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Question

Calculate the electron affinity of oxygen atom in gaining two electrons to give the oxide ion O2 from the following data.
Standard heat of formation of MgO(s)=600 kJ mol1
Lattice energy of MgO(s)=3860 kJ mol1
lonization energy of Mg(g) to give Mg2+ = +2170 kJ mol1
Dissociation energy of O2(g) = +494 kJ mol1
Heat of sublimation of Mg(s) = +150 kJ mol1

A
493 kJ mol1
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B
+493 kJ mol1
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C
693 kJ mol1
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D
+693 kJ mol1
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Solution

The correct option is D +693 kJ mol1
Given, ΔfHMgO=600 kJmol1
Mg(s)Mg(g) ΔsH=+150 kJmol1
Mg(g)Mg2+(g) I.E.=+2170 kJmol1
12O2(g)O(g) B.E.=+494 kJmol1 Lattice energy of MgO(s)=3860 kJmol1
ΔfHMgO=ΔsH+I.E.+12B.E.+E.A.+ Lattice energy
600=150+2170+247+E.A.3860
E.A.=6001502170247+3860
EA=+693 kJmol1

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