) against only 158.8 kJ mol\(^{-1}\) for F–F. Small size, which makes fluorine so reactive towards other elements, is exactly what makes it hold its own twin loosely.

Arrange \(\mathrm{F_2}\), \(\mathrm{Cl_2}\), \(\mathrm{Br_2}\) and \(\mathrm{I_2}\) in decreasing order of bond dissociation enthalpy. Justify the position of \(\mathrm{F_2}\) in your order, and state why \(\mathrm{I_2}\) occupies the last place for a completely different reason.

c. Group 18 Elements (Noble Gases)

Noble gases are monoatomic. They exist as single atoms, not as \(\mathrm{X_2}\) molecules, because each atom already has a complete valence shell and gains nothing by pairing up.

They are sparingly soluble in water.

Noble gases have very low melting and boiling points. Helium has the lowest boiling point (4.2 K) of any known substance.

Noble gases have very low melting and boiling points. Why?

Noble gases are monoatomic, and the only type of interatomic interactions which exist between them are van der Waals forces. Therefore they can be liquefied at very low temperatures and have very low melting or boiling points.

physical character of group 16 vs group 17 vs group 18

Feature Group 16 Group 17 Group 18
State at room temperature O gas; S, Se, Te, Po solids F, Cl gases; Br liquid; I solid All gases
Atomicity of the element Diatomic (\(\mathrm{O_2}\)) and polyatomic (\(\mathrm{S_8}\)) Diatomic (\(\mathrm{X_2}\)) Monoatomic
Metallic character O, S nonmetals; Se, Te metalloids; Po metal All nonmetals All nonmetals
Forces between particles van der Waals forces between molecules van der Waals forces between molecules van der Waals forces between atoms only
Melting and boiling points Increase with atomic number Increase down the group Very low; He has the lowest b.p. known (4.2 K)
Allotropy Shown by all members Not a feature of the group Not a feature of the group

The first member of a group usually differs in properties from the rest of the members of the group. Why?

7.5 Anomalous Behaviour

In every group of the p-block, the first member behaves differently from the rest of the family. It is much smaller, it holds electrons far more tightly, and — most importantly — its valence shell has no d-orbitals available. Oxygen in group 16 and fluorine in group 17 are the two clearest examples, and both are worth studying closely because examiners return to them again and again.

7.5.1 Anomalous Behaviour of Oxygen

Oxygen shows the following anomalous properties compared to the other members of group 16.

Atomicity

Oxygen is a diatomic molecule (\(\mathrm{O_2}\)) while the others are polyatomic molecules — for example \(\mathrm{P_4}\), \(\mathrm{S_8}\). Being small, two oxygen atoms can form a strong double bond between themselves; the larger atoms of sulfur and selenium overlap poorly in that way and prefer to link into rings and chains of single bonds instead.

Magnetic Property

Oxygen is paramagnetic while the others are diamagnetic. A paramagnetic substance contains unpaired electrons and is drawn into a magnetic field; a diamagnetic one has all its electrons paired and is weakly pushed out. The reason behind the paramagnetism of \(\mathrm{O_2}\) is taken up in section 7.9.

Oxidation State

Oxygen shows −2, −1 and +2 oxidation states, while the other elements of the group show −2, +2, +4 and +6 oxidation states.

Oxygen cannot exhibit a higher oxidation state due to the absence of vacant d orbitals. To reach +4 or +6, an atom has to unpair its valence electrons and place them in empty orbitals of the same shell; sulfur has empty 3d orbitals for this, but the second shell of oxygen has only 2s and 2p orbitals, and no 2d orbitals exist at all.

Q. What exactly does “expanding the octet” mean?

A. It means holding more than eight electrons in the valence shell. This is possible only when the shell contains empty d orbitals into which paired electrons can be promoted, so it begins from the third shell onwards. Elements of the second period — oxygen, fluorine, nitrogen, carbon — can never expand the octet.

Oxygen forms only \(\mathrm{OF_2}\) with fluorine while sulfur forms \(\mathrm{SF_6}\). Explain. Why?

Diagram

Fig. 7.4 – Two orbital box diagrams side by side: oxygen 2s²2p⁴ showing only two unpaired electrons and no 2d boxes available, and sulfur 3s²3p⁴ with an empty 3d set, arrows showing promotion of paired 3s and 3p electrons into 3d to give six unpaired electrons; label them “O: covalency 2” and “S: covalency up to 6”.

Q3. Sulfur combines with fluorine to give \(\mathrm{SF_6}\), but oxygen manages only \(\mathrm{OF_2}\). What stops oxygen from taking on six fluorine atoms?

A. To bond with six fluorine atoms, the central atom must supply six unpaired electrons. Sulfur can do this because its valence shell is the third shell, which contains empty 3d orbitals; two of its paired electrons are promoted into these orbitals, giving six unpaired electrons and a covalency of six, as shown in Fig. 7.4. Oxygen’s valence shell is the second shell, and there are no 2d orbitals in existence for it to use. Its two unpaired 2p electrons are all it has, so its covalency stops at two and \(\mathrm{OF_2}\) is as far as it can go. The limit is not a matter of size alone — it is the plain absence of d orbitals in the second shell.

Nature of Hydrides

The hydride of oxygen (\(\mathrm{H_2O}\)) is a liquid at room temperature while the hydrides of the other members of the group are gases. Oxygen is small and highly electronegative, so the O–H bond is strongly polar and each water molecule is held to its neighbours by hydrogen bonds. Breaking that network needs far more energy than separating \(\mathrm{H_2S}\) molecules, which are held only by weak van der Waals forces.

Which of the following possess hydrogen bonding: \(\mathrm{H_2S}\), \(\mathrm{H_2O}\), \(\mathrm{H_2Se}\), \(\mathrm{H_2Te}\)? Show hydrogen bonding in that molecule with the help of a diagram.

Diagram

Fig. 7.3 – Three water molecules drawn with the bent H–O–H shape, partial charges δ⁻ on O and δ⁺ on H marked, and dotted lines from the H of one molecule to the lone pair of O on the next labelled “hydrogen bond”.

Q2. Oxygen and sulfur belong to the same family, but water is a drinkable liquid while hydrogen sulfide is a foul-smelling poisonous gas. Why does the first member of a group behave so unlike the rest?

A. Both molecules have the same bent shape and the same \(\mathrm{H_2E}\) formula, but oxygen is by far the smaller and more electronegative atom. That makes the O–H bond strongly polar, and it lets each water molecule form hydrogen bonds with its neighbours, as drawn in Fig. 7.3. Sulfur is bigger and much less electronegative, so \(\mathrm{H_2S}\) molecules cannot hydrogen bond and are held together only by weak van der Waals forces. Water therefore stays liquid at room temperature while \(\mathrm{H_2S}\) boils away as a gas. The same smallness and high electronegativity of the first member — together with its lack of d orbitals — is what sets oxygen apart from the whole of the rest of group 16.

Covalency

The common covalency of oxygen is 2, and in rare cases it is four. But for the other members of group 16 the covalency can exceed four.

NoteDefinition

Covalency is the number of covalent bonds that an atom of an element forms in a compound, and it is limited by the number of unpaired electrons the atom can make available in its valence shell.

The anomalous behaviour of oxygen is due to the following reasons: small atomic size, high electronegativity, and absence of inner d-orbitals.

Assign the oxidation number of oxygen in (i) \(\mathrm{H_2O}\), (ii) \(\mathrm{H_2O_2}\) and (iii) \(\mathrm{OF_2}\).

The working rule is that the oxidation numbers of all atoms in a neutral molecule add up to zero, hydrogen is taken as +1 when joined to a nonmetal, and fluorine, being the most electronegative element of all, is always −1.

  1. In \(\mathrm{H_2O}\): let the oxidation number of O be \(x\). Then

\[2(+1) + x = 0 \quad \Rightarrow \quad x = -2\]

  1. In \(\mathrm{H_2O_2}\) there are two oxygen atoms, each with oxidation number \(x\):

\[2(+1) + 2x = 0 \quad \Rightarrow \quad x = -1\]

This is the peroxide value, and it arises because one of the bonds is an O–O bond, which contributes nothing to the oxidation number of either oxygen.

  1. In \(\mathrm{OF_2}\), fluorine is −1:

\[x + 2(-1) = 0 \quad \Rightarrow \quad x = +2\]

Oxygen is positive here only because it is bonded to fluorine, the one element more electronegative than itself. These three values, −2, −1 and +2, are exactly the oxidation states listed for oxygen above.

oxygen vs the other members of group 16

Property Oxygen S, Se, Te, Po
Atomicity of the element Diatomic, \(\mathrm{O_2}\) Polyatomic, e.g. \(\mathrm{S_8}\)
Magnetic nature Paramagnetic Diamagnetic
Oxidation states shown −2, −1, +2 −2, +2, +4, +6
State of the hydride \(\mathrm{H_2E}\) Liquid (\(\mathrm{H_2O}\)) Gases
Common covalency 2 (rarely 4) Can exceed 4
d orbitals in valence shell Absent Present

7.5.2 Anomalous Behaviour of Fluorine

Fluorine, the first member of group 17, differs in properties from the other members of the group. The anomalous behaviour of fluorine is due to the following reasons: its small atomic size, its high electronegativity, the absence of d-orbitals in its valence shell, and its low F–F bond dissociation enthalpy.

Some anomalous properties of fluorine are listed below.

Ionisation enthalpy, electronegativity and electrode potential are all higher for fluorine than the expected trends shown by the other halogens.

Ionic and covalent radii, melting point, boiling point and electron gain enthalpy are quite lower than expected — the values in Table 7.3 bear this out.

Most of the reactions of fluorine are exothermic. This is due to the short and strong bond formed by it with other elements: little energy is spent breaking the weak F–F bond, while a great deal is released when the strong bonds to the new partner are formed.

It forms only one oxoacid (HOF), while the other halogens form a number of oxoacids. Without d orbitals fluorine cannot expand its octet to hold several oxygen atoms, and being the most electronegative element it cannot take a positive oxidation state in the way chlorine and bromine do.

Hydrogen fluoride is a liquid (b.p. 293 K) due to strong hydrogen bonding, while the other hydrogen halides are gases — the same hydrogen bonding argument that explained liquid water above.

fluorine vs the other halogens

Property Fluorine Cl, Br, I
Oxidation states Only −1 −1 and positive states such as +1, +3, +5, +7
d orbitals in valence shell Absent Present
Number of oxoacids formed One only (HOF) Several
Bond dissociation enthalpy of \(\mathrm{X_2}\) Abnormally low (158.8 \(\text{kJ mol}^{-1}\)) Higher, then falling down the group
State of the hydride HX Liquid, b.p. 293 K Gases
Nature of its reactions Mostly exothermic Exothermic, but less markedly so

Which of the following species cannot exist, and why: \(\mathrm{OF_6}\), \(\mathrm{SF_6}\), \(\mathrm{TeF_6}\), \(\mathrm{OCl_4}\)? For each species you reject, name the property of the central atom that rules it out.

7.6 Chemical Properties of Elements of Groups 16, 17 and 18

Having seen how these atoms are built and how big they are, we can now ask what they actually do. The chemistry of all three families follows from one question: how far is the valence shell from being complete, and can that shell be expanded?

7.6.1 Oxidation State

i. The Group 16 Elements

The group 16 elements have the valence shell electronic configuration \(\mathrm{ns^2np^4}\). They attain a noble gas configuration either by gaining two electrons, forming \(\mathrm{E^{2-}}\) ions, or by sharing two electrons, forming two covalent bonds. These elements, thus, show −2 and +2 oxidation states in their compounds.

Oxygen, being highly electronegative, shows a common oxidation state of −2 except in two cases. In the case of \(\mathrm{OF_2}\) its oxidation state is +2, and in peroxides it shows the oxidation state −1 (\(\mathrm{H_2O_2}\), \(\mathrm{Na_2O_2}\)).

The other elements of the group exhibit +2, +4 and +6 oxidation states, with +4 and +6 being more common. The stability of the higher (+6) oxidation state decreases down the group while the stability of the lower (+4) oxidation state increases down the group, due to the inert pair effect.

NoteDefinition

The inert pair effect is the reluctance of the two electrons of the outermost s-orbital to take part in bonding, which grows stronger as we move down a group; the heavier elements therefore favour an oxidation state two units lower than the group would predict.

Bonding in the +4 and +6 oxidation states is primarily covalent. An atom cannot realistically lose four or six electrons, so these states are reached by sharing rather than by transfer.

Flowchart

FC 7.1 – Deciding the oxidation states available to an element of groups 16–18: start → count valence electrons from the group → is the element the first member of its group (no d orbitals in the valence shell)? → if yes: only the negative state and, for oxygen, the special cases OF₂ (+2) and peroxides (−1); if no: d orbitals available → octet can expand → positive states up to the group maximum (+6 for group 16, +7 for group 17, +8 for xenon) → check inert pair effect for the heaviest members → lower state (+4) becomes the more stable one.

Assign the oxidation number of the underlined element in (i) \(\mathrm{H_2\underline{S}}\), (ii) \(\mathrm{\underline{Se}O_2}\) and (iii) \(\mathrm{HO\underline{Cl}}\).

In each case the oxidation numbers of all the atoms in the neutral molecule must add up to zero. Hydrogen is +1 when bonded to a nonmetal, and oxygen is −2 except in peroxides and in compounds with fluorine.

  1. \(\mathrm{H_2S}\): \(\quad 2(+1) + x = 0 \quad \Rightarrow \quad x = -2\)

Sulfur takes the same −2 state that oxygen takes in water — the group 16 shortfall of two electrons, filled.

  1. \(\mathrm{SeO_2}\): \(\quad x + 2(-2) = 0 \quad \Rightarrow \quad x = +4\)

  2. \(\mathrm{HOCl}\): \(\quad (+1) + (-2) + x = 0 \quad \Rightarrow \quad x = +1\)

Chlorine is positive here because it is bonded to oxygen, which is more electronegative than itself.

Complete the following tables.

Element O O S F
Compound \(\mathrm{H_2O}\) \(\mathrm{OF_2}\) \(\mathrm{H_2S}\) \(\mathrm{HF}\)
Oxidation state −2 …… …… ……
Element Se Se Te Cl
Compound \(\mathrm{SeO_2}\) \(\mathrm{SeO_3}\) \(\mathrm{TeF_6}\) \(\mathrm{HOCl}\)
Oxidation state …… +6 …… ……
ii. The Group 17 Elements

The group 17 elements are represented by their valence shell electronic configuration \(\mathrm{ns^2np^5}\). They attain a noble gas configuration either by gaining one electron, forming \(\mathrm{E^{-}}\) ions, or by sharing one electron, forming one covalent bond.

All halogens exhibit the −1 oxidation state. However Cl, Br and I exhibit +1, +3, +5 and +7 oxidation states as well. This is because they are less electronegative than F and possess empty d-orbitals in the valence shell, and therefore can expand the octet. The oxidation states +4 and +6 occur in the oxides and oxoacids of Cl and Br.

The fluorine atom has no d-orbitals in its valence shell and therefore cannot expand its octet. Thus fluorine, being the most electronegative element, exhibits mostly the −1 oxidation state.

Q. Why do the positive oxidation states of the halogens go up in steps of two — +1, +3, +5, +7 — rather than one at a time?

A. Each positive state is reached by unpairing one more pair of valence electrons and promoting one of them to an empty d orbital. Every promotion frees two electrons for bonding at once, so the number of bonds, and hence the oxidation state, rises two units at a time.

iii. The Group 18 Elements

Group 18 elements (noble gases) have a stable valence shell electronic configuration \(\mathrm{ns^2np^6}\) with completely filled orbitals. Thus they have no tendency to gain or lose electrons, that is, they are zero valent and mostly exist as monoatomic gases.

However, xenon has a large atomic size and lower ionisation enthalpy compared to He, Ne, Ar and Kr. Hence xenon exhibits higher oxidation states.

Its outermost shell has d-orbitals. The paired electrons of the valence shell can be unpaired and promoted to the empty d-orbitals. The unpaired electrons are then shared with fluorine or oxygen atoms, and covalent compounds showing higher oxidation states — such as \(\mathrm{XeF_2}\) (+2), \(\mathrm{XeF_4}\) (+4), \(\mathrm{XeF_6}\) (+6), \(\mathrm{XeO_3}\) (+6) and \(\mathrm{XeOF_4}\) (+6) — are formed.

Notice that the partners are always fluorine or oxygen. Only the two most electronegative elements pull hard enough to draw electrons out of a noble gas atom at all.

Q4. Noble gases were believed to be completely unreactive for over sixty years. Why did xenon, and not helium or neon, finally break that record?

A. Because reactivity in this group depends on how tightly the filled shell is held. Helium and neon are small atoms with very high ionisation enthalpies — 2372 and 2080 \(\text{kJ mol}^{-1}\) in Table 7.4 — and their valence shells contain no d orbitals, so their electrons can neither be pulled away nor promoted. Xenon is much larger, its outermost electrons are far from the nucleus and well shielded, and its ionisation enthalpy has fallen to 1170 \(\text{kJ mol}^{-1}\). Its valence shell also has empty d orbitals into which paired electrons can be promoted. Given a partner as electronegative as fluorine or oxygen, xenon can therefore be persuaded to share electrons and form real compounds, while helium and neon still cannot.

oxidation states in groups 16, 17 and 18

Feature Group 16 Group 17 Group 18
Negative state shown −2 −1 none
Positive states shown +2, +4, +6 +1, +3, +5, +7 (not F) +2, +4, +6 (Xe only)
First member’s limitation O shows only −2, −1, +2 F shows only −1 He, Ne form no compounds
Reason for the limitation No d orbitals in the second shell No d orbitals in the valence shell Small size, very high ionisation enthalpy
Effect down the group +6 becomes less stable, +4 more stable (inert pair effect) Higher states become more stable Reactivity increases towards Xe and Rn

7.6.2 Chemical Reactivity Towards Hydrogen

i. Group 16 Elements

The elements of group 16 react with hydrogen to form hydrides of the type \(\mathrm{H_2E}\), where E = O, S, Se, Te, Po — for example \(\mathrm{H_2O}\), \(\mathrm{H_2S}\), \(\mathrm{H_2Se}\), \(\mathrm{H_2Te}\) and \(\mathrm{H_2Po}\). Some properties of the hydrides of group 16 are given in Table 7.5.

\(\mathrm{H_2O}\) is a colourless, odourless liquid, while \(\mathrm{H_2S}\), \(\mathrm{H_2Se}\), \(\mathrm{H_2Te}\) and \(\mathrm{H_2Po}\) are colourless, bad smelling, poisonous gases at ambient conditions.

All the hydrides have angular structures, which involve \(\mathrm{sp^3}\) hybridisation of the central atom (E). Four electron pairs surround the central atom — two bond pairs and two lone pairs — and the two lone pairs push the bonds closer together, which is why the molecule is bent rather than straight, as drawn in Fig. 7.5.

Diagram

Fig. 7.5 – Bent H–E–H molecule with the central atom E carrying two lone pairs drawn above it; mark the bond angle and show the tetrahedral arrangement of the four electron pairs in faint outline.

Table 7.5 : Properties of hydrides of group 16 elements

Property \(\mathrm{H_2O}\) \(\mathrm{H_2S}\) \(\mathrm{H_2Se}\) \(\mathrm{H_2Te}\)
m.p (K) 273 188 208 222
b.p (K) 373 213 232 269
H–E bond length (pm) 96 134 146 169
\(\Delta_{\text{diss}}H\) (H–E) kJ/mol 463 347 276 238
\(\Delta_f H\) kJ/mol −286 −20 73 100
HEH angle (°) 104 92 91 90
\(\mathrm{p}K_a\) 14.0 7.0 3.8 2.6

The hydrides of group 16 elements are weakly acidic. The acidic character of the hydrides increases, while the thermal stability decreases, from \(\mathrm{H_2O}\) to \(\mathrm{H_2Te}\). This is due to the decrease in the bond dissociation enthalpy of the H–E bond down the group (Table 7.5).

Both changes come from the same cause. A weaker H–E bond releases its hydrogen more readily as \(\mathrm{H^+}\), which means greater acidity; and a weaker bond also comes apart more easily on heating, which means lower thermal stability. Follow the \(\mathrm{p}K_a\) row of Table 7.5 and you can watch acidity rise as the number falls from 14.0 to 2.6.

All the hydrides except \(\mathrm{H_2O}\) possess reducing property, which increases in the order \(\mathrm{H_2S} < \mathrm{H_2Se} < \mathrm{H_2Te}\).

properties of the group 16 hydrides (down the group ↓: H–E bond length increases and bond dissociation enthalpy falls, because the central atom grows larger and overlaps less well with hydrogen. As a consequence — acidic character increases, thermal stability decreases, and reducing power increases. The bond angle also falls from 104° to 90° as the central atom grows larger and its lone pairs press the bonds together.)

ii. Group 17 Elements

The elements of group 17 react with hydrogen to give hydrogen halides.

\[\mathrm{H_2} + \mathrm{X_2} \rightarrow 2\,\mathrm{HX} \qquad (\text{where } \mathrm{X} = \mathrm{F},\ \mathrm{Cl},\ \mathrm{Br},\ \mathrm{I})\]

Some of the properties of the hydrogen halides are given in Table 7.6.

