Atoms of metals x ,   y and z form face-centred cubic (fcc) unit cell of edge length L x , body-centred…

Atoms of metals x, y and z form face-centred cubic (fcc) unit cell of edge length Lx, body-centred cubic (bcc) unit cell of edge length Ly, and simple cubic unit cell of edge length Lz, respectively. If rz=32ry;ry=83rx;Mz=32My and Mz=3Mx, then the correct statement(s) is(are)

[Given: Mx,My, and Mz are molar masses of metals x, y, and z, respectively. rx,ry, and rz are atomic radii of metals x,y, and z, respectively.]

  1. Packing efficiency of unit cell of x> Packing efficiency of unit cell of y> Packing efficiency of unit cell of z
  2. Ly>Lz
  3. Lx>Ly
  4. Density of x> Density of y

Solution

Metal x forms FCC (edge length Lx )

Metal y forms BCC (edge length Ly )

Metal z forms SC (edge length Lz )

Given rz=32ry and ry=83rx

  rz=32×83rx=4rx

Mz=32My  Mz=3Mx

My=2Mx

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 Lx=4rxLx=22rx

In BCC unit cell: atoms along the body diagonal are

3Ly=4ryLy=43ry=43×83rx=323rx

Ly=323rx

In SC unit cell, atoms along the edge are in contact

Lz=2rz

=2×4rx=8rx

Lx=22rx

Ly=323rx

Lz=8rx

Ly>Lz>Lx

Density of x (Number of atoms of x per unit cell (z)=4)

dx=zMxLx3 NA=4×Mx22rx3×NA

=4Mx162rx3NA=Mx42rx3NA

Density of y : (Number of atoms of y per unit cell (z)=2 )

dy=zMyLy3NA=2×2Mx323rx3NA=108Mx32768rx3NA

Density of x> density of y.

Asked in: JEE Advanced 2023 (Paper 2)

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