According to Born-Haber's cycle, the enthalpy of formation of ionic compounds can be determined. The formation of NaCl involves following steps – (i) Na(s)S−−−−−−−−−−−−−→Energy of subtimationNa(g) (ii) Na(g)I−−−−−−−−−−−→Ionisation potenitalNa+(g)+e− (iii) Cl2(g)D−−−−−−−−−−−−−−−→Bond dissociation energy2Cl(g) (iv) Cl(g)+e−E−−−−−−−−−−→Electron - affinityCl−(g) (v) Cl−(g)+Na+(g)U−−−−−−−−→Lattice energyNaCl(s) The enthalpy of formation of NaCl will be -

According to Born-Haber's cycle, the enthalpy of formation of ionic co

1.	According to Born-Haber's cycle, the enthalpy of formation of ionic compounds can be determined. The formation of NaCl involves following steps – (i) Na(s)S−−−−−−−−−−−−−→Energy of subtimationNa(g) (ii) Na(g)I−−−−−−−−−−−→Ionisation potenitalNa+(g)+e− (iii) Cl2(g)D−−−−−−−−−−−−−−−→Bond dissociation energy2Cl(g) (iv) Cl(g)+e−E−−−−−−−−−−→Electron - affinityCl−(g) (v) Cl−(g)+Na+(g)U−−−−−−−−→Lattice energyNaCl(s) The enthalpy of formation of NaCl will be -
Answer» According to Born-Haber's cycle, the enthalpy of formation of ionic compounds can be determined. The formation of NaCl involves following steps – (i) $N a (s) S - ------------ \to Energy of subtimation N a (g)$ (ii) $N a (g) I - ---------- \to Ionisation potenital N a^{+} (g) + e^{-}$ (iii) $C l_{2} (g) D - -------------- \to Bond dissociation energy 2 C l (g)$ (iv) $C l (g) + e^{-} E - --------- \to Electron - affinity C l^{-} (g)$ (v) $C l^{-} (g) + N a^{+} (g) U - ------- \to Lattice energy N a C l (s)$ The enthalpy of formation of NaCl will be -