What is the bond order of the F_2^+ ion?

• A. 1
• B. 1.5
• C. 2
• D. 0.5

Answer: (B)

Verified step by step guidance

1. Recall that bond order is calculated using the formula: \(\text{Bond order} = \frac{(\text{number of bonding electrons}) - (\text{number of antibonding electrons})}{2}\).
2. Write the electron configuration for the neutral F\(_2\) molecule. Each fluorine atom has 9 electrons, so F\(_2\) has a total of 18 electrons.
3. Fill the molecular orbitals for F\(_2\) with 18 electrons according to the molecular orbital theory, noting the order of orbitals for second period homonuclear diatomic molecules: \(\sigma_{2s}\), \(\sigma^*_{2s}\), \(\sigma_{2p_z}\), \(\pi_{2p_x} = \pi_{2p_y}\), \(\pi^*_{2p_x} = \pi^*_{2p_y}\), \(\sigma^*_{2p_z}\).
4. For the F\(_2^+\) ion, remove one electron from the highest occupied molecular orbital (HOMO) of F\(_2\). Identify which orbital loses an electron.
5. Count the number of electrons in bonding and antibonding orbitals after removing one electron, then apply the bond order formula to find the bond order of F\(_2^+\).