The melting point of RbBr is 955 K while that of NaF is 1260 K.... (1) Melting point of NaF is much larger that that of RbBr...why?(it is mainly due to.. .)

1 Answer
Truong-Son N.
Feb 19, 2018

Due to the comparative ionic radii.

We know that #"Na"^(+)# is much smaller than #"Rb"^(+)#, since #"Na"# is much smaller than #"Rb"#. Similarly, #"F"^(-)# is much smaller than #"Br"^(-)#, since #"F"# is much smaller than #"Br"#.

Smaller ionic radii

#-># shorter internuclear bonding distance

#-># stronger ion-pairing interaction

#-># higher melting and boiling point

Their charges don't matter... they're the same charges.

Perhaps a more advanced explanation is that because #"Na"^(+)# and #"F"^(-)# are smaller, their electron density is more concentrated, making each of them a harder acid and base, respectively.

That means #"Na"^(+)# is more polarizing than #"Rb"^(+)# and #"F"^(-)# is more polarizing than #"Br"^(-)#... in a flow chart,

Smaller ion

#-># more concentrated electron density

#-># harder acid (cation) or base (anion)

#-># less polarizable and more polarizing

#-># more tightly-held ionic interaction is made

#-># higher melting and boiling point

That again leads to the conclusion that a stronger ion-pairing interaction is made and thus a higher melting and boiling point are achieved.

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