Problem 50 - Chemistry and Chemical Reactivity

Question

Nitrogen, \(\mathbf{N}_{2}\), can ionize to form \(\mathbf{N}_{2}^{+}\) or add an electron to give \(\mathrm{N}_{2}\). . Using molecular orbital theory, compare these species with regard to (a) their magnetic character, (b) net number of \(\pi\) bonds, (c) bond order, (d) bond length, and (e) bond strength.

Step-by-Step Solution

Answer

\(\mathrm{N}_2\) and \(\mathrm{N}_2^+\) are diamagnetic, \(\mathrm{N}_2^-\) is paramagnetic; \(\mathrm{N}_2\) has 2 \(\pi\) bonds, \(\mathrm{N}_2^+\) has 1, \(\mathrm{N}_2^-\) has 2; bond orders are 3, 2.5, and 2.5, respectively; bond lengths are shortest in \(\mathrm{N}_2\) and longest in \(\mathrm{N}_2^-\) and \(\mathrm{N}_2^+\); bond strength follows the same order as bond order.

Step 1: Determining Magnetic Character

Nitrogen \( \mathbf{N}_2 \) molecule is diatomic with a bond order of 3 (filled \pi-bonding and \sigma-bonding). Both \textbf{\(\mathrm{N}_2\)} and \textbf{\(\mathrm{N}_2^+\)} are diamagnetic because all electrons are paired. For \textbf{\(\mathrm{N}_2^-\)}, adding an electron to the \pi-antibonding orbital results in unpaired electrons, making it paramagnetic.

Step 2: Calculating Net Number of \(\pi\) Bonds

For \textbf{\(\mathbf{N}_2\)}, there are two \pi-bonds (from filled \pi-bonding orbitals). The \(\mathbf{N}_2^+\) loses one electron from the \pi-bonding orbitals, resulting in one \pi-bond. In \(\mathbf{N}_2^-\), the added electron goes into an antibonding orbital, still leaving two \pi-bonds.

Step 3: Determining Bond Order

The bond order is calculated as \(\frac{1}{2}(\text{number of bonding electrons} - \text{number of antibonding electrons})\). For \(\mathbf{N}_2\), the bond order is 3. For \(\mathbf{N}_2^+\), the bond order is 2.5. For \(\mathbf{N}_2^-\), the bond order is 2.5 as well.

Step 4: Comparing Bond Length

Bond length is inversely related to bond order. Therefore, \(\mathbf{N}_2\) has the shortest bond length with a bond order of 3. The length increases in \(\mathbf{N}_2^+\) and \(\mathbf{N}_2^-\) because they have bond orders of 2.5.

Step 5: Comparing Bond Strength

Bond strength is directly related to bond order. Hence, \(\mathbf{N}_2\) has the strongest bond. \(\mathbf{N}_2^+\) and \(\mathbf{N}_2^-\) both have weaker bonds than \(\mathbf{N}_2\) due to lower bond order.

Key Concepts

Nitrogen Ionization

When looking at molecular behavior, ionization refers to the loss or gain of electrons in a molecule. This results in the formation of ions. In the case of nitrogen molecules (_2), there are two notable forms: _2^+ and _2^-. These forms are derived by either removing or adding an electron, respectively.
\(N_2^+\) forms when _2 loses an electron. This results in a positively charged ion, with one less electron occupying the bonding or antibonding orbitals. Conversely, \(N_2^-\) results from the addition of an extra electron, yielding a negative charge.
Understanding these ionized forms is vital as they influence the molecule's properties, such as magnetic character and bond characteristics.

Magnetic Character

Molecular magnetic character depends on the electron configuration and pairing within the molecule.
Diamagnetism and paramagnetism are two critical types of magnetic behavior.

• **Diamagnetic molecules** contain only paired electrons. These molecules do not respond to a magnetic field. <\li>
• **Paramagnetic molecules** feature unpaired electrons, which causes them to be attracted to a magnetic field. <\li>

In terms of nitrogen and its ions, we note:

• \(_2\) and \(_2^+\) are both diamagnetic due to having all electrons paired.<\li>
• In contrast, \(_2^-\) is paramagnetic because the additional electron funds itself unpaired in the \(\pi^*\) orbital. <\li>

These differences in magnetic character arise as the balance of electrons in periodic orbitals shifts due to ionization.

Bond Order

Bond order is a fundamental concept in molecular orbital theory. It is a measure of the number of bonds between a pair of atoms. A higher bond order suggests stronger and shorter bonds.
The bond order for a molecule is calculated using the formula: \[\text{Bond Order} = \frac{1}{2}(\text{number of bonding electrons} - \text{number of antibonding electrons})\]

• For \(N_2\), the bond order is 3. This indicates a strong triple bond.<\li>
• \(N_2^+\) results in a bond order of 2.5 due to the loss of an electron, weakening the bond slightly. <\li>
• \(N_2^-\), with an extra electron in the antibonding orbital, also has a bond order of 2.5, emphasizing less stability than \(N_2\).<\li>

Changes in bond order shed light on how the molecular structure responds to ionization and provide insight into bond strength and length.

Bond Strength

Bond strength is closely tied to bond order. It refers to how much energy is needed to break the bond between two atoms.
Higher bond orders usually lead to stronger bonds because more electron sharing occurs between atoms. In the nitrogen molecule and its ions:

• \(N_2\) has the highest bond strength due to its bond order of 3. This strong triple bond requires a significant amount of energy to break. <\li>
• Both \(N_2^+\) and \(N_2^-\) have weaker bonds with bond orders of 2.5. This lower bond order implies less energy needed to break these bonds, making them weaker than \(N_2\).<\li>

Understanding bond strength in relation to bond order offers valuable insights into the reactivity and stability of molecular ions, particularly in chemical reactions.