Chlorine dioxide is a somewhat exceptional molecule because it contains an unpaired electron. — Analytical Chemistry Chemistry Question
Chlorine dioxide photochemistry
Chlorine dioxide is a somewhat exceptional molecule because it contains an unpaired electron.
Draw the Lewis structure of chlorine dioxide.
Model Answer
[VISUAL] (Two resonance structures of chlorine dioxide showing the unpaired electron on the chlorine atom and on the oxygen atom, respectively.)
Name at least two other stable molecules which do not contain metals but have an odd number of electrons.
Model Answer
NO, NO2
Chlorine dioxide is increasingly used in water treatment. In one study, the reaction of chlorine dioxide with iodine was studied in aqueous solution. Light accelerated the process greatly. Chloride and iodate ions were detected as final products. Write the expected balanced equation for the process.
Model Answer
2 ClO2 + I2 + 2 H2O → 2 Cl− + 2 IO3− + 4 H+
The ratio of the decreases in chlorine dioxide and iodine concentration was experimentally determined to be 2.3. What side reactions can cause this deviation from the expected stoichiometry?
Model Answer
The side reaction taking place is the disproportionation of ClO2:
6 ClO2 + 3 H2O → Cl− + 5 ClO3− + 6 H+
Other possibilities include a different disproportionation of ClO2:
8 ClO2 + 4 H2O → 3 Cl− + 5 ClO4− + 8 H+
or the formation of periodate ion in the redox reaction:
14 ClO2 + 5 I2 + 12 H2O → 14 Cl− + 10 IO4− + 24 H+
Suggest a method to detect the possible side reactions.
Model Answer
Qualitative analysis for chlorate, perchlorate, and periodate ions can be used to differentiate between possible side reactions.
Quantitative measurements on the photochemical reaction were carried out using a very intense halogen lamp. Between the lamp and the photoreactor, an interference filter was also used, which excluded all light with the exception of the 455 – 465 nm wavelength region. This wavelength band is sufficiently narrow to consider the light beam as monochromatic 460 nm light. Iodine has a molar absorption of ε = 740 dm3 mol–1 cm–1 at 460 nm, whereas the molar absorption of chlorine dioxide is immeasurably low at this wavelength. A 25.00 cm3 photoreactor with a 5.00 cm long light path length was used in all of the experiments. The reaction was studied in an acidic solution with initial concentrations [I2]o = 5.1·10^–4 mol dm–3 and [ClO2]o = 4.0·10^–4 mol dm–3. Reference experiments were also carried out in the absence of selected reagents. When the solution contained chlorine dioxide but no iodine, no change was observable. Does this prove that chlorine dioxide does not photo-decompose?
Model Answer
No. At 460 nm, chlorine dioxide does not absorb light; therefore a 460 nm light beam cannot cause the photodecomposition. Lower wavelength light actually induces such a photodecomposition.
When the solution contained iodine but no chlorine dioxide, a very slow decay of iodine was observed, but it was orders of magnitude smaller than in the presence of chlorine dioxide. In the remaining experiments, the solution contained both reagents. In each experiment, the initial rate of the loss of chlorine dioxide (v0) was determined. The first experiment was carried out using the described experimental setup. In later experiments, a grey filter was inserted into the light beam before the photoreactor. The absorbance of this grey filter at 460 nm was measured in an independent experiment. In the final experiment, a steel sheet was inserted into the light beam that did not let any light through. The initial rates were determined in all of these experiments:
no filter
filter 1 A460 = 0.125
filter 2 A460 = 0.316
filter 3 A460 = 0.582
steel sheet
[VISUAL]
Why doesn’t the rate fall to 0 in the experiment with the metal sheet?
Model Answer
The reaction also takes place in the absence of light, so there are simultaneous photochemical and thermal pathways.
