The standard enthalpy of neutralisation for the reaction between a strong acid and a strong base in dilute aqueous solution is approximately –57.33 kJ mol⁻¹, representing the formation of one mole of water from $H^{+}(aq)$ and $OH^{-}(aq)$ ions [Source knowledge]. However, for the reaction involving a solid basic oxide like MgO, we apply Hess's Law using the standard enthalpies of formation ($\Delta_{f}H^{\circ}$) found in the sources:

1. Identify Enthalpy Values (at 298 K): $\Delta_{f}H^{\circ} [Mg^{2+}(aq)] = -466.85 \text{ kJ mol}^{-1}$ $\Delta_{f}H^{\circ} [H_{2}O(l)] = -285.83 \text{ kJ mol}^{-1}$ $\Delta_{f}H^{\circ} [MgO(s)] = -601.70 \text{ kJ mol}^{-1}$ $\Delta_{f}H^{\circ} [H^{+}(aq)] = 0 \text{ kJ mol}^{-1}$ (by convention)

2. Calculate Reaction Enthalpy ($\Delta_{r}H^{\circ}$): The net ionic equation is $MgO(s) + 2H^{+}(aq) \rightarrow Mg^{2+}(aq) + H_{2}O(l)$. $\Delta_{r}H^{\circ} = [\Delta_{f}H^{\circ}(Mg^{2+}, aq) + \Delta_{f}H^{\circ}(H_{2}O, l)] - [\Delta_{f}H^{\circ}(MgO, s) + 2 \times \Delta_{f}H^{\circ}(H^{+}, aq)]$ $\Delta_{r}H^{\circ} = [(-466.85) + (-285.83)] - [(-601.70) + 0]$ $\Delta_{r}H^{\circ} = -752.68 - (-601.70) = -150.98 \text{ kJ mol}^{-1}$.

Since –150.98 kJ mol⁻¹ is numerically less than –57.33 kJ mol⁻¹ (more exothermic), the correct option is A.