I thought that when compounds have similar forces and charges, the one with the higher molecular weight has the higher melting point. If that is true, then why is the melting point of $\ce{KBr}$ higher than that of $\ce{CsCl}$?
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2$\begingroup$ Why, the charges are indeed similar, but the distances between them are not. Hence the forces are different, too. $\endgroup$– Ivan NeretinSep 8, 2016 at 21:17
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1$\begingroup$ see this:-chemistry.stackexchange.com/questions/17064/… $\endgroup$– Nilay GhoshSep 10, 2016 at 9:59
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1 Answer
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It may be helpful considering molecular weight for say $\ce{KBr}$ vs $\ce{KCl}$ or $\ce{CsCl}$ vs $\ce{CsBr}$, and actually melting point would go down with increasing molecular weight, but it actually has nothing to do with molecular weight despite the trend. It would be more helpful to look at lattice energy, electronegativity, electron affinity, and atomic radii, but its a little more complicated to compare a $\ce{Cs}$ salt of a halide with a $\ce{K}$ salt of a different halide as you've changed more factors. Although chlorine has a higher electronegativity and smaller atomic radius than bromine, caesium has an even larger atomic radius than potassium (relative to the size difference between chlorine and bromine) as well as a lower electronegativity than potassium. All of these factors will affect the lattice energy and therefore the melting points. But, like I said, when comparing two similar salts, make sure one of the elements stays constant. Its much more tricky to compare $\ce{KBr}$ with $\ce{CsCl}$ than it is to compare ($\ce{KBr}$ with $\ce{KCl}$) or ($\ce{CsBr}$ with $\ce{CsCl}$), or even ($\ce{KBr}$ with $\ce{CsBr}$) or ($\ce{KCl}$ with $\ce{CsCl}$). When comparing $\ce{KBr}$ with $\ce{CsCl}$, you must determine what will affect melting point more: the ($\ce{Cs}$ vs $\ce{K}$) difference or the ($\ce{Cl}$ vs $\ce{Br}$) difference.
