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Chapter 9: Problem 85

Consider the molecular orbital electron configurations for \(\mathrm{N}_{2}\), \(\mathrm{N}_{2}^{+}\), and \(\mathrm{N}_{2}^{-}\). For each compound or ion, fill in the table below with the correct number of electrons in each molecular orbital.

Short Answer

Expert verified
The molecular orbital electron configurations for \(\mathrm{N}_{2}\), \(\mathrm{N}_{2}^{+}\), and \(\mathrm{N}_{2}^{-}\) are as follows: | MO | \(\mathrm{N}_{2}\) | \(\mathrm{N}_{2}^{+}\) | \(\mathrm{N}_{2}^{-}\) | |------------|--------|-------------|-------------| | \(\mathrm{1\sigma_{g}}\) | 2 | 2 | 2 | | \(\mathrm{1\sigma_{u}^{*}}\) | 0 | 0 | 0 | | \(\mathrm{2\sigma_{g}}\) | 2 | 2 | 2 | | \(\mathrm{2\sigma_{u}^{*}}\) | 0 | 0 | 0 | | \(\mathrm{3\sigma_{g}}\) | 2 | 2 | 2 | | \(\mathrm{3\sigma_{u}^{*}}\) | 0 | 0 | 0 | | \(\mathrm{1\pi_{u}}\) | 6 | 5 | 6 | | \(\mathrm{1\pi_{g}^{*}}\) | 2 | 0 | 3 | | \(\mathrm{4\sigma_{g}}\) | 0 | 0 | 0 | | \(\mathrm{4\sigma_{u}^{*}}\) | 0 | 0 | 0 |

Step by step solution

01

Determine the number of electrons for each compound

For \(\mathrm{N}_{2}\), each nitrogen atom contributes 7 electrons, so total electrons = 7 * 2 = 14 electrons. For \(\mathrm{N}_{2}^{+}\), one electron is removed; hence total electrons = 14 - 1 = 13 electrons. For \(\mathrm{N}_{2}^{-}\), one electron is added; hence total electrons = 14 + 1 = 15 electrons. Step 2: Fill in the molecular orbitals
02

Fill electrons in the molecular orbitals for each compound

For \(\mathrm{N}_{2}\), we have 14 electrons to fill: \(\mathrm{1\sigma_{g}}\)(2), \(\mathrm{1\sigma_{u}^{*}}\)(0), \(\mathrm{2\sigma_{g}}\)(2), \(\mathrm{2\sigma_{u}^{*}}\)(0), \(\mathrm{3\sigma_{g}}\)(2), \(\mathrm{3\sigma_{u}^{*}}\)(0), \(\mathrm{1\pi_{u}}\)(6), \(\mathrm{1\pi_{g}^{*}}\)(2), \(\mathrm{4\sigma_{g}}\)(0), and \(\mathrm{4\sigma_{u}^{*}}\)(0). For \(\mathrm{N}_{2}^{+}\), we have 13 electrons to fill: \(\mathrm{1\sigma_{g}}\)(2), \(\mathrm{1\sigma_{u}^{*}}\)(0), \(\mathrm{2\sigma_{g}}\)(2), \(\mathrm{2\sigma_{u}^{*}}\)(0), \(\mathrm{3\sigma_{g}}\)(2), \(\mathrm{3\sigma_{u}^{*}}\)(0), \(\mathrm{1\pi_{u}}\)(5), \(\mathrm{1\pi_{g}^{*}}\)(0), \(\mathrm{4\sigma_{g}}\)(0), and \(\mathrm{4\sigma_{u}^{*}}\)(0). For \(\mathrm{N}_{2}^{-}\), we have 15 electrons to fill: \(\mathrm{1\sigma_{g}}\)(2), \(\mathrm{1\sigma_{u}^{*}}\)(0), \(\mathrm{2\sigma_{g}}\)(2), \(\mathrm{2\sigma_{u}^{*}}\)(0), \(\mathrm{3\sigma_{g}}\)(2), \(\mathrm{3\sigma_{u}^{*}}\)(0), \(\mathrm{1\pi_{u}}\)(6), \(\mathrm{1\pi_{g}^{*}}\)(3), \(\mathrm{4\sigma_{g}}\)(0), and \(\mathrm{4\sigma_{u}^{*}}\)(0). Step 3: Complete the table
03

Fill in the table with the molecular orbital electron configurations

| MO | \(\mathrm{N}_{2}\) | \(\mathrm{N}_{2}^{+}\) | \(\mathrm{N}_{2}^{-}\) | |------------|--------|-------------|-------------| | \(\mathrm{1\sigma_{g}}\) | 2 | 2 | 2 | | \(\mathrm{1\sigma_{u}^{*}}\) | 0 | 0 | 0 | | \(\mathrm{2\sigma_{g}}\) | 2 | 2 | 2 | | \(\mathrm{2\sigma_{u}^{*}}\) | 0 | 0 | 0 | | \(\mathrm{3\sigma_{g}}\) | 2 | 2 | 2 | | \(\mathrm{3\sigma_{u}^{*}}\) | 0 | 0 | 0 | | \(\mathrm{1\pi_{u}}\) | 6 | 5 | 6 | | \(\mathrm{1\pi_{g}^{*}}\) | 2 | 0 | 3 | | \(\mathrm{4\sigma_{g}}\) | 0 | 0 | 0 | | \(\mathrm{4\sigma_{u}^{*}}\) | 0 | 0 | 0 |

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