The standard heat of formation, in kcal/mol of ${\mathrm{Ba}}^{2+}$ is :

[Given : standard heat of formation of ${\mathrm{SO}}_4^{2-}$ ion (aq) = $-216$ kcal/mol, standard heat of crystallisation of $\mathrm{BaSO}\mathrm{₄}$ (s) = $-4.5$ kcal/mol, standard heat of formation of $\mathrm{BaSO}\mathrm{₄}$ (s) = $-349$ kcal/mol ]

Given thermodynamic data:

Δ_fH° ( ${\mathrm{SO}}_4^{2-}$ (aq) ) = -216 kcal·mol⁻¹

Δ_cH° ( BaSO₄ (s) ) = -4.5 kcal·mol⁻¹

Δ_fH° ( BaSO₄ (s) ) = -349 kcal·mol⁻¹

Thermodynamic cycle for forming solid barium sulfate from its aqueous ions:

$\Delta f_H⁰\left(\mathrm{BaSO}{4}_{}\right(s\left)\right)=\Delta f_H⁰\left({\mathrm{Ba}}^{2+}\right(\mathrm{aq}\left)\right)+\Delta f_H⁰\left({\mathrm{SO}}_4^{2-}\right(\mathrm{aq}\left)\right)+\Delta c_H⁰\left(\mathrm{BaSO}{4}_{}\right(s\left)\right)$

Re‑arranging to isolate the formation enthalpy of the barium ion:

$\Delta f_H⁰\left({\mathrm{Ba}}^{2+}\right(\mathrm{aq}\left)\right)=\Delta f_H⁰\left(\mathrm{BaSO}{4}_{}\right(s\left)\right)-\Delta f_H⁰\left({\mathrm{SO}}_4^{2-}\right(\mathrm{aq}\left)\right)-\Delta c_H⁰\left(\mathrm{BaSO}{4}_{}\right(s\left)\right)$

Substituting the numerical values:

$\Delta f_H⁰\left({\mathrm{Ba}}^{2+}\right(\mathrm{aq}\left)\right)=-349+-216--4.5$

Carrying out the arithmetic step‑by‑step:

-349 kcal·mol⁻¹ + 216 kcal·mol⁻¹ + 4.5 kcal·mol⁻¹ = -128.5 kcal·mol⁻¹

Therefore, the standard heat of formation of Ba²⁺ (aq) is -128.5 kcal·mol⁻¹, which matches the given correct option.