The bond dissociation enthalpy of $$X_{2}\Delta H_{bond}$$ calculated from the given data is_____$$kJmol^{-1}$$.(Nearest integer)
$$M^{+}X^{-}(s)\rightarrow M^{+}(g)+X^{-}(g)\Delta H_{lattice}^{*}=800kJmol^{-1}\\M(s)\rightarrow M(g)\Delta H_{sub}^{\circ}=100kJmol^{-1}\\M(g)\rightarrow M^{+}(g)+e^{-}(g)\Delta H_{i}=500kJmol^{-1}X(g)+e^{-}(g)\rightarrow X^{-}(g)\Delta H_{eg}^{*}=-300kJmol^{-1}\\M(g)+\frac{1}{2}X_{2}(g)\rightarrow M^{+}X^{-}(s)\Delta H_{f}^{\circ}=-400kJmol^{-1}$$
[Given : $$M^{+}X^{-}$$ is a pure ionic compound and X forms a diatomic molecule $$X_{2}$$ in gaseous state]

To determine the bond dissociation enthalpy of $$X_2$$, we apply the Born-Haber cycle to the formation reaction of the ionic solid from its elements. The relevant enthalpy changes are as follows: the lattice enthalpy for the process $$M^+X^-(s) \rightarrow M^+(g) + X^-(g)$$ is $$\Delta H_{lattice} = 800$$ kJ/mol; the sublimation enthalpy for $$M(s) \rightarrow M(g)$$ is $$\Delta H_{sub} = 100$$ kJ/mol; the ionization enthalpy for $$M(g) \rightarrow M^+(g) + e^-$$ is $$\Delta H_i = 500$$ kJ/mol; the electron gain enthalpy for $$X(g) + e^- \rightarrow X^-(g)$$ is $$\Delta H_{eg} = -300$$ kJ/mol; and the standard enthalpy of formation $$M(s) + \frac{1}{2}X_2(g) \rightarrow M^+X^-(s)$$ is $$\Delta H_f = -400$$ kJ/mol.

The formation of the ionic compound can be dissected into individual steps as follows:
Sublimation of metal: $$M(s) \rightarrow M(g),\;\Delta H_{sub} = +100$$ kJ/mol;
Dissociation of $$X_2$$: $$\frac{1}{2}X_2(g) \rightarrow X(g),\;\Delta H = \frac{1}{2}\Delta H_{bond}$$;
Ionization of metal: $$M(g) \rightarrow M^+(g) + e^-,\;\Delta H_i = +500$$ kJ/mol;
Electron gain by halogen: $$X(g) + e^- \rightarrow X^-(g),\;\Delta H_{eg} = -300$$ kJ/mol;
Formation of lattice: $$M^+(g) + X^-(g) \rightarrow M^+X^-(s),\;\Delta H = -\Delta H_{lattice} = -800$$ kJ/mol.

By Hess's law, the sum of these steps equals the enthalpy of formation:

$$\Delta H_f = \Delta H_{sub} + \frac{1}{2}\Delta H_{bond} + \Delta H_i + \Delta H_{eg} - \Delta H_{lattice}$$

Substituting the given values into this expression gives:

$$-400 = 100 + \frac{1}{2}\Delta H_{bond} + 500 + (-300) - 800$$
$$-400 = 100 + \frac{1}{2}\Delta H_{bond} + 500 - 300 - 800$$
$$-400 = -500 + \frac{1}{2}\Delta H_{bond}$$
$$\frac{1}{2}\Delta H_{bond} = -400 + 500 = 100$$
$$\Delta H_{bond} = 200 \text{ kJ/mol}$$

Therefore, the bond dissociation enthalpy of $$X_2$$ is 200 kJ/mol.