# Q43. Ans : option b. How? Q43. Which of the following formula has involved all the energy terms used to calculated  ($∆$Hsb: Sublimation energy ; LE1 : First ionisation energy; LE2 : Second ionisation energy ; B.D.E, ; Bond dissosiation energy ; E. G.E1: First electron gain enthalpy ; E.G.E2 : Second electron gain enthalpy ; U : Lattice energy)           (A) + 2$∆$Hsb + LE1 + LE2 +            (B) + 2$∆$Hsb +2LE1 +            (C) + 2$∆$Hsb +2LE1  + $+\frac{\mathrm{B}.\mathrm{D}.\mathrm{E}}{2}+2\mathrm{E}.\mathrm{G}.{\mathrm{E}}_{1}+\mathrm{U}$           (D) + 2$∆$Hsb + LE1 +

Reaction of formation: 2Na(g) + 0.5O2(g) → Na2O(s)
Note that 0.5O2 represents half mole of O2, and not half molecule, wihch is meaningless.

Now, Na, which is a solid has to be sublimated for the reaction to take place; there are 2 moles of Na: 2∆HSb
Now, Na has to be ionized once to form Na+; there are 2 moles of Na: 2IE1
Now, O2 has to be broken into 2O, that is, the O=O bond has to be broken (dissociated); there are 0.5 moles of O2: (1/2) BDE
Note that when 0.5 moles O2 is dissociated, we will have 0.5 * 2= 1 mole of O.
Now, 2 electrons have to be added successively (one after the other) to O to form O2-; there is 1 mole of O: 1(EGE1 + EGE2)
Now, Na2O, which is formed in gaseuos phase has to be convereted into a solid lattice; there is 1 mole of Na2O: 1(U)

Hence, total ∆Hf= 2∆Hsb + 2IE1 + (1/2)BDE + EGE1 + EGE2 + U
Hence, correct option is (B).

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