UNIT 11

THE p -BLOCK ELEMENTS

“The variation in properties of the p-block elements due to the influence of $d$ and $f$ electrons in the inner core of the heavier elements makes their chemistry interesting”

In $p$-block elements the last electron enters the outermost $p$ orbital. As we know that the number of $p$ orbitals is three and, therefore, the maximum number of electrons that can be accommodated in a set of $p$ orbitals is six. Consequently there are six groups of $p$-block elements in the periodic table numbering from 13 to 18 . Boron, carbon, nitrogen, oxygen, fluorine and helium head the groups. Their valence shell electronic configuration is $\mathit{n}{\mathit{s}}^2\mathit{n}{\mathit{p}}^{\mathbf{1}\mathbf{-}\mathbf{6}}$ (except for $\mathrm{He}$ ). The inner core of the electronic configuration may, however, differ. The difference in inner core of elements greatly influences their physical properties (such as atomic and ionic radii, ionisation enthalpy, etc.) as well as chemical properties. Consequently, a lot of variation in properties of elements in a group of $p$-block is observed. The maximum oxidation state shown by a $p$-block element is equal to the total number of valence electrons (i.e., the sum of the $s^{-}$ and $p$-electrons). Clearly, the number of possible oxidation states increases towards the right of the periodic table. In addition to this so called group oxidation state, $p$-block elements may show other oxidation states which normally, but not necessarily, differ from the total number of valence electrons by unit of two. The important oxidation states exhibited by $p$-block elements are shown in Table 11.1. In boron, carbon and nitrogen families the group oxidation state is the most stable state for the lighter elements in the group. However, the oxidation state two unit less than the group oxidation state becomes progressively more stable for the heavier elements in each group. The occurrence of oxidation states two unit less than the group oxidation states are sometime attributed to the ‘inert pair effect’.

Table 11.1 General Electronic Configuration and Oxidation States of p-Block Elements

Group	$\mathbf{1}\mathbf{3}$	$\mathbf{1}\mathbf{4}$	$\mathbf{1}\mathbf{5}$	$\mathbf{1}\mathbf{6}$	$\mathbf{1}\mathbf{7}$	$\mathbf{1}\mathbf{8}$
General electronic configuration	$n{s}^{2}n{p}^1$	$n{s}^{2}n{p}^2$	$n{s}^{2}n{p}^3$	$n{s}^{2}n{p}^4$	$n{s}^{2}n{p}^5$	$n{s}^{2}n{p}^6$ $(1s^2$ for $\mathrm{He})$
First member of the group	$\mathrm{B}$	$\mathrm{C}$	$\mathrm{N}$	$\mathrm{O}$	$\mathrm{F}$	$\mathrm{He}$
Group oxidation state	+3	+4	+5	+6	+7	+8
Other oxidation states	+1	+2, -4	+3, -3	+4, +2, -2	+5, +3, +1, -1	+6, +4, +2

The relative stabilities of these two oxidation states - group oxidation state and two unit less than the group oxidation state - may vary from group to group and will be discussed at appropriate places.

It is interesting to note that the non-metals and metalloids exist only in the $p$-block of the periodic table. The non-metallic character of elements decreases down the group. In fact the heaviest element in each $p$-block group is the most metallic in nature. This change from nonmetallic to metallic character brings diversity in the chemistry of these elements depending on the group to which they belong.

In general, non-metals have higher ionisation enthalpies and higher electronegativities than the metals. Hence, in contrast to metals which readily form cations, non-metals readily form anions. The compounds formed by highly reactive non-metals with highly reactive metals are generally ionic because of large differences in their electronegativities. On the other hand, compounds formed between non-metals themselves are largely covalent in character because of small differences in their electronegativities. The change of non-metallic to metallic character can be best illustrated by the nature of oxides they form. The non-metal oxides are acidic or neutral whereas metal oxides are basic in nature. The first member of p-block differs from the remaining members of their corresponding group in two major respects. First is the size and all other properties which depend on size. Thus, the lightest $p$-block elements show the same kind of differences as the lightest $s$-block elements, lithium and beryllium. The second important difference, which applies only to the $p$-block elements, arises from the effect of $d$ orbitals in the valence shell of heavier elements (starting from the third period onwards) and their lack in second period elements. The second period elements of $p$-groups starting from boron are restricted to a maximum covalence of four (using $2s$ and three $2p$ orbitals). In contrast, the third period elements of $p$-groups with the electronic configuration $3s^{2}3p^n$ have the vacant $3d$ orbitals lying between the $3p$ and the $4s$ levels of energy. Using these $d$-orbitals the third period elements can expand their covalence above four. For example, while boron forms only ${\left[{\mathrm{BF}}_4\right]}^{-}$, aluminium gives ${\left[{\mathrm{AlF}}_6\right]}^{3-}$ ion. The presence of these $d$-orbitals influences the chemistry of the heavier elements in a number of other ways. The combined effect of size and availability of $d$ orbitals considerably influences the ability of these elements to form $\pi$ bonds. The first member of a group differs from the heavier members in its ability to form $p\pi -p\pi$ multiple bonds to itself ( e.g., $\mathrm{C}=\mathrm{C},\mathrm{C}\equiv \mathrm{C}$, $\mathrm{N}\equiv \mathrm{N}$ ) and to other second row elements (e.g., $\mathrm{C}=\mathrm{O},\mathrm{C}=\mathrm{N},\mathrm{C}\equiv \mathrm{N},\mathrm{N}=\mathrm{O}$ ). This type of $\pi$ - bonding is not particularly strong for the heavier $p$-block elements. The heavier elements do form $\pi$ bonds but this involves $d$ orbitals $(d\pi -p\pi$ or $d\pi -d\pi$ ). As the $d$ orbitals are of higher energy than the $p$ orbitals, they contribute less to the overall stability of molecules than does $\mathrm{p}\pi -\mathrm{p}\pi$ bonding of the second row elements. However, the coordination number in species of heavier elements may be higher than for the first element in the same oxidation state. For example, in +5 oxidation state both $\mathrm{N}$ and $\mathrm{P}$ form oxoanions : ${\mathrm{NO}}_3^{-}$(three-coordination with $\pi$ - bond involving one nitrogen $p$-orbital) and ${\mathrm{PO}}_4^{3-}$ (four-coordination involving $s,p$ and $d$ orbitals contributing to the $\pi$-bond). In this unit we will study the chemistry of group 13 and 14 elements of the periodic table.

11.1 GROUP 13 ELEMENTS: THE BORON FAMILY

This group elements show a wide variation in properties. Boron is a typical non-metal, aluminium is a metal but shows many chemical similarities to boron, and gallium, indium and thallium are almost exclusively metallic in character.

Boron is a fairly rare element, mainly occurs as orthoboric acid, $\left({\mathrm{H}}_3{\mathrm{BO}}_3\right)$, borax, ${\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7\cdot 10{\mathrm{H}}_2\mathrm{O}$, and kernite, ${\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7\cdot 4{\mathrm{H}}_2\mathrm{O}$. In India borax occurs in Puga Valley (Ladakh) and Sambhar Lake (Rajasthan). The abundance of boron in earth crust is less than $0.0001$ by mass. There are two isotopic forms of boron ${}^{10}\mathrm{B}(19$ and ${}^{11}\mathrm{B}(81$. Aluminium is the most abundant metal and the third most abundant element in the earth’s crust ( $8.3$ by mass) after oxygen ( $45.5$ ) and $\mathrm{Si}(27.7$. Bauxite, ${\mathrm{Al}}_2{\mathrm{O}}_3\cdot 2{\mathrm{H}}_2\mathrm{O}$ and cryolite, ${\mathrm{Na}}_3{\mathrm{AlF}}_6$ are the important minerals of aluminium. In India it is found as mica in Madhya Pradesh, Karnataka, Orissa and Jammu. Gallium, indium and thallium are less abundant elements in nature.

The atomic, physical and chemical properties of these elements are discussed below.

11.1.1 Electronic Configuration

The outer electronic configuration of these elements is $n{s}^{2}n{p}^1$. A close look at the electronic configuration suggests that while boron and aluminium have noble gas core, gallium and indium have noble gas plus $10d$-electrons, and thallium has noble gas plus $14f$-electrons plus $10d$-electron cores. Thus, the electronic structures of these elements are more complex than for the first two groups of elements discussed in unit 10. This difference in electronic structures affects the other properties and consequently the chemistry of all the elements of this group.

11.1.2 Atomic Radii

On moving down the group, for each successive member one extra shell of electrons is added and, therefore, atomic radius is expected to increase. However, a deviation can be seen. Atomic radius of Ga is less than that of Al. This can be understood from the variation in the inner core of the electronic configuration. The presence of additional $10d$-electrons offer only poor screening effect (Unit 2) for the outer electrons from the increased nuclear charge in gallium. Consequently, the atomic radius of gallium (135 pm) is less than that of aluminium (143 pm).

11.1.3 Ionization Enthalpy

The ionisation enthalpy values as expected from the general trends do not decrease smoothly down the group. The decrease from $\mathrm{B}$ to $\mathrm{Al}$ is associated with increase in size. The observed discontinuity in the ionisation enthalpy values between $\mathrm{Al}$ and $\mathrm{Ga}$, and between In and Tl are due to inability of $d$ - and $f$-electrons, which have low screening effect, to compensate the increase in nuclear charge.

The order of ionisation enthalpies, as expected, is ${\mathrm{\Delta }}_i{\mathrm{H}}_1<{\mathrm{\Delta }}_i{\mathrm{H}}_2<{\mathrm{\Delta }}_i{\mathrm{H}}_3$. The sum of the first three ionisation enthalpies for each of the elements is very high. Effect of this will be apparent when you study their chemical properties.

11.1.4 Electronegativity

Down the group, electronegativity first decreases from $\mathrm{B}$ to $\mathrm{Al}$ and then increases marginally (Table 11.2). This is because of the discrepancies in atomic size of the elements.

11.1.5 Physical Properties

Boron is non-metallic in nature. It is extremely hard and black coloured solid. It exists in many allotropic forms. Due to very strong crystalline lattice, boron has unusually high melting point. Rest of the members are soft metals with low melting point and high electrical conductivity. It is worthwhile to note that gallium with unusually low melting point (303K), could exist in liquid state during summer. Its high boiling point $(2676\mathrm{K})$ makes it a useful material for measuring high temperatures. Density of the elements increases down the group from boron to thallium.

11.1.6 Chemical Properties

Oxidation state and trends in chemical reactivity

Due to small size of boron, the sum of its first three ionization enthalpies is very high. This prevents it to form +3 ions and forces it to form only covalent compounds. But as we move from $\mathrm{B}$ to $\mathrm{Al}$, the sum of the first three ionisation enthalpies of Al considerably decreases, and is therefore able to form ${\mathrm{Al}}^{3+}$ ions. In fact, aluminium is a highly electropositive metal. However, down the group, due to poor shielding effect of intervening $d$ and $f$ orbitals, the increased effective nuclear charge holds $ns$ electrons tightly (responsible for inert pair effect) and thereby, restricting their participation in bonding. As a result of this, only $p$-orbital electron may be involved in bonding. In fact in Ga, In and Tl, both +1 and +3 oxidation states are observed. The relative stability of +1 oxidation state progressively increases for heavier elements: $\mathrm{Al}<\mathrm{Ga}<\mathrm{In}<\mathrm{Tl}$. In thallium +1 oxidation state is predominant whereas the +3 oxidation state is highly oxidising in character. The compounds in +1 oxidation state, as expected from energy considerations, are more ionic than those in +3 oxidation state.

