04-12-2018, 03:35 AM

1. The problem statement, all variables and given/known data

Below is an image of a unit cell for MgO. Does MgO have the lattice structure of NaCl or ZnS? If the density of MgO, ρρ = 3,58g/cm3, evaluate the sizes of the radii of Mg2+- and O2--ions.

Answers: 149pm and 62pm

2. Relevant equations

3. The attempt at a solution

Looking at the picture, MgO seems to have the lattice structure of NaCl. Therefore as an ionic solid it has octahedral holes in the middle of the larger O2−O2−-ions, where the Mg2+Mg2+-ions are located. A relationship between the radii of ions in an octahedral arrangement can be derived to be r=2–√R−R≈0.414Rr=2R−R≈0.414R, where r is the radius of the smaller ion and R the radius of the larger ion.

We know ρρ = 3,58g/cm3. We also know from looking at the picture, that the unit cell contains 4 whole O2−O2−-ions and 4 whole Mg2+Mg2+-ions, since there is 1 whole Mg-ion in the middle of the cell (the O-ions are cut either in half or into eights by the edges of the cube if it is drawn, and the Mg-ions on the edges of the cube are cut into quarters).

Therefore if we take 1cm3 block of MgO, it's mass m=nM=ρVim=nM=ρVi, where ViVi is the total volume of the ions in the unit cell.

Now VionsVunit cell=4(43πR3)+4(43πr3)(2R+2r)3=r=2√R−R(192√−27)π62√−9≈0,7931VionsVunit cell=4(43πR3)+4(43πr3)(2R+2r)3=r=2R−R(192−27)π62−9≈0,7931.

If we then take 1cm31cm3 block of MgO, its mass m=0,7931ρV=0,7931(3,58g/cm3)(1cm3)=2,8393gm=0,7931ρV=0,7931(3,58g/cm3)(1cm3)=2,8393g.

Now the amount of MgO in this sample is n=mM=2.8393g40,3044g/mol=0,07447moln=mM=2.8393g40,3044g/mol=0,07447mol, and since the lattice structure of MgO is stoichiometric, meaning the ratios of ions is the lattice match the chemical formula, each unit cell contains 4 MgOs.

Since there are N=nNA=(0,07447mol)(6,022⋅10231/mol)≈4,2423⋅1022N=nNA=(0,07447mol)(6,022⋅10231/mol)≈4,2423⋅1022 MgOs in the sample and each unit cell contains 4 MgOs, the amount of unit cells in the sample is N4=1,0606⋅1022N4=1,0606⋅1022.

This number is equal to the ratio VsampleVunit cell=1cm3e3=N4⟺e=1cm3N/4−−−−√3=4,55148⋅10−8VsampleVunit cell=1cm3e3=N4⟺e=1cm3N/43=4,55148⋅10−8.

Since the edge of the unit cell e=2R+2r=2R+2(2–√R−R)=R(2+2(√2)−2)⟺R=e22√=4,55148⋅10−822√=1,60919⋅10−8cme=2R+2r=2R+2(2R−R)=R(2+2(2)−2)⟺R=e22=4,55148⋅10−822=1,60919⋅10−8cm.

Therefore r=0,414R=0,414(1,60919⋅10−8cm)=6,66548⋅10−9cmr=0,414R=0,414(1,60919⋅10−8cm)=6,66548⋅10−9cm.

Answer:

{R≈161pmr≈66,7pm{R≈161pmr≈66,7pm

This is not what the book says. What am I doing wrong?

Below is an image of a unit cell for MgO. Does MgO have the lattice structure of NaCl or ZnS? If the density of MgO, ρρ = 3,58g/cm3, evaluate the sizes of the radii of Mg2+- and O2--ions.

Answers: 149pm and 62pm

2. Relevant equations

3. The attempt at a solution

Looking at the picture, MgO seems to have the lattice structure of NaCl. Therefore as an ionic solid it has octahedral holes in the middle of the larger O2−O2−-ions, where the Mg2+Mg2+-ions are located. A relationship between the radii of ions in an octahedral arrangement can be derived to be r=2–√R−R≈0.414Rr=2R−R≈0.414R, where r is the radius of the smaller ion and R the radius of the larger ion.

We know ρρ = 3,58g/cm3. We also know from looking at the picture, that the unit cell contains 4 whole O2−O2−-ions and 4 whole Mg2+Mg2+-ions, since there is 1 whole Mg-ion in the middle of the cell (the O-ions are cut either in half or into eights by the edges of the cube if it is drawn, and the Mg-ions on the edges of the cube are cut into quarters).

Therefore if we take 1cm3 block of MgO, it's mass m=nM=ρVim=nM=ρVi, where ViVi is the total volume of the ions in the unit cell.

Now VionsVunit cell=4(43πR3)+4(43πr3)(2R+2r)3=r=2√R−R(192√−27)π62√−9≈0,7931VionsVunit cell=4(43πR3)+4(43πr3)(2R+2r)3=r=2R−R(192−27)π62−9≈0,7931.

If we then take 1cm31cm3 block of MgO, its mass m=0,7931ρV=0,7931(3,58g/cm3)(1cm3)=2,8393gm=0,7931ρV=0,7931(3,58g/cm3)(1cm3)=2,8393g.

Now the amount of MgO in this sample is n=mM=2.8393g40,3044g/mol=0,07447moln=mM=2.8393g40,3044g/mol=0,07447mol, and since the lattice structure of MgO is stoichiometric, meaning the ratios of ions is the lattice match the chemical formula, each unit cell contains 4 MgOs.

Since there are N=nNA=(0,07447mol)(6,022⋅10231/mol)≈4,2423⋅1022N=nNA=(0,07447mol)(6,022⋅10231/mol)≈4,2423⋅1022 MgOs in the sample and each unit cell contains 4 MgOs, the amount of unit cells in the sample is N4=1,0606⋅1022N4=1,0606⋅1022.

This number is equal to the ratio VsampleVunit cell=1cm3e3=N4⟺e=1cm3N/4−−−−√3=4,55148⋅10−8VsampleVunit cell=1cm3e3=N4⟺e=1cm3N/43=4,55148⋅10−8.

Since the edge of the unit cell e=2R+2r=2R+2(2–√R−R)=R(2+2(√2)−2)⟺R=e22√=4,55148⋅10−822√=1,60919⋅10−8cme=2R+2r=2R+2(2R−R)=R(2+2(2)−2)⟺R=e22=4,55148⋅10−822=1,60919⋅10−8cm.

Therefore r=0,414R=0,414(1,60919⋅10−8cm)=6,66548⋅10−9cmr=0,414R=0,414(1,60919⋅10−8cm)=6,66548⋅10−9cm.

Answer:

{R≈161pmr≈66,7pm{R≈161pmr≈66,7pm

This is not what the book says. What am I doing wrong?