Explain the Lewis version of acid-base chemistryWrite equations because that the development of adducts and complex ionsPerform equilibrium calculations involving formation constants

In 1923, G. N. Lewis suggest a generalized an interpretation of acid-base actions in i beg your pardon acids and also bases are determined by their capability to accept or come donate a pair of electrons and type a coordinate covalent bond.

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A coordinate covalent bond (or datil bond) occurs when one of the atoms in the bond offers both bonding electrons. For example, a coordinate covalent link occurs once a water molecule combines with a hydrogen ion to kind a hydronium ion. A name: coordinates covalent bond additionally results when an ammonia molecule combines through a hydrogen ion to kind an ammonium ion. Both of these equations are shown here.


A Lewis acid is any varieties (molecule or ion) that can accept a pair the electrons, and a Lewis base is any varieties (molecule or ion) that can donate a pair of electrons.

A Lewis acid-base reaction occurs as soon as a basic donates a pair of electron to one acid. A Lewis acid-base adduct, a link that has a coordinate covalent bond in between the Lewis acid and the Lewis base, is formed. The complying with equations show the basic application of the Lewis concept.

The boron atom in boron trifluoride, BF3, has only six electrons in the valence shell. Being brief of the wanted octet, BF3 is a very good Lewis acid and also reacts with many Lewis bases; a fluoride ion is the Lewis base in this reaction, donating among its lone pairs:


In the adhering to reaction, every of 2 ammonia molecules, Lewis bases, donates a pair of electron to a silver ion, the Lewis acid:


Nonmetal oxides act together Lewis acids and react with oxide ions, Lewis bases, to type oxyanions:


Many Lewis acid-base reactions are displacement reaction in i m sorry one Lewis basic displaces another Lewis basic from an acid-base adduct, or in which one Lewis mountain displaces one more Lewis acid:


The critical displacement reaction shows just how the reaction the a Brønsted-Lowry acid with a base fits into the Lewis concept. A Brønsted-Lowry acid such as HCl is one acid-base adduct according to the Lewis concept, and proton transfer occurs because a more stable acid-base adduct is formed. Thus, return the definitions of acids and also bases in the two theories are quite different, the theories overlap considerably.

Many slightly dissolve ionic solids dissolve once the concentration the the metal ion in systems is decreased through the development of complex (polyatomic) ion in a Lewis acid-base reaction. For example, silver- chloride dissolves in a systems of ammonia because the silver- ion reacts through ammonia to kind the complex ion extAg(NH_3)_2^;;+. The Lewis framework of the extAg(NH_3)_2^;;+ ion is:


The equations for the dissolved of AgCl in a solution of NH3 are:

extAg^+(aq);+;2 extNH_3(aq);longrightarrow; extAg(NH_3)_2^;;+(aq)
extNet:;AgCl(s);+;2 extNH_3(aq);longrightarrow; extAg(NH_3)_2^;;+(aq);+; extCl^-(aq)

Aluminum hydroxide disappear in a systems of salt hydroxide or another strong base since of the development of the complex ion extAl(OH)_4^;;-. The Lewis structure of the extAl(OH)_4^;;- ion is:


The equations for the resolution are:

extAl^3+(aq);+;4 extOH^-(aq);longrightarrow; extAl(OH)_4^;;-(aq)
extNet:;Al(OH)_3(s);+; extOH^-(aq);longrightarrow; extAl(OH)_4^;;-(aq)
extHg^2+(aq);+;2 extS^2-(aq);longrightarrow; extHgS_2^;;2-(aq)
extNet:;HgS(s);+; extS^2-(aq);longrightarrow; extHgS_2^;;2-(aq)

A facility ion is composed of a main atom, commonly a transition metal cation, surrounded by ions, or molecules dubbed ligands. This ligands deserve to be neutral molecules choose H2O or NH3, or ion such together CN– or OH–. Often, the ligands act together Lewis bases, donating a pair of electrons to the main atom. The ligands accumulation themselves approximately the central atom, creating a new ion with a charge equal come the amount of the fees and, many often, a transitional steel ion. This more complicated arrangement is why the result ion is called a complex ion. The facility ion created in these reactions cannot be predicted; it must be identified experimentally. The types of bonds formed in complex ions are called coordinate covalent bonds, as electrons from the ligands space being common with the main atom. Since of this, complex ions are periodically referred to together coordination complexes. This will certainly be studied further in upcoming chapters.

