Question

If ${Al}^{3+}$ ions replace ${Na}^{+}$ ions at the edge centres of $NaCl$ lattice, then the number of vacancies in 1 mole of $NaCl$ will be:

• A. $3.01{\times 10}^{23}$
• B. $6.02{\times 10}^{23}$
• C. $12.04{\times 10}^{23}$
• D. $12.04{\times 10}^{21}$

Answer 1

The correct option is A $3.01{\times 10}^{23}$

In $NaCl$ lattice, $C{l}^{-}$ ions forms FCC and $N{a}^{+}$ ions are present at edge centers and body center.

Number of $N{a}^{+}$ ions present $=\frac{1}{4}\times 12=3$

Now, charge on $A{l}^{3+}=+3$ and charge on $N{a}^{+}=1$.

So, each $A{l}^{3+}$ ion will replace $3N{a}^{+}$ ions and two vacant sites will be created.

Now, $3/4$ moles of $N{a}^{+}$ ions will be replaced by $1/4$ moles of $A{l}^{3+}$ ions to maintain electrical neutrality. Now, remaining $1/4$ moles of $N{a}^{+}$ ions and $1/4$ moles of $A{l}^{3+}$ ion will occupy $1/2$ moles of lattice sites.

So, Number of vacancies $=\frac{1}{2}moles=\frac{1}{2}\times 6.022\times {10}^{23}=3.0125\times {10}^{23}$