Problem Set 5 Solutions Chemistry 104a

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1 Chem 104a omework 5 Solutions all 2011 Problem Set 5 Solutions Chemistry 104a Problem 1 (a) Prepare a molecular orbital energy level diagram for the cyanide ion. Use sketches to show clearly how the atomic orbitals interact to form MOs. C CN - N 4σ 2π σ ,2 (b) What is the bond order, and how many unpaired electrons does cyanide have? We can say that the bond order is approximately 3 because the and the two orbitals are filled; and 3σ are considered non-bonding. 1 (6 0) = B.O. 2 Looking at the molecular orbital diagram, you can see that there are no unpaired electrons. (c) Which molecular orbital of CN - would you predict to interact most strongly with a hydrogen 1s orbital to form an -C bond in the reaction CN CN? Explain. 1

2 Chem 104a omework 5 Solutions all 2011 The 3σ (OMO) of CN - would interact most with the LUMO of + 1s orbital, which has an energy of ev, because they match in symmetry and in energy. Problem 2 Although Kr + and Xe + have been studied, KrBr + has not yet been prepared. or KrBr + : (a) Propose a molecular orbital diagram, showing the interactions of the valence shell s and p orbitals to form molecular orbitals. B N Kr 4σ 2π σ , (b) Toward which atom would the OMO be polarized? Why? The OMO = 2π. The OMO is more polarized toward the Br 4p x and 4p y. These Br 4p x and 4p y orbitals are closer in energy to the 2π MO than the Kr 4p x and 4p y. 2

3 Chem 104a omework 5 Solutions all 2011 (c) Predict the bond order. Bond order is approximately 1 since the effects of the and 3σ cancel each other out just like the and the 2π. (d) Which is more electronegative, Kr or Br? Explain your reasoning. Kr is more electronegative because its atomic orbitals are lower in energy than the corresponding Br atomic orbitals. As a result, you can see more filled orbitals have Kr character than Br character. Problem 3 Consider the molecule N and the ions N + and N -. Write the Lewis structure and the molecular orbital description of the ground state for each species. Determine which of the three species would be paramagnetic, and tell how many unpaired electrons there would be in each paramagnetic molecule. Predict the bond orders for all three species. N N 4σ 2π 11, ,10 3σ 5,7 6, , N

4 , , Chem 104a omework 5,7 Solutions 3σ 6,8 all π 3σ 5,7 6, N + N 1, electrons 1, electrons are bonding and 3 electrons are antibonding. The bond order is 5/2, and the molecule is paramagnetic due to one unpaired electron. N N N 12 electrons 8 electrons are bonding and 4 electrons are antibonding. The bond order is 2, and the molecule is paramagnetic N due to 2 unpaired electrons. N - N 13 electrons 8 electrons are bonding and 5 electrons are antibonding. The bond order is 3 /2, and the molecule is paramagnetic due to one unpaired electron. 4

5 Chem 104a omework 5 Solutions all 2011 Problem 4 The first ionization energies of B, CO, and N 2 are ev, ev, and ev, respectively. Explain the increase in ionization energy for this isoelectronic series on the basis of atomic-orbital composition of the highest occupied molecular orbital. B B C CO O N N σ 2π 3σ σ σ 2π N 2 4σ 2π 3σ Note: Not showing s-p mixing for all 3 molecules. We can answer this question simply by inspecting the OMO levels of each molecule because the OMO level is a minimum value for ionization energy (IE). The OMO level of B is the highest since it is higher than ev. The OMO level of CO is slightly higher than the O orbital (-15.8 ev) from s-p mixing; therefore, its OMO level is lower than the OMO level of B, so its IE is larger. The OMO level of N 2 is the lowest because its 3σ bond is strongly bonding despite weak s-p mixing. Therefore, diatomic nitrogen has the largest IE. 5

6 nb 3nb Chem 104a omework 5 Solutions all 2011 Problem nb The species of - 2 is linear and symmetrical, 1nb -- -, and has one of the strongest known systems of hydrogen bonding. (a) Draw atomic overlap representations for each of the σ and π orbitals of 2 -. σ π π (b) Draw a molecular-orbital energy-level scheme for 2 -. Place the atom 1s orbital on the left, the fluorine and atomic orbitals (or the appropriate SALCs) on the right, and the orbitals of 2 - in the center nb 4nb 3nb nb 1nb 6 σ

7 Chem 104a omework 5 Solutions all 2011 (c) Given the energy-level scheme derived in part (b), would you expect 2 - to be diamagnetic or paramagnetic? diamagnetic (all electrons are paired up) (d) What is the - bond order in 2 -? The total bond order = 1 (2 0) = 1, which represents the bond order 2 over two - bonds. A single BO - would then be 1 /2 Problem 6 Consider the borohydride anion, B 4 -, to be formed in the reaction - + B 3 B 4 - D 3h B B 3 3 2E 2E 2A 1 2A p x p y p z E E A 2 1A 2 1A s A 1 1E A 1 1E E 1A 1 1A 1 Which orbital will the - nucleophile attack on the B 3 molecule? ollowing the symmetry rules for chemical reactions laid out in pg of DG, we can argue that - will act as a nucleophile and attack the LUMO on B 3. The LUMO is a non bonding a 2 orbital, which corresponds to the z orbital of boron. 7

