Problem 37
Question
Consider the molecules \(\mathrm{NO}^{+}\) and \(\mathrm{N}_{2}^{+}\) and use molecular orbital theory to answer the following: (a) Write the molecular orbital configuration of each ion (ignore the 1 s electrons). (b) Predict the bond order of each ion. (c) Which of these ions is paramagnetic? Which is diamagnetic? (d) Which of these ions do you think has the greater bond length? Explain.
Step-by-Step Solution
Verified Answer
\(\mathrm{NO}^{+}\) has a bond order of 3 and is diamagnetic, while \(\mathrm{N}_{2}^{+}\) has a bond order of 3.5 and is paramagnetic. Therefore, \(\mathrm{N}_{2}^{+}\) has a shorter bond length.
1Step 1: Molecular Orbital Configuration
We first need to determine the electron configuration. Nitrogen atom has 7 electrons, Oxygen atom has 8. \(\mathrm{NO}^{+}\) loses one electron, so it has 14 electrons. The molecular orbital configuration for 14 electrons is: \(\sigma_{1s}^{2}\), \(\sigma_{2s}^{2}\), \(\sigma_{2p}^{2}\), \(\pi_{2p}^{4}\), \(\sigma_{2p}^{2}\), \(\pi_{2p}^{2}\). \(\mathrm{N}_{2}^{+}\) has 13 electrons. The molecular orbital configuration for 13 electrons is: \(\sigma_{1s}^{2}\), \(\sigma_{2s}^{2}\), \(\sigma_{2p}^{2}\), \(\pi_{2p}^{4}\), \(\sigma_{2p}^{2}\), \(\pi_{2p}^{1}\).
2Step 2: Predict the Bond Order
Next, we have to calculate the bond order. The formula for bond order is \(\frac{1}{2}(n_{b} - n_{a})\), where \(n_{b}\) is the number of electrons in bonding orbitals and \(n_{a}\) is the number of electrons in antibonding orbitals. For \(\mathrm{NO}^{+}\), it has 10 bonding electrons and 4 antibonding electrons, giving a bond order of 3. For \(\mathrm{N}_{2}^{+}\), it has 10 bonding electrons and 3 antibonding electrons, hence bond order of 3.5.
3Step 3: Check for Paramagnetism or Diamagnetism
A molecule is paramagnetic if it has unpaired electrons, and diamagnetic if all electrons are paired. In \(\mathrm{NO}^{+}\), all electrons are paired, so it is diamagnetic. In \(\mathrm{N}_{2}^{+}\), there is one unpaired electron, hence it is paramagnetic.
4Step 4: Determine the Bond Length
Lastly, when comparing bond lengths, it’s useful to remember that, all else being equal, a greater bond order generally indicates a shorter bond length. Therefore, \(\mathrm{N}_{2}^{+}\), with a bond order of 3.5, likely has a shorter bond length than \(\mathrm{NO}^{+}\) with a bond order of 3.
Key Concepts
Bond OrderParamagnetism and DiamagnetismMolecule Bond Length
Bond Order
Bond order is a concept used in molecular orbital theory to predict the strength and length of a bond in a molecule. It provides an insight into how many bonds exist between two atoms. Think of it as a way to quantify bonding interactions.
The formula for bond order is given by \( \frac{1}{2}(n_{b} - n_{a}) \), where \( n_{b} \) is the number of electrons in bonding molecular orbitals and \( n_{a} \) is the number of electrons in antibonding molecular orbitals. This formula helps calculate the net bonding effect.
A higher bond order suggests a stronger, more stable bond. For example, a bond order of 3, as in NO\(^+\), typically indicates a triple bond, whereas a bond order of 3.5, as in N\(_2^+\), suggests a bond that's even stronger than a typical triple bond.
The formula for bond order is given by \( \frac{1}{2}(n_{b} - n_{a}) \), where \( n_{b} \) is the number of electrons in bonding molecular orbitals and \( n_{a} \) is the number of electrons in antibonding molecular orbitals. This formula helps calculate the net bonding effect.
A higher bond order suggests a stronger, more stable bond. For example, a bond order of 3, as in NO\(^+\), typically indicates a triple bond, whereas a bond order of 3.5, as in N\(_2^+\), suggests a bond that's even stronger than a typical triple bond.
- Greater bond order implies more electron density between atoms.
- Stronger bonds usually result in shorter distances between bonded atoms.
- Increased bond order often accompanies increased bond energy.
Paramagnetism and Diamagnetism
Paramagnetism and diamagnetism are properties of molecules that describe their magnetic behavior based on their electronic configuration.
In simple terms, these properties are determined by the presence or absence of unpaired electrons in a molecule:
In simple terms, these properties are determined by the presence or absence of unpaired electrons in a molecule:
- Paramagnetism: If a molecule has one or more unpaired electrons, it is paramagnetic. Such molecules exhibit attraction towards a magnetic field. This occurs because unpaired electrons create a magnetic moment. For instance, N\(_2^+\) is paramagnetic due to its one unpaired electron.
- Diamagnetism: Conversely, if all electrons in a molecule are paired, the molecule is considered diamagnetic. These molecules are slightly repelled by magnetic fields. NO\(^+\), for instance, is diamagnetic as all its electrons are paired.
Molecule Bond Length
Molecular bond length is the distance between the nuclei of two bonded atoms in a molecule. It is a critical factor that influences a molecule's properties and behavior.
Bond length is inherently linked to bond order. Typically, as bond order increases, bond length decreases. This is because a higher bond order means more electrons are shared between atoms, pulling them closer together, which results in a shorter bond length.
Bond length is inherently linked to bond order. Typically, as bond order increases, bond length decreases. This is because a higher bond order means more electrons are shared between atoms, pulling them closer together, which results in a shorter bond length.
- For example, N\(_2^+\) with a bond order of 3.5, is likely to have a shorter bond length than NO\(^+\), which has a bond order of 3.
- Bonds involving multiple pairs of electrons, like double or triple bonds, tend to be shorter and stronger than single bonds.
- It's important to consider that bond length can also depend on other factors such as the size of the atoms involved and the presence of any electronegative groups.
Other exercises in this chapter
Problem 34
For each of the species \(\mathrm{C}_{2}^{+}, \mathrm{O}_{2}^{-}, \mathrm{F}_{2}^{+},\) and \(\mathrm{NO}^{+}\) (a) Write the molecular orbital diagram (as in E
View solution Problem 35
Write plausible molecular orbital diagrams for the following heteronuclear diatomic species: (a) \(\mathrm{NO} ;\) (b) \(\mathrm{NO}^{+}\) (c) \(\mathrm{CO} ;\)
View solution Problem 38
Consider the molecules \(\mathrm{CO}^{+}\) and \(\mathrm{CN}^{-}\) and use molecular orbital theory to answer the following: (a) Write the molecular orbital con
View solution Problem 39
Construct the molecular orbital diagram for CF. Would you expect the bond length of \(\mathrm{CF}^{+}\) to be longer or shorter than that of CF?
View solution