In trivalent state, the number of electrons around the central atom in a molecule

Table 11.2 Atomic and Physical Properties of Group 13 Elements

of the compounds of these elements (e.g., boron in ${\mathrm{BF}}_3$ ) will be only six. Such electron deficient molecules have tendency to accept a pair of electrons to achieve stable electronic configuration and thus, behave as Lewis acids. The tendency to behave as Lewis acid decreases with the increase in the size down the group. ${\mathrm{BCl}}_3$ easily accepts a lone pair of electrons from ammonia to form ${\mathrm{BCl}}_3\cdot {\mathrm{NH}}_3$.

In trivalent state most of the compounds being covalent are hydrolysed in water. For example, the trichlorides on hyrolysis in water form tetrahedral ${[\mathrm{M}(\mathrm{OH}{)}_4]}^{-}$species; the hybridisation state of element $\mathrm{M}$ is $s{p}^3$. Aluminium chloride in acidified aqueous solution forms octahedral ${\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}$ ion. In this complex ion, the $3d$ orbitals of $\mathrm{Al}$ are involved and the hybridisation state of $\mathrm{Al}$ is $s{p}^3{d}^2$.

Problem 11.1

Standard electrode potential values, ${\mathrm{E}}^{\ominus }$ for ${\mathrm{Al}}^{3+}/\mathrm{Al}$ is $-1.66\mathrm{V}$ and that of ${\mathrm{Tl}}^{3+}/\mathrm{Tl}$ is $+1.26\mathrm{V}$. Predict about the formation of ${\mathrm{M}}^{3+}$ ion in solution and compare the electropositive character of the two metals.

Solution

Standard electrode potential values for two half cell reactions suggest that aluminium has high tendency to make ${\mathrm{Al}}^{3+}(\mathrm{aq})$ ions, whereas ${\mathrm{Tl}}^{3+}$ is not only unstable in solution but is a powerful oxidising agent also. Thus ${\mathrm{Tl}}^{+}$is more stable in solution than ${\mathrm{Tl}}^{3+}$. Aluminium being able to form +3 ions easily, is more electropositive than thallium.

(i) Reactivity towards air

Boron is unreactive in crystalline form. Aluminium forms a very thin oxide layer on the surface which protects the metal from further attack. Amorphous boron and aluminium metal on heating in air form ${\mathrm{B}}_2{\mathrm{O}}_3$ and ${\mathrm{Al}}_2{\mathrm{O}}_3$ respectively. With dinitrogen at high temperature they form nitrides.

$$\begin{array}{rl}& 2\mathrm{E}(\mathrm{s})+3{\mathrm{O}}_2(\mathrm{g})\stackrel{\mathrm{\Delta }}{\to }2{\mathrm{E}}_2{\mathrm{O}}_3(\mathrm{s})\\ & 2\mathrm{E}(\mathrm{s})+{\mathrm{N}}_2(\mathrm{g})\stackrel{\mathrm{\Delta }}{\to }2\mathrm{EN}(\mathrm{s})\\ & & \mathrm{E}=\text{ element}\end{array}$$

The nature of these oxides varies down the group. Boron trioxide is acidic and reacts with basic (metallic) oxides forming metal borates. Aluminium and gallium oxides are amphoteric and those of indium and thallium are basic in their properties.

(ii) Reactivity towards acids and alkalies

Boron does not react with acids and alkalies even at moderate temperature; but aluminium dissolves in mineral acids and aqueous alkalies and thus shows amphoteric character.

Aluminium dissolves in dilute $\mathrm{HCl}$ and liberates dihydrogen.

$2\mathrm{Al}(\mathrm{s})+6\mathrm{HCl}(\mathrm{aq})\to 2{\mathrm{Al}}^{3+}(\mathrm{aq})+6{\mathrm{Cl}}^{-}(\mathrm{aq})+3{\mathrm{H}}_2(\mathrm{g})$

However, concentrated nitric acid renders aluminium passive by forming a protective oxide layer on the surface.

Aluminium also reacts with aqueous alkali and liberates dihydrogen.

$$\begin{array}{cc}& 2\mathrm{Al}(\mathrm{s})+2\mathrm{NaOH}(\mathrm{aq})+6{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\\ & \downarrow \\ & \underset{\begin{array}{c}\text{ Sodium }\\ \text{ tetrahydroxoaluminate(III) }\end{array}}{2{\mathrm{Na}}^{+}{[\mathrm{Al}(\mathrm{OH}{)}_4]}^{-}(\mathrm{aq})}+3{\mathrm{H}}_2(\mathrm{g})\end{array}$$

(iii) Reactivity towards halogens

These elements react with halogens to form trihalides (except ${\mathrm{TlI}}_3$ ).

$2\mathrm{E}(\mathrm{s})+3{\mathrm{X}}_2(\mathrm{g})\to 2{\mathrm{EX}}_3(\mathrm{s})(\mathrm{X}=\mathrm{F},\mathrm{Cl},\mathrm{Br},\mathrm{I})$

Problem 11.2

White fumes appear around the bottle of anhydrous aluminium chloride. Give reason.

Solution

Anhydrous aluminium chloride is partially hydrolysed with atmospheric moisture to liberate $\mathrm{HCl}$ gas. Moist $\mathrm{HCl}$ appears white in colour.

11.2 IMPORTANT TRENDS ANOMALOUS PROPERTIES OF BORON

Certain important trends can be observed in the chemical behaviour of group 13 elements. The tri-chlorides, bromides and iodides of all these elements being covalent in nature are hydrolysed in water. Species like tetrahedral ${[\mathrm{M}(\mathrm{OH}{)}_4]}^{-}$ and octahedral ${\left[\mathrm{M}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}$, except in boron, exist in aqueous medium.

The monomeric trihalides, being electron deficient, are strong Lewis acids. Boron trifluoride easily reacts with Lewis bases such as ${\mathrm{NH}}_3$ to complete octet around boron.

$${\mathrm{F}}_3\mathrm{B}+:{\mathrm{NH}}_3\to {\mathrm{F}}_3\mathrm{B}\leftarrow {\mathrm{NH}}_3$$

It is due to the absence of $d$ orbitals that the maximum covalence of $B$ is 4 . Since the $d$ orbitals are available with $\mathrm{Al}$ and other elements, the maximum covalence can be expected beyond 4. Most of the other metal halides (e.g., ${\mathrm{AlCl}}_3$ ) are dimerised through halogen bridging (e.g., ${\mathrm{Al}}_2{\mathrm{Cl}}_6$ ). The metal species completes its octet by accepting electrons from halogen in these halogen bridged molecules.

Problem 11.3

Boron is unable to form ${\mathrm{BF}}_6{}^{3-}$ ion. Explain.

Solution

Due to non-availability of $d$ orbitals, boron is unable to expand its octet. Therefore, the maximum covalence of boron cannot exceed 4.

11.3 SOME IMPORTANT COMPOUNDS OF BORON

Some useful compounds of boron are borax, orthoboric acid and diborane. We will briefly study their chemistry.

11.3.1 Borax

It is the most important compound of boron. It is a white crystalline solid of formula ${\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7\cdot 10{\mathrm{H}}_2\mathrm{O}$. In fact it contains the tetranuclear units ${[{\mathrm{B}}_4{\mathrm{O}}_5(\mathrm{OH}{)}_4]}^{2-}$ and correct formula; therefore, is ${\mathrm{Na}}_2[{\mathrm{B}}_4{\mathrm{O}}_5(\mathrm{OH}{)}_4]\cdot 8{\mathrm{H}}_2\mathrm{O}$. Borax dissolves in water to give an alkaline solution.

$$\begin{array}{r}{\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7+7{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{NaOH}+4{\mathrm{H}}_3{\mathrm{BO}}_3\\ \text{ Orthoboric acid }\end{array}$$

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

$${\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7\cdot 10{\mathrm{H}}_2\mathrm{O}\stackrel{\mathrm{\Delta }}{\to }\underset{\begin{array}{c}\text{Sodium}\\ \text{metaborate}\end{array}}{{\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7}\stackrel{\mathrm{\Delta }}{\to }2{\mathrm{NaBO}}_2+\underset{\begin{array}{c}\text{Boric}\\ \text{anhydride}\end{array}}{{\mathrm{B}}_2{\mathrm{O}}_3}$$

The metaborates of many transition metals have characteristic colours and, therefore, borax bead test can be used to identify them in the laboratory. For example, when borax is heated in a Bunsen burner flame with $\mathrm{CoO}$ on a loop of platinum wire, a blue coloured $\mathrm{Co}{\left({\mathrm{BO}}_2\right)}_2$ bead is formed.

11.3.2 Orthoboric acid

Orthoboric acid, ${\mathrm{H}}_3{\mathrm{BO}}_3$ is a white crystalline solid, with soapy touch. It is sparingly soluble in water but highly soluble in hot water. It can be prepared by acidifying an aqueous solution of borax.

${\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7+2\mathrm{HCl}+5{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{NaCl}+4\mathrm{B}(\mathrm{OH}{)}_3$

It is also formed by the hydrolysis (reaction with water or dilute acid) of most boron compounds (halides, hydrides, etc.). It has a layer structure in which planar ${\mathrm{BO}}_3$ units are joined by hydrogen bonds as shown in Fig. 11.1.

Boric acid is a weak monobasic acid. It is not a protonic acid but acts as a Lewis acid by accepting electrons from a hydroxyl ion:

$\mathrm{B}(\mathrm{OH}{)}_3+2\mathrm{HOH}\to {[\mathrm{B}(\mathrm{OH}{)}_4]}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}$

On heating, orthoboric acid above $370\mathrm{K}$ forms metaboric acid, ${\mathrm{HBO}}_2$ which on further heating yields boric oxide, ${\mathrm{B}}_2{\mathrm{O}}_3$.

${\mathrm{H}}_3{\mathrm{BO}}_3\stackrel{\mathrm{\Delta }}{\to }{\mathrm{HBO}}_2\stackrel{\mathrm{\Delta }}{\to }{\mathrm{B}}_2{\mathrm{O}}_3$

Problem 11.4

Why is boric acid considered as a weak acid?

Solution

Because it is not able to release ${\mathrm{H}}^{+}$ions on its own. It receives ${\mathrm{OH}}^{-}$ions from water molecule to complete its octet and in turn releases ${\mathrm{H}}^{+}$ions.

11.3.3 Diborane, ${\mathrm{B}}_2{\mathrm{H}}_6$

The simplest boron hydride known, is diborane. It is prepared by treating boron trifluoride with ${\mathrm{LiAlH}}_4$ in diethyl ether.

$4{\mathrm{BF}}_3+3{\mathrm{LiAlH}}_4\to 2{\mathrm{B}}_2{\mathrm{H}}_6+3\mathrm{LiF}+3{\mathrm{AlF}}_3$

A convenient laboratory method for the preparation of diborane involves the oxidation of sodium borohydride with iodine.

