An understanding of the trends in Lewis acidity and basicity enables us to predict the outcome of many reactions of the s- and p-block elements.
(a) Lewis acids and bases of the s-block elements
KEY POINT Alkali metal ions act as Lewis acids with water, forming hydrated ions.
The existence of hydrated alkali metal ions in water can be regarded as an aspect of their Lewis acid character, with H2O the Lewis base. In practice, alkali metal ions do not act as Lewis bases but may do so indirectly—an example being that their fluorides act as a source of the uncom- plexed Lewis base F− and form fluoride complexes with Lewis acids, such as SF4:
CsF + SF4 → Cs+[SF5]−
The Be atom in beryllium dihalides acts as a Lewis acid by forming a polymeric chain structure in the solid state (14). In this structure, a σ bond is formed when a lone pair of elec- trons of a halide ion, acting as a Lewis base, is donated into an empty sp3 hybrid orbital on the Be atom (two in total).
The Lewis acidity of beryllium chloride is also demonstrated by the formation of tetrahedral adducts such as BeCl42−.
Be
Hal 14 Be(Hal)2
(b) Group 13 Lewis acids
KEY POINTS The ability of boron trihalides to act as Lewis acids generally increases in the order BF3 < BCl3 < BBr3; aluminium halides are dimeric in the gas phase and are used as catalysts in solution.
The planar molecules BX3 and AlX3 have incomplete octets, and the vacant p orbital perpendicular to the plane can accept a lone pair from a Lewis base:
X
X B X B N
X X Me
MeMe N X
Me
MeMe
The acid molecule becomes pyramidal as the complex is formed and the B–X bonds bend away from their new neighbours.
The order of thermodynamic stability of complexes of :N(CH3)3 with BX3 is BF3 < BCl3 < BBr3. This order is opposite to that expected on the basis of the relative electro negativities of the halogens: an electronegativity argument would suggest that F, the most electronegative halogen, ought to leave the B atom in BF3 most electron deficient and hence able to form the strongest bond to the incoming base. The currently accepted explanation is that the halogen atoms in the BX3 molecule can form π bonds with the empty B2p orbital (15), and that these
π bonds must be disrupted to make the acceptor orbital available for complex formation. The π bond also favours the planar structure of the molecule, a structure that must be converted into tetrahedral in the adduct. The small F atom forms the strongest π bonds with the B2p orbital: recall that p–p π bonding is strongest for Period 2 elements, largely on account of the small atomic radii of these elements and the significant overlap of their compact 2p orbitals (Section 2.5). Thus, the BF3 molecule has the strongest π bond to be broken when the amine forms an N–B bond.
X
Empty
B X
X
15 p–p π bonding in BX3
Boron trifluoride is widely used as an industrial catalyst.
Its role there is to extract bases bound to carbon and hence to generate carbocations:
F
F B F B X
F F C F
R RR X
R R C R
+ +
– +
Boron trifluoride is a gas at room temperature and pressure, but it dissolves in diethyl ether to give a solution that is convenient to use and commercially available. This dissolu- tion is also an aspect of Lewis acid character because, as BF3 dissolves, it forms a complex with the :O atom of a solvent molecule (16).
B F
O
C2H5
16 Adduct of BF3 with diethyl ether
Aluminium halides are dimers in the gas phase;
aluminium chloride, for example, has the molecular formula Al2Cl6 in the vapour (17). Each Al atom acts as an acid towards a Cl atom initially belonging to the other Al atom. Aluminium chloride is widely used as a Lewis acid catalyst for organic reactions. The classic examples are Friedel–Crafts alkylation (the attachment of R+ to an aro- matic ring) and acylation (the attachment of RCO) during which AlCl4− is formed. The catalytic cycle is shown in Fig. 5.9.
Al
Cl
17 Al2Cl6
(c) Group 14 Lewis acids
KEY POINTS Group 14 elements other than carbon exhibit hyperva- lence and act as Lewis acids by becoming five- or six-coordinate; tin(II) chloride is both a Lewis acid and a Lewis base.
Unlike carbon, a Si atom can expand its valence shell (or is simply large enough) to become hypervalent. Stable struc- tures with five-coordinate trigonal bipyramidal geometry can be isolated (18) and a six-coordinate adduct is formed when the Lewis acid SiF4 reacts with two F− ions:
F
F Si FF+ 2 F–
F Si F F F
F
F 2–
Germanium and tin fluorides can react similarly. Because the Lewis base F−, aided by a proton, can displace O2− from silicates, hydrofluoric acid is corrosive towards glass (SiO2).
The trend in acidity for SiX4, which follows the order SiF4 >
SiCl4 > SiBr4 > SiI4, correlates with the decrease in the elec- tron-withdrawing power of the halogen from F to I and is the reverse of that for BX3.
AlCl3 RCH Cl:2
RCH Cl–AlCl2 3
AlCl4– CH R2
H + + HCl
CH2R
FIGURE 5.9 Catalytic cycle for the Friedel–Crafts alkylation reaction.
Si O
C6H5
–
18 [Si(C6H5)(OC6H4O)2]−
Tin(II) chloride is both a Lewis acid and a Lewis base. As an acid, SnCl2 combines with Cl− to form SnCl3− (19). This complex retains a lone pair, and it is sometimes more reveal- ing to write its formula as :SnCl3–. It acts as a base to give metal–metal bonds, as in the complex (CO)5Mn–SnCl3 (20).
