Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals

Transcription

1 Chemical Bonding II: and Hybridization of Atomic Orbitals Chapter 10 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Valence shell electron pair repulsion (VSEPR) model: Predict the geometry of the molecule from the electrostatic repulsions between the electron (bonding and nonbonding) pairs. Class AB linear linear B B 2 0 lone Cl Be Cl 2 atoms 3 1

2 Class AB linear linear AB VSEPR 4 Boron Trifluoride 5 Class AB linear linear AB VSEPR AB tetrahedral tetrahedral 6 2

3 Methane 7 Class AB linear linear AB VSEPR AB tetrahedral tetrahedral AB Phosphorus Pentachloride 9 3

4 Class AB linear linear AB VSEPR AB tetrahedral tetrahedral AB AB octahedral octahedral 10 Sulfur Hexafluoride

5 lone-pair vs. lone-pair repulsion lone-pair vs. bondingpair > > repulsion bonding-pair vs. bondingpair repulsion 13 VSEPR Class AB AB 2 E 2 1 bent 14 VSEPR Class AB tetrahedral tetrahedral AB 3 E 3 1 tetrahedral pyramidal 15 5

6 VSEPR Class AB tetrahedral tetrahedral AB 3 E 3 1 tetrahedral pyramidal AB 2 E tetrahedral bent 16 VSEPR Class AB AB 4 E 4 1 distorted tetrahedron 17 VSEPR Class AB AB 4 E 4 1 AB 3 E distorted tetrahedron T-shaped 18 6

7 Class VSEPR AB AB 4 E 4 1 AB 3 E AB 2 E distorted tetrahedron T-shaped linear 19 VSEPR Class AB octahedral octahedral AB 5 E 5 1 octahedral square pyramidal 20 VSEPR Class AB octahedral octahedral AB 5 E 5 1 octahedral AB 4 E octahedral square pyramidal square 21 7

8 22 Predicting 1. Draw Lewis structure for molecule. 2. Count number of lone the and number of atoms the. 3. Use VSEPR to predict the geometry of the molecule. 23 Example 10.1 Use the VSEPR model to predict the geometry of the following molecules and ions: (a) AsH 3 (b) OF 2 (c) (d) (e) C 2 H 4 8

9 Example 10.1 Strategy The sequence of steps in determining molecular geometry is as follows: Solution (a) The Lewis structure of AsH 3 is There are four around the ; therefore, the electron pair arrangement is tetrahedral (see Table 10.1). Example 10.1 Recall that the geometry of a molecule is determined only by the arrangement of atoms (in this case the As and H atoms). Thus, removing the lone pair leaves us with three bonding pairs and a pyramidal geometry, like NH 3. We cannot predict the HAsH angle accurately, but we know that it is less than because the repulsion of the bonding in the As H bonds by the lone pair on As is greater than the repulsion between the bonding pairs. (b) The Lewis structure of OF 2 is There are four around the ; therefore, the electron pair arrangement is tetrahedral. Example 10.1 Recall that the geometry of a molecule is determined only by the arrangement of atoms (in this case the O and F atoms). Thus, removing the two lone pairs leaves us with two bonding pairs and a bent geometry, like H 2 O. We cannot predict the FOF angle accurately, but we know that it must be less than because the repulsion of the bonding in the O F bonds by the lone pairs on O is greater than the repulsion between the bonding pairs. (c) The Lewis structure of is 9