$2{\mathrm{NaBH}}_4+{\mathrm{I}}_2\to {\mathrm{B}}_2{\mathrm{H}}_6+2\mathrm{NaI}+{\mathrm{H}}_2$

Diborane is produced on an industrial scale by the reaction of ${\mathrm{BF}}_3$ with sodium hydride.

$$2{\mathrm{BF}}_3+6\mathrm{NaH}\stackrel{450\mathrm{K}}{\to }{\mathrm{B}}_2{\mathrm{H}}_6+6\mathrm{NaF}$$

Diborane is a colourless, highly toxic gas with a b.p. of $180\mathrm{K}$. Diborane catches fire spontaneously upon exposure to air. It burns in oxygen releasing an enormous amount of energy.

$$\begin{array}{rl}& {\mathrm{B}}_2{\mathrm{H}}_6+3{\mathrm{O}}_2\to {\mathrm{B}}_2{\mathrm{O}}_3+3{\mathrm{H}}_2\mathrm{O};\\ & {\mathrm{\Delta }}_c{H}^{\ominus }=-1976\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$$

Most of the higher boranes are also spontaneously flammable in air. Boranes are readily hydrolysed by water to give boric acid. ${\mathrm{B}}_2{\mathrm{H}}_6(\mathrm{g})+6{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 2\mathrm{B}(\mathrm{OH}{)}_3(\mathrm{aq})+6{\mathrm{H}}_2(\mathrm{g})$

Diborane undergoes cleavage reactions with Lewis bases(L) to give borane adducts, ${\mathrm{BH}}_3\cdot \mathrm{L}$

$$\begin{array}{rl}& {\mathrm{B}}_2{\mathrm{H}}_6+2{\mathrm{NMe}}_3\to 2{\mathrm{BH}}_3\cdot {\mathrm{NMe}}_3\\ & {\mathrm{B}}_2{\mathrm{H}}_6+2\mathrm{CO}\to 2{\mathrm{BH}}_3\cdot \mathrm{CO}\end{array}$$

Reaction of ammonia with diborane gives initially ${\mathrm{B}}_2{\mathrm{H}}_6.2{\mathrm{NH}}_3$ which is formulated as ${\left[{\mathrm{BH}}_2{\left({\mathrm{NH}}_3\right)}_2\right]}^{+}{\left[{\mathrm{BH}}_4\right]}^{-}$; further heating gives borazine, ${\mathrm{B}}_3{\mathrm{N}}_3{\mathrm{H}}_6$ known as “inorganic benzene” in view of its ring structure with alternate $\mathrm{BH}$ and $\mathrm{NH}$ groups.

$$3{\mathrm{B}}_2{\mathrm{H}}_6+6{\mathrm{NH}}_3\to 3{\left[{\mathrm{BH}}_2{\left({\mathrm{NH}}_3\right)}_2\right]}^{+}{\left[{\mathrm{BH}}_4\right]}^{-}\stackrel{\text{Heat}}{\to }2{\mathrm{B}}_3{\mathrm{N}}_3{\mathrm{H}}_6+12{\mathrm{H}}_2$$

The structure of diborane is shown in Fig.11.2(a). The four terminal hydrogen atoms and the two boron atoms lie in one plane. Above and below this plane, there are two bridging hydrogen atoms. The four terminal $\mathrm{B}-\mathrm{H}$ bonds are regular two centre-two electron bonds while the two bridge ( $\mathrm{B}-\mathrm{H}-\mathrm{B})$ bonds are different and can be described in terms of three

centre-two electron bonds shown in Fig.11.2 (b).

Boron also forms a series of hydridoborates; the most important one is the tetrahedral ${\left[{\mathrm{BH}}_4\right]}^{-}$ ion. Tetrahydridoborates of several metals are known. Lithium and sodium tetrahydridoborates, also known as borohydrides, are prepared by the reaction of metal hydrides with ${\mathrm{B}}_2{\mathrm{H}}_6$ in diethyl ether.

$2\mathrm{MH}+{\mathrm{B}}_2{\mathrm{H}}_6\to 2{\mathrm{M}}^{+}{\left[{\mathrm{BH}}_4\right]}^{-}(\mathrm{M}=\mathrm{Li}$ or $\mathrm{Na})$

Both ${\mathrm{LiBH}}_4$ and ${\mathrm{NaBH}}_4$ are used as reducing agents in organic synthesis. They are useful starting materials for preparing other metal borohydrides.

11.4 USES OF BORON AND ALUMINIUM AND THEIR COMPOUNDS

Boron being extremely hard refractory solid of high melting point, low density and very low electrical conductivity, finds many applications. Boron fibres are used in making bullet-proof vest and light composite material for aircraft. The boron-10 $\left({}^{10}\mathrm{B}\right)$ isotope has high ability to absorb neutrons and, therefore, metal borides are used in nuclear industry as protective shields and control rods. The main industrial application of borax and boric acid is in the manufacture of heat resistant glasses (e.g., Pyrex), glass-wool and fibreglass. Borax is also used as a flux for soldering metals, for heat, scratch and stain resistant glazed coating to earthenwares and as constituent of medicinal soaps. An aqueous solution of orthoboric acid is generally used as a mild antiseptic.

Aluminium is a bright silvery-white metal, with high tensile strength. It has a high electrical and thermal conductivity. On a weight-to-weight basis, the electrical conductivity of aluminium is twice that of copper. Aluminium is used extensively in industry and every day life. It forms alloys with $\mathrm{Cu},\mathrm{Mn},\mathrm{Mg},\mathrm{Si}$ and $\mathrm{Zn}$. Aluminium and its alloys can be given shapes of pipe, tubes, rods, wires, plates or foils and, therefore, find uses in packing, utensil making, construction, aeroplane and transportation industry. The use of aluminium and its compounds for domestic purposes is now reduced considerably because of their toxic nature.

11.5 GROUP 14 ELEMENTS: THE CARBON FAMILY

Carbon (C), silicon (Si), germanium (Ge), tin (Sn) and lead $(\mathrm{Pb})$ are the members of group 14. Carbon is the seventeenth most abundant element by mass in the earth’s crust. It is widely distributed in nature in free as well as in the combined state. In elemental state it is available as coal, graphite and diamond; however, in combined state it is present as metal carbonates, hydrocarbons and carbon dioxide gas $(0.03$ in air. One can emphatically say that carbon is the most versatile element in the world. Its combination with other elements such as dihydrogen, dioxygen, chlorine and sulphur provides an astonishing array of materials ranging from living tissues to drugs and plastics. Organic chemistry is devoted to carbon containing compounds. It is an essential constituent of all living organisms. Naturally occurring carbon contains two stable isotopes: ${}^{12}\mathrm{C}$ and ${}^{13}\mathrm{C}$. In addition to these, third isotope, ${}^{14}\mathrm{C}$ is also present. It is a radioactive isotope with halflife 5770 years and used for radiocarbon dating. Silicon is the second ( $27.7$ by mass) most abundant element on the earth’s crust and is present in nature in the form of silica and silicates. Silicon is a very important component of ceramics, glass and cement.

Germanium exists only in traces. Tin occurs mainly as cassiterite, ${\mathrm{SnO}}_2$ and lead as galena, $\mathrm{PbS}$.

Ultrapure form of germanium and silicon are used to make transistors and semiconductor devices.

The important atomic and physical properties of the group 14 elements along with their electronic configuration are given in Table 11.3 Some of the atomic, physical and chemical properties are discussed below:

11.5.1 Electronic Configuration

The valence shell electronic configuration of these elements is $n{s}^{2}n{p}^2$. The inner core of the electronic configuration of elements in this group also differs.

11.5.2 Covalent Radius

There is a considerable increase in covalent radius from $\mathrm{C}$ to $\mathrm{Si}$, thereafter from $\mathrm{Si}$ to $\mathrm{Pb}$ a small increase in radius is observed. This is due to the presence of completely filled $d$ and $f$ orbitals in heavier members.

11.5.3 Ionization Enthalpy

The first ionization enthalpy of group 14 members is higher than the corresponding members of group 13. The influence of inner core electrons is visible here also. In general the ionisation enthalpy decreases down the group. Small decrease in ${\mathrm{\Delta }}_{i}H$ from $\mathrm{Si}$ to Ge to $\mathrm{Sn}$ and slight increase in ${\mathrm{\Delta }}_{i}H$ from $\mathrm{Sn}$ to $\mathrm{Pb}$ is the consequence of poor shielding effect of intervening $d$ and $f$ orbitals and increase in size of the atom.

11.5.4 Electronegativity

Due to small size, the elements of this group are slightly more electronegative than group 13 elements. The electronegativity values for elements from $\mathrm{Si}$ to $\mathrm{Pb}$ are almost the same.

11.5.5 Physical Properties

All group 14 members are solids. Carbon and siliconarenon-metals, germanium is a metalloid,

Table 11.3 Atomic and Physical Properties of Group 14 Elements

whereas tin and lead are soft metals with low melting points. Melting points and boiling points of group 14 elements are much higher than those of corresponding elements of group 13.

11.5.6 Chemical Properties

Oxidation states and trends in chemical reactivity

The group 14 elements have four electrons in outermost shell. The common oxidation states exhibited by these elements are +4 and +2 . Carbon also exhibits negative oxidation states. Since the sum of the first four ionization enthalpies is very high, compounds in +4 oxidation state are generally covalent in nature. In heavier members the tendency to show +2 oxidation state increases in the sequence $\mathrm{Ge}<\mathrm{Sn}<\mathrm{Pb}$. It is due to the inability of $n{s}^2$ electrons of valence shell to participate in bonding. The relative stabilities of these two oxidation states vary down the group. Carbon and silicon mostly show +4 oxidation state. Germanium forms stable compounds in +4 state and only few compounds in +2 state. Tin forms compounds in both oxidation states ( $\mathrm{Sn}$ in +2 state is a reducing agent). Lead compounds in +2 state are stable and in +4 state are strong oxidising agents. In tetravalent state the number of electrons around the central atom in a molecule (e.g., carbon in ${\mathrm{CCl}}_4$ ) is eight. Being electron precise molecules, they are normally not expected to act as electron acceptor or electron donor species. Although carbon cannot exceed its covalence more than 4 , other elements of the group can do so. It is because of the presence of $d$ orbital in them. Due to this, their halides undergo hydrolysis and have tendency to form complexes by accepting electron pairs from donor species. For example, the species like, ${{\mathrm{SiF}}_6}^{2-},{\left[{\mathrm{GeCl}}_6\right]}^{2-}$, ${[\mathrm{Sn}(\mathrm{OH}{)}_6]}^{2-}$ exist where the hybridisation of the central atom is $s{p}^3{d}^2$.

(i) Reactivity towards oxygen

All members when heated in oxygen form oxides. There are mainly two types of oxides, i.e., monoxide and dioxide of formula $\mathrm{MO}$ and ${\mathrm{MO}}_2$ respectively. $\mathrm{SiO}$ only exists at high temperature. Oxides in higher oxidation states of elements are generally more acidic than those in lower oxidation states. The dioxides - ${\mathrm{CO}}_2,{\mathrm{SiO}}_2$ and ${\mathrm{GeO}}_2$ are acidic, whereas ${\mathrm{SnO}}_2$ and ${\mathrm{PbO}}_2$ are amphoteric in nature. Among monoxides, $\mathrm{CO}$ is neutral, GeO is distinctly acidic whereas $\mathrm{SnO}$ and $\mathrm{PbO}$ are amphoteric.

Problem 11.5

Select the member(s) of group 14 that (i) forms the most acidic dioxide, (ii) is commonly found in +2 oxidation state, (iii) used as semiconductor.

Solution

(i) carbon

(ii) lead

(iii) silicon and germanium

(ii) Reactivity towards water

Carbon, silicon and germanium are not affected by water. Tin decomposes steam to form dioxide and dihydrogen gas.

$$\mathrm{Sn}+2{\mathrm{H}}_2\mathrm{O}\stackrel{\mathrm{\Delta }}{\to }{\mathrm{SnO}}_2+2{\mathrm{H}}_2$$

Lead is unaffected by water, probably because of a protective oxide film formation.

(iii) Reactivity towards halogen

These elements can form halides of formula ${\mathrm{MX}}_2$ and ${\mathrm{MX}}_4$ (where $\mathrm{X}=\mathrm{F},\mathrm{Cl},\mathrm{Br},\mathrm{I}$ ). Except carbon, all other members react directly with halogen under suitable condition to make halides. Most of the ${\mathrm{MX}}_4$ are covalent in nature. The central metal atom in these halides undergoes $s{p}^3$ hybridisation and the molecule is tetrahedral in shape. Exceptions are ${\mathrm{SnF}}_4$ and ${\mathrm{PbF}}_4$, which are ionic in nature. ${\mathrm{PbI}}_4$ does not exist because $\mathrm{Pb}-\mathrm{I}$ bond initially formed during the reaction does not release enough energy to unpair $6s^2$ electrons and excite one of them to higher orbital to have four unpaired electrons around lead atom. Heavier members Ge to $\mathrm{Pb}$ are able to make halides of formula ${\mathrm{MX}}_2$. Stability of dihalides increases down the group. Considering the thermal and chemical stability, ${\mathrm{GeX}}_4$ is more stable than ${\mathrm{GeX}}_2$, whereas ${\mathrm{PbX}}_2$ is more than ${\mathrm{PbX}}_4$. Except ${\mathrm{CCl}}_4$, other tetrachlorides are easily hydrolysed by water because the central atom can accommodate the lone pair of electrons from oxygen atom of water molecule in $d$ orbital.

