Home
Class 12
CHEMISTRY
The melting points RbBr is 682^(@)C, whi...

The melting points RbBr is `682^(@)C`, while that of NaF is `988^(@)C`. The principal reason of this fact is :

Text Solution

Verified by Experts

Both are crystalline solids and the melting points are directly linked to the respective lattice energies. Now, the magnitude of lattice energy is directly proporational to charge on ions and inversely to the interionic distance. In the above case both the cation shave +1 unit charge while anions have -1 unit charge. Since `Br^(-)` ions is bigger than `F^(-)` ion, the lattice energy for NaF crystals is more and therefore, the melting point is higher than of RbBr crystals.
Promotional Banner

Similar Questions

Explore conceptually related problems

The melting point of RbBr is 682^(@)C , while that of NaF is 988^(@)C . The principla reason that melting point of NaF is much higher than that of RbBr is that :

The melting point of MgBr_(2) is 700^(@)C while that of AlBr_(3) is only 97^(@)C . Give reason.

Melting of DNA at 70^(@)C is due to breakdown of

Why is the melting point of phosphorous higher than that of nitrogen while the melting point of Bi is lower than that of Sb?

Find the principal value of the following : cos e c^(-1)(2)

The melting point and boiling point of sulphur dioxide are -75^(@)C and -10^(@)C respectively. If the change in enthalpy of fusion and vaporisation of sulphur dioxide are 7.4 " kJ mol"^(-1) and 25.2 " kJ mol"^(-1) respectively, calculate the change in entropy of fusion and vaporisation of SO_(2) ?

The temperature of a piece of lead is 27^(@)C . Find out the minimum velocity of its impact with a wall so that it melts completely. Suppose 58% of the heat generated is dissipated. Given, J = 4.2 J cdot cal^(-1) , melting point, specific heat and latent heat of fusion of lead are 327^(@)C, 0.03 cal cdot g^(-1) cdot^(@)C^(-1) and 5 cal cdot g^(-1) , respectively.