The intensity of the light beam was determined by ferrioxalate actinometry. A solution of K3[Fe(C2O4)3] (c = 0.00600 mol dm–3) was prepared in a solution of H2SO4 (c = 0.05 mol dm–3). A volume of 25.00 cm3 of this solution was measured into the photoreactor. Previously, the absorbance of this solution was measured to be 1.41 at 460 nm in a 1.000 cm quartz cell. The sample was illuminated for 30.00 minutes. The following process takes place in the solution:
2 [Fe(C2O4)3]3- + hν → 2 Fe2+ + 2 CO2 + 5 C2O42-
The quantum yield for the formation of iron(II) is 1.12.
After illumination, a 1.000 cm3 sample from the solution was measured into a 5.000 cm3 volumetric flask, which was then filled with a solution that contained 0.0100 mol dm–3 1,10-phenanthroline and 0.50 mol dm–3 1 : 1 acetate : acetic acid buffer. The absorbance of this solution at 510 nm was measured in a 1.000 cm cell, the absorbance reading was 0.3823. The molar absorption coefficient of the complex Fe(phen32+ is ε = 1.10·10^4 dm3 mol–1 cm–1 at 510 nm and nothing else absorbs light in this solution at this wavelength. What is the concentration of the iron(II) complex in the cell?
Model Answer
cFe(II) = 0.3823 / (1.10·10^4 dm3 mol−1 cm−1 * 1.000 cm) = 3.48·10^−5 mol dm–3
How much iron(II) is formed during the illumination?
Model Answer
The amount of iron(II) in the 5.000 cm3 volumetric flask is 1.74·10^−7 mol. This amount of iron(II) was present in 1.000 cm3 of the illuminated solution whose overall volume was 25.00 cm3. Therefore, the total amount of iron(II) formed is 4.35·10^−6 mol.
What is the intensity of the light at 460 nm in mol photon/s and W units?
Model Answer
The quantum yield of the process producing iron(II) is 1.12. Therefore, in the 30.00 minute illumination period:
4.35·10^−6 mol / 1.12 = 3.88·10^−6 mol photon caused photoreaction.
The absorbance of the solution in a 1.00 cm cell is 1.41. The photoreactor has an optical path length of 5.00 cm, therefore the absorbance is expected to be 7.05. This means the solution absorbs (1 − 10^−7.05) * 100 = 100.00 % of the light (practically all of it). The light intensity from this and the illumination time is:
3.88·10^−6 mol / 1800 s = 2.16·10^−9 mol/s.
The energy of a 460-nm photon is hν, where h is Planck’s constant and the frequency of the photon can be calculated using the speed of light c as ν = c/λ:
I = 2.16·10^−9 mol s−1 * NA * h * c / λ = 5.62·10^−4 J s−1 = 0.562 mW.
Determine the quantum yield in the reaction of chlorine dioxide with iodine for both the loss of chlorine dioxide and iodine.
Model Answer
The light intensity after the filter can be calculated as I = Io * 10^-A. The absorbance of iodine in the photoreactor:
5.1·10^−4 mol dm−3 * 740 dm3 mol−1 cm−1 * 5.00 cm = 1.887
therefore (1 − 10^−1.887) * 100 = 98.7 % of the light is absorbed by this solution at 460 nm.
Calculation summary across filters:
- no filter: I = 2.16·10^-9 mol/s, absorbed = 2.13·10^-9 mol/s, v0 = 2.51·10^-9 mol dm-3 s-1
- filter 1: I = 1.62·10^-9 mol/s, absorbed = 1.60·10^-9 mol/s, v0 = 1.97·10^-9 mol dm-3 s-1
- filter 2: I = 1.04·10^-9 mol/s, absorbed = 1.03·10^-9 mol/s, v0 = 1.40·10^-9 mol dm-3 s-1
- filter 3: I = 5.66·10^-10 mol/s, absorbed = 5.59·10^-10 mol/s, v0 = 9.3·10^-10 mol dm-3 s-1
- steel sheet: I = 0, absorbed = 0, v0 = 3.7·10^-10 mol dm-3 s-1
A plot of initial rate as a function of light intensity:
[VISUAL]
The slope of this plot is the quantum yield for the loss of ClO2 (1.00). The intercept is the initial rate of the dark reaction.
The quantum yield for the loss of iodine is obtained by taking the stoichiometry into account: 1.00 / 2.30 = 0.43.