Chapter 3: Problem 12

Use MO diagram to predict the magnetic behaviour of the following molecules: (a) \(\mathrm{H}_{2}^{+}\) (b) \(\mathrm{B}_{2}\) (c) \(\mathrm{O}_{2}^{+}\) (d) \(\mathrm{N}_{2}^{2-}\)

Short Answer

Expert verified

\(\mathrm{H}_{2}^{+}\) and \(\mathrm{N}_{2}^{2-}\) are diamagnetic, while \(\mathrm{B}_{2}\) and \(\mathrm{O}_{2}^{+}\) are paramagnetic.

Step by step solution

- \(\mathrm{H}_{2}^{+}\) MO Diagram

Starting with \(\mathrm{H}_{2}^{+}\), which has 1 electron. That electron will be placed into the \(\sigma_{1s}\) bonding molecular orbital. Since there's no any unpaired electron, \(\mathrm{H}_{2}^{+}\) is diamagnetic.

- \(\mathrm{B}_{2}\) MO Diagram

For \(\mathrm{B}_{2}\), there are 6 valence electrons in total. They will fill the \(\sigma_{2s}\), \(\sigma_{2s}^{*}\), and two out of three \(2p\) orbitals. There are two unpaired electrons, one in each of the \(2p\) orbitals. This means that the compound will be paramagnetic.

- \(\mathrm{O}_{2}^{+}\) MO Diagram

\(\mathrm{O}_{2}^{+}\) has 11 valence electrons: 4 in the \(2s\) orbitals, and 7 in the \(2p\) orbitals. After filling the \(\sigma_{2s}\), \(\sigma_{2s}^{*}\), and \(\sigma_{2p}\) orbitals, 2 electrons remain. They fill two out of three \(\pi_{2p}\) orbitals. Again, the molecule is paramagnetic because of the unpaired electrons.

- \(\mathrm{N}_{2}^{2-}\) MO Diagram

\(\mathrm{N}_{2}^{2-}\) has 12 valence electrons: 4 in the \(2s\) orbitals, and 8 in the \(2p\) orbitals. They fill all the orbitals up to \(\sigma_{2p}\). The molecule is diamagnetic as all electrons are paired.

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Use MO diagram to predict the magnetic behaviour of the following molecules: (a) \(\mathrm{H}_{2}^{+}\) (b) \(\mathrm{B}_{2}\) (c) \(\mathrm{O}_{2}^{+}\) (d) \(\mathrm{N}_{2}^{2-}\)

Short Answer

Step by step solution

- \(\mathrm{H}_{2}^{+}\) MO Diagram

- \(\mathrm{B}_{2}\) MO Diagram

- \(\mathrm{O}_{2}^{+}\) MO Diagram

- \(\mathrm{N}_{2}^{2-}\) MO Diagram

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