Table 7.6 : Properties of hydrides of group 17 elements

Property HF HCl HBr HI
m.p (K) 190 159 185 222
b.p (K) 293 189 206 238
Bond length (H–X) pm 91.7 127.4 141.4 160.9
\(\Delta_{\text{diss}}H^{\circ}\) kJ/mol 574 432 363 295
\(\mathrm{p}K_a\) 3.2 −7.0 −9.5 −10.0

The acidic strength of the halogen acids increases in the order

\[\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}\]

It is due to the decreasing bond dissociation enthalpy of the H–X bond in the order \(\mathrm{HF} > \mathrm{HCl} > \mathrm{HBr} > \mathrm{HI}\).

The thermal stability of the hydrogen halides decreases in the order \(\mathrm{HF} > \mathrm{HCl} > \mathrm{HBr} > \mathrm{HI}\). It is due to the decrease in bond dissociation enthalpy of the H–X bond down the group.

Q. Fluorine is the most electronegative halogen, so why is HF the weakest acid of the four?

A. Acid strength here is decided by how easily the H–X bond breaks, not by how polar it is. The H–F bond is the shortest and strongest of the four (574 \(\text{kJ mol}^{-1}\) in Table 7.6), and hydrogen bonding between HF molecules holds the acid together as well. Both effects make it reluctant to release \(\mathrm{H^+}\), so HF is the weakest acid even though it is the most polar.

iii. Group 18 Elements (Noble Gases)

Noble gases are chemically inert towards hydrogen due to their stable electronic configuration.

7.6.3 Reactivity Towards Oxygen

i. Group 16 Elements

All the elements of group 16 form oxides of the type \(\mathrm{EO_2}\) and \(\mathrm{EO_3}\), where E = S, Se, Te, Po.

\(\mathrm{EO_2}\) type oxides: ozone (\(\mathrm{O_3}\)) and sulfur dioxide (\(\mathrm{SO_2}\)) are gases, while selenium dioxide (\(\mathrm{SeO_2}\)) is a solid. They are acidic in nature and react with water to form acids.

\[\mathrm{SO_2} + \mathrm{H_2O} \rightarrow \mathrm{H_2SO_3}\ (\text{sulfurous acid})\]

\[\mathrm{SeO_2} + \mathrm{H_2O} \rightarrow \mathrm{H_2SeO_3}\ (\text{selenious acid})\]

The reducing property of the dioxides decreases from \(\mathrm{SO_2}\) to \(\mathrm{TeO_2}\). \(\mathrm{SO_2}\) is reducing while \(\mathrm{TeO_2}\) serves as an oxidising agent. This follows the stability pattern noted earlier: for sulfur the +6 state is easily reached, so \(\mathrm{SO_2}\) readily gives up electrons to move up to it, whereas for tellurium the +4 state is the comfortable one and \(\mathrm{TeO_2}\) would rather take electrons in.

\(\mathrm{EO_3}\) type oxides, \(\mathrm{SO_3}\), \(\mathrm{SeO_3}\) and \(\mathrm{TeO_3}\), are also acidic in nature. They dissolve in water to form acids.

\[\mathrm{SeO_3} + \mathrm{H_2O} \rightarrow \mathrm{H_2SeO_4}\ (\text{selenic acid})\]

\[\mathrm{TeO_3} + 3\,\mathrm{H_2O} \rightarrow \mathrm{H_6TeO_6}\ (\text{telluric acid})\]

ii. Group 17 Elements

Elements of group 17 (halogens) form many oxides with oxygen, but most of them are unstable.

Fluorine forms two oxides, \(\mathrm{OF_2}\) and \(\mathrm{O_2F_2}\). However, only \(\mathrm{OF_2}\) is thermally stable at 298 K. Both are strong fluorinating agents. \(\mathrm{O_2F_2}\) oxidises plutonium to \(\mathrm{PuF_6}\), and the reaction is used in removing plutonium as \(\mathrm{PuF_6}\) from spent nuclear fuel.

Chlorine oxides, \(\mathrm{Cl_2O}\), \(\mathrm{ClO_2}\), \(\mathrm{Cl_2O_6}\) and \(\mathrm{Cl_2O_7}\), are highly reactive oxidising agents and tend to explode.

\(\mathrm{ClO_2}\) is used as a bleaching agent for paper pulp and textiles, and in water treatment.

Bromine oxides, \(\mathrm{Br_2O}\), \(\mathrm{BrO_2}\) and \(\mathrm{BrO_3}\), are the least stable halogen oxides (middle row anomaly). They are very powerful oxidising agents.

Iodine oxides, \(\mathrm{I_2O_4}\), \(\mathrm{I_2O_5}\) and \(\mathrm{I_2O_7}\), are insoluble solids and decompose on heating. \(\mathrm{I_2O_5}\) is a very good oxidising agent and is used for the estimation of carbon monoxide.

The higher oxides of halogens are more stable than the lower ones.

iodine pentoxide, \(\mathrm{I_2O_5}\) (what it does → it is a solid oxidising agent that oxidises carbon monoxide to carbon dioxide, and is itself reduced to iodine → the conditions it needs → dry gas passed over the solid at a moderate temperature → what it is used for → estimating carbon monoxide in a gas sample, since the iodine liberated can be measured and corresponds to the amount of CO present.)

Chlorine dioxide, \(\mathrm{ClO_2}\), is one of the workhorses of a modern water treatment plant. It kills bacteria and viruses like chlorine does, but it also destroys the compounds that give water an earthy taste and smell. Paper mills use it to bleach wood pulp white without weakening the fibre. Because the gas is explosive when concentrated, it is never transported in cylinders — it is generated on site, in small quantities, exactly as it is needed.

iii. Group 18 Elements

Noble gas elements are chemically inert and do not directly react with oxygen.

7.6.4 Reactivity Towards Halogens

i. Group 16 Elements

Elements of group 16 react with halogens to give a large number of halides of the types \(\mathrm{EX_6}\), \(\mathrm{EX_4}\) and \(\mathrm{EX_2}\), where E = S, Se, Te.

Hexahalides: \(\mathrm{SF_6}\), \(\mathrm{SeF_6}\) and \(\mathrm{TeF_6}\) are formed by direct combination. They are colourless gases. They have \(\mathrm{sp^3d^2}\) hybridisation and possess octahedral structure. \(\mathrm{SF_6}\) is an exceptionally stable halide for steric reasons — the six fluorine atoms wrap the sulfur atom so completely that no reagent can reach it.

The stability of the halides decreases in the order

\[\text{fluorides} > \text{chlorides} > \text{bromides} > \text{iodides}\]

Tetrahalides: \(\mathrm{SF_4}\), \(\mathrm{SeF_4}\), \(\mathrm{TeF_4}\) and \(\mathrm{TeCl_4}\) have \(\mathrm{sp^3d}\) hybridisation and thus a trigonal bipyramidal geometry, with one equatorial position occupied by a lone pair.

Dihalides: \(\mathrm{SCl_2}\), \(\mathrm{SeCl_2}\) and \(\mathrm{TeCl_2}\) have \(\mathrm{sp^3}\) hybridisation, so their four electron pairs take up a tetrahedral arrangement in which two positions are occupied by lone pairs.

Diagram

Fig. 7.6 – Three structures in a row: SF₆ octahedral with six F atoms; SF₄ trigonal bipyramidal with a lone pair in an equatorial position, giving the see-saw shape; SCl₂ with a tetrahedral arrangement of four electron pairs, two of them lone pairs, giving a bent molecule. Label hybridisation under each: sp³d², sp³d, sp³.

Monohalides are dimeric in nature. For example, \(\mathrm{S_2F_2}\), \(\mathrm{S_2Cl_2}\), \(\mathrm{Se_2Cl_2}\) and \(\mathrm{SeBr_2}\). These dimeric halides undergo disproportionation.

\[2\,\mathrm{Se_2Cl_2} \rightarrow \mathrm{SeCl_4} + 3\,\mathrm{Se}\]

Disproportionation means that one and the same element is oxidised and reduced in the same reaction: selenium goes from +1 in \(\mathrm{Se_2Cl_2}\) up to +4 in \(\mathrm{SeCl_4}\) and down to 0 in free selenium.

Predict the hybridisation and the shape of the \(\mathrm{SF_4}\) molecule.

Sulfur has six electrons in its valence shell. Four of them are used to form four S–F bonds, and the remaining two stay together as one lone pair.

Total electron pairs around sulfur \(= 4\) bond pairs \(+\ 1\) lone pair \(= 5\).

Five electron pairs require five hybrid orbitals, which means one s, three p and one d orbital are mixed:

\[\text{hybridisation} = \mathrm{sp^3d}\]

Five electron pairs arrange themselves in a trigonal bipyramid. The lone pair, being bulkier than a bond pair, takes an equatorial position where it has more room.

Removing the lone pair from the picture leaves four bonded atoms in a shape like a see-saw. So \(\mathrm{SF_4}\) is \(\mathrm{sp^3d}\) hybridised with a see-saw shape, based on a trigonal bipyramidal arrangement of electron pairs.

Find and draw the structures of \(\mathrm{SeF_4}\) and \(\mathrm{SCl_2}\).

ii. Group 17 Elements

Halogens (group 17 elements) combine amongst themselves to form a number of compounds known as interhalogen compounds.

These are of the following types: \(\mathrm{XX'}\), \(\mathrm{XX'_3}\), \(\mathrm{XX'_5}\) and \(\mathrm{XX'_7}\), where X is the halogen atom with the larger size and X’ is the halogen atom with the smaller size. More details of interhalogen compounds are included in section 7.12.

iii. Group 18 Elements

Group 18 elements (noble gases) are chemically inert. Krypton and xenon, however, react directly with fluorine to give their fluorides. For example,

\[\mathrm{Xe}\text{(g)} + \mathrm{F_2}\text{(g)} \xrightarrow{\ 673\ \text{K},\ 1\ \text{atm}\ } \mathrm{XeF_2}\text{(s)}\]

Xenon fluorides \(\mathrm{XeF_2}\), \(\mathrm{XeF_4}\) and \(\mathrm{XeF_6}\) are crystalline and colourless, and they sublime readily at 298 K. They are powerful fluorinating agents.

7.6.5 Reactivity Towards Metals

i. Group 16 Elements

Elements of group 16 react with metals to form the corresponding compounds — oxides, sulfides and selenides in which the group 16 element takes its −2 state.

\[4\,\mathrm{Al} + 3\,\mathrm{O_2} \rightarrow 2\,\mathrm{Al_2O_3}\]

\[\mathrm{Cu} + \mathrm{S} \rightarrow \mathrm{CuS}\]

\[\mathrm{Mg} + \mathrm{Se} \rightarrow \mathrm{MgSe}\ (\text{magnesium selenide})\]

Gold is famous for combining with almost nothing, yet tellurium has the unusual property of combining with gold metal to form a telluride.

\[2\,\mathrm{Au} + 3\,\mathrm{Te} \rightarrow \mathrm{Au_2Te_3}\ (\text{gold telluride})\]

Gold telluride minerals are worth knowing about because prospectors have been fooled by them: the ore carries no golden glint at all, and gold has to be recovered from it by roasting off the tellurium.

ii. Group 17 Elements

Elements of group 17 (halogens) react with metals instantly to give metal halides.

\[2\,\mathrm{Na}\text{(s)} + \mathrm{Cl_2}\text{(g)} \rightarrow 2\,\mathrm{NaCl}\text{(s)}\]

\[\mathrm{Mg}\text{(s)} + \mathrm{Br_2}\text{(l)} \rightarrow \mathrm{MgBr_2}\text{(s)}\ (\text{magnesium bromide})\]

The ionic character of the halides decreases in the order \(\mathrm{MF} > \mathrm{MCl} > \mathrm{MBr} > \mathrm{MI}\), where M is a monovalent metal. Fluorine is the smallest and most electronegative halogen, so it takes the electron most completely and gives the most ionic bond; iodide is large and easily distorted, so its bond to the metal has more covalent character.

The metal halides having metals in their higher oxidation states are more covalent than the ones having metals in lower oxidation states. For example, \(\mathrm{SnCl_4}\), \(\mathrm{PbCl_4}\), \(\mathrm{SbCl_5}\) and \(\mathrm{UF_6}\) are more covalent than \(\mathrm{SnCl_2}\), \(\mathrm{PbCl_2}\), \(\mathrm{SbCl_3}\) and \(\mathrm{UF_4}\) respectively. A metal ion in a high oxidation state is small and highly charged, so it pulls the electron cloud of the halide ion towards itself and the bond becomes more covalent.

iii. Group 18 Elements

Noble gases do not directly react with metals.

Determine the oxidation state of the halogen atom in each of the following: \(\mathrm{HClO_4}\), \(\mathrm{KBrO_3}\), \(\mathrm{IF_7}\) and \(\mathrm{CaF_2}\). State which one of these values fluorine could never show, and why.

Complete and balance the following, naming the product in each case: (i) \(\mathrm{SO_2} + \mathrm{H_2O} \rightarrow\) (ii) \(\mathrm{Mg} + \mathrm{Se} \rightarrow\) (iii) \(\mathrm{Xe} + \mathrm{F_2} \xrightarrow{\ 673\ \text{K},\ 1\ \text{atm}\ }\) (iv) \(\mathrm{TeO_3} + \mathrm{H_2O} \rightarrow\). Then state which of these four reactions would not occur at all if xenon were replaced by neon, and give the reason.

7.7 Allotropy

What is allotropy? What is the difference between allotropy and polymorphism?

Elements of the group 16 exhibit allotropy. The same element can be packaged in more than one way — as molecules of different size, or as rings stacked differently in a crystal — and each package has its own colour, melting point and density even though the atoms in it are identical.

NoteDefinition

Allotropy is the existence of an element in two or more different physical forms in the same physical state, the forms differing in the way their atoms are bonded or arranged. These different forms are called allotropes.

It is worth separating this from a term that sounds similar. Polymorphism is the existence of a compound in more than one crystalline form, whereas allotropy refers to different forms of a single element. Calcium carbonate occurring as calcite and as aragonite is polymorphism; oxygen occurring as \(\mathrm{O_2}\) and as \(\mathrm{O_3}\) is allotropy.

Oxygen has two allotropes, \(\mathrm{O_2}\) and \(\mathrm{O_3}\) (ozone). Here the two forms differ in the number of atoms in the molecule itself, and their properties differ sharply, as section 7.9 will show.

Sulfur exists in a number of allotropic forms. Rhombic and monoclinic sulfur are the important allotropes of sulfur. Both are non metallic.

Selenium exists in two allotropic forms, red (non metallic) and grey (metallic).

Tellurium exists in two allotropic forms, (i) crystalline and (ii) amorphous.

Polonium reveals two allotropic forms, \(\alpha\) and \(\beta\), both metallic.

Read down that list and a pattern appears. In oxygen the allotropes are simply different molecules; by the time we reach tellurium and polonium, the difference is one of metallic character and crystal packing. This mirrors the shift from nonmetal to metal down the group.

The grey allotrope of selenium is a photoconductor: it hardly conducts electricity in the dark, but conducts as soon as light falls on it. That single property made the office photocopier possible. A selenium-coated rotating drum is given a uniform positive charge and is then exposed to the image to be copied. Wherever light strikes, the coating conducts and the charge leaks away; the dark regions of the image keep their charge. Negatively charged toner particles stick only to those charged regions, and the powder pattern is then pressed onto paper and fixed with heat. The steps are set out in FC 7.2.

Flowchart

FC 7.2 – Photocopying with a selenium drum: selenium-coated drum given a uniform positive charge (step 1) → drum exposed to the image; light discharges the lit areas, charge remains on the dark areas (step 2) → negatively charged toner particles attracted to the charged areas (step 3) → image transferred from drum to sheet of paper (step 4) → heating fixes the image; drum flooded with light and cleaned (step 5) → returns to step 1 for the next cycle.

7.7.1 Allotropes of Sulfur

Sulfur exhibits numerous allotropic forms. However, rhombic sulfur (\(\alpha\)-sulfur) and monoclinic sulfur (\(\beta\)-sulfur) are the most important allotropes of sulfur, and they are compared in Table 7.7.

Table 7.7 : Allotropes of sulfur

Rhombic Sulfur (\(\alpha\)-Sulfur) Monoclinic Sulfur (\(\beta\)-Sulfur)
Color Pale Yellow Bright yellow solid
Shape Orthorhombic crystals Needle shaped monoclinic crystals
M. P. 385.8 K 393 K
Density 2.069 g/cm³ 1.989 g/cm³
Solubility Insoluble in water and soluble in \(\mathrm{CS_2}\) Soluble in \(\mathrm{CS_2}\)
Stability Stable below 369 K and transforms to \(\beta\)-sulphur above this temperature Stable above 369 K and transforms into \(\alpha\)-sulphur below this temperature
Structure \(\mathrm{S_8}\) molecules having puckered ring structure \(\mathrm{S_8}\) molecules with puckered ring structure
Method of preparation It is prepared by evaporation of roll sulphur in \(\mathrm{CS_2}\) Rhombic sulphur melted in a dish and cooled till crust is formed. Two holes are made in the crust and the remaining liquid is poured out to give needle shaped crystals of \(\beta\)-sulphur

The temperature 369 K is worth remembering, because it is the point at which the two forms change places: below it the rhombic form is the stable one, above it the monoclinic form takes over.

Q. Both allotropes are built from the same puckered \(\mathrm{S_8}\) rings. Then why do they differ in colour, density and melting point at all?

A. Because the difference is not in the molecule but in how the molecules are stacked. The same \(\mathrm{S_8}\) rings pack into an orthorhombic lattice in one case and a monoclinic lattice in the other. Different packing means different distances between the rings, and therefore different density and melting point.

The Sulfur Rings

Several modifications of sulfur containing 6 to 20 sulfur atoms per ring have been synthesised. In the \(\mathrm{S_8}\) molecule the ring is puckered and has a crown shape, with an S–S bond length of 204 pm and a bond angle of about 107°, as shown in Fig. 7.7.

Diagram

Fig. 7.7 – Puckered crown-shaped S₈ ring drawn in perspective, with eight S atoms alternately above and below the mean plane; mark one S–S bond length as 204 pm and one S–S–S bond angle as 107°.

In cyclo-\(\mathrm{S_6}\) the ring adopts the chair form instead, with a bond length of 205.7 pm and a bond angle of 102.2°, drawn in Fig. 7.8.

Diagram

Fig. 7.8 – Chair form of the cyclo-S₆ ring shown in perspective like a cyclohexane chair; mark the S–S bond length 205.7 pm and the bond angle 102.2°.

At elevated temperature (about 1000 K), \(\mathrm{S_2}\) is the dominant species, which like \(\mathrm{O_2}\) is paramagnetic. The heat has broken the rings apart into small diatomic molecules, and those molecules turn out to have the same electronic peculiarity as dioxygen.

Which form of sulfur shows paramagnetic behaviour?

In the vapour state, sulfur partly exists as the \(\mathrm{S_2}\) molecule.

The \(\mathrm{S_2}\) molecule has two unpaired electrons in the antibonding \(\pi^*\) orbitals, exactly like \(\mathrm{O_2}\).

A species with unpaired electrons is attracted into a magnetic field. Hence \(\mathrm{S_2}\) exhibits paramagnetism, while \(\mathrm{S_8}\), in which every electron is paired, does not.

A sample of rhombic sulfur is heated slowly to 380 K and then allowed to stand at that temperature. Using Table 7.7, state (i) which allotrope will be present after some time and why, (ii) whether the density of the sample will rise or fall during the change, and (iii) whether the \(\mathrm{S_8}\) molecule itself is altered by the change. Give a reason for each answer.

7.8 Oxoacids

NoteDefinition

An oxoacid is an acid in which the acidic hydrogen is attached to oxygen, and that oxygen is in turn attached to a central atom — here sulfur or a halogen. Its general skeleton is therefore H–O–E, where E is the central atom that may carry further oxygen atoms.

7.8.1 Oxoacids of Sulfur

Sulfur forms a number of oxoacids. Some of them are unstable and cannot be isolated. They are known to exist in aqueous solutions or in the form of their salts.

Some important oxoacids of sulfur, together with their structures, are the following.

  1. Sulfurous acid, \(\mathrm{H_2SO_3}\)

  2. Sulfuric acid, \(\mathrm{H_2SO_4}\)

  3. Di or pyrosulfuric acid, \(\mathrm{H_2S_2O_7}\)

  4. Peroxy monosulfuric acid, \(\mathrm{H_2SO_5}\)

  5. Peroxy disulfuric acid, \(\mathrm{H_2S_2O_8}\)

  6. Thiosulfuric acid, \(\mathrm{H_2S_2O_3}\)

The structures of all six are collected in Fig. 7.9. Notice while reading them that sulfur sits at the centre of every one, holding some combination of –OH groups, doubly bonded oxygen atoms, and — in three cases — an extra link that gives the acid its name.

