ASR Basic Principles of Spectroscopy Tutorial

2025 JC2 H3 Chemistry ©2025ASRJC/CHEM WE ARE ASR 1 Anderson Serangoon Junior College H3 Chemistry Basic Principles Of Spectroscopy Tutorial 1 When electrons are being removed from or gained by molecular oxygen, the following species could form: peroxide O22–, superoxide O2– and dioxygenyl ion O2+. Draw an MO diagram for each and rank them in order of increasing bond length, and find the number of unpaired electrons in each. MO diagram for O22– O: 1s2 2s2 2p4 O2 has 16 electrons Therefore, O22– has 18 electrons (9 electrons for each atomic orbital) O O22– O With reference to the O22– MO diagram, Bond order of O22– = ½(10 – 8) = 1 (lowest bond order), 0 unpaired electrons Note: the electrons in the MO diagram can also be written just above the line. Energy

2025 JC2 H3 Chemistry ©2025ASRJC/CHEM WE ARE ASR 2 MO diagram for O2– O2– has 17 electrons (one O AO has 8 electrons, the other has 9) O O2– O MO diagram for O2+ O2+ has 15 electrons (one O AO has 8 electrons, the other has 7) O O2+ O  rank in increasing bond length would be O2+ < O2– < O22–. Bond order of O2– = ½ (10 – 7) = 1.5, 1 unpaired electron Bond order of O2+ = ½ (10 – 5) = 2.5 (highest bond order), 1 unpaired electron Energy Energy Bond length is defined as the distance between the centers of two covalently bonded atoms. The number of bonded electrons (the bond order) determines the length of the bond. The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length

2025 JC2 H3 Chemistry ©2025ASRJC/CHEM WE ARE ASR 3 2 The bond strength in the oxygen molecule, 496 kJ mol–1, is similar to that of the N=N in diazo compounds, 410 kJ mol–1. Its bonding is therefore often represented by structure I. However, the oxygen molecule is paramagnetic, showing that it contains unpaired electrons. This can be represented by the alternative structure II. This would predict that it contains a O–O single bond, similar in strength to the O–O bond in hydrogen peroxide, H2O2, 150 kJ mol–1. O=O .O–O. I II Draw a molecular orbital diagram for the O2 molecule and use it to explain why it contains unpaired electrons but has a bond order of 2. MO diagram for O2 O O2 O As the two *2p orbitals are degenerate, the last two electrons occupy them singly, resulting in 2 unpaired electrons. Bond order = ½ (10 – 6) = 2 Energy

2025 JC2 H3 Chemistry ©2025ASRJC/CHEM WE ARE ASR 4 3 Molecular Orbital (MO) Theory was introduced by Mulliken in the 1940’s and 1950’s for which he won the 1964 Nobel Prize in Chemistry. It allows for the prediction of bond orders and paramagnetism of simple molecules. (a) Give a relative energy diagram for the MO’s of diatomic molecules which possess only 1s, 2s and 2p electrons. (b) G