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 crystallisation of
BaSO₄(s) = –4.5 kcal/mol, standard heat of formation of BaSO₄(s) = – 349 kcal/mol]

The correct option is – 128.5.

Step-by-step Explanation:

We are given the following thermodynamic values:
1. Standard heat of formation of ${\text{SO}}_4^{2-}\left(\text{aq}\right)$:
$\Delta H_{\text{f}}({\text{SO}}_4^{2-},\text{aq})=-216\text{kcal/mol}$

2. Standard heat of crystallisation of ${\text{BaSO}}_4\left(\text{s}\right)$:
This corresponds to the reaction where gaseous/aqueous ions form the solid crystal:
${\text{Ba}}^{2+}\left(\text{aq}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)\to {\text{BaSO}}_4\left(\text{s}\right)$
The enthalpy change for this crystallization process is:
$\Delta H_{\text{cryst}}=-4.5\text{kcal/mol}$

3. Standard heat of formation of ${\text{BaSO}}_4\left(\text{s}\right)$:
$\Delta H_{\text{f}}({\text{BaSO}}_4,\text{s})=-349\text{kcal/mol}$

According to Hess's Law, the enthalpy change of the crystallization reaction can be expressed in terms of the standard enthalpies of formation of the products and reactants:
$\Delta H_{\text{cryst}}=\Delta H_{\text{f}}({\text{BaSO}}_4,\text{s})-\left[\Delta H_{\text{f}}\right({\text{Ba}}^{2+},\text{aq})+\Delta H_{\text{f}}({\text{SO}}_4^{2-},\text{aq}\left)\right]$

Substituting the given values into the equation:
$-4.5=-349-\left[\Delta H_{\text{f}}\right({\text{Ba}}^{2+})+(-216\left)\right]$

Rearranging the equation to solve for $\Delta H_{\text{f}}\left({\text{Ba}}^{2+}\right)$:
$-4.5=-349-\Delta H_{\text{f}}\left({\text{Ba}}^{2+}\right)+216$
$-4.5=-133-\Delta H_{\text{f}}\left({\text{Ba}}^{2+}\right)$
$\Delta H_{\text{f}}\left({\text{Ba}}^{2+}\right)=-133+4.5$
$\Delta H_{\text{f}}\left({\text{Ba}}^{2+}\right)=-128.5\text{kcal/mol}$

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