NEET Chemistry The Solid State Questions Solved


Compute the percentage void space per unit volume of unit cell in zinc fluoride structure.

(A) 15.03%                                             

(B) 22.18%

(C) 18.23%                                             

(D) 25.07%

(D). Anions occupy fcc positions and half of the tetrahedral holes are occupied by cations. Since there are four anions and 8 tetrahedral holes per unit cell, the fraction of volume occupied by spheres per unit volume of the unit cell is 

=4×43πra3+12×8×43πrc3162ra3=π321+rcra3

 for tetrahedral holes, rcra=0.225

=π321+0.2253=0.7493

  Void volume = 1 – 0.7493 = 0.2507/unit volume of unit cell.
% void space = 25.07%

Difficulty Level:

  • 7%
  • 28%
  • 18%
  • 50%