Exercises 2, 4, 8, 10, 12, 16, 18, 24, 26, 32, 38, 40, 44, 46, 52, 58, 60

2. The bond angles in a tetrahedral molecule are equal to 109.5°.

4. (a) Electron domains are determined by counting the number of unshared pairs of electrons plus the number of single, double, or triple bonds around the central atom of the molecule or ion.

(b) A bonding electron domain is a single, double, or triple bond. A nonbonding electron domain is an unshared pair of electrons (a lone pair).

8. A trigonal bipyramidal electron domain is associated with 5 electron domains. An AB₃ molecule will have 3 bonding electron domains. In order to have a total of 5 electron domains, there must be 2 nonbonding electron domains.

10. (a) There are a total of three electron domains (three bonding electron domains plus zero nonbonding electron domains). The trigonal planar electron domain geometry is associated with three electron domains. With three bonding electron domains, the molecular geometry is also trigonal planar.

(b) There are a total of four electron domains (three bonding plus one nonbonding). The tetrahedral electron-domain geometry is associated with 4 electron domains. With three bonding electron domains, the molecular geometry is called trigonal pyramidal.

(c) There are a total of five electron domains (two bonding plus three nonbonding). The trigonal bipyramidal electon-domain geometry is associated with 5 electron domains. With two bonding electron domains, the molecular geometry is linear.

12. (a) The Lewis structure of N₂O is:

There are a total of two electron domains about the central N atom. The electron-domain geometry and molecular geometry is linear.

(b) The Lewis structure of SO₃ is:

There are a total of three electron domains about the central S atom. The electron-domain geometry and the molecular geometry is trigonal planar.

(c) The Lewis structure of PCl₃ is:

There are a total of 4 electron domains around the central P atom. The electron domain geometry for 4 pairs is tetrahedral. With 3 bonding electron domains, the molecular geometry is trigonal pyramidal.

(d) NH₂Cl

The Lewis structure of NH₂Cl is:

There are a total of 4 electron domains around the central P atom. The electron domain geometry for 4 pairs is tetrahedral. With 3 bonding electron domains, the molecular geometry is trigonal pyramidal.

(e) The Lewis structure of BrF₅ is:

There are a total of 6 electron domains around the central Br atom. The electron domain geometry for 6 pairs is octahedral. With 5 bonding electron domains, the molecular geometry is square pyramidal.

(f) The Lewis structure of KrF₂ is:

There are a total of 5 electron domains around the central Kr atom. The electron domain geometry for 5 pairs is trigonal bipyramidal. With 2 bonding electron domains, the molecular geometry is linear.

14. (a) There are 4 electron domains around the O atom associated with angle 1. A tetrahedral electron-domain geometry is assigned to 4 pairs. Since two of the pairs are nonbonding, the bond angle will be less than 109.5°. There are 3 electron domains around the N atom associated with angle 2. A trigonal planar electron-domain geometry is assigned to 3 pairs. Since one pair is nonbonding, the bond angle will be slightly less than 120°.

(b) There are 4 electron domains about the C atom associated with angle 3. A tetrahedral electron-domain geometry is associated with 4 pairs. Since all the electron pairs are bonding, the bond angle should be 109.5°. There are 3 electron domains about the C atom associated with angle 4. A trigonal planar geometry is associated with 3 pairs. Since this bond angle involves a double bond, the bond angle will be slightly greater than 120°.

(c) There are 4 electron domains about the N atom associated with angle 5. A tetrahedral electron-domain geometry is associated with 4 pairs. Since one of the pairs is nonbonding, the bond angle will be slightly less than 109.5°. ) There are 4 electron domains around the O atom associated with angle 6. A tetrahedral electron-domain geometry is assigned to 4 pairs. Since two of the pairs are nonbonding, the bond angle will be less than 109.5°.

(d) There are 4 pairs about the C atom associated with angle 8. A tetrahedral geometry is associated with 4 pairs. The bond angle is 109.5°. There are two pairs about the C atom associated with angle 7. A linear geometry is associated with two pairs. The bond angle is 180°.

18. (a) The Lewis structures of ClO₂¯ and NO₂¯ are:

The structures of the two ions are different because there are a different number of nonbonding electron pairs on each ion. NO₂¯ has three electron domains about the central N atom,which is associated with a trigonal planar electron domain geometry. Since one of the electron domains is non-bonding, the bond angle is less than 120°. ClO₂¯ has 4 electron domains about the central Cl atom, which is associated with a tetrahedral geometry. Since two of the electron domains are nonbonding, the bond angle is less than 109.5°.

