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The stoichiometry of a complex can be determined by various methods [1]. One of these is Job's methoPhysical Chemistry — Electrochemistry Chemistry Question

Structural study of copper (II) complexes

The stoichiometry of a complex can be determined by various methods [1]. One of these is Job's method of continuous variation, which involves measuring the absorbance of a series of mixtures with varying mole fractions of metal and ligand while keeping the total concentration constant [1, 2]. Additionally, the electronic structure of transition metal complexