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Chapter 9: Elements of Groups 13, 14 and 15

In the late 1660s, a German alchemist named Hennig Brand was working in Hamburg on the problem that had obsessed alchemists for centuries — finding a way to turn ordinary metal into gold. His chosen raw material was urine, which he collected in enormous quantities. He let it stand, boiled it down to a thick paste, and then heated the residue strongly in a retort.

No gold appeared. What collected instead was a pale, waxy solid, and in the darkness of his workroom it glowed with a faint, cold light. Left standing in open air, it did something even more alarming — it caught fire on its own, without a flame being brought near it.

Brand had isolated phosphorus, and the name later given to the element comes from the Greek word for “light-bearer.” Without knowing it, he had also walked straight into the central theme of this chapter. Phosphorus is a single element, yet it exists in more than one form: the substance Brand made is the dangerously reactive white form, while a red form of the very same element is stable enough to be coated on the side of a matchbox and handled safely.

One group to the left, carbon does something similar — appearing both as diamond, the hardest natural substance known, and as graphite, soft enough to leave a grey mark on paper. Groups 13, 14 and 15 are full of such surprises.

Diagram

Fig. 9.0 – Opening illustration: a strip of the periodic table with groups 13, 14 and 15 highlighted, and beside each group a real specimen labelled — borax crystals (group 13), diamond and a graphite pencil tip (group 14), and a piece of white phosphorus glowing under water (group 15)

NoteCuriosity Corner

Q1. White phosphorus must be kept stored under water, while red phosphorus can sit safely on a shelf. If both are made of nothing but phosphorus atoms, why do they behave so differently? Q2. Diamond and graphite are both pure carbon, yet one is the hardest natural substance and the other is soft and slippery. Is the difference in the atoms, or in the way the atoms are joined? Q3. Nitrogen and phosphorus belong to the same group, and phosphorus readily forms \(\mathrm{PCl_5}\) — but nitrogen never forms \(\mathrm{NCl_5}\). What does the nitrogen atom simply not possess? Q4. Thallium lies at the bottom of group 13, where +3 is the expected oxidation state, yet its +1 compounds are the more stable ones. Why should a heavier member of a group prefer a lower oxidation state than its own group number suggests? Q5. Boron oxide is acidic, aluminium oxide reacts with both acids and bases, and thallium oxide is basic — and all three are oxides of group 13. What changes as we move down a group that can reverse the very nature of its oxide?

As you work through this chapter, you will gather exactly the ideas needed to reason out each of these questions for yourself.

9.1 Introduction

In Chapter 7 you learnt that the periodic table is divided into blocks according to the orbital that receives the last electron of an atom. In the p-block elements, the differentiating electron — the last electron to be filled in — enters the outermost p orbital. This one structural fact is what decides the chemistry of every element in this chapter.

If the valence shell electronic configuration of an element is \(3\mathrm{s}^2\,3\mathrm{p}^1\), in which block of the periodic table is it placed?

Why the p-Block Has Six Groups

A p-subshell is made up of three p orbitals, and each orbital can hold at most two electrons. A p-subshell can therefore accommodate a maximum of six electrons. Since these six electrons are filled in one at a time as we move from element to element, six successive columns are needed to complete the process.

These six columns are groups 13 to 18 of the periodic table, and together they form the p-block. Each group in this block differs from the next by just one p electron.

Q. If the p-block begins right after the s-block, why are its groups numbered from 13 and not from 3?

A. Because the ten groups of the d-block (groups 3 to 12) lie in between. The s-block occupies groups 1 and 2, the d-block takes up groups 3 to 12, and only then does the p-block begin — so its first group carries the number 13.

Greater Variation Than in the s-Block

In the previous chapter you studied the s-block elements, whose properties change fairly smoothly from one member of a group to the next. The p-block behaves very differently. Here the properties show much greater variation, and a single group can contain a nonmetal at the top, a metalloid in the middle and a metal at the bottom.

A metalloid is an element whose properties lie in between those of metals and nonmetals — it may look metallic and yet conduct electricity poorly. It is precisely this mixture of behaviour within one group that makes the p-block worth studying closely. In this chapter we shall study the elements of groups 13, 14 and 15 in some detail.

The Three Families

Each of these three groups is named after its first, lightest member, and that first member sets the family name.

The elements boron (\(_5\mathrm{B}\)), aluminium (\(_{13}\mathrm{Al}\)), gallium (\(_{31}\mathrm{Ga}\)), indium (\(_{49}\mathrm{In}\)) and thallium (\(_{81}\mathrm{Tl}\)) constitute group 13, called the boron family.

The elements carbon (\(_6\mathrm{C}\)), silicon (\(_{14}\mathrm{Si}\)), germanium (\(_{32}\mathrm{Ge}\)), tin (\(_{50}\mathrm{Sn}\)) and lead (\(_{82}\mathrm{Pb}\)) form group 14, called the carbon family.

The elements nitrogen (\(_7\mathrm{N}\)), phosphorous (\(_{15}\mathrm{P}\)), arsenic (\(_{33}\mathrm{As}\)), antimony (\(_{51}\mathrm{Sb}\)) and bismuth (\(_{83}\mathrm{Bi}\)) belong to group 15 of the periodic table, called the nitrogen family.

Notice the atomic numbers as you read down any one of these families. They rise steeply — from 5 to 81 in group 13, for example — because each step down a group adds a whole new shell along with the d and f electrons that fill in on the way. That steep rise in atomic number is the seed of every trend you will meet in the sections that follow.

Group 13 vs Group 14 vs Group 15 — the three families at a glance | Feature | Group 13 | Group 14 | Group 15 | |—|—|—|—| | Family name | Boron family | Carbon family | Nitrogen family | | First (lightest) member | Boron, \(_5\mathrm{B}\) | Carbon, \(_6\mathrm{C}\) | Nitrogen, \(_7\mathrm{N}\) | | Other members | \(_{13}\mathrm{Al}\), \(_{31}\mathrm{Ga}\), \(_{49}\mathrm{In}\), \(_{81}\mathrm{Tl}\) | \(_{14}\mathrm{Si}\), \(_{32}\mathrm{Ge}\), \(_{50}\mathrm{Sn}\), \(_{82}\mathrm{Pb}\) | \(_{15}\mathrm{P}\), \(_{33}\mathrm{As}\), \(_{51}\mathrm{Sb}\), \(_{83}\mathrm{Bi}\) | | Number of members | 5 | 5 | 5 | | Position in the p-block | First group of the p-block | Second group of the p-block | Third group of the p-block |

Several names in these three families carry their origin inside them. Boron takes its name from borax, the mineral in which the element was found, whose own name travelled into European languages from Arabic and Persian words for the same white salt. Gallium was named after Gallia, the old name for France, by its French discoverer Lecoq de Boisbaudran in the 1870s, and germanium was named soon after by Winkler in honour of Germany. Element names, in other words, are often a record of where the chemistry was done.

You meet these three families every day without naming them. Silicon from group 14 is the material of computer chips and solar panels, because it is a semiconductor — it conducts far better than an insulator but far worse than a metal. Aluminium from group 13 wraps food in kitchens and forms the body panels of aircraft, since it is light and resists corrosion. Nitrogen from group 15 makes up most of the air around you, and food packets are often flushed with nitrogen gas precisely because it is so unreactive that it will not spoil the contents.

(question only — no solution) The valence shell electronic configuration of an element is \(4\mathrm{s}^2\,4\mathrm{p}^3\). State the block, the group and the family to which this element belongs, and name the element.

9.2 Electronic Configuration of Elements of Groups 13, 14 and 15

The General Outer Configuration

The general outer electronic configuration of the group 13 elements is \(n\mathrm{s}^2\,n\mathrm{p}^1\), that of the group 14 elements is \(n\mathrm{s}^2\,n\mathrm{p}^2\), while the group 15 elements are shown as \(n\mathrm{s}^2\,n\mathrm{p}^3\).

Here \(n\) is the principal quantum number of the outermost shell — the shell number. So \(n = 2\) for the first member of each family, \(n = 3\) for the second, and so on down the group. Only the value of \(n\) changes as we descend; the pattern of the outer electrons stays the same, which is exactly why the members of one group resemble one another chemically. The way these outer electrons occupy the three p orbitals is drawn in Fig. 9.6.

Diagram

Fig. 9.6 – Orbital-box diagram of the outer shell for the three families: ns and np boxes filled as ns²np¹ (group 13), ns²np² (group 14) and ns²np³ (group 15), showing single electrons placed in separate p orbitals with parallel spins, and labelling group 15 as a half-filled p subshell

Distance From a Noble-Gas Arrangement

These electronic configurations differ from their nearest inert gas by 3 or 4 electrons. Counting carefully makes this clear: a group 13 atom carries three electrons more than the noble gas that comes just before it, a group 14 atom carries four more, and a group 15 atom is three electrons short of the noble gas that comes just after it.

This distance matters. An atom that is three or four electrons away from a noble-gas arrangement cannot reach that arrangement by simply gaining or losing a single electron, as the s-block metals and the halogens do. Such large gains or losses of charge would cost far too much energy, so these elements mostly reach stability by sharing electrons instead — a point that will return again and again in this chapter.

The noble gases were called “inert” and “noble” for the same reason a noble was thought to keep aloof from ordinary company — these gases were found to combine with almost nothing, standing apart from the busy chemistry going on around them. The name has stuck even though a few compounds of the heavier members are now known.

Why These Elements Are Never Found Alone

Because their atoms are so far from a stable arrangement, these elements do not occur in the free monoatomic state. They are found as compounds with other elements, or as polyatomic molecules (such as \(\mathrm{N_2}\), \(\mathrm{P_4}\), \(\mathrm{C_{60}}\)), or as polyatomic covalent arrays (such as graphite, diamond).

The difference between the last two is worth fixing in your mind now. A polyatomic molecule is a separate unit containing a definite number of atoms — \(\mathrm{P_4}\) always means four phosphorus atoms bonded into one small molecule. A polyatomic covalent array has no such fixed size: the covalent bonding simply continues from atom to atom throughout the whole crystal, so a single molecule cannot be picked out of it at all.

Lead is a good illustration of “never found alone.” In nature it occurs chiefly combined with sulphur as the mineral galena, and never as the free metal lying about in the ground. Once extracted, however, lead spreads easily through the environment and is a cumulative poison that affects the nervous system, especially in children. This is why leaded petrol and lead-based house paints were phased out worldwide, and why lead–acid car batteries are collected for recycling rather than thrown away with ordinary waste.

Reading the Condensed Configurations

Table 9.1 shows the condensed electronic configuration of the elements of group 13, group 14 and group 15. A condensed configuration is a shorthand: the completely filled inner shells are replaced by the symbol of the preceding noble gas written in square brackets, so that only the electrons added after that noble gas have to be written out.

Table 9.1 : Condensed electronic configuration of elements of groups 13, 14 and 15

Group 13 (Boron family) Condensed electronic configuration Group 14 (Carbon family) Condensed electronic configuration Group 15 (Nitrogen family) Condensed electronic configuration
\(_5\mathrm{B}\) [He]\(2\mathrm{s}^2 2\mathrm{p}^1\) \(_6\mathrm{C}\) [He]\(2\mathrm{s}^2 2\mathrm{p}^2\) \(_7\mathrm{N}\) [He]\(2\mathrm{s}^2 2\mathrm{p}^3\)
\(_{13}\mathrm{Al}\) [Ne]\(3\mathrm{s}^2 3\mathrm{p}^1\) \(_{14}\mathrm{Si}\) [Ne]\(3\mathrm{s}^2 3\mathrm{p}^2\) \(_{15}\mathrm{P}\) [Ne]\(3\mathrm{s}^2 3\mathrm{p}^3\)
\(_{31}\mathrm{Ga}\) [Ar]\(3\mathrm{d}^{10} 4\mathrm{s}^2 4\mathrm{p}^1\) \(_{32}\mathrm{Ge}\) [Ar]\(3\mathrm{d}^{10} 4\mathrm{s}^2 4\mathrm{p}^2\) \(_{33}\mathrm{As}\) [Ar]\(3\mathrm{d}^{10} 4\mathrm{s}^2 4\mathrm{p}^3\)
\(_{49}\mathrm{In}\) [Kr]\(4\mathrm{d}^{10} 5\mathrm{s}^2 5\mathrm{p}^1\) \(_{50}\mathrm{Sn}\) [Kr]\(4\mathrm{d}^{10} 5\mathrm{s}^2 5\mathrm{p}^2\) \(_{51}\mathrm{Sb}\) [Kr]\(4\mathrm{d}^{10} 5\mathrm{s}^2 5\mathrm{p}^3\)
\(_{81}\mathrm{Tl}\) [Xe]\(4\mathrm{f}^{14} 5\mathrm{d}^{10} 6\mathrm{s}^2 6\mathrm{p}^1\) \(_{82}\mathrm{Pb}\) [Xe]\(4\mathrm{f}^{14} 5\mathrm{d}^{10} 6\mathrm{s}^2 6\mathrm{p}^2\) \(_{83}\mathrm{Bi}\) [Xe]\(4\mathrm{f}^{14} 5\mathrm{d}^{10} 6\mathrm{s}^2 6\mathrm{p}^3\)

Read the table down any one column and you will see the outer pattern repeat unchanged, while the inner part grows steadily — first a d subshell appears from the fourth member onwards, and then an f subshell as well in the last member.

Q. In gallium, the configuration is [Ar]3d¹⁰4s²4p¹. Are those ten 3d electrons valence electrons?

A. No. The \(3\mathrm{d}\) subshell is complete and lies inside the outermost shell, so its electrons do not take part in bonding — gallium’s valence electrons are only the three in \(4\mathrm{s}\) and \(4\mathrm{p}\). But the filled d subshell is not without effect: it screens the outer electrons from the nucleus rather poorly, and that single fact explains some surprising sizes and energies in the next section.

Atomic numbers of the group 13 elements are in the order B < Al < Ga < In < Tl. Arrange these elements in increasing order of ionic radii of \(\mathrm{M^{3+}}\).

Solution: The given elements are in an increasing order of atomic number. As we go down the group 13, their general outer electronic configuration is \(n\mathrm{s}^2\,n\mathrm{p}^1\). The \(\mathrm{M^{3+}}\) ion is formed by removal of three electrons from the outermost shell \(n\). In the \(\mathrm{M^{3+}}\) ion the \((n-1)\) shell becomes the outermost. Size of the added \((n-1)\) shell increases down the group. Therefore the ionic radii of \(\mathrm{M^{3+}}\) also increase down the group as follows:

\[\mathrm{B^{3+}} < \mathrm{Al^{3+}} < \mathrm{Ga^{3+}} < \mathrm{In^{3+}} < \mathrm{Tl^{3+}}\]

Notice the reasoning step that does the real work here: once all three outer electrons are stripped away, the ion’s size is no longer decided by shell \(n\) at all, but by the shell just below it. Since that shell is a larger one for each element further down the group, the ionic radius must increase in the same order as the atomic number.

(question only — no solution) An element of groups 13–15 has the condensed electronic configuration [Ar]\(3\mathrm{d}^{10}4\mathrm{s}^2 4\mathrm{p}^2\). Identify the element, state its group and family, and give the number of its valence electrons.

9.4 Chemical Properties of the Elements of the Groups 13, 14 and 15

9.4.1 Oxidation State

Oxidation state is the primary chemical property of all elements. It tells us how many electrons an atom has effectively given away or taken on in a compound, and once you know it you can usually predict what formulae that element will form.

The highest oxidation state exhibited by the p-block elements is equal to the total number of valence electrons, that is, the sum of the s-electrons and the p-electrons of the outer shell. This is sometimes called the group oxidation state.

NoteDefinition

The group oxidation state of a p-block element is the highest oxidation state it can show, and it equals the total number of electrons in its valence shell (s-electrons + p-electrons).

For group 13, with the outer configuration \(n\mathrm{s}^2\,n\mathrm{p}^1\), that total is three, so the group oxidation state is \(+3\). For group 14 it is \(+4\) and for group 15 it is \(+5\). The logic is simply that an atom cannot lose or share more electrons than it actually has available in its valence shell.

oxidation state (Oxidation state is a number, not a physical quantity, so it carries no unit; the sign is written before the digit, as \(+3\) or \(-3\). In a neutral compound all the oxidation numbers add up to zero; in an ion they add up to the charge on the ion. By convention oxygen is taken as \(-2\) and hydrogen as \(+1\) in ordinary compounds.)

Oxidation States Two Units Lower

In boron, carbon and nitrogen families the group oxidation state is the most stable oxidation state for the lighter elements. Besides this, the elements of groups 13, 14 and 15 exhibit other oxidation states which are lower than the group oxidation state by two units.

The lower oxidation states become increasingly stable as we move down to the heavier elements in the groups. So carbon at the top of group 14 is happiest at \(+4\), while lead at the bottom of the same group is happiest at \(+2\). Table 9.5 shows various oxidation states exhibited by elements belonging to these groups.

Table 9.5 : Oxidation states of the elements of groups 13, 14 and 15

Group 13 14 15
Outer electronic configuration \(n\mathrm{s}^2\,n\mathrm{p}^1\) \(n\mathrm{s}^2\,n\mathrm{p}^2\) \(n\mathrm{s}^2\,n\mathrm{p}^3\)
Group oxidation state \(+3\) \(+4\) \(+5\)
Other oxidation states \(+1\) \(+2\), \(-4\) \(+3\), \(-3\)
Examples of stable compounds \(\mathrm{BF_3}\), \(\mathrm{AlCl_3}\), \(\mathrm{GaCl_3}\), \(\mathrm{InCl_3}\), \(\mathrm{TlBr}\) \(\mathrm{CH_4}\), \(\mathrm{CO_2}\), \(\mathrm{CCl_4}\), \(\mathrm{SiCl_4}\), \(\mathrm{GeCl_4}\), \(\mathrm{SnCl_2}\), \(\mathrm{PbCl_2}\) \(\mathrm{N_2O_5}\), \(\mathrm{NH_3}\), \(\mathrm{NF_3}\), \(\mathrm{PH_3}\), \(\mathrm{PCl_5}\), \(\mathrm{AsH_3}\), \(\mathrm{SbH_3}\), \(\mathrm{SbCl_3}\), \(\mathrm{BiCl_3}\)

Look closely at the last row and you will notice that the negative oxidation states appear only for groups 14 and 15. Those are the groups whose atoms are close enough to a noble-gas arrangement to gain electrons instead of losing them — four in the case of carbon in \(\mathrm{CH_4}\), three in the case of nitrogen in \(\mathrm{NH_3}\). Group 13 is three electrons past its noble gas, so a negative state does not appear in its ordinary chemistry.

The Inert Pair Effect

Why should the lower oxidation state become steadily more stable as we go down a group? The answer lies in the two s-electrons of the valence shell.

The increased stability of the oxidation state lowered by 2 units than the group oxidation state in heavier p-block elements is due to the inert pair effect. In these elements, the two s-electrons are involved less readily in chemical reactions. This is because the s-electrons of the valence shell in p-block elements experience poor shielding than the valence p-electron, due to the ten inner d-electrons.

NoteDefinition

Inert pair effect — in heavier p-block elements the two valence s-electrons are held so firmly that they take little part in bonding, so the oxidation state two units below the group oxidation state becomes the more stable one.

Notice that this is the same poor-shielding idea that made gallium smaller than aluminium in the previous section, used here for a different purpose. Ten inner d-electrons screen the nucleus badly, the s-pair therefore feels an unusually strong pull, and electrons that are held tightly are electrons that will not be shared or lost. Only the single p-electron (in group 13) or the p-electrons alone remain chemically available, which is exactly a state two units lower.

stability of the lower oxidation state (down a group ↓: the lower oxidation state — \(+1\) in group 13, \(+2\) in group 14, \(+3\) in group 15 — becomes increasingly stable, because the inert pair effect grows stronger as more inner d, and later f, electrons shield the nucleus poorly and lock the valence s-pair in place; correspondingly the group oxidation state becomes a stronger oxidising state in the heaviest members)

Why is the \(\mathrm{Tl^+}\) ion more stable than \(\mathrm{Tl^{3+}}\)?

Solution: Tl is a heavy element which belongs to group 13 of the p-block. The common oxidation state for this group is 3. In the p-block, the lower oxidation state is more stable for heavier elements due to the inert pair effect. Therefore the \(\mathrm{Tl^+}\) ion is more stable than the \(\mathrm{Tl^{3+}}\) ion.

Q4. Thallium lies at the bottom of group 13, where +3 is the expected oxidation state, yet its +1 compounds are the more stable ones. Why should a heavier member of a group prefer a lower oxidation state than its own group number suggests?

A. Because of the inert pair effect. In thallium the valence shell is \(6\mathrm{s}^2\,6\mathrm{p}^1\), and the nucleus is screened very poorly by the inner filled \(5\mathrm{d}\) and \(4\mathrm{f}\) subshells. The \(6\mathrm{s}\) pair is therefore gripped tightly and does not take part in bonding, leaving only the single \(6\mathrm{p}\) electron chemically usable. Using one electron gives the \(+1\) state, which is two units below the group oxidation state of \(+3\) — and that is why \(\mathrm{TlBr}\), not a thallium(III) salt, appears among the stable compounds in Table 9.5.

Before working the next example, recall how an oxidation number is actually assigned. Take oxygen as \(-2\) and hydrogen as \(+1\), let the unknown be \(x\), and then set the sum of all oxidation numbers equal to zero for a neutral compound.

Assign the oxidation number of nitrogen in (i) \(\mathrm{N_2O_5}\) and (ii) \(\mathrm{NH_3}\), and state in each case whether it is the group oxidation state.

Solution:

  1. In \(\mathrm{N_2O_5}\), let the oxidation number of nitrogen be \(x\). There are two N atoms and five O atoms, each oxygen being \(-2\):

\[2x + 5(-2) = 0\]

\[2x - 10 = 0 \quad \Rightarrow \quad x = +5\]

Nitrogen is in the \(+5\) state, which for group 15 is the group oxidation state.

  1. In \(\mathrm{NH_3}\), let the oxidation number of nitrogen be \(x\). There are three H atoms, each \(+1\):

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

Nitrogen is in the \(-3\) state here, a negative state and not the group oxidation state. Both compounds appear in the group 15 row of Table 9.5, which is a useful reminder that one element can be stable in two very different oxidation states depending on what it is bonded to — to oxygen, which pulls electrons away from nitrogen, and to hydrogen, from which nitrogen pulls electrons towards itself.

(question only — no solution) Assign the oxidation number of the group 15 element in each of the following, and state whether it is the group oxidation state or a lower one: \(\mathrm{PH_3}\), \(\mathrm{PCl_5}\), \(\mathrm{SbCl_3}\), \(\mathrm{BiCl_3}\).

9.4.2 Bonding in Compounds of Group 13, 14 and 15 Elements

Why the Lighter Members Bond Covalently

The lighter elements in groups 13, 14 and 15 have small atomic radii and high ionization enthalpy values. Both facts point the same way: a small atom holds its electrons tightly, and high ionization enthalpy means it costs a great deal of energy to pull an electron off altogether. Such atoms therefore form covalent bonds with other atoms by overlapping their valence shell orbitals, sharing electrons rather than surrendering them.

As we move down the group, the ionization enthalpies are lowered. The atomic radii increase, since the valence shell orbitals are more diffused — that is, the outer electron cloud is spread over a larger, looser volume. The heavier elements in these groups therefore tend to form ionic bonds, because losing an electron outright has become affordable.

lighter members vs heavier members of groups 13, 14 and 15 | Feature | Lighter members (top of the group) | Heavier members (bottom of the group) | |—|—|—| | Atomic radius | Small | Large | | Ionization enthalpy | High | Lowered | | Valence shell orbitals | Compact | More diffused | | Preferred bond type | Covalent, by orbital overlap | Ionic | | Typical example | \(\mathrm{BF_3}\), \(\mathrm{CCl_4}\) | \(\mathrm{TlBr}\), \(\mathrm{BiCl_3}\) |

Octet Expansion and the Missing d Orbitals

The first member of each of these groups belongs to the second period and does not have d orbitals. Consequently B, C and N cannot expand their octet — they can never hold more than eight electrons in their valence shell, because there is simply no extra subshell available at that energy level to put more electrons into.

The subsequent elements in each group possess vacant d orbitals in their valence shell. These vacant orbitals can accept additional electron pairs, so these elements can expand their octet, forming a variety of compounds that the first member cannot. The contrast is drawn in Fig. 9.8.

Diagram

Fig. 9.8 – Side-by-side orbital sketch: nitrogen’s valence shell (2s, 2p only, no 3d box) limited to three bonds as in NCl₃, and phosphorus’s valence shell (3s, 3p plus empty 3d boxes) allowing five bonds as in PCl₅ with a trigonal bipyramidal shape

Q. Aluminium is in period 3 — does it have vacant d orbitals in its valence shell?

A. Yes. Aluminium’s valence shell is the third shell, which contains empty \(3\mathrm{d}\) orbitals alongside the filled \(3\mathrm{s}\) and \(3\mathrm{p}\). This is why aluminium can accept a lone pair from a neighbouring atom, a possibility closed to boron above it. You will see exactly this happening when we look at the structure of aluminium chloride later in the chapter.

(question only — no solution) Silicon forms the ion \([\mathrm{SiF_6}]^{2-}\), but carbon does not form \([\mathrm{CF_6}]^{2-}\). Explain this difference in terms of the valence shell of the two elements.

9.4.3 Reactivity Towards Air/Oxygen

The oxide an element forms is one of the quickest ways to judge its character, so all three families are compared here on the same test — heating in air or oxygen. The branching pattern is summarised in FC 9.1.

Flowchart

FC 9.1 – Heating an element E of groups 13–15 in air/oxygen: E + O₂ (Δ) → branch by group: group 13 → E₂O₃ only (and with N₂ → EN); group 14 → EO (limited O₂, lower oxidation state) or EO₂ (excess O₂, group oxidation state); group 15 → E₂O₃ (limited O₂) or E₂O₅ (excess O₂) → then judge the oxide: acidic (top of group) → amphoteric (middle) → basic (bottom)

Group 13 Elements

Elements of group 13 on heating with air or oxygen produce oxide of the type \(\mathrm{E_2O_3}\), where E stands for the element:

\[4\,\mathrm{E}\text{(s)} + 3\,\mathrm{O_2}\text{(g)} \xrightarrow{\ \Delta\ } 2\,\mathrm{E_2O_3}\text{(s)}\]

On heating in air these elements also combine with the nitrogen present, forming nitrides:

\[2\,\mathrm{E}\text{(s)} + \mathrm{N_2}\text{(g)} \xrightarrow{\ \Delta\ } 2\,\mathrm{EN}\text{(s)}\]

Only one oxide formula appears here because group 13 has only one important positive state in ordinary chemistry, \(+3\); three electrons shared with oxygen give \(\mathrm{E_2O_3}\) every time.

Boron nitride, the group 13 nitride formed in the second reaction, is also called inorganic graphite. The nickname comes from its structure: like graphite it is built of flat hexagonal sheets, only with boron and nitrogen atoms alternating in place of carbon, and the sheets slide over one another in the same way. It is white rather than grey, and unlike graphite it does not conduct electricity.

Aluminium foil burns in oxygen when strongly heated. Using the general equation for group 13, write the balanced equation for this reaction and calculate the mass of the oxide formed from 5.40 g of aluminium. (Atomic masses: Al \(= 26.98\), O \(= 16.00\))

Solution: Substituting \(\mathrm{E} = \mathrm{Al}\) in the general equation:

\[4\,\mathrm{Al}\text{(s)} + 3\,\mathrm{O_2}\text{(g)} \xrightarrow{\ \Delta\ } 2\,\mathrm{Al_2O_3}\text{(s)}\]

Moles of aluminium taken:

\[n_{\mathrm{Al}} = \frac{5.40\ \text{g}}{26.98\ \text{g mol}^{-1}} = 0.200\ \text{mol}\]

From the balanced equation, 4 mol Al give 2 mol \(\mathrm{Al_2O_3}\), that is, the oxide is formed in half the number of moles:

\[n_{\mathrm{Al_2O_3}} = \frac{0.200}{2} = 0.100\ \text{mol}\]

Molar mass of \(\mathrm{Al_2O_3} = 2(26.98) + 3(16.00) = 101.96\ \text{g mol}^{-1}\).

\[m_{\mathrm{Al_2O_3}} = 0.100\ \text{mol} \times 101.96\ \text{g mol}^{-1} = 10.2\ \text{g}\]

So 5.40 g of aluminium yields about 10.2 g of aluminium oxide. The mass grows because oxygen from the air has been built into the product — a point worth remembering whenever a metal appears to “gain weight” on burning.

Group 14 Elements

The elements of group 14 on heating in air or oxygen form oxides of the type EO and \(\mathrm{EO_2}\), in accordance with the stable oxidation state and the availability of oxygen:

\[\mathrm{E}\text{(s)} + \tfrac{1}{2}\,\mathrm{O_2}\text{(g)} \xrightarrow{\ \Delta\ } \mathrm{EO}\]

\[\mathrm{E}\text{(s)} + \mathrm{O_2}\text{(g)} \xrightarrow{\ \Delta\ } \mathrm{EO_2}\]

Two oxides are possible here because group 14 has two accessible oxidation states, \(+2\) and \(+4\). Which one you get depends on both factors named above: how much oxygen is available, and which oxidation state that particular element prefers.

Group 15 Elements

The elements of group 15 on heating in air or oxygen form two types of oxide, \(\mathrm{E_2O_3}\) and \(\mathrm{E_2O_5}\), corresponding to the \(+3\) and \(+5\) states:

\[\mathrm{P_4} + 3\,\mathrm{O_2} \rightarrow \mathrm{P_4O_6}\]

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

\[\mathrm{As_4} + 3\,\mathrm{O_2} \rightarrow \mathrm{As_4O_6}\]

\[2\,\mathrm{Bi} + 3\,\mathrm{O_2} \rightarrow \mathrm{Bi_2O_3}\]

The formulae of the first three products look unfamiliar only because phosphorus and arsenic exist as \(\mathrm{P_4}\) and \(\mathrm{As_4}\) molecules rather than as single atoms. \(\mathrm{P_4O_6}\) is simply the \(\mathrm{E_2O_3}\) type written for a four-atom molecule, and \(\mathrm{P_4O_{10}}\) is the \(\mathrm{E_2O_5}\) type written the same way.

Nature of the Oxides

The increase in metallic character down all these groups 13, 14 and 15 is reflected in their oxides, which gradually vary from acidic through amphoteric to basic. An amphoteric oxide is one that can react with an acid as well as with a base, behaving either way depending on what it meets. The nature of the stable oxides of groups 13, 14 and 15 is given in Table 9.6.

Table 9.6 : Nature of stable oxides of groups 13, 14 and 15 elements

Group Element Oxide Nature
13 B \(\mathrm{B_2O_3}\) Acidic
13 Al \(\mathrm{Al_2O_3}\) Amphoteric
13 Ga \(\mathrm{Ga_2O_3}\) Amphoteric
13 In \(\mathrm{In_2O_3}\) Basic
13 Tl \(\mathrm{Tl_2O_3}\) Basic
14 C \(\mathrm{CO_2}\) Acidic
14 Si \(\mathrm{SiO_2}\) Acidic
14 Ge \(\mathrm{GeO_2}\) Acidic
14 Sn \(\mathrm{SnO_2}\) Amphoteric
14 Pb \(\mathrm{PbO_2}\) Amphoteric
15 N \(\mathrm{N_2O_5}\) Acidic
15 P \(\mathrm{P_2O_5}\) Acidic
15 As \(\mathrm{As_4O_6}\) Amphoteric
15 Sb \(\mathrm{Sb_2O_3}\) Amphoteric
15 Bi \(\mathrm{Bi_2O_3}\) Basic

acidic to basic nature of oxides (down a group ↓: acidic → amphoteric → basic, because metallic character increases as atomic radius grows and ionization enthalpy falls; a nonmetal holds oxygen covalently and the oxide reacts with water or bases as an acid, while a metal transfers electrons to oxygen and the resulting oxide reacts with acids as a base — the middle members, being borderline, manage both and are amphoteric)

Q5. Boron oxide is acidic, aluminium oxide reacts with both acids and bases, and thallium oxide is basic — and all three are oxides of group 13. What changes as we move down a group that can reverse the very nature of its oxide?

A. Metallic character. Going down group 13 the atomic radius increases and ionization enthalpy falls, so the element gives up its electrons more and more readily. Boron, still essentially a metalloid, keeps its bonding electrons and forms a covalent, acidic oxide, \(\mathrm{B_2O_3}\). Thallium, a heavy metal, hands its electrons over to oxygen and forms a basic oxide, \(\mathrm{Tl_2O_3}\). Aluminium and gallium lie in between and are amphoteric, reacting with acids and with bases alike. The oxide is really a report on how metallic the element has become.

(question only — no solution) Using the general equations of this subsection, write the balanced equation for the reaction of boron with dinitrogen on heating, name the product, and predict whether the oxide \(\mathrm{B_2O_3}\) will dissolve in aqueous sodium hydroxide, in hydrochloric acid, or in both.

9.4.4 Reaction With Water

Most of the elements of groups 13, 14 and 15 are unaffected by water. Aluminium reacts with water on heating and forms hydroxide, while tin reacts with steam to form oxide:

\[2\,\mathrm{Al}\text{(s)} + 6\,\mathrm{H_2O}\text{(l)} \xrightarrow{\ \Delta\ } 2\,\mathrm{Al(OH)_3}\text{(s)} + 3\,\mathrm{H_2}\text{(g)}\]

\[\mathrm{Sn}\text{(s)} + 2\,\mathrm{H_2O}\text{(g)} \xrightarrow{\ \Delta\ } \mathrm{SnO_2}\text{(s)} + 2\,\mathrm{H_2}\text{(g)}\]

Notice that both reactions release dihydrogen. In each case the metal takes the oxygen of water for itself and leaves the hydrogen behind — the difference is only that aluminium in liquid water ends up with hydroxide, while tin in steam, at a higher temperature, ends up with the oxide.

Lead is unaffected by water, due to the formation of a protective film of oxide. Such a film is invisible and extremely thin, but once formed it seals the metal beneath from further attack, so the reaction stops almost as soon as it begins.

Phosphorous is stored under water because it catches fire when exposed to air. This makes it one of the substances handled with the greatest care in a school or college laboratory: a piece of white phosphorus must be cut while still submerged, moved with forceps and never with bare fingers, and returned under water at once. The water is not there to dissolve it — phosphorus is insoluble in water — but simply to keep air away from its surface.

Q. If aluminium reacts with water, how can aluminium vessels and window frames be used safely for years?

A. For the same reason lead resists water — an oxide film. Aluminium exposed to air acquires a thin, tightly adhering coat of its own oxide, and it is that coat, not the metal, which faces the water. The reaction written above needs a heated, freshly exposed metal surface, which everyday aluminium never has.

9.4.5 Reaction With Halogens

Halides of Group 13

All the elements of group 13 react directly with halogens to form trihalides (\(\mathrm{EX_3}\)). Thallium is an exception, which forms mono halides (TlX):

\[2\,\mathrm{E}\text{(s)} + 3\,\mathrm{X_2}\text{(g)} \rightarrow 2\,\mathrm{EX_3}\text{(s)}\]

Thallium’s exception should no longer surprise you. It is the heaviest member of the group, its inert pair is firmly locked, and so it uses only one electron — giving TlX rather than \(\mathrm{TlX_3}\).

Halides of Group 14

All the elements of group 14 (except carbon) react directly with halogens to form tetrahalides (\(\mathrm{EX_4}\)). The heavy elements Ge and Pb form dihalides as well. The stability of the dihalides increases down the group, once again because of the inert pair effect. The ionic character of the halides also increases steadily down the group, since the heavier elements are more metallic and part with their electrons more completely.

\(\mathrm{GeCl_4}\) is more stable than \(\mathrm{GeCl_2}\), while \(\mathrm{PbCl_2}\) is more stable than \(\mathrm{PbCl_4}\). Explain.

Solution: Ge and Pb are the 4th and 5th period elements down group 14. The group oxidation state of group 14 is 4, and the stability of the other oxidation state, lower by 2 units, increases down the group due to the inert pair effect. The stability of the oxidation state 2 is more in Pb than in Ge.

In other words, germanium is still high enough in the group to prefer its group oxidation state of \(+4\), giving the stable \(\mathrm{GeCl_4}\), whereas lead at the bottom prefers \(+2\), giving the stable \(\mathrm{PbCl_2}\). The same reasoning tells you which of a pair of halides will act as the stronger oxidising agent: \(\mathrm{PbCl_4}\), holding lead in a state it would rather leave, gives up chlorine readily.

Halides of Group 15

Elements of the group 15 react with halogens to form two series of halides: \(\mathrm{EX_3}\) and \(\mathrm{EX_5}\). The pentahalides possess more covalent character due to the availability of vacant d orbitals of the valence shell for bonding. Nitrogen, being a second period element, does not have d orbitals in its valence shell, and therefore does not form pentahalides.

Trihalides of the group 15 elements are predominantly covalent, except \(\mathrm{BiF_3}\). This single exception fits the pattern of the group: bismuth is the most metallic member and fluorine the most electronegative halogen, so this is precisely the combination in which electrons are transferred outright rather than shared. The only stable halide of nitrogen is \(\mathrm{NF_3}\).

trihalides vs pentahalides of group 15 | Feature | Trihalides, \(\mathrm{EX_3}\) | Pentahalides, \(\mathrm{EX_5}\) | |—|—|—| | Oxidation state of E | \(+3\) | \(+5\) (group oxidation state) | | Formed by nitrogen? | Yes — \(\mathrm{NF_3}\) is its only stable halide | No — no d orbitals available | | Bonding character | Predominantly covalent, except \(\mathrm{BiF_3}\) | More covalent, using vacant d orbitals | | Example | \(\mathrm{SbCl_3}\), \(\mathrm{BiCl_3}\) | \(\mathrm{PCl_5}\) |

Nitrogen does not form \(\mathrm{NCl_5}\) or \(\mathrm{NF_5}\), but phosphorous can. Explain.

Solution: Phosphorous and other members of the group can make use of d-orbitals in their bonding, and thus compounds \(\mathrm{MX_3}\) as well as \(\mathrm{MX_5}\) are formed. Nitrogen cannot form \(\mathrm{NCl_5}\) or \(\mathrm{NF_5}\) since it is void of d-orbitals in its second shell.

Q3. Nitrogen and phosphorus belong to the same group, and phosphorus readily forms \(\mathrm{PCl_5}\) — but nitrogen never forms \(\mathrm{NCl_5}\). What does the nitrogen atom simply not possess?

A. Vacant d orbitals in its valence shell. Nitrogen belongs to the second period, whose valence shell contains only \(2\mathrm{s}\) and \(2\mathrm{p}\) orbitals — there is no \(2\mathrm{d}\) subshell in existence. With eight electrons its valence shell is full, so nitrogen can form at most three bonds to a halogen and its octet cannot be expanded. Phosphorus, in the third period, has empty \(3\mathrm{d}\) orbitals available alongside \(3\mathrm{s}\) and \(3\mathrm{p}\); it can accommodate ten electrons around itself and so forms \(\mathrm{PCl_5}\) as well as \(\mathrm{PCl_3}\). The limit on nitrogen is set by which orbitals exist, not by how reactive the halogen is.

(question only — no solution) Predict which member of each pair is more stable and give the reason: (i) \(\mathrm{GeCl_2}\) or \(\mathrm{PbCl_2}\), (ii) \(\mathrm{TlCl}\) or \(\mathrm{TlCl_3}\). Also write the formula of the halide formed when thallium reacts directly with chlorine.

9.5 Catenation

The property of self linking of atoms of an element by covalent bonds to form chains and rings is called catenation. In a catenated compound, atoms of the same element are joined to one another — a carbon bonded to a carbon bonded to another carbon — instead of each atom bonding only to atoms of some other element.

NoteDefinition

Catenation is the self linking of atoms of an element with one another through covalent bonds, forming chains and rings.

The catenation tendency of an element depends upon the strength of the bond formed. This is a simple survival condition: a chain can exist only if the bond holding two like atoms together is strong enough not to break in preference to the other bonds in the molecule.

Q. Why should the strength of the E–E bond decide whether an element catenates?

A. Because the chain is always in competition. If the E–E bond is weaker than the bond E can form with oxygen, hydrogen or a halogen, the chain will simply come apart and the atoms will bond to those partners instead. Only when the E–E bond is comparably strong does a long chain of like atoms survive long enough to be a real compound.

Catenation Among the Group 14 Elements

Among the group 14 elements, carbon shows the maximum tendency for catenation because of the stronger C–C bond, whose bond enthalpy is \(348\ \text{kJ mol}^{-1}\).

Bond enthalpy is the energy needed to break one mole of a given bond in the gaseous state. A larger value means a bond that is harder to pull apart, so bond enthalpy is exactly the right quantity to compare when asking which element will hold a chain together.

bond enthalpy (SI-based unit: \(\text{kJ mol}^{-1}\). The value is always positive, because breaking a bond absorbs energy; forming the same bond releases the same amount. Comparisons are only fair between bonds of the same type — here, E–E single bonds.)

The order of catenation of group 14 elements is \(\mathrm{C} \gg \mathrm{Si} > \mathrm{Ge} = \mathrm{Sn}\). Lead does not show catenation. This can be clearly seen from the bond enthalpy values listed in Table 9.7.

Table 9.7 : Bond enthalpies of E–E bonds in group 14

Bond Bond enthalpy (kJ mol\(^{-1}\))
C – C 348
Si – Si 297
Ge – Ge 260
Sn – Sn 240

Read the column downwards and the reason for the order is plain. As we move down group 14 the atoms get larger, so the two bonding orbitals meet further apart and overlap less effectively. Weaker overlap means a weaker E–E bond, and a weaker bond means less catenation — until, at lead, the bond is too feeble to hold any chain at all.

catenation tendency in group 14 (down a group ↓: catenation falls sharply, \(\mathrm{C} \gg \mathrm{Si} > \mathrm{Ge} = \mathrm{Sn}\), with none for Pb, because atomic radius increases down the group, orbital overlap between two like atoms becomes poorer, and the E–E bond enthalpy drops from 348 to \(240\ \text{kJ mol}^{-1}\))

the E–E bond enthalpies of Table 9.7 plotted as a bar graph against the element (By how many kJ mol⁻¹ does the bond enthalpy fall from C–C to Si–Si, and by how much from Ge–Ge to Sn–Sn? Which single step in the graph is the largest, and how does that step explain the “much greater than” sign in the order \(\mathrm{C} \gg \mathrm{Si}\)? Lead has no bar in this graph — what would you predict about the strength of a Pb–Pb bond, and what does that predict about catenation in lead?)

Diagram

Fig. 9.9 – Bar graph of E–E bond enthalpy (kJ mol⁻¹) on the y-axis against C, Si, Ge, Sn on the x-axis, using the Table 9.7 values 348, 297, 260, 240; leave a blank labelled position for Pb and draw beside each bar a small sketch of two atoms of increasing size showing overlap becoming poorer

The C–C bond enthalpy is \(348\ \text{kJ mol}^{-1}\). Calculate the total energy required to break all the carbon–carbon bonds in one mole of propane, \(\mathrm{CH_3CH_2CH_3}\).

Solution: First count the C–C bonds. Propane is a chain of three carbon atoms, \(\mathrm{CH_3\!-\!CH_2\!-\!CH_3}\), so there are two C–C bonds in each molecule.

Energy required per mole of propane:

\[E = 2 \times 348\ \text{kJ mol}^{-1} = 696\ \text{kJ mol}^{-1}\]

So \(696\ \text{kJ}\) must be supplied to break the carbon skeleton of one mole of propane. Note what the answer implies: a chain of only three carbon atoms already demands nearly 700 kJ per mole to dismantle. This is why carbon chains, once formed, persist through so many reactions — and why catenation in carbon is the structural basis of an entire branch of chemistry.

Catenation is the reason organic chemistry exists as a subject of its own. Because C–C bonds are strong, carbon builds chains and rings of almost any length, and the millions of known carbon compounds — from the fuels in a vehicle’s tank to the proteins in your body — are all built on catenated carbon skeletons. The same property is exploited deliberately in plastics: polyethene, used for carry bags and bottles, is nothing but an extremely long catenated carbon chain with hydrogen atoms attached along its length. It is exactly this stability of the C–C backbone that makes such plastics durable in use and, unfortunately, slow to break down once discarded.

(question only — no solution) Using the bond enthalpy values of Table 9.7, calculate the total energy needed to break all the Si–Si bonds in one mole of \(\mathrm{Si_4H_{10}}\), in which four silicon atoms are joined in a chain. Then state, with a reason drawn from the same table, why no lead analogue \(\mathrm{Pb_4H_{10}}\) is known.

9.6 Allotropy

What is common between diamond and graphite?

When a solid element exists in different crystalline forms with different physical properties such as colour, density, melting point, etc., the phenomenon is called allotropy, and the individual crystalline forms are called allotropes.

NoteDefinition

Allotropy is the existence of a solid element in two or more different crystalline forms, called allotropes, which differ in physical properties such as colour, density and melting point while remaining chemically the same element.

Read that definition carefully, because the whole idea rests on one point: the allotropes contain nothing but atoms of the same element. What differs is the arrangement of those atoms, and arrangement alone is enough to change hardness, colour, density and electrical behaviour beyond recognition.

Diamond and graphite are well known allotropes of carbon. Fullerenes, graphene and carbon nanotubes are other allotropes of carbon. Elements such as boron, bismuth, silicon, etc. of groups 13, 14 and 15 also exhibit allotropy. In this chapter we are going to look at some aspects of the allotropes of carbon and phosphorous.

9.6.1 Allotropes of Carbon

Diamond

In diamond each carbon atom is linked to four other carbon atoms, through \(\mathrm{sp}^3\) hybrid orbitals, in a tetrahedral manner. The \(\mathrm{sp}^3\) label means that one s and three p orbitals of the carbon atom have mixed to give four equivalent orbitals pointing towards the corners of a tetrahedron, so each carbon sits at the centre of a tetrahedron of four neighbours, as seen in Fig. 9.1.

Diagram

Fig. 9.1 – Structure of diamond: three-dimensional network of carbon atoms, each bonded tetrahedrally to four others by single bonds, with one tetrahedral unit highlighted and the C–C distance of 154 pm marked

The C–C bond distance is 154 pm. The tetrahedra are linked together forming a three dimensional network structure (Fig. 9.1) involving strong C–C single bonds.

Because the whole crystal is one continuous network of these strong bonds, deforming or scratching diamond means breaking covalent bonds rather than merely sliding atoms past one another. Thus diamond is the hardest natural substance, with an abnormally high melting point of \(3930\ ^\circ\text{C}\).

Diamond is a bad conductor of electricity. All four valence electrons of every carbon atom are used up in the four sigma bonds, so no electron is left free to carry a current.

Uses: Diamond is used

  • for cutting glass and in drilling tools.
  • for making dies for drawing thin wire from metal.
  • for making jewellery.

The first two uses follow directly from its hardness — only a substance harder than the material being cut can cut it — and the third from the way its high refractive index returns light to the eye.

Graphite

Graphite is composed of layers of two dimensional sheets of carbon atoms (Fig. 9.2), each sheet being made up of a hexagonal net of \(\mathrm{sp}^2\) carbons bonded to three neighbours, forming three sigma bonds.

Diagram

Fig. 9.2 – Structure of graphite: parallel flat sheets of hexagonal carbon rings, with the interlayer separation of 335 pm and the in-plane C–C bond length of 141.5 pm both marked with arrows

Since \(\mathrm{sp}^2\) hybridisation uses only three of the four valence orbitals, the fourth electron is left in the unhybridised p-orbital of each carbon. The p-orbitals on all the carbons are parallel to each other. These overlap laterally to form \(\pi\) bonds, and the \(\pi\) electrons are delocalised over the whole layer — that is, they belong to the sheet as a whole rather than to any single pair of atoms, and can therefore travel through it.

The C–C bond length in graphite is 141.5 pm, shorter than the 154 pm of diamond because of this additional \(\pi\) bonding. The individual layers are held by weak van der Waals forces and are separated by 335 pm. This makes graphite soft and slippery: one sheet can slide over the next without any covalent bond being broken.

Graphite is the thermodynamically most stable form (allotrope) of carbon.

