The order of stability of metal oxides is

{A l}_{2} O_{3} < M g O < {F e}_{2} O_{3} < {C r}_{2} O_{3}

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{C r}_{2} O_{3} < M g O < {A l}_{2} O_{3} < {F e}_{2} O_{3}

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{F e}_{2} O_{3} < {C r}_{2} O_{3} < {A l}_{2} O_{3} < M g O

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{F e}_{2} O_{3} < {A l}_{2} O_{3} < {C r}_{2} O_{3} < M g O

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Solution

The correct option is D ${F e}_{2} O_{3} < {C r}_{2} O_{3} < {A l}_{2} O_{3} < M g O$
The order of stability of metal oxide is as :
${F e}_{2} O_{3} < {C r}_{2} O_{3} < {A l}_{2} O_{3} < M g O$

This can be explained on the basis of Gibb's free energy anaylsed through Ellingham diagram. According to Ellingham diagram the lower the position of a metal line in the Ellingham diagram, the greater is the stability of its oxide i.e lower metal line will have most negative Gibbs free energy due to which metal will be most stable in the form of its oxide.

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