The formula of a metal oxide with a metal def | vedclass

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(C) The formula is $A_{0.8}O$. This implies that for every $1$ mole of $O^{2-}$ ions,there are $0.8$ moles of $A$ ions.Let the number of $A^{2+}$ ions

Vedclass · Accepted Answer

$0.50$

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(C) The formula is $A_{0.8}O$. This implies that for every $1$ mole of $O^{2-}$ ions,there are $0.8$ moles of $A$ ions.Let the number of $A^{2+}$ ions be $x$ and the number of $A^{3+}$ ions be $(0.8 - x)$.Since the crystal is electrically neutral,the total positive charge must equal the total negative charge:$2x + 3(0.8 - x) = 2$$2x + 2.4 - 3x = 2$$-x = -0.4$$x = 0.4$Thus,the number of $A^{2+}$ ions is $0.4$.The fraction of metal existing as $A^{2+}$ ions is given by:$	ext{Fraction} = \frac{	ext{Number of } A^{2+} 	ext{ ions}}{	ext{Total number of } A 	ext{ ions}} = \frac{0.4}{0.8} = 0.5$.

The formula of a metal oxide with a metal deficiency defect in its crystal is $A_{0.8}O$. The crystal contains $A^{2+}$ and $A^{3+}$ ions. The fraction of metal existing as $A^{2+}$ ions in the crystal is -

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