Q. Bond order of \( \mathrm{F_2^+} \).

We want the bond order of the ion \( \mathrm{F_2^+} \). The key idea is to compare how many electrons occupy the bonding and antibonding molecular orbitals for fluorine.

Step 1: Write the electronic configuration for the molecule

Each fluorine atom has 7 valence electrons. Molecular orbital theory uses all valence electrons when finding bond order for main-group diatomics.

So for \( \mathrm{F_2} \):

• \(7 + 7 = 14\) valence electrons

For \( \mathrm{F_2^+} \), we remove one electron (because of the \(+\) charge):

• \(14 – 1 = 13\) valence electrons

Step 2: Determine the molecular orbital filling (for \( \mathrm{F_2} \) framework)

The ordering for \( \mathrm{F_2} \) (and lighter halogens) is typically:

• \(\sigma 2s\)
• \(\sigma^{\ast} 2s\)
• \(\pi 2p\)
• \(\sigma 2p\)
• \(\pi^{\ast} 2p\)
• \(\sigma^{\ast} 2p\)

We can summarize the electron count by placing electrons into the standard valence MO set for \( \mathrm{F_2} \), then remove one for \( \mathrm{F_2^+} \).

For \( \mathrm{F_2} \) (14 valence electrons)

• \(\sigma 2s\): 2 electrons
• \(\sigma^{\ast} 2s\): 2 electrons
• \(\pi 2p\): 4 electrons
• \(\sigma 2p\): 2 electrons
• \(\pi^{\ast} 2p\): 4 electrons

That accounts for \(2 + 2 + 4 + 2 + 4 = 14\).

Now create \( \mathrm{F_2^+} \) (13 valence electrons)

Remove one electron from the highest-energy occupied orbital in \( \mathrm{F_2} \).

In \( \mathrm{F_2} \), the highest-energy occupied set is \( \pi^{\ast} 2p \) (it has 4 electrons). Removing one gives:

• \(\pi^{\ast} 2p\): 3 electrons

All lower orbitals stay the same.

Step 3: Use the bond order formula

The bond order is calculated by:

\[
\text{Bond order} = \frac{N_b – N_a}{2}
\]

where \(N_b\) is the total number of electrons in bonding MOs and \(N_a\) is the total number of electrons in antibonding MOs.

Step 4: Count bonding and antibonding electrons for \( \mathrm{F_2^+} \)

Bonding molecular orbitals (bonding set):

• \(\sigma 2s\) (2 electrons)
• \(\pi 2p\) (4 electrons)
• \(\sigma 2p\) (2 electrons)

So:

\[
N_b = 2 + 4 + 2 = 8
\]

Antibonding molecular orbitals (antibonding set):

• \(\sigma^{\ast} 2s\) (2 electrons)
• \(\pi^{\ast} 2p\) (3 electrons in \( \mathrm{F_2^+} \))

So:

\[
N_a = 2 + 3 = 5
\]

Step 5: Compute the bond order

\[
\text{Bond order} = \frac{8 – 5}{2} = \frac{3}{2} = 1.5
\]

Final Answer

The bond order of \( \mathrm{F_2^+} \) is \(1.5\).