Compounds containing metal–metal bonds are currently the focus of much attention in inorganic chemistry, as we see later in the text (Section 19.6). Tin(IV) halides are Lewis acids. They react with halide ions to form SnX62−:
SnCl4 + 2 Cl– → SnCl62–
The strength of the Lewis acidity again follows the order SnF4 > SnCl4 > SnBr4 > SnI4.
Sn Cl
–
19 SnCl3−
Mn Sn
Cl
CO
20 [Mn(CO)5(SnCl3)]
EXAMPLE 5.8 Predicting the relative Lewis basicity of compounds
Rationalize the following relative Lewis basicities: (a) (H3Si)2O <
(H3C)2O; (b) (H3Si)3N < (H3C)3N.
Answer Nonmetallic elements in Period 3 and below can expand their valence shells by delocalization of the O or N lone pairs
to create multiple bonds (O and N are thus acting as π electron donors). So the silyl ether and silyl amine are the weaker Lewis bases in each pair.
Self-test 5.8 Given that π bonding between Si and the lone pairs of N is important, what difference in structure between (H3Si)3N and (H3C)3N do you expect?
(d) Group 15 Lewis acids
KEY POINT Oxides and halides of the heavier Group 15 elements act as Lewis acids.
Phosphorus pentafluoride is a strong Lewis acid and forms complexes with ethers and amines. The heavier elements of the nitrogen group (Group 15) form very important Lewis acids, SbF5 being one of the most widely studied compounds. The reaction with HF produces a superacid (Section 5.16).
F Sb F
F
F + 2 HF
F Sb F
F F
F F F
+ H2F+ –
(e) Group 16 Lewis acids
KEY POINTS Sulfur dioxide can act as a Lewis acid by accepting an electron pair at the S atom; to act as a Lewis base, the SO2 molecule can donate either its S or its O lone pair to a Lewis acid.
Sulfur dioxide is both a Lewis acid and a Lewis base. Its Lewis acidity is illustrated by the formation of a complex with a trialkylamine acting as a Lewis base:
S N O O R
RR N
R RR O
S O
To act as a Lewis base, the SO2 molecule can donate either its S or its O lone pair to a Lewis acid. When SbF5 is the acid, the O atom of SO2 acts as the electron-pair donor, but when Ru(II) is the acid, the S atom acts as the donor (21).
Ru
SO2
NH3 Cl
21 [RuCl(NH3)4(SO2)]+
Sulfur trioxide is a strong Lewis acid and a very weak (O donor) Lewis base. Its acidity is illustrated by the reaction
O
O S O S N
O O R
RR N O
R
RR
A classic aspect of the acidity of SO3 is its highly exothermic reaction with water in the formation of sulfuric acid. The resulting problem of having to remove large quantities of heat from the reactor used for the commercial production of sulfuric acid is alleviated by exploiting the Lewis acidity of sulfur trioxide further to carry out the hydration by a two- stage process (Section 16.13). Before dilution, sulfur triox- ide is dissolved in sulfuric acid to form the mixture known as oleum. This reaction is an example of Lewis acid–base complex formation:
S O S O O
OH O HO O O
O S O
S O O
OH HO
The resulting H2S2O7 can then be hydrolysed in a less exo- thermic reaction:
H2S2O7 + H2O → 2 H2SO4
(f) Group 17 Lewis acids
KEY POINT Bromine and iodine molecules act as mild Lewis acids.
Lewis acidity is expressed in an interesting and subtle way by Br2 and I2, which are both strongly coloured. The strong visible absorption spectra of Br2 and I2 arise from transitions to low-lying unfilled antibonding orbitals. The colours of the species therefore suggest that the empty orbitals may be low enough in energy to serve as acceptor orbitals in Lewis acid–
base complex formation.5 Iodine is violet in the solid and gas phases and in non-donor solvents such as trichloromethane. In water, propanone (acetone), or ethanol, all of which are Lewis bases, iodine is brown. The colour changes because a solvent–
solute complex is formed from the lone pair of donor molecule O atoms and a low-lying σ* orbital of the dihalogen.
The weak interaction between halogen atoms and Lewis bases is known as halogen bonding. Use of the empty σ*
orbital on the halogen in compound AX (A is typically alkyl or halogen) leads to a tendency for linear arrangement along an axis A−X−B. Like hydrogen bonding that is dis- cussed next, intermolecular halogen bonding is responsible for the arrangement of molecules as they appear in crystals, such as the infinite chains formed by pyrazine and iodine
5 The terms donor–acceptor complex and charge-transfer complex were at one time used to denote these complexes. However, the distinc- tion between these complexes and the more familiar Lewis acid–base complexes is arbitrary and in the current literature the terms are used more or less interchangeably.
(22). Halogen bonding may also be important in the micro- biological degradation of environmentally harmful halocar- bons by Co-containing enzymes (Chapter 26).
281.7 pm 273.3 pm
22 Linear chains formed by I2 and pyrazine
The interaction of Br2 with the carbonyl group of pro- panone is shown in Fig. 5.10. The illustration also shows the transition responsible for the new absorption band observed when a complex is formed. The orbital from which the elec- tron originates in the transition is predominantly the lone pair orbital of the base (the ketone). The orbital to which the transition occurs is predominantly the LUMO of the acid (the dihalogen). Thus, to a first approximation, the transition transfers an electron from the base to the acid and is therefore called a charge-transfer transition.
The triiodide ion, I3−, is another example of a complex between a halogen acid (I2) and a halide base (I−). One of the applications of its formation is to render molecular iodine soluble in water so that it can be used as a titration reagent:
I2(s) + I−(aq) → I3−(aq) K = 725
The triiodide ion is one example of a large class of polyhal- ide ions (Section 17.10).