Use $Δ_{i} H_{1} (Na) = 496 kJ mo l^{- 1}$ and the electron gain enthalpy of $Cl$ ( $- 349 kJ mo l^{- 1}$ ). The enthalpy change for $Na (g) + Cl (g) \to Na X^{+} (g) + Cl X^{-} (g)$ is

A $+ 845 kJ mo l^{- 1}$
B $- 147 kJ mo l^{- 1}$
C $- 845 kJ mo l^{- 1}$
D $+ 147 kJ mo l^{- 1}$

Answer: D

The process is ionisation of Na (+496) followed by gain of an electron by Cl (-349). Net = 496 - 349 = +147 kJ/mol, so it is endothermic; the ionic solid forms only because of lattice energy.

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Use Δi​H1​(Na)=496 kJ mol−1 and the electron gain enthalpy of Cl (−349 kJ mol−1). The enthalpy change for Na(g)+Cl(g)→NaX+(g)+ClX−(g) is

Answer: D

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Use $Δ_{i} H_{1} (Na) = 496 kJ mo l^{- 1}$ and the electron gain enthalpy of $Cl$ ( $- 349 kJ mo l^{- 1}$ ). The enthalpy change for $Na (g) + Cl (g) \to Na X^{+} (g) + Cl X^{-} (g)$ is