[stevemont7]

Hello,

I had a homework question that asked whether NaCl or CaCl₂ has a higher melting point. I said NaCl, as I believed that the smaller radius of the Na⁺ ion would increase the attraction between the ions. However, CaCl₂ is the correct answer. I thought that this might be due to the increased magnitude of charge on the cation, but still, sodium chloride's melting point is 801°C according to Wikipedia and calcium chloride's is 772°C.

I'm confused.

[Borek]

However, CaCl₂ is the correct answer.

You have done the right thing - you have checked the primary data and you can be sure CaCl₂ is not the right answer. There is not much more than can be said.

[stevemont7]

However, CaCl₂ is the correct answer.

You have done the right thing - you have checked the primary data and you can be sure CaCl₂ is not the right answer. There is not much more than can be said.

Hi Borek, thank you for the response. I'm afraid I misrepresented the question.  It actually says, "For each of the following pairs of substances, predict which will have the higher melting point." Does magnitude of ionic charge influence the attraction between ions more than ionic radius? Because, if that is true, the attraction between ions would be greater in CaCl₂, and one could assume that the melting point should actually be higher too, right? Then CaCl₂ would be the correct answer.

[Borek]

This is tricky. Such predictions/comparisons work only when you assume "all other things kept constant". Between other things that means identical crystal structure and it is obvious structures of NaCl and CaCl₂ have to be different (because of different cation charges; NaCl structure is possible only when both anion and cation have the same charge so it occurs for many alkali halides and numerous sulfides and oxides of divalent metals), so the prediction based on just comparing the charges is guaranteed to be problematic.

Asking a question of "predict" type and expecting an answer that is incorrect is a rather poor (pedagogically) idea if you ask me.

[stevemont7]

I see. Thank you for your explanation, Borek.