Hydrolysis can be understood by taking the example of ${\mathrm{SiCl}}_4$. It undergoes hydrolysis by initially accepting lone pair of electrons from water molecule in $d$ orbitals of $\mathrm{Si}$, finally leading to the formation of $\mathrm{Si}(\mathrm{OH}{)}_4$ as shown below:

Problem 11. 6

${\left[{\mathrm{SiF}}_6\right]}^{2-}$ is known whereas ${\left[{\mathrm{SiCl}}_6\right]}^{2-}$ not. Give possible reasons.

Solution

The main reasons are :

(i) six large chloride ions cannot be accommodated around ${\mathrm{Si}}^{4+}$ due to limitation of its size.

(ii) interaction between lone pair of chloride ion and ${\mathrm{Si}}^{4+}$ is not very strong.

11.6 IMPORTANT TRENDS AND ANOMALOUS BEHAVIOUR OF CARBON

Like first member of other groups, carbon also differs from rest of the members of its group. It is due to its smaller size, higher electronegativity, higher ionisation enthalpy and unavailability of $d$ orbitals.

In carbon, only $s$ and $p$ orbitals are available for bonding and, therefore, it can accommodate only four pairs of electrons around it. This would limit the maximum covalence to four whereas other members can expand their covalence due to the presence of $d$ orbitals. Carbon also has unique ability to form $p\pi -p\pi$ multiple bonds with itself and with other atoms of small size and high electronegativity. Few examples of multiple bonding are: $\mathrm{C}=\mathrm{C}$, $\mathrm{C}\equiv \mathrm{C},\mathrm{C}=\mathrm{O},\mathrm{C}=\mathrm{S}$, and $\mathrm{C}\equiv \mathrm{N}$. Heavier elements do not form $p\pi -p\pi$ bonds because their atomic orbitals are too large and diffuse to have effective overlapping.

Carbon atoms have the tendency to link with one another through covalent bonds to form chains and rings. This property is called catenation. This is because $\mathrm{C}-\mathrm{C}$ bonds are very strong. Down the group the size increases and electronegativity decreases, and, thereby, tendency to show catenation decreases. This can be clearly seen from bond enthalpies values. The order of catenation is $\mathrm{C}»\mathrm{Si}>$ $\mathrm{Ge}\approx \mathrm{Sn}$. Lead does not show catenation.

Bond	Bond enthalpy / kJ ${\mathbf{m}\mathbf{o}\mathbf{l}}^{\mathbf{-}\mathbf{1}}$
$\mathrm{C}-\mathrm{C}$	348
$\mathrm{Si}-\mathrm{Si}$	297
$\mathrm{Ge}-\mathrm{Ge}$	260
$\mathrm{Sn}-\mathrm{Sn}$	240

Due to property of catenation and $p\pi -p\pi$ bond formation, carbon is able to show allotropic forms.

11.7 ALLOTROPES OF CARBON

Carbon exhibits many allotropic forms; both crystalline as well as amorphous. Diamond and graphite are two well-known crystalline forms of carbon. In 1985, third form of carbon known as fullerenes was discovered by H.W.Kroto, E.Smalley and R.F.Curl. For this discovery they were awarded the Nobel Prize in 1996.

11.7.1 Diamond

It has a crystalline lattice. In diamond each carbon atom undergoes $s{p}^3$ hybridisation and linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion. The $\mathrm{C}-\mathrm{C}$ bond length is $154\mathrm{pm}$. The structure extends in space and produces a rigid threedimensional network of carbon atoms. In this

structure (Fig. 11.3) directional covalent bonds are present throughout the lattice.

It is very difficult to break extended covalent bonding and, therefore, diamond is a hardest substance on the earth. It is used as an abrasive for sharpening hard tools, in making dyes and in the manufacture of tungsten filaments for electric light bulbs.

Problem 11.7

Diamond is covalent, yet it has high melting point. Why?

Solution

Diamond has a three-dimensional network involving strong $\mathrm{C}-\mathrm{C}$ bonds, which are very difficult to break and, in turn has high melting point.

11.7.2 Graphite

Graphite has layered structure (Fig.11.4). Layers are held by van der Waals forces and distance between two layers is 340 pm. Each layer is composed of planar hexagonal rings of carbon atoms. $\mathrm{C}-\mathrm{C}$ bond length within the layer is $141.5\mathrm{pm}$. Each carbon atom in hexagonal ring undergoes $s{p}^2$ hybridisation and makes three sigma bonds with three neighbouring carbon atoms. Fourth electron forms a $\pi$ bond. The electrons are delocalised over the whole sheet. Electrons are mobile and,

therefore, graphite conducts electricity along the sheet. Graphite cleaves easily between the layers and, therefore, it is very soft and slippery. For this reason graphite is used as a dry lubricant in machines running at high temperature, where oil cannot be used as a lubricant.

11.7.3 Fullerenes

Fullerenes are made by the heating of graphite in an electric arc in the presence of inert gases such as helium or argon. The sooty material formed by condensation of vapourised ${\mathrm{C}}^{\mathrm{n}}$ small molecules consists of mainly ${\mathrm{C}}_{6}0$ with smaller quantity of ${\mathrm{C}}_{7}0$ and traces of fullerenes consisting of even number of carbon atoms up to 350 or above. Fullerenes are the only pure form of carbon because they have smooth structure without having ‘dangling’ bonds. Fullerenes are cage like molecules. ${\mathrm{C}}_{6}0$ molecule has a shape like soccer ball and called Buckminsterfullerene (Fig. 11.5).

It contains twenty six- membered rings and twelve five membered rings. A six membered ring is fused with six or five membered rings but a five membered ring can only fuse with six membered rings. All the carbon atoms are equal and they undergo $s{p}^2$ hybridisation. Each carbon atom forms three sigma bonds with other three carbon atoms. The remaining electron at each carbon is delocalised in molecular orbitals, which in turn give aromatic character to molecule. This ball shaped molecule has 60 vertices and each one is occupied by one carbon atom and it also contains both single and double bonds with $\mathrm{C}-\mathrm{C}$ distances of $143.5\mathrm{pm}$ and $138.3\mathrm{pm}$ respectively. Spherical fullerenes are also called bucky balls in short.

It is very important to know that graphite is thermodynamically most stable allotrope of carbon and, therefore, ${\mathrm{\Delta }}_f{H}^{\ominus }$ of graphite is taken as zero. ${\mathrm{\Delta }}_f{H}^{\ominus }$ values of diamond and fullerene, ${\mathrm{C}}_{6}0$ are 1.90 and $38.1\mathrm{kJ}{\mathrm{mol}}^{-1}$, respectively.

Other forms of elemental carbon like carbon black, coke, and charcoal are all impure forms of graphite or fullerenes. Carbon black is obtained by burning hydrocarbons in a limited supply of air. Charcoal and coke are obtained by heating wood or coal respectively at high temperatures in the absence of air.

11.7.4 Uses of Carbon

Graphite fibres embedded in plastic material form high strength, lightweight composites. The composites are used in products such as tennis rackets, fishing rods, aircrafts and canoes. Being good conductor, graphite is used for electrodes in batteries and industrial electrolysis. Crucibles made from graphite are inert to dilute acids and alkalies. Being highly porous, activated charcoal is used in adsorbing poisonous gases; also used in water filters to remove organic contaminators and in airconditioning system to control odour. Carbon black is used as black pigment in black ink and as filler in automobile tyres. Coke is used as a fuel and largely as a reducing agent in metallurgy. Diamond is a precious stone and used in jewellery. It is measured in carats ( 1 carat $=200\mathrm{mg}$ ).

11.8 SOME IMPORTANT COMPOUNDS OF CARBON AND SILICON

Oxides of Carbon

Two important oxides of carbon are carbon monoxide, $\mathrm{CO}$ and carbon dioxide, ${\mathrm{CO}}_2$.

11.8.1 Carbon Monoxide

Direct oxidation of $\mathrm{C}$ in limited supply of oxygen or air yields carbon monoxide.

$$2\mathrm{C}(\mathrm{s})+{\mathrm{O}}_2(\mathrm{g})\stackrel{\mathrm{\Delta }}{\to }2\mathrm{CO}(\mathrm{g})$$

On small scale pure $\mathrm{CO}$ is prepared by dehydration of formic acid with concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$ at $373\mathrm{K}$

$$\mathrm{HCOOH}\underset{\text{ conc. }{\mathrm{H}}_2{\mathrm{SO}}_4}{\overset{373\mathrm{K}}{\to }}{\mathrm{H}}_2\mathrm{O}+\mathrm{CO}$$

On commercial scale it is prepared by the passage of steam over hot coke. The mixture of $\mathrm{CO}$ and ${\mathrm{H}}_2$ thus produced is known as water gas or synthesis gas.

$\mathrm{C}(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{g})\stackrel{473-1273\mathrm{K}}{\to }\underset{\text{Water gas}}{\mathrm{CO}(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})}$

When air is used instead of steam, a mixture of $\mathrm{CO}$ and ${\mathrm{N}}_2$ is produced, which is called producer gas.

$$\begin{array}{rl}2\mathrm{C}(\mathrm{s})+{\mathrm{O}}_2(\mathrm{g})+4{\mathrm{N}}_2(\mathrm{g})\stackrel{1273\mathrm{K}}{\to }& \underset{\text{Producer Gas}}{2\mathrm{CO}(\mathrm{g})+4{\mathrm{N}}_2(\mathrm{g})}\end{array}$$

Water gas and producer gas are very important industrial fuels. Carbon monoxide in water gas or producer gas can undergo further combustion forming carbon dioxide with the liberation of heat.

Carbon monoxide is a colourless, odourless and almost water insoluble gas. It is a powerful reducing agent and reduces almost all metal oxides other than those of the alkali and alkaline earth metals, aluminium and a few transition metals. This property of $\mathrm{CO}$ is used in the extraction of many metals from their oxides ores.

$$\begin{array}{rl}& {\mathrm{Fe}}_2{\mathrm{O}}_3(\mathrm{s})+3\mathrm{CO}(\mathrm{g})\stackrel{\mathrm{\Delta }}{\to }2\mathrm{Fe}(\mathrm{s})+3{\mathrm{CO}}_2(\mathrm{g})\\ & \mathrm{ZnO}(\mathrm{s})+\mathrm{CO}(\mathrm{g})\stackrel{\mathrm{\Delta }}{\to }\mathrm{Zn}(\mathrm{s})+{\mathrm{CO}}_2(\mathrm{g})\end{array}$$

In $\mathrm{CO}$ molecule, there are one sigma and two $\pi$ bonds between carbon and oxygen, :C $\equiv \mathrm{O}:$. Because of the presence of a lone pair on carbon, $\mathrm{CO}$ molecule acts as a donor and reacts with certain metals when heated to form metal carbonyls. The highly poisonous nature of $\mathrm{CO}$ arises because of its ability to form a complex with haemoglobin, which is about 300 times more stable than the oxygen-haemoglobin complex. This prevents haemoglobin in the red blood corpuscles from carrying oxygen round the body and ultimately resulting in death.

