The principal reason that the melting point of $NaF$ is much higher than that of $RbBr$ is that

• A
  the two crystals are not isomorphs
• B
  the molar mass of $NaF$ is smaller than that of $RbBr$
• C
  the bond in $RbBr$ has more covalent character than the bond in $NaF$
• D
  the internuclear distance $(r^+ + r^-)$ is greater for $RbBr$ than for $NaF$