Metal deficiency defect is shown by Fe_{0.93} | vedclass

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(A) Let the number of $Fe^{2+}$ ions be $x$ and the number of $Fe^{3+}$ ions be $y$.Total number of $Fe$ ions is $x + y = 0.93$.Since the crystal is e

Vedclass · Accepted Answer

$85$

Vedclass · Answer

(A) Let the number of $Fe^{2+}$ ions be $x$ and the number of $Fe^{3+}$ ions be $y$.Total number of $Fe$ ions is $x + y = 0.93$.Since the crystal is electrically neutral,the total positive charge must equal the total negative charge ($-2$ for $O^{2-}$).$2x + 3y = 2$.Substituting $y = 0.93 - x$ into the charge equation:$2x + 3(0.93 - x) = 2$$2x + 2.79 - 3x = 2$$-x = -0.79$$x = 0.79$.Thus,the number of $Fe^{2+}$ ions is $0.79$.The percentage of $Fe^{2+}$ ions is $\frac{0.79}{0.93} 	imes 100 \approx 84.94\%$.The nearest integer is $85\%$.

Metal deficiency defect is shown by $Fe_{0.93}O$. In the crystal,some $Fe^{2+}$ cations are missing and loss of positive charge is compensated by the presence of $Fe^{3+}$ ions. The percentage of $Fe^{2+}$ ions in the $Fe_{0.93}O$ crystals is $....$ (Nearest integer)

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