(b) The Lewis structure of XeF₂ is:

There are 5 electron domains about the central Xe atom. A trigonal bipyramidal electron domain geometry is associated with 5 pairs. The equatorial positions in the trigonal bipyramidal geometry are roomier spatially than the axial positions. Therefore, the more spatially demanding nonbonding pairs occupy the three equatorial positions, leaving the bonding domains in the axial positions. This will give a linear geometry for the molecule.

24. (a) The vector sum of the bond dipoles in the polar bonds must be equal to zero in order for a molecule with polar bonds to be nonpolar.

(b) A linear geometry for an AB₂ molecule, a trigonal planar geometry for an AB₃, and a tetrahedral geometry for an AB₄ molecule will all give nonpolar molecules.

26. (a) IF is linear with a polar bond. It is a polar molecule.

(b) Since C and S possess the same electronegativity value (2.5), CS₂ does not have any polar bonds so it can't be a polar molecule.

(c) SO₃ has a trigonal planar geometry (see Lewis structure in problem 12 (b)). It has polar bonds but because the bond dipoles exactly cancel, the molecule is not polar.

(d) PCl₃ is trigonal pyramidal (see Lewis structure in problem 12 (c)). It has polar bonds. The partially negatively charged Cl atoms lie one end of the molecule, while the partially positively charged P atom lies at the other end. Therefore, this is a polar molecule.

(e) SF₆ contains polar bonds. The Lewis structure consists of a central S atom connected to 6 F atoms by single bonds in an octahedral geometry. Since the bond dipoles cancel out, this molecule is not polar.

(f) IF₅ contains polar bonds. The Lewis structure consists of a central I atom connected to 5 F atoms by single bonds with a additional nonbonding pair on the I atom. The molecular geometry of this molecule is square pyramidal. This molecule is polar because the bond dipole opposite the lone pair is not cancelled out.

32. (a) One s and two p orbitals combine to form 3 sp² hybrid orbitals. They point at 120° angles.

(b) One s, three p, and one d orbital combine to form 5 sp³d hybrid orbitals. They point at 120° and 90° angles.

(c) One s, three p, and two d orbitals combind to form 6 sp³d² hybrid orbitals. They point at 90° angles.

38. (a) The Lewis structure of SiCl₄ is:

There are 4 electron domains about the central Si atom. The electron-domain geometry is tetrahedral. sp³ hybrid orbitals are associated with tetrahedral geometries.

(b) The Lewis structure of HCN is:

There are 2 electron domains about the central C atom. The electron-domain geometry is linear. sp hydrid orbitals are associated with linear geometries.

(c) The Lewis structure of SO₃ is:

There are 3 electron domains about the central S atom. The electron-domain geometry is trigonal planar. sp² hybrid orbitals are associated with trigonal planar geometries.

(d) The Lewis structure is ICl₂¯:

There are 5 electron domains about the central I atom. The electron-domain geometry is trigonal bipyramidal, sp³d hybrid orbitals are associated with trigonal bipyramidal geometries.

(e) The Lewis structure of BrF₄¯ is:

There are 6 electron domains about the central Br atom. The electron-domain geometry associated with 6 domains is octahedral. sp³d² hybrid orbitals are associated with octahedral geometries.

40. (a) If one s and one p orbital is used to form 2 sp hybrid orbitals, that leaves 2 unhybridized p orbitals (remember there are 3 p orbitals a p subshell). Two p bonds can be formed from these unhybridized p orbitals.

(b) A triple bond consists of one s bond and two p bonds.

(c) Multiple bonds do not allow rotation about the bond axis because rotation breaks the p bonds.

44. Rewriting the Lewis structure:

(a) The carbon atom attached to the hydrogen atoms and one carbon atom has three electron domains around it. It is assigned a trigonal planar geometry. sp² hybrid orbitals are associated with trigonal planar geometries. The carbon atom attached to the oxygen and the other carbon has two electron domain geometries about it. It has a linear geometry. sp hybrid orbitals are associated with linear geometries.

(b) There are a total of 16 valence electrons (1 for each H, 4 for each C, 6 for O).

(c) Each single and double bond has one s bond. There are a total of 4 s bonds in ketene. 4 s bonds would use 8 valence electrons.