The equilibrium continuous for the reaction that the components of a complicated ion to type the complicated ion in equipment is referred to as a formation constant (Kf) (sometimes called a security constant). Because that example, the complex ion extCu(CN)_2^;;- is displayed here:


It creates by the reaction:

extCu^+(aq);+;2 extCN^-(aq); ightleftharpoons; extCu(CN)_2^;;-(aq)
K_ extf = Q = frac< extCu(CN)_2^;;->< extCu^+>< extCN^->^2

The train station of the formation continuous is the dissociation continuous (Kd), the equilibrium consistent for the decomposition the a complex ion right into its materials in solution. We will work with dissociation constants more in the exercises for this section. Attachment K and Table 2 space tables of formation constants. In general, the bigger the development constant, the more stable the complex; however, together in the situation of Ksp values, the stoichiometry of the compound should be considered.

SubstanceKf in ~ 25 °C
< extCd(CN)_4>^2-3 × 1018
extAg(NH_3)_2^;;+1.7 × 107
< extAlF_6>^3-7 × 1019
Table 2. Common complex Ions by to decrease Formulation Constants

As an instance of dissolved by complex ion formation, permit us think about what happens once we add aqueous ammonia come a mixture of silver chloride and also water. Silver- chloride dissolves slightly in water, offering a little concentration the Ag+ ( = 1.3 × 10–5M):

However, if NH3 is current in the water, the complicated ion, extAg(NH_3)_2^;;+, can kind according come the equation:

extAg^+(aq);+;2 extNH_3(aq); ightleftharpoons; extAg(NH_3)_2^;;+(aq)
K_ extf = frac< extAg(NH_3)_2^;;+>< extAg^+>< extNH_3>^2 = 1.7; imes;10^7

The big size the this formation continuous indicates that most of the cost-free silver ions created by the dissolution of AgCl incorporate with NH3 to form extAg(NH_3)_2^;;+. Together a consequence, the concentration of silver ions, , is reduced, and also the reaction quotient for the dissolved of silver chloride, , falls listed below the solubility product of AgCl:

More silver- chloride then dissolves. If the concentration that ammonia is an excellent enough, all of the silver chloride dissolves.

Example 1

Dissociation of a complex IonCalculate the concentration that the silver ion in a systems that originally is 0.10 M with respect to extAg(NH_3)_2^;;+.

SolutionWe usage the acquainted path to deal with this problem:


Determine x and also equilibrium concentrations. we let the adjust in concentration of Ag+ be x. Dissociation of 1 mol that extAg(NH_3)_2^;;+ offers 1 mol of Ag+ and 2 mol the NH3, so the adjust in is 2x and that of extAg(NH_3)_2^;;+ is –x. In summary:
Solve for x and also the equilibrium concentrations. at equilibrium:

Both Q and also Kf room much bigger than 1, therefore let united state assume that the changes in concentrations necessary to with equilibrium room small. Therefore 0.10 – x is approximated together 0.10:

Because just 1.1% that the extAg(NH_3)_2^;;+ dissociates right into Ag+ and NH3, the presumption that x is little is justified.

Now we identify the equilibrium concentrations:

The concentration of cost-free silver ion in the equipment is 0.0011 M.

Check the work. The worth of Q calculated utilizing the equilibrium concentration is same to Kf within the error linked with the significant figures in the calculation.

