Solution Key, Problemset 2

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1 Solution Ke, Problemset Use VSEPR theor to obtain the molecular structures. After ou have the shape of the molecule, if necessar, use Pauling electronegativit values to obtain the bond dipole moments and then consider the molecular dipole moment. a. b. B non polar S c. Br P Br Br d. S C S non polar e. C 3.2 a. E is the identit operator. It effectivel identifies the molecular configuration. The operator E leaves the molecule unchanged. All objects can be operated upon b the identit operator E. b. A plane of smmetr (mirror plane) is denoted b σ. c. The smmetr operation of rotation about an n-fold ais (the smmetr element) is denoted b C n, in which the angle of rotation is 360 /n, where n = 1, 2, 3, 4. d. An S n ais is an n-fold improper rotation ais: rotation through 360 /n about the ais is followed b reflection through a plane perpendicular to the same ais. If the mirror plane lies perpendicular to the principal ais, it is denoted σ h. If the plane contains the principal ais, it is denoted σ v. C 2 σv σ v '

2 To see the difference between σ v and σ v, it is best to use an eample: 2 is a simple eample. The figure above shows the principal rotation ais (C 2 ) in a molecule of 2, and the two mirror planes, both of which contain the principal rotation ais. The plane which bisects the molecule is labeled σ v, and the plane in which the molecule lies is labeled σ v. A σ d plane contains the principal rotation ais, and also bisects the angle between two adjacent 2-fold aes. 3.4 S 2 is a bent molecule (see 3.1 a) The smmetr operators are E, C 2, σ v and σ v. 3.9 irst, draw out the structure of the molecules and ions using the VSEPR model. C I S The onl two species with a C 4 ais and a σ h plane are the square planar ones: [I 4 ] - and Xe Xe 3.11 a. is in the same group as C, so 2 6 is epected to have the same structure as ethane. b. A staggered conformer will be the most favored in terms of steric energ. a c. The midpoint of the - bond is an inversion center. This point is shown in the structure above. To confirm that this is a center of inversion, reflect each point of the molecule through center I and show that an identical point is generated. or eample, take the atom marked a reflect through I. ou end up at atom a which is indistinguishable from atom a. d. The eclipsed conformer is the least favored in terms of steric energ. i a' e. This conformer does not contain a center of inversion.

3 3.17 If Br 3 were trigonal planar or trigonal pramidal, the molecule would have a C 3 ais. The fact that it belongs to the C 2 point group means that the principal ais is a C 2 ais. The other 3-coordinate structure possibilit is T- shaped, so the net step in the answer is to work out the smmetr properties of a T-shaped Br 3 molecule. Apart from the E operator, T- shaped Br 3 contains a C 2 ais, a σ v plane, and a σ v plane. These are consistent with the C 2v point group. Now see if this agrees with VSEPR theor: Central atom is Br Br is in Group 17, number of VE = 7 Number of bonding pairs (3 Br- bonds) = 3 Number of lone pairs (for Br) = 2 Total number of electron pairs (for Br) = 5 Parent shape = trigonal bipramidal Molecular shape = T-shape a. 2 is polar and possesses a bent molecular shape. All three modes of vibration are IR active b. 4 is tetrahedral and non-polar; T d smmetr. 4 modes of vibrational freedom (2 degenerate pairs) are IR active and give rise to 2 bands in the IR spectrum. c. P 3 has a trigonal pramidal structure and is polar. Both smmetric stretch and smmetric deformation are IR active, the doubl degenerate (asmmetric stretch and the doubl degenerate deformation are also IR active. In summar, there are 6 modes of vibrational freedom giving rise to 4 bands in the IR spectrum. d. Al 3 has a trigonal planar structure and is non-polar. Therefore, the smmetric stretch is IR inactive. The IR active modes are the smmetric deformation, doubl degenerate stretch and doubl degenerate deformation 5 modes of vibrational freedom giving rise to 3 bands in the IR spectrum. e. CS 2 is linear and polar. Thus, the asmmetric stretch and the doubl degenerate deformation are IR active, giving rise to two bands in the IR spectrum. f. CN is linear and polar. Thus, both smmetric and asmmetric stretches as well as the doubl degenerate deformation are IR active. Three fundamental absorptions are seen in the IR spectrum.

