7.14 Linear triatomic: A-----B-----C. Bond angles = 180 degrees. Trigonal planar: Bond angles = 120 degrees. B < B A B = 120

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1 APTER SEVEN Molecular Geometry 7.13 Molecular geometry may be defined as the three-dimensional arrangement of atoms in a molecule. The study of molecular geometry is important in that a molecule s geometry affects its physical and chemical properties, such as melting point, boiling point, density, and the types of reactions it undergoes Linear triatomic: A-----B Bond angles = 180 degrees. Trigonal planar: Bond angles = 120 degrees. B < B A B = 120 A / \ B B Tetrahedral: Bond angles = degrees B < B A B = A B / \ B B Trigonal bipyramidal: Bond angles = 90 or 120 degrees (depends on angle in question) B < A B = 90 / < B A B = 120 B A / \ B B ctahedral: Bond angles = 90 degrees. B B < B A B = 90 / B A B / B B 7.15 Tetrahedral: 4 atoms Trigonal bipyramidal: 5 atoms ctahedral: 6 atoms 7.16 The VESPR model, or the valence-shell electron pair repulsion model, accounts for the geometric arrangements of electron pairs around a central atom in terms of the electrostatic repulsion between electron pairs. Using this model we can predict the geometry of molecules (and ions) in a systematic way. With this model double bonds and triple bonds can be treated like single bonds, which is good for qualitative purposes. Also, if a molecule has two or more resonance structures, we can apply the VESPR model to any one of them. ormal charges are usually not shown. The magnitude of repulsion decreases in that order because electrons in a bond are held by the attractive forces exerted by the nuclei of the two bonded atoms. These electrons have less spatial distribution than lone pairs; that is, they take up less space than lone-pair electrons, which are associated with only one particular atom. Because lone-pair electrons in a molecule occupy more space, they experience greater repulsion from neighboring lone pairs and bonding pairs By the VSEPR theory, lone pair bonding pair repulsion is stronger than bonding pair bonding pair repulsion. Therefore, the lone pair occupies a spot where it has two bonds 120 degrees away

2 and two bonds 90 degrees away, instead of the spot which has three bonds 90 degrees away and one bond 180 degrees away In the proposed arrangement each - bond has two - bonds 90 degrees away and one - bond 180 degrees away. This is not as stable as the arrangement which puts three - bonds degrees away, which is the observed geometry. < = 90 / \ Square Planar (proposed) Tetrahedral (actual) 7.19 (a) The Lewis structure of P 3 is shown below. Since in the VSEPR method the number of bonding pairs and lone pairs of electrons around the central atom (phosphorus, in this case) is important in determining the structure, the lone pairs of electrons around the chlorine atoms have been omitted for simplicity. There are three bonds and one lone electron pair around the central atom, phosphorus, which makes this an AB 3 E case. The information in Table 10.2 shows that the structure is a trigonal pyramid like ammonia. P What would be the structure of the molecule if there were no lone pairs and only three bonds? (b) The Lewis structure of 3 is shown below. There are four bonds and no lone pairs around carbon which makes this an AB 4 case. The molecule should be tetrahedral like methane (Table 10.1). (c) (d) The Lewis structure of Si 4 is shown below. Like part (b), it is a tetrahedral AB 4 molecule. Si The Lewis structure of Te 4 is shown below. There are four bonds and one lone pair which make this an AB 4 E case. onsulting Table 10.2 shows that the structure should be that of a distorted tetrahedron like S 4. Te

3 Are Te 4 and S 4 isoelectronic? Should isoelectronic molecules have similar VSEPR structures? 7.20 In each case, you should use the following approach. Step 1: Write the Lewis structure of the molecule. The lone pairs on the terminal atoms have been omitted. Step 2: ount the number of electron pairs around the central atom. Step 3: Build a model, or consult Table 10.1 of your text to predict the geometry of the molecule. Lewis structure Electron pairs Electron Lone pairs Geometry on central atom arrangement (a) Al 3 trigonal planar 0 trigonal planar, AB 3 (b) Zn 2 linear 0 linear, AB 2 2 Zn (c) 4 tetrahedral 0 tetrahedral, AB The lone pairs of electrons on the bromine atoms have been omitted for simplicity. Br g linear Br N + N linear S linear N 7.22 (a) AB 4 tetrahedral (f) AB 4 tetrahedral (b) AB 2 E 2 bent (g) AB 5 trigonal bipyramid (c) AB 3 trigonal planar (h) AB 3 E trigonal pyramid (d) AB 2 E 3 linear (i) AB 4 tetrahedral (e) AB 4 E 2 square planar 7.23 The Lewis structure is:

