Explain the following:
a. The formation of `Cs_(2)O` from its element is less exothermic than the formation of `ZnO` from its element.
b. On the basis of appropriate Born-Haber cycle, state what factor `(s)` is (are)responsible for the fact that lithium nitride `(Li_(3)N)` is more stable while potassium nitride `(K_(3)N)` is unstable.
c. Why `Al^(3+)` is the only stable oxidation state of `Al` in its compounds while `TI` has `+1` and `+3` oxidation states?
d. `Pb^(4+)` is a powerful oxidising agent. What is the reducing ability of `Pb^(2+)`
e. Which is more soluble in water `LiI` or `KI`?

Explanation

a. The lattice energy of `ZnO` must be greater since it is the major contribution to the exhothermic nature of each reaction.
b. The lattice energy of `Li_(3)N` is higher than that of `K_(3)N`, due to the large difference in ionic size between `Li^(o=)` and `K^(o+)` inons.
Note: Comparable size of the ions forms the strongest lattice, so size of `Li^(o+)` and `N^(3-)` are comparable.
The difference in `IE`, sublimation energy and other factor of the Born-Haber cycle are relatively small between the two compounds.
c. `TI` has an inert pair of electron `(6s^(2)), Al` has no corresponding inert pair.
d. Due to the inert pair effect `Pb^(2+)` is more stable than `Pb^(4+)`. Therefore, `Pb^(4+)` can undergo reduction easily and acts as powerful oxidising agnet.
` Pb^(4+)+2e^(-)rarrPb^(2+)`(reduction and acts as oxidising agent)
Therefore, `Pb^(2+)` cannot undergo oxidation easily and hence acts as a relatively poor reducing agent.
`P b^(2+) cancel to Pb^(4+) +2e` [does not undergo oxidation]
e. `LiI`. According to Fajans'rule, `LiI` is less ionic than `KI` and therefore `KI` shold be more soluble in `H_(2)O` than `LiI`.
But the lattice energy of `LiI` ( small cation and large anion) is less than that of `KI`, due to comparable sizes of `K^(o+)` and `I^(o+)` ions. Thus the energy required to break up the lattice of `LiI` is lower and is easily provided by hydration energy. So, `LiI` is more soluble in `H_(2)O` than `KI`.