Question

In $NaCl$ lattice structure, if one of the sodium ion is removed from the corner, then find the formula of the resulting compound.

• A. $N{a}_{7}C{l}_8$
• B. $N{a}_{3}C{l}_8$
• C. $N{a}_{7}C{l}_5$
• D. $N{a}_{3}C{l}_4$

Answer 1

The correct option is A $N{a}_{7}C{l}_8$

In NaCl structure $N{a}^{+}$ ion will occupy the tetrahedral voids while the $C{l}^{-}$ ion will occupy the FCC lattice

At the corners $=\frac{1}{2}\times 6=3$

And at the octahedral body diagonal $=1$

Total number of octahedral voids in FCC lattice is $=4$

Thus, if the $N{a}^{+}$ ion is lost from one of the corner then the total number of $N{a}^{+}$ ion $=\frac{1}{2}(6-1)+1=\frac{7}{2}$

And Number of $C{l}^{-}$ ions $=4$

Therefore the formula is $N{a}_{7}C{l}_8$