Question

The stability of ${{K}_{2}}O,{{K}_{2}}{{O}_{2}}$ and $K{{O}_{2}}$ is inorder ${{\operatorname{K}}_{2}}O\,<\,{{K}_{2}}{{O}_{2}}\,<\,K{{O}_{2}}$. The increasing stability at the size of metal ion increase is due to stabilization of:
A) Large cations by smaller anions through lattice energy effects
B) Larger cations by larger anions through lattice energy effect.
C) Smaller cations by smaller anions through m.p
D) Smaller cations by larger anions through m.p

Answer 1

Hint The lattice energy of the molecule is proportional to the charge the metal carries. As we move down the group the size of the atom increases.

Complete step by step solution:
First let’s check the data in the question, in the question the increasing order of stability of various oxides of potassium is provided. They have asked to give the factor that helps the stabilization order of oxides of Potassium. In the question the potassium oxide, potassium superoxide and potassium peroxide is give and we know the formulae of,
Oxide ion -${{O}^{2-}}$, which contains one oxygen atom.
Peroxide ion-$O_{2}^{2-}$, which contains 2 oxygen atoms and is much bigger in size than oxide ion.
Superoxide ion -$O_{2}^{-}$, which contains 2 oxygen atoms and a negative charge and is larger than ${{\operatorname{O}}^{2-}}\,<\,O_{2}^{2-}\,<\,O_{2}^{-}$ peroxide.
Here superoxide and peroxide require a larger cation to form a compound so that it gets stability through lattice energy effects. The ions having equivalent size to superoxide and peroxide is of s block elements i.e. Alkali metals ions and alkaline metal ions, so they only form the superoxide and peroxides.
So the increasing order of size of oxide ions are:
${{\operatorname{O}}^{2-}}\,<\,O_{2}^{2-}\,<\,O_{2}^{-}$
In the three given oxides, the metal ion, cation is same i.e. ${{K}^{+}}$. Only there is change in the anions.
One of the main factors that stabilizes compounds like oxides of potassium is the lattice energy effect. Lattice energy is the amount of energy released during the bond formation or we can say that it is the energy required to break the bonds and break the ionic compound to its constituents.
Lattice energy mainly depends on charge and size of the ions.
As the size of ions increases, the attraction between nuclei and electrons will be less and hence attraction is less and the lattice energy released decreases. As lattice energy decreases, stability also decreases.
For the above compounds the cations and anions are large enough and their stability is mainly depending on the lattice energy effect.

So the correct answer from the options given is Option (B).

Note: Difference between the superoxide ion and peroxide ion ,even though both contain 2 oxygen atoms, peroxide has -2 charge for 2 oxygen whereas for superoxide there is only one negative charge for 2 oxygen atoms. Lattice energy increases with increase in charge and hence the stability increases.