Check your LearningCalculate the silver ion concentration, , that a solution ready by dissolve 1.00 g that AgNO3 and 10.0 g the KCN in adequate water to make 1.00 together of solution. (Hint: due to the fact that Q f, assume the reaction goes to completion then calculate the developed by dissociation that the complex.)

Key Concepts and Summary

G.N. Lewis propose a an interpretation for acids and bases that relies on one atom’s or molecule’s capacity to accept or donate electron pairs. A Lewis acid is a types that can accept an electron pair, whereas a Lewis base has an electron pair easily accessible for donation to a Lewis acid. Complicated ions are examples of Lewis acid-base adducts. In a complex ion, we have a central atom, regularly consisting that a shift metal cation, which acts as a Lewis acid, and also several neutral molecule or ions surrounding them referred to as ligands that act together Lewis bases. Complex ions kind by sharing electron pairs to form coordinate covalent bonds. The equilibrium reaction the occurs when developing a facility ion has actually an equilibrium constant associated with it referred to as a formation constant, Kf. This is regularly referred to together a stability constant, as it to represent the security of the facility ion. Formation of complicated ions in solution deserve to have a profound impact on the solubility the a shift metal compound.

Chemistry finish of thing Exercises

Under what circumstances, if any, go a sample of hard AgCl fully dissolve in pure water?Explain why the enhancement of NH3 or HNO3 come a saturated systems of Ag2CO3 in call with heavy Ag2CO3 increases the solubility that the solid.Calculate the cadmium ion concentration, , in a solution ready by mix 0.100 l of 0.0100 M Cd(NO3)2 v 1.150 together of 0.100 NH3(aq).Explain why addition of NH3 or HNO3 come a saturated solution of Cu(OH)2 in contact with hard Cu(OH)2 boosts the solubility that the solid.Sometimes equilibria for complicated ions are defined in terms of dissociation constants, Kd. Because that the facility ion extAlF_6^;;3- the dissociation reaction is:

extAlF_6^;;3-; ightleftharpoons; extAl^3+;+;6 extF^- and K_ extd = frac< extAl^3+>< extF^->^6< extAlF_6^;;3-> = 2; imes;10^-24

Calculate the worth of the formation constant, Kf, because that extAlF_6^;;3-.

Using the worth of the formation constant for the complicated ion extCo(NH_3)_6^;;2+, calculation the dissociation constant.Using the dissociation constant, Kd = 7.8 × 10–18, calculate the equilibrium concentration of Cd2+ and CN– in a 0.250-M solution of extCd(CN)_4^;;2-.Using the dissociation constant, Kd = 3.4 × 10–15, calculate the equilibrium concentrations of Zn2+ and OH– in a 0.0465-M systems of extZn(OH)_4^;;2-.Using the dissociation constant, Kd = 2.2 × 10–34, calculate the equilibrium concentration of Co3+ and also NH3 in a 0.500-M equipment of extCo(NH_3)_6^;;3+.Using the dissociation constant, Kd = 1 × 10–44, calculation the equilibrium concentrations of Fe3+ and also CN– in a 0.333 M systems of extFe(CN)_6^;;3-.Calculate the massive of potassium cyanide ion that need to be added to 100 mL of solution to dissolve 2.0 × 10–2 mol of silver cyanide, AgCN.Calculate the minimum concentration the ammonia needed in 1.0 l of systems to dissolve 3.0 × 10–3 mol of silver- bromide.A role of 35-mm black and also white photographic movie contains about 0.27 g the unexposed AgBr prior to developing. What massive of Na2S2O3·5H2O (sodium thiosulfate pentahydrate or hypo) in 1.0 l of developer is required to dissolve the AgBr together extAg(S_2 extO_3)_2^;;3- (Kf = 4.7 × 1013)?We have actually seen one introductory definition of an acid: An mountain is a compound the reacts v water and increases the lot of hydronium ion present. In the thing on acids and bases, we observed two an ext definitions the acids: a compound that donates a proton (a hydrogen ion, H+) to another compound is called a Brønsted-Lowry acid, and a Lewis mountain is any species that can accept a pair the electrons. Define why the introductory definition is a macroscopic definition, while the Brønsted-Lowry meaning and the Lewis an interpretation are microscope definitions.Write the Lewis structures of the reactants and also product of each of the adhering to equations, and identify the Lewis acid and also the Lewis basic in each:

