So, as $r_c+r_a$ is greater for $RbBr$ than for $NaF$ so $NaF$ have higher melting point.
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(image) $\xrightarrow{{B{r_2}/{H_2}O}}$ product
| $List-I$ | $List-II$ |
| $A$ $\left( NH _{4}\right)_{2} Cr _{2} O _{7} \stackrel{\Delta}{\longrightarrow}$ | $I$ $H _{2}$ |
| $B$ $KMnO _{4}+ HCl \rightarrow$ | $II$ $N _{2}$ |
| $C$ $Al + NaOH + H _{2} O \rightarrow$ | $III$ ${ O _{2}}$ |
| $D$ $NaNO _{3} \stackrel{\Delta}{\longrightarrow}$ | $IV$ ${ Cl _{2}}$ |
Choose the correct answer from the options given below.
${N_2} + {O_2}$ $\rightleftharpoons$ $2NO - 43,200\,kcal$