Diagram

Fig. 7.9 – Six structural formulae of the oxoacids of sulfur drawn side by side and labelled: (i) H₂SO₃, S with two OH, one S=O and a lone pair; (ii) H₂SO₄, S with two OH and two S=O; (iii) H₂S₂O₇, two SO₃H units joined through a bridging O; (iv) H₂SO₅, S with one OH, two S=O and one O–OH peroxo group; (v) H₂S₂O₈, two SO₃H units joined through an O–O peroxo link; (vi) H₂S₂O₃, like H₂SO₄ but with one doubly bonded O replaced by a terminal S.

the six oxoacids of sulfur at a glance

Acid Formula Distinguishing feature of its structure
Sulfurous acid \(\mathrm{H_2SO_3}\) One S=O bond and a lone pair on sulfur
Sulfuric acid \(\mathrm{H_2SO_4}\) Two S=O bonds and two –OH groups
Di or pyrosulfuric acid \(\mathrm{H_2S_2O_7}\) Two sulfur atoms joined through a bridging oxygen, S–O–S
Peroxy monosulfuric acid \(\mathrm{H_2SO_5}\) Contains a peroxo (–O–O–) linkage
Peroxy disulfuric acid \(\mathrm{H_2S_2O_8}\) Two sulfur atoms joined through a peroxo (–O–O–) linkage
Thiosulfuric acid \(\mathrm{H_2S_2O_3}\) One oxygen of sulfuric acid replaced by sulfur

Determine the oxidation state of sulfur in peroxy disulfuric acid, \(\mathrm{H_2S_2O_8}\).

A quick count that treats every oxygen as −2 gives the wrong answer here, so the structure has to be used.

Fig. 7.9 shows that two of the eight oxygen atoms in \(\mathrm{H_2S_2O_8}\) form a peroxo linkage, –O–O–. Oxygen in a peroxo linkage has the oxidation number −1, not −2; the remaining six oxygen atoms are −2 as usual. Each hydrogen is +1.

Let the oxidation state of each sulfur atom be \(x\). The molecule is neutral, so all the oxidation numbers add up to zero:

\[2(+1) + 2x + 6(-2) + 2(-1) = 0\]

\[2 + 2x - 12 - 2 = 0\]

\[2x = 12 \quad \Rightarrow \quad x = +6\]

So sulfur is in the +6 state, the same as in sulfuric acid. Had we carelessly taken all eight oxygen atoms as −2, we would have obtained the impossible value +7 — impossible because sulfur has only six valence electrons to give. Whenever a formula carries the word “peroxy”, look for the –O–O– link before counting.

Determine the oxidation state of sulfur in (i) peroxy monosulfuric acid, \(\mathrm{H_2SO_5}\), and (ii) di or pyrosulfuric acid, \(\mathrm{H_2S_2O_7}\). State clearly, for each, which oxygen atoms you have taken as −1 and which as −2, and why.

7.8.2 Oxoacids of Halogens

Halogens form several oxoacids, which are set out in Table 7.8. Only four oxoacids have been isolated in pure form: hypofluorous acid (HOF), perchloric acid (\(\mathrm{HClO_4}\)), iodic acid (\(\mathrm{HIO_3}\)) and metaperiodic acid (\(\mathrm{H_2IO_6}\)). The others are stable only in aqueous solutions or in the form of their salts.

Read Table 7.8 across the rows and the pattern is clear at once: fluorine appears in one row only, while chlorine, bromine and iodine appear in several. This is the consequence of the point made in section 7.6 — fluorine has no d orbitals, cannot expand its octet, and cannot take a positive oxidation state beyond +1.

Table 7.8 : Oxoacids of halogens

Oxidation state of X Generic name Oxoacids of fluorine Oxoacids of chlorine Oxoacids of bromine Oxoacids of iodine
+1 Hypohalous acid (HXO) HOF HOCl HOBr HOI
+3 Halous acid (\(\mathrm{HXO_2}\)) − HOClO − −
+5 Halic acid (\(\mathrm{HXO_3}\)) − \(\mathrm{HOClO_2}\) \(\mathrm{HOBrO_2}\) \(\mathrm{HOIO_2}\)
+7 Perhalic acid (\(\mathrm{HXO_4}\)) − \(\mathrm{HOClO_3}\) \(\mathrm{HOBrO_3}\) \(\mathrm{HOIO_3}\)

The acid strength of the halogen oxoacids increases with the increasing oxidation state of the halogen. For example, acid strength increases from \(\mathrm{HClO}\), a weak acid (\(K_a = 3.5 \times 10^{-8}\)), to \(\mathrm{HClO_4}\), a very strong acid (\(K_a \gg 1\)).

Q. Why should adding more oxygen atoms to the same central chlorine atom make the acid stronger?

A. An acid is strong when the anion it leaves behind is stable. Each extra oxygen atom is highly electronegative and pulls electron density away from the central atom, spreading the negative charge of the anion over more atoms. The \(\mathrm{ClO_4^{-}}\) ion spreads its charge over four oxygen atoms and is very stable, so \(\mathrm{HClO_4}\) lets its proton go readily; \(\mathrm{ClO^{-}}\) has only one oxygen to hold the charge, so \(\mathrm{HClO}\) holds on to its proton and stays weak.

acid strength of the halogen oxoacids (with rising oxidation state of the halogen, \(\mathrm{HXO} < \mathrm{HXO_2} < \mathrm{HXO_3} < \mathrm{HXO_4}\): acid strength increases, because each additional oxygen withdraws more electron density and stabilises the anion left behind after the proton departs. Across the halogens at the same oxidation state, acid strength falls as the halogen becomes less electronegative, so \(\mathrm{HClO_4} > \mathrm{HBrO_4} > \mathrm{HIO_4}\).)

Structures of the Oxoacids of Chlorine

The four oxoacids of chlorine are built on the same H–O–Cl skeleton, with oxygen atoms added one at a time to the chlorine. Their structures are drawn in Fig. 7.10.

  1. Hypochlorous acid, HOCl

  2. Chlorous acid, HOClO or \(\mathrm{HClO_2}\)

  3. Chloric acid, \(\mathrm{HClO_3}\)

  4. Perchloric acid, \(\mathrm{HClO_4}\)

Diagram

Fig. 7.10 – Four structural formulae in a row: HOCl drawn as H–O–Cl bent at oxygen; HClO₂ as H–O–Cl with one Cl=O; HClO₃ as H–O–Cl with two Cl=O; HClO₄ as H–O–Cl with three Cl=O. Label the oxidation state of chlorine beneath each as +1, +3, +5 and +7.

Hypochlorous acid may be the weakest of the four, but it is the one you meet most often. When bleaching powder or a hypochlorite solution is added to water, it is HOCl that forms and does the disinfecting, and it is HOCl that keeps swimming pool water free of bacteria. Its weakness is exactly what makes it useful: a strong acid would attack skin, fabric and pipes, while HOCl stays mild enough to handle and still oxidises the organisms it meets.

7.9 Oxygen and Compounds of Oxygen

7.9.1 Dioxygen

a. Preparation
i. Laboratory methods

Oxygen is prepared in the laboratory by taking a compound that already contains oxygen and persuading it to let go.

By heating oxygen containing salts such as chlorates, nitrates and permanganates:

\[2\,\mathrm{KClO_3}\text{(s)} \xrightarrow{\ \text{Heat},\ \mathrm{MnO_2}\ } 2\,\mathrm{KCl}\text{(s)} + 3\,\mathrm{O_2}\text{(g)}\]

By thermal decomposition of oxides of metals:

\[2\,\mathrm{Ag_2O}\text{(s)} \xrightarrow{\ \Delta\ } 4\,\mathrm{Ag}\text{(s)} + \mathrm{O_2}\text{(g)}\]

\[2\,\mathrm{HgO}\text{(s)} \xrightarrow{\ \Delta\ } 2\,\mathrm{Hg}\text{(l)} + \mathrm{O_2}\text{(g)}\]

\[2\,\mathrm{PbO_2}\text{(s)} \xrightarrow{\ \Delta\ } 2\,\mathrm{PbO}\text{(s)} + \mathrm{O_2}\text{(g)}\]

By decomposition of hydrogen peroxide in presence of a catalyst such as finely divided metals and manganese dioxide:

\[2\,\mathrm{H_2O_2}\text{(aq)} \xrightarrow{\ \Delta,\ \mathrm{MnO_2}\ } 2\,\mathrm{H_2O}\text{(l)} + \mathrm{O_2}\text{(g)}\]

manganese dioxide, \(\mathrm{MnO_2}\) (what it does → it acts as a catalyst, lowering the temperature at which potassium chlorate or hydrogen peroxide gives up its oxygen, and is recovered unchanged at the end → the conditions it needs → simply mixed as a fine black powder with the substance being decomposed, with gentle heating for \(\mathrm{KClO_3}\) → what it is used for → the standard laboratory preparation of dioxygen; note that it appears above the arrow, never as a reactant, so it is not balanced into the equation.)

ii. Electrolysis

Dioxygen can be prepared on a large scale by the electrolysis of water, when hydrogen is liberated at the cathode and oxygen at the anode.

\[2\,\mathrm{H_2O} \xrightarrow{\ \text{Electrolysis}\ } 2\,\mathrm{H_2} + \mathrm{O_2}\]

iii. Industrial method

Dioxygen is obtained from air, by first removing carbon dioxide and water vapour. The remaining gases are liquified subsequently. This is followed by fractional distillation, which gives dinitrogen and dioxygen. The separation works because liquid nitrogen boils at a lower temperature than liquid oxygen, so nitrogen distils off first. The steps are summarised in FC 7.3.

Flowchart

FC 7.3 – Industrial manufacture of dioxygen: air → removal of CO₂ and water vapour → compression and cooling → liquefaction of the remaining gases → fractional distillation → dinitrogen distils off first (lower boiling point) → dioxygen left behind and collected.

b. Physical properties

Dioxygen is a colourless and odourless gas.

Dioxygen is sparingly soluble in water: 30.8 cm³ of \(\mathrm{O_2}\) dissolves in 1000 cm³ of water at 293 K. A small amount of dissolved dioxygen is sufficient to sustain marine and aquatic life.

It liquifies at 90 K and freezes at 55 K.

Oxygen has three stable isotopes, \(^{16}\mathrm{O}\), \(^{17}\mathrm{O}\) and \(^{18}\mathrm{O}\). All three have eight protons and differ only in the number of neutrons, so they are chemically alike.

Molecular oxygen, \(\mathrm{O_2}\), exhibits paramagnetism.

Why does the water in a fish pot need to be changed from time to time?

Q. If only about 30 cm³ of oxygen dissolves in a litre of water, how do fish manage at all?

A. Because that small dissolved amount is continuously replaced. Oxygen passes into the water from the air at the surface and from aquatic plants, and moving water mixes it through. Fish gills are efficient enough to extract what they need from this thin supply — which is exactly why still, crowded water goes stale and has to be changed.

c. Chemical Properties
i. Reaction with metals

Dioxygen directly reacts with almost all metals except Au and Pt to form their oxides.

\[2\,\mathrm{Ca} + \mathrm{O_2} \rightarrow 2\,\mathrm{CaO}\]

\[4\,\mathrm{Al} + 3\,\mathrm{O_2} \rightarrow 2\,\mathrm{Al_2O_3}\]

Gold and platinum resist this attack, which is one reason they survive in nature as free metals and are used for jewellery and laboratory ware.

ii. Reaction with nonmetals

Dioxygen reacts with nonmetals (except noble gases) to form their oxides.

\[\mathrm{C} + \mathrm{O_2} \rightarrow \mathrm{CO_2}\]

\[\mathrm{P_4} + 5\,\mathrm{O_2} \rightarrow \mathrm{P_4O_{10}}\]

iii. Reaction with some compounds

\[2\,\mathrm{ZnS} + 3\,\mathrm{O_2} \xrightarrow{\ \Delta\ } 2\,\mathrm{ZnO} + 2\,\mathrm{SO_2}\]

\[\mathrm{CH_4} + 2\,\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\,\mathrm{H_2O}\]

\[2\,\mathrm{SO_2} + \mathrm{O_2} \xrightarrow{\ \mathrm{V_2O_5}\ } 2\,\mathrm{SO_3}\]

\[4\,\mathrm{HCl} + \mathrm{O_2} \xrightarrow{\ \mathrm{CuCl_2},\ \Delta\ } 2\,\mathrm{Cl_2} + 2\,\mathrm{H_2O}\]

The first of these is the roasting of a sulfide ore, and the last two will return later in this chapter as steps in the manufacture of sulfuric acid and of chlorine.

d. Uses

Dioxygen is important for respiration, to sustain animal and aquatic life.

It is used in the manufacture of steel.

It is used in the oxyacetylene flame for welding and cutting of metals.

Oxygen cylinders are widely used in hospitals, in high altitude flying and in mountaineering.

It is used in the combustion of fuels; for example, hydrazine in liquid oxygen provides tremendous thrust (energy) in rockets.

Dioxygen is paramagnetic in spite of having an even number of electrons. Explain.

Dioxygen is a covalently bonded molecule, and its Lewis structure is written as

\[:\!\ddot{\mathrm{O}}::\ddot{\mathrm{O}}\!: \quad \text{or} \quad :\!\ddot{\mathrm{O}}\!=\!\ddot{\mathrm{O}}\!:\]

This structure shows every electron paired, and so it cannot account for paramagnetism at all. The paramagnetic behaviour of \(\mathrm{O_2}\) can be explained with the help of molecular orbital theory.

Electronic configuration of \(\mathrm{O_2}\):

\[\mathrm{KK}\ \sigma(2s)^2\ \sigma^*(2s)^2\ \sigma(2p_z)^2\ \pi(2p_x)^2\ \pi(2p_y)^2\ \pi^*(2p_x)^1\ \pi^*(2p_y)^1\]

The last two electrons occupy two different antibonding \(\pi^*\) orbitals of equal energy, and by Hund’s rule they stay unpaired with parallel spins. The presence of these two unpaired electrons explains the paramagnetic nature of dioxygen, as shown in Fig. 7.11.

Diagram

Fig. 7.11 – Molecular orbital energy level diagram for O₂ with atomic orbitals of the two O atoms on either side; fill σ(2s)², σ(2s)², σ(2p_z)², π(2p_x)², π(2p_y)², and show single arrows of the same spin in π(2p_x) and π*(2p_y); label these two “unpaired electrons — paramagnetic”.

Calculate the volume of dioxygen, measured at STP, obtained by the complete decomposition of 24.52 g of potassium chlorate according to the equation \(2\,\mathrm{KClO_3} \rightarrow 2\,\mathrm{KCl} + 3\,\mathrm{O_2}\). (Molar masses in \(\text{g mol}^{-1}\): K = 39.1, Cl = 35.5, O = 16.0; molar volume of a gas at STP = 22.4 L \(\text{mol}^{-1}\))

7.9.2 Simple Oxides

NoteDefinition

A binary compound of oxygen with another element is called an oxide.

Oxides can be classified into (a) acidic oxides, (b) basic oxides, (c) amphoteric oxides and (d) neutral oxides. The classification is based on one simple test: what does the oxide do when it meets water, an acid or a base?

a. Acidic oxides

An oxide which dissolves in water to give an acid, or reacts with a base to give a salt, is called an acidic oxide. For example, \(\mathrm{SO_2}\), \(\mathrm{SO_3}\), \(\mathrm{CO_2}\), \(\mathrm{N_2O_5}\), \(\mathrm{Cl_2O_7}\) etc.

\[\mathrm{SO_2} + \mathrm{H_2O} \rightarrow \mathrm{H_2SO_3}\]

\[\mathrm{SO_3} + 2\,\mathrm{NaOH} \rightarrow \mathrm{Na_2SO_4} + \mathrm{H_2O}\]

Generally, oxides of nonmetals are acidic oxides.

b. Basic oxides

An oxide which dissolves in water to give a base, or reacts with an acid to give a salt, is called a basic oxide. For example, \(\mathrm{Na_2O}\), \(\mathrm{CaO}\), \(\mathrm{BaO}\) etc.

\[\mathrm{CaO} + \mathrm{H_2O} \rightarrow \mathrm{Ca(OH)_2}\]

\[\mathrm{BaO} + 2\,\mathrm{HCl} \rightarrow \mathrm{BaCl_2} + \mathrm{H_2O}\]

These are oxides of metals, in which the metal has given its electrons away fairly completely, so the oxide ion is free to accept a proton from water or acid.

c. Amphoteric oxides

The oxide which reacts with a base as well as with an acid to give a salt is called an amphoteric oxide. For example, \(\mathrm{Al_2O_3}\).

Acting as an acidic oxide, towards a base:

\[\mathrm{Al_2O_3}\text{(s)} + 6\,\mathrm{NaOH}\text{(aq)} + 3\,\mathrm{H_2O}\text{(l)} \rightarrow 2\,\mathrm{Na_3[Al(OH)_6]}\text{(aq)}\]

Acting as a basic oxide, towards an acid:

\[\mathrm{Al_2O_3}\text{(s)} + 6\,\mathrm{HCl}\text{(aq)} + 9\,\mathrm{H_2O}\text{(l)} \rightarrow 2\,\mathrm{[Al(H_2O)_6]^{3+}}\text{(aq)} + 6\,\mathrm{Cl^{-}}\text{(aq)}\]

d. Neutral oxides

The oxides which are neither acidic nor basic are called neutral oxides. For example, \(\mathrm{CO}\), \(\mathrm{NO}\), \(\mathrm{N_2O}\) etc.

the four classes of oxides

Class Behaviour with water Behaviour with acid Behaviour with base Examples
Acidic Gives an acid No reaction Gives a salt and water \(\mathrm{SO_2}\), \(\mathrm{SO_3}\), \(\mathrm{CO_2}\), \(\mathrm{N_2O_5}\), \(\mathrm{Cl_2O_7}\)
Basic Gives a base Gives a salt and water No reaction \(\mathrm{Na_2O}\), \(\mathrm{CaO}\), \(\mathrm{BaO}\)
Amphoteric Generally insoluble Gives a salt Gives a salt \(\mathrm{Al_2O_3}\)
Neutral No reaction No reaction No reaction \(\mathrm{CO}\), \(\mathrm{NO}\), \(\mathrm{N_2O}\)

Classify each of the following as an acidic, basic, amphoteric or neutral oxide, and for the first three write one balanced equation that justifies your choice: \(\mathrm{CaO}\), \(\mathrm{Cl_2O_7}\), \(\mathrm{Al_2O_3}\), \(\mathrm{N_2O}\).

7.9.3 Ozone

Ozone (\(\mathrm{O_3}\)) is an allotrope of oxygen. Oxygen in the upper atmosphere absorbs energy in the form of ultra-violet light and changes to atomic oxygen, which combines with molecular oxygen to form \(\mathrm{O_3}\).

\[\mathrm{O_2} \xrightarrow{\ \text{U.V. light}\ } \mathrm{O} + \mathrm{O}\]

\[\mathrm{O_2} + \mathrm{O} \rightarrow \mathrm{O_3}\]

The layer of ozone protects the earth’s surface from harmful ultraviolet (U.V.) radiations. Hence it is called the ‘ozone umbrella’.

The name ozone was coined from the Greek word ozein, meaning to smell. The gas was noticed by its sharp odour near electrical machines long before its formula was established — the same smell that hangs in the air after a lightning storm.

a. Preparation of Ozone

Ozone is prepared in the laboratory by passing a silent electric discharge through pure and dry oxygen in an apparatus called an ozoniser, drawn in Fig. 7.12. As the conversion of oxygen to ozone is only 10 %, the product is known as ozonised oxygen. It is an endothermic process.

\[3\,\mathrm{O_2}\text{(g)} \rightleftharpoons 2\,\mathrm{O_3} \qquad \Delta H = +142\ \text{kJ mol}^{-1}\]

The discharge is called silent because it passes as a diffuse glow between the glass walls rather than as a spark; a spark would supply so much heat that the ozone formed would break down again.

Diagram

Fig. 7.12 – Ozoniser: two concentric glass tubes with tin foil coatings connected to a high voltage induction coil, dry oxygen entering at one end and ozonised oxygen leaving at the other; label “silent electric discharge”.

b. Physical properties of ozone

Pure ozone is a pale-blue gas, a dark blue liquid and a violet-black solid.

Ozone has a characteristic smell. When inhaled in concentrations above 100 ppm, it causes nausea and headache.

It is diamagnetic in nature — unlike dioxygen, all the electrons in the ozone molecule are paired.

Gibbs energy change, \(\Delta G\) (SI unit: \(\text{kJ mol}^{-1}\); sign convention — \(\Delta G\) negative means the change is spontaneous, and the more negative it is the stronger the driving force. It is built from two other quantities through \(\Delta G = \Delta H - T\Delta S\): a negative \(\Delta H\) (heat given out) and a positive \(\Delta S\) (disorder increasing) both push \(\Delta G\) downward.)