11.8.2 Carbon Dioxide

It is prepared by complete combustion of carbon and carbon containing fuels in excess of air.

$$\begin{array}{rl}& \mathrm{C}(\mathrm{s})+{\mathrm{O}}_2(\mathrm{g})\stackrel{\mathrm{\Delta }}{\to }{\mathrm{CO}}_2(\mathrm{g})\\ & {\mathrm{CH}}_4(\mathrm{g})+2{\mathrm{O}}_2(\mathrm{g})\stackrel{\mathrm{\Delta }}{\to }{\mathrm{CO}}_2(\mathrm{g})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{g})\end{array}$$

In the laboratory it is conveniently prepared by the action of dilute $\mathrm{HCl}$ on calcium carbonate.

${\mathrm{CaCO}}_3(\mathrm{s})+2\mathrm{HCl}(\mathrm{aq})\to {\mathrm{CaCl}}_2(\mathrm{aq})+{\mathrm{CO}}_2(\mathrm{g})+$ ${\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

On commercial scale it is obtained by heating limestone.

It is a colourless and odourless gas. Its low solubility in water makes it of immense biochemical and geo-chemical importance. With water, it forms carbonic acid, ${\mathrm{H}}_2{\mathrm{CO}}_3$ which is a weak dibasic acid and dissociates in two steps:

${\mathrm{H}}_2{\mathrm{CO}}_3(\mathrm{aq})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to {{\mathrm{HCO}}_3}^{-}(\mathrm{aq})+{\mathrm{H}}_3{\mathrm{O}}^{+}(\mathrm{aq})$

${{\mathrm{HCO}}_3}^{-}(\mathrm{aq})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to {{\mathrm{CO}}_3}^{2-}(\mathrm{aq})+{\mathrm{H}}_3{\mathrm{O}}^{+}(\mathrm{aq})$

${\mathrm{H}}_2{\mathrm{CO}}_3/{{\mathrm{HCO}}_3}^{-}$buffer system helps to maintain pH of blood between 7.26 to 7.42. Being acidic in nature, it combines with alkalies to form metal carbonates.

Carbon dioxide, which is normally present to the extent of $\sim 0.03$ by volume in the atmosphere, is removed from it by the process known as photosynthesis. It is the process by which green plants convert atmospheric ${\mathrm{CO}}_2$ into carbohydrates such as glucose. The overall chemical change can be expressed as:

$$6{\mathrm{CO}}_2+12{\mathrm{H}}_2\mathrm{O}\underset{\text{Chlorophyll}}{\overset{hv}{\to }}{\mathrm{C}}_6{H}_{12}{\mathrm{O}}_6+6{\mathrm{O}}_2+6{\mathrm{HO}}_2$$

By this process plants make food for themselves as well as for animals and human beings. Unlike CO, it is not poisonous. But the increase in combustion of fossil fuels and decomposition of limestone for cement manufacture in recent years seem to increase the ${\mathrm{CO}}_2$ content of the atmosphere. This may lead to increase in green house effect and thus, raise the temperature of the atmosphere which might have serious consequences.

Carbon dioxide can be obtained as a solid in the form of dry ice by allowing the liquified ${\mathrm{CO}}_2$ to expand rapidly. Dry ice is used as a refrigerant for ice-cream and frozen food. Gaseous ${\mathrm{CO}}_2$ is extensively used to carbonate soft drinks. Being heavy and non-supporter of combustion it is used as fire extinguisher. A substantial amount of ${\mathrm{CO}}_2$ is used to manufacture urea.

In ${\mathrm{CO}}_2$ molecule carbon atom undergoes $sp$ hybridisation. Two $sp$ hybridised orbitals of carbon atom overlap with two $p$ orbitals of oxygen atoms to make two sigma bonds while other two electrons of carbon atom are involved in $p\pi -p\pi$ bonding with oxygen atom. This results in its linear shape [with both $\mathrm{C}-\mathrm{O}$ bonds of equal length (115 pm)] with no dipole moment. The resonance structures are shown below:

Resonance structures of carbon dioxide

11.8.3 Silicon Dioxide, ${\mathrm{SiO}}_2$

$95\mathrm{\%}$ of the earth’s crust is made up of silica and silicates. Silicon dioxide, commonly known as silica, occurs in several crystallographic forms. Quartz, cristobalite and tridymite are some of the crystalline forms of silica, and they are interconvertable at suitable temperature. Silicon dioxide is a covalent, three-dimensional network solid in which each silicon atom is covalently bonded in a tetrahedral manner to four oxygen atoms. Each oxygen atom in turn covalently bonded to another silicon atoms as shown in diagram (Fig 11.6). Each corner is shared with another tetrahedron. The entire crystal may be considered as giant molecule in which eight membered rings are formed with alternate silicon and oxygen atoms.

Silica in its normal form is almost nonreactive because of very high $\mathrm{Si}-\mathrm{O}$ bond enthalpy. It resists the attack by halogens, dihydrogen and most of the acids and metals even at elevated temperatures. However, it is attacked by $\mathrm{HF}$ and $\mathrm{NaOH}$.

$$\begin{array}{rl}& {\mathrm{SiO}}_2+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{SiO}}_3+{\mathrm{H}}_2\mathrm{O}\\ & {\mathrm{SiO}}_2+4\mathrm{HF}\to {\mathrm{SiF}}_4+2{\mathrm{H}}_2\mathrm{O}\end{array}$$

Quartz is extensively used as a piezoelectric material; it has made possible to develop extremely accurate clocks, modern radio and television broadcasting and mobile radio communications. Silica gel is used as a drying agent and as a support for chromatographic materials and catalysts. Kieselghur, an amorphous form of silica is used in filtration plants.

11.8.4 Silicones

They are a group of organosilicon polymers, which have $\left({\mathrm{R}}_2\mathrm{SiO}\right)$ as a repeating unit. The starting materials for the manufacture of silicones are alkyl or aryl substituted silicon chlorides, ${\mathrm{R}}_{\mathrm{n}}{\mathrm{SiCl}}_{(}4-\mathrm{n})$, where $\mathrm{R}$ is alkyl or aryl group. When methyl chloride reacts with silicon in the presence of copper as a catalyst at a temperature $573\mathrm{K}$ various types of methyl substituted chlorosilane of formula ${\mathrm{MeSiCl}}_3$, ${\mathrm{Me}}_2{\mathrm{SiCl}}_2,{\mathrm{Me}}_3\mathrm{SiCl}$ with small amount of ${\mathrm{Me}}_4\mathrm{Si}$ are formed. Hydrolysis of dimethyldichlorosilane, ${\left({\mathrm{CH}}_3\right)}_2{\mathrm{SiCl}}_2$ followed by condensation polymerisation yields straight chain polymers.

$2{\mathrm{CH}}_3\mathrm{Cl}+\mathrm{Si}\underset{\text{570 K}}{\overset{\text{Cu powder}}{\to }}({\mathrm{CH}}_3{)}_2{\mathrm{SiCl}}_2\underset{-2\mathrm{HCl}}{\overset{+2{\mathrm{H}}_2\mathrm{O}}{\to }}({\mathrm{CH}}_3{)}_2\mathrm{Si}(\mathrm{OH}{)}_2$

The chain length of the polymer can be controlled by adding ${\left({\mathrm{CH}}_3\right)}_3\mathrm{SiCl}$ which blocks the ends as shown below :

Silicones being surrounded by non-polar alkyl groups are water repelling in nature. They have in general high thermal stability, high dielectric strength and resistance to oxidation and chemicals. They have wide applications. They are used as sealant, greases, electrical insulators and for water proofing of fabrics. Being biocompatible they are also used in surgical and cosmetic plants.

Problem: 11.8

What are silicones?

Solution

Simple silicones consist of ${(-\begin{array}{l}|\\ \mathrm{Si}\\ |\end{array}-\mathrm{O}-)}_{\mathrm{n}}$ chains in which alkyl or phenyl groups occupy the remaining bonding positions on each silicon. They are hydrophobic (water repellant) in nature.

11.8.5 Silicates

A large number of silicates minerals exist in nature. Some of the examples are feldspar, zeolites, mica and asbestos. The basic structural unit of silicates is ${\mathrm{SiO}}_4{}^{4-}$ (Fig. 11.7) in which silicon atom is bonded to four oxygen atoms in tetrahedron fashion. In silicates either the discrete unit is present or a number of such units are joined together via corners by sharing $1,2,3$ or 4 oxygen atoms per silicate units. When silicate units are linked together, they form chain, ring, sheet or three-dimensional structures. Negative charge on silicate structure is

neutralised by positively charged metal ions. If all the four corners are shared with other tetrahedral units, three-dimensional network is formed.

Two important man-made silicates are glass and cement.

11.8.6 Zeolites

If aluminium atoms replace few silicon atoms in three-dimensional network of silicon dioxide, overall structure known as aluminosilicate, acquires a negative charge. Cations such as ${\mathrm{Na}}^{+},{\mathrm{K}}^{+}$or ${\mathrm{Ca}}^{2+}$ balance the negative charge. Examples are feldspar and zeolites. Zeolites are widely used as a catalyst in petrochemical industries for cracking of hydrocarbons and isomerisation, e.g., ZSM-5 (A type of zeolite) used to convert alcohols directly into gasoline. Hydrated zeolites are used as ion exchangers in softening of “hard” water.

SUMMARY

p-Block of the periodic table is unique in terms of having all types of elements - metals, non-metals and metalloids. There are six groups of $p$-block elements in the periodic table numbering from 13 to 18 . Their valence shell electronic configuration is $\mathit{n}{\mathit{s}}^{\mathbf{2}}\mathit{n}{\mathit{p}}^{\mathbf{1}\mathbf{-}\mathbf{6}}$ (except for He). Differences in the inner core of their electronic configuration greatly influence their physical and chemical properties. As a consequence of this, a lot of variation in properties among these elements is observed. In addition to the group oxidation state, these elements show other oxidation states differing from the total number of valence electrons by unit of two. While the group oxidation state is the most stable for the lighter elements of the group, lower oxidation states become progressively more stable for the heavier elements. The combined effect of size and availability of $d$ orbitals considerably influences the ability of these elements to form $\pi$-bonds. While the lighter elements form $\mathit{p}\mathit{\pi }-\mathit{p}\mathit{\pi }$ bonds, the heavier ones form $\mathit{d}\mathit{\pi }\mathit{p}\mathit{\pi }$ or $\mathit{d}\mathit{\pi }\mathit{d}\mathit{\pi }$ bonds. Absence of $\mathit{d}$ orbital in second period elements limits their maximum covalence to 4 while heavier ones can exceed this limit.

Boron is a typical non-metal and the other members are metals. The availability of 3 valence electrons $\left(2s^{2}2p^1\right)$ for covalent bond formation using four orbitals $(2s,2p_{\mathrm{x},}2p_{\mathrm{y}}$ and $2p_z$ ) leads to the so called electron deficiency in boron compounds. This deficiency makes them good electron acceptor and thus boron compounds behave as Lewis acids. Boron forms covalent molecular compounds with dihydrogen as boranes, the simplest of which is diborane, ${\mathrm{B}}_2{\mathrm{H}}_6$. Diborane contains two bridging hydrogen atoms between two boron atoms; these bridge bonds are considered to be three-centre two-electron bonds. The important compounds of boron with dioxygen are boric acid and borax. Boric acid, $\mathrm{B}(\mathrm{OH}{)}_3$ is a weak monobasic acid; it acts as a Lewis acid by accepting electrons from hydroxyl ion. Borax is a white crystalline solid of formula ${\mathrm{Na}}_2[{\mathrm{B}}_4{\mathrm{O}}_5(\mathrm{OH}{)}_4]\cdot 8{\mathrm{H}}_2\mathrm{O}$. The borax bead test gives characteristic colours of transition metals.

Aluminium exhibits +3 oxidation state. With heavier elements +1 oxidation state gets progressively stabilised on going down the group. This is a consequence of the so called inert pair effect.

