A) 4y pm
B) y/4 pm
C) y/2 pm
D) 2y pm
Correct Answer: D
Solution :
The distance between sodium and chloride ion = y pm The distance between two chloride ions \[=\sqrt{{{y}^{2}}+{{y}^{2}}}=\sqrt{2{{y}^{2}}}=y\sqrt{2}\] \[{{r}_{C{{l}^{-}}}}=\frac{y\sqrt{2}}{2}\] In a face-centred cubic lattice, anions touch each other along the face diagonal of the cube \[\therefore \] \[a=\frac{4r_{Cl}^{-}}{\sqrt{2}}=\frac{4y\sqrt{2}}{\sqrt{2}.2}=2ypm\]
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