If the distance between Na^{+} and Cl^{-} ion | vedclass

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(B) In a sodium chloride $(NaCl)$ crystal,the structure is $fcc$ (face-centered cubic).$Cl^{-}$ ions are present at the corners and face centers,while

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$2 Y \ pm$

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(B) In a sodium chloride $(NaCl)$ crystal,the structure is $fcc$ (face-centered cubic).$Cl^{-}$ ions are present at the corners and face centers,while $Na^{+}$ ions occupy all octahedral voids.The distance between the nearest $Na^{+}$ and $Cl^{-}$ ions is equal to half of the edge length $(a)$ of the unit cell.Given,distance between $Na^{+}$ and $Cl^{-} = Y \ pm$.Therefore,$Y = \frac{a}{2}$.This implies that the edge length $a = 2 Y \ pm$.Thus,option $(b)$ is the correct answer.

If the distance between $Na^{+}$ and $Cl^{-}$ ions in sodium chloride crystal is '$Y$' $pm$,the length of the edge of the unit cell will be:

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