High concentrations of ozone can be dangerously explosive. Explain.

Consider the thermal stability of ozone. Ozone is thermodynamically less stable than oxygen, and decomposes into \(\mathrm{O_2}\).

\[\mathrm{O_3} \rightarrow \mathrm{O_2} + \mathrm{O}\]

The decomposition is exothermic and results in the liberation of heat, so \(\Delta H\) is negative. It also increases the number of independent particles, so the entropy increases and \(\Delta S\) is positive.

Putting both into \(\Delta G = \Delta H - T\Delta S\): a negative \(\Delta H\) and a positive \(\Delta S\) together give a large negative Gibbs energy change.

A large negative \(\Delta G\) means the decomposition is strongly spontaneous, and the heat it releases speeds up further decomposition. Therefore a high concentration of ozone can be dangerously explosive.

c. Chemical Properties
i. Oxidising property

Ozone is a powerful oxidising agent, as it easily decomposes to liberate nascent oxygen.

\[\mathrm{O_3} \rightarrow \mathrm{O_2} + \mathrm{O}\]

Nascent oxygen means a single, freshly formed oxygen atom, which is far more reactive than the paired-up \(\mathrm{O_2}\) molecule; this atom is what does the oxidising.

Ozone oxidises lead sulfide to lead sulfate, and iodide ions to iodine.

\[\mathrm{PbS}\text{(s)} + 4\,\mathrm{O_3}\text{(g)} \rightarrow \mathrm{PbSO_4}\text{(s)} + 4\,\mathrm{O_2}\text{(g)}\]

\[2\,\mathrm{KI}\text{(aq)} + \mathrm{H_2O}\text{(l)} + \mathrm{O_3}\text{(g)} \rightarrow 2\,\mathrm{KOH}\text{(aq)} + \mathrm{I_2}\text{(g)} + \mathrm{O_2}\text{(g)}\]

Ozone oxidises nitrogen oxide and gives nitrogen dioxide.

\[\mathrm{NO}\text{(g)} + \mathrm{O_3}\text{(g)} \rightarrow \mathrm{NO_2}\text{(g)} + \mathrm{O_2}\text{(g)}\]

Hence the nitrogen oxide emitted from the exhaust systems of supersonic jet aeroplanes can bring forth depletion of the ozone layer in the upper atmosphere.

ii. Bleaching property

Ozone acts as a good bleaching agent due to its oxidising nature.

\[\mathrm{O_3} \rightarrow \mathrm{O} + \mathrm{O_2}\]

\[\text{Coloured matter} + \mathrm{O} \rightarrow \text{colourless matter}\]

Ozone bleaches in the absence of moisture, so it is also known as dry bleach. This is a useful contrast to keep in mind, because chlorine — as section 7.11 will show — bleaches only when moisture is present.

iii. Reducing property

Ozone reduces peroxides to oxides.

\[\mathrm{H_2O_2} + \mathrm{O_3} \rightarrow \mathrm{H_2O} + 2\,\mathrm{O_2}\]

\[\mathrm{BaO_2} + \mathrm{O_3} \rightarrow \mathrm{BaO} + 2\,\mathrm{O_2}\]

  1. Ozone is used as a bleaching agent. Explain. (b) Why does ozone act as a powerful oxidising agent?

ozone in ozonolysis (what it does → ozone adds across a carbon–carbon double bond of an unsaturated compound to give an addition product called an ozonide → the conditions it needs → the ozonide is then decomposed by water or dilute acids → what it is used for → the products are aldehydes or ketones, and identifying them tells a chemist exactly where the double bond was located in the original molecule. The whole sequence is called ozonolysis, and you will meet it again in organic chemistry.)

iv. Ozone depletion

The thinning of the ozone layer in the upper atmosphere is called ozone depletion.

The ozone (\(\mathrm{O_3}\)) layer in the upper atmosphere absorbs harmful UV radiations from the sun, thus protecting people on the earth.

Depletion of the ozone layer in the upper atmosphere is caused by nitrogen oxide released from the exhaust systems of cars or supersonic jet aeroplanes.

\[\mathrm{NO}\text{(g)} + \mathrm{O_3}\text{(g)} \rightarrow \mathrm{NO_2}\text{(g)} + \mathrm{O_2}\text{(g)}\]

Depletion (thinning) of the ozone layer can also be caused by chlorofluoro carbons (freons), used in aerosols and refrigerators, and their subsequent escape into the atmosphere.

The depletion of the ozone layer has been most pronounced in the polar regions, especially over Antarctica.

Ozone depletion is a major environmental problem because it increases the amount of ultraviolet (UV) radiation that reaches the earth’s surface, thus causing an increase in the rate of skin cancer, eye cataracts, and genetic as well as immune system damage among people. The cycle of formation and destruction is set out in FC 7.4.

Flowchart

FC 7.4 – Ozone in the stratosphere: O₂ absorbs UV → splits into O + O → O + O₂ → O₃ (ozone umbrella formed); destruction branch: O₃ + NO from jet and car exhausts → NO₂ + O₂, and O₃ attacked by chlorine atoms released from CFCs → ozone layer thins → more UV reaches the earth’s surface → skin cancer, cataracts, genetic and immune damage.

The Antarctic ozone hole, first reported in the mid-1980s, showed how a gas released in one part of the world could damage the atmosphere over another. Chlorofluorocarbons were prized precisely because they were unreactive and non-toxic, which is what allowed them to drift intact all the way to the stratosphere before UV light broke them apart. The nations of the world agreed in the Montreal Protocol to phase them out, and refrigerants and aerosol propellants had to be redesigned. Measurements since then show the layer slowly recovering — a reminder that a chemical property as innocent-looking as “unreactive” has to be judged over the whole life of a substance.

Q5. Oxygen and ozone are built from the same kind of atom, yet one sustains life at ground level and the other protects it from far above — while ozone near the ground is treated as a pollutant. How can one element behave in two such different ways?

A. Because allotropes are different molecules, and the molecule decides the chemistry. \(\mathrm{O_2}\) is a stable, moderately reactive molecule that our cells can use for respiration. \(\mathrm{O_3}\) is thermodynamically less stable and readily gives up an oxygen atom, which makes it a powerful oxidising agent. High in the atmosphere that instability is exactly what protects us: ozone absorbs harmful UV radiation and re-forms in the cycle shown in FC 7.4. At ground level the same eagerness to release nascent oxygen means ozone attacks lung tissue, fabrics and rubber, and above 100 ppm it causes nausea and headache. Same element, two molecules, two entirely different roles.

d. Structure of Ozone

Ozone is an angular molecule. The two O—O bond lengths in the ozone molecule are identical, 128 pm, and the O—O—O bond angle is about 117°. It is a resonance hybrid of two canonical forms, drawn in Fig. 7.13.

The word identical is the key to the structure. If ozone had one single bond and one double bond, the two lengths would differ; because they are equal, neither canonical form is the real molecule. The true molecule is the resonance hybrid, in which the extra pair of electrons is spread evenly over both bonds.

Diagram

Fig. 7.13 – Two canonical forms of ozone drawn side by side with a double-headed resonance arrow between them, each showing one O=O and one O–O with lone pairs and formal charges; to the right, the resonance hybrid with both bond lengths marked 128 pm and the bond angle 117°.

dioxygen vs ozone

Feature Dioxygen, \(\mathrm{O_2}\) Ozone, \(\mathrm{O_3}\)
Molecular shape Linear (diatomic) Angular, bond angle about 117°
Colour Colourless gas Pale-blue gas
Smell Odourless Characteristic sharp smell
Magnetic behaviour Paramagnetic Diamagnetic
Thermal stability Stable Less stable; decomposes to \(\mathrm{O_2}\)
Oxidising power Moderate Powerful, through nascent oxygen
Formation — \(3\,\mathrm{O_2} \rightleftharpoons 2\,\mathrm{O_3}\), \(\Delta H = +142\ \text{kJ mol}^{-1}\)
e. Uses of ozone

Ozone is used for air purification at crowded places like cinema halls, tunnels and railways.

It is used in sterilizing drinking water by oxidising all germs and bacteria.

It is used for bleaching ivory, oils, starch, wax and delicate fabrics such as silk.

It is used in the manufacture of synthetic camphor, potassium permanganate, etc.

7.10 Compounds of Sulfur

7.10.1 Sulfur Dioxide

a. Preparation
i. From sulfur

Sulfur dioxide gas can be prepared by burning sulfur in air.

\[\mathrm{S}\text{(s)} + \mathrm{O_2}\text{(g)} \rightarrow \mathrm{SO_2}\text{(g)}\]

ii. From sulfite

In the laboratory, sulfur dioxide is prepared by treating an aqueous solution of sodium sulfite with dilute sulfuric acid.

\[\mathrm{Na_2SO_3} + \mathrm{H_2SO_4}\text{(aq)} \rightarrow \mathrm{Na_2SO_4} + \mathrm{H_2O}\text{(l)} + \mathrm{SO_2}\text{(g)}\]

This is the usual bench method because the gas comes off steadily and needs no heating — the stronger, less volatile acid simply displaces the weaker, volatile one from its salt.

iii. From sulfides (industrial method)

Sulfur dioxide can be prepared by roasting zinc sulfide and iron pyrites.

\[2\,\mathrm{ZnS}\text{(s)} + 3\,\mathrm{O_2}\text{(g)} \xrightarrow{\ \Delta\ } 2\,\mathrm{ZnO}\text{(s)} + 2\,\mathrm{SO_2}\text{(g)}\]

\[4\,\mathrm{FeS_2}\text{(s)} + 11\,\mathrm{O_2}\text{(g)} \xrightarrow{\ \Delta\ } 2\,\mathrm{Fe_2O_3}\text{(s)} + 8\,\mathrm{SO_2}\text{(g)}\]

Roasting means heating a sulfide ore strongly in a plentiful supply of air. Industry gets its sulfur dioxide almost free this way, as a by-product of extracting zinc and iron.

b. Physical properties of \(\mathrm{SO_2}\)

Sulfur dioxide is a colourless gas with a pungent smell.

It is poisonous in nature.

\(\mathrm{SO_2}\) is highly soluble in water, and its solution in water is called sulfurous acid.

It liquifies at room temperature under a pressure of 2 atm and boils at 263 K.

c. Chemical Properties
i. Reaction with \(\mathrm{Cl_2}\)

Sulfur dioxide reacts with chlorine in the presence of charcoal (catalyst) to form sulfuryl chloride.

\[\mathrm{SO_2}\text{(g)} + \mathrm{Cl_2}\text{(g)} \xrightarrow{\ \text{charcoal}\ } \mathrm{SO_2Cl_2}\text{(l)}\]

ii. Reaction with \(\mathrm{O_2}\)

Sulfur dioxide is oxidised by dioxygen in the presence of vanadium(V) oxide to sulfur trioxide.

\[2\,\mathrm{SO_2}\text{(g)} + \mathrm{O_2}\text{(g)} \xrightarrow{\ \mathrm{V_2O_5}\ } 2\,\mathrm{SO_3}\text{(g)}\]

iii. Reaction with NaOH

Sulfur dioxide readily reacts with sodium hydroxide solution to form sodium sulfite.

\[2\,\mathrm{NaOH} + \mathrm{SO_2} \rightarrow \mathrm{Na_2SO_3} + \mathrm{H_2O}\]

iv. Reaction with \(\mathrm{Na_2SO_3}\)

Sulfur dioxide reacts with sodium sulfite solution to form sodium hydrogen sulfite.

\[\mathrm{Na_2SO_3} + \mathrm{H_2O}\text{(l)} + \mathrm{SO_2} \rightarrow 2\,\mathrm{NaHSO_3}\]

These last two reactions together confirm that \(\mathrm{SO_2}\) is an acidic oxide: it neutralises a base, and then reacts further with its own salt.

v. Reducing property

Sulfur dioxide acts as a reducing agent in the presence of moisture. In each of the reactions below, sulfur moves from the +4 state to the +6 state, giving up two electrons to whatever it reduces.

Moist sulfur dioxide reduces ferric salts into ferrous salts.

\[2\,\mathrm{Fe^{3+}} + \mathrm{SO_2} + 2\,\mathrm{H_2O} \rightarrow 2\,\mathrm{Fe^{2+}} + \mathrm{SO_4^{2-}} + 4\,\mathrm{H^{+}}\]

Moist sulfur dioxide decolourises acidified potassium permanganate(VII) solution.

\[2\,\mathrm{KMnO_4} + 5\,\mathrm{SO_2} + 2\,\mathrm{H_2O} \rightarrow \mathrm{K_2SO_4} + 2\,\mathrm{MnSO_4} + 2\,\mathrm{H_2SO_4}\]

The loss of the purple colour is a standard laboratory test for a reducing gas, since the deep purple \(\mathrm{MnO_4^{-}}\) ion is reduced to the almost colourless \(\mathrm{Mn^{2+}}\) ion.

Moist sulfur dioxide reduces halogens to halogen acids.

\[\mathrm{I_2} + \mathrm{SO_2} + 2\,\mathrm{H_2O} \rightarrow \mathrm{H_2SO_4} + 2\,\mathrm{HI}\]

In the reaction \(2\,\mathrm{KMnO_4} + 5\,\mathrm{SO_2} + 2\,\mathrm{H_2O} \rightarrow \mathrm{K_2SO_4} + 2\,\mathrm{MnSO_4} + 2\,\mathrm{H_2SO_4}\), identify the substance oxidised and the substance reduced. Show the change in oxidation number of sulfur and of manganese, and state the number of electrons transferred per mole of \(\mathrm{KMnO_4}\).

d. Structure of \(\mathrm{SO_2}\)

Sulfur dioxide is angular, with an O—S—O bond angle of 119.5°. The S—O double bond arises from \(\mathrm{d\pi\!-\!p\pi}\) bonding, in which a filled p orbital of oxygen overlaps sideways with an empty d orbital of sulfur — an option open to sulfur precisely because its valence shell has d orbitals, and closed to oxygen. It is a resonance hybrid of two canonical forms, shown in Fig. 7.14.

Diagram

Fig. 7.14 – Two canonical forms of SO₂ side by side with a double-headed resonance arrow between them; each drawn bent at sulfur with one S=O and one S–O, lone pairs marked on S and on both O atoms; mark the O–S–O bond angle 119.5°.

e. Uses

Sulfur dioxide is used in refining of petroleum and sugar, in bleaching wool and silk, as an anti-chlor and disinfectant, and as a preservative.

It is used in the manufacture of \(\mathrm{H_2SO_4}\) and \(\mathrm{NaHSO_3}\).

Liquid \(\mathrm{SO_2}\) is used as a solvent to dissolve a number of organic and inorganic chemicals.

7.10.2 Sulfuric Acid, \(\mathrm{H_2SO_4}\)

a. Preparation

Sulfuric acid is manufactured by the Contact process, which involves the following three steps. The plant is drawn in Fig. 7.15, and the chemical sequence is summarised in FC 7.5.

Diagram

Fig. 7.15 – Flow diagram for the manufacture of sulfuric acid: sulfur burner with air inlet → dust precipitator → washing and cooling tower with water spray → drying tower with conc. H₂SO₄ spray → arsenic purifier containing gelatinous hydrated ferric oxide → preheater → catalytic converter packed with V₂O₅ → absorption tower with conc. H₂SO₄ spray → oleum outlet; mark waste water and waste acid outlets.

Flowchart

FC 7.5 – Contact process: S or FeS₂ roasted in air → SO₂ → purification of the gas (dust, moisture and arsenic removed, since arsenic poisons the catalyst) → catalytic oxidation over V₂O₅ at 720 K and 2 bar → SO₃ → absorbed in concentrated H₂SO₄ → oleum (H₂S₂O₇) → diluted with water → H₂SO₄ of the desired concentration, 96–98 % pure.

i. Formation of sulfur dioxide

Sulfur or sulfide ore (iron pyrites) on burning or roasting in air produces sulfur dioxide.

\[\mathrm{S}\text{(s)} + \mathrm{O_2}\text{(g)} \xrightarrow{\ \Delta\ } \mathrm{SO_2}\text{(g)}\]

\[4\,\mathrm{FeS_2}\text{(s)} + 11\,\mathrm{O_2}\text{(g)} \xrightarrow{\ \Delta\ } 2\,\mathrm{Fe_2O_3}\text{(s)} + 8\,\mathrm{SO_2}\text{(g)}\]

ii. Catalytic oxidation to sulfur trioxide

Sulfur dioxide is oxidised catalytically with oxygen to sulfur trioxide, in the presence of \(\mathrm{V_2O_5}\) catalyst.

\[2\,\mathrm{SO_2}\text{(g)} + \mathrm{O_2} \xrightarrow{\ \mathrm{V_2O_5}\ } 2\,\mathrm{SO_3}\text{(g)}\]

The reaction is exothermic and reversible, and the forward reaction leads to a decrease in volume. Therefore low temperature (720 K) and high pressure (2 bar) are favourable conditions for the maximum yield of \(\mathrm{SO_3}\).

Count the molecules on each side and the reasoning becomes clear: three molecules of gas become two, so squeezing the mixture pushes the equilibrium forward. Since heat is given out, keeping the temperature down also favours the product. In practice the temperature cannot be lowered too far, because the reaction would then become impractically slow — which is why a catalyst is used to obtain a workable rate at a moderate temperature.

NotePrinciple / Law

(Le Chatelier’s principle — if a system at equilibrium is disturbed by a change of concentration, pressure or temperature, the equilibrium shifts in the direction that opposes the change.)

iii. Absorption to oleum, and dilution

Sulfur trioxide gas (from the catalytic converter) is absorbed in concentrated \(\mathrm{H_2SO_4}\) to produce oleum. Dilution of oleum with water gives sulfuric acid of the desired concentration.

\[\mathrm{SO_3}\text{(g)} + \mathrm{H_2SO_4} \rightarrow \mathrm{H_2S_2O_7}\ (\text{oleum})\]

\[\mathrm{H_2S_2O_7} + \mathrm{H_2O} \rightarrow 2\,\mathrm{H_2SO_4}\]

The sulfuric acid obtained by the contact process is 96–98 % pure.

Q. Sulfur trioxide reacts with water to give sulfuric acid directly. Why does the plant take the roundabout route through oleum instead?

A. Because \(\mathrm{SO_3}\) reacts with water so violently that it produces a dense mist of fine sulfuric acid droplets, which is difficult to condense and escapes from the plant. Absorbing the gas in concentrated \(\mathrm{H_2SO_4}\) gives oleum smoothly, and the oleum can then be diluted with water in a controlled way.

Calculate the mass of sulfuric acid that can be obtained from 6.40 kg of sulfur by the contact process, assuming complete conversion at every stage. (Molar masses in \(\text{g mol}^{-1}\): S = 32.06, H = 1.008, O = 16.00)

Follow the sulfur atom through the three steps of FC 7.5:

\[\mathrm{S} \rightarrow \mathrm{SO_2} \rightarrow \mathrm{SO_3} \rightarrow \mathrm{H_2SO_4}\]

At no stage is a sulfur atom lost or doubled up, so one mole of sulfur finally gives one mole of sulfuric acid.

Moles of sulfur taken:

\[n(\mathrm{S}) = \frac{6400\ \text{g}}{32.06\ \text{g mol}^{-1}} = 199.6\ \text{mol}\]

Therefore moles of \(\mathrm{H_2SO_4}\) formed \(= 199.6\ \text{mol}\).

Molar mass of \(\mathrm{H_2SO_4} = 2(1.008) + 32.06 + 4(16.00) = 98.08\ \text{g mol}^{-1}\)

Mass of sulfuric acid:

\[m = 199.6 \times 98.08 = 1.958 \times 10^{4}\ \text{g} \approx 19.6\ \text{kg}\]

So about 19.6 kg of sulfuric acid comes from every 6.40 kg of sulfur. Notice that the answer needed no balancing of the individual equations — tracking the one atom that is conserved is quicker and less error-prone, and this trick works for any multi-step industrial process.

b. Physical properties of \(\mathrm{H_2SO_4}\)

Sulfuric acid is a colourless, dense, oily liquid.

It has a density (specific gravity) of 1.84 g/cm³ at 298 K.

It freezes at 283 K and boils at 611 K.

It is highly corrosive and produces severe burns on the skin.

Sulfuric acid dissolves in water with the evolution of a large quantity of heat. Hence care must be taken while preparing a solution of sulfuric acid from the concentrated acid: concentrated \(\mathrm{H_2SO_4}\) must be added slowly to water with constant stirring, keeping the beaker in a water bath. Never the reverse. If water is poured onto the concentrated acid, the heat is released in the small volume of water sitting on top, which boils instantly and throws concentrated acid out of the vessel.

c. Chemical Properties
i. Acidic property

Sulfuric acid ionises in aqueous solution in two steps.

\[\mathrm{H_2SO_4}\text{(aq)} + \mathrm{H_2O}\text{(l)} \rightarrow \mathrm{H_3O^{+}}\text{(aq)} + \mathrm{HSO_4^{-}}\text{(aq)} \qquad K_a > 10\]

\[\mathrm{HSO_4^{-}}\text{(aq)} + \mathrm{H_2O}\text{(l)} \rightarrow \mathrm{H_3O^{+}}\text{(aq)} + \mathrm{SO_4^{2-}}\text{(aq)} \qquad K_a = 1.2 \times 10^{-2}\]

The greater value of \(K_a\) (\(K_a > 10\)) means that \(\mathrm{H_2SO_4}\) is largely dissociated into \(\mathrm{H^{+}}\) and \(\mathrm{HSO_4^{-}}\) ions. Thus \(\mathrm{H_2SO_4}\) is a strong acid.

