Atoms of element X form hcp lattice and those of element Y occupy 2/3 of its tetrahedral voids. The percentage of element X in the lattice is ____________. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Every atom in the lattice brings two tetrahedral voids with it, so start the count from there.

1. Count the voids. If X forms the lattice with N atoms, there are 2N tetrahedral voids and N octahedral voids. In one hcp unit cell, that means 6 atoms of X and 12 tetrahedral voids.
2. Fill two-thirds. Y takes 2/3 of the 12 tetrahedral voids, which is 8 atoms of Y for every 6 atoms of X.

   X : Y = 6 : 8 = 3 : 4 ⇒ X₃Y₄

3. Find the share. Out of every 7 atoms, 3 are X. So the percentage of X is 3/7 × 100 = 42.86%.
4. So. The percentage of element X in the lattice is about 42.86%, which rounds to 43.

The common slip: 60. Taking the number of tetrahedral voids as N instead of 2N makes the ratio 3 : 2 and gives 60, not 43.

The idea to keep. For N lattice atoms there are 2N tetrahedral and N octahedral voids. Multiply by the fraction filled, get the atom ratio, then take each element's share of the total.

Answer from the NTA final answer key. Solution by PrepWiser.