Chemistry - PERIODIC TABLE Question with Solution | TestHub

The question provides first ionization energy (IE₁) data for several anions and asks for the electron gain enthalpy ($\Delta H_{eg}$) for a fluorine atom.

The given values are IE₁ for:

I⁻, O⁻, Br⁻, F⁻, Cl⁻, Na⁻, and O²⁻.

The values are: +53, –798, +141, +295, +325, +328, +349 kJ/mol.

The first ionization energy (IE₁) of an anion X⁻ is the energy required to remove an electron from X⁻ to form X.

$$X^{-}(g)\to X(g)+e^{-}$$

By definition, this process is the reverse of the electron gain by the neutral atom X to form X⁻.

$$X(g)+e^{-}\to X^{-}(g)$$

Therefore, the first ionization energy of an anion X⁻ is equal in magnitude but opposite in sign to the electron gain enthalpy ($\Delta H_{eg}$) of the neutral atom X.

$$I{E}_1(X^{-})=-\Delta H_{eg}(X)$$

We need to find $\Delta H_{eg}$ for a fluorine atom (F). This means we need to find IE₁ for F⁻.

The halogens (F, Cl, Br, I) have high electron affinities. Their electron gain enthalpies are highly negative.

Among the given anions, F⁻, Cl⁻, Br⁻, I⁻ are halide ions.

The IE₁ values for these anions will correspond to the negative of the electron gain enthalpy of the respective neutral halogen atoms.

The given values are:

+53 (I⁻)

+141 (Br⁻)

+295 (O⁻)

+325 (Cl⁻)

+328 (F⁻)

+349 (Na⁻)

–798 (O²⁻)

The value +328 kJ/mol corresponds to the IE₁ of F⁻.

Therefore, $\Delta H_{eg}$ for a fluorine atom is $-328$ kJ/mol.

Final Answer: -328 kJ/mol