The conversion of gaseous atoms `K` and `F` to `K^(o+)` and `F^(ɵ)` absorbs `0.85 eV` of energy. If the `IE` and `Delta_("eg")H^(ɵ)` of `K` and `F` have magnitudes in the ratio of `7:6`, what is the electron gain enthalpy `(Delta_("eg")H^(ɵ))` of fluorine ?

Explanation

To solve the problem, we need to find the electron gain enthalpy (ΔegH°) of fluorine given the following information:

1. The conversion of gaseous potassium (K) and fluorine (F) to their respective ions (K⁺ and F⁻) absorbs 0.85 eV of energy.
2. The ionization energy (IE) and electron gain enthalpy (ΔegH°) of potassium and fluorine are in the ratio of 7:6.

### Step-by-Step Solution:

**Step 1: Define the reactions and the energy change.**
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