Which is not correct about $F e O_{0.94}$ ?

It is non-stoichiometric compound

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Some of

F e^{2 +}

ions get replaced by as many two third

F e^{3 +}

ions

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It is metal excess solid

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It is metal deficient solid

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Solution

The correct option is D It is metal excess solid
$F e O_{0.94}$ is not metal excess solid but a metal deficient solid because here some of $F e^{2 +}$ ions get replaced by as many two third $F e^{3 +}$ ions, hence forming a non-stoichiometric compound.

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Similar questions

F e_{0.94} O

x %

F e^{2 +}

F e^{3 +}

x

The non-stoichiometric compound $F e_{0.94} O$ is formed when x% of $F e^{2 +}$ ions are replaced by as many $\frac{2}{3} F e^{3 +}$ ions, x is ?

The non-stoichiometric compound $F e_{0.94} O$ is formed when x% of $F e^{2 +}$ ions are replaced by $F e^{3 +}$ ions then x is

The non-stoichiometric compound $F e_{0.94} O$ is formed when x% of $F e^{2 +}$ ions are replaced by as many $\frac{2}{3} F e^{3 +}$ ions, x is ?

{F e}_{0.94} O

x

{F e}^{2 +}

\frac{2}{3} {F e}^{3 +}

x

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