(a) extCO_2;+; extOH^-;longrightarrow; extHCO_3^;;-

(b) extB(OH)_3;+; extOH^-;longrightarrow; extB(OH)_4^;;-

(c) extI^-;+; extI_2;longrightarrow; extI_3^;;-

(d) extAlCl_3;+; extCl^-;longrightarrow; extAlCl_4^;;- (use Al-Cl solitary bonds)

(e) extO^2-;+; extSO_3;longrightarrow; extSO_4^;;2-

Write the Lewis frameworks of the reactants and product of each of the following equations, and also identify the Lewis acid and the Lewis base in each:

(a) extCS_2;+; extSH^-;longrightarrow; extHCS_3^;;-

(b) extBF_3;+; extF^-;longrightarrow; extBF_4^;;-

(c) extI^-;+; extSnI_2;longrightarrow; extSnI_3^;;-

(d) extAl(OH)_3;+; extOH^-;longrightarrow; extAl(OH)_4^;;-

(e) extF^-;+; extSO_3;longrightarrow; extSFO_3^;;-

Using Lewis structures, write balanced equations for the complying with reactions:

(a) extHCl(g);+; extPH_3(g);longrightarrow

(b) extH_3 extO^+;+; extCH_3^;;-;longrightarrow

(c) extCaO;+; extSO_3;longrightarrow

(d) extNH_4^;;+;+; extC_2 extH_5 extO^-;longrightarrow

Calculate < extHgCl_4^;;2-> in a solution all set by adding 0.0200 mol of NaCl to 0.250 l of a 0.100-M HgCl2 solution.In a titration the cyanide ion, 28.72 mL of 0.0100 M AgNO3 is added before precipitation begins. extAg(CN)_2^;;- complex.> Precipitation of heavy AgCN takes place when excess Ag+ is included to the solution, above the amount needed to complete the formation of extAg(CN)_2^;;-. How numerous grams that NaCN to be in the original sample?What space the concentrations of Ag+, CN–, and also extAg(CN)_2^;;- in a saturated equipment of AgCN?In dilute aqueous equipment HF acts as a weak acid. However, pure liquid HF (boiling suggest = 19.5 °C) is a solid acid. In fluid HF, HNO3 acts prefer a base and accepts protons. The mountain of fluid HF have the right to be boosted by including one of several not natural fluorides that are Lewis acids and accept F– ion (for example, BF3 or SbF5). Write balanced chemical equations because that the reaction the pure HNO3 through pure HF and also of pure HF v BF3.The simplest amino mountain is glycine, H2NCH2CO2H. The typical feature of amino acids is that they save the useful groups: one amine group, –NH2, and a carboxylic mountain group, –CO2H. One amino mountain can role as one of two people an mountain or a base. Because that glycine, the acid stamin of the carboxyl team is around the same as that of acetic acid, CH3CO2H, and the base stamin of the amino team is slightly higher than the of ammonia, NH3.

(a) write the Lewis structures of the ion that form when glycine is dissolved in 1 M HCl and also in 1 M KOH.

(b) compose the Lewis structure of glycine as soon as this amino acid is dissolved in water. (Hint: think about the family member base staminas of the –NH2 and also - extCO_2^;;- groups.)

Boric acid, H3BO3, is no a Brønsted-Lowry acid yet a Lewis acid.

(a) create an equation for its reaction with water.

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(b) suspect the shape of the anion thus formed.

(c) What is the hybridization top top the boron continuous with the shape you have predicted?