The melting point of \( \mathrm{RbBr} \) is \( 682^{\circ} \mathrm{...
The melting point of \( \mathrm{RbBr} \) is \( 682^{\circ} \mathrm{C} \) while that of \( \mathrm{NaF} \) is \( 988^{\circ} \mathrm{C} \). The principal reason for this fact is :
\( \mathrm{P} \)
(a) the molar mass of \( \mathrm{NaF} \) is smaller than that of \( \mathrm{RbBr} \)
\( \mathrm{W} \)
(b) the bond in \( \mathrm{RbBr} \) has more covalent character than the bond in \( \mathrm{NaF} \)
(c) the difference in electronegativity between \( \mathrm{Rb} \) and \( \mathrm{Br} \) is smaller than the difference between \( \mathrm{Na} \) and \( \mathrm{F} \)
(d) the internuclear distance, \( \mathrm{r}_{\mathrm{c}}+\mathrm{r}_{\mathrm{a}} \) is greater for \( \mathrm{RbBr} \) than for \( \mathrm{NaF} \)
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