Rank the following ionic compounds by lattice energy. rank from highest to lowest lattice energy.

a) Rank the following ionic compounds by the magnitude of their
lattice energy. Rank from highest to lowest magnitude of lattice
energy. Highest magnitude to Lowest Magnitude LiCl, MgO, Na2O, BeO,
Na2s b) Given the following thermodynamic data, calculate the
lattice energy of CaBr2(s). Term Value (kJ/mol) ΔH∘f[CaBr2(s)] -675
ΔH∘f[Ca(g)] 179 ΔH∘f[Br(g)] 112 I1(Ca) 590. I2(Ca) 1145 E(Br) -325
Express your answer as an integer, and include the appropriate
units.

Answer

General guidance

Concepts and reason
The problem is based on the concept of lattice energy. Lattice energy is defined as the amount of energy released when gaseous ions combine to form ionic solids.

Fundamentals

Lattice energy is the energy released when a crystal lattice is formed. It can also be defined as the energy required to break the crystal lattice into gaseous ions. Usually, it is the energy released so it is taken as negative. Lattice energy depends on the distance between the ions and charge on the ions. As the ionic charge increases, the lattice energy also increases and becomes more negative. And as the ions get closer, then also there is an increase in lattice energy.

Step-by-step

Step 1 of 2

(A)

Lattice energy depends on the distance between ions and charge on ions.

Depending on this condition the order is as follows:

BeO > MgO > Na,O > Na S> Lici

Part A

The decreasing order of the ionic compounds by the magnitude of their lattice energies is as follows:

BeO > MgO > Na,O > Na S> Lici

Since Beryllium has the smallest ionic radius amongst the given ionic compounds, its lattice energy is the highest. Lithium ion and chloride ion carry the lowest charge; therefore, the lattice energy of lithium chloride is the lowest.

Step 2 of 2

(2)

Calculate the lattice energy as follows:

H,(CaBr) = Datomisation(Ca)+1, (Ca)+1,(Ca)+2x Datomisation(Bry)+2xEA(Br)
+LE(CaBr)

Here, and represent the ionization energy, EA represents the electron affinity, LE represents lattice energy.

Substituting the values in the above equation:

-675 kJmol =179 kJmol- +590 kJmol- +1145 kJmol + (2x112) kJmol
+(2x-325) kJmol- +LE(CaBr)
LE(CaBr) = -675 kJmol--179

Therefore, the lattice energy of CaBr is -2163 kJmol.

Part 2

The lattice energy of CaBr is -2163 kJmol.

Enthalpy of formation is the sum of all the enthalpies that includes enthalpy of atomization, electron affinity and lattice energy.

Answer

Part A

The decreasing order of the ionic compounds by the magnitude of their lattice energies is as follows:

BeO > MgO > Na,O > Na S> Lici

Part 2

The lattice energy of CaBr is -2163 kJmol.

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