The metallic radius of chromium is estimated to be 126 pm. The density of chromium is 7.14 g cm –3. — Inorganic Chemistry — Solid State Chemistry Question
Chromium Chemistry and Latimer Diagrams
The metallic radius of chromium is estimated to be 126 pm. The density of chromium is 7.14 g cm –3. Solid chromium belongs to the regular cubic crystal system.
Determine the lattice type of chromium using only the data given above.
Model Answer
In the cubic system we can have simple cubic, face-centred cubic (fcc) and body-centred cubic (bcc) lattices. These unit cells contain 1, 4 or 2 atoms, respectively. We can determine how the lattice constant depends on the metallic radius utilizing the fact that in each lattice the atoms touch each other. We obtain for simple cubic a = 2R; for fcc: a = 2√2R ; for bcc a = 4/√3R.
If the lattice is simple cubic, then we have 1 atom in a unit cell. Using the density value we obtain M(Cr)/(ρ NA) = 1.209×10 7 pm3 for the volume of a unit cell. The lattice constant is therefore (1.209×10 7 pm3) 1/3 = 229.5 pm. However, this value does not agree with the 252 pm (a = 2 R) expected for a simple cubic lattice.
If the lattice is fcc, then we have 4 atoms in a unit cell. Using the density value we obtain 4M(Cr) / (ρ NA) = 4.838×10 7 pm3 for the volume of a unit cell. The lattice constant is therefore (4.838×10 7 pm3) 1/3 = 364.4 pm. For fcc, a should be 2√2R = 356 pm. Again, this does not agree with the density.
If the lattice is bcc, then we have 2 atoms in a unit cell. Using the density value we obtain 2M(Cr)/(ρ NA) = 2.419×10 7 pm3 for the volume of a unit cell. The lattice constant is therefore (2.418×10 7 pm3) 1/3 = 289.2 pm. For bcc, a = 4/√3R = 291 pm, which is reasonably close. Hence the lattice of chromium is body-centred cubic.
A test for the presence of Cl – ions used to be the following: a dry mixture of the unknown material and potassium dichromate is heated with concentrated H2SO4. The gases produced are passed into NaOH solution, where the appearance of a yellow color indicates the presence of chlorine.
3.2 What is the volatile chromium compound produced in the reaction? Draw its structure. Note that neither Cr nor Cl changes oxidation state during the reaction.
Model Answer
CrO2Cl2, chromyl chloride. Its structure is:
Acidifying a solution of potassium chromate gives rise to the formation of the orange dichromate ion, then the deeper red tri- and tetrachromate ions. Using concentrated sulfuric acid we obtain a red precipitate not containing potassium.
3.3 Write the equations and draw the structure of the ions. Can you propose a structure for the precipitate?
Model Answer
The reaction equations:
2 CrO4 2− + 2 H + = Cr2O7 2− + H2O
3 CrO4 2− + 4 H + = Cr3O10 2− + 2 H2O
4 CrO4 2− + 6 H + = Cr4O13 2− + 3 H2O
The structure of the polychromate ions:
CrO3 is a polymeric chain derived from the polychromate ions.
The Latimer diagrams for a series of chromium species in acidic (pH=0) and basic (pH=14) media is given below:
3.4 Find the missing three values.
Model Answer
The formula to be used is
Eº = (n1Eº1 + n2Eº2) / (n1 + n2)
From this we obtain Eº = 1.35 V for the Cr(V) – Cr(IV) process, Eº = 1.33 V for the Cr2O7 2– – Cr 3+ process, Eº = –0.90 V and for the Cr 2+ – Cr process.
Are Cr(V) and Cr(IV) stable with respect to disproportionation? Identify a simple criterion based on the Latimer diagram. What is the equilibrium constant for the disproportionation of Cr 2+?
Model Answer
The criterion for disproportionation is: E ox < E red, where E ox is the redox potential for the oxidation process and E red is the redox potential for the reduction. A species is prone to disproportionate if there is a larger number on the Latimer diagram on its right side than on its left side. It follows that both Cr(V) and Cr(IV) are unstable with respect to disproportionation.
For the 3 Cr 2+ = 2 Cr 3+ + Cr process the equilibrium constant can be calculated from standard redox potential values: ∆rGº = –2 F × (–0.90 V – (–0.42 V)) = 93 kJ mol – 1. From ∆rGº we obtain the equilibrium constant by using the formula K = exp(–∆rGº/RT) = 5.91·10 –17
Calculate the solubility constant of chromium(III) hydroxide and the overall stability constant of tetrahydroxo-chromate(III) anion.
Model Answer
In order to calculate the solubility constant of Cr(OH)3 we turn to the Latimer diagram for pH = 14. Comparing the standard potentials in volts for the Cr(III) → Cr(0) processes for pH = 0 and pH = 14 we can write:
0.059/3 lg Ksp = 0.74 - 1.33
This gives pKsp = 30.
Similarly for the overall stability constant of the tetrahydroxo-chromate(III) anion we can write:
0.059/3 log(1/K[OH–]4) = 0.74 - 1.33
Substituting 1.0 for [Cr(OH)4 – ] and [OH – ] we obtain: pK = –30.
The Latimer diagram of a series of oxygen-related species in acidic (pH = 0) and basic (pH = 14) media is the following:
3.7 What will happen if the pH of a solution containing chromate(VI), Cr(III) and hydrogen peroxide is set to 0? What will happen if we set the pH to 14? Write down the reactions and the corresponding standard cell potential.
Model Answer
The highest possible standard cell potential belonging to these species in acidic conditions is 1.33 V – 0.695 V = 0.635 V. Thus in acidic solution hydrogen peroxide is oxidized to O2 and Cr(VI) is reduced to Cr(III). The reaction equation is:
Cr2O7 2− + 3 H2O2 + 8 H + 2 Cr 3+ + 3 O2 + 7 H2O
In basic solution OH – formation and the oxidation of Cr(III) to Cr(VI) is associated with the highest standard cell potential: 0.87 V – (–0.72 V) = 1.59 V. The reaction equation is:
2 [Cr(OH)4 ] − + 3 HO2 − 2 CrO4 2− + 5 H2O + OH −