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.. .)

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.