Q. Both are pure carbon, so why does graphite conduct electricity while diamond does not?

A. It is a question of whether any electron is free to move. In diamond all four valence electrons of every carbon are locked into \(\mathrm{sp}^3\) sigma bonds, leaving nothing mobile. In graphite only three are used in sigma bonds; the fourth sits in an unhybridised p orbital, and lateral overlap of these orbitals spreads the electrons over the entire layer. Delocalised electrons can drift when a voltage is applied, so graphite conducts along its sheets.

Q2. Diamond and graphite are both pure carbon, yet one is the hardest natural substance and the other is soft and slippery. Is the difference in the atoms, or in the way the atoms are joined?

A. Entirely in the way the atoms are joined. In diamond every carbon is \(\mathrm{sp}^3\) hybridised and bonded to four neighbours in a rigid three-dimensional network of C–C single bonds, so no part of the crystal can move without covalent bonds being broken — hence its extreme hardness. In graphite every carbon is \(\mathrm{sp}^2\) hybridised and bonded to only three neighbours within a flat sheet; the sheets themselves are held to one another only by weak van der Waals forces across a gap of 335 pm. Those weak forces give way easily, letting the layers slide, which is why graphite feels soft and slippery. Same atoms, different architecture, opposite properties.

Diamond vs Graphite vs Fullerene (\(\mathrm{C_{60}}\)) | Feature | Diamond | Graphite | Fullerene \(\mathrm{C_{60}}\) | |—|—|—|—| | Hybridisation of carbon | \(\mathrm{sp}^3\) | \(\mathrm{sp}^2\) | \(\mathrm{sp}^2\)-type, in a curved cage | | Neighbours per carbon | 4 | 3 within a sheet | 3 within the cage | | Structural type | Three-dimensional covalent network | Layers of two-dimensional sheets | Discrete covalent molecule | | C–C distance | 154 pm | 141.5 pm (layers 335 pm apart) | 143.5 pm and 138.3 pm | | Hardness / feel | Hardest natural substance | Soft and slippery | Soft molecular solid | | Electrical conduction | Bad conductor | Conducts along the layers | \(\mathrm{K_3C_{60}}\) is a superconductor below 18 K | | Solubility | Insoluble | Insoluble | Soluble in organic solvents |

Graphite’s two odd properties are put to work separately. Because it conducts, it is shaped into electrodes for electrolysis and for electric-arc furnaces, where a metal electrode would melt or react. Because its layers slide, powdered graphite serves as a dry lubricant in places where oil would fail — in very hot machinery, where oil would burn away, and in locks and hinges, where oil would collect dust and jam the mechanism.

Fullerene

Fullerenes are allotropes of carbon in which carbon molecules are formed by linking a definite number of carbon atoms — for example \(\mathrm{C_{60}}\). This “definite number” is what sets fullerenes apart from diamond and graphite: a fullerene is a genuine molecule of fixed size, not an endless array.

Fullerenes are produced when an electric arc is struck between graphite electrodes in an inert atmosphere of argon or helium. The soot formed contains a significant amount of \(\mathrm{C_{60}}\) fullerene and smaller amounts of other fullerenes, \(\mathrm{C_{32}}\), \(\mathrm{C_{50}}\), \(\mathrm{C_{70}}\) and \(\mathrm{C_{84}}\). The steps are set out in FC 9.2.

Flowchart

FC 9.2 – Production of fullerenes: graphite electrodes → electric arc struck between them → inert atmosphere of argon or helium (no oxygen, so the carbon cannot burn) → soot deposited → soot contains mainly C₆₀ with smaller amounts of C₃₂, C₅₀, C₇₀, C₈₄ → separation to obtain C₆₀

\(\mathrm{C_{60}}\) has a shape like a soccer ball and is called Buckminsterfullerene or bucky ball (Fig. 9.3). It contains twenty hexagonal and twelve pentagonal fused rings of carbons.

Diagram

Fig. 9.3 – Buckminsterfullerene: closed cage of 60 carbon atoms made of 20 hexagonal and 12 pentagonal fused rings, drawn like a football, with one pentagon and one hexagon shaded to distinguish them

The name honours the American architect Buckminster Fuller, who designed geodesic domes built from a network of hexagons and pentagons — exactly the pattern the \(\mathrm{C_{60}}\) cage turned out to have. The molecule was identified in the mid-1980s by Kroto, Smalley and Curl, work for which they received the Nobel Prize in Chemistry in 1996. The shorter nickname, bucky ball, came from the same source.

Unlike diamond and graphite, the \(\mathrm{C_{60}}\) fullerene structure exhibits two distinct distances between the neighbouring carbons, 143.5 pm and 138.3 pm. The two values arise because the bonds shared between two hexagons are not identical to those shared between a hexagon and a pentagon.

Fullerenes are covalent molecules and are soluble in organic solvents. Fullerene \(\mathrm{C_{60}}\) reacts with group 1 metals, forming solids such as \(\mathrm{K_3C_{60}}\). The compound \(\mathrm{K_3C_{60}}\) behaves as a superconductor below 18 K, which means that it conducts electric current with zero resistance.

Q. Diamond and graphite dissolve in nothing, yet \(\mathrm{C_{60}}\) dissolves in organic solvents. Why the difference?

A. Because dissolving requires separating one unit of the solid from the next. In \(\mathrm{C_{60}}\) the units are individual cage molecules held to their neighbours only by weak van der Waals forces, which an organic solvent can overcome. In diamond and graphite there are no separate molecules at all — the covalent bonding runs through the entire crystal, and no solvent can dismantle a covalent network.

Buckminsterfullerene contains twenty hexagonal and twelve pentagonal fused rings, and each carbon atom in the cage is shared between three rings. Show that these figures are consistent with the molecular formula \(\mathrm{C_{60}}\).

Solution: Count the corners contributed by all the rings, then correct for sharing.

Corners of the hexagons: \(20 \times 6 = 120\)

Corners of the pentagons: \(12 \times 5 = 60\)

Total corner positions: \(120 + 60 = 180\)

Each carbon atom sits at a corner where three rings meet, so every atom has been counted three times in that total. The actual number of carbon atoms is therefore

\[\frac{180}{3} = 60\]

which matches the formula \(\mathrm{C_{60}}\). The same counting method — total contributions divided by the sharing factor — is the one you use for atoms shared between unit cells in a crystal, so it is worth being fluent in it.

(question only — no solution) Taking the atomic mass of carbon as 12.01, calculate the molar mass of \(\mathrm{C_{60}}\) and the number of \(\mathrm{C_{60}}\) molecules present in 1.441 g of buckminsterfullerene. (\(N_A = 6.022\times10^{23}\ \text{mol}^{-1}\))

Carbon Nanotubes

Carbon nanotubes are cylindrical in shape, consisting of a rolled-up graphite sheet (Fig. 9.4). Nanotubes can be single-walled (SWNTs), with a diameter of less than 1 nm, or multi-walled (MWNTs), with diameter reaching more than 100 nm. Their lengths range from several micrometres to millimetres.

Diagram

Fig. 9.4 – Carbon nanotube: a hexagonal graphite sheet rolled into a cylinder, shown alongside a magnified portion of the wall labelling a carbon atom and a covalent bond, with single-walled and multi-walled forms drawn side by side

Note the extraordinary proportions implied by those numbers: a tube barely a nanometre across may be a million times longer than it is wide.

Carbon nanotubes are robust. They can be bent, and when released, they will spring back to their original shape. Carbon nanotubes have high electrical and heat conductivities, and the highest strength to weight ratio of any known material to date. The conductivity is inherited from the graphite sheet they are rolled from, with its delocalised \(\pi\) electrons.

The researchers of NASA are working on combining carbon nanotubes with other materials to obtain composites that can be used to build light weight space craft. Every kilogram saved on a spacecraft is fuel saved, which is why a material that is both strong and light is so valuable there.

Graphene

An isolated layer of graphite is called graphene (Fig. 9.5). A graphene sheet is a two dimensional solid — a solid that is only one atom thick, yet extends indefinitely in the other two directions. Graphene has unique electronic properties.

Diagram

Fig. 9.5 – Graphene: a single one-atom-thick sheet of hexagonally arranged carbon atoms, shown slightly rippled, with a note that stacking many such sheets 335 pm apart gives graphite

The discovery of graphene was awarded the Nobel Prize to Geim and Novoselov in 2010. What makes the story memorable is how simple the successful method was: a single atomic layer was lifted from ordinary graphite, showing that a two-dimensional solid, long thought impossible to isolate, had been sitting inside pencil-grade graphite all along.

9.6.2 Allotropes of Phosphorus

Phosphorus is found in different allotropic forms, the important ones being white and red phosphorus.

White (Yellow) Phosphorus

White (yellow) phosphorus consists of discrete tetrahedral \(\mathrm{P_4}\) molecules. The P–P–P bond angle is \(60^\circ\), as shown in Fig. 9.10.

Diagram

Fig. 9.10 – Structure of white phosphorus: a discrete P₄ tetrahedron with the four P atoms at the corners, every P joined to the other three, and one P–P–P bond angle marked 60°

White phosphorus is less stable and hence more reactive, because of the angular strain in the \(\mathrm{P_4}\) molecule. A phosphorus atom would “prefer” its bonds to be spread much wider apart than \(60^\circ\); forcing them into the corners of a small tetrahedron leaves the bonds strained, and strained bonds are easier to break — which is precisely what high reactivity means.

Properties of white phosphorus:

  • It is a translucent white waxy solid.
  • It glows in dark (chemiluminescence) — it emits light as a result of a chemical reaction, not because it has been heated.
  • It is insoluble in water but dissolves in boiling NaOH solution.
  • It is poisonous.
Red Phosphorus

Red phosphorus consists of chains of \(\mathrm{P_4}\) tetrahedra linked together by covalent bonds. Thus it is polymeric in nature, as drawn in Fig. 9.11.

Diagram

Fig. 9.11 – Structure of red phosphorus: several P₄ tetrahedra opened up and joined end to end by covalent P–P bonds into a continuous chain, showing the polymeric linkage

Properties of red phosphorus:

  • It is stable and less reactive.
  • It is odourless. It possesses an iron grey lustre.
  • It does not glow in dark.
  • It is insoluble in water.
  • It is non poisonous.

Red phosphorus is prepared by heating white phosphorus at 573 K in an inert atmosphere. The inert atmosphere is essential — heating white phosphorus in air would simply set it on fire. The conversion is summarised in FC 9.3.

Flowchart

FC 9.3 – Conversion of white to red phosphorus: white phosphorus (discrete strained P₄ tetrahedra) → heat at 573 K in an inert atmosphere → tetrahedra open and link end to end → red phosphorus (polymeric chains of P₄ units, stable and less reactive)

White phosphorus vs Red phosphorus | Feature | White (yellow) phosphorus | Red phosphorus | |—|—|—| | Structure | Discrete tetrahedral \(\mathrm{P_4}\) molecules | Chains of \(\mathrm{P_4}\) tetrahedra linked covalently; polymeric | | P–P–P bond angle | \(60^\circ\), with angular strain | Strain relieved on linking into chains | | Stability and reactivity | Less stable, more reactive | Stable, less reactive | | Appearance | Translucent white waxy solid | Iron grey lustre, odourless | | Glow in dark | Glows (chemiluminescence) | Does not glow | | Solubility in water | Insoluble; dissolves in boiling NaOH solution | Insoluble | | Toxicity | Poisonous | Non poisonous |

Q1. White phosphorus must be kept stored under water, while red phosphorus can sit safely on a shelf. If both are made of nothing but phosphorus atoms, why do they behave so differently?

A. Because the atoms are joined differently. White phosphorus is made of separate \(\mathrm{P_4}\) tetrahedra in which the bond angle is squeezed to \(60^\circ\); the resulting angular strain makes those bonds easy to break, so the solid is highly reactive and ignites on contact with air — hence the layer of water kept over it. In red phosphorus the tetrahedra have opened out and linked to one another by covalent bonds into long polymeric chains. Bonding to more neighbours relieves the strain, and the stable chain structure is far less reactive: red phosphorus does not catch fire in air, does not glow and is not poisonous. Identical atoms, a different arrangement, and a completely different level of danger.

The difference between these two allotropes once had grim consequences. In the nineteenth century match heads were made with white phosphorus, and workers in match factories — many of them very young — developed severe, disfiguring damage to the jaw bone from long exposure to its vapour. The condition, known at the time as phossy jaw, led eventually to international agreement to ban white phosphorus in matches. Modern safety matches use the harmless red allotrope, which is coated on the striking surface of the box rather than on the match head. It is a reminder that in chemistry, “the same element” is not the same as “the same substance”.

(question only — no solution) The P–P–P bond angle in white phosphorus is \(60^\circ\). State what this small angle implies about the strength of the P–P bonds in \(\mathrm{P_4}\), and use your answer to explain why red phosphorus, and not white phosphorus, is the allotrope used in safety matches.

9.7 Molecular Structures of Some Important Compounds of the Group 13, 14 and 15 Elements

  • Which element from the following pairs has higher ionization enthalpy? B and Tl; N and Bi.
  • Does boron form covalent compounds or ionic?

Sections 9.2 and 9.3 established that the lighter elements of groups 13, 14 and 15 have higher ionization enthalpy and, because of their smaller atomic radius, do not form cations readily. These elements form covalent compounds.

Covalent molecules have a definite shape, which is described with the help of bond lengths and bond angles. Inorganic molecules are often represented by molecular formulae indicating their elemental composition — but a formula such as \(\mathrm{B_2H_6}\) tells you only what is present, not how it is put together.

In the case of covalent inorganic molecules, the reactivity is better understood from their structures. A vacant orbital, an incomplete octet or a lone pair in the right place is what decides how a molecule will behave, and none of these is visible in the formula. In this section we will consider molecular structures of common compounds of elements of groups 13, 14 and 15.

9.7.1 Boron Trichloride (\(\mathrm{BCl_3}\))

Boron trichloride (\(\mathrm{BCl_3}\)) is covalent. Here the boron atom is \(\mathrm{sp}^2\) hybridised, having one vacant unhybridised p orbital. Three \(\mathrm{sp}^2\) orbitals lie in one plane, pointing to the corners of an equilateral triangle, and each overlaps with an orbital of a chlorine atom.

B in \(\mathrm{BCl_3}\) has an incomplete octet. Counting shows why: boron contributes three electrons and each of the three B–Cl bonds places two electrons around it, giving only six electrons in the valence shell of boron instead of eight.

\(\mathrm{BCl_3}\) is a nonpolar trigonal planar molecule, as shown in Fig. 9.12.

Diagram

Fig. 9.12 – Structure of BCl₃: boron at the centre with three Cl atoms at the corners of an equilateral triangle in one plane, all Cl–B–Cl angles marked 120°, with the vacant unhybridised p orbital shown perpendicular to the plane

Although each B–Cl bond is polar, the three bond dipoles are equal and point symmetrically outwards at \(120^\circ\) to one another in the same plane, so they cancel exactly and the molecule as a whole is nonpolar.

Q. Boron in \(\mathrm{BCl_3}\) has only six electrons around it. What does the molecule do about the missing pair?

A. It looks for one. The vacant unhybridised p orbital, standing perpendicular to the molecular plane, is exactly the right size and shape to accept a lone pair offered by another species. A molecule that accepts an electron pair in this way is behaving as an electron-pair acceptor, and this single structural feature — a vacant orbital on an atom with an incomplete octet — is the source of most of the chemistry of boron compounds.

9.7.2 Aluminium Chloride (\(\mathrm{AlCl_3}\))

The aluminium atom in aluminium chloride is \(\mathrm{sp}^2\) hybridised, with one vacant unhybrid p orbital. So far the picture matches \(\mathrm{BCl_3}\) exactly. What happens next does not.

Aluminium chloride (\(\mathrm{AlCl_3}\)) exists as the dimer (\(\mathrm{Al_2Cl_6}\)), formed by overlap of the vacant 3d orbital of Al with a lone pair of electrons of Cl. Two \(\mathrm{AlCl_3}\) units join together, each aluminium accepting a lone pair from a chlorine atom belonging to the other unit, as shown in Fig. 9.13.

