Question

(a) The nitric oxide molecule, NO, readily loses one electron to form the \(\mathrm{NO}^{+}\) ion. Why is this consistent with the electronic structure of \(\mathrm{NO} ?\) (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in \(\mathrm{NO}, \mathrm{NO}^{+},\) and \(\mathrm{NO}^{-},\) and describe the magnetic properties of each. (c) With what neutral homonuclear diatomic molecules are the \(\mathrm{NO}^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?

Short answer

The electronic structure of NO consists of 11 valence electrons, making it paramagnetic due to one unpaired electron. In contrast, \(\mathrm{NO^{+}}\) has 10 valence electrons, resulting in a completely filled molecular orbitals and diamagnetic behavior. On the other hand, \(\mathrm{NO^{-}}\) has 12 valence electrons and is paramagnetic like NO. The bond strength order for these molecules and ions is NO+ > NO > NO-, due to their respective bond orders of 3, 2.5, and 2. Finally, the ions are isoelectronic with neutral homonuclear diatomic molecules: \(\mathrm{NO^{+}}\) with N2 and \(\mathrm{NO^{-}}\) with O2.

Step-by-step solution

Examine the electronic structure of NO

To determine the electronic structure, we first need to find the number of valence electrons in the molecule. Nitrogen (N) has 5 valence electrons, and oxygen (O) has 6 valence electrons. Thus, the nitric oxide (NO) molecule has a total of 11 valence electrons. To find which orbitals are filled, we can start filling them using the mnemonic (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 5d...): - \(1\sigma_{g} (2e^{⁻})\) - \(1\sigma_{u}^{*} (2e^{⁻})\) - \(2\sigma_{g} (2e^{⁻})\) - \(1\pi_{u} (4e^{⁻})\) - two degenerate orbitals - \(1\pi_{g}^{*} (1e^{⁻})\) - two degenerate orbitals, but only one has an electron

Electronic structure of NO+

In the case of NO+, one electron is removed from the molecule, resulting in 10 valence electrons. So, the removal of an electron will be from the highest-energy orbital, which is \(1\pi_{g}^{*}\), thus fully filling all the molecular orbitals.

Predict N-O bond strengths in NO, NO+, and NO-

Bond strength is inversely proportional to bond order. For each molecule/ion, we can calculate bond order by considering half the difference of electrons in bonding and anti-bonding orbitals: NO: Bond order = (8 - 3) / 2 = 2.5 NO+: Bond order = (8 - 2) / 2 = 3 NO-: Bond order = (8 - 4) / 2 = 2 Thus, the bond strength follows the order: NO+ > NO > NO-.

Describe the magnetic properties of each

A molecule is paramagnetic if it has unpaired electrons and diamagnetic if it has no unpaired electrons. In the case of the electronic structure of NO, there is one unpaired electron present in the highest-energy orbital. Therefore, NO is paramagnetic. For NO+, all molecular orbitals are filled, so there are no unpaired electrons and it is diamagnetic. For NO-, since it has an extra electron, it will also have one unpaired electron and it will be paramagnetic as well.

Identify isoelectronic neutral diatomic molecules

To find neutral homonuclear diatomic molecules isoelectronic to the NO+ and NO- ions, we need to find molecules with the same number of valence electrons: NO+: 10 valence electrons - Molecule: N2 (nitrogen) - 2 atoms of nitrogen each with 5 valence electrons -> total of 10 valence electrons NO-: 12 valence electrons - Molecule: O2 (oxygen) - 2 atoms of oxygen each with 6 valence electrons -> total of 12 valence electrons So, the \(\mathrm{NO^{+}}\) ion is isoelectronic with the N2 molecule, and the \(\mathrm{NO^{-}}\) ion is isoelectronic with the O2 molecule.

Key concepts

Electronic Structure

The electronic structure of a molecule reveals how electrons are distributed in its molecular orbitals. For nitric oxide (NO), we first examine its valence electrons. Nitrogen has 5 valence electrons and oxygen has 6, giving NO a total of 11 valence electrons. In molecular orbital theory, these electrons fill the orbitals according to their energy levels, starting with the lowest.

For NO, the orbitals are filled as follows:

• The lowest energy levels fill first: \(1\sigma_{g} (2e^{-})\), \(1\sigma_{u}^{*} (2e^{-})\), \(2\sigma_{g} (2e^{-})\)
• Next, the electrons occupy: \(1\pi_{u} (4e^{-})\) - which is a degenerate orbital, meaning identical in energy.
• Finally, one electron sits in a the higher energy orbital: \(1\pi_{g}^{*} (1e^{-})\).

In the case of NO⁺, which loses one electron to have 10 valence electrons, the electron is removed from the highest-energy orbital, \(1\pi_{g}^{*}\), resulting in complete filling of all lower energy orbitals.

Bond Strength

The concept of bond strength in molecular orbital theory is closely related to the bond order, which is calculated as half the difference between the number of electrons in bonding and antibonding orbitals. For NO, NO⁺, and NO^- the bond strengths can be determined by evaluating their bond orders:

- **NO:** Bond Order = \((8 - 3)/2 = 2.5\)- **NO⁺:** Bond Order = \((8 - 2)/2 = 3\)- **NO^-:** Bond Order = \((8 - 4)/2 = 2\)

• Higher bond order corresponds to stronger bond strength.
• Therefore, the bond strength order is NO⁺ > NO > NO^-.

The removal or addition of electrons in forming NO⁺ and NO^- affects the electron distribution between bonding and antibonding orbitals, thus influencing the bond strength.

Magnetic Properties

Magnetic properties of molecules depend on the presence of unpaired electrons. These properties categorize molecules as either paramagnetic or diamagnetic.

A molecule is **paramagnetic** if it contains unpaired electrons, causing it to be attracted to magnetic fields. NO exhibits paramagnetism due to its unpaired electron in the \(1\pi_{g}^{*}\) orbital. Conversely, a molecule is **diamagnetic** if all its electrons are paired, resulting in no attraction to magnetic fields. NO⁺ is diamagnetic because removing one electron results in fully filled orbitals with all paired electrons.

Lastly, NO^- gains an electron, which sits as an unpaired electron in one of its molecular orbitals, making it paramagnetic like NO.

Isoelectronic Species

Isoelectronic species are atoms, molecules, or ions that have the same number of electrons. This concept helps in identifying molecules with similar electronic configurations and properties. For instance, NO⁺ and N₂ both have 10 electrons, making them isoelectronic. The NO molecule loses an electron to form NO⁺, matching the electron count in the nitrogen molecule (N₂).

Similarly, when NO^- adds an electron, it becomes isoelectronic with O₂, as both have 12 electrons. Understanding which species are isoelectronic aids in predicting reactivity and chemical behavior, as similar electron counts often lead to analogous chemical properties across different molecules.