NaCl is a FCC lattice where Na^{+} ions are a | vedclass

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(B) In a standard $NaCl$ crystal structure,$Na^{+}$ ions occupy the octahedral voids (edge centres and body centre) and $Cl^{-}$ ions occupy the $FCC$

Vedclass · Accepted Answer

$4$

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(B) In a standard $NaCl$ crystal structure,$Na^{+}$ ions occupy the octahedral voids (edge centres and body centre) and $Cl^{-}$ ions occupy the $FCC$ lattice points (corners and face centres). However,based on the specific distribution provided in the question: $Na^{+}$ ions at corners $(8 	imes \frac{1}{8} = 1)$ and face centres $(6 	imes \frac{1}{2} = 3)$,total $Na^{+} = 1 + 3 = 4$. $Cl^{-}$ ions at edge centres $(12 	imes \frac{1}{4} = 3)$ and body centre $(1 	imes 1 = 1)$,total $Cl^{-} = 3 + 1 = 4$. Since there are $4$ $Na^{+}$ ions and $4$ $Cl^{-}$ ions per unit cell,the number of $NaCl$ formula units is $4$.

$NaCl$ is a $FCC$ lattice where $Na^{+}$ ions are at corner and face centre positions. Chloride ions are at edge centres and body centre positions. How many $NaCl$ formula units will be in a unit cell?

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