Information was always regarded as the most valuable product resulting from mankind activity. It is — Physical Chemistry — Thermodynamics Chemistry Question
Chameleonic cobalt
Information was always regarded as the most valuable product resulting from mankind activity. It is not striking that recognition of this fact was followed by numerous efforts aimed at information safety. Cryptography seemed to be a convenient way to reach such safety from unrecorded time. Cryptography cannot be detached from sympathetic ink that becomes visible only after special treatment, for instance, heating. History knows a number of recipes of such ink, among them that based on salts of cobalt(II). Being pale-pink in color, cobalt ink is virtually invisible when dried on paper. However, once heated with a candle flame, a letter written with such ink reveals hidden text colored in bright-blue.
We know other applications of cobalt(II) salts, less secret, but dependent on the color transition described above. Blue granules of silica-gel doped with Co(II) salt and placed into a desiccators to dry some product, become pink at last. This is the signal to regenerate silica-gel (just to dry, since it accumulates too much water). Similarly, a paper soaked with saturated solution of CoCl2 turns blue in dry air due to formation of CoCl2·4 H2O, and changes its color back to pink CoCl2·6 H2O in a humid environment. Apparently, the paper works as a humidity meter, hygrometer.
Using the thermodynamic data below, determine the threshold of air humidity (in %) specific to the response of such a hygrometer.
[VISUAL]
| Compound | \Delta_f H^\circ (298) / kJ mol-1 | S^\circ (298) / J mol-1 K-1 |
| :--- | :---: | :---: |
| CoCl2·6 H2O(s) | -2113.0 | 346.0 |
| CoCl2·4 H2O(s) | -1538.6 | 211.4 |
| H2O (l) | -285.8 | 70.1 |
| H2O(g) | -241.8 | 188.7 |
Model Answer
The reaction corresponding to the hygrometer response is:
CoCl2·6 H2O(s) <=> CoCl2·4 H2O(s) + 2 H2O(g)
Reaction thermodynamics calculations:
\Delta_r H^\circ_{298} = -1538.6 + 2 \times (-241.8) - (-2113.0) = 90.8 kJ
\Delta_r S^\circ_{298} = 211.4 + 2 \times 188.7 - 346.0 = 242.8 J K-1
\Delta_r G^\circ_{298} = 90800 - 298 \times 242.8 = 18450 J = 18.45 kJ
From the relationship -ln K_p = \Delta_r G^\circ / RT:
lg p_{H2O} = - \Delta_r G^\circ / (2 \times 2.3 \times R \times T) = -18450 / (2 \times 2.3 \times 298 \times 8.31) = -1.62
p_{H2O} = 10^{-1.62} = 0.024 atm
At 298 K, the pressure of saturated water vapor is estimated from the vaporization equilibrium:
H2O(l) <=> H2O(g)
\Delta_r H^\circ_{298} = -241.8 - (-285.8) = 44.0 kJ
\Delta_r S^\circ_{298} = 188.7 - 70.1 = 118.6 J K-1
\Delta_r G^\circ_{298} = 44000 - 298 \times 118.6 = 8.66 kJ
lg p^\circ_{H2O} = -8660 / (2.3 \times 298 \times 8.31) = -1.52
p^\circ_{H2O} = 0.030 atm
The threshold of relative humidity of air specific to the hygrometer response is:
RH = p_{H2O} / p^\circ_{H2O} = 0.024 / 0.030 = 0.80 or 80%
The “pink (sometimes, violet) ↔ blue” color transition described above is related to the reconstruction of the coordination sphere of Co2+ ion: octahedron ↔ tetrahedron. The examples discussed in a previous section deal with the transition [Co(H2O)6]oct 2+ ↔ [Co(H2O)4]tetr 2+. As a rule, coordination compounds with tetrahedral geometry are less abundant compared to octahedral ones. However, in particular case of Co2+ tetrahedral complexes competes with octahedral compounds.
To understand the reason behind such behavior, consider the following octahedral and tetrahedral complexes:
a) [Cr(H2O)6] 3+ and [Cr(H2O)4] 3+,
b) [Co(H2O)6] 2+ and [Co(H2O)4] 2+.
Draw diagrams for the case of an octahedral and a tetrahedral ligand field showing clearly the energy levels of all metal 3d-orbitals; indicate the d-orbital splitting parameter ∆. For each of the ions above use the appropriate diagram and fill it in with the electrons available in the metal d-subshell. Calculate the Crystal Field Stabilization Energy (CFSE) for each of the ions.
Compare the results and draw a conclusion.
Model Answer
In a weak Crystal Field (where ligands are water molecules):
[VISUAL]
a) Cr3+ (d3):
- Octahedral [Cr(H2O)6]3+: t_2g^3 e_g^0 configuration.
CFSE = 3 \times (-2/5 \Delta_o) = -1.2 \Delta_o
- Tetrahedral [Cr(H2O)4]3+: e^2 t_2^1 configuration.
CFSE = 2 \times (-3/5 \Delta_t) + 1 \times (2/5 \Delta_t) = -4/5 \Delta_t
Assuming \Delta_t \approx 4/9 \Delta_o, CFSE \approx -16/45 \Delta_o = -0.36 \Delta_o
b) Co2+ (d7):
- Octahedral [Co(H2O)6]2+: t_2g^5 e_g^2 configuration.
CFSE = 5 \times (-2/5 \Delta_o) + 2 \times (3/5 \Delta_o) = -4/5 \Delta_o = -0.80 \Delta_o
- Tetrahedral [Co(H2O)4]2+: e^4 t_2^3 configuration.
