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For a reaction $A (g) + B (g) \to A B (s)$ the enthalpy change for the reaction will be (assuming AB(s) to be an ionic compound) if $I. E_{A} + E \cdot A_{A} < I \cdot E_{B} + E.A A_{B}$

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Δ H_{LEA} + Δ H_{e_{E_{A}}} + Δ H_{{LE}_{A B (0)}}

Δ H_{e g_{A}} + Δ H_{{LE}_{B}} + Δ H_{{LE}_{A B (0)}}

I. E_{A} - E \cdot A_{B} - Δ H_{LE} (E_{A B (s)}

I. E_{A} + E \cdot A_{B} + Δ H_{LE} (E_{A B (s)}

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detailed solution

Correct option is C

As $I. E_{A} + E \cdot A_{A} < I \cdot E_{B} + E.A A_{B}$ the compound AB(s) will be $A^{+} B^{-} (s)$ and not $A^{-} B^{+} (s)$

From Hess's Law

$Δ H = Δ H_{LE} + Δ H_{eg} + Δ H_{LE . (s)}$ .

Hence, option C is correct.

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For a reaction A(g)+B(g)→AB(s) the enthalpy change for the reaction will be (assuming AB(s) to be an ionic compound) if I. EA+E⋅AA<I⋅EB+ E.A AB