Question

An example of non-stoichiometric compound is:
A.${\text{PbO}}$
B.${\text{Ni}}{{\text{O}}_{\text{2}}}$
C.${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$
D.${\text{F}}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}$

Answer 1

Non-stoichiometric compounds are solid chemical inorganic compounds, mostly the proportions of their elemental compositions cannot be represented in small natural numbers.

Complete step by step answer:
So let’s start our discussion with the basic definition of the non-stoichiometric compound, these are the solid chemical inorganic compounds, mostly the proportions of their elemental compositions cannot be represented in small natural numbers.
The majority of these compounds have some missing atoms or either too many atoms are closely packed in an otherwise perfect lattice structure.
These compounds with help of modern science are being identified as homogenous contrary to earlier claims saying them to be a mixture of stoichiometric chemical compounds.
Being of pervasive nature in metal oxides in which the metal does not tends to be in its highest oxidation state.
This can be very well proved with the example of ferrous oxide whose stoichiometric formula is considered to be an ideal one that is ${\text{FeO}}$, but the reality seems to be far away from the visual claim.
In reality, ferrous oxide tends to have the formula close to ${\text{F}}{{\text{e}}_{{\text{0}}{\text{.95}}}}{\text{O}}$.
Coming back to our question we find out that ${\text{F}}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}$ is very close to the definition of being non-stoichiometric and the simple reason being the number of different cations and anions that can be perceived from the formula.
Hence per the stated definition of non-stoichiometric compounds, ${\text{F}}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}$ is an example of a non-stoichiometric compound.

Note:
Along with iron oxides, some iron sulphides are also non stoichiometric in nature. These Iron (II) Sulphides include ${\text{F}}{{\text{e}}_{\text{7}}}{{\text{S}}_{\text{8}}}$, ${\text{F}}{{\text{e}}_{\text{9}}}{{\text{S}}_{{\text{10}}}}$, ${\text{F}}{{\text{e}}_{{\text{11}}}}{{\text{S}}_{{\text{12}}}}$, All these compounds are iron deficient which leads to the presence of lattice defects. Despite the presence of these defects the ratio is expressed as a large number and the crystal stability is also high for these compounds.