Compare the two constants and you can see that the second proton is released far less readily than the first. Pulling a positive proton away from an already negative \(\mathrm{HSO_4^{-}}\) ion is much harder than pulling it from the neutral molecule.

ii. Reaction with metals and nonmetals (oxidising property)

Metals and nonmetals are both oxidised by hot, concentrated sulfuric acid, which itself gets reduced to \(\mathrm{SO_2}\).

\[\mathrm{Cu} + 2\,\mathrm{H_2SO_4}\ (\text{Conc.}) \rightarrow \mathrm{CuSO_4} + \mathrm{SO_2} + 2\,\mathrm{H_2O}\]

\[\mathrm{S} + 2\,\mathrm{H_2SO_4}\ (\text{Conc.}) \rightarrow 3\,\mathrm{SO_2} + 2\,\mathrm{H_2O}\]

\[\mathrm{C} + 2\,\mathrm{H_2SO_4}\ (\text{Conc.}) \rightarrow \mathrm{CO_2} + 2\,\mathrm{SO_2} + 2\,\mathrm{H_2O}\]

The oxidising properties of sulfuric acid depend on its concentration and temperature. In dilute solutions at room temperature, \(\mathrm{H_2SO_4}\) behaves like HCl, oxidising only those metals that stand above hydrogen in the e.m.f. series.

\[\mathrm{Fe}\text{(s)} + 2\,\mathrm{H^{+}}\text{(aq)} \rightarrow \mathrm{Fe^{2+}}\text{(aq)} + \mathrm{H_2}\text{(g)}\]

Hot, concentrated \(\mathrm{H_2SO_4}\) is a better oxidising agent than the dilute, cold acid. It oxidises metals like copper, which dilute acid leaves untouched.

iii. Dehydrating property

Concentrated sulfuric acid is a strong dehydrating agent.

Sulfuric acid removes water from sugar and carbohydrates. The carbon left behind is called sugar charcoal, and the process is called charring.

\[\mathrm{C_{12}H_{22}O_{11}} \xrightarrow{\ \text{conc. } \mathrm{H_2SO_4}\ } 12\,\mathrm{C} + 11\,\mathrm{H_2O}\]

The acid does not simply drive off water that was already present — it strips hydrogen and oxygen atoms out of the sugar molecule itself in the ratio in which they occur in water, which is why a white solid turns into a rising black column of carbon.

iv. Reaction with salts

Concentrated sulfuric acid decomposes the salts of more volatile acids to the corresponding acid.

\[\mathrm{NaCl} + \mathrm{H_2SO_4} \rightarrow \mathrm{NaHSO_4} + \mathrm{HCl}\]

\[\mathrm{KNO_3} + \mathrm{H_2SO_4} \rightarrow \mathrm{KHSO_4} + \mathrm{HNO_3}\]

\[\mathrm{CaF_2} + \mathrm{H_2SO_4} \rightarrow \mathrm{CaSO_4} + 2\,\mathrm{HF}\]

Sulfuric acid succeeds here not because it is the stronger acid in every sense, but because it has a very high boiling point (611 K). On heating, the volatile acid escapes as a gas and cannot come back, so the reaction is driven to completion.

concentrated sulfuric acid (what it does → three distinct jobs: it oxidises metals and nonmetals when hot, it dehydrates carbohydrates and other organic matter, and it displaces volatile acids from their salts → the conditions it needs → the oxidising role needs the acid hot and concentrated, while dilute cold acid behaves merely like HCl → what it is used for → preparing HCl, \(\mathrm{HNO_3}\) and HF from their salts, drying gases that do not react with it, and as the standard laboratory oxidising acid. Handle only with the water-bath dilution rule above.)

dilute vs hot concentrated sulfuric acid

Feature Dilute \(\mathrm{H_2SO_4}\) (cold) Hot concentrated \(\mathrm{H_2SO_4}\)
Chief role Ordinary strong acid, behaves like HCl Powerful oxidising agent
Action on copper No reaction \(\mathrm{Cu} \rightarrow \mathrm{CuSO_4}\), acid reduced to \(\mathrm{SO_2}\)
Action on iron \(\mathrm{Fe} + 2\,\mathrm{H^{+}} \rightarrow \mathrm{Fe^{2+}} + \mathrm{H_2}\) Oxidises the metal, no \(\mathrm{H_2}\) liberated
Gas evolved with metals Hydrogen Sulfur dioxide
Dehydrating action Negligible Strong; chars sugar to carbon

What is the action of concentrated \(\mathrm{H_2SO_4}\) on (a) HBr and (b) HI?

In both cases the concentrated acid acts as an oxidising agent, and the halide ion is oxidised to the free halogen.

  1. Concentrated sulfuric acid oxidises hydrobromic acid to bromine.

\[2\,\mathrm{HBr} + \mathrm{H_2SO_4} \rightarrow \mathrm{Br_2} + \mathrm{SO_2} + 2\,\mathrm{H_2O}\]

  1. It oxidises hydroiodic acid to iodine.

\[2\,\mathrm{HI} + \mathrm{H_2SO_4} \rightarrow \mathrm{I_2} + \mathrm{SO_2} + 2\,\mathrm{H_2O}\]

This is also the reason why HBr and HI cannot be prepared from their salts using concentrated sulfuric acid, although HCl can — the acid destroys the very product it liberates.

97.46 g of zinc sulfide is roasted completely in air according to \(2\,\mathrm{ZnS} + 3\,\mathrm{O_2} \rightarrow 2\,\mathrm{ZnO} + 2\,\mathrm{SO_2}\). Calculate (i) the volume of \(\mathrm{SO_2}\) produced at STP and (ii) the mass of sulfuric acid that this \(\mathrm{SO_2}\) could yield if it were passed on to a contact process plant working with complete conversion. (Molar masses in \(\text{g mol}^{-1}\): Zn = 65.4, S = 32.06, O = 16.00, H = 1.008; molar volume at STP = 22.4 L \(\text{mol}^{-1}\))

d. Uses

Sulfuric acid is a very important industrial chemical. It is used in the manufacture of fertilizers — for example, ammonium sulfate and superphosphate.

It is used in the manufacture of pigments, paints and dyestuff intermediates.

It is used in petroleum refining and in the detergent industry.

It is used in metallurgy, for cleaning of metals, electroplating and galvanising.

It is used in storage batteries, as a laboratory reagent, and in the manufacture of nitrocellulose products.

7.11 Chlorine and Compounds of Chlorine

7.11.1 Chlorine

Chlorine was discovered by Scheele, a German-Swedish chemist, in 1774, by the action of HCl on \(\mathrm{MnO_2}\). In 1810, Davy established its elementary nature and suggested the name chlorine on account of its colour, from the Greek chloros, meaning greenish yellow.

Scheele obtained the greenish gas in 1774 but did not claim it as a new element; the chemistry of his day assumed that every acid contained oxygen, so the gas was long taken to be a compound of oxygen. It fell to Humphry Davy, more than thirty years later, to show that no amount of effort could extract oxygen from it, and that the substance was an element in its own right. He named it for the one property nobody could miss — its colour.

a. Preparation

1. Chlorine can be prepared by the oxidation of hydrochloric acid with any of the following oxidising agents.

i. Manganese dioxide

\[\mathrm{MnO_2} + 4\,\mathrm{HCl} \rightarrow \mathrm{MnCl_2} + \mathrm{Cl_2} + 2\,\mathrm{H_2O}\]

ii. Potassium permanganate

\[2\,\mathrm{KMnO_4} + 16\,\mathrm{HCl} \rightarrow 2\,\mathrm{KCl} + 2\,\mathrm{MnCl_2} + 8\,\mathrm{H_2O} + 5\,\mathrm{Cl_2}\]

In both reactions the chloride ion loses an electron and is oxidised to free chlorine, while manganese is reduced — from +4 to +2 in the first case and from +7 to +2 in the second. Potassium permanganate is the more powerful oxidant, which is why it needs no heating.

2. Chlorine can also be prepared by the action of concentrated sulfuric acid on a mixture of sodium chloride (common salt) and manganese dioxide. The reaction takes place in two steps.

\[4\,\mathrm{NaCl} + 4\,\mathrm{H_2SO_4} \rightarrow 4\,\mathrm{NaHSO_4} + 4\,\mathrm{HCl}\]

\[\mathrm{MnO_2} + 4\,\mathrm{HCl} \rightarrow \mathrm{MnCl_2} + 2\,\mathrm{H_2O} + \mathrm{Cl_2}\]

Adding the two steps gives the overall change:

\[4\,\mathrm{NaCl} + \mathrm{MnO_2} + 4\,\mathrm{H_2SO_4} \rightarrow 4\,\mathrm{NaHSO_4} + \mathrm{MnCl_2} + 2\,\mathrm{H_2O} + \mathrm{Cl_2}\]

The sulfuric acid first sets HCl free from common salt, and the manganese dioxide then oxidises that HCl — so the second method is really the first method with the hydrochloric acid generated in the flask.

Calculate the volume of chlorine, measured at STP, obtained when 8.69 g of manganese dioxide reacts completely with excess concentrated hydrochloric acid. (Molar masses in \(\text{g mol}^{-1}\): Mn = 54.94, O = 16.00; molar volume at STP = 22.4 L \(\text{mol}^{-1}\))

The balanced equation is

\[\mathrm{MnO_2} + 4\,\mathrm{HCl} \rightarrow \mathrm{MnCl_2} + \mathrm{Cl_2} + 2\,\mathrm{H_2O}\]

One mole of \(\mathrm{MnO_2}\) gives one mole of \(\mathrm{Cl_2}\).

Molar mass of \(\mathrm{MnO_2} = 54.94 + 2(16.00) = 86.94\ \text{g mol}^{-1}\)

Moles of \(\mathrm{MnO_2}\) taken:

\[n = \frac{8.69}{86.94} = 0.100\ \text{mol}\]

Since the mole ratio is 1 : 1, moles of \(\mathrm{Cl_2}\) formed \(= 0.100\ \text{mol}\).

Volume at STP:

\[V = 0.100 \times 22.4 = 2.24\ \text{L}\]

Hydrochloric acid is in excess, so manganese dioxide is the limiting reagent and decides the yield. Always check which reactant runs out first before using a mole ratio.

b. Manufacture of Chlorine
i. Deacon’s process

Chlorine is manufactured by the oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of \(\mathrm{CuCl_2}\) as catalyst at 723 K.

\[4\,\mathrm{HCl} + \mathrm{O_2} \xrightarrow{\ \mathrm{CuCl_2}\ } 2\,\mathrm{Cl_2} + 2\,\mathrm{H_2O}\]

ii. Electrolytic process

By the electrolysis of brine (concentrated NaCl solution), chlorine is liberated at the anode.

\[\mathrm{NaCl} \rightleftharpoons \mathrm{Na^{+}} + \mathrm{Cl^{-}}\]

At the cathode:

\[2\,\mathrm{H_2O} + 2\,\mathrm{e^{-}} \rightarrow \mathrm{H_2} + 2\,\mathrm{OH^{-}}\]

\[\mathrm{Na^{+}} + \mathrm{OH^{-}} \rightarrow \mathrm{NaOH}\]

At the anode:

\[\mathrm{Cl^{-}} \rightarrow \mathrm{Cl} + \mathrm{e^{-}}\]

\[\mathrm{Cl} + \mathrm{Cl} \rightarrow \mathrm{Cl_2}\]

Three saleable products come out of this one cell — chlorine at the anode, hydrogen at the cathode and sodium hydroxide left in the solution — which is why the process is the backbone of the chemical industry. The arrangement is shown in Fig. 7.16 and the products are traced in FC 7.6.

Diagram

Fig. 7.16 – Electrolytic cell for brine: tank of concentrated NaCl solution with an anode and a cathode, chlorine gas bubbling off at the anode, hydrogen gas at the cathode, and NaOH accumulating in the solution; label the ion movements Cl⁻ → anode and Na⁺ → cathode.

Flowchart

FC 7.6 – Electrolysis of brine: concentrated NaCl solution → dissociates into Na⁺ and Cl⁻ → branch 1: Cl⁻ migrates to anode → loses an electron → Cl atoms pair up → Cl₂ gas collected; branch 2: at the cathode water is reduced → H₂ gas + OH⁻ → OH⁻ combines with Na⁺ → NaOH remains in solution.

c. Physical Properties of Chlorine

Chlorine is a greenish-yellow gas having a pungent and suffocating odour.

It is poisonous in nature.

It can be easily liquified into a greenish yellow liquid, which boils at 239 K.

It dissolves in water to give chlorine water.

It is about 2.5 times heavier than air, which is why the gas collects near the floor and is always handled in a fume cupboard.

d. Chemical Properties of Chlorine
i. Reaction with metals

Chlorine reacts with metals to form chlorides.

\[2\,\mathrm{Al} + 3\,\mathrm{Cl_2} \rightarrow 2\,\mathrm{AlCl_3}\]

\[2\,\mathrm{Na} + \mathrm{Cl_2} \rightarrow 2\,\mathrm{NaCl}\]

\[2\,\mathrm{Fe} + 3\,\mathrm{Cl_2} \rightarrow 2\,\mathrm{FeCl_3}\]

Notice that iron is taken to its higher (+3) state. Chlorine is a strong enough oxidant to push a metal to its upper oxidation state, which is a useful point of contrast with dilute acids, where iron stops at +2.

ii. Reaction with nonmetals

Chlorine reacts with nonmetals to form their chlorides.

\[\mathrm{P_4} + 6\,\mathrm{Cl_2} \rightarrow 4\,\mathrm{PCl_3}\]

\[\mathrm{S_8} + 4\,\mathrm{Cl_2} \rightarrow 4\,\mathrm{S_2Cl_2}\]

iii. Affinity for hydrogen

Chlorine has a great affinity for hydrogen. It reacts with hydrogen, and with compounds containing hydrogen, to form HCl.

\[\mathrm{H_2} + \mathrm{Cl_2} \rightarrow 2\,\mathrm{HCl}\]

\[\mathrm{H_2S} + \mathrm{Cl_2} \rightarrow 2\,\mathrm{HCl} + \mathrm{S}\]

In the second reaction chlorine strips the hydrogen out of hydrogen sulfide and leaves sulfur behind — a direct demonstration of that affinity.

iv. Reaction with \(\mathrm{NH_3}\)

Chlorine, when reacted with an excess of ammonia, gives ammonium chloride and nitrogen.

\[8\,\mathrm{NH_3}\ (\text{Excess}) + 3\,\mathrm{Cl_2} \rightarrow 6\,\mathrm{NH_4Cl} + \mathrm{N_2}\]

An excess of chlorine reacts with ammonia to give nitrogen trichloride, which is explosive.

\[\mathrm{NH_3} + 3\,\mathrm{Cl_2}\ (\text{Excess}) \rightarrow \mathrm{NCl_3} + 3\,\mathrm{HCl}\]

Q. Why does the same pair of reactants give two completely different products depending on which one is in excess?

A. Because chlorine attacks the hydrogen atoms of ammonia one at a time. When ammonia is plentiful, the HCl formed is immediately mopped up by spare ammonia as \(\mathrm{NH_4Cl}\), and the nitrogen released pairs up as \(\mathrm{N_2}\). When chlorine is plentiful, there is no spare ammonia to take up the HCl, and the chlorine keeps replacing hydrogen until all three have gone, leaving \(\mathrm{NCl_3}\).

v. Reaction with alkali

Chlorine reacts with cold and dilute alkali to produce a mixture of chloride and hypochlorite. When reacted with hot concentrated alkali, chloride and chlorate are produced.

\[2\,\mathrm{NaOH}\ (\text{Cold and dilute}) + \mathrm{Cl_2} \rightarrow \mathrm{NaCl} + \mathrm{NaOCl} + \mathrm{H_2O}\]

\[6\,\mathrm{NaOH}\ (\text{Hot and conc.}) + 3\,\mathrm{Cl_2} \rightarrow 5\,\mathrm{NaCl} + \mathrm{NaClO_3} + 3\,\mathrm{H_2O}\]

Both are disproportionation reactions: the chlorine that entered as \(\mathrm{Cl_2}\) (oxidation state 0) leaves partly as chloride (−1) and partly as hypochlorite (+1) or chlorate (+5).

Chlorine, when reacted with dry slaked lime, gives bleaching powder.

\[2\,\mathrm{Ca(OH)_2} + 2\,\mathrm{Cl_2} \rightarrow \mathrm{Ca(OCl)_2} + \mathrm{CaCl_2} + 2\,\mathrm{H_2O}\]

chlorine with cold dilute alkali vs hot concentrated alkali

Feature Cold, dilute alkali Hot, concentrated alkali
Products Chloride + hypochlorite Chloride + chlorate
Example \(2\,\mathrm{NaOH} + \mathrm{Cl_2} \rightarrow \mathrm{NaCl} + \mathrm{NaOCl} + \mathrm{H_2O}\) \(6\,\mathrm{NaOH} + 3\,\mathrm{Cl_2} \rightarrow 5\,\mathrm{NaCl} + \mathrm{NaClO_3} + 3\,\mathrm{H_2O}\)
Oxidation states of Cl formed −1 and +1 −1 and +5
Ratio of the two products 1 : 1 5 : 1
Type of reaction Disproportionation Disproportionation
vi. Reaction with hydrocarbons

Chlorine reacts with saturated hydrocarbons to give substitution products, and with unsaturated hydrocarbons gives addition products.

\[\mathrm{CH_4} + \mathrm{Cl_2} \xrightarrow{\ \text{U.V.}\ } \mathrm{CH_3Cl} + \mathrm{HCl}\]

Methane gives methyl chloride, and a hydrogen atom has been replaced — that is substitution.

\[\mathrm{H_2C}\!=\!\mathrm{CH_2} + \mathrm{Cl_2} \rightarrow \mathrm{CH_2Cl\!-\!CH_2Cl}\]

Ethene gives 1,2-dichloroethane, and nothing has been lost — the double bond has simply opened to accept both chlorine atoms. That is addition.

vii. Oxidising property

Chlorine oxidises ferrous salts to ferric salts and sulfites to sulfates.

\[2\,\mathrm{FeSO_4} + \mathrm{H_2SO_4} + \mathrm{Cl_2} \rightarrow \mathrm{Fe_2(SO_4)_3} + 2\,\mathrm{HCl}\]

\[\mathrm{Na_2SO_3} + \mathrm{Cl_2} + \mathrm{H_2O} \rightarrow \mathrm{Na_2SO_4} + 2\,\mathrm{HCl}\]

It oxidises sulfur dioxide to sulfur trioxide, and iodine to iodate. In the presence of water they form sulfuric acid and iodic acid respectively.

\[\mathrm{SO_2} + 2\,\mathrm{H_2O} + \mathrm{Cl_2} \rightarrow \mathrm{H_2SO_4} + 2\,\mathrm{HCl}\]

\[\mathrm{I_2} + 6\,\mathrm{H_2O} + 5\,\mathrm{Cl_2} \rightarrow 2\,\mathrm{HIO_3} + 10\,\mathrm{HCl}\]

The last reaction is worth pausing over: chlorine, being the more electronegative halogen, forces iodine all the way up to the +5 state. A halogen higher in the group can always oxidise one below it.

viii. Bleaching property

Chlorine requires the presence of moisture (water) for bleaching. It liberates nascent oxygen from water, which is responsible for its oxidising and bleaching property.

\[\mathrm{Cl_2} + \mathrm{H_2O} \rightarrow \mathrm{HCl} + \mathrm{HOCl}\]

\[\mathrm{HOCl} \rightarrow \mathrm{HCl} + [\mathrm{O}]\]

Chlorine bleaches vegetable matter or coloured organic matter in the presence of moisture to colourless matter.

\[\text{Coloured organic matter} + [\mathrm{O}] \rightarrow \text{Colourless organic matter}\]

Compare this with ozone, which bleaches without moisture and is therefore called a dry bleach. Chlorine has to make its oxidising agent out of water first.

  1. Give the reasons for the bleaching action of chlorine. (ii) Name the two gases used in war.

Bleaching by chlorine is permanent — the colour does not return on standing, because the coloured compound has been oxidised and destroyed rather than merely covered up. It bleaches cotton fabrics, wood pulp, litmus and the like. However, chlorine is not used to bleach delicate materials such as silk and wool, because it is a strong bleaching and oxidising agent, and this dual action will damage the base material along with the stain.

e. Uses

Chlorine is used for the purification (sterilizing) of drinking water.

