Magnesium sulfide (molar mass 56.37 g/mol)has a face-centered cubic unit cell with magnesium ions in octahedral holes.The ionic radii of magnesium ions and sulfide ions are 79 pm and 184 pm,respectively.What is the density of MgS (in g/cm³)?

Question

Quizplus · Accepted Answer

First, we need to calculate the edge length (a) of the unit cell:
a = 2r√2
where r is the radius of the sulfide ion; since the sulfide ion is assumed to occupy the corners of the unit cell in a face-centered cubic arrangement, r is half the length of the diagonal across a face of the cube:
r = 1/2 × √2 × 184 pm = 260 pm = 2.60 Å

a = 2(2.60 Å) = 5.20 Å

The volume of the unit cell is then:
V = a^3 = (5.20 Å)^3 = 143.75 Å^3

There are four sulfide ions and eight magnesium ions in each unit cell (since each octahedral hole is shared by 6 unit cells, each Mg ion contributes 1/6 to each of the 6 adjacent cells, and there are 4 such holes per cell). Therefore, the total mass of the unit cell is:
(8 Mg ions × 79 g/mol) + (4 S ions × 32.06 g/mol) = 984.48 g/mol

The density is given by the ratio of mass to volume:
density = (984.48 g/mol) / (143.75 Å^3 × 1.6605 × 10^-24 g/Å^3) ≈ 2.57 g/cm3
Therefore, the best choice is B.

Magnesium Sulfide (Molar Mass 56