Use Delta_iH_1(Mg) = 738, Delta_iH_2(Mg) = 1451 and Delta_egH(Cl) = -349 kJ mol^-1. The enthalpy change for Mg(g) + 2 Cl(g) -> Mg^2+(g) + 2 Cl-(g) is

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JEE Advanced levelSingle correct optionDifficulty 3 of 3pt157

Use $Δ_{i} H_{1} (Mg) = 738$ , $Δ_{i} H_{2} (Mg) = 1451$ and $Δ_{e g} H (Cl) = - 349$ $kJ mo l^{- 1}$ . The enthalpy change for $Mg (g) + 2 Cl (g) \to Mg X^{2 +} (g) + 2 Cl X^{-} (g)$ is

A $+ 2189 kJ mo l^{- 1}$
B $+ 1491 kJ mo l^{- 1}$
C $+ 1840 kJ mo l^{- 1}$
D $+ 2887 kJ mo l^{- 1}$

Answer: B

Total ionisation of Mg = 738 + 1451 = 2189. Two Cl atoms release 2 x 349 = 698. Net = 2189 - 698 = +1491 kJ/mol. Using only one Cl gives 1840, omitting Cl gives 2189.

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Use Δi​H1​(Mg)=738, Δi​H2​(Mg)=1451 and Δeg​H(Cl)=−349 kJ mol−1. The enthalpy change for Mg(g)+2 Cl(g)→MgX2+(g)+2 ClX−(g) is

Answer: B

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Use $Δ_{i} H_{1} (Mg) = 738$ , $Δ_{i} H_{2} (Mg) = 1451$ and $Δ_{e g} H (Cl) = - 349$ $kJ mo l^{- 1}$ . The enthalpy change for $Mg (g) + 2 Cl (g) \to Mg X^{2 +} (g) + 2 Cl X^{-} (g)$ is