10 Example 10.1 There are four around the ; therefore, the electron pair arrangement is tetrahedral. Because there are no lone pairs present, the arrangement of the bonding pairs is the same as the electron pair arrangement. Therefore, has a tetrahedral geometry and the ClAlCl angles are all (d) The Lewis structure of is There are five around the central I atom; therefore, the electron pair arrangement is. Of the five, three are lone pairs and two are bonding pairs. Example 10.1 Recall that the lone pairs preferentially occupy the equatorial positions in a bipyramid (see Table 10.2). Thus, removing the lone pairs leaves us with a linear geometry for, that is, all three I atoms lie in a straight line. (e) The Lewis structure of C 2 H 4 is The C=C bond is treated as though it were a single bond in the VSEPR model. Because there are three around each C atom and there are no lone pairs present, the arrangement around each C atom has a shape like BF 3, discussed earlier. Example 10.1 Thus, the predicted bond angles in C 2 H 4 are all 120. Comment (1) The ion is one of the few structures for which the bond angle (180 ) can be predicted accurately even though the contains lone pairs. (2) In C 2 H 4, all six atoms lie in the same plane. The overall geometry is not predicted by the VSEPR model, but we will see why the molecule prefers to be later. In reality, the angles are close, but not equal, to 120 because the bonds are not all equivalent. 10

11 Dipole Moments and Polar Molecules electron poor region H electron rich region F d+ d- m = Q x r Q is the charge r is the distance between charges 1 D = 3.36 x C m 31 Behavior of Polar Molecules field off field on 32 Bond moments and resultant dipole moments in NH 3 and NF

12 34 Example 10.2 Predict whether each of the following molecules has a dipole moment: (a) BrCl (b) BF 3 ( ) (c) CH 2 Cl 2 (tetrahedral) Example 10.2 Strategy Keep in mind that the dipole moment of a molecule depends on both the difference in electronegativities of the elements present and its geometry. A molecule can have polar bonds (if the bonded atoms have different electronegativities), but it may not possess a dipole moment if it has a highly symmetrical geometry. 12

13 Example 10.2 Solution (a) Because bromine chloride is diatomic, it has a linear geometry. Chlorine is more electronegative than bromine (see Figure 9.5), so BrCl is polar with chlorine at the negative end Thus, the molecule does have a dipole moment. In fact, all diatomic molecules containing different elements possess a dipole moment. Example 10.2 (b) Because fluorine is more electronegative than boron, each B F bond in BF 3 (boron trifluoride) is polar and the three bond moments are equal. However, the symmetry of a shape means that the three bond moments exactly cancel one another: An analogy is an object that is pulled in the directions shown by the three bond moments. If the forces are equal, the object will not move. Consequently, BF 3 has no dipole moment; it is a nonpolar molecule. Example 10.2 (c) The Lewis structure of CH 2 Cl 2 (methylene chloride) is This molecule is similar to CH 4 in that it has an overall tetrahedral shape. However, because not all the bonds are identical, there are three different bond angles: HCH, HCCl, and ClCCl. These bond angles are close to, but not equal to,

14 Example 10.2 Because chlorine is more electronegative than carbon, which is more electronegative than hydrogen, the bond moments do not cancel and the molecule possesses a dipole moment: Thus, CH 2 Cl 2 is a polar molecule. Change in Potential Energy of Two Hydrogen Atoms as a Function of Their Distance of Separation 41 Change in electron density as two hydrogen atoms approach each other

15 Hybridization mixing of two or more atomic orbitals to form a new set of hybrid orbitals 1. Mix at least 2 nonequivalent atomic orbitals (e.g. s and p). Hybrid orbitals have very different shape from original atomic orbitals. 2. Number of hybrid orbitals is equal to number of pure atomic orbitals used in the hybridization process. 3. Covalent bonds are formed by: a. Overlap of hybrid orbitals with atomic orbitals b. Overlap of hybrid orbitals with other hybrid orbitals 43 Formation of sp 3 Hybrid Orbitals 44 Formation of Covalent Bonds in CH

16 sp 3 -Hybridized N Atom in NH 3 Predict correct bond angle 46 Formation of sp Hybrid Orbitals 47 Formation of sp 2 Hybrid Orbitals 48 16

17 How do I predict the hybridization of the? 1. Draw the Lewis structure of the molecule. 2. Count the number of lone pairs AND the number of atoms the # of Lone Pairs + # of Bonded Atoms Hybridization Examples 2 sp BeCl sp 2 sp 3 sp 3 d BF 3 CH 4, NH 3, H 2 O PCl 5 6 sp 3 d 2 SF Example 10.3 Determine the hybridization state of the central (underlined) atom in each of the following molecules: (a) BeH 2 (b) AlI 3 (c) PF 3 Describe the hybridization process and determine the molecular geometry in each case. 17