4 AB 4 tetrahedral AB 3 trigonal planar AB 2 E 2 bent 7.24 nly molecules with four bonds to the central atom and no lone pairs are tetrahedral (AB 4 ). 2 I Si I I d I What are the Lewis structures and shapes for Xe 4 and Se 4? 7.25 nly (c) will not be tetrahedral. All the others have AB 4 type Lewis structures and will therefore be tetrahedral. or S 4 the Lewis structure is of the AB 4 E type which gives rise to a distorted tetrahedral geometry (Table 10.2 of the text) Step 1: Write the Lewis structure of the molecule. Br g Br Step 2: ount the number of electron pairs around the central atom. There are two electron pairs around g. Step 3: Since there are two electron pairs around g, the electron-pair arrangement that minimizes electron-pair repulsion is linear. In addition, since there are no lone pairs around the central atom, the geometry is also linear (AB 2 ). You could establish the geometry of gbr 2 by measuring its dipole moment. If mercury(ii) bromide were bent, it would have a measurable dipole moment. Experimentally, it has no dipole moment and therefore must be linear Geometry: bent; hybridization: sp 3.

5 7.28 The Lewis structures and VSEPR geometries of these species are shown below. The three nonbonding pairs of electrons on each fluorine atom have been omitted for simplicity. + + Xe Xe Sb AB 3 E 2 AB 5 E AB 6 T-shaped Square Pyramid ctahedral 7.29 (a) The Lewis structure is: B The shape will be trigonal planar (AB 3 ) (b) The Lewis structure is: - The molecule will be a trigonal pyramid (nonplanar). (e) The Lewis structure is: N The nitrogen atom is of the AB 2 E type, but there is only one unshared electron rather than the usual pair. As a result, the repulsion will not be as great and the N angle will be greater than 120 expected for AB 2 E geometry. Experiment shows the angle to be around 135. Which of the species in this problem has resonance structures? 7.30 To predict the bond angles for the molecules, you would have to draw the Lewis structure and determine the geometry using the VSEPR model. rom the geometry, you can predict the bond angles. (a) Be 2 : AB 2 type, 180 (linear). (b) B 3 : AB 3 type, 120 (trigonal planar). (c) 4 : AB 4 type, (tetrahedral). (d) 3 : AB 4 type, (tetrahedral with a possible slight distortion resulting from the different sizes of the chlorine and hydrogen atoms).

6 (e) g 2 2 : Each mercury atom is of the AB 2 type. The entire molecule is linear, 180 bond angles. (f) Sn 2 : AB 2 E type, roughly 120 (bent). (g) 2 2 : The atom arrangement is. Each oxygen atom is of the AB 2 E 2 type and the angles will be roughly (h) Sn 4 : AB 4 type, (tetrahedral) (d) The Lewis structure is: Si The ion has a trigonal pyramidal geometry (AB 3 E) nly I 2 and dbr 2 will be linear. The rest are bent The Lewis structure is shown below. 2 Be The molecule is of the AB 4 type and should therefore be tetrahedral. The hybridization of the Be atom should be sp or an octahedral AX 4 Y 2 molecule only two different structures are possible: one with the two Y s next to each other like (b) and (d), and one with the two Y s on opposite sides of the molecule like (a) and (c). The different looking drawings simply depict the same molecule seen from a different angle or side. It would help to develop your power of spatial visualization to make some simple models and convince yourself of the validity of these answers. ow many different structures are possible for octahedral AX 5 Y or AX 3 Y 3 molecules? Would an octahedral AX 2 Y 4 molecule have a different number of structures from AX 4 Y 2? Ask your instructor if you aren t sure.

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