Question

The melting point of RbBr is ${682}^{\circ }C$, while that of NaF is ${988}^{\circ }C$. The principal reason that melting point of $NaF$ is much higher than that of $RbBr$ is that :

• A. the two crystals are not isomorphous
• B. the molar mass of $NaF$ is smaller than that of $RbBr$
• C. the inter-nuclear distance, $r_c+r_a$ is lesser for $NaF$ than for $RbBr$
• D. the bond in $RbBr$ has more covalent character than the bond in $NaF$

Answer 1

Answer: (C)

the inter-nuclear distance, $r_c+r_a$ is lesser for $NaF$ than for $RbBr$

The melting point of $RbBr$ is ${682}^{\circ }C$, while that of $\text{NaF}$ is ${988}^{\circ }C$. The principal reason that melting point of $\text{NaF}$ is much higher than that of $\text{RbBr}$ is that the inter-nuclear distance $r_c+r_a$ is lesser for $\text{NaF}$ than for $\text{RbBr}$. This leads to stronger Coulombic forces of attractions in $\text{NaF}$.

Hence, the correct option is $\text{C}$