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Question

Z2O3(s) is the ionic compound formed from oxygen and a metal with the form Z(s) at 1.00 bar and 298 K.

(0.5 points) (a) Draw the Lewis structure for Z2O3. Assume that all the valence electrons from Z are required.

(4.5 points) (b) Use the following information to determine the enthalpy of formation for Z2O3(s). Express your answer in kJ/(mol Z2O3(s)). 

Lattice energy for Z2O3(s) = -15100 kJ mol-1

ΔHsub for Z(s) = 485 kJ mol-1

First ionization energy for Z(g) = 590 kJ mol-1

Second ionization energy for Z(g) = 1710 kJ mol-1

Third ionization energy for Z(g) = 3948 kJ mol-1

Bond dissociation energy for O2(g) = 498 kJ mol-1

First electron affinity for O = -141 kJ mol-1

Second electron affinity for O = 744 kJ mol-1

  

As part of your problem work you must show the full equations with phases for all steps in the Born-Haber process and you must label each step.

Answer 1

Z₂O₃ suggests that Z has 3 valence electrons (3+ ion) similar to aluminum (Al).

Can't really draw Lewis structures on this Wyzant platform, but I'll describe it so you can draw it.

We have 2 Z atoms each with 3 electrons which they will DONATE to the 3 oxygen atoms. Each O atom starts with 6 valence electrons. We end up with 2 Z atoms with no valence electrons and a 3+ charge, and 3 O atoms each with eight valence electrons and a charge of 2-.

[O]^2- [Z]³⁺ (put 4 pairs of electrons around the O)

[O]^2- (put 4 pairs of electrons around the O)

[O]^2- [Z]³⁺ (put 4 pairs of electrons around the O)

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Born-Haber process

∆H_formation = Hsub + IE + Dissoc E + EA + LE

2 Z(s) --> 2Z(g) ∆Hsub = 2x485 = 970 kJ SUBLIMATION

2Z(g) --> 2Z(g)⁺ IE1 =2x590 = 1180 kJ FIRST IONIZATION ENERGY

2Z(g)⁺ --> 2Z(g)²⁺ IE2 = 2x1710 = 3420 kJ SECOND IONIZATION ENERGY

2Z(g)²⁺ --> 2Z(g)³⁺ IE3 = 2x3948 = 7896 kJ THIRD IONIZATION ENERGY

1 1/2 O₂(g) --> 3O(g) dissoc. energy = 1.5x498 = 747 kJ BOND DISSOCIATION ENERGY

3O(g) --> 3O^- EA₁ = 3x-141 = -423 kJ FIRST ELECTRON AFFINITY

3O(g)- --> 3O(g)^2- EA₂ = 3x-744 = -2232 kJ SECOND ELECTRON AFFINITY

2Z(g)³⁺ + 3O(g)^2- --> Z₂O₃(s) LE = -15,100 LATTICE ENERGY

∆H_formation = Hsub + IE + Dissoc E + EA + LE

∆H_formation = 970 + 1180 + 3420 + 7986 + 747 -423 -2232 - 15,100

∆H_formation = -3452 kJ/mol