In NaCl type structure, the shortest distance | vedclass

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(B) In an $NaCl$ type structure, the nearest neighbours are the cation and anion along the edge, separated by a distance of $d_1 = a / 2$.Given $d_1 =

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$100\sqrt{2} \ pm$

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(B) In an $NaCl$ type structure, the nearest neighbours are the cation and anion along the edge, separated by a distance of $d_1 = a / 2$.Given $d_1 = 100 \ pm$, we have $a / 2 = 100 \ pm$, which implies the edge length $a = 200 \ pm$.The next nearest neighbours are the ions of the same type (e.g., $Na^+$ to $Na^+$ or $Cl^-$ to $Cl^-$) located at the corners and face centers, which are separated by the face diagonal distance $d_2 = a / \sqrt{2}$.Substituting $a = 200 \ pm$, we get $d_2 = 200 / \sqrt{2} = 100\sqrt{2} \ pm$.

In $NaCl$ type structure, the shortest distance between two nearest neighbours is $100 \ pm$. Then, the distance between two next nearest neighbours in the same unit cell will be:

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