JEE Advanced levelSingle correct optionDifficulty 3 of 3pt158
Use ΔiH1=738 and ΔiH2=1451 for Mg, and electron gain enthalpies −141 (first) and +744 (second) for O, all in kJmol−1. The enthalpy change for Mg(g)+O(g)→MgX2+(g)+OX2−(g) is
A+2792kJmol−1
B+2189kJmol−1
C+1998kJmol−1
D+1586kJmol−1
Answer: A
Ionisation of Mg: 738 + 1451 = 2189. Formation of OX2−: -141 + 744 = +603. Total = 2189 + 603 = +2792 kJ/mol. Taking only the first gain gives 2048 and a wrong sign for the second gives 1586.