Carbon is a typical non-metal forming covalent bonds employing all its four valence electrons $\left(2s^{2}2p^2\right)$. It shows the property of catenation, the ability to form chains or rings, not only with $\mathrm{C}-\mathrm{C}$ single bonds but also with multiple bonds $(\mathrm{C}=\mathrm{C}$ or $\mathrm{C}\equiv \mathrm{C})$. The tendency to catenation decreases as $\mathrm{C}»\mathrm{Si}>\mathrm{Ge}\simeq \mathrm{Sn}>\mathrm{Pb}$. Carbon provides one of the best examples of allotropy. Three important allotropes of carbon are diamond, graphite and fullerenes. The members of the carbon family mainly exhibit +4 and +2 oxidation states; compouds in +4 oxidation states are generally covalent in nature. The tendency to show +2 oxidation state increases among heavier elements. Lead in +2 state is stable whereas in +4 oxidation state it is a strong oxidising agent. Carbon also exhibits negative oxidation states. It forms two important oxides: $\mathrm{CO}$ and ${\mathrm{CO}}_2$. Carbon monoxide is neutral whereas ${\mathrm{CO}}_2$ is acidic in nature. Carbon monoxide having lone pair of electrons on $\mathrm{C}$ forms metal carbonyls. It is deadly poisonous due to higher stability of its haemoglobin complex as compared to that of oxyhaemoglobin complex. Carbon dioxide as such is not toxic. However, increased content of ${\mathrm{CO}}_2$ in atmosphere due to combustion of fossil fuels and decomposition of limestone is feared to cause increase in ‘green house effect’. This, in turn, raises the temperature of the atmosphere and causes serious complications. Silica, silicates and silicones are important class of compounds and find applications in industry and technology.

EXERCISES

11.1 Discuss the pattern of variation in the oxidation states of (i) $\mathrm{B}$ to $\mathrm{Tl}$ and (ii) $\mathrm{C}$ to $\mathrm{Pb}$.

Show Answer

Answer

(i) B to TI

The electric configuration of group 13 elements is $n{s}^{2}n{p}^1$. Therefore, the most common oxidation state exhibited by them should be +3 . However, it is only boron and aluminium which practically show the +3 oxidation state. The remaining elements, i.e., $Ga$, In, $TI$, show both the +1 and +3 oxidation states. On moving down the group, the +1 state becomes more stable. For example, $TI(+1)$ is more stable than $TI(+3)$. This is because of the inert pair effect. The two electrons present in the s-shell are strongly attracted by the nucleus and do not participate in bonding. This inert pair effect becomes more and more prominent on moving down the group. Hence, Ga (+1) is unstable, In (+1) is fairly stable, and $TI(+1)$ is very stable.

Group 13 element	Oxidation state
$B$	+3
$Al$	+3
$Ga,In,Tl$	$+1,+3$

The stability of the +3 oxidation state decreases on moving down the group.

(ii) $C$ to $Pb$

The electronic configuration of group 14 elements is $n{s}^{2}n{p}^2$. Therefore, the most common oxidation state exhibited by them should be +4 . However, the +2 oxidation state becomes more and more common on moving down the group. $C$ and Si mostly show the +4 state. On moving down the group, the higher oxidation state becomes less stable. This is because of the inert pair effect. Thus, although $Ge,Sn$, and $Pb$ show both the +2 and +4 states, the stability of the lower oxidation state increases and that of the higher oxidation state decreases on moving down the group.

Group 14 element	Oxidation state
$C$	+4
$Si$	+4
$Ge,Sn,Pb$	$+2,+4$

C $SiGeSnPb$

stability of +4 state decreases

11.2 How can you explain higher stability of ${\mathrm{BCl}}_3$ as compared to ${\mathrm{TlCl}}_3$ ?

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Answer

Boron and thallium belong to group 13 of the periodic table. In this group, the +1 oxidation state becomes more stable on moving down the group. $BC{l}_3$ is more stable than $TIC{l}_3$ because the +3 oxidation state of $B$ is more stable than the +3 oxidation state of $TI$. In $TI$, the +3 state is highly oxidising and it reverts back to the more stable +1 state.

11.3 Why does boron triflouride behave as a Lewis acid ?

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Answer

The electric configuration of boron is $n{s}^{2}n{p}^1$. It has three electrons in its valence shell. Thus, it can form only three covalent bonds. This means that there are only six electrons around boron and its octet remains incomplete. When one atom of boron combines with three fluorine atoms, its octet remains incomplete. Hence, boron trifluoride remains electron-deficient and acts as a Lewis acid.

11.4 Consider the compounds, ${\mathrm{BCl}}_3$ and ${\mathrm{CCl}}_4$. How will they behave with water? Justify.

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Answer

Being a Lewis acid, $BC{l}_3$ readily undergoes hydrolysis. Boric acid is formed as a result.

$BC{l}_3+3H_{2}O⟶3HCl+B(OH{)}_3$

$CC{l}_4$ completely resists hydrolysis. Carbon does not have any vacant orbital. Hence, it cannot accept electrons from water to form an intermediate. When $CC{l}_4$ and water are mixed, they form separate layers.

$CC{l}_4+H_{2}O⟶$ No reaction

11.5 Is boric acid a protic acid ? Explain.

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Answer

Boric acid is not a protic acid. It is a weak monobasic acid, behaving as a Lewis acid.

$B(OH{)}_3+2HOH⟶[B(OH{)}_4{]}^{-}+H_3{O}^{+}$

It behaves as an acid by accepting a pair of electrons from ${}^{\text{âe" }}OH$ ion.

11.6 Explain what happens when boric acid is heated .

Show Answer

Answer

On heating orthoboric acid $(H_{3}B{O}_3)$ at $370K$ or above, it changes to metaboric acid $(HB{O}_2)$. On further heating, this yields boric oxide $B_2{O}_3$.

$H_{3}B{O}_3\underset{370K}{\overset{\mathrm{\Delta }}{\to }}HB{O}_2\underset{\text{ red hot }}{\overset{\mathrm{\Delta }}{\to }}{B}_2{O}_3$

Metaboric acid Boric oxide

11.7 Describe the shapes of ${\mathrm{BF}}_3$ and ${{\mathrm{BH}}_4}^{-}$. Assign the hybridisation of boron in these species.

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Answer

(i) $B{F}_3$

As a result of its small size and high electronegativity, boron tends to form monomeric covalent halides. These halides have a planar triangular geometry. This triangular shape is formed by the overlap of three $s{p}^2$ hybridised orbitals of boron with the sporbitals of three halogen atoms. Boron is $s{p}^2$ hybridised in $B{F}_3$.

(ii) $B{H}_4{}^{-}$

Boron-hydride ion $(B{H}_4)$ is formed by the $s{p}^3$ hybridisation of boron orbitals. Therefore, it is tetrahedral in structure.

11.8 Write reactions to justify amphoteric nature of aluminium.

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Answer

A substance is called amphoteric if it displays characteristics of both acids and bases. Aluminium dissolves in both acids and bases, showing amphoteric behaviour.

(i) $2A{l}_{(s)}+6HC{l}_{(oq)}⟶2A{l}_{(aq)}^{3+}+6C{l}_{(\alpha q)}^{-}+3H_{2(g)}$

(ii) $2A{l}_{(s)}+2NaO{H}_{(xq)}+6H_2{O}_{(l)}⟶2N{a}^{+}[Al(OH{)}_4{]}_{(aq)}^{-}+3H_{2(g)}$

11.9 What are electron deficient compounds? Are ${\mathrm{BCl}}_3$ and ${\mathrm{SiCl}}_4$ electron deficient species? Explain.

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Answer

In an electron-deficient compound, the octet of electrons is not complete, i.e., the central metal atom has an incomplete octet. Therefore, it needs electrons to complete its octet.

(i) $BC{l}_3$

$BC{l}_3$ is an appropriate example of an electron-deficient compound. $B$ has 3 valence electrons. After forming three covalent bonds with chlorine, the number of electrons around it increases to 6 . However, it is still short of two electrons to complete its octet. (ii) $SiC{l}_4$

The electronic configuration of silicon is $n{s}^{2}n{p}^2$. This indicates that it has four valence electrons. After it forms four covalent bonds with four chlorine atoms, its electron count increases to eight. Thus, $SiC{l}_4$ is not an electron-deficient compound.

11.10 Write the resonance structures of ${{\mathrm{CO}}_3}^{2-}$ and ${{\mathrm{HCO}}_3}^{-}$.

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Answer

(a) $C{O}_3^{2-}$

(b) $HC{O}_3^{-}$ion

(A)

(B)

There are only two resonating structures for the bicarbonate ion.

11.11 What is the state of hybridisation of carbon in (a) ${{\mathrm{CO}}_3}^{2-}$ (b) diamond (c) graphite?

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Answer

The state of hybridisation of carbon in: (a) $C{O}_3^{2-}$

$C$ in $C{O}_3^{2-}$ is $s{p}^2$ hybridised and is bonded to three oxygen atoms.

(b) Diamond

Each carbon in diamond is $s{p}^3$ hybridised and is bound to four other carbon atoms.

(c) Graphite

11.12 Explain the difference in properties of diamond and graphite on the basis of their structures.

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Answer

Diamond	Graphite
It has a crystalline lattice.	It has a layered structure.
In diamond, each carbon atom is $s{p}^3$ hybridised and is bonded to four other carbon atoms through a $\tilde{A}\AE$ ’ bond.	In graphite, each carbon atom is $s{p}^2$ hybridised and is bonded to three other carbon atoms through a $\tilde{A}\AE$ ’ bond. The fourth electron forms a $\pi$ bond.
It is made up of tetrahedral units.	It has a planar geometry.
The C‑C bond length in diamond is 154 pm.	The C–C bond length in graphite is 141.5 pm.
It has a rigid covalent bond network which is difficult to break.	It is quite soft and its layers can be separated easily.
It acts as an electrical insulator.	It is a good conductor of electricity.

11.13 Rationalise the given statements and give chemical reactions :

• Lead(II) chloride reacts with ${\mathrm{Cl}}_2$ to give ${\mathrm{PbCl}}_4$.
• Lead(IV) chloride is highly unstable towards heat.
• Lead is known not to form an iodide, ${\mathrm{PbI}}_4$.

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Answer

(a) Lead belongs to group 14 of the periodic table. The two oxidation states displayed by this group is +2 and +4 . On moving down the group, the +2 oxidation state becomes more stable and the +4 oxidation state becomes less stable.

This is because of the inert pair effect. Hence, $PbC{l}_4$ is much less stable than $PbC{l}_2$. However, the formation of $PbC{l}_4$ takes place when chlorine gas is bubbled through a saturated solution of $PIC{l}_2$.

$PbC{l}_{2(s)}+C{l}_{2(g)}⟶PbC{l}_{4(l)}$

(b) On moving down group IV, the higher oxidation state becomes unstable because of the inert pair effect. $Pb(IV)$ is highly unstable and when heated, it reduces to $Pb(II)$.

$PbC{l}_{4(l)}\stackrel{\mathrm{\Delta }}{\to }PbC{l}_{2(s)}+C{l}_{2(g)}$

(c) Lead is known not to form $Pb{l}_4$. $Pb(+4)$ is oxidising in nature and lis reducing in nature. A combination of $Pb(IV)$ and iodide ion is not stable. lodide ion is strongly reducing in nature. $Pb(IV)$ oxidises $I^a$ to $I^2$ and itself gets reduced to $Pb(II)$.

$Pb{I}_4⟶Pb{I}_2+I_2$

11.14 Suggest reasons why the $\mathrm{B}-\mathrm{F}$ bond lengths in ${\mathrm{BF}}_3(130\mathrm{pm})$ and ${{\mathrm{BF}}_4}^{-}$ (143 pm) differ.

