See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Y(CCP) = 4 X(Oct. void) = 4 Atoms along one B. D removed  15 Y 4  , X = 3 Z  introduced in available voids = 2 Td. Voids + 1 oct. void = 3 3 15 3 4 5 4 4 X Y Z X Y Z 