It is used for bleaching wood pulp required for the manufacture of paper and rayon, and for bleaching cotton and textiles.

It is used for the extraction of metals like gold and platinum.

It is used in the manufacture of dyes, drugs and organic compounds such as \(\mathrm{CCl_4}\), \(\mathrm{CHCl_3}\), DDT, refrigerants, etc.

It is used in the preparation of poisonous gases such as phosgene (\(\mathrm{COCl_2}\)), tear gas (\(\mathrm{CCl_3NO_2}\)) and mustard gas (\(\mathrm{ClCH_2CH_2SCH_2CH_2Cl}\)).

Chlorine is passed separately into (i) cold dilute sodium hydroxide and (ii) hot concentrated sodium hydroxide. Write the balanced equation in each case, assign the oxidation number of chlorine in every chlorine-containing species, and explain why both reactions are described as disproportionation.

7.11.2 Hydrogen Chloride

Hydrogen chloride was prepared by Glauber in 1648 by heating common salt with concentrated sulfuric acid. Davy, in 1810, showed that it is a compound of hydrogen and chlorine.

The gap between those two dates is worth noticing. Glauber made hydrogen chloride more than a century and a half before anyone knew what it was made of, and for most of that time it was assumed — like chlorine itself — to contain oxygen, since every acid was believed to. Davy’s work settled both questions at once: chlorine was an element, and the acid built from it contained no oxygen at all.

a. Preparation

In the laboratory, hydrogen chloride is prepared by heating sodium chloride (common salt) with concentrated sulfuric acid.

\[\mathrm{NaCl} + \mathrm{H_2SO_4} \xrightarrow{\ 420\ \text{K}\ } \mathrm{NaHSO_4} + \mathrm{HCl}\]

\[\mathrm{NaHSO_4} + \mathrm{NaCl} \xrightarrow{\ 420\ \text{K}\ } \mathrm{Na_2SO_4} + \mathrm{HCl}\]

HCl gas can be dried by passing it through concentrated sulfuric acid. This works because sulfuric acid absorbs water strongly but does not react with hydrogen chloride — recall from section 7.10 that concentrated \(\mathrm{H_2SO_4}\) oxidises HBr and HI, but not HCl, which is why the same drying trick cannot be used for those two gases.

b. Physical properties of HCl

Hydrogen chloride is a colourless and pungent smelling gas.

It can be easily liquified to a colourless liquid (b.p. 189 K), which freezes to a white crystalline solid (m.p. 159 K).

It is highly soluble in water.

c. Chemical properties
i. Acidic property

Hydrogen chloride is highly soluble in water and ionises as follows:

\[\mathrm{HCl}\text{(g)} + \mathrm{H_2O}\text{(l)} \rightarrow \mathrm{H_3O^{+}}\text{(aq)} + \mathrm{Cl^{-}}\text{(aq)} \qquad K_a = 10^{7}\]

The aqueous solution of HCl gas is called hydrochloric acid. The high value of the dissociation constant (\(K_a\)) indicates that it is a strong acid in water — the equilibrium lies so far to the right that practically every molecule has given up its proton.

ii. Reaction with \(\mathrm{NH_3}\)

Hydrochloric acid reacts with ammonia and gives white fumes of ammonium chloride.

\[\mathrm{NH_3} + \mathrm{HCl} \rightarrow \mathrm{NH_4Cl}\]

The dense white cloud formed when the two gases meet is the standard test for either of them.

iii. Reaction with noble metals

When three parts of concentrated HCl and one part of concentrated \(\mathrm{HNO_3}\) are mixed, aqua regia is formed.

Noble metals like gold and platinum get dissolved in aqua regia.

\[\mathrm{Au} + 4\,\mathrm{H^{+}} + \mathrm{NO_3^{-}} + 4\,\mathrm{Cl^{-}} \rightarrow \mathrm{AuCl_4^{-}} + \mathrm{NO} + 2\,\mathrm{H_2O}\]

\[3\,\mathrm{Pt} + 16\,\mathrm{H^{+}} + 4\,\mathrm{NO_3^{-}} + 18\,\mathrm{Cl^{-}} \rightarrow 3\,\mathrm{PtCl_6^{2-}} + 4\,\mathrm{NO} + 8\,\mathrm{H_2O}\]

aqua regia (what it does → dissolves gold and platinum, which neither concentrated hydrochloric acid nor concentrated nitric acid can dissolve on its own → the conditions it needs → freshly mixed in the ratio of three parts concentrated HCl to one part concentrated \(\mathrm{HNO_3}\); it loses strength on standing → what it is used for → refining and assaying precious metals, and recovering gold from electronic scrap. The trick is teamwork: nitric acid oxidises a little of the metal, and the chloride ions immediately lock the metal ion away as \(\mathrm{AuCl_4^{-}}\) or \(\mathrm{PtCl_6^{2-}}\), so more metal keeps dissolving.)

d. Uses

Hydrochloric acid is used as an important laboratory reagent, in cleaning and pickling metal surfaces, and in the preparation of chlorides and of aqua regia as described above.

Concentrated sulfuric acid can be used to prepare HCl from sodium chloride and to dry the gas afterwards, but it cannot be used to prepare HBr from sodium bromide. Explain the difference, writing the equation that shows what happens in the bromide case, and suggest what property an alternative acid would need in order to work.

7.12 Interhalogen Compounds

Which type of bonds do halogens form with other elements? Does \(\mathrm{BrF_5}\) obey the octet rule? What is the oxidation state of Br in \(\mathrm{BrF_5}\)? How many electrons do halogens require to complete their octet? What is the shape of \(\mathrm{ClF_3}\)?

We have seen that all halogen molecules are diatomic. They form binary compounds with hydrogen; with oxygen they form oxyacids; and with halogens they form interhalogen compounds.

Although all halogens belong to the same group, they have different electronegativities. Due to this difference in electronegativity, two or more halogen atoms combine to form species which may be ionic or neutral. The neutral molecules are called interhalogen compounds — for example, ClF and \(\mathrm{BrF_3}\).

NoteDefinition

An interhalogen compound is a compound formed by the combination of atoms of different halogens. It is regarded as the halide of the more electropositive halogen, and a given halogen forms an interhalogen compound only with a halogen of lesser electronegativity.

The ionic species are worth noting alongside. Ions formed by the combination of different halogens are called polyhalide ions, or interhalogen ions — for example \(\mathrm{K^{+}[Cl_3^{-}]}\) and \(\mathrm{[NH_4]^{+}[I_5]^{-}}\), which contain the \(\mathrm{Cl_3^{-}}\) and \(\mathrm{I_5^{-}}\) ions.

Table 7.9 : Various types of interhalogen compounds

Element Fluoride Chloride Bromide Iodide
Chlorine ClF, \(\mathrm{ClF_3}\), \(\mathrm{ClF_5}\) − − −
Bromine BrF, \(\mathrm{BrF_3}\), \(\mathrm{BrF_5}\) BrCl − −
Iodine IF, \(\mathrm{IF_3}\), \(\mathrm{IF_5}\), \(\mathrm{IF_7}\) ICl, \(\mathrm{ICl_3}\) IBr −

Read Table 7.9 down the left column and across the top and the rule stated in the definition shows itself. Every entry pairs a larger halogen with a smaller one; there is no fluorine row and no iodide column, because fluorine is never the central atom and iodine is never the attached one.

Chlorine and fluorine combine to form interhalogen compounds. Will the halide ion be of chlorine or of fluorine? Why does fluorine combine with other halogens to form the maximum number of fluorides?

7.12.1 Classification

Depending on their composition, interhalogen compounds are classified into four types.

Type Example
\(\mathrm{XX'}\) ClF, BrF, BrCl, ICl, IBr
\(\mathrm{XX'_3}\) \(\mathrm{ClF_3}\), \(\mathrm{BrF_3}\), \(\mathrm{IF_3}\)
\(\mathrm{XX'_5}\) \(\mathrm{ClF_5}\), \(\mathrm{BrF_5}\), \(\mathrm{IF_5}\)
\(\mathrm{XX'_7}\) \(\mathrm{IF_7}\)

In the general formula \(\mathrm{XX'_n}\), X is the halogen having the larger size and is more electropositive. X’ is the halogen having the smaller size and more electronegativity.

7.12.2 General Characteristics of Interhalogen Compounds

1. The compound is considered as the halide of X. For example, in ClF the halogen having the larger size is chlorine, which is more electropositive than F, and hence the interhalogen compound is named as chlorine monofluoride. Here \(n\) is the number of atoms of X’ attached to X.

As the ratio [radius of X : radius of X’] increases, the value of \(n\) also increases. This is simply a matter of room — a large central atom with small partners can fit more of them around itself, which is why \(\mathrm{IF_7}\) exists while \(\mathrm{ClF_7}\) does not.

2. Interhalogen compounds have an even number of atoms — 2, 4, 6 or 8. For example, \(\mathrm{ClF_3}\) has 4 atoms.

3. The properties of interhalogen compounds are generally intermediate between those of the halogens from which they are made.

4. The central halogen exhibits different oxidation states in different interhalogen compounds.

5. The number of X’ atoms in the compounds is always odd.

6. They are all diamagnetic — every electron in the molecule is paired.

Q. Why must the number of attached atoms X’ always be odd, and the total number of atoms therefore even?

A. Because the central halogen has seven valence electrons, an odd number. Each bond it forms uses up one of them, and the electrons left over must still pair up among themselves as lone pairs. That is possible only if an odd number of electrons has been used in bonding — so \(n\) is 1, 3, 5 or 7, and the molecule as a whole contains an even number of atoms.

What will be the names of the following compounds: ICl, BrF? Which halogen (X) will have the maximum number of the other halogen (X’) attached? Which halogen has a tendency to form more interhalogen compounds?

Which will be more reactive? (a) \(\mathrm{ClF_3}\) or ClF, (b) \(\mathrm{BrF_5}\) or BrF. Then complete the table.

Formula Name
ClF Chlorine monofluoride
\(\mathrm{ClF_3}\) ……
…… Chlorine pentafluoride
BrF ……
…… Bromine pentafluoride
ICl ……
\(\mathrm{ICl_3}\) ……

Q. Why are interhalogen compounds more reactive than the halogens they are made from?

A. In an X–X’ bond the two atoms differ in electronegativity, so the bond is polar and weaker than the bond in \(\mathrm{X_2}\) or \(\mathrm{X'_2}\), where both atoms pull equally. A weaker bond breaks more readily, so XX’ compounds are more reactive than either \(\mathrm{X_2}\) or \(\mathrm{X'_2}\).

Table 7.10 : States of interhalogen compounds at 25 °C

Type Compound State and appearance
\(\mathrm{XX'}\) ClF Colorless gas
BrF Pale brown gas
BrCl Gas
ICl Ruby red solid (\(\alpha\)-form); brown red solid (\(\beta\)-form)
IBr Black solid
\(\mathrm{XX'_3}\) \(\mathrm{ClF_3}\) Colorless gas
\(\mathrm{BrF_3}\) Yellow green liquid
\(\mathrm{IF_3}\) Yellow powder
\(\mathrm{ICl_3}\) Orange solid; dimerises to form \(\mathrm{I_2Cl_6}\) having Cl-bridges
\(\mathrm{XX'_5}\) \(\mathrm{IF_5}\) Colorless gas at R.T. but solid below 77 K
\(\mathrm{BrF_5}\) Colorless liquid
\(\mathrm{ClF_5}\) Colorless liquid
\(\mathrm{XX'_7}\) \(\mathrm{IF_7}\) Colorless gas
Methods of Preparation

Table : Methods of preparation of interhalogen compounds

Method \(\mathrm{XX'}\) \(\mathrm{XX'_3}\)
1. Direct combination Both in equal volumes: \(\mathrm{Cl_2} + \mathrm{F_2} \xrightarrow{\ 523\ \text{K},\ \text{Cu-vessel}\ } 2\,\mathrm{ClF}\); \(\mathrm{I_2} + \mathrm{Cl_2} \xrightarrow{\ \mathrm{HCl} + \mathrm{HNO_3},\ \text{boil}\ } 2\,\mathrm{ICl}\); \(\mathrm{Br_2} + \mathrm{Cl_2} \rightarrow 2\,\mathrm{BrCl}\) \(\mathrm{Cl_2} + 3\,\mathrm{F_2}\ (\text{excess}) \rightarrow 2\,\mathrm{ClF_3}\); \(\mathrm{Br_2} + 3\,\mathrm{F_2} \rightarrow 2\,\mathrm{BrF_3}\); \(\mathrm{I_2} + 3\,\mathrm{Cl_2}\ (\text{excess}) \rightarrow 2\,\mathrm{ICl_3}\)
2. Reaction of a halogen with an interhalogen compound \(\mathrm{Br_2} + \mathrm{BrF_3} \rightarrow 3\,\mathrm{BrF}\) \(\mathrm{Br_2} + \mathrm{ClF_3} \rightarrow 2\,\mathrm{BrF_3} + \mathrm{BrCl}\)
Special reaction for ICl \(\mathrm{I_2} + \mathrm{KClO_3} \xrightarrow{\ \Delta\ } \mathrm{ICl} + \mathrm{KIO_3}\) ——-

Look closely at the first row and you will see how the product is controlled: equal volumes of the two halogens give the 1 : 1 compound, while an excess of the smaller halogen drives the reaction on to \(\mathrm{XX'_3}\).

In the special reaction for ICl above, identify the oxidant and the reductant. Denote the oxidation states of the species.

Some Properties

Table : Some properties of interhalogen compounds

Reaction/Property \(\mathrm{XX'}\) \(\mathrm{XX'_3}\)
1 Thermal stability \(\mathrm{ClF} > \mathrm{ICl} > \mathrm{IBr} > \mathrm{BrCl} > \mathrm{BrF}\)
2 Hydrolysis gives oxoacids \(\mathrm{BrCl} + \mathrm{H_2O} \rightarrow \mathrm{HOBr} + \mathrm{HCl}\); \(5\,\mathrm{ICl} + 3\,\mathrm{H_2O} \rightarrow \mathrm{HIO_3} + 5\,\mathrm{HCl} + 2\,\mathrm{I_2}\) \(2\,\mathrm{ICl_3} + 3\,\mathrm{H_2O} \rightarrow \mathrm{HIO_3} + 5\,\mathrm{HCl} + \mathrm{ICl}\)
3 Disproportionation / autoionisation \(5\,\mathrm{BrF} \rightarrow 2\,\mathrm{Br_2} + \mathrm{BrF_5}\) \(2\,\mathrm{ClF_3} \rightarrow \mathrm{ClF_2^{+}} + \mathrm{ClF_4^{-}}\); \(\mathrm{ICl_3} \xrightarrow{\ 341\ \text{K}\ } \mathrm{ICl} + \mathrm{Cl_2}\)
4 Fluorination ——— \(\mathrm{U} + \mathrm{ClF_3} \rightarrow \mathrm{UF_6}\text{(l)} + \mathrm{ClF}\text{(g)}\)
5 Addition across olefins \(\mathrm{H_2C}\!=\!\mathrm{CH_2} + \mathrm{ICl} \rightarrow \mathrm{ICH_2\!-\!CH_2Cl}\)

In hydrolysis the more electropositive halogen ends up in the oxoacid and the more electronegative one leaves as the halide acid — which is exactly what the definition of an interhalogen compound predicts.

Liquid ICl also undergoes autoionization, like water, to form a cation and an anion:

\[2\,\mathrm{ICl} \rightleftharpoons \mathrm{I^{+}}\ (\text{solvent cation}) + \mathrm{ICl_2^{-}}\ (\text{solvent anion})\]

This is what makes some interhalogen compounds useful as nonaqueous solvents: like water, they can supply their own ions and dissolve salts.

Uses

Table : Uses of interhalogen compounds

\(\mathrm{XX'}\) \(\mathrm{XX'_3}\)
ICl is used to determine the iodine value of oils For preparation of polyhalides
As catalyst for oxidation of As(III) As fluorinating agent
For preparation of polyhalides As nonaqueous solvent

The first of those uses turns up in every food-testing laboratory. Iodine monochloride dissolved in glacial acetic acid is called Wijs solution, and it is used to determine the iodine value of an oil — the mass of halogen the oil will absorb, which measures how many carbon–carbon double bonds it contains. A high iodine value means a highly unsaturated oil, so the same reagent that tells a chemist about bonding also tells a manufacturer whether an oil will stay liquid, turn rancid quickly, or set into a film.

Structures

The shapes of these molecules follow straight from VSEPR theory, once the lone pairs on the central atom are counted. Table 7.11 lists the structures, which are drawn in Fig. 7.17.

Table 7.11 : Structures of some interhalogen compounds

Formula Name Shape
ICl Iodine monochloride Linear
\(\mathrm{ClF_3}\) Chlorine trifluoride Bent T-shaped (bond angle 89°)
\(\mathrm{BrF_3}\) Bromine trifluoride Bent T-shaped (bond angle 86°)
\(\mathrm{BrF_5}\) Bromine pentafluoride Square pyramidal
\(\mathrm{ClF_5}\) Chlorine pentafluoride Square pyramidal

Diagram

Fig. 7.17 – Five structures in a row with lone pairs shown on the central atom: ICl linear; ClF₃ bent T-shape with the 89° angle marked; BrF₃ bent T-shape with the 86° angle marked; BrF₅ square pyramidal with one lone pair above the Br; ClF₅ square pyramidal with one lone pair above the Cl.

Table 7.12 : Oxidation states of the central halogen atom in interhalogen compounds

O.S. of central halogen No. of lone pairs of electrons Examples
+7 0 \(\mathrm{IF_7}\)
+5 1 \(\mathrm{ClF_5}\), \(\mathrm{BrF_5}\), \(\mathrm{IF_5}\)
+3 2 \(\mathrm{ClF_3}\), \(\mathrm{BrF_3}\), \(\mathrm{IF_3}\), \(\mathrm{I_2Cl_6}\)
+1 3 ClF, BrF, IF, BrCl, ICl, IBr

Table 7.12 is more useful than it looks. Read it as a rule: each step down in oxidation state adds one lone pair to the central atom, because the two electrons that are no longer used in bonding stay behind as a pair. Once you know the oxidation state, you know the number of lone pairs, and once you know that, VSEPR gives you the shape.

shape and hybridisation of the four interhalogen types

Type Bond pairs Lone pairs Total electron pairs Hybridisation Shape of the molecule
\(\mathrm{XX'}\) 1 3 4 \(\mathrm{sp^3}\) Linear
\(\mathrm{XX'_3}\) 3 2 5 \(\mathrm{sp^3d}\) Bent T-shaped
\(\mathrm{XX'_5}\) 5 1 6 \(\mathrm{sp^3d^2}\) Square pyramidal
\(\mathrm{XX'_7}\) 7 0 7 \(\mathrm{sp^3d^3}\) Pentagonal bipyramidal

Predict the shape of the \(\mathrm{ClF_3}\) molecule, and state the oxidation state of chlorine in it.

Chlorine, the central atom, has seven valence electrons.

Three of these are used in forming three Cl–F bonds. The remaining four stay on chlorine as two lone pairs.

Total electron pairs around chlorine \(= 3\) bond pairs \(+\ 2\) lone pairs \(= 5\).

Five electron pairs require \(\mathrm{sp^3d}\) hybridisation and a trigonal bipyramidal arrangement. The two lone pairs, being bulkier than bond pairs, occupy two of the three equatorial positions, where they are furthest apart.

What remains visible is one equatorial and two axial fluorine atoms, which gives a bent T-shape — matching Table 7.11, where the bond angle is slightly squeezed to 89° by the lone pairs.

For the oxidation state, fluorine is always −1, and the molecule is neutral:

\[x + 3(-1) = 0 \quad \Rightarrow \quad x = +3\]

Chlorine is therefore in the +3 state, which Table 7.12 correctly pairs with two lone pairs.

For the molecule \(\mathrm{BrF_5}\): (i) count the bond pairs and lone pairs on bromine, (ii) give the hybridisation and predict the shape, (iii) state the oxidation state of bromine, and (iv) explain whether the molecule obeys the octet rule.

Explain each of the following in one or two sentences: (i) \(\mathrm{IF_7}\) exists but \(\mathrm{ClF_7}\) does not; (ii) fluorine never appears as the central atom in any interhalogen compound; (iii) \(\mathrm{BrF_3}\) is more reactive than either \(\mathrm{Br_2}\) or \(\mathrm{F_2}\); (iv) all interhalogen compounds are diamagnetic.

7.13 Compounds of Xenon

What is the correlation between ionization energies and reactivity of elements? What is the trend in ionization energy down a group?

We have studied earlier in this chapter that group 18 elements have very high ionisation energies, and due to this property they are unreactive. Each noble gas atom has a completely filled valence electron shell, which makes it inert.

The first ionization potential decreases down the group. Hence the heavier noble gases Kr, Xe and Rn can form compounds, because of their low ionization energy.

