To determine the correct answer for the conversion of \( \text{ICl} \) to \( \text{ICl}^+ \), we will consider each option and use molecular orbital theory to explain which is correct:
1. Removal of an electron from a \( \pi^* \) molecular orbital of \( \text{ICl} \):
Molecular orbital theory suggests that the \( \pi^* \) antibonding orbital is typically higher in energy. Removal of an electron from this orbital would stabilize the molecule. Given that \( \text{ICl} \) transitions to \( \text{ICl}^+ \), the most plausible scenario involves removing an electron from the \( \pi^* \) orbital.
2. Increase in bond order from 1 in \( \text{ICl} \) to 1.5 in \( \text{ICl}^+ \):
The bond order is calculated as \( \text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} \). Removing an electron from an antibonding orbital (\( \pi^* \)) increases the bond order, which increases the stability and bond strength, making this statement correct.
3. Formation of a paramagnetic species:
If the electron is removed from a singly occupied molecular orbital, the resultant \( \text{ICl}^+ \) would indeed be paramagnetic due to the presence of an unpaired electron.
4. Removal of an electron from a molecular orbital localized predominantly on Cl:
This option implies that the electron is removed specifically from an orbital mostly associated with chlorine. While it could happen depending on the specific energy levels, molecular orbital diagrams for diatomic halogen compounds often involve interaction of orbitals across both atoms rather than one element exclusively.
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