# Suggest why melting point of sodium iodide is lower than the melting point of sodium bromide [duplicate]

This question already has an answer here:

I would assume this has something to do with bonding and perhaps density, but I haven't the faintest where I'd begin.

## marked as duplicate by Karl, Mithoron, pentavalentcarbon, airhuff, Todd MinehardtNov 25 '17 at 23:54

• – Ivan Neretin Nov 25 '17 at 18:05
• The charges are further separated in the iodide, so ... – Karl Nov 25 '17 at 18:12

Reactivity decreases as you go down the halogen group. This is because when you go down the halogen group, you add another shell of electrons and increase the amount of electron shielding. So the valence electrons feel less of a pull from the nucleus, and are not tightly bound. Note that this is the same logic for why reactivity increases as you go down the alkali metal and alkaline earth metals group.

So, $Br$ is more reactive than $I$ with $Na$, because $Br$ is higher up the group (the atomic number of $Br$ is $35$, while the atomic number of $I$ is $53$). Since $Br$ is more readily reactive with $Na$ than $I$, the bond between $Na$ and $Br$ is stronger than the bond between $Na$ and $I$. Thus, the boiling point of $NaBr$ is higher than the boiling point of $NaI$ .

To check, the boiling point of sodium bromide ($NaBr$) is $~1,396^{o}C$, while the boiling point of sodium iodide ($NaI$) is $~1,304^{o}C$.