Question

If all $F{e}^{2+}$ ions are replaced by $S{r}^{4+}$ ions in 0.1 mole of $F{e}_{0.9}O,$ then the number of vacancies created are-

• A. $0.035N_A$
• B. $0.01N_A$
• C. $0.07N_A$
• D. $0.1N_A$

Answer 1

Answer: (A)

$F{e}_{0.9}O$

Let no. of $F{e}^{2+}=x$

No, of $F{e}^{3+}=0.9-x$

$2x+3(0.9-x)=+2$

$2x+2.7-3x=+2$

$-x=-0.7\Rightarrow x=0.7$

No. of $F{e}^{2+}=0.7$

No. of $F{e}^{3+}=0.9-0.7=0.2$

$1$ molecule of $F{e}_{0,9}{O}_{1.00}$ has $F{e}^{2+}=0.7$

$0.1\times N_A$ molecule of $F{e}_{0.9}{O}_{1.00}$ has

$F{e}^{2+}=0.4\times 0.1\times N_A$

$=0.07\times N_A$

Now, $2F{e}^{2+}$ substitute $S{r}^{4+}=1$

$0.07\times N_A$ substitute

$S{r}^{4+}=\frac{1}{2}\times 0.07\times N_{A}S{r}^{4+}$



$1S{r}^{4+}$ creates, cation vacancy $=1$

$\frac{0.07}{2}\times N_{A}S{r}^{4+}$ creates, cation vacancy

$=\frac{0.070}{2}\times N_A$

$=0.035\times N_A$