LECTURE 13 1. Draw a molecular orbital diagram and determine the bond order expected for the molecule B2. For full credit on MO diagrams, • label increasing energy with an arrow next to the diagram. • pay attention to whether the question asks for valence electrons or all electrons. • for any bonding orbital drawn, include the corresponding anti-bonding orbital, even if it is not filled with any electrons. • Label each atomic orbital (1s, 2s, 2px, 2py, 2pz etc.) and each molecular orbital (s2s, p2px, p2py, etc.) that you draw. • Fill in the electrons for both the atomic and molecular orbitals. Bond order = ½(6-4) = 1 2. (a) Write the valence electron configuration (from lowest to highest orbital energies) for the ion N2-1. Your answer should be in a form similar to (s2s)2, which is the valence configuration for Li2. (b) What is the bond order of N2-1? (c) Which has a longer bond, N2-1 or N2? Justify your answer using bond order. (a) (s2s)2(s2s*)2(p2px)2(p2py)2(s2pz)2(p2px*)1 (b) 2.5 (c) The N-N bond is stronger in N2 since an electron is removed from an antibonding orbital, increasing the bond order from 2.5 to 3. LECTURE 13 3. (a) Draw a MO diagram for the valence electrons of BC. Label all atomic and molecular orbitals. (b) Write the molecular orbital configuration for the valence electrons in BC and in BC1-. (c) Which of the molecular orbitals in BC do not have a planar node along the internuclear axis? (d) Which has the stronger B–C bond, BC or BC1-? Justify your answer using bond order. (a) (b) BC: 7 valence electrons (s2s)2(s2s*)2(p2px)2(p2py)1 BC1-: 8 valence electrons (s2s)2(s2s*)2(p2px)2(p2py)2 (c) Only p orbitals have planar nodes at the internuclear (bonding) axis. The following orbitals do not have nodal planes along the bonding axis: s2s, s2s*, s2pz, and s2pz* (d) The B-C bond is stronger in BC1- since an electron is added to a bonding orbital, increasing the bond order from 1.5 to 2. 4. For each of the following molecules, (i) write the valence electron configuration (Your answer should be in a form similar to (s2s)2, which is the valence configuration for Li2) and (ii) determine if the molecule is paramagnetic (has unpaired electrons) or diamagnetic (does not have unpaired electrons). If the species is paramagnetic, identify the number of unpaired electrons. (a) Cl21+; (b) O21+ (a) Cl2+: 13 valence electrons (s3s)2(s3s*)2(s3pz)2(p3px)2(p3py)2(p3px*)2(p3py*)1 paramagnetic: 1 unpaired electron LECTURE 13 (b) O2+: 11 valence electrons (s2s)2(s2s*)2(s2pz)2(p2px)2(p2py)2(p2px*)1 paramagnetic: 1 unpaired electron