JEE Advanced Chemistry Question (2023) — Solution

Step-by-step solution

Metal x forms FCC (edge length L x ) Metal y forms BCC (edge length L y ) Metal z forms SC (edge length L z ) Given r z = 3 2 r y and r y = 8 3 r x ∴    r z = 3 2 × 8 3 r x = 4 r x M z = 3 2 M y    M z = 3 M x ∴ M y = 2 M x Packing efficiency FCC   >   BCC   >   SC Packing efficiency unit cell x > y > z In FCC unit cell:- atoms along the face diagonals are in contact. ∴ 2   L x = 4 r x ⇒ L x = 2 2 r x In BCC unit cell: atoms along the body diagonal are ∴ 3 L y = 4 r y ⇒ L y = 4 3 r y = 4 3 × 8 3 r x = 32 3 r x L y = 32 3 r x In SC unit cell, atoms along the edge are in contact ∴ L z = 2 r z = 2 × 4 r x = 8 r x L x = 2 2 r x L y = 32 3 r x L z = 8 r x ∴ L y > L z > L x Density of x  (Number of atoms of x per unit cell ( z ) = 4 ) d x = zM x L x 3   N A = 4 × M x 2 2 r x 3 × N A = 4 M x 16 2 r x 3 N A = M x 4 2 r x 3 N A Density of y : (Number of atoms of y per unit cell ( z ) = 2  ) d y = zM y L y 3 N A = 2 × 2 M x 32 3 r x 3 N A = 108 M x 32768 r x 3 N A ∴ Density of x > density of y .