Assume that in each pair of compounds be- low the type of crystal structure is similar. Consider the strength of the attractive force
bonding the ions together in each compound.

In which case would the relative strengths be ranked correctly?

1. MgCl2 < CaBr2
2. LiCl > LiBr
3. KF < KCl
4. NaI > Li

Respuesta :

The answer is three or two I'm pretty sure

The attraction between ions in a crystal depends on the size of the ions. Since chloride ion is smaller than bromide ion, the statement LiCl > LiBr is correct.

The attraction between oppositely charged ions in the crystal lattice of an ionic compound has a lot to do with the size of the ions.

For a given crystal lattice, smaller ions implies greater attraction and higher lattice energy.

LiCl and LiBr contains the same cation which is the lithium ion so we turn our attention to the anions.

Chloride ion is smaller than bromide ion, the statement LiCl > LiBr is correct based on the strength of the attractive force  bonding the ions together in each compound .

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