Which of the following explains why the entropy change is greater for the dissolution of nai compared to the dissolution of nabr?

a. iodide has weaker ion-dipole interactions with water than bromide.
b. the interactions between bromide ions with other bromide ions is stronger than the interactions between iodide ions with other iodide ions.
c. the more negative change in enthalpy observed with nai implies greater dissociation and hence greater entropy.
d. the bromide ion has more negative charge than the iodide ion. therefore, because of the greater charge, it forms a stronger ion-dipole network with water.
e. the cation forms stronger ion-dipole networks with water in nabr than nai because of the weaker bond to br.

Iodide has weaker ion-dipole interactions with water than bromide.

Explanation:

The electronegativity difference between sodium and iodine in sodium iodide is about 1.73. This shows that the compound is not composed of purely ionic bonds. Electro negativity decreases down the group hence iodine is far less electronegative than bromine and is thus ineffective in forming strong dipole interactions with water hence the higher entropy due to much less association of ions in solution.

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.