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

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

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

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 $~1,304^{o}C$.