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Chapter 8: Problem 170

Explain the molecular orbital structure of \(\mathrm{O}_{2}\). Calculate the bond order.

Short Answer

Expert verified
The molecular orbital structure of O₂ is derived by considering the energy levels of atomic orbitals and filling the molecular orbitals according to the aufbau principle. The molecular orbitals are arranged in increasing energy order as σ(2s), σ*(2s), σ(2p), π(2p), π(2p), π*(2p), π*(2p), and σ*(2p). The 12 valence electrons are distributed accordingly, and the bond order is calculated using the formula \(Bond\, Order = \frac{1}{2} (Number\, of\, electrons\, in\, bonding\, orbitals - Number\, of\, electrons\, in\, antibonding\, orbitals)\). The bond order of O₂ is found to be 2, indicating a double bond between the two oxygen atoms.

Step by step solution

01

Identify atomic orbitals of oxygen

Since oxygen has an atomic number of 8, its electronic configuration is given by 1s² 2s² 2p⁴. The valence orbitals are the 2s and 2p orbitals. There are two oxygen atoms in O₂, so we have a total of 2(2s² + 2p⁴) = 12 valence electrons.
02

Arrange atomic orbitals into molecular orbitals

For the molecular orbitals, we will consider the energy levels of different atomic orbitals and use the aufbau principle to fill them with electrons. The molecular orbitals are arranged in increasing energy order as follows: 1. σ(2s) 2. σ*(2s) 3. σ(2p) 4. π(2p) and π(2p) - these two orbitals are degenerate (having the same energy) 5. π*(2p) and π*(2p) - these two orbitals are degenerate (having the same energy) 6. σ*(2p)
03

Fill molecular orbitals with electrons

We have a total of 12 valence electrons in O₂. We will now fill the molecular orbitals using the aufbau principle. 1. σ(2s) – 2 electrons 2. σ*(2s) – 2 electrons 3. σ(2p) – 2 electrons 4. π(2p) – 2 electrons (one in each degenerate orbital) 5. π*(2p) – 4 electrons (two in each degenerate orbital) 6. σ*(2p) – 0 electrons
04

Calculate bond order

Now that we have filled the molecular orbitals with electrons, we can calculate the bond order. Bond order is calculated using the formula: \(Bond\, Order = \frac{1}{2} (Number\, of\, electrons\, in\, bonding\, orbitals - Number\, of\, electrons\, in\, antibonding\, orbitals)\) In our case, Bond order = \(\frac{1}{2}(8 - 4) = \frac{1}{2}(4) = 2\) The molecular orbital structure of O₂ has a bond order of 2, indicating a double bond between the two oxygen atoms.

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Most popular questions from this chapter

(a) Define the following terms, and draw a diagram showing their relationship to each other. Amplitude Nodes Nodal plane Phase "Out of phase" Wave equation (b) How do they pertain to molecular orbitals?

In 1,3 -butadiene, the observed length of the \(\mathrm{C}_{2}-\mathrm{C}_{3}\) bond is \(1.48 \AA\), while the "standard" single bond distance is generally accepted to be \(1.54 \AA\). Explain this discrepancy and calculate the \(\pi\) bond order of the \(\mathrm{C}_{2}-\mathrm{C}_{3}\) butadiene bond. (Hint, consider the resonance effects and the molecular orbital structure).

Discuss the direct, concerted addition of \(\mathrm{H}_{2}\) to an alkene from the standpoint of orbital symmetry. (a) In the absence of catalyst, and (b) in the presence of photochemical stimulation.

The commonly observed conversion of cyclopropy1 cations into allyl cations is considered to be an example of an electrocyclic reaction. (a) What is the HOMO of the ally1 cation? How many \(\pi\) electrons has it? (b) Would you expect conrotatory or disrotatory motion? (c) What prediction would you make about interconversion of allyl and cyclopropy1 anions? (d) About the interconversion of pentadienyl cations and cyclopentenyl cations?

Draw the LCAO model of ethylene in the bonding and antibonding orbital. Distinguish the ground state of ethylene from its excited state. Distinguish \(\pi^{2}\) from \(\pi^{*} \pi\).
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