Give the decreasing order of melting points of the following compounds:
i. NaF ii. BeO iii. MgO iv. SrO
Given: The inter-ionic distance in `Å` as
`NaF=2.31, BeO=1.65, MgO=2.106, SrO=2.58`
b. Given the decreasing order of hardness of the following compounds:
i. CaO ii. BeO iii. TiC
Given : The inter-ionic distances in `Å` as
`CaO=2.405, BaO=2.762, TiC=2.159`

Explanation

a. `U_(0) prop (|Z^(o+)||Z^(ɵ)|)/r_(0)`
i. Smaller the inter-ionic distance and high charges on ions, higher the lattice energy and high is the melting point.
ii. Increasing order of inert-ionic distance with same
`|Z^(+)|=2` and `|Z^(ɵ)|=2` is `overset(+2)(Be)overset(-2)(O) lt overset(+2)(Mg)overset(-2)(O)lt overset(+2)(Sr)overset(-2)(O)`.
iii. Charge in NaF is `|Z^(+)|=1, |Z^(ɵ)|=1`.
So, the decreasing order of melting point is
`BeO gt MgO gt SrO gt NaF (ii gt iii gt iv gt i)`
m.pt. `(.^(@)C) 2930 gt 2800 gt 2430 gt 990`
b. i. Higher the lattice energy, higher is the hardness.
Charge in `overset(+4)(Ti)overset(-4)(C)` is the highest, `|Z^(o+)|=4, |Z^(ɵ)|=4`
ii. Increasing order of inter-ionic distance with same charge, `|Z^(o+)|=2`, and `|Z^(ɵ)|=2` is `CaO lt BaO`.
So, the decreasing order of hardness is