Diagram

Fig. 9.13 – Structure of the Al₂Cl₆ dimer: two AlCl₃ units joined through two bridging Cl atoms, with arrows drawn from the lone pair of each bridging Cl towards the Al of the other unit to show the coordinate overlap, and the four terminal Cl atoms labelled

The arrow in such a structure is not an ordinary shared-pair bond. It shows that both electrons of the pair come from the same atom — the chlorine — while the aluminium contributes only an empty orbital.

\(\mathrm{BCl_3}\) vs \(\mathrm{AlCl_3}\) | Feature | \(\mathrm{BCl_3}\) | \(\mathrm{AlCl_3}\) | |—|—|—| | Hybridisation of central atom | \(\mathrm{sp}^2\) | \(\mathrm{sp}^2\) | | Vacant orbital present | One unhybridised p orbital | One unhybrid p orbital, plus vacant 3d orbitals | | Octet of central atom | Incomplete | Incomplete in the monomer | | Existing form | Monomer, trigonal planar and nonpolar | Dimer, \(\mathrm{Al_2Cl_6}\) | | How the deficiency is met | Retained; the molecule remains an electron-pair acceptor | Satisfied by accepting a lone pair from Cl of the other unit |

The difference traces back to a point made in section 9.4.2. Boron belongs to the second period and has no d orbitals at all, so it cannot use the route aluminium uses; aluminium, in the third period, has vacant 3d orbitals available in its valence shell and dimerises readily.

(question only — no solution) \(\mathrm{AlCl_3}\) exists as the dimer \(\mathrm{Al_2Cl_6}\), but \(\mathrm{BCl_3}\) does not dimerise. Explain this difference in terms of the orbitals available on the central atom of each molecule.

9.7.3 Orthoboric Acid / Boric Acid (\(\mathrm{H_3BO_3}\))

The orthoboric acid molecule has a central boron atom bound to three –OH groups. Each of the three \(\mathrm{sp}^2\) orbitals of boron therefore holds an –OH group instead of a chlorine, giving a flat, triangular \(\mathrm{B(OH)_3}\) unit.

The solid orthoboric acid has a layered crystal structure in which trigonal planar \(\mathrm{B(OH)_3}\) units are joined together by hydrogen bonds, as shown in Fig. 9.14.

Diagram

Fig. 9.14 – Layered structure of solid orthoboric acid: several planar B(OH)₃ units in one plane, with dotted lines showing hydrogen bonds between the –OH group of one unit and the oxygen of a neighbouring unit, forming a two-dimensional sheet

A hydrogen bond is the attraction between a hydrogen atom already bonded to a strongly electronegative atom, such as oxygen, and a lone pair on the oxygen of a neighbouring molecule. Being much weaker than a covalent bond, these links let one layer separate from the next fairly easily — which is why solid boric acid is a soft, flaky solid rather than a hard crystal.

9.7.4 Diborane (\(\mathrm{B_2H_6}\))

Boron has only three valence electrons. In diborane each boron atom is \(\mathrm{sp}^3\) hybridized. Three of these hybrid orbitals are half filled, while the fourth \(\mathrm{sp}^3\) hybrid orbital is vacant.

The two half filled \(\mathrm{sp}^3\) hybrid orbitals of each B atom overlap with the 1s orbitals of two H atoms and form four B–H covalent bonds. These hydrogen atoms are located at the terminal positions, and each of these four bonds is an ordinary 2-centre – 2-electron bond: two atomic centres sharing two electrons between them.

Besides these, the ‘1s’ orbital of each of the remaining two H atoms simultaneously overlaps with the half filled hybrid orbital of one B atom and the vacant hybrid orbital of the other B atom. This type of overlap produces two three centred – two electron bonds (3c – 2e), or banana bonds. The hydrogen atoms involved in the (3c – 2e) bonds are the bridging H-atoms. The overlap is drawn in Fig. 9.15.

Diagram

Fig. 9.15 – Orbital picture of the 3-centre–2-electron bonds in diborane: two B atoms with their sp³ orbitals, four terminal H atoms in ordinary 2c–2e overlaps, and the two bridging H atoms each overlapping simultaneously with a half-filled orbital of one B and the vacant orbital of the other B; alongside it, the same molecule drawn with two curved “banana” bonds joining B–H–B

The name comes from the shape: because a single pair of electrons has to spread over three atomic centres, the region of overlap curves outward like a banana instead of lying straight along a line between two atoms.

In diborane the two B atoms and the four terminal H atoms lie in one plane, while the two bridging H atoms lie symmetrically above and below this plane. The full geometry, with its bond lengths and angles, is given in Fig. 9.16.

Diagram

Fig. 9.16 – Bonding and structure of diborane: two B atoms with four terminal H atoms in one plane and two bridging H atoms above and below it, marking the bridging H–B–H angle 97°, the terminal H–B–H angle 120°, the terminal B–H bond length 119 pm and the bridging B–H bond length 134 pm

Notice that the bridging B–H distance of 134 pm is longer than the terminal one of 119 pm. That is exactly what a shared-out electron pair should give: two electrons stretched over three centres hold each pair of atoms less tightly than two electrons concentrated between two centres.

Diborane, \(\mathrm{B_2H_6}\), is called an electron-deficient molecule. Using an electron count, show that its twelve valence electrons are not enough for eight ordinary two-centre bonds, and account for the bonding actually present.

Solution: First count the valence electrons available. Boron has 3 valence electrons and hydrogen has 1:

\[2 \times 3 + 6 \times 1 = 6 + 6 = 12\ \text{valence electrons, that is, 6 pairs}\]

A structure with eight ordinary 2-centre – 2-electron bonds (six B–H bonds plus a B–B bond, for instance) would need more pairs than the molecule possesses, so such a structure is impossible.

Now account for the 12 electrons as the actual structure does. Four terminal B–H bonds, each a 2c – 2e bond, use

\[4 \times 2 = 8\ \text{electrons}\]

The electrons left over are

\[12 - 8 = 4\ \text{electrons, that is, 2 pairs}\]

These two pairs form the two three-centre – two-electron bonds, one pair holding together the B–H–B bridge on each side of the plane. The count balances exactly, and the “deficiency” is not a shortage of bonding but a different way of spreading the same electrons over more atoms.

(question only — no solution) For one molecule of \(\mathrm{B_2H_6}\), state (i) the number of terminal hydrogen atoms, (ii) the number of bridging hydrogen atoms, (iii) the number of 2-centre – 2-electron bonds, and (iv) the number of 3-centre – 2-electron bonds. Then name the type of hybridisation of each boron atom.

9.7.5 Silicon Dioxide (\(\mathrm{SiO_2}\))

Silicon dioxide is commonly known as silica. Quartz, cristobalite and tridymite are different crystalline forms of silica. They are inter-convertible at a suitable temperature — heating one form past the right temperature rearranges the same \(\mathrm{Si\!-\!O}\) framework into another, which is allotropy of the kind you met in section 9.6, here shown by a compound rather than an element.

Silicon dioxide (silica) is a covalent three dimensional network solid, in which each silicon atom is covalently bound in a tetrahedral manner to four oxygen atoms. The crystal contains eight membered rings having alternate silicon and oxygen atoms, as drawn in Fig. 9.17.

Diagram

Fig. 9.17 – Structure of SiO₂: extended network of alternating Si and O atoms, each Si bonded tetrahedrally to four O atoms and each O bridging two Si atoms, with one eight-membered Si–O–Si–O ring outlined and the bonds continuing beyond the edges of the figure

Compare this with carbon dioxide, the oxide of the element directly above silicon. Carbon forms strong \(\pi\) bonds to oxygen and so exists as small \(\mathrm{O=C=O}\) molecules, a gas at room temperature. Silicon, being larger, forms \(\pi\) bonds to oxygen poorly and instead makes four single bonds, building an endless network — which is why silica is a hard, high-melting solid while \(\mathrm{CO_2}\) is a gas.

Silica is the chief constituent of ordinary sand, and two of its uses depend on the very purity that sand lacks. Fibres drawn from extremely pure silica form the core of optical fibre cables, carrying telephone and internet signals as pulses of light over long distances with very little loss. A thin slice cut from a quartz crystal, meanwhile, vibrates at a fixed frequency when a voltage is applied across it, and it is this steady vibration that keeps a quartz wristwatch accurate.

9.7.6 Nitric Acid (\(\mathrm{HNO_3}\))

Nitric acid is a strong, oxidising mineral acid. The central nitrogen atom is \(\mathrm{sp}^2\) hybridised, so the \(\mathrm{NO_3}\) part of the molecule is planar.

\(\mathrm{HNO_3}\) exhibits the resonance phenomenon. Two equivalent structures can be written for the molecule, differing only in which of the two terminal oxygen atoms carries the double bond, as shown in Fig. 9.18.

Diagram

Fig. 9.18 – Resonance in nitric acid: two resonance structures of HNO₃ connected by a double-headed arrow, showing the N=O double bond on either terminal oxygen with the formal ⊕ on N and ⊖ on the singly bonded O; beside them the resonance hybrid with both terminal N–O bonds shown identical and each oxygen carrying δ⊖, and the bond angles 120° and 130° and the bond lengths 96 pm and 140.6 pm marked

NoteDefinition

Resonance is the representation of a molecule by two or more structures that differ only in the placement of electrons; the real molecule is a single resonance hybrid of all of them, and does not switch back and forth between the structures.

The measurable consequence is important. Because the true molecule is one hybrid and not an alternating pair, the two terminal nitrogen–oxygen bonds are identical to each other and each oxygen carries an equal partial negative charge, marked \(\delta^-\) in the hybrid. Spreading the electrons over more than one bond in this way also lowers the energy of the molecule, which is why resonance is always stabilising.

Nitric acid was known to early chemists as aqua fortis, “strong water”, for its power to attack metals that ordinary acids left alone. Mixed with hydrochloric acid it forms a liquid that will dissolve even gold, and that mixture earned the still grander name aqua regia — “royal water” — because gold was the royal metal.

concentrated nitric acid, \(\mathrm{HNO_3}\) (What it does → it acts both as a strong acid, supplying \(\mathrm{H^+}\), and as a strong oxidising agent, in which role the nitrogen in the \(+5\) state is reduced. Conditions it needs → the concentrated, and often warm, acid for oxidation; the dilute acid where only its acidic behaviour is wanted. Typically used for → preparing nitrates, oxidising metals and non-metals, and, mixed with concentrated sulphuric acid, as the standard nitrating mixture. Handling → it stains and destroys skin on contact, so it is always poured with the vessel held away from the face, and dilution is done by adding the acid to water.)

(question only — no solution) In the resonance hybrid of nitric acid, both terminal oxygen atoms are shown carrying a partial negative charge \(\delta^-\), and the two N–O bonds are shown as identical. Explain both of these observations, and state the hybridisation of the central nitrogen atom.

9.7.7 Orthophosphoric Acid / Phosphoric Acid (\(\mathrm{H_3PO_4}\))

Phosphorus forms a number of oxyacids. Orthophosphoric acid is a strong, non toxic mineral acid. It contains three ionizable acidic hydrogens — the three that are attached to oxygen atoms, and which can therefore be released into water as \(\mathrm{H^+}\).

The central phosphorous atom is tetrahedral, with four oxygen atoms arranged around it, as shown in Fig. 9.19.

Diagram

Fig. 9.19 – Structure of orthophosphoric acid: central P atom bonded tetrahedrally to four O atoms — one by a double bond (P=O) and three as –OH groups — with the three ionizable hydrogens labelled

Notice why exactly three of the four hydrogens’ positions matter here: the molecule has three –OH groups, and only a hydrogen attached to oxygen is acidic enough to be released. This is also why phosphoric acid can be neutralised in three separate stages, giving three different series of salts.

Because orthophosphoric acid is a strong acid and yet non toxic, it can be used where other mineral acids cannot. A dilute solution of it is the acidity regulator that gives cola drinks their sharp taste, and much larger quantities go into the manufacture of phosphate fertilizers, which return to the soil the phosphorus that crops remove from it. The same acid is also used to treat rusted iron before painting, where it converts the loose rust into a firmer surface layer that paint can grip.

Assign the oxidation state of phosphorus in \(\mathrm{H_3PO_4}\), state whether it is the group oxidation state, and give the number of ionizable hydrogen atoms in the molecule.

Solution: Let the oxidation number of phosphorus be \(x\). Take hydrogen as \(+1\) and oxygen as \(-2\), and set the sum equal to zero for the neutral molecule:

\[3(+1) + x + 4(-2) = 0\]

\[3 + x - 8 = 0 \quad \Rightarrow \quad x = +5\]

Phosphorus is in the \(+5\) state. Since phosphorus belongs to group 15, whose group oxidation state is \(+5\), this is indeed the group oxidation state — consistent with phosphorus being a lighter member of the group, for which the group oxidation state is the stable one.

The number of ionizable hydrogen atoms is three, because all three hydrogens are held in –OH groups attached to oxygen. The structure, not the formula, is what tells you this: a formula like \(\mathrm{H_3PO_4}\) would look the same whether or not the hydrogens were attached to oxygen.

Find out the structural formulae of various oxyacids of phosphorus.

9.8 Chemistry of Notable Compounds of Elements of Groups 13, 14 and 15

9.8.1 Borax (\(\mathrm{Na_2B_4O_7}\))

Borax is one of the most important boron compounds. The crystalline borax has the formula \(\mathrm{Na_2B_4O_7\cdot 10H_2O}\), or, written to show its actual structure, \(\mathrm{Na_2[B_4O_5(OH)_4]\cdot 8H_2O}\).

The two formulae describe the same substance. The first simply counts up all ten water molecules; the second recognises that four of those ten hydrogens and oxygens are in fact held as –OH groups built into the borate anion itself, leaving only eight molecules as water of crystallisation.

Occurrence

Borax naturally occurs as tincal, which contains about 90% borax, in certain inland lakes in India, Tibet and California (U.S.A.). Such lakes have no outlet to the sea, so dissolved salts brought in by streams stay behind and concentrate as the water evaporates.

Preparation

Borax is prepared from the mineral colemanite by boiling it with a solution of sodium carbonate:

\[\mathrm{Ca_2B_6O_{11}} + 2\,\mathrm{Na_2CO_3} \xrightarrow{\ \Delta\ } \mathrm{Na_2B_4O_7} + 2\,\mathrm{NaBO_2} + 2\,\mathrm{CaCO_3}\downarrow\]

The downward arrow marks calcium carbonate as an insoluble precipitate. This is what makes the preparation practical: the calcium leaves the solution as a solid that can be filtered off, while the borax stays dissolved and is recovered by crystallising the filtrate.

Properties

i. Borax is a white crystalline solid.

ii. Borax dissolves in water and gives an alkaline solution due to its hydrolysis:

\[\mathrm{Na_2B_4O_7} + 7\,\mathrm{H_2O} \rightarrow 2\,\mathrm{NaOH} + 4\,\mathrm{H_3BO_3}\]

The second product is ortho boric acid, whose structure you met in section 9.7.3.

Q. Borax is a salt, so why is its solution alkaline rather than neutral?

A. Because the two products of hydrolysis are not equally strong. Sodium hydroxide is a strong base and dissociates fully, whereas orthoboric acid is a very weak acid and barely dissociates at all. The \(\mathrm{OH^-}\) ions released therefore have no matching supply of \(\mathrm{H^+}\) to neutralise them, and the solution turns alkaline. This is the general rule for the salt of a weak acid and a strong base.

iii. On heating, borax first loses its water molecules and swells. On further heating it turns into a transparent liquid, which solidifies into a glass-like material known as the borax bead:

\[\mathrm{Na_2B_4O_7\cdot 10H_2O} \xrightarrow[\ -10\,\mathrm{H_2O}\ ]{\ \Delta\ } \mathrm{Na_2B_4O_7} \xrightarrow{\ \Delta\ } 2\,\mathrm{NaBO_2} + \mathrm{B_2O_3}\]

The borax bead, consisting of sodium metaborate and boric anhydride, is used to detect coloured transition metal ions, under the name borax bead test. For example, when borax is heated in a Bunsen burner flame with CoO on a loop of platinum wire, a blue coloured \(\mathrm{Co(BO_2)_2}\) bead is formed. The whole sequence is set out in FC 9.4.

