Question

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

• A. $0.035\, N _{ A }$
• B. $0.01\, N _{ A }$
• C. $0.07\, N _{ A }$
• D. $0.1\, N _{ A }$

Answer 1

Answer: (A)

$Fe _{0.9} O$

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

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

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

$2 x+2.7-3 x=+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 $Fe _{0,9} O _{1.00}$ has $Fe ^{2+}=0.7$

$0.1 \times N _{ A }$ molecule of $Fe _{0.9} O _{1.00}$ has

$Fe ^{2+}=0.4 \times 0.1 \times N _{ A }$

$=0.07 \times N _{ A }$

Now, $2 Fe ^{2+}$ substitute $Sr ^{4+}=1$

$0.07 \times N _{ A }$ substitute

$Sr ^{4+}=\frac{1}{2} \times 0.07 \times N _{ A }\, Sr ^{4+}$



$1\, Sr ^{4+}$ creates, cation vacancy $=1$

$\frac{0.07}{2} \times N _{ A }\, Sr ^{4+}$ creates, cation vacancy

$=\frac{0.070}{2} \times N _{ A }$

$=0.035 \times N _{ A }$