Only xenon reacts directly with fluorine to form xenon fluorides, and the stable fluorides are \(\mathrm{XeF_2}\), \(\mathrm{XeF_4}\) and \(\mathrm{XeF_6}\). Xenon also forms compounds with oxygen, such as \(\mathrm{XeO_3}\), \(\mathrm{XeOF_2}\), \(\mathrm{XeOF_4}\) and \(\mathrm{XeO_2F_2}\).

The first true compound of a noble gas was made in 1962 by Neil Bartlett, working with Lohman. Bartlett had noticed that platinum hexafluoride was powerful enough to tear an electron away from oxygen, and he realised that the ionisation enthalpy of xenon was close to that of the oxygen molecule. If \(\mathrm{PtF_6}\) could oxidise one, it should oxidise the other. He tried it, and an orange-yellow solid formed at once.

\[\mathrm{Xe} + 2\,\mathrm{[PtF_6]} \rightleftharpoons \mathrm{[XeF]^{+}[Pt_2F_{11}]^{-}}\]

Sixty years of textbooks calling this group the “inert gases” ended with that experiment, and the family was renamed the noble gases instead.

7.13.1 Xenon Fluorides

a. Preparation of Xenon Fluorides

Xenon fluorides are generally prepared by the direct reaction of xenon and fluorine in different ratios and conditions, such as temperature, electric discharge and photochemical reaction.

  1. \[\mathrm{Xe} + \mathrm{F_2} \xrightarrow{\ \text{sealed Ni tube},\ 400\ ^\circ\text{C}\ } \mathrm{XeF_2}\]

  2. \[\mathrm{Xe} + \mathrm{F_2}\ (1:5) \xrightarrow{\ \text{Ni tube},\ 400\ ^\circ\text{C},\ 5\text{–}6\ \text{atm}\ } \mathrm{XeF_4}\]

  3. \[\mathrm{Xe} + 2\,\mathrm{F_2} \xrightarrow{\ \text{electric discharge},\ -80\ ^\circ\text{C}\ } \mathrm{XeF_4}\]

  4. \[\mathrm{Xe} + 3\,\mathrm{F_2} \xrightarrow{\ \text{electric discharge},\ \text{low temp.}\ } \mathrm{XeF_6}\]

The pattern is worth reading as a whole: the more fluorine that is supplied, and the harsher the conditions, the further the reaction is pushed — \(\mathrm{XeF_2}\), then \(\mathrm{XeF_4}\), then \(\mathrm{XeF_6}\). Controlling the ratio of the two gases is what controls the product.

b. Important chemical reactions of \(\mathrm{XeF_2}\)

i. Hydrolysis: \(\mathrm{XeF_2}\) undergoes hydrolysis to form HF.

\[2\,\mathrm{XeF_2} + 2\,\mathrm{H_2O} \rightarrow 4\,\mathrm{HF} + 2\,\mathrm{Xe} + \mathrm{O_2}\]

Water pulls the compound apart completely here, releasing the xenon as free gas. This is why xenon fluorides must be prepared and stored under strictly dry conditions.

ii. Reaction with \(\mathrm{PF_5}\): \(\mathrm{XeF_2}\) forms adducts on reaction with \(\mathrm{PF_5}\).

\[\mathrm{XeF_2} + \mathrm{PF_5} \rightarrow \mathrm{XeF_2{\cdot}PF_5}\]

7.13.2 Xenon Trioxide

a. Preparation of xenon trioxide (\(\mathrm{XeO_3}\))

Fluorides of xenon react with water to form \(\mathrm{XeO_3}\).

\[3\,\mathrm{XeF_4} + 6\,\mathrm{H_2O} \rightarrow 2\,\mathrm{Xe} + \mathrm{XeO_3} + 12\,\mathrm{HF} + 1\tfrac{1}{2}\,\mathrm{O_2}\]

\[\mathrm{XeF_6} + 3\,\mathrm{H_2O} \rightarrow \mathrm{XeO_3} + 6\,\mathrm{HF}\]

Notice the difference between the two. Complete hydrolysis of \(\mathrm{XeF_6}\) replaces every fluorine by oxygen in a single clean step, while \(\mathrm{XeF_4}\) partly falls apart to free xenon at the same time.

Xenon trioxide is a colourless solid, and it is a dangerous one: it is thermodynamically unstable with respect to xenon and oxygen, and can decompose explosively. The reasoning is the same as for concentrated ozone in section 7.9 — a strongly exothermic decomposition that also increases the number of gas molecules gives a large negative \(\Delta G\). In laboratories the compound is handled only in milligram quantities, behind a shield, and never allowed to dry out in bulk.

7.13.3 Oxyfluorides of Xenon

Xenon forms the following oxyfluorides: \(\mathrm{XeOF_2}\), \(\mathrm{XeO_2F_2}\), \(\mathrm{XeO_3F_2}\), \(\mathrm{XeO_2F_4}\) and \(\mathrm{XeOF_4}\).

a. Preparation

i. Partial hydrolysis of xenon fluorides yields different oxyfluorides.

\[\mathrm{XeF_4} + \mathrm{H_2O} \xrightarrow{\ 80\ ^\circ\text{C}\ } \mathrm{XeOF_2} + 2\,\mathrm{HF}\]

\[\mathrm{XeF_6} + \mathrm{H_2O} \rightarrow \mathrm{XeOF_4} + 2\,\mathrm{HF}\]

The word partial is doing the work here. One molecule of water replaces just two fluorine atoms by one oxygen atom; with three molecules of water the replacement goes all the way and \(\mathrm{XeO_3}\) is obtained instead, as in section 7.13.2.

ii. Reaction of a xenon oxyfluoride with \(\mathrm{SiO_2}\), or its hydrolysis, yields xenon dioxydifluoride.

\[2\,\mathrm{XeOF_4} + \mathrm{SiO_2} \rightarrow 2\,\mathrm{XeO_2F_2} + \mathrm{SiF_4}\]

\[\mathrm{XeOF_4} + \mathrm{H_2O} \rightarrow \mathrm{XeO_2F_2} + 2\,\mathrm{HF}\]

The reaction with silica is a warning as much as a preparation: it means xenon oxyfluorides attack ordinary glassware, and so must be handled in nickel or fluoropolymer vessels. The whole hydrolysis family is collected in FC 7.7.

Flowchart

FC 7.7 – Hydrolysis routes of xenon fluorides: XeF₆ → with 1 H₂O (partial) → XeOF₄ → with more H₂O or with SiO₂ → XeO₂F₂ → further hydrolysis → XeO₃; parallel branch: XeF₆ + 3H₂O (complete) → XeO₃ + 6HF; second branch: XeF₄ + H₂O at 80 °C → XeOF₂; third branch: XeF₄ + excess H₂O → Xe + XeO₃ + HF + O₂; fourth branch: XeF₂ + H₂O → Xe + HF + O₂ (complete breakdown, no oxide formed).

What are the missing entries?

Formula Name
\(\mathrm{XeOF_2}\) Xenon mono oxyfluoride
…… Xenon dioxydifluoride
\(\mathrm{XeO_3F_2}\) ……
\(\mathrm{XeO_2F_4}\) ……
Structures of the Xenon Compounds

The shapes follow from VSEPR theory once the lone pairs left on xenon are counted, exactly as they did for the interhalogen compounds. The structures are drawn in Fig. 7.18.

Table 7.13 : Structure of xenon compounds

Sr. No. Formula Oxidation state of Xe Structure
1. \(\mathrm{XeF_2}\) +2 Linear
2. \(\mathrm{XeF_4}\) +4 Square planar
3. \(\mathrm{XeF_6}\) +6 Distorted octahedral
4. \(\mathrm{XeO_3}\) +6 Trigonal pyramidal
5. \(\mathrm{XeOF_4}\) +6 Square pyramidal

Diagram

Fig. 7.18 – Five structures in a row with lone pairs on xenon shown: XeF₂ linear with three lone pairs in the equatorial plane; XeF₄ square planar with one lone pair above and one below; XeF₆ distorted octahedral with a lone pair pushing one face out of shape; XeO₃ trigonal pyramidal with one lone pair on top; XeOF₄ square pyramidal with the oxygen at the apex and a lone pair below.

electron pairs, hybridisation and shape of the xenon compounds

Compound Bond pairs Lone pairs on Xe Total electron pairs Hybridisation Shape
\(\mathrm{XeF_2}\) 2 3 5 \(\mathrm{sp^3d}\) Linear
\(\mathrm{XeF_4}\) 4 2 6 \(\mathrm{sp^3d^2}\) Square planar
\(\mathrm{XeF_6}\) 6 1 7 \(\mathrm{sp^3d^3}\) Distorted octahedral
\(\mathrm{XeO_3}\) 3 1 4 \(\mathrm{sp^3}\) Trigonal pyramidal
\(\mathrm{XeOF_4}\) 5 1 6 \(\mathrm{sp^3d^2}\) Square pyramidal

Q. Why is \(\mathrm{XeF_6}\) described as distorted octahedral rather than simply octahedral?

A. Because xenon in \(\mathrm{XeF_6}\) has six bond pairs and one lone pair — seven electron pairs in all. Six bond pairs alone would give a regular octahedron, but the extra lone pair has to be accommodated somewhere, and it pushes the fluorine atoms nearest to it out of their symmetrical positions. The result is an octahedron that has been visibly bent out of shape.

Predict the hybridisation and shape of \(\mathrm{XeF_4}\), and confirm the oxidation state of xenon in it.

Xenon has eight electrons in its valence shell.

Four of them are used in forming four Xe–F bonds. The remaining four stay on xenon as two lone pairs.

Total electron pairs around xenon \(= 4\) bond pairs \(+\ 2\) lone pairs \(= 6\).

Six electron pairs require six hybrid orbitals, so the hybridisation is \(\mathrm{sp^3d^2}\) and the electron pairs point towards the corners of an octahedron.

The two lone pairs take up positions opposite each other, one above and one below the plane, since that keeps them as far apart as possible. The four fluorine atoms are then left in one plane at 90° to each other.

Ignoring the lone pairs, the shape of the molecule is square planar, in agreement with Table 7.13.

For the oxidation state, each fluorine is −1 and the molecule is neutral:

\[x + 4(-1) = 0 \quad \Rightarrow \quad x = +4\]

Table 7.14 : Uses of helium, neon and argon

Element | Uses |

|—|—|

Helium | (i) A mixture of He and \(\mathrm{O_2}\) is used for artificial breathing of asthma patients. (ii) A mixture of He and \(\mathrm{O_2}\) is used for respiration by sea divers. (iii) For filling balloons, a mixture of helium (85 %) and hydrogen (15 %) is used. (iv) Helium is used for producing the inert atmosphere required for welding and metallurgy of some metals. (v) Liquid helium is used for producing the low temperatures required for research. (vi) In low temperature gas thermometry, and for production of lasers. (vii) Used to pressurize fuel tanks of liquid fuelled rockets. (viii) Used as shielding gas for arc welding. (ix) In supersonic wind tunnels. (x) The helium nucleus is used as a bombarding particle for disintegration of atoms. (xi) Used for magnetic resonance imaging. |
Neon | (i) In neon discharge lamps and signs — these signs are visible from long distances and also in mist or fog. (ii) A mixture of Ne and He is used in certain protective electrical devices such as voltage stabilizers and current rectifiers. (iii) For production of lasers. (iv) In fluorescent tubes. |
Argon | (i) For producing an inert atmosphere in welding and steel production. (ii) A mixture of 85 % Ar and 15 % \(\mathrm{N_2}\) is filled in electric bulbs to increase the life of the filament. (iii) In filling fluorescent tubes and radio valves. (iv) It is mixed with neon to get lights of various colors. |

Notice how many of these uses depend on the very unreactivity that once made the group look chemically uninteresting. An inert atmosphere protects hot metal from oxidation; an inert gas around a filament stops it from burning away; and helium, being both inert and the least soluble of gases in blood, is what lets a diver breathe safely at depth.

For \(\mathrm{XeO_3}\): (i) count the bond pairs and lone pairs on xenon, (ii) give the hybridisation and predict the shape, and (iii) determine the oxidation state of xenon. Then explain why \(\mathrm{XeO_3}\) and \(\mathrm{XeF_6}\) have the same oxidation state but different shapes.

24.53 g of xenon hexafluoride is hydrolysed completely according to \(\mathrm{XeF_6} + 3\,\mathrm{H_2O} \rightarrow \mathrm{XeO_3} + 6\,\mathrm{HF}\). Calculate (i) the number of moles of \(\mathrm{XeF_6}\) taken, (ii) the mass of HF produced and (iii) the mass of \(\mathrm{XeO_3}\) formed. (Molar masses in \(\text{g mol}^{-1}\): Xe = 131.3, F = 19.00, O = 16.00, H = 1.008)

7.14 Summary

  1. In the p-block elements the differentiating electron enters a p-orbital of the outermost shell. A p-subshell holds a maximum of six electrons, which is why the p-block is six columns wide, giving groups 13 to 18.

  2. Group 16 (O, S, Se, Te, Po) is the oxygen family, also called the chalcogens or ore-forming elements. Group 17 (F, Cl, Br, I, At) are the halogens, or salt producers. Group 18 (He, Ne, Ar, Kr, Xe, Rn) are the noble gases.

  3. Oxygen is the most abundant element on earth — 20.95 % by volume of air and about 46.6 % by mass of the crust. Sulfur (0.034 % by mass of the crust) occurs as sulfates such as gypsum, epsom salt and baryte, and as sulfides such as galena, zinc blende and copper pyrites.

  4. Halogens are too reactive to occur free. Fluorine occurs as fluorspar, cryolite and fluorapatite; sea water supplies chloride, bromide and iodide, and contains 2.5 % NaCl by mass. Astatine is radioactive with a half life of 8.1 hours.

  5. All noble gases except radon occur in the atmosphere, about 1 % by volume of dry air, with argon as the major constituent. Helium comes commercially from natural gas.

  6. General valence configurations: group 16 is \(\mathrm{ns^2np^4}\), group 17 is \(\mathrm{ns^2np^5}\), group 18 is \(\mathrm{ns^2np^6}\). Groups 16 and 17 are two and one electrons short of the nearest noble gas respectively.

  7. Atomic and ionic radii increase down each group as quantum shells are added, and decrease across a period as effective nuclear charge rises. Halogens are the smallest atoms in their periods.

  8. Ionisation enthalpy is high for all three groups; it decreases down a group with increasing size and increases across a period. Group 16 values fall below those of group 15 in the same period because the half filled \(\mathrm{ns^2np^3}\) configuration of group 15 is extra stable.

  9. Electronegativity decreases down a group. Fluorine is the most electronegative element, and oxygen is next after it.

  10. Electron gain enthalpy becomes less negative down groups 16 and 17. Oxygen is less negative than sulfur, and fluorine less negative than chlorine, because the first member is so small that the incoming electron meets strong repulsion. Noble gases have large positive values.

  11. Physically: oxygen is a gas and the rest of group 16 are solids; O and S are nonmetals, Se and Te metalloids, Po a metal. In group 17, \(\mathrm{F_2}\) and \(\mathrm{Cl_2}\) are gases, \(\mathrm{Br_2}\) a liquid and \(\mathrm{I_2}\) a solid. Noble gases are monoatomic with very low melting and boiling points; helium has the lowest boiling point known, 4.2 K.

  12. Bond dissociation enthalpy of the halogens runs \(\mathrm{Cl\!-\!Cl} > \mathrm{Br\!-\!Br} > \mathrm{F\!-\!F} > \mathrm{I\!-\!I}\). \(\mathrm{F_2}\) is out of place because lone pair–lone pair repulsion in the tiny molecule weakens the bond.

  13. Oxygen behaves anomalously: it is diatomic and paramagnetic, shows only −2, −1 and +2 oxidation states, its hydride is a liquid, and its covalency is limited to 2. The causes are its small size, high electronegativity and absence of d-orbitals.

  14. Fluorine behaves anomalously for the same three reasons together with its low F–F bond dissociation enthalpy. It shows only the −1 state, forms just one oxoacid (HOF), reacts exothermically, and HF is a liquid because of hydrogen bonding.

  15. Group 16 shows −2 and +2 states, and also +4 and +6 for the heavier members; the +6 state becomes less stable and +4 more stable down the group because of the inert pair effect. Halogens all show −1, while Cl, Br and I also show +1, +3, +5 and +7 by expanding the octet into empty d orbitals.

  16. Xenon is the noble gas that reacts, because it is large and its ionisation enthalpy is low enough for paired electrons to be promoted into empty d orbitals.

  17. Hydrides \(\mathrm{H_2E}\) are angular and \(\mathrm{sp^3}\) hybridised. Down the group the H–E bond weakens, so acidic character and reducing power increase while thermal stability decreases, from \(\mathrm{H_2O}\) to \(\mathrm{H_2Te}\).

  18. Hydrogen halides follow the same logic: acid strength increases \(\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}\) while thermal stability decreases in the reverse order, both because the H–X bond dissociation enthalpy falls down the group.

  19. Group 16 forms acidic oxides of types \(\mathrm{EO_2}\) and \(\mathrm{EO_3}\) that give acids with water. \(\mathrm{SO_2}\) is reducing while \(\mathrm{TeO_2}\) is oxidising. Halogen oxides are mostly unstable, and the higher oxides are more stable than the lower ones.

  20. Group 16 forms hexahalides (\(\mathrm{sp^3d^2}\), octahedral), tetrahalides (\(\mathrm{sp^3d}\), trigonal bipyramidal with one lone pair) and dihalides (\(\mathrm{sp^3}\), two lone pairs). Stability falls in the order fluorides > chlorides > bromides > iodides.

  21. Group 16 elements combine with metals to give oxides, sulfides and selenides; halogens give metal halides whose ionic character falls in the order \(\mathrm{MF} > \mathrm{MCl} > \mathrm{MBr} > \mathrm{MI}\), and which become more covalent as the oxidation state of the metal rises.

  22. All group 16 elements show allotropy. Oxygen has \(\mathrm{O_2}\) and \(\mathrm{O_3}\); sulfur has rhombic (\(\alpha\)) and monoclinic (\(\beta\)) forms, both built from puckered \(\mathrm{S_8}\) rings and interconverting at 369 K. Allotropy applies to elements, polymorphism to compounds.

  23. Sulfur forms the oxoacids \(\mathrm{H_2SO_3}\), \(\mathrm{H_2SO_4}\), \(\mathrm{H_2S_2O_7}\), \(\mathrm{H_2SO_5}\), \(\mathrm{H_2S_2O_8}\) and \(\mathrm{H_2S_2O_3}\); several exist only in solution or as salts.

  24. Halogen oxoacid strength increases with the oxidation state of the halogen, from the weak \(\mathrm{HClO}\) (\(K_a = 3.5 \times 10^{-8}\)) to the very strong \(\mathrm{HClO_4}\).

  25. Dioxygen is prepared in the laboratory from chlorates, from metal oxides and from hydrogen peroxide, and industrially by the fractional distillation of liquefied air. It is colourless, sparingly soluble in water, and paramagnetic — a fact only molecular orbital theory explains, through two unpaired electrons in \(\pi^*\) orbitals.

  26. Oxides are classified as acidic (mostly of nonmetals), basic (of metals), amphoteric (such as \(\mathrm{Al_2O_3}\)) and neutral (CO, NO, \(\mathrm{N_2O}\)).

  27. Ozone is prepared by silent electric discharge through dry oxygen in an ozoniser, only 10 % conversion being achieved: \(3\,\mathrm{O_2} \rightleftharpoons 2\,\mathrm{O_3}\), \(\Delta H = +142\ \text{kJ mol}^{-1}\). It is angular, a resonance hybrid with equal O–O bond lengths of 128 pm and a bond angle of about 117°.

  28. Ozone is a powerful oxidising agent because it releases nascent oxygen; it bleaches without moisture (dry bleach) and reduces peroxides to oxides. Its layer shields the earth from UV radiation, and it is depleted by nitrogen oxides and by chlorofluorocarbons.

  29. Sulfur dioxide is prepared by burning sulfur, by acidifying a sulfite, and industrially by roasting sulfide ores. It is angular (119.5°), a resonance hybrid with \(\mathrm{d\pi\!-\!p\pi}\) bonding, and a reducing agent in moist conditions.

  30. Sulfuric acid is manufactured by the contact process: \(\mathrm{SO_2}\) is oxidised over \(\mathrm{V_2O_5}\) at 720 K and 2 bar, the \(\mathrm{SO_3}\) is absorbed in concentrated acid to give oleum, and the oleum is diluted. The product is 96–98 % pure.

\[2\,\mathrm{SO_2}\text{(g)} + \mathrm{O_2} \xrightarrow{\ \mathrm{V_2O_5}\ } 2\,\mathrm{SO_3}\text{(g)}\]

  1. Sulfuric acid is a strong dibasic acid, an oxidising agent when hot and concentrated, a powerful dehydrating agent that chars sugar, and it displaces more volatile acids from their salts.

  2. Chlorine, discovered by Scheele in 1774 and named by Davy in 1810, is prepared by oxidising HCl with \(\mathrm{MnO_2}\) or \(\mathrm{KMnO_4}\), and manufactured by Deacon’s process and by the electrolysis of brine.