CFSE = 4 \times (-3/5 \Delta_t) + 3 \times (2/5 \Delta_t) = -6/5 \Delta_t
Assuming \Delta_t \approx 4/9 \Delta_o, CFSE \approx -24/45 \Delta_o = -0.53 \Delta_o
Conclusion:
The absolute difference |CFSE(octahedral) - CFSE(tetrahedral)| becomes minimal for the d7 configuration of Co2+ (0.80 \Delta_o - 0.53 \Delta_o = 0.27 \Delta_o) compared to the d3 configuration of Cr3+ (1.20 \Delta_o - 0.36 \Delta_o = 0.84 \Delta_o).
Furthermore, according to the Hard and Soft Acids and Bases (HSAB) concept, the Crystal Field Theory's assumption of purely ionic bonding holds best for hard acid-hard base interactions. Co2+ is an intermediate acid close to a soft one, and its covalent contribution to chemical bonding with large polarizable ligands acts as an additional stabilizing factor for tetrahedral complexes.
The following reaction
[Co(H2O)6] 2+ + 4 X– = [CoX4] 2– + 6 H2O, (1)
where X– = Cl–, Br–, I–, SCN–, is used in some textbooks to illustrate Le Chatelier’s principle related to equilibrium shifting. If one adds an excess of salt containing X–, the solution becomes blue, and under dilution with water it turns back pale-pink.
a) Predict the signs of the enthalpy (\Delta_r H^\circ_{298}) and entropy (\Delta_r S^\circ_{298}) changes for the reaction (1).
b) What effect does temperature produce on the equilibrium (1)?
c) Consider reaction (1) and KCl and KSCN as a source of ions X– for it. Which salt present in the same molar concentration shifts the equilibrium (1) to the right in a greater extent? Explain using the principle of Hard and Soft Acids and Bases (HSAB).
Model Answer
a) We expect \Delta_r S^\circ_{298} > 0 because the reaction is accompanied by an increase in the number of species (from 5 reactant particles to 7 product particles). Since \Delta_r G^\circ_{298} is slightly above zero (as the reaction requires high concentrations of X- to proceed), \Delta_r H^\circ_{298} must be greater than T \times \Delta_r S^\circ_{298} > 0. Thus, \Delta_r H^\circ_{298} > 0 (endothermic).
b) Heating shifts the equilibrium to the right since the reaction is endothermic (\Delta_r H^\circ_{298} > 0), turning the pink solution deep blue.
c) Since Co2+ is an intermediate acid (close to a soft one), it forms more stable complexes with softer bases. Thiocyanate (SCN-) is a softer base than chloride (Cl-), so SCN- shifts the equilibrium (1) to the right to a greater extent. This reaction is used in qualitative analysis to identify Co2+.
Consider a similar equilibrium (2):
[CoX2L4] = [CoX2L2] + 2 L. (2)
a) If L = pyridine (py), which ligand X (Cl– or I–) helps better shift the equilibrium (2) to the right? Explain using the principle of Hard and Soft Acids and Bases (HSAB).
b) If L = PH3, which ligand X (Cl– or I–) helps better shift the equilibrium (2) to the right? Explain using the HSAB principle.
c) The coordination compound with the formula [CoX2L2], where L = py, X = Cl– exists in two forms colored blue and violet. The structure of the former is quite apparent, whereas that of the latter is less obvious. For the violet form, draw a fragment of its structure large enough to show clearly the coordination mode of the cobalt ion.
Model Answer
a) X = I–. According to HSAB, I– is a softer base than Cl–, which coordinates better to the intermediate-to-soft Co2+ center.
b) For L = PH3, the tetrahedral coordination compounds [CoX2L2] are highly stable for both X = I– and X = Cl– because PH3 is a much softer base compared to pyridine, meaning that the softness of the secondary ligand X is no longer the determining factor in stabilizing the tetrahedral complex.
c) Violet color in Co(II) complexes corresponds to an octahedral coordination environment. To achieve this, the compound [CoCl2py2] must possess a polymeric structure bridged by chlorine atoms:
[VISUAL]
Structure description:
Each Co2+ center resides in an octahedral geometry coordinated by two trans-pyridine ligands and bridged by four equatorial chlorine atoms forming an infinite chain (-Cl-Co(py2-Cl-Co(py2-Cl-).
With some knowledge of coordination chemistry of Co(II) described above, you may be able to account for the transformations described below.
NaOH solution is added dropwise to a solution of Co(II) under cooling (0 °C), which results in a precipitate of blue color. If the precipitate is left at room temperature (25 °C) for a while, it becomes pink. If an excess of alkali is further added to the precipitate, it dissolves giving blue solution.
Write down equations corresponding to the transformations described above.
Model Answer
The reactions correspond to the following coordination transitions:
1) Under cooling (0 °C), a blue tetrahedral hydroxide precipitate is formed:
CoCl2 + 2 NaOH → [Co(H2O)2(OH)2]↓ + 2 NaCl
2) Standing at room temperature (25 °C) leads to coordination sphere expansion, yielding a pink octahedral precipitate:
[Co(H2O)2(OH)2] + 2 H2O → [Co(H2O)4(OH)2]↓
3) Adding excess NaOH dissolves the precipitate to form a blue tetrahedral tetrahydroxocobaltate(II) complex:
[Co(H2O)4(OH)2] + 2 NaOH → Na2[Co(OH)4] + 4 H2O