Problem 105 - Chemistry: An Atoms-Focused Approach

Question

Which of the following molecular ions are paramagnetic? (a) \(\mathrm{N}_{2}^{+} ;\) (b) \(\mathrm{O}_{2}^{+} ;\) (c) \(\mathrm{C}_{2}^{2+} ;\) (d) \(\mathrm{Br}_{2}^{2-}\)

Step-by-Step Solution

Answer

The ions \(\mathrm{N}_{2}^{+}\) and \(\mathrm{O}_{2}^{+}\) are paramagnetic, as they have unpaired electrons in their molecular orbital configurations.

Step 1: Recall the molecular orbital configuration for \(\mathrm{N}_2\), \(\mathrm{O}_2\), \(\mathrm{C}_2\), and \(\mathrm{Br}_2\)

Before discussing the ions, let's recall the molecular orbital configurations for the parent molecules - \(\mathrm{N}_2\), \(\mathrm{O}_2\), \(\mathrm{C}_2\), and \(\mathrm{Br}_2\). The molecular orbitals in these diatomic molecules can be represented using the following notation: 1sσ, 1sσ*, 2sσ, 2sσ*, 2pσ, 2pπ, 2pπ*, 2pσ*, 3sσ, 3pσ, and so on. Molecular orbitals with an asterisk (*) indicate anti-bonding orbitals.

Step 2: Determine the electron configuration for each ion

After recalling the molecular orbitals, now determine the electron configuration for each ion. (a) \(\mathrm{N}_{2}^+\): The parent molecule \(\mathrm{N}_2\) has 14 electrons; the positive charge indicates one electron has been removed, resulting in 13 electrons. (b) \(\mathrm{O}_{2}^+\): The parent molecule \(\mathrm{O}_2\) has 16 electrons; the positive charge indicates one electron has been removed, resulting in 15 electrons. (c) \(\mathrm{C}_{2}^{2+}\): The parent molecule \(\mathrm{C}_2\) has 12 electrons; the double positive charge indicates two electrons have been removed, resulting in 10 electrons. (d) \(\mathrm{Br}_{2}^{2-}\): The parent molecule \(\mathrm{Br}_2\) has 70 electrons; the double negative charge indicates two electrons have been added, resulting in 72 electrons.

Step 3: Fill the molecular orbitals with electrons

Now, fill the molecular orbitals with the correct number of electrons for each ion: (a) \(\mathrm{N}_{2}^+\): 1sσ(2), 1sσ*(2), 2sσ(2), 2sσ*(2), 2pσ(2), 2pπ(3). The 2pπ molecular orbital has an unpaired electron. (b) \(\mathrm{O}_{2}^+\): 1sσ(2), 1sσ*(2), 2sσ(2), 2sσ*(2), 2pσ(2), 2pπ(4), 2pπ*(1). The 2pπ* molecular orbital has an unpaired electron. (c) \(\mathrm{C}_{2}^{2+}\): 1sσ(2), 1sσ*(2), 2sσ(2), 2sσ*(2), 2pσ(2). All electrons are paired. (d) \(\mathrm{Br}_{2}^{2-}\): The molecular orbitals are filled up to 3pπ(4), 3pπ*(2). All electrons are paired.

Step 4: Determine if the ions are paramagnetic or not

Finally, determine if the ions are paramagnetic based on the presence of unpaired electrons: (a) \(\mathrm{N}_{2}^+\): Paramagnetic (1 unpaired electron in 2pπ) (b) \(\mathrm{O}_{2}^+\): Paramagnetic (1 unpaired electron in 2pπ*) (c) \(\mathrm{C}_{2}^{2+}\): Diamagnetic (no unpaired electrons) (d) \(\mathrm{Br}_{2}^{2-}\): Diamagnetic (no unpaired electrons) So, the ions \(\mathrm{N}_{2}^{+}\) and \(\mathrm{O}_{2}^{+}\) are paramagnetic, while \(\mathrm{C}_{2}^{2+}\) and \(\mathrm{Br}_{2}^{2-}\) are diamagnetic.