  3. Chlorine reacts with metals, nonmetals, hydrogen, ammonia, alkalis and hydrocarbons. With cold dilute alkali it gives chloride and hypochlorite; with hot concentrated alkali, chloride and chlorate; with dry slaked lime, bleaching powder. It bleaches only in the presence of moisture, through nascent oxygen.

  4. Hydrogen chloride, first made by Glauber in 1648, is prepared by heating NaCl with concentrated \(\mathrm{H_2SO_4}\) and dried over the same acid. Its aqueous solution is a strong acid, and mixed 3 : 1 with concentrated \(\mathrm{HNO_3}\) it forms aqua regia, which dissolves gold and platinum.

  5. Interhalogen compounds \(\mathrm{XX'_n}\) form because halogens differ in electronegativity; X is the larger, more electropositive atom and \(n\) is 1, 3, 5 or 7. They are diamagnetic, more reactive than the parent halogens, and hydrolyse to oxoacids.

  6. Their shapes follow VSEPR: \(\mathrm{XX'}\) linear, \(\mathrm{XX'_3}\) bent T-shaped, \(\mathrm{XX'_5}\) square pyramidal, \(\mathrm{XX'_7}\) pentagonal bipyramidal, with the number of lone pairs rising as the oxidation state of the central halogen falls.

  7. The first noble gas compound was made by Neil Bartlett in 1962. Xenon forms \(\mathrm{XeF_2}\), \(\mathrm{XeF_4}\) and \(\mathrm{XeF_6}\) by direct combination with fluorine under different ratios and conditions, and these hydrolyse to \(\mathrm{XeO_3}\) and, partially, to oxyfluorides such as \(\mathrm{XeOF_4}\) and \(\mathrm{XeO_2F_2}\).

  8. Xenon compound shapes: \(\mathrm{XeF_2}\) linear, \(\mathrm{XeF_4}\) square planar, \(\mathrm{XeF_6}\) distorted octahedral, \(\mathrm{XeO_3}\) trigonal pyramidal, \(\mathrm{XeOF_4}\) square pyramidal.

  9. Helium, neon and argon are used wherever chemical inertness or extreme cold is needed — breathing mixtures, welding atmospheres, discharge lamps, filament bulbs, lasers, cryogenic research and magnetic resonance imaging.

Flowchart

FC 7.8 – Chapter overview: valence configuration (ns²np⁴ / ns²np⁵ / ns²np⁶) → decides electrons short of the octet → decides atomic properties (radius, ionisation enthalpy, electronegativity, electron gain enthalpy) → decides oxidation states, with the branch “first member: no d orbitals → anomalous behaviour of O and F” → leads to compound classes: hydrides H₂E and HX → oxides and oxoacids → halides and interhalogens → xenon fluorides, oxides and oxyfluorides → industrial routes: contact process for H₂SO₄, Deacon’s process and brine electrolysis for Cl₂.

7.15 Objective Questions

  1. Which of the following has highest electron gain enthalpy?
    1. Fluorine
    2. Chlorine
    3. Bromine
    4. Iodine
  2. Hydrides of group 16 are weakly acidic. The correct order of acidity is
    1. \(\mathrm{H_2O} > \mathrm{H_2S} > \mathrm{H_2Se} > \mathrm{H_2Te}\)
    2. \(\mathrm{H_2Te} > \mathrm{H_2O} > \mathrm{H_2S} > \mathrm{H_2Se}\)
    3. \(\mathrm{H_2Te} > \mathrm{H_2Se} > \mathrm{H_2S} > \mathrm{H_2O}\)
    4. \(\mathrm{H_2Te} > \mathrm{H_2Se} > \mathrm{H_2O} > \mathrm{H_2S}\)
  3. Which of the following elements does not show oxidation state of +4?
    1. O
    2. S
    3. Se
    4. Te
  4. HI acid when heated with conc. \(\mathrm{H_2SO_4}\) forms
    1. \(\mathrm{HIO_3}\)
    2. \(\mathrm{KIO_3}\)
    3. \(\mathrm{I_2}\)
    4. KI
  5. Ozone layer is depleted by
    1. NO
    2. \(\mathrm{NO_2}\)
    3. \(\mathrm{NO_3}\)
    4. \(\mathrm{N_2O_5}\)
  6. Which of the following occurs in liquid state at room temperature?
    1. \(\mathrm{HIO_3}\)
    2. HBr
    3. HCl
    4. HF
  7. In pyrosulfurous acid the oxidation state of sulfur is
    1. Only +2
    2. Only +4
    3. +2 and +6
    4. Only +6
  8. Stability of interhalogen compounds follows the order
    1. \(\mathrm{BrF} > \mathrm{IBr} > \mathrm{ICl} > \mathrm{ClF} > \mathrm{BrCl}\)
    2. \(\mathrm{IBr} > \mathrm{BeF} > \mathrm{ICl} > \mathrm{ClF} > \mathrm{BrCl}\)
    3. \(\mathrm{ClF} > \mathrm{ICl} > \mathrm{IBr} > \mathrm{BrCl} > \mathrm{BrF}\)
    4. \(\mathrm{ICl} > \mathrm{ClF} > \mathrm{BrCl} > \mathrm{IBr} > \mathrm{BrF}\)
  9. BrCl reacts with water to form
    1. HBr
    2. \(\mathrm{Br_2} + \mathrm{Cl_2}\)
    3. HOBr
    4. HOBr + HCl
  10. Chlorine reacts with excess of fluorine to form
  1. ClF
  2. \(\mathrm{ClF_3}\)
  3. \(\mathrm{ClF_2}\)
  4. \(\mathrm{Cl_2F_3}\)
  1. In interhalogen compounds, which of the following halogens is never the central atom?
  1. I
  2. Cl
  3. Br
  4. F
  1. Which of the following has one lone pair of electrons?
  1. \(\mathrm{IF_3}\)
  2. ICl
  3. \(\mathrm{IF_5}\)
  4. \(\mathrm{ClF_3}\)
  1. In which of the following pairs are molecules paired with their correct shapes?
  1. \(\mathrm{[I_3]}\) : bent
  2. \(\mathrm{BrF_5}\) : trigonal bipyramid
  3. \(\mathrm{ClF_3}\) : trigonal planar
  4. \(\mathrm{[BrF_4]}\) : square planar
  1. Among the known interhalogen compounds, the maximum number of atoms is
  1. 3
  2. 6
  3. 7
  4. 8

7.16 Check Your Concepts

  1. Assertion: The first ionisation enthalpy of sulfur is lower than that of phosphorus. Reason: Sulfur has a larger atomic radius than phosphorus. State whether the assertion and the reason are true, and whether the reason correctly explains the assertion.

  2. Why does the electron gain enthalpy of fluorine turn out to be less negative than that of chlorine, even though fluorine is the more electronegative of the two?

  3. Arrange \(\mathrm{F_2}\), \(\mathrm{Cl_2}\), \(\mathrm{Br_2}\) and \(\mathrm{I_2}\) in decreasing order of bond dissociation enthalpy. Justify the position of \(\mathrm{F_2}\), and explain why \(\mathrm{I_2}\) occupies its position for an entirely different reason.

  4. Oxygen and sulfur belong to the same group, yet \(\mathrm{H_2O}\) is a liquid and \(\mathrm{H_2S}\) is a gas at room temperature. Account for this difference, and predict where \(\mathrm{H_2Se}\) would fit.

  5. Explain why sulfur forms \(\mathrm{SF_6}\) while oxygen forms only \(\mathrm{OF_2}\) with fluorine. Which species out of \(\mathrm{OF_6}\), \(\mathrm{SeF_6}\) and \(\mathrm{OCl_4}\) cannot exist, and why?

  6. What happens if chlorine is passed into (a) cold dilute NaOH and (b) hot concentrated NaOH? Write the balanced equation in each case and state why both are called disproportionation reactions.

  7. Complete the following reactions and name the main product in each case:

    1. \(\mathrm{SeO_3} + \mathrm{H_2O} \rightarrow\)
    2. \(2\,\mathrm{Ca(OH)_2} + 2\,\mathrm{Cl_2} \rightarrow\)
    3. \(\mathrm{Na_2SO_3} + \mathrm{H_2O} + \mathrm{SO_2} \rightarrow\)
    4. \(\mathrm{XeF_2} + \mathrm{PF_5} \rightarrow\)
  8. Assertion: Ozone acts as a powerful oxidising agent. Reason: Ozone is thermodynamically more stable than dioxygen. Comment on the truth of each statement and on whether the reason explains the assertion.

  9. Ozone is called a dry bleach while chlorine bleaches only in the presence of moisture. Explain the difference using the chemistry of each bleaching action.

  10. Why is concentrated sulfuric acid always added slowly to water, and never water to the acid? Explain what would physically happen if the order were reversed.

  11. Concentrated \(\mathrm{H_2SO_4}\) can be used to prepare HCl from NaCl but not HBr from NaBr. Explain, writing the equation that shows what goes wrong in the bromide case.

  12. Arrange HF, HCl, HBr and HI in increasing order of (a) acid strength and (b) thermal stability. Explain why the two orders are the reverse of each other and why one single property accounts for both.

  13. State what you would observe, and write the equation, when moist \(\mathrm{SO_2}\) is passed through acidified potassium permanganate solution. Which property of \(\mathrm{SO_2}\) does this demonstrate?

  14. In the contact process, the oxidation of \(\mathrm{SO_2}\) to \(\mathrm{SO_3}\) is favoured by low temperature and high pressure. Explain both conditions from the equation, and state why in practice the temperature cannot be lowered indefinitely.

  15. Why is sulfur trioxide absorbed in concentrated sulfuric acid rather than in water during the manufacture of sulfuric acid?

  16. Predict the shape and hybridisation of each of the following, showing the count of bond pairs and lone pairs: \(\mathrm{ClF_3}\), \(\mathrm{BrF_5}\), \(\mathrm{XeF_4}\) and \(\mathrm{XeO_3}\).

  17. Why is \(\mathrm{XeF_6}\) described as distorted octahedral while \(\mathrm{SF_6}\) is a regular octahedron?

  18. Explain why interhalogen compounds are more reactive than the halogens from which they are made, and why the number of atoms in an interhalogen molecule is always even.

  19. Why does iodine form \(\mathrm{IF_7}\) while chlorine forms no compound beyond \(\mathrm{ClF_5}\)? Refer to the ratio of the radii of the two halogens in your answer.

  20. Match the following.

A B
a. \(\mathrm{I_2O_5}\) i. Bleaching agent for paper pulp
b. \(\mathrm{ClO_2}\) ii. Estimation of carbon monoxide
c. ICl in glacial acetic acid iii. Removal of plutonium from spent nuclear fuel
d. \(\mathrm{O_2F_2}\) iv. Determination of iodine value of an oil
  1. Helium and xenon are both noble gases, yet only xenon forms compounds. Give two reasons, and explain why fluorine and oxygen are the only partners xenon accepts.

  2. Both rhombic and monoclinic sulfur are built from \(\mathrm{S_8}\) rings. Then why do they differ in colour, density and melting point? What happens to a sample of rhombic sulfur kept at 380 K?

  3. Explain why (a) polonium has no ore of its own, (b) astatine cannot be collected in weighable amounts, and (c) helium is obtained from natural gas rather than from air.

  4. The stability of the +6 oxidation state falls down group 16 while that of the +4 state rises. Name the effect responsible and predict, with reasoning, which of \(\mathrm{SO_2}\) and \(\mathrm{TeO_2}\) acts as a reducing agent and which as an oxidising agent.

  5. Write the structural difference between peroxy monosulfuric acid and pyrosulfuric acid, and explain how that difference changes the way the oxidation state of sulfur must be calculated in each.

  6. Write the order of the thermal stability of the hydrides of group 16 elements.

  7. Name two gases which deplete the ozone layer.

  8. Give two uses of \(\mathrm{ClO_2}\).

  9. What is the action of bromine on magnesium metal?

  10. Write the names of the allotropic forms of selenium.

  11. The \(\mathrm{p}K_a\) value of HCl is −7.0 and that of HI is −10.0. Which is the stronger acid?

  12. Give one example showing the reducing property of ozone.

  13. Write the reaction of conc. \(\mathrm{H_2SO_4}\) with sugar.

  14. Give two uses of chlorine.

  15. Complete the following.

  1. \(\mathrm{ICl_3} + \mathrm{H_2O} \rightarrow\) …… + …… + ICl
  2. \(\mathrm{I_2} + \mathrm{KClO_3} \rightarrow\) …… + \(\mathrm{KIO_2}\)
  3. \(\mathrm{BrCl} + \mathrm{H_2O} \rightarrow\) …… + HCl
  4. \(\mathrm{Cl_2} + \mathrm{ClF_3} \rightarrow\) ……
  5. \(\mathrm{H_2C}\!=\!\mathrm{CH_2} + \mathrm{ICl} \rightarrow\) ……
  6. \(\mathrm{XeF_4} + \mathrm{SiO_2} \rightarrow\) …… + \(\mathrm{SiF_4}\)
  7. \(\mathrm{XeF_6} + 6\,\mathrm{H_2O} \rightarrow\) …… + HF
  8. \(\mathrm{XeOF_4} + \mathrm{H_2O} \rightarrow\) …… + HF
  1. Match the following.
A B
a. \(\mathrm{XeOF_2}\) i. Xenon trioxydifluoride
b. \(\mathrm{XeO_2F_2}\) ii. Xenon monooxydifluoride
c. \(\mathrm{XeO_3F_2}\) iii. Xenon dioxytetrafluoride
d. \(\mathrm{XeO_2F_4}\) iv. Xenon dioxydifluoride
  1. The first ionisation enthalpies of S, Cl and Ar are 1000, 1256 and 1520 \(\text{kJ mol}^{-1}\) respectively. Explain the observed trend.

  2. “Acidic character of hydrides of group 16 elements increases from \(\mathrm{H_2O}\) to \(\mathrm{H_2Te}\).” Explain.

  3. How is dioxygen prepared in the laboratory from \(\mathrm{KClO_3}\)?

  4. What happens when (a) lead sulfide reacts with ozone (\(\mathrm{O_3}\)), (b) nitric oxide reacts with ozone?

  5. Give two chemical reactions to explain the oxidizing property of concentrated \(\mathrm{H_2SO_4}\).

  6. Discuss the structure of sulfur dioxide.

  7. Fluorine shows only −1 oxidation state while other halogens show −1, +1, +3, +5 and +7 oxidation states. Explain.

  8. What is the action of chlorine on the following: (a) Fe (b) excess of \(\mathrm{NH_3}\)?

  9. How is hydrogen chloride prepared from sodium chloride?

  10. Draw the structures of \(\mathrm{XeF_6}\), \(\mathrm{XeO_3}\), \(\mathrm{XeOF_4}\) and \(\mathrm{XeF_2}\).

  11. What are inter-halogen compounds? Give two examples.

  12. What is the action of hydrochloric acid on the following: (a) \(\mathrm{NH_3}\) (b) \(\mathrm{Na_2CO_3}\)?

  13. Give two uses of HCl.

  14. Write the names and structural formulae of the oxoacids of chlorine.

  15. What happens when (a) \(\mathrm{Cl_2}\) reacts with \(\mathrm{F_2}\) in equal volume at 437 K, (b) \(\mathrm{Br_2}\) reacts with excess of \(\mathrm{F_2}\)?

  16. How are xenon fluorides \(\mathrm{XeF_2}\), \(\mathrm{XeF_4}\) and \(\mathrm{XeF_6}\) obtained? Give suitable reactions.

  17. How are \(\mathrm{XeO_3}\) and \(\mathrm{XeOF_4}\) prepared?

  18. Give two uses of neon and argon.

  19. Describe the structure of ozone. Give two uses of ozone.

  20. Explain the trend in the following atomic properties of group 16 elements: (i) atomic radii (ii) ionisation enthalpy (iii) electronegativity.

  21. Distinguish between rhombic sulfur and monoclinic sulfur.

  22. How is \(\mathrm{SO_2}\) prepared in the laboratory from sodium sulfite? Give two physical properties of \(\mathrm{SO_2}\).

  23. Describe the manufacturing of \(\mathrm{H_2SO_4}\) by the contact process.

7.17 Practice With Numericals

  1. Calculate the volume of dioxygen, measured at STP, liberated by the complete decomposition of 12.26 g of potassium chlorate: \(2\,\mathrm{KClO_3} \rightarrow 2\,\mathrm{KCl} + 3\,\mathrm{O_2}\). (Molar masses in \(\text{g mol}^{-1}\): K = 39.1, Cl = 35.5, O = 16.0; molar volume at STP = 22.4 L \(\text{mol}^{-1}\))

  2. 3.20 kg of sulfur is converted to sulfuric acid by the contact process. Assuming complete conversion at every stage, calculate the mass of \(\mathrm{H_2SO_4}\) obtained. (Molar masses in \(\text{g mol}^{-1}\): S = 32.06, H = 1.008, O = 16.00)

  3. The bond dissociation enthalpy of \(\mathrm{F_2}\) is 158.8 \(\text{kJ mol}^{-1}\). Calculate the energy required to dissociate all the molecules in 3.80 g of fluorine gas into atoms. (Molar mass of F = 19.00 \(\text{g mol}^{-1}\))

  4. Calculate the percentage of sulfur by mass in copper pyrites, \(\mathrm{CuFeS_2}\). (Molar masses in \(\text{g mol}^{-1}\): Cu = 63.5, Fe = 55.85, S = 32.06)

  5. Determine the oxidation state of the central atom in each of the following: \(\mathrm{HIO_3}\), \(\mathrm{KBrO_3}\), \(\mathrm{XeO_2F_2}\), \(\mathrm{H_2S_2O_7}\) and \(\mathrm{ClF_5}\).

  6. 15.80 g of potassium permanganate reacts with excess concentrated hydrochloric acid according to \(2\,\mathrm{KMnO_4} + 16\,\mathrm{HCl} \rightarrow 2\,\mathrm{KCl} + 2\,\mathrm{MnCl_2} + 8\,\mathrm{H_2O} + 5\,\mathrm{Cl_2}\). Calculate the volume of chlorine obtained at STP. (Molar masses in \(\text{g mol}^{-1}\): K = 39.1, Mn = 54.94, O = 16.00; molar volume at STP = 22.4 L \(\text{mol}^{-1}\))

  7. 0.30 mol of \(\mathrm{XeF_4}\) is hydrolysed completely according to \(3\,\mathrm{XeF_4} + 6\,\mathrm{H_2O} \rightarrow 2\,\mathrm{Xe} + \mathrm{XeO_3} + 12\,\mathrm{HF} + 1\tfrac{1}{2}\,\mathrm{O_2}\). Calculate (i) the moles of HF formed, (ii) the moles of free xenon released and (iii) the volume of \(\mathrm{O_2}\) obtained at STP. (Molar volume at STP = 22.4 L \(\text{mol}^{-1}\))

  8. 178.14 g of oleum, \(\mathrm{H_2S_2O_7}\), is diluted with the required quantity of water: \(\mathrm{H_2S_2O_7} + \mathrm{H_2O} \rightarrow 2\,\mathrm{H_2SO_4}\). Calculate the mass of sulfuric acid produced and the mass of water consumed. (Molar masses in \(\text{g mol}^{-1}\): H = 1.008, S = 32.06, O = 16.00)

  9. 6.41 g of sulfur dioxide is passed into acidified potassium permanganate solution: \(2\,\mathrm{KMnO_4} + 5\,\mathrm{SO_2} + 2\,\mathrm{H_2O} \rightarrow \mathrm{K_2SO_4} + 2\,\mathrm{MnSO_4} + 2\,\mathrm{H_2SO_4}\). Calculate the mass of \(\mathrm{KMnO_4}\) required for complete reaction. (Molar masses in \(\text{g mol}^{-1}\): S = 32.06, O = 16.00, K = 39.1, Mn = 54.94)

  10. At 293 K, 30.8 cm³ of dioxygen dissolves in 1000 cm³ of water. Calculate the mass of dissolved dioxygen present in a 5.00 L aquarium filled with water saturated at this temperature. (Molar volume of a gas at 293 K and 1 atm = 24.0 L \(\text{mol}^{-1}\); molar mass of O = 16.00 \(\text{g mol}^{-1}\))

  11. A sample of gypsum, \(\mathrm{CaSO_4{\cdot}2H_2O}\), weighing 8.61 g is heated until all the water of crystallisation is driven off. Calculate the mass of the residue obtained. (Molar masses in \(\text{g mol}^{-1}\): Ca = 40.08, S = 32.06, O = 16.00, H = 1.008)

  12. What is the oxidation state of Te in \(\mathrm{TeO_3}\)?

  13. What is the oxidation state of S in \(\mathrm{H_2SO_4}\)?

  14. What is the oxidation state of xenon in the following compounds: \(\mathrm{XeOF_4}\), \(\mathrm{XeO_3}\), \(\mathrm{XeF_6}\), \(\mathrm{XeF_4}\), \(\mathrm{XeF_2}\)?