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Answer

The B–F bond length in $B{F}_3$ is shorter than the B— $F$ bond length in $B{F}_4^{-}.B{F}_3$ is an electron-deficient species. With a vacant $p$-orbital on boron, the fluorine and boron atoms undergo $p$ тâ€" $p\pi$ mack-bonding to remove this deficiency. This imparts a double-bond character to the B–F bond.

This double-bond character causes the bond length to shorten in $B{F}_3(130pm)$. However, when $B{F}_3$ coordinates with

the double-bond character is lost. This accounts for a B— $F$ bond length of 143 pm in $B{F}_4^{-}$ion.

Tetrahedral structure

11.15 If $\mathrm{B}-\mathrm{Cl}$ bond has a dipole moment, explain why ${\mathrm{BCl}}_3$ molecule has zero dipole moment.

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Answer

As a result of the difference in the electronegativities of $B$ and $Cl$, the $B-Cl$ bond is polar in nature. However, the $BC{l}_3$ molecule is non-polar. This is because $BC{l}_3$ is trigonal planar in shape. It is a symmetrical molecule. Hence, the respective dipole-moments of the $B-Cl$ bond cancel each other, thereby causing a zero-dipole moment.

11.16 Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous ${\mathrm{BF}}_3$ is bubbled through. Give reasons.

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Answer

Hydrogen fluoride (HF) is a covalent compound and has a very strong intermolecular hydrogen-bonding. Thus, it does not provide ions and aluminium fluoride (AIF) does not dissolve in it. Sodium fluoride (NaF) is an ionic compound and when it is added to the mixture, AIF dissolves. This is because of the availability of free $F^{\widehat{ϵ}{ϵ}^{\ast }}$. The reaction involved in the process is: $Al{F}_3+3NaF$ $N{a}_3[Al{F}_6]$

Sodium hexafluoroaluminate(III)

When boron trifluoride $(B{F}_3)$ is added to the solution, aluminium fluoride precipitates out of the solution. This happens because the tendency of boron to form complexes is much more than that of aluminium. Therefore, when $B{F}_3$ is added to the solution, $B$ replaces $Al$ from the complexes according to the following reaction:

$N{a}_3[Al{F}_6]+3B{F}_3⟶3Na[B{F}_4]+Al{F}_3$

11.17 Suggest a reason as to why $\mathrm{CO}$ is poisonous.

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Answer

Carbon monoxide is highly-poisonous because of its ability to form a complex with haemoglobin. The CO-Hb complex is more stable than the $O_2-Hb$ complex. The former prevents $Hb$ from binding with oxygen. Thus, a person dies because of suffocation on not receiving oxygen. It is found that the $CO-Hb$ complex is about 300 times more stable than the $O_2-Hb$ complex.

11.18 How is excessive content of ${\mathrm{CO}}_2$ responsible for global warming ?

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Answer

Carbon dioxide is a very essential gas for our survival. However, an increased content of $C{O}_2$ in the atmosphere poses a serious threat. An increment in the combustion of fossil fuels, decomposition of limestone, and a decrease in the number of trees has led to greater levels of carbon dioxide. Carbon dioxide has the property of trapping the heat provided by sunrays. Higher the level of carbon dioxide, higher is the amount of heat trapped. This results in an increase in the atmospheric temperature, thereby causing global warming.

11.19 Explain structures of diborane and boric acid.

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Answer

(a) Diborane

$B_2{H}_6$ is an electron-deficient compound. $B_2{H}_6$ has only 12 electrons - 6 efrom $6H$ atoms and 3 eeach from $2B$ atoms. Thus, after combining with $3H$ atoms, none of the boron atoms has any electrons left. X-ray diffraction studies have shown the structure of diborane as:

2 boron and 4 terminal hydrogen atoms $(H_t)$ lie in one plane, while the other two bridging hydrogen atoms $(H_b)$ lie in a plane perpendicular to the plane of boron atoms. Again, of the two bridging hydrogen atoms, one $H$ atom lies above the plane and the other lies below the plane. The terminal bonds are regular two-centre two-electron $(2c-2e)$ bonds, while the two bridging ( $B-H-B)$ bonds are three-centre two-electron $(3c-2e)$ bonds.

(b) Boric acid

Boric acid has a layered structure. Each planar $B{O}_3$ unit is linked to one another through $H$ atoms. The $H$ atoms form a covalent bond with $B{O}_3$ unit, while a hydrogen bond is formed with another $B{O}_3$ unit. In the given figure, the dotted lines represent hydrogen bonds.

11.20 What happens when

(a) Borax is heated strongly,

(b) Boric acid is added to water,

(c) Aluminium is treated with dilute $\mathrm{NaOH}$,

(d) ${\mathrm{BF}}_3$ is reacted with ammonia?

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Answer

(a) When heated, borax undergoes various transitions. It first loses water molecules and swells. Then, it turns into a transparent liquid, solidifying to form a glass-like material called borax bead.

$\underset{\text{ Borax }}{N{a}_2{B}_4{O}_7\cdot 10H_{2}O}\stackrel{\text{ Sodium }}{\to }\underset{\text{ metaborate }}{N{a}_2{B}_4{O}_7}\stackrel{\mathrm{\Delta }}{\to }2NaB{O}_2+\underset{\text{ Boric }}{B_2{O}_3}$

(b) When boric acid is added to water, it accepts electrons from ${}^{at}OH$ ion.

$B(OH{)}_3+2HOH⟶[B(OH{)}_4{]}^{-}+H_3{O}^{+}$

(c) Al reacts with dilute $NaOH$ to form sodium tetrahydroxoaluminate(III). Hydrogen gas is liberated in the process.

$2A{l}_{(s)}+2NaO{H}_{(aq)}+6H_2{O}_{(l)}⟶2N{a}^{+}[Al(OH{)}_4{]}_{(oq)}^{-}+3H_{2(g)}$

(d) $B{F}_3$ (a Lewis acid) reacts with $N{H}_3$ (a Lewis base) to form an adduct. This results in a complete octet around $B$ in $B{F}_3$.

$F_{3}B+:N{H}_3⟶F_{3}B\to :N{H}_3$

11.21 Explain the following reactions

(a) Silicon is heated with methyl chloride at high temperature in the presence of copper;

(b) Silicon dioxide is treated with hydrogen fluoride;

(c) $\mathrm{CO}$ is heated with $\mathrm{ZnO}$;

(d) Hydrated alumina is treated with aqueous $\mathrm{NaOH}$ solution.

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Answer

(a) When silicon reacts with methyl chloride in the presence of copper (catalyst) and at a temperature of about $537K$, a class of organosilicon polymers called methyl-substituted chlorosilanes ( $MeSiC{l}_3,M{e}_{2}SiC{l}_2,M{e}_{3}SiCl$, and $M{e}_{4}Si$ ) are formed.

(b) When silicon dioxide $(Si{O}_2)$ is heated with hydrogen fluoride $(HF)$, it forms silicon tetrafluoride $(Si{F}_4)$. Usually, the Si–O bond is a strong bond and it resists any attack by halogens and most acids, even at a high temperature. However, it is attacked by HF.

$Si{O}_2+4HF⟶Si{F}_4+2H_{2}O$

The $Si{F}_4$ formed in this reaction can further react with $HF$ to form hydrofluorosilicic acid.

$Si{F}_4+2HF⟶H_{2}Si{F}_6$

(c) When $CO$ reacts with $ZnO$, it reduces $ZnO$ to $Zn$. $CO$ acts as a reducing agent.

$Zn{O}_{(s)}+C{O}_{(g)}\stackrel{\mathrm{\Delta }}{\to }Z{n}_{(s)}+C{O}_{2(g)}$

(d) When hydrated alumina is added to sodium hydroxide, the former dissolves in the latter because of the formation of sodium meta-aluminate.

$A{l}_2{O}_3\cdot 2H_{2}O+2NaOH⟶2NaAl{O}_2+3H_{2}O$

11.22 Give reasons :

(i) Conc. ${\mathrm{HNO}}_3$ can be transported in aluminium container.

(ii) A mixture of dilute $\mathrm{NaOH}$ and aluminium pieces is used to open drain.

(iii) Graphite is used as lubricant.

(iv) Diamond is used as an abrasive.

(v) Aluminium alloys are used to make aircraft body.

(vi) Aluminium utensils should not be kept in water overnight.

(vii) Aluminium wire is used to make transmission cables.

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Answer

(i) Concentrated $HN{O}_3$ can be stored and transported in aluminium containers as it reacts with aluminium to form a thin protective oxide layer on the aluminium surface. This oxide layer renders aluminium passive.

(ii) Sodium hydroxide and aluminium react to form sodium tetrahydroxoaluminate(III) and hydrogen gas. The pressure of the produced hydrogen gas is used to open blocked drains.

$2Al+2NaOH+6H_{2}O⟶2N{a}^{+}[Al(OH{)}_4{]}^{-}+3H_2$

(iii) Graphite has a layered structure and different layers of graphite are bonded to each other by weak van der Waals’ forces. These layers can slide over each other. Graphite is soft and slippery. Therefore, graphite can be used as a lubricant.

(iv) In diamond, carbon is $s{p}^3$ hybridised. Each carbon atom is bonded to four other carbon atoms with the help of strong covalent bonds. These covalent bonds are present throughout the surface, giving it a very rigid 3-D structure. It is very difficult to break this extended covalent bonding and for this reason, diamond is the hardest substance known. Thus, it is used as an abrasive and for cutting tools.

(v) Aluminium has a high tensile strength and is very light in weight. It can also be alloyed with various metals such as $Cu,Mn,Mg,Si$, and $Zn$. It is very malleable and ductile. Therefore, it is used in making aircraft bodies.

(vi) The oxygen present in water reacts with aluminium to form a thin layer of aluminium oxide. This layer prevents aluminium from further reaction. However, when water is kept in an aluminium vessel for long periods of time, some amount of aluminium oxide may dissolve in water. As aluminium ions are harmful, water should not be stored in aluminium vessels overnight.

(vii) Silver, copper, and aluminium are among the best conductors of electricity. Silver is an expensive metal and silver wires are very expensive. Copper is quite expensive and is also very heavy. Aluminium is a very ductile metal. Thus, aluminium is used in making wires for electrical conduction.

11.23 Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?

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Answer

Ionisation enthalpy of carbon (the first element of group 14) is very high (1086 kJ/mol). This is expected owing to its small size. However, on moving down the group to silicon, there is a sharp decrease in the enthalpy ( $786kJ)$. This is because of an appreciable increase in the atomic sizes of elements on moving down the group.

11.24 How would you explain the lower atomic radius of $\mathrm{Ga}$ as compared to $\mathrm{Al}$ ?

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Answer

Atomic radius (in pm)
Aluminium	143
Gallium	135

Although Ga has one shell more than AI, its size is lesser than Al. This is because of the poor shielding effect of the $3d$-electrons. The shielding effect of $d$-electrons is very poor and the effective nuclear charge experienced by the valence electrons in gallium is much more than it is in the case of $Al$.

11.25 What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and graphite. What is the impact of structure on physical properties of two allotropes?

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Answer

Allotropy is the existence of an element in more than one form, having the same chemical properties but different physical properties. The various forms of an element are called allotropes.

Diamond:

The rigid 3-D structure of diamond makes it a very hard substance. In fact, diamond is one of the hardest naturallyoccurring substances. It is used as an abrasive and for cutting tools.

Graphite:

It has $s{p}^2$ hybridised carbon, arranged in the form of layers. These layers are held together by weak van der Walls’ forces. These layers can slide over each other, making graphite soft and slippery. Therefore, it is used as a lubricant.

11.26 (a) Classify following oxides as neutral, acidic, basic or amphoteric: $\mathrm{CO},{\mathrm{B}}_2{\mathrm{O}}_3,{\mathrm{SiO}}_2,{\mathrm{CO}}_2,{\mathrm{Al}}_2{\mathrm{O}}_3,{\mathrm{PbO}}_2,{\mathrm{Tl}}_2{\mathrm{O}}_3$

(b) Write suitable chemical equations to show their nature.

