The standard heat of formation, in kcal/mol of Ba²⁺ is: [Given : standard heat of formation of SO₄^2- ion (aq)= –216 kcal/mol, standard heat of crystallization of BaSO4(s) = –4.5 kcal/mol, standard heat of formation of BaSO4(s) = –349 kcal/mol]

The correct option is -128.5.

Let us solve the problem step-by-step using thermochemical principles.

First, let us write down the given thermochemical data:
1. Standard heat of formation of ${\text{SO}}_4^{2-}\left(\text{aq}\right)$:
${\Delta }_{\text{f}}{H}^{⦵}\left[{\text{SO}}_4^{2-}\right(\text{aq}\left)\right]=-216\text{kcal/mol}$
2. Standard heat of formation of ${\text{BaSO}}_4$:
${\Delta }_{\text{f}}{H}^{⦵}\left[{\text{BaSO}}_4\right]=-349\text{kcal/mol}$
3. Standard heat of crystallization of ${\text{BaSO}}_4$ from its constituent aqueous ions:
The crystallization process is represented by the following chemical equation:
${\text{Ba}}^{2+}\left(\text{aq}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)\to {\text{BaSO}}_4$
The standard enthalpy change for this crystallization process is:
${\Delta }_{\text{cryst}}{H}^{⦵}=-4.5\text{kcal/mol}$

According to the principles of thermochemistry, the enthalpy change of a reaction is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants:
${\Delta }_{\text{cryst}}{H}^{⦵}={\Delta }_{\text{f}}{H}^{⦵}\left[{\text{BaSO}}_4\right]-\left({\Delta }_{\text{f}}{H}^{⦵}\right[{\text{Ba}}^{2+}\left(\text{aq}\right)]+{\Delta }_{\text{f}}{H}^{⦵}[{\text{SO}}_4^{2-}\left(\text{aq}\right)\left]\right)$

Now, let us substitute the given values into the equation:
$-4.5=-349-\left({\Delta }_{\text{f}}{H}^{⦵}\right[{\text{Ba}}^{2+}\left(\text{aq}\right)]+(-216\left)\right)$

Rearranging the equation to solve for the standard heat of formation of ${\text{Ba}}^{2+}\left(\text{aq}\right)$:
${\Delta }_{\text{f}}{H}^{⦵}\left[{\text{Ba}}^{2+}\right(\text{aq}\left)\right]-216=-349-(-4.5)$
${\Delta }_{\text{f}}{H}^{⦵}\left[{\text{Ba}}^{2+}\right(\text{aq}\left)\right]-216=-349+4.5$
${\Delta }_{\text{f}}{H}^{⦵}\left[{\text{Ba}}^{2+}\right(\text{aq}\left)\right]-216=-344.5$
${\Delta }_{\text{f}}{H}^{⦵}\left[{\text{Ba}}^{2+}\right(\text{aq}\left)\right]=-344.5+216$
${\Delta }_{\text{f}}{H}^{⦵}\left[{\text{Ba}}^{2+}\right(\text{aq}\left)\right]=-128.5\text{kcal/mol}$

Thus, the standard heat of formation of ${\text{Ba}}^{2+}$ is -128.5 kcal/mol.