Question

An example of a non-stoichiometric compound is:
(a) $PbO$
(b) $\text {Ni}{{\text{O}} _ {2}} $
(c) $\text{A}{{\text{l}} _ {2}} {{\text{O}} _ {3}} $
(d) $\text{F}{{\text{e}} _ {3}} {{\text{O}} _ {4}} $

Answer 1

Hint: Non- stoichiometric compounds consist of an unequal ratio of the elements in their compounds and from the above compounds, the one having an unequal ratio of elements in the compound would be a non-stoichiometric compound. Identify that.

Complete step by step solution:
-By compounds we mean, two or more different elements combine resulting in the formation of compounds. E.g. water, ammonia, carbon dioxide etc.
Like:
$C\text{ + }{{\text{O}}_{2}}\to \text{C}{{\text{O}}_{2}}$
-Carbon and oxygen two different elements combine to form the compound molecule \i.e. Carbon dioxide.
-The word stoichiometric deals with the calculation of the masses of the reactants and the products involved in chemical reaction.
-Stoichiometric compounds are those compounds which contain all its component elements in the same proportion as indicated by their chemical formula i.e. they follow the law of constant or definite composition. According to this law, a compound always contains exactly the same proportion of elements by weight.
-On the other hand, non- stoichiometric compounds are those compounds which don't contain all its component elements in the same proportion as indicated by their chemical formula i.e. they don’t follow the law of constant or definite composition. They are mostly the transition metal oxides but can also be hydrides, carbides, nitrides, sulphides etc.
-From the above given compounds $\text{F}{{\text{e}}_{3}}{{\text{O}}_{4}}$, is a non-stoichiometric compound because in this the elements are in different proportion i.e. the ratio of cations (Fe) to the anions (O) is different in the $\text{F}{{\text{e}}_{3}}{{\text{O}}_{4}}$ as indicated by their chemical formula.
-Fe can lose 3 electrons to acquire the stable electronic configuration and O can gain 2 electrons to attain nearest noble gas configuration.
${Fe} \to {3}$
${O} \to {2}$
So, their chemical formula should be $\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}$ but it is not so and its actual chemical formula is $\text{F}{{\text{e}}_{3}}{{\text{O}}_{4}}$ due to the difference in proportion of elements in the compound.

So, the answer is (C).
Note: The non-stoichiometric compounds mostly result due to the missing of atoms from their lattice due to the defects in the crystal lattice (i.e. may be Schottky defect (atoms are missing) or the Frenkel defect (atoms occupy the interstitial sites)).