(d) Each double bond has one p bond. There are a total of 2 p bonds in ketene using 4 electrons.

(e) There are two nonbonding pairs on the O atom using 4 electrons.

46. Rewriting the structure:

(a) The central carbon atom in angle 1 has three electron domains about it for a trigonal planar geometry. The bond angle should be about 120°. The central carbon atom in angle 2 has three electron domains about it for a trigonal planar geometry. The bond angle should be about 120°. The central oxygen angle in angle 3 has 4 electron domains for a tetrahedral electron-domain geometry. Having only 2 bonding electron domains, the geometry is bent and the bond angle will be less than 109.5°.

(b) The central carbon atom in angle 1 and the central carbon atom in angle 2 will have sp² hybridization. The central oxygen atom in angle 3 has sp³.

(c) Each single and double bond will have 1 s bond each. There are 21 s bonds in the molecule.

52. (a) The molecular orbitals for the H₂¯ ion are s_1s and s_1s^* orbitals. They look like the following:

The energy level diagram is:

(b) The molecular orbital electron configuration is (s_1s)²(s_1s^*)¹.

(c) The bond order is equal to:

(d) Since the bond order is quite small (less than 1), the bond energy holding the ion together should be rather small. If you excite the electron in the antibonding molecular orbital to a higher energy molecular orbital, this excess energy will probably exceed the bond energy, which will break the ion apart. Therefore, the excited H₂¯ ion is not expected to be stable.

56. (a) The molecular orbital electron configuration for O₂^2- (with 14 valence electrons) is (s_2s)²(s_2s^*)²(s_2p)²(p_2p)⁴(p_2p^*)⁴. It will have a bond order of :

The molecular orbital electron configuration for O₂^¯ (with 13 valence electrons) is (s_2s)²(s_2s^*)²(s_2p)²(p_2p)⁴(p_2p^*)³. It will have a bond order of :

Since the peroxide ion, O₂^2-, has a lower bond order than the superoxide ion, O₂¯, the peroxide ion has a longer bond.

(b) For boron, the p_2p molecular orbitals are lower in energy than the s_2p molecular orbital because of interactions between the 2s and 2p orbitals in B. Therefore, the p_2p molecular orbitals are filled first. The molecular orbital electron configuration is: (s_2s)²(s_2s^*)²(p_2p)² (there are 6 valence electrons in B₂). This electron configuration will have 2 unpaired electrons in the p_2p.

58. (a) Paramagnetism is an attraction to a magnetic field.

(b) Any species with unpaired electrons will be paramagnetic and exhibit paramagnetism.

(c) Any odd electron species will be paramagnetic. This includes O₂⁺ (11 valence electrons) and Li₂⁺ (1 valence electron). N₂^2- has 12 valence electrons with a molecular orbital electron configuration of (s_2s)²(s_2s^*)²(p_2p)⁴(s_2p)² (p_2p^*)². It will have two unpaired electrons in the p_2p^* molecular orbital. O₂^2- has 14 valence electrons with a molecular orbital electron configuration of (s_2s)²(s_2s^*)²(s_2p)²(p_2p)⁴(p_2p^*)⁴. It has no unpaired electrons and is not paramagnetic. O₂⁺ and Li₂⁺ each have one unpaired electron. N₂^2- has two unpaired electrons.

60. (a) CO has a total of 10 valence electrons. It will have a molecular orbital electron configuration of (s_2s)²(s_2s^*)²(s_2p)²(p_2p)⁴. There are no unpaired electrons so the molecule is diamagnetic. The bond order is:

(b) NO¯ has a total of 12 valence electrons. It will have a molecular orbital electron configuration of (s_2s)²(s_2s^*)²(p_2p)⁴(s_2p)² (p_2p^*)². It has two unpaired electrons in the p_2p^* molecular orbital. The bond order is:

(c) OF⁺ has a total of 12 valence electrons. It will have a molecular orbital electron configuration of (s_2s)²(s_2s^*)²(p_2p)⁴(s_2p)² (p_2p^*)². It has two unpaired electrons in the p_2p^* molecular orbital. The bond order is:

(d) NeF⁺ has a total of 14 valence electrons. It will have a molecular orbital electron configuration of (s_2s)²(s_2s^*)²(p_2p)⁴(s_2p)² (p_2p^*)⁴. It has no unpaired electrons so the molecule is diamagnetic. The bond order is: