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3 years ago

Arrange the following substances in order of decreasing magnitude of lattice energy.

2 answers:

5 0

The formula is F = ( q1 * q2 ) / r ^ 2
where: q is the individual charges of each ion 
r is the distance between the nuclei 
The formula is not important but to explain the relationship between the atoms in the compounds and their lattice energy. 

From the formula we can first conclude that compounds of ions with greater charges will have a greater lattice energy. This is a direct relationship. 
For example, the compounds BaO and SrO, whose ions' charges are ( + 2 ) and ( - 2 ) respectively for each, will have greater lattice energies that the compounds NaF and KCl, whose ions' charges are ( + 1 ) and ( - 1 ) respectively for each. 

So Far: ( BaO and SrO ) > ( NaF and KCl ) 

The second part required you find the relative distance between the atoms of the compounds. Really, the lattice energy is stronger with smaller atoms, an indirect relationship. 
For example, in NaF the ions are smaller than the ions in KCl so it has a greater lattice energy. Because Sr is smaller than Ba, SrO has a greater lattice energy than BaO. 

Therefore: 
Answer: SrO > BaO > NaF > KCl

qwelly [4]3 years ago

3 0

The following substances in order of decreasing magnitude of lattice energy are as follows: SrO > BaO > NaF > KCl

<h3>Explanation: </h3>

Ionic compounds are more stable because their have an electrostatic force between the two opposite ions. After the formation of ions, they combine together to form the ionic compound. The lattice energy or lattice enthalpy is the energy released in this process. It means the energy released when a cation and a anion combine together to form one mole of an ionic compound that known as the lattice energy or lattice enthalpy.

To calculation of lattice energy we can done by using Hess’s law (or called Born Haber cycle). Hess' law of constant or the heat summation, also known as Hess' law is a relationship in physical chemistry that named after Germain Hess. Hess's Law states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.

The formula is $F = \frac{ ( q1 * q2 ) }{ r ^ 2 }$

The compounds BaO and SrO, have ions' charges ( + 2 ) and ( - 2 ) respectively will have greater lattice energies that the compounds NaF and KCl that have charges ( + 1 ) and ( - 1 ) respectively

( BaO and SrO ) > ( NaF and KCl )

In NaF, the ions are smaller than the KCl ions so it has the greater lattice energy. Sr is smaller than Ba therefore SrO has a greater lattice energy than BaO.

Therefore SrO > BaO > NaF > KCl

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