18 Example 10.3 Strategy The steps for determining the hybridization of the in a molecule are: Solution (a) The ground-state electron configuration of Be is 1s 2 2s 2 and the Be atom has two valence electrons. The Lewis structure of BeH 2 is H Be H Example 10.3 There are two bonding pairs around Be; therefore, the electron pair arrangement is linear. We conclude that Be uses sp hybrid orbitals in bonding with H, because sp orbitals have a linear arrangement (see Table 10.4). The hybridization process can be imagined as follows. First, we draw the orbital diagram for the ground state of Be: By promoting a 2s electron to the 2p orbital, we get the excited state: Example 10.3 The 2s and 2p orbitals then mix to form two hybrid orbitals: The two Be H bonds are formed by the overlap of the Be sp orbitals with the 1s orbitals of the H atoms. Thus, BeH 2 is a linear molecule. 18

19 Example 10.3 (b) The ground-state electron configuration of Al is [Ne]3s 2 3p 1. Therefore, the Al atom has three valence electrons. The Lewis structure of AlI 3 is There are three pairs of electrons around Al; therefore, the electron pair arrangement is. We conclude that Al uses sp 2 hybrid orbitals in bonding with I because sp 2 orbitals have a arrangement. The orbital diagram of the ground-state Al atom is Example 10.3 By promoting a 3s electron into the 3p orbital we obtain the following excited state: The 3s and two 3p orbitals then mix to form three sp 2 hybrid orbitals: The sp 2 hybrid orbitals overlap with the 5p orbitals of I to form three covalent Al I bonds. We predict that the AlI 3 molecule is and all the IAlI angles are 120. Example 10.3 (c) The ground-state electron configuration of P is [Ne]3s 2 3p 3. Therefore, P atom has five valence electrons. The Lewis structure of PF 3 is There are four pairs of electrons around P; therefore, the electron pair arrangement is tetrahedral. We conclude that P uses sp 3 hybrid orbitals in bonding to F, because sp 3 orbitals have a tetrahedral arrangement. The hybridization process can be imagined to take place as follows. The orbital diagram of the ground-state P atom is 19

20 Example 10.3 By mixing the 3s and 3p orbitals, we obtain four sp 3 hybrid orbitals. As in the case of NH 3, one of the sp 3 hybrid orbitals is used to accommodate the lone pair on P. The other three sp 3 hybrid orbitals form covalent P F bonds with the 2p orbitals of F. We predict the geometry of the molecule to be pyramidal; the F F angle should be somewhat less than Example 10.4 Describe the hybridization state of phosphorus in phosphorus pentabromide (PBr 5 ). Example 10.4 Strategy Follow the same procedure shown in Example Solution The ground-state electron configuration of P is [Ne]3s 2 3p 3. Therefore, the P atom has five valence electrons. The Lewis structure of PBr 5 is There are five pairs of electrons around P; therefore, the electron pair arrangement is. We conclude that P uses sp 3 d hybrid orbitals in bonding to Br, because sp 3 d hybrid orbitals have a arrangement. 20

21 Example 10.4 The hybridization process can be imagined as follows. The orbital diagram of the ground-state P atom is Promoting a 3s electron into a 3d orbital results in the following excited state: Example 10.4 Mixing the one 3s, three 3p, and one 3d orbitals generates five sp 3 d hybrid orbitals: These hybrid orbitals overlap the 4p orbitals of Br to form five covalent P Br bonds. Because there are no lone the P atom, the geometry of PBr 5 is. sp 2 Hybridization of Carbon 63 21

22 Unhybridized 2p z orbital (gray), which is perpendicular to the plane of the hybrid (green) orbitals. 64 sp Hybridization of Carbon 65 22