NiO has a rock-salt structure. The ionic radius of Ni2+ is 0.69 Å, an…
NiO has a rock-salt structure. The ionic radius of Ni2+ is 0.69 Å, and the radius of O2- is 1.32 Å. Determine the lattice parameter, density, and ionic packing factor of NiO. The atomic mass of Ni and O are 58.71 and 16.00 g/mol, respectively. Please explain to me step by step as i have a test coming up
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Technifi Expert’s Answer:
Let, the length of one side of the cube is a. At the limiting condition the atoms along the side of the cube touch each other. If the radius of the anion is r- and radius of the cation is r+, we can write
lattice parameter, a = (r-) + (2r+) + (r-) = 2 [(r+) + ( r-)]
= 2(0.69 + 1.32)Ao
= 4.02 Ao
Now As, Density = (mass of molecules in one unit cell)/(volume of one unit cell)
We have to determine no of molecules per cell first. There are 8 anions at 8 corners. Each corner ion is shared by 8 adjacent cells. There are 6 anions at the centre of six faces, each of which shared by two adjacent cells.
So no of anions per cell = 8*(1/8) + 6*(1/2) = 4
There are 12 cations at the center of 12 sides of the cell, each of which is shared by 4 adjacent cells. There is one cation at the centre of the cell.
So no of cations per cell = 12*(1/4) + 1 = 4
no of molecules per cell = 4 (Ni4O4 or 4NiO)
Mass of 4 molecules = [4(58.71 + 16)/(6.023*1023)] g = 4.96*10-22 g
Volume of the cell
= (4.02 * 10-8 cm)3
Density = (mass of molecules in one unit cell)/(volume of one unit cell)
= 4.96*10-22 g/[(4.02 * 10-8 cm)3]
= 7.635 g/cm3
packing factor = % of space occupied
= [ volume of 4 anions + volume of 4 cations]/volume of the cell
= [ 4(4/3)π(2.01)3 + 4(4/3)π(0.69)3 ] /[(4.02)3]
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