8 Chem 104a omework 5 Solutions all 2011 Problem 7 Diborane, B 2 6, has the structure shown. Using molecular orbitals (and showing appropriate orbitals on B and from which the MOs are formed), explain how hydrogen can form bridges between two B atoms. (This type of bonding is discussed in Chapter 8 of MT) B B B B You can look at this problem as 2 molecules of B 3, labeled (1) and (2), reacting with each other to make B 2 6. B B (1) (2) B (1) (2) B B B B Problem #6 shows us that the OMO and B LUMO of each B 3 molecule will look like: (1) (2) B OMO B LUMO OMO 2 interactions LUMO (1) (1) The goal now is to orient the OMO on (1) with the LUMO on (2) to create one interaction. To create another 2 interactions interaction, we can orient the (1) OMO on (2) with the (1) LUMO on (1) to create another interaction while the molecules areomo still in the same relative LUMO (2) position. This can(2) create a bridged molecule. 2 interactions (2) (2) (1) (1) (2) (2) 8

9 Chem 104a omework 5 Solutions all 2011 Problem 8 In the bent geometry, we should consider s-p mixing. The Walsh diagram is made with s-p mixing. Please note that the position of 2a 1 is system dependent. (a) N N (c) PES 2b 2 2b 2 3a 1 3a (p x ) b 2 (p y ) b 1 (p z ) a 1b1 1 b 2 a 1 1b nonbonding (p y ) 2a 1 2a 1 2a 1 (slightly bonding) 1b 2 a 1 (s) 1b 2 1b 2 1a 1 1a 1 1a 1 (b) u 2b 2 3a 1 g 1b 1 p x p y 2a 1 1b 2 u g 1a 1 C 2v D h C 2v is more stable because it provides the greater energy stabilization for 8 valence electrons Problem 9 Consider the hypothetical square-planar molecule Xe 4 in the following coordinate system: 9

10 Chem 104a omework 5 Solutions all 2011 Z c d Xe b Y a X (a) Write appropriate SALCs of the hydrogen 1s atomic orbitals which will overlap with each of the following xenon orbitals that can form molecular orbitals: 5s, 5p x, 5p y, and 5d x 2 y 2. D 4h E 2C 4 C 2 2C 2 2C 2 i 2S 4 σ h v d Γ red Applying the reduction formula: Z Γ red = + + c We can now apply the Projection operator to generate LGOs. d Xe b Don t pay too much attention to the shading. The sign is arbitrary. Just know when the sign flips on a symmetry. a X Y ˆP ( a ) = 4( a + b + c + d ) ˆP ( a ) = 4( a b + c d ) 10 5p y with 5p x with

11 c d Xe b Y a X Z Chem 104a omework 5 Solutions all 2011 c b Y Z X c ˆP Eu ( a ) = 4( a c ) d Xe b Y a X c Z 5p x with 5p y with 5d x^2-y^2 with d Xe b Y a X ˆP Eu ( b ) = 4( b d ) 5d z^2 witha 5p 1g y with 5p x withe 5d x^2-y^2 withb (b) Draw u atomic-orbital overlaps with the appropriate 1g Xe orbital for each of these SALCs. 5p y with 5p y with with 5p x withe u 5p x with 5d x^2-y^2 with 5d x^2-y^2 with 5d z^2 with 5d z^2 witha (c) 1g Sketch a molecular-orbital energy-level diagram for Xe 4, showing the Xe atomic orbitals on the left, the 1s orbitals on the right, and the Xe 4 orbitals in the center. 11

12 Chem 104a omework 5 Solutions all 2011 D 4h Xe Xe x 2 -y 2 5d xy yz xz z 2 B 2g E g E g 1B 2g 1E g 2 5p p x p y p z A 2u 1A 2u s 1 1 (d) Indicate which molecular orbitals are occupied, and calculate the net Xe- bond order, based on the molecular-orbital model. a b 1g, e b u, b b 1g and a 1g are occupied. Bond Order = (8 2) = 3 /4 Problem 10 Consider the molecule Si 4 : (a) Draw the MO diagram for the molecule in a planar configuration. Use isolobal analogy so that we can use only s-orbitals for our LGO s. This will make life easier. These orbitals will be take on the energy of the cholorine 3p (13.7 ev) orbitals. D 4h E 2C 4 C 2 2C 2 2C 2 i 2S 4 σ h v d Γ red

13 Si Chem 104a omework 5 Solutions all 2011 Γ red = + + Don t pay too much attention to the shading. The sign is arbitrary. Just know when the sign flips on a symmetry. Si ˆP (σ 1 ) = 4(σ 1 + σ 2 + σ 3 + σ 4 ) Si ˆP ( 1 ) = 4(σ 1 σ 2 + σ 3 σ 4 ) Si 5p y with 5p x with ˆP Eu ( 1 ) = 4(σ 1 σ 3 ) 5p y with ˆP Eu ( 2 ) = 4(σ 2 σ 4 ) 5p x with 5p y with 5p x with 13 5p y with 5p x with

14 Chem 104a omework 5 Solutions all 2011 D 4h Si Si 4 4 D 4h -7.7 Si Si p x p y p z A 2u 1A 2u p x p y p z A 2u 1A 2u s s 1 1 (b) It will be tetrahedral T d 1A 2u 2T d 1A 2u 2 2 2A 1 2A T d 1T d 1 1A D 1g 4h 1A 1 T d 1A 1 D 4h 14 T d