The relative electronegativity of iodine is 2.5, and its ionization potential is 10.45 V. Determine the electron affinity of iodine.

To determine the electron affinity of iodine using its electronegativity and ionization potential, we can apply the following relationship: \[ \text{Electron Affinity (EA)} = \text{Electronegativity (EN)} + \text{Ionization Potential (IP)} \] However, this specific relationship holds well primarily for elements when considering how easily they can gain electrons to form negative ions, and it's also applied generally to predict ionic behavior in compounds. Given: - Electronegativity (EN) of iodine = 2.5 - Ionization Potential (IP) of iodine = 10.45 V First, we need to ensure that the values we use are in consistent units. The electronegativity can be treated in relative terms and might not convert directly to energy units like voltage (eV). So here we will convert the ionization potential from volts to eV directly in order to assess electron affinity properly. Assuming that the effective units are consistent, let's rearrange the terms to find the electron affinity: \[ \text{EA} = 2.5 + 10.45 \, \text{eV} \] However, it's important to note that electronegativity itself does not have the same units as electron affinity. Therefore, it's more rigorous to conclude that although we can describe the general trend of electronegativity influencing how easily an atom acquires an electron, the standard direct addition isn’t fully compliant with the units. Instead, we can look for experimental values. For iodine: The known electron affinity of iodine is approximately **-3.06 eV**. This discrepancy from the intended calculation may arise from the simplified model used, but in practice, the electron affinity signifies the energy change that occurs when an electron is added to a neutral atom to form a negative ion. Thus, the electron affinity of iodine is best taken from experimental data as approximately **-3.06 eV**.