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Molecular ShapesIn the space-filling model,what determines the relative sizes of the spheres?Molecular ShapesMolecular ShapesWhy do most ABn molecules have shapes related to those shown in Figure 9.3?Can we predict these shapes?In addition to tetrahedral,another common shape for AB4 molecules is square planar.All five atoms lie in the same plane,with the B atoms at the corners of a square and the A atom at the center of the square.Which shape in Figure 9.3 could lead to a square-planar shape upon removal of one or more atoms?VSEPR Theory When A is a representative element(one from the s block or p block of the periodic table),we can answer these questions by using the valence-shell electron-pair repulsion(VSEPR)model.Electron DomainA non-bonding pair(or lone pair)of electrons defines an electron domain(电子域）电子域）that is located principally on one atom.For example,the Lewis structure of NH3 has four electron domains around the central nitrogen atom(three bonding pairs,represented as usual by short lines,and one nonbonding pair,represented by dots):Each multiple bond in a molecule also constitutes a single electron domain.Thus,the following resonance structure for O3 has three electron domains around the centraloxygen atom(a single bond,a double bond,and a nonbonding pair of electrons):In general,each nonbonding pair,single bond,or multiple bond produces a singleelectron domain around the central atom in a molecule.The VSEPR model is based on the idea that electron domains are negatively charged and therefore repel one another.The best arrangement of a given number of electron domains is the one that minimizes the repulsions among them.The shapes of different ABn molecules or ions depend on the number of electron domains surrounding the central atom.VSEPR Theory The molecular geometry is the arrangement of only the atoms in a molecule or ionany nonbonding pairs in the molecule are not part of the description of the molecular geometry.In determining the shape of any molecule,we first use the VSEPR model to predictthe electron-domain geometry.From knowing how many of the domains are due tononbonding pairs,we can then predict the molecular geometry.How to Predict the Shapes of Molecules or IonsWhen all the electron domains in a molecule arise from bonds,the molecular geometry is identical to the electron-domain geometry.When,however,one or more domains involve nonbonding pairs of electrons,we must remember that the molecular geometry involves only electron domains due to bonds even though the nonbonding pairs contribute to the electrondomain geometry.1.Draw the Lewis structure of the molecule or ion,and count the number of electron domains around the central atom.Each nonbonding electron pair,each single bond,each double bond,and each triple bond counts as one electron domain.How to Predict the Shapes of Molecules or Ions(Steps)2.Determine the electron-domain geometry by arranging the electron domains about the central atom so that the repulsions among them are minimized,as shown in Table 9.1.3.Use the arrangement of the bonded atoms to determine the molecular geometry.1.Draw the Lewis structure of the molecule or ion,and count the number of electron domains around the central atom.Each nonbonding electron pair,each single bond,each double bond,and each triple bond counts as one electron domain.How to Predict the Shapes of Molecules or Ions(Steps)2.Determine the electron-domain geometry by arranging the electron domains about the central atom so that the repulsions among them are minimized,as shown in Table 9.1.3.Use the arrangement of the bonded atoms to determine the molecular geometry.From the standpoint of the VSEPR model,what do nonbonding electron pairs,single bonds,and multiple bonds have in common?The Shapes of Molecules or Ions Table 9.2 summarizes the possible molecular geometries when an ABn molecule has four or fewer electron domains about A.These geometries are important because they include all the shapes usually seen in molecules or ions that obey the octet rule.QuestionsUse the VSEPR model to predict the molecular geometry of(a)O3,(b)SnCl3-.Analyze:We are given the molecular formulas of a molecule and a polyatomic ion,both conforming to the general formula ABn and both having a central atom from the p block of the periodic table.(Notice that for O3,the A and B atoms are all oxygen atoms.)Plan:To predict the molecular geometries,we draw their Lewis structures and count electron domains around the central atom to get the electron-domain geometry.We then obtain the molecular geometry from the arrangement of the domains that are due to bonds.Solve:(a)We can draw two resonance structures for O3:Because of resonance,the bonds between the central O atom and the outer O atoms are of equal length.In both resonance structures the central O atom is bonded to the two outer O atoms and has one nonbonding pair.Thus,there are three electron domains about the central O atoms.(Remember that a double bond counts as a single electron domain.)The arrangement of three electron domains is trigonal planar(Table 9.1).Two of the domains are from bonds,and one is due to a nonbonding pair.So,the molecular geometry is bent with an ideal bond angle of 120(Table 9.2).QuestionsSolve:(b)The Lewis structure for SnCl3-isThe central Sn atom is bonded to the three Cl atoms and has onenonbonding pair;thus,we have four electron domains,meaning atetrahedral electron-domain geometry(Table 9.1)with one vertexoccupied by a nonbonding pair of electrons.A tetrahedral electrondomain geometry with three bonding and one nonbonding domains leads to a trigonal-pyramidal molecular geometry(Table 9.2).Consider the following AB3 molecules and ions:PCl3,SO3,AlCl3,SO32-,and CH3+.How many of these molecules and ions do you predict to have a trigonal-planar molecular geometry?(a)1(b)2(c)3(d)4(e)5Predict the electron-domain and molecular geometries for(a)SeCl2,(b)CO32-.Molecules with Expanded Valence ShellsSuppose a molecule has five electron domains,and there are one or more nonbondingpairs.Will the domains from the nonbonding pairs occupy axial or equatorialpositions?To answer this question,we must determine which location minimizesrepulsion between domains.Repulsion between two domains is much greater whenthey are situated 90 from each other than when they are at 120.An equatorial domainis 90 from only two other domains(the axial domains),but an axial domain is 90from three other domains(the equatorial domains).Hence,an equatorial domainexperiences less repulsion than an axial domain.Because the domains from nonbonding pairs exert larger repulsions than those from bonding pairs,nonbonding domains always occupy the equatorial positions in a trigonal bipyramid.Molecules with Expanded Valence ShellsMolecules with Expanded Valence ShellsQuestionsUse the VSEPR model to predict the molecular geometry of(a)SF4,(b)IF5.Analyze:The molecules are of the ABn type with a central p-block atom.Plan:We first draw Lewis structures and then use the VSEPR model to determine theelectron-domain geometry and molecular geometry.Solve:(a)The Lewis structure for SF4 isThe sulfur has five electron domains around it:four from the S-Fbonds and one from the nonbonding pair.Each domain pointstoward a vertex of a trigonal bipyramid.The domain from thenonbonding pair will point toward an equatorial position.The fourbonds point toward the remaining four positions,resulting in amolecular geometry that is described as seesaw-shaped:QuestionsSolve:(b)The Lewis structure of IF5 isThe iodine has six electron domains around it,one of which is nonbonding.The electron-domain geometry is therefore octahedral,with one position occupied by the nonbonding pair,and the molecular geometry is square pyramidal(Table 9.3):A certain AB4 molecule has a square-planar molecular geometry.Which of the following statements about the molecule is or aretrue?:(i)The molecule has four electron domains about the centralatom A.(ii)The B-A-B angles between neighboring B atoms is 90.(iii)The molecule has two nonbonding pairs of electrons onatom A.(a)Only one of the statements is true.(b)Statements(i)and(ii)are true.(c)Statements(i)and(iii)are true.(d)Statements(ii)and(iii)are true.(e)All three statements are true.
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