4 4.4 a. Take the shaded lobes of the p and d 2-2 orbital to point along the + ais, and the shaded lob of the p orbital to point along the + ais. In the plane, the orbital combinations to give 4 sp 2 d hbrid orbital are: b. Available for hbridization are one s, two p and one d orbital. Each hbrid orbital must contain the same amount of s character; since there are 4 hbrid orbitals, each contains 25% s character. Each hbrid orbital also must contain the same amount of p character, i.e. 50% p character. Each hbrid orbital contains 25% d character. 4.5 a. sp 3 ; b. sp 3 d 2 ; c. sp 3 ; d. sp 3 ; e. sp 3 d; f. sp 3 d 2 ; g. sp 3 d; h. sp a. The B 3 molecule is defined as ling in the plane. The atoms lie in the nodal plane of the B 2p z orbitals; there is no net overlap between the 1s and the B 2p z orbitals, so the B 2p z orbital becomes a non-bonding M in B 3. b. Schematic representations are: Bonding M's Antibonding M's

5 4.14 a. Schematic representations are: ψ 3 ψ 5 ψ 6 ψ 7 b. A small degree of N- bonding character The structure of [B 2 7 ] - is shown below: a. 2 B = 6 VE; 7 = 7 VE; negative charge. Total = 14 electrons. b. Each B has left one sp 3 hbrid orbital. c. See M diagram below; - provides 2 electrons, but all electrons in each B 3 unit are used for terminal B- bonding (M s for this are not shown). The B- -B bridge is therefore a 3c-2e interaction the σ-m is delocalized over three atoms. σ* Energ non-bonding σ 3 B B 3-3 B B 3

6 Additional Problems: 1. a. p has C v smmetr. ignoring the difference in sign between the two lobes, it is D h. b. d has D 2h smmetr. Ignoring the signs, it is D 4h. c. d 2-2 has D 2h smmetr. Ignoring the signs, it is D 4h. d. d z 2 has D h smmetr. 2. The superimposed octahedron and cube show the matching smmetr elements. C 2, C 4 The descriptions below are for the elements of a cube; each element also applies to the octahedron. E Ever object has an identit operation 8C 3 Diagonals through opposite corners of the cube are C 3 aes 6C 2 Lines bisecting opposite edges are C 2 aes 6C 4 Lines through the centers of opposite faces are C 4 aes. Although there are onl three such lines, there are si aes, counting C C 2 (=C 2 4 ) The lines through the center of opposite faces are C 4 aes as well as C 2 aes. I The center of the cube is the inversion center. 6S 4 The C 4 aes are also S 4 aes. 8S 6 The C 3 aes are also S 6 aes. 3σ h These mirror planes are parallel to the faces of the cube. 6σ d These mirror planes are through opposite edges. C 3

7 3. As a cclic (triangular) ion 3 + has a pair of electrons in a bonding orbital and two vacant orbitals that are slightl antibonding: node LG(2) LG(3) LG(1) a. KrBr + energ level diagram is below. Energ σ 4p * π 4p * 4p 4p π 4p σ 4p σ 4s * 4s Kr KrBr + σ 4s 4s Br b. The M is polarized toward Br, since its energ is closer to that of the Br 4p orbital. c. Bond order = 1. d. Kr is more electronegative. Its greater nuclear charge eerts a stronger pull on the shared electrons.

Chapter 9. Chemical reactivity of molecules depends on the nature of the bonds between the atoms as well on its 3D structure

Chapter 9. Chemical reactivity of molecules depends on the nature of the bonds between the atoms as well on its 3D structure Chapter 9 Molecular Geometry & Bonding Theories I) Molecular Geometry (Shapes) Chemical reactivity of molecules depends on the nature of the bonds between the atoms as well on its 3D structure Molecular