Chapter 10: Problem 61

In 2009 the ion \(\mathrm{N}_{2}^{3-}\) was isolated. Use a molecular orbital diagram to compare its properties (bond order and magnetism) with the isoelectronic ion \(\mathrm{O}_{2}^{-}\)

Short Answer

Expert verified

Both \(\mathrm{N}_{2}^{3-}\) and \(\mathrm{O}_{2}^{-}\) are paramagnetic due to the presence of unpaired electrons. \(\mathrm{N}_{2}^{3-}\) has greater bond order (2.5) compared to \(\mathrm{O}_{2}^{-}\) (2), indicating stronger bonding in \(\mathrm{N}_{2}^{3-}\).

Step by step solution

Understanding Molecular Orbital Theory

Molecular orbital (MO) Theory suggests that atomic orbitals combine to form new molecular orbitals which belong to the molecule as a whole. Electrons fill the MOs just as they would in atomic orbitals, with the lowest energy orbitals filling first.

Constructing the MO Diagram for \(\mathrm{N}_{2}^{3-}\)

The \(\mathrm{N}_{2}^{3-}\) ion has 17 electrons (14 from the two nitrogen atoms and 3 from the charge). The MO diagram for nitrogen has a total of 10 electrons populating the orbitals. The 3 additional electrons for \(\mathrm{N}_{2}^{3-}\) would populate the next highest orbitals, leaving two unpaired electrons making \(\mathrm{N}_{2}^{3-}\) paramagnetic.

Determining the Bond Order for \(\mathrm{N}_{2}^{3-}\)

Bond order is the difference between the number of bonding electrons and the number of antibonding electrons, divided by 2. After populating the \(\mathrm{N}_{2}^{3-}\) MO diagram, count the number of bonding and antibonding electrons to find a bond order of 2.5.

Constructing the MO Diagram for \(\mathrm{O}_{2}^{-}\)

The \(\mathrm{O}_{2}^{-}\) ion has 17 electrons (16 from the two oxygen atoms and 1 from the charge). The MO diagram for oxygen has a total of 12 electrons populating the orbitals. The 5 additional electrons for \(\mathrm{O}_{2}^{-}\) would populate the next highest orbitals, leaving one unpaired electron making \(\mathrm{O}_{2}^{-}\) paramagnetic.

Determining the Bond Order for \(\mathrm{O}_{2}^{-}\)

Use the same formula to determine the bond order for \(\mathrm{O}_{2}^{-}\) as used for \(\mathrm{N}_{2}^{3-}\). After populating the \(\mathrm{O}_{2}^{-}\) MO diagram, count the number of bonding and antibonding electrons to find a bond order of 2.

Comparing the MO Properties

Both \(\mathrm{N}_{2}^{3-}\) and \(\mathrm{O}_{2}^{-}\) are paramagnetic due to the presence of unpaired electrons. The bond order of \(\mathrm{N}_{2}^{3-}\) is greater than that of \(\mathrm{O}_{2}^{-}\), making it more stable.

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In 2009 the ion \(\mathrm{N}_{2}^{3-}\) was isolated. Use a molecular orbital diagram to compare its properties (bond order and magnetism) with the isoelectronic ion \(\mathrm{O}_{2}^{-}\)

Short Answer

Step by step solution

Understanding Molecular Orbital Theory

Constructing the MO Diagram for \(\mathrm{N}_{2}^{3-}\)

Determining the Bond Order for \(\mathrm{N}_{2}^{3-}\)

Constructing the MO Diagram for \(\mathrm{O}_{2}^{-}\)

Determining the Bond Order for \(\mathrm{O}_{2}^{-}\)

Comparing the MO Properties

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