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Answer

(1) $CO=$ Neutral

(2) $B_2{O}_3=$ Acidic

Being acidic, it reacts with bases to form salts. It reacts with $NaOH$ to form sodium metaborate.

$B_2{O}_3+2NaOH⟶2NaB{O}_2+H_{2}O$

(3) $Si{O}_2=$ Acidic

Being acidic, it reacts with bases to form salts. It reacts with $NaOH$ to form sodium silicate.

$Si{O}_2+2NaOH⟶2N{a}_{2}Si{O}_3+H_{2}O$

(4) $C{O}_2=$ Acidic

Being acidic, it reacts with bases to form salts. It reacts with $NaOH$ to form sodium carbonate.

$C{O}_2+2NaOH⟶N{a}_{2}C{O}_3+H_{2}O$

(5) $A{l}_2{O}_3=$ Amphoteric

Amphoteric substances react with both acids and bases. $A{l}_2{O}_3$ reacts with both $NaOH$ and $H_{2}S{O}_4$.

$A{l}_2{O}_3+2NaOH⟶NaAl{O}_2$

$A{l}_2{O}_3+3H_{2}S{O}_4⟶A{l}_2(S{O}_4{)}_3+3H_{2}O$

(6) $Pb{O}_2=$ Amphoteric

Amphoteric substances react with both acids and bases. $Pb{O}_2$ reacts with both $NaOH$ and $H_{2}S{O}_4$.

$Pb{O}_2+2NaOH⟶N{a}_{2}Pb{O}_3+H_{2}O$

$2Pb{O}_2+2H_{2}S{O}_4⟶2PbS{O}_4+2H_{2}O+O_2$

(7) $T{l}_2{O}_3=$ Basic

Being basic, it reacts with acids to form salts. It reacts with $HCl$ to form thallium chloride.

$T{l}_2{O}_3+6HCl⟶2TlC{l}_3+3H_{2}O$

11.27 In some of the reactions thallium resembles aluminium, whereas in others it resembles with group I metals. Support this statement by giving some evidences.

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Answer

Thallium belongs to group 13 of the periodic table. The most common oxidation state for this group is +3 . However, heavier members of this group also display the +1 oxidation state. This happens because of the inert pair effect. Aluminium displays the +3 oxidation state and alkali metals display the +1 oxidation state. Thallium displays both the oxidation states. Therefore, it resembles both aluminium and alkali metals.

Thallium, like aluminium, forms compounds such as $TIC{l}_3$ and $T{l}_2{O}_3$. It resembles alkali metals in compounds $T{l}_{2}O$ and $TICl$.

11.28 When metal $\mathrm{X}$ is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of $\mathrm{NaOH}$ to give soluble complex (B). Compound (A) is soluble in dilute $\mathrm{HCl}$ to form compound $(\mathrm{C})$. The compound (A) when heated strongly gives (D), which is used to extract metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identities.

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Answer

The given metal $X$ gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, $X$ must be aluminium.

The white precipitate (compound A) obtained is aluminium hydroxide. The compound $B$ formed when an excess of the base is added is sodium tetrahydroxoaluminate(III).

$2Al+3NaOH⟶Al(OH{)}_3\downarrow +3N{a}^{+}$

$\text{ Aluminium }(X)\text{ Sodium hydroxide }\text{ White ppt.(A) }$

$$\begin{array}{}\text{(A)}& Al(OH{)}_3+NaOH⟶N{a}^{+}[Al(OH{)}_4{]}^{-}\end{array}$$

(Soluble complex B)

Now, when dilute hydrochloric acid is added to aluminium hydroxide, aluminium chloride (compound C) is obtained.

$$\begin{array}{}\text{(A)}& Al(OH{)}_3+3HCl⟶AlC{l}_3+3H_{2}O\end{array}$$

Also, when compound $A$ is heated strongly, it gives compound $D$. This compound is used to extract metal X. Aluminium metal is extracted from alumina. Hence, compound $D$ must be alumina.

$2Al(OH{)}_3\stackrel{\mathrm{\Delta }}{\to }A{l}_2{O}_3+3H_{2}O$

(A) (D)

11.29 What do you understand by (a) inert pair effect (b) allotropy and (c) catenation?

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Answer

(a) Inert pair effect

As one moves down the group, the tendency of s-block electrons to participate in chemical bonding decreases. This effect is known as inert pair effect. In case of group 13 elements, the electronic configuration is $n{s}^{2}n{p}^1$ and their group valency is +3 . However, on moving down the group, the +1 oxidation state becomes more stable. This happens because of the poor shielding of the $n{s}^2$ electrons by the $d$ - and $f$ - electrons. As a result of the poor shielding, the $n{s}^2$ electrons are held tightly by the nucleus and so, they cannot participate in chemical bonding.

(b) Allotropy

Allotropy is the existence of an element in more than one form, having the same chemical properties but different physical properties. The various forms of an element are called allotropes. For example, carbon exists in three allotropic forms: diamond, graphite, and fullerenes.

(c) Catenation

The atoms of some elements (such as carbon) can link with one another through strong covalent bonds to form long chains or branches. This property is known as catenation. It is most common in carbon and quite significant in $Si$ and $S$.

11.30 A certain salt $\mathrm{X}$, gives the following results.

(i) Its aqueous solution is alkaline to litmus.

(ii) It swells up to a glassy material Y on strong heating.

(iii) When conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ is added to a hot solution of $\mathrm{X}$, white crystal of an acid $Z$ separates out.

Write equations for all the above reactions and identify $\mathrm{X},\mathrm{Y}$ and $\mathrm{Z}$.

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Answer

The given salt is alkaline to litmus. Therefore, $X$ is a salt of a strong base and a weak acid. Also, when Xis strongly heated, it swells to form substance $Y$. Therefore, Xmust be borax.

When borax is heated, it loses water and swells to form sodium metaborate. When heating is continued, it solidifies to form a glassy material $Y$. Hence, $Y$ must be a mixture of sodium metaborate and boric anhydride.

$\begin{array}{rl}& N{a}_2{B}_4{O}_7+7H_{2}O\stackrel{\text{ water }}{\to }2NaOH+4H_{3}B{O}_3\\ & \text{ Borax }(X)\text{ Sodium hydroxide }\text{ Orthoboric acid }\end{array}$

$\underset{\text{ Borax }(X)}{N{a}_2{B}_4{O}_7\cdot 10H_{2}O\stackrel{\mathrm{\Delta }}{\to }\underset{\text{ Sodium metaborate }}{N{a}_2{B}_4{O}_7}\stackrel{\mathrm{\Delta }}{\to }\underset{\begin{array}{c}\text{ Boric anhydride }\\ \text{ (Glassy material) }\end{array}}{B_2{O}_3}+2NaB{O}_2}$

When concentrated acid is added to borax, white crystals of orthoboric acid $(Z)$ are formed.

$N{a}_2{B}_4{O}_7\cdot 10H_{2}O+H_{2}S{O}_4\stackrel{\mathrm{\Delta }}{\to }N{a}_{2}S{O}_4+4H_{3}B{O}_3+5H_{2}O$

Borax $(X)$

Orthoboric acid (Z)

11.31 Write balanced equations for:

(i) ${\mathrm{BF}}_3+\mathrm{LiH}\to$

(ii) ${\mathrm{B}}_2{\mathrm{H}}_6+{\mathrm{H}}_2\mathrm{O}\to$

(iii) $\mathrm{NaH}+{\mathrm{B}}_2{\mathrm{H}}_6\to$

(iv) ${\mathrm{H}}_3{\mathrm{BO}}_3\stackrel{\mathrm{\Delta }}{\to }$

(v) $\mathrm{Al}+\mathrm{NaOH}\to$

(vi) ${\mathrm{B}}_2{\mathrm{H}}_6+{\mathrm{NH}}_3\to$

Show Answer

Answer

(i)

$2B{F}_3+6LiH⟶B_2{H}_6+6LiF$

Boron trifluoride Lithium hydride Diborane Lithium fluoride

(ii) $B_2{H}_6+6H_{2}O⟶2H_{3}B{O}_3+6H_2$

Diborane Water Orthoboric acid Hydrogen

(iii) $B_2{H}_6+2NaH\stackrel{\text{ ether }}{\to }2NaB{H}_4$

Diborane Sodium hydride Sodium borohydride

(iv)

Boron trioxide

(v) $2Al+2NaOH+6H_{2}O⟶2N{a}^{+}[Al(OH{)}_4{]}_{(aq)}^{-}+3H_2$

Sodium tetrahydroxoaluminate(III)

(vi) $3B_2{H}_6+6N{H}_3⟶3[B{H}_2(N{H}_3{)}_2{]}^{+}[B{H}_4{]}^{-}⟶2B_3{N}_3{H}_6+12H_2$

Borazene

11.32. Give one method for industrial preparation and one for laboratory preparation of $\mathrm{CO}$ and ${\mathrm{CO}}_2$ each.

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Answer

Caron dioxide

In the laboratory, $C{O}_2$ can be prepared by the action of dilute hydrochloric acid on calcium carbonate. The reaction involved is as follows:

$CaC{O}_3+2HC{l}_{(aq)}⟶CaC{l}_{2(aq)}+C{O}_{2(g)}+H_2{O}_{(j)}$

$C{O}_2$ is commercially prepared by heating limestone. The reaction involved is as follows: $CaC{O}_3\stackrel{\mathrm{\Delta }}{\to }CaO+C{O}_2\uparrow$

Caron monoxide

In the laboratory, $CO$ is prepared by the dehydration of formic acid with conc. $H_{2}S{O}_4$, at $373K$. The reaction involved is as follows:

$HCOOH\underset{\text{ conc. }{H}_{2}S{O}_4}{\overset{373K}{\to }}{H}_{2}O+CO\uparrow$

$CO$ is commercially prepared by passing steam over hot coke. The reaction involved is as follows:

$C_{(s)}+H_2{O}_{(g)}\stackrel{473-1273K}{\to }\underset{\text{water gas }}{\underbrace{C{O}_{(g)}+H_{2(g)}}}$

11.33 An aqueous solution of borax is (a) neutral (b) amphoteric (c) basic (d) acidic

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Answer

(c) Borax is a salt of a strong base $(NaOH)$ and a weak acid $(H_{3}B{O}_3)$. It is, therefore, basic in nature.

11.34 Boric acid is polymeric due to (a) its acidic nature (b) the presence of hydrogen bonds (c) its monobasic nature (d) its geometry

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Answer

(b) Boric acid is polymeric because of the presence of hydrogen bonds. In the given figure, the dotted lines represent hydrogen bonds.

11.35 The type of hybridisation of boron in diborane is (a) $sp$ (b) $s{p}^2$ (c) $s{p}^3$ (d) $ds{p}^2$

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Answer

(c) Boron in diborane is $s{p}^3$ hybridised.

11.36 Thermodynamically the most stable form of carbon is (a) diamond (b) graphite (c) fullerenes (d) coal

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

11.37 Elements of group 14

(a) exhibit oxidation state of +4 only

(b) exhibit oxidation state of +2 and +4

(c) form ${\mathrm{M}}^{2-}$ and ${\mathrm{M}}^{4+}$ ions

(d) form ${\mathrm{M}}^{2+}$ and ${\mathrm{M}}^{4+}$ ions

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(b)The elements of group 14 have 4 valence electrons. Therefore, the oxidation state of the group is +4 . However, as a result of the inert pair effect, the lower oxidation state becomes more and more stable and the higher oxidation state becomes less stable. Therefore, this group exhibits +4 and +2 oxidation states.

Group 14 element	Oxidation state
$C$	+4
$Si$	+4
$Ge,Sn,Pb$	$+2,+4$