Flowchart

FC 9.4 – Borax bead test: borax·10H₂O on a platinum wire loop → heat in Bunsen flame → loses 10H₂O and swells → anhydrous Na₂B₄O₇ → further heating → transparent liquid → cools to a clear glassy bead of NaBO₂ + B₂O₃ → touch the hot bead to the metal oxide (e.g. CoO) and reheat → coloured metaborate bead (blue Co(BO₂)₂) → identify the metal ion from the colour

Diagram

Fig. 9.20 – Borax bead test: a platinum wire with a small loop at the end held in the non-luminous Bunsen flame, showing the glassy transparent bead in the loop, with a labelled inset of the coloured bead after the metal oxide has been added

Platinum is chosen for the wire because it neither melts nor reacts at flame temperature, so it contributes no colour of its own to the test.

iv. Borax, when heated with ethyl alcohol and concentrated \(\mathrm{H_2SO_4}\) acid, gives volatile vapours of triethyl borate which burn with a green edged flame:

\[\mathrm{Na_2B_4O_7} + \mathrm{H_2SO_4} + 5\,\mathrm{H_2O} \rightarrow \mathrm{Na_2SO_4} + 4\,\mathrm{H_3BO_3}\]

\[\mathrm{H_3BO_3} + 3\,\mathrm{C_2H_5OH} \rightarrow \mathrm{B(OC_2H_5)_3} + 3\,\mathrm{H_2O}\]

The above reaction is used as a test for detection and removal of borate in qualitative analysis. Notice that it performs both jobs at once: the green-edged flame identifies borate, and because triethyl borate is volatile, the boron leaves the mixture as vapour and stops interfering with the tests that follow.

two tests for the borate radical | Feature | Borax bead test | Triethyl borate flame test | |—|—|—| | Reagents needed | Borax on a platinum loop, plus the metal oxide | Ethyl alcohol and concentrated \(\mathrm{H_2SO_4}\) | | What is observed | A coloured glassy bead | A flame with a green edge | | What it identifies | The coloured transition metal ion | The borate radical | | Additional purpose | — | Removes borate from the mixture as volatile vapour |

Calculate the percentage by mass of boron in anhydrous borax, \(\mathrm{Na_2B_4O_7}\). (Atomic masses: Na \(= 22.99\), B \(= 10.81\), O \(= 16.00\))

Solution: First find the molar mass of \(\mathrm{Na_2B_4O_7}\) by adding the contribution of each element:

Na: \(2 \times 22.99 = 45.98\)

B: \(4 \times 10.81 = 43.24\)

O: \(7 \times 16.00 = 112.00\)

\[M = 45.98 + 43.24 + 112.00 = 201.22\ \text{g mol}^{-1}\]

The percentage of boron is the mass of boron in one mole expressed as a fraction of the whole:

\[\%\,\mathrm{B} = \frac{43.24}{201.22} \times 100 = 21.49\ \%\]

So a little over one-fifth of the mass of anhydrous borax is boron. Note that the crystalline form, \(\mathrm{Na_2B_4O_7\cdot 10H_2O}\), would give a smaller percentage, because the ten water molecules add mass without adding any boron — a reminder to check whether a formula includes water of crystallisation before starting such a calculation.

Uses

Borax is used

i. in the manufacture of optical and hard borosilicate glass.

ii. as a flux for soldering and welding.

iii. as a mild antiseptic in the preparation of medical soaps.

iv. in qualitative analysis for the borax bead test.

v. as a brightener in washing powder.

A flux, in the second use, is a substance that dissolves the oxide film on a metal surface so that the solder can wet the clean metal beneath; without it the joint will not hold.

(question only — no solution) Write the balanced equation for the hydrolysis of borax in water, name both products, and state, with a reason, whether the resulting solution will turn red litmus blue or blue litmus red.

9.8.2 Silicones

Silicones represent organo-silicon polymers where \(\{\mathrm{R_2SiO}\}\) is a repeating unit. These are held together by the \(-\mathrm{Si}-\mathrm{O}-\mathrm{Si}-\) linkage. Silicones have the empirical formula \(\mathrm{R_2SiO}\), where R is a \(\mathrm{CH_3}\) or \(\mathrm{C_6H_5}\) group.

NoteDefinition

Silicones are organo-silicon polymers built from the repeating unit \(\{\mathrm{R_2SiO}\}\), in which silicon atoms are joined to one another through oxygen atoms by \(-\mathrm{Si}-\mathrm{O}-\mathrm{Si}-\) linkages, R being an alkyl or aryl group.

This empirical formula is similar to that of a ketone, \(\mathrm{R_2CO}\), and hence the name silicones. The resemblance is only in the formula, however: a ketone has a genuine \(\mathrm{C{=}O}\) double bond, whereas silicon prefers single bonds to oxygen and so builds a long chain instead of a small molecule — the same preference that made silica a network solid in section 9.7.5.

Preparation

The starting materials for the manufacture of silicones are alkyl or aryl substituted silicon chlorides, \(\mathrm{R_nSiCl_{(4-n)}}\), where R is an alkyl or aryl group.

When methyl chloride reacts with silicon in the presence of a copper catalyst at a temperature of 573 K, various types of methyl substituted chlorosilane of the formulae \(\mathrm{MeSiCl_3}\), \(\mathrm{Me_2SiCl_2}\) and \(\mathrm{Me_3SiCl}\), with small amounts of \(\mathrm{Me_4Si}\), are formed. (Here Me is the usual short form for the methyl group, \(\mathrm{CH_3}\).)

\[2\,\mathrm{CH_3Cl} + \mathrm{Si} \xrightarrow{\ \mathrm{Cu\ powder},\ 573\ \text{K}\ } \mathrm{(CH_3)_2SiCl_2}\]

Hydrolysis of dimethyl dichlorosilane, \(\mathrm{(CH_3)_2SiCl_2}\), followed by condensation polymerisation, yields straight chain silicone polymers. Water replaces the two chlorine atoms by two –OH groups, hydrogen chloride being eliminated:

\[\mathrm{(CH_3)_2SiCl_2} + 2\,\mathrm{H_2O} \xrightarrow{\ -2\,\mathrm{HCl}\ } \mathrm{(CH_3)_2Si(OH)_2}\]

The resulting silanediol then joins to its neighbours, each new \(-\mathrm{Si}-\mathrm{O}-\mathrm{Si}-\) linkage being formed with the loss of one water molecule:

\[n\ \mathrm{(CH_3)_2Si(OH)_2} \xrightarrow{\ \text{polymerisation},\ -\mathrm{H_2O}\ } [-\mathrm{Si(CH_3)_2}-\mathrm{O}-]_n\]

The three stages are shown as structural formulae in Fig. 9.21, and summarised in FC 9.5.

Diagram

Fig. 9.21 – Preparation of a silicone in three structural stages: (a) Cl–Si(CH₃)₂–Cl with the two methyl groups drawn above and below the Si; (b) HO–Si(CH₃)₂–OH after hydrolysis; (c) the straight-chain polymer –[Si(CH₃)₂–O]ₙ– with brackets and subscript n, showing the repeating –Si–O–Si– backbone

Flowchart

FC 9.5 – Manufacture of silicones: CH₃Cl + Si → (Cu powder, 573 K) → mixture of MeSiCl₃, Me₂SiCl₂, Me₃SiCl with a little Me₄Si → separate Me₂SiCl₂ → hydrolysis with H₂O (−2HCl) → (CH₃)₂Si(OH)₂ → condensation polymerisation (−H₂O) → straight-chain silicone polymer; branch: add (CH₃)₃SiCl at the ends → chain growth stopped → polymer of controlled length

Controlling the Chain Length

The chain length of the polymer can be controlled by adding \(\mathrm{(CH_3)_3SiCl}\) at the end, as shown in Fig. 9.22.

Diagram

Fig. 9.22 – Chain-length control in silicones: the polymer written as CH₃–Si(CH₃)₂–O–[Si(CH₃)₂–O]ₙ–Si(CH₃)₂–CH₃, with the two terminal (CH₃)₃Si– groups circled and labelled as chain-stopping end units

The reason this works is worth seeing. A molecule of \(\mathrm{(CH_3)_2SiCl_2}\) can be linked at two positions, so it lets the chain continue growing at both ends. A molecule of \(\mathrm{(CH_3)_3SiCl}\) has only one position available, so once it attaches, that end of the chain is closed and can grow no further. By choosing how much of it to add, a manufacturer decides how long the chains will be, and therefore whether the product is a thin oil, a thick grease or a rubbery solid.

(question only — no solution) Predict the product formed when \(\mathrm{(CH_3)_3SiCl}\) is hydrolysed, and explain why adding this compound to the polymerisation mixture shortens the silicone chains while adding \(\mathrm{(CH_3)_2SiCl_2}\) lengthens them.

Properties of Silicones

i. They are water repellant.

ii. They have high thermal stability.

iii. They are good electrical insulators.

iv. They are resistant to oxidation and chemicals.

These four properties all follow from the same structural facts. The backbone is made of strong \(\mathrm{Si}-\mathrm{O}\) bonds, which resist heat and oxidation far better than the \(\mathrm{C}-\mathrm{C}\) backbone of an ordinary organic polymer, while the methyl or phenyl groups form a non-polar outer surface that water cannot wet and current cannot pass through.

Uses of Silicones

They are used as

i. insulating material for electrical appliances.

ii. water proofing of fabrics.

iii. sealant.

iv. high temperature lubricants.

v. for mixing in paints and enamels to make them resistant to high temperature, sunlight and chemicals.

The thermal stability of the \(\mathrm{Si}-\mathrm{O}\) backbone is why silicone bakeware and silicone spatulas can go into a hot oven or rest against a hot pan without softening, charring or giving off fumes, where an ordinary plastic would fail at once. The same stability, combined with chemical resistance, is why silicone rubber is chosen for gaskets and seals around ovens and engines — parts that must stay flexible through thousands of heating and cooling cycles.

9.8.3 Ammonia (\(\mathrm{NH_3}\))

Preparation

1. Ammonia is present in small quantities in air and soil, where it is formed by the decomposition of nitrogenous organic matter such as urea:

\[\mathrm{NH_2CONH_2} + 2\,\mathrm{H_2O} \rightarrow \mathrm{(NH_4)_2CO_3} \rightleftharpoons 2\,\mathrm{NH_3} + \mathrm{H_2O} + \mathrm{CO_2}\]

Urea first takes up water to give ammonium carbonate, which is itself unstable and breaks down to release ammonia, water and carbon dioxide. This is the reaction responsible for the sharp smell of stale urine and of decaying organic matter.

2. Ammonia is prepared on the laboratory scale by decomposition of ammonium salts with calcium hydroxide or caustic soda:

\[2\,\mathrm{NH_4Cl} + \mathrm{Ca(OH)_2} \rightarrow 2\,\mathrm{NH_3} + \mathrm{CaCl_2} + 2\,\mathrm{H_2O}\]

\[\mathrm{(NH_4)_2SO_4} + 2\,\mathrm{NaOH} \rightarrow 2\,\mathrm{NH_3} + \mathrm{Na_2SO_4} + 2\,\mathrm{H_2O}\]

In both reactions a strong base takes the extra proton away from the ammonium ion, converting \(\mathrm{NH_4^{+}}\) back into free \(\mathrm{NH_3}\), which escapes as a gas.

3. On the large scale, ammonia is prepared by direct combination of dinitrogen and dihydrogen. This is Haber’s process:

\[\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \rightleftharpoons 2\,\mathrm{NH_3}\text{(g)}; \qquad \Delta_f H^\circ = -46.1\ \text{kJ mol}^{-1}\]

\(\Delta_f H^\circ\), standard enthalpy of formation (Unit: \(\text{kJ mol}^{-1}\). The superscript \(^\circ\) means standard conditions and the subscript f means formation from the elements. A negative value, as here, means the formation is exothermic — heat is released to the surroundings; a positive value would mean heat is absorbed.)

three routes to ammonia | Feature | In air and soil | Laboratory scale | Haber’s process | |—|—|—|—| | Starting material | Nitrogenous matter such as urea | An ammonium salt | \(\mathrm{N_2}\) and \(\mathrm{H_2}\) | | Other reagent | Water | \(\mathrm{Ca(OH)_2}\) or NaOH | Catalyst | | Conditions | Ordinary temperature, natural decay | Gentle heating | \(200 \times 10^5\ \text{Pa}\), \(\sim 700\ \text{K}\) | | Scale and purpose | Traces only, natural | Small quantities for tests | Industrial, for fertilizers |

Optimum Conditions for Haber’s Process

High pressure favours the formation of ammonia. The reason can be read off the equation: four molecules of gas on the left produce only two on the right, so squeezing the mixture pushes the reaction towards the side that occupies less volume — the ammonia side.

The optimum conditions for the production of ammonia are a high pressure of \(200 \times 10^5\ \text{Pa}\) (200 atm), a temperature of about 700 K, and the use of a catalyst such as iron oxide with trace amounts of \(\mathrm{K_2O}\) and \(\mathrm{Al_2O_3}\). Under these conditions equilibrium is attained rapidly. The industrial sequence is given in FC 9.6.

Flowchart

FC 9.6 – Haber’s process: purified N₂ (from air) + H₂ → compressed to 200 × 10⁵ Pa → heated to ~700 K → passed over iron oxide catalyst with traces of K₂O and Al₂O₃ → equilibrium mixture of N₂, H₂ and NH₃ → NH₃ liquefied and separated → unreacted N₂ and H₂ returned to the catalyst chamber (cycle)

Q. The formation of ammonia is exothermic, so why is the mixture heated to about 700 K instead of being kept cool?

A. Because two different things matter in a manufacturing plant — how much ammonia can form, and how fast it forms. A lower temperature would allow a greater yield at equilibrium, but the reaction would be far too slow to be useful. The chosen temperature, together with the catalyst, is a working compromise: it sacrifices a little yield so that equilibrium is reached rapidly, and the ammonia is removed as it forms so that more can be made.

Haber’s process is named after Fritz Haber, who in the years just before the First World War showed that atmospheric nitrogen could be combined with hydrogen over a catalyst under high pressure; Carl Bosch then solved the formidable engineering problem of doing this on an industrial scale. Both received Nobel Prizes for the work — Haber in Chemistry for the synthesis, and Bosch later for high-pressure methods. Until then, crops depended almost entirely on nitrogen already present in the soil, so this single reaction changed how much food the world could grow.

In Haber’s process, calculate the mass of ammonia obtained from 28.0 g of dinitrogen, assuming complete conversion, and the mass of dihydrogen required for it. (Atomic masses: N \(= 14.0\), H \(= 1.0\))

Solution: The balanced equation is

\[\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \rightleftharpoons 2\,\mathrm{NH_3}\text{(g)}\]

Molar mass of \(\mathrm{N_2} = 2 \times 14.0 = 28.0\ \text{g mol}^{-1}\), so

\[n_{\mathrm{N_2}} = \frac{28.0\ \text{g}}{28.0\ \text{g mol}^{-1}} = 1.00\ \text{mol}\]

From the equation, 1 mol \(\mathrm{N_2}\) gives 2 mol \(\mathrm{NH_3}\) and requires 3 mol \(\mathrm{H_2}\).

Molar mass of \(\mathrm{NH_3} = 14.0 + 3(1.0) = 17.0\ \text{g mol}^{-1}\):

\[m_{\mathrm{NH_3}} = 2.00\ \text{mol} \times 17.0\ \text{g mol}^{-1} = 34.0\ \text{g}\]

Molar mass of \(\mathrm{H_2} = 2.0\ \text{g mol}^{-1}\):

\[m_{\mathrm{H_2}} = 3.00\ \text{mol} \times 2.0\ \text{g mol}^{-1} = 6.0\ \text{g}\]

So 34.0 g of ammonia is obtained, using 6.0 g of dihydrogen. As a check on the arithmetic, notice that \(28.0 + 6.0 = 34.0\) — mass is conserved, as it must be.

Physical Properties

i. Ammonia is a colourless gas with a pungent odour.

ii. It has a freezing point of 198.4 K and a boiling point of 239.7 K.

iii. It is highly soluble in water. The concentrated aqueous solution of \(\mathrm{NH_3}\) is called liquor ammonia.

Ammonia has a higher melting point and boiling point than its small molecular mass would suggest, because in the solid and liquid state \(\mathrm{NH_3}\) molecules get associated together through hydrogen bonding. Thus more energy is required to break such intermolecular hydrogen bonds before the substance can melt or boil. The same hydrogen bonding, formed now with water molecules, explains why the gas is so highly soluble in water.

Chemical Properties

i. Basic nature: The aqueous solution of ammonia is basic in nature, due to the formation of \(\mathrm{OH^-}\) ions:

\[\mathrm{NH_3}\text{(g)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{NH_4^{+}}\text{(aq)} + \mathrm{OH^{-}}\text{(aq)}\]

The lone pair on nitrogen takes a proton from a water molecule, and the \(\mathrm{OH^-}\) left behind is what makes the solution basic. Since the reaction is reversible and does not go to completion, ammonia is a weak base.

ii. Ammonia reacts with acids to form ammonium salts:

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

\[2\,\mathrm{NH_3} + \mathrm{H_2SO_4} \rightarrow \mathrm{(NH_4)_2SO_4}\]

iii. The aqueous solution of ammonia precipitates out hydroxides (or hydrated oxides) of metals from their salt solutions:

\[\mathrm{ZnSO_4}\text{(aq)} + 2\,\mathrm{NH_4OH}\text{(aq)} \rightarrow \mathrm{Zn(OH)_2}\text{(s)} + \mathrm{(NH_4)_2SO_4}\text{(aq)}\]

Here \(\mathrm{Zn(OH)_2}\) appears as a white precipitate.

\[\mathrm{FeCl_3}\text{(aq)} + 2\,\mathrm{NH_4OH}\text{(aq)} \rightarrow \mathrm{Fe_2O_3\cdot xH_2O} + \mathrm{NH_4Cl}\text{(aq)}\]

The formula \(\mathrm{Fe_2O_3\cdot xH_2O}\) is written with an \(x\) because the number of water molecules held in the gelatinous brown precipitate is not fixed — this is what “hydrated oxide” means.

iv. Complex formation: The lone pair of electrons on the nitrogen atom facilitates complexation of ammonia with transition metal ions:

\[\mathrm{Cu^{2+}}\text{(aq)} + 4\,\mathrm{NH_3}\text{(aq)} \rightarrow [\mathrm{Cu(NH_3)_4}]^{2+}\text{(aq)}\]

The product is deep blue.

\[\mathrm{AgCl}\text{(s)} + 2\,\mathrm{NH_3}\text{(aq)} \rightarrow [\mathrm{Ag(NH_3)_2}]\mathrm{Cl}\text{(aq)}\]

The product here is colourless. This reaction is used for the detection of metal ions such as \(\mathrm{Cu^{2+}}\) and \(\mathrm{Ag^{+}}\).

In each case the ammonia molecule donates its lone pair to the metal ion, forming a coordinate bond — the same kind of bond that held the \(\mathrm{Al_2Cl_6}\) dimer together in section 9.7.2, with ammonia now playing the part of the donor. Note also what the second reaction achieves: solid silver chloride, insoluble in water, dissolves to give a clear solution, which is why the test is so easy to see.

aqueous ammonia, \(\mathrm{NH_4OH}\) (What it does → supplies a small, controlled concentration of \(\mathrm{OH^-}\), being a weak base, and supplies \(\mathrm{NH_3}\) molecules with a lone pair to donate. Conditions → added drop by drop to precipitate a metal hydroxide; added in excess to dissolve the precipitate again as a complex where one can form. Typically used for → precipitating hydroxides such as \(\mathrm{Zn(OH)_2}\), and identifying \(\mathrm{Cu^{2+}}\) by the deep blue \([\mathrm{Cu(NH_3)_4}]^{2+}\) and \(\mathrm{Ag^{+}}\) by the colourless \([\mathrm{Ag(NH_3)_2}]\mathrm{Cl}\). Handling → the pungent vapour irritates the eyes and airways, so the bottle is opened and the solution added in a well-ventilated place, never with the face over the mouth of the vessel.)

Aqueous ammonia is added in excess to a pale blue solution of copper(II) sulphate. State what is observed, write the equation for the reaction, and explain which feature of the ammonia molecule makes it possible.

Solution: On adding excess aqueous ammonia, the solution turns a deep blue colour. The colour is due to the formation of a complex ion:

\[\mathrm{Cu^{2+}}\text{(aq)} + 4\,\mathrm{NH_3}\text{(aq)} \rightarrow [\mathrm{Cu(NH_3)_4}]^{2+}\text{(aq)}\]

The feature responsible is the lone pair of electrons on the nitrogen atom of \(\mathrm{NH_3}\). Nitrogen uses three of its five valence electrons in the three N–H bonds, leaving one pair unshared. Four such molecules each donate their lone pair into vacant orbitals of the \(\mathrm{Cu^{2+}}\) ion, forming four coordinate bonds, and the resulting complex ion is deep blue. This is why the test identifies \(\mathrm{Cu^{2+}}\): the change from pale blue to deep blue is the signal.

Uses

Ammonia is used in

i. the manufacture of fertilizers such as urea, diammonium phosphate, ammonium nitrate, ammonium sulphate, etc.

ii. the manufacture of some inorganic compounds like nitric acid.

iii. as a refrigerant (liquid ammonia).

iv. as a laboratory reagent in qualitative and quantitative analysis (aqueous solution of ammonia).

Liquid ammonia works as a refrigerant because it absorbs a large quantity of heat when it evaporates — the same hydrogen bonding that raises its boiling point must be overcome for it to turn into vapour.

Ammonia also gives a brown precipitate with Nessler’s reagent, an alkaline solution of \(\mathrm{K_2HgI_4}\). The reagent is prepared in two steps, and the third equation shows the test itself:

\[2\,\mathrm{KI} + \mathrm{HgCl_2} \rightarrow \mathrm{HgI_2} + 2\,\mathrm{KCl}\]

\[2\,\mathrm{KI} + \mathrm{HgI_2} \rightarrow \mathrm{K_2HgI_4}\]

\[2\,\mathrm{K_2HgI_4} + \mathrm{NH_3} + 3\,\mathrm{KOH} \rightarrow \mathrm{H_2N\!-\!HgO\!-\!HgI} + 7\,\mathrm{KI} + 2\,\mathrm{H_2O}\]

The brown precipitate formed is Millon’s base. Because the colour appears even with very little ammonia present, this test is used to detect traces of ammonia — for example, in checking water samples.

Nessler’s reagent (What it is → an alkaline solution of potassium tetraiodomercurate(II), \(\mathrm{K_2HgI_4}\), prepared by adding KI to \(\mathrm{HgCl_2}\) and then dissolving the \(\mathrm{HgI_2}\) formed in more KI. What it does → reacts with ammonia in alkaline medium to give the brown precipitate of Millon’s base. Typically used for → detecting small amounts of ammonia or ammonium ion. Handling → it contains mercury compounds, which are toxic, so the used solution is collected for proper disposal and never washed down the sink.)

Two names in this single test belong to nineteenth-century chemists rather than to any substance — the reagent is named after Julius Nessler and the brown product after Auguste Millon. Such reagent names are common in qualitative analysis, and they are worth noticing: a name that looks like a technical term is often just the surname of the person who first described the reaction.

Almost all the nitrogen fertilizer used on the world’s farms begins as ammonia from Haber’s process, so this one reaction supports a large share of the food supply. It comes with two environmental costs worth knowing. First, the process is energy-intensive: high pressure and high temperature must be maintained, and the hydrogen is usually obtained from fossil fuels, so fertilizer manufacture accounts for a noticeable fraction of industrial energy use. Second, fertilizer that is applied in excess washes off fields into rivers and lakes, where the extra nitrogen makes algae grow explosively; when the algae die and decay they strip the water of dissolved oxygen and fish suffocate. Applying the right quantity at the right time is therefore as much a part of good chemistry as making the ammonia in the first place.

(question only — no solution) \(5.35\ \text{g}\) of ammonium chloride is heated with excess calcium hydroxide. Write the balanced equation and calculate the volume of ammonia gas liberated at STP, taking the molar volume of a gas at STP as \(22.4\ \text{L mol}^{-1}\). (Molar mass of \(\mathrm{NH_4Cl} = 53.5\ \text{g mol}^{-1}\))

9.9 Summary

  1. The p-block consists of six groups, 13 to 18, because a p-subshell has three orbitals and can hold a maximum of six electrons. In these elements the differentiating electron enters the outermost p orbital.

  2. Group 13 is the boron family (B, Al, Ga, In, Tl), group 14 the carbon family (C, Si, Ge, Sn, Pb) and group 15 the nitrogen family (N, P, As, Sb, Bi).

  3. The general outer electronic configurations are \(n\mathrm{s}^2\,n\mathrm{p}^1\) for group 13, \(n\mathrm{s}^2\,n\mathrm{p}^2\) for group 14 and \(n\mathrm{s}^2\,n\mathrm{p}^3\) for group 15 — differing from the nearest inert gas by 3 or 4 electrons.

  4. Because of this, none of these elements occurs free as single atoms. They are found as compounds, as polyatomic molecules such as \(\mathrm{N_2}\), \(\mathrm{P_4}\) and \(\mathrm{C_{60}}\), or as polyatomic covalent arrays such as graphite and diamond.

  5. Atomic radius increases down each group, but the increase is broken at gallium, which is smaller than aluminium because the ten inner \(3\mathrm{d}\) electrons shield the outer electrons poorly against a nuclear charge greater by 18 units.

  6. Ionization enthalpy increases across a period, as the nuclear charge rises while the valence shell stays the same, and decreases down a group.

  7. All three families contain more than one type of element: boron is a metalloid and the rest of group 13 are metals; group 14 has the nonmetal carbon, the metalloids silicon and germanium, and the metals tin and lead; group 15 has the nonmetals nitrogen and phosphorous, the metalloids arsenic and antimony, and the metal bismuth. Metallic character increases down every group.

  8. The group oxidation state equals the total number of valence electrons: \(+3\) for group 13, \(+4\) for group 14, \(+5\) for group 15. Each group also shows a state two units lower, and group 14 and 15 elements show negative states as well.

  9. The lower oxidation state becomes more stable down a group because of the inert pair effect: the two valence s-electrons, poorly shielded by ten inner d-electrons, are held too tightly to take part in bonding. This is why \(\mathrm{Tl^{+}}\) is more stable than \(\mathrm{Tl^{3+}}\) and \(\mathrm{PbCl_2}\) more stable than \(\mathrm{PbCl_4}\).

  10. Lighter members, with small radii and high ionization enthalpy, form covalent bonds; heavier members, with lower ionization enthalpy and diffused orbitals, tend to form ionic bonds.

  11. B, C and N belong to the second period, have no d orbitals and cannot expand their octet. Later members have vacant valence-shell d orbitals and can expand their octet — which is why phosphorus forms \(\mathrm{PCl_5}\) but nitrogen forms no pentahalide, \(\mathrm{NF_3}\) being its only stable halide.

  12. On heating in air or oxygen, group 13 gives \(\mathrm{E_2O_3}\) (and nitrides EN with dinitrogen), group 14 gives EO and \(\mathrm{EO_2}\), and group 15 gives \(\mathrm{E_2O_3}\) and \(\mathrm{E_2O_5}\):

\[4\,\mathrm{E}\text{(s)} + 3\,\mathrm{O_2}\text{(g)} \xrightarrow{\ \Delta\ } 2\,\mathrm{E_2O_3}\text{(s)} \qquad 2\,\mathrm{E}\text{(s)} + \mathrm{N_2}\text{(g)} \xrightarrow{\ \Delta\ } 2\,\mathrm{EN}\text{(s)}\]

  1. As metallic character increases down a group, the nature of the stable oxide changes from acidic through amphoteric to basic — for example \(\mathrm{B_2O_3}\) is acidic, \(\mathrm{Al_2O_3}\) amphoteric and \(\mathrm{Tl_2O_3}\) basic.

  2. Most of these elements are unaffected by water. Aluminium reacts on heating to give the hydroxide and tin reacts with steam to give the oxide, while lead is protected by an oxide film:

\[2\,\mathrm{Al}\text{(s)} + 6\,\mathrm{H_2O}\text{(l)} \xrightarrow{\ \Delta\ } 2\,\mathrm{Al(OH)_3}\text{(s)} + 3\,\mathrm{H_2}\text{(g)}\]

  1. With halogens, group 13 gives trihalides \(\mathrm{EX_3}\) (thallium gives TlX), group 14 gives tetrahalides \(\mathrm{EX_4}\) with dihalides also for Ge and Pb, and group 15 gives both \(\mathrm{EX_3}\) and \(\mathrm{EX_5}\). Ionic character of the halides increases down each group, and group 15 trihalides are predominantly covalent except \(\mathrm{BiF_3}\).

  2. Catenation is the self linking of atoms of an element by covalent bonds into chains and rings, and it depends on the strength of the E–E bond. In group 14 the order is \(\mathrm{C} \gg \mathrm{Si} > \mathrm{Ge} = \mathrm{Sn}\), with no catenation in lead, matching the fall in bond enthalpy from \(348\) to \(240\ \text{kJ mol}^{-1}\).

  3. Allotropy is the existence of a solid element in different crystalline forms with different physical properties. Carbon and phosphorus both show it.

  4. In diamond every carbon is \(\mathrm{sp}^3\) hybridised and bonded tetrahedrally to four others in a three-dimensional network, with a C–C distance of 154 pm. It is the hardest natural substance, melts at \(3930\ ^\circ\text{C}\) and does not conduct electricity.

  5. In graphite \(\mathrm{sp}^2\) carbons form hexagonal sheets held by three sigma bonds, the fourth electron occupying an unhybridised p orbital; lateral overlap gives \(\pi\) bonds with electrons delocalised over the layer. The C–C length is 141.5 pm, the layers are 335 pm apart and held by weak van der Waals forces, making graphite soft, slippery and conducting. Graphite is the thermodynamically most stable allotrope of carbon.

  6. Fullerenes are molecular allotropes containing a definite number of carbon atoms, made by striking an electric arc between graphite electrodes in argon or helium. \(\mathrm{C_{60}}\), or Buckminsterfullerene, has 20 hexagonal and 12 pentagonal fused rings, two C–C distances of 143.5 pm and 138.3 pm, dissolves in organic solvents, and gives \(\mathrm{K_3C_{60}}\), a superconductor below 18 K.

  7. Carbon nanotubes are rolled-up graphite sheets, single-walled below 1 nm or multi-walled above 100 nm in diameter, with lengths from micrometres to millimetres; they are robust and springy, and have high electrical and heat conductivity and the highest known strength-to-weight ratio. Graphene is a single isolated layer of graphite, a two-dimensional solid with unique electronic properties.

  8. White phosphorus consists of discrete tetrahedral \(\mathrm{P_4}\) molecules with a P–P–P bond angle of \(60^\circ\); the angular strain makes it less stable and more reactive. It is a translucent white waxy solid, glows in the dark, is insoluble in water but dissolves in boiling NaOH solution, and is poisonous.

  9. Red phosphorus consists of \(\mathrm{P_4}\) tetrahedra linked covalently into polymeric chains. It is stable and less reactive, odourless with an iron grey lustre, does not glow, is insoluble in water and is non poisonous. It is made by heating white phosphorus at 573 K in an inert atmosphere.

  10. \(\mathrm{BCl_3}\) has an \(\mathrm{sp}^2\) boron with one vacant unhybridised p orbital and an incomplete octet; it is a nonpolar trigonal planar molecule. \(\mathrm{AlCl_3}\), also \(\mathrm{sp}^2\), exists as the dimer \(\mathrm{Al_2Cl_6}\), formed by overlap of a vacant \(3\mathrm{d}\) orbital of Al with a lone pair of Cl.

  11. Orthoboric acid, \(\mathrm{H_3BO_3}\), has boron bound to three –OH groups; in the solid, trigonal planar \(\mathrm{B(OH)_3}\) units are joined into layers by hydrogen bonds.

  12. In diborane, \(\mathrm{B_2H_6}\), each boron is \(\mathrm{sp}^3\) hybridised with three half-filled and one vacant hybrid orbital. Four terminal B–H bonds are ordinary 2-centre – 2-electron bonds; the two bridging hydrogens form two 3-centre – 2-electron (banana) bonds. The two B atoms and four terminal H atoms lie in one plane, with the bridging H atoms symmetrically above and below it.

  13. Silica, \(\mathrm{SiO_2}\), is a covalent three-dimensional network solid with each silicon bonded tetrahedrally to four oxygens and eight-membered rings of alternating Si and O atoms. Quartz, cristobalite and tridymite are its crystalline forms, interconvertible at a suitable temperature.

  14. Nitric acid, \(\mathrm{HNO_3}\), is a strong oxidising mineral acid with an \(\mathrm{sp}^2\) nitrogen, and it exhibits resonance. Orthophosphoric acid, \(\mathrm{H_3PO_4}\), is a strong non-toxic mineral acid with a tetrahedral phosphorus and three ionizable acidic hydrogens.

  15. Borax, \(\mathrm{Na_2B_4O_7\cdot 10H_2O}\) or \(\mathrm{Na_2[B_4O_5(OH)_4]\cdot 8H_2O}\), occurs as tincal and is prepared by boiling colemanite with sodium carbonate solution:

\[\mathrm{Ca_2B_6O_{11}} + 2\,\mathrm{Na_2CO_3} \xrightarrow{\ \Delta\ } \mathrm{Na_2B_4O_7} + 2\,\mathrm{NaBO_2} + 2\,\mathrm{CaCO_3}\downarrow\]

  1. Borax hydrolyses to give an alkaline solution, \(\mathrm{Na_2B_4O_7} + 7\,\mathrm{H_2O} \rightarrow 2\,\mathrm{NaOH} + 4\,\mathrm{H_3BO_3}\); on heating it swells, loses water and forms the borax bead of \(\mathrm{NaBO_2}\) and \(\mathrm{B_2O_3}\), used to detect coloured transition metal ions. Heated with ethyl alcohol and concentrated \(\mathrm{H_2SO_4}\) it gives triethyl borate, which burns with a green edged flame — the test used to detect and remove borate.

  2. Borax is used in optical and borosilicate glass, as a flux for soldering and welding, as a mild antiseptic in medical soaps, in the borax bead test, and as a brightener in washing powder.

  3. Silicones are organo-silicon polymers with the repeating unit \(\{\mathrm{R_2SiO}\}\) joined by \(-\mathrm{Si}-\mathrm{O}-\mathrm{Si}-\) linkages, named for the resemblance of the formula \(\mathrm{R_2SiO}\) to a ketone \(\mathrm{R_2CO}\). They are made from \(\mathrm{R_nSiCl_{(4-n)}}\); methyl chloride and silicon over copper at 573 K give the chlorosilanes, and hydrolysis of \(\mathrm{(CH_3)_2SiCl_2}\) followed by condensation polymerisation gives straight-chain silicones, the chain length being controlled with \(\mathrm{(CH_3)_3SiCl}\).

  4. Silicones are water repellant, thermally stable, good electrical insulators and resistant to oxidation and chemicals, and are used as insulation, waterproofing for fabrics, sealants, high-temperature lubricants and additives in paints and enamels.

  5. Ammonia occurs in air and soil from the decomposition of nitrogenous matter such as urea, is prepared in the laboratory by warming an ammonium salt with \(\mathrm{Ca(OH)_2}\) or NaOH, and is manufactured by Haber’s process:

\[\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \rightleftharpoons 2\,\mathrm{NH_3}\text{(g)}; \qquad \Delta_f H^\circ = -46.1\ \text{kJ mol}^{-1}\]

  1. High pressure favours ammonia, since four molecules of gas give two. The optimum conditions are \(200 \times 10^5\ \text{Pa}\) (200 atm), about 700 K, and an iron oxide catalyst with traces of \(\mathrm{K_2O}\) and \(\mathrm{Al_2O_3}\).

  2. Ammonia is a colourless gas of pungent odour, freezing at 198.4 K and boiling at 239.7 K, highly soluble in water; the concentrated solution is liquor ammonia. Hydrogen bonding between \(\mathrm{NH_3}\) molecules raises its melting and boiling points.

  3. Chemically, ammonia is a weak base giving \(\mathrm{OH^-}\) in water, forms ammonium salts with acids, precipitates metal hydroxides or hydrated oxides from salt solutions, and forms complexes such as the deep blue \([\mathrm{Cu(NH_3)_4}]^{2+}\) and colourless \([\mathrm{Ag(NH_3)_2}]\mathrm{Cl}\) using the lone pair on nitrogen. It gives a brown precipitate of Millon’s base with Nessler’s reagent, an alkaline solution of \(\mathrm{K_2HgI_4}\).

  4. Ammonia is used to make fertilizers such as urea, diammonium phosphate, ammonium nitrate and ammonium sulphate, to make inorganic compounds such as nitric acid, as a refrigerant in the liquid form, and as a laboratory reagent in qualitative and quantitative analysis.

Flowchart

FC 9.7 – The chapter’s chain of reasoning: outer electronic configuration ns²np¹⁻³ → atomic radius, ionization enthalpy, electronegativity trends → metallic character increasing down each group → oxidation states (group state and the state two units lower, inert pair effect) → bond type: covalent for lighter members, ionic for heavier → reactions with oxygen, water and halogens, and the acidic → amphoteric → basic change in oxides → catenation and allotropy → molecular structures (BCl₃, Al₂Cl₆, H₃BO₃, B₂H₆, SiO₂, HNO₃, H₃PO₄) → notable compounds (borax, silicones, ammonia)

9.10 Check Your Concepts

  1. Aluminium lies above gallium in group 13, yet the atomic radius of gallium is smaller than that of aluminium. Explain, and state whether the same kind of irregularity is possible between boron and aluminium.

  2. Assertion (A): \(\mathrm{TlBr}\) is a stable compound while thallium(III) bromide is not. Reason (R): The inert pair effect becomes more prominent on moving down group 13. Choose the correct option and justify: both A and R are true and R explains A; both are true but R does not explain A; A is true and R is false; A is false and R is true.

  3. Each B–Cl bond in boron trichloride is polar, and yet the molecule as a whole is nonpolar. Explain this using the shape of the molecule.

  4. \(\mathrm{AlCl_3}\) exists as \(\mathrm{Al_2Cl_6}\), whereas \(\mathrm{BCl_3}\) remains a monomer. Explain the difference, and name the type of bond that holds the dimer together.

  5. Explain why the bridging B–H bond in diborane is longer than the terminal B–H bond.

  6. Distinguish between a 2-centre – 2-electron bond and a 3-centre – 2-electron bond, giving one example of each from this chapter.

  7. Both diamond and graphite are pure carbon. Explain why diamond is used in drilling tools while graphite is used as a lubricant.

  8. Graphite conducts electricity along its layers but diamond does not conduct at all. Account for this difference in terms of hybridisation and the availability of electrons.

  9. \(\mathrm{C_{60}}\) dissolves in organic solvents, while graphite and diamond are insoluble in every solvent. Explain.

  10. What happens when white phosphorus is heated to 573 K in an inert atmosphere? Name the product and state why an inert atmosphere is essential.

  11. Explain why white phosphorus is more reactive than red phosphorus, and state which of the two must be stored under water.

  12. Arrange the following in the order asked and give a reason for the order in each case.

    1. C, Si, Ge, Sn — decreasing tendency for catenation
    2. \(\mathrm{B_2O_3}\), \(\mathrm{Al_2O_3}\), \(\mathrm{Tl_2O_3}\) — decreasing acidic character
    3. N, P, Bi — decreasing first ionization enthalpy

  13. Complete the following reactions and name the main product in each case.

    1. \(2\,\mathrm{E}\text{(s)} + 3\,\mathrm{X_2}\text{(g)} \rightarrow\)
    2. \(\mathrm{Sn}\text{(s)} + 2\,\mathrm{H_2O}\text{(g)} \xrightarrow{\ \Delta\ }\)
    3. \(\mathrm{Na_2B_4O_7} + 7\,\mathrm{H_2O} \rightarrow\)
    4. \(\mathrm{H_3BO_3} + 3\,\mathrm{C_2H_5OH} \rightarrow\)
    5. \(\mathrm{(CH_3)_2SiCl_2} + 2\,\mathrm{H_2O} \rightarrow\)

  14. What would you observe in each of the following, and what is the reason for it?

    1. Aqueous ammonia is added to a solution of zinc sulphate.
    2. Excess aqueous ammonia is added to a solution of a copper(II) salt.
    3. Borax is heated on a platinum loop with cobalt oxide in a Bunsen flame.
    4. Borax is warmed with ethyl alcohol and concentrated sulphuric acid, and the vapour is ignited.

  15. Why is a platinum wire, and not a copper or an iron wire, used in the borax bead test?

  16. Silicon dioxide is a hard, high-melting solid whereas carbon dioxide is a gas at room temperature, although both are dioxides of group 14 elements. Explain.

  17. Explain why silicones withstand high temperatures better than ordinary organic polymers, and relate this to two of their listed uses.

  18. Explain why the chain length of a silicone polymer can be controlled by adding \(\mathrm{(CH_3)_3SiCl}\) to the reaction mixture.

  19. The trihalides of group 15 elements are predominantly covalent, but \(\mathrm{BiF_3}\) is an exception. Account for this exception.

  20. Nitrogen forms \(\mathrm{NF_3}\) but no pentahalide, while phosphorus forms both \(\mathrm{PCl_3}\) and \(\mathrm{PCl_5}\). Explain, and state what this tells you about the valence shell of a second-period element.

  21. High pressure favours the formation of ammonia in Haber’s process, yet the reaction is carried out at a high temperature of about 700 K even though it is exothermic. Explain both facts.

  22. Ammonia has an unusually high boiling point for such a small molecule and is highly soluble in water. Explain both observations using the same idea.

  23. Match the pairs.

A B
a. Bucky ball i. Isolated single layer of graphite
b. Graphene ii. Rolled-up graphite sheet
c. Carbon nanotube iii. Twenty hexagonal and twelve pentagonal rings
iv. Discrete tetrahedral \(\mathrm{P_4}\) molecules

  1. Identify the substance in each case and give one reason for your choice.

    1. A group 13 nitride called inorganic graphite.
    2. A compound of \(\mathrm{C_{60}}\) with a group 1 metal that conducts with zero resistance below 18 K.
    3. A mineral containing about 90% borax.
    4. The concentrated aqueous solution of ammonia.

  2. Identify the group 14 element that best fits each of the following descriptions.

    1. Non metallic element
    2. Forms the most acidic oxide
    3. They prefer \(+2\) oxidation state
    4. Forms strong \(\pi\) bonds

  3. Give reasons.

    1. \(\mathrm{Ga^{3+}}\) salts are better reducing agents while \(\mathrm{Tl^{3+}}\) salts are better oxidising agents.
    2. \(\mathrm{PbCl_4}\) is less stable than \(\mathrm{PbCl_2}\).

  4. Give the formula of a compound in which carbon exhibits an oxidation state of

    1. \(+4\)
    2. \(+2\)
    3. \(-4\)

  5. Explain the trend of the following in group 13 elements:

    1. atomic radii
    2. ionization enthalpy
    3. electron affinity

  6. What is the hybridization of Al in \(\mathrm{AlCl_3}\)?

  7. Name a molecule having a banana bond.

  8. Draw the structure of the following.

    1. Orthophosphoric acid
    2. Resonance structures of nitric acid

  9. Find out the difference between

    1. Diamond and graphite
    2. White phosphorus and red phosphorus

  10. What are silicones? Where are they used?

  11. Explain the trend in oxidation state of elements from nitrogen to bismuth.

  12. Give the test that is used to detect the borate radical in qualitative analysis.

  13. Explain the structure and bonding of diborane.

  14. Ammonia is a good complexing agent. Explain.

  15. Give the reactions supporting the basic nature of ammonia.

9.11 Practice With Numericals

  1. Calculate the percentage by mass of boron in orthoboric acid, \(\mathrm{H_3BO_3}\). (Atomic masses: H \(= 1.008\), B \(= 10.81\), O \(= 16.00\))

  2. Calculate the percentage by mass of sodium in crystalline borax, \(\mathrm{Na_2B_4O_7\cdot 10H_2O}\). (Atomic masses: Na \(= 22.99\), B \(= 10.81\), O \(= 16.00\), H \(= 1.008\))

  3. Assign the oxidation number of the underlined element in each of the following: \(\mathrm{Na_2\underline{B}_4O_7}\), \(\mathrm{H\underline{N}O_3}\), \(\mathrm{\underline{Sn}Cl_2}\), \(\mathrm{\underline{As}_4O_6}\).

  4. In Haber’s process, calculate the mass of ammonia that can be obtained from 1.40 kg of dinitrogen, assuming complete conversion, and the mass of dihydrogen consumed. (Atomic masses: N \(= 14.0\), H \(= 1.0\))

  5. Calculate the volume of dihydrogen, measured at STP, required to produce 3.40 g of ammonia by Haber’s process. (Molar volume at STP \(= 22.4\ \text{L mol}^{-1}\))

  6. What mass of ammonium chloride must be heated with excess calcium hydroxide to obtain 4.48 L of ammonia at STP? Write the balanced equation first. (Molar mass of \(\mathrm{NH_4Cl} = 53.5\ \text{g mol}^{-1}\))

  7. Aluminium is heated in oxygen. Write the balanced equation and calculate the mass of the oxide formed from 10.8 g of aluminium. (Atomic masses: Al \(= 26.98\), O \(= 16.00\))

  8. Phosphorus burns in excess oxygen according to \(\mathrm{P_4} + 5\,\mathrm{O_2} \rightarrow \mathrm{P_4O_{10}}\). Calculate the mass of \(\mathrm{P_4O_{10}}\) formed from 12.4 g of \(\mathrm{P_4}\). (Atomic masses: P \(= 30.97\), O \(= 16.00\))

  9. Using the bond enthalpy values given in the chapter, calculate the total energy required to break all the Si–Si bonds in one mole of \(\mathrm{Si_6H_{14}}\), in which six silicon atoms are joined in a straight chain.

  10. Taking the atomic mass of carbon as 12.01, calculate the molar mass of the fullerene \(\mathrm{C_{70}}\) and the number of \(\mathrm{C_{70}}\) molecules present in 4.204 g of it. (\(N_A = 6.022\times10^{23}\ \text{mol}^{-1}\))

  11. In graphite the C–C bond length within a layer is 141.5 pm and the separation between two layers is 335 pm. Calculate how many times greater the interlayer separation is than the in-layer bond length, and comment on what this ratio suggests about the strength of the forces between the layers.

  12. Using the ionization enthalpy data for group 13, calculate the difference between the third and the first ionization enthalpy for boron and for thallium. State which element shows the larger difference and suggest a reason.

  13. Using the density values given in the chapter, calculate the mass of a cube of germanium of edge 2.00 cm, and the volume that the same mass of silicon would occupy.

9.12 Case-Based Questions

  1. A compound is prepared from the mineral colemanite by boiling it with a solution of sodium carbonate. It is a white crystalline solid and is used for inorganic qualitative analysis.
    1. Name the compound produced.
    2. Write the reaction that explains its formation.
  2. Shravani was performing inorganic qualitative analysis of a salt. To an aqueous solution of that salt, she added silver nitrate, and a white precipitate was formed. On adding ammonium hydroxide to this, she obtained a clear solution. Comment on her observations and write the chemical reactions involved.

9.13 Objective Questions

  1. Which of the following is not an allotrope of carbon?
    1. bucky ball
    2. diamond
    3. graphite
    4. emerald
  2. ………… is inorganic graphite.
    1. borax
    2. diborane
    3. boron nitride
    4. colemanite
  3. Haber’s process is used for the preparation of …………
    1. \(\mathrm{HNO_3}\)
    2. \(\mathrm{NH_3}\)
    3. \(\mathrm{NH_2CONH_2}\)
    4. \(\mathrm{NH_4OH}\)
  4. Thallium shows different oxidation states because …………
    1. of inert pair effect
    2. it is an inner transition element
    3. it is a metal
    4. of its high electronegativity
  5. Which of the following shows the most prominent inert pair effect?
    1. C
    2. Si
    3. Ge
    4. Pb
  6. State true or false. Correct the false statement.
    1. The acidic nature of oxides of group 13 increases down the group.
    2. The tendency for catenation is much higher for C than for Si.
  7. Match the pairs from column A and column B.
A B
a. \(\mathrm{BCl_3}\) i. Angular molecule
b. \(\mathrm{SiO_2}\) ii. Linear covalent molecule
c. \(\mathrm{CO_2}\) iii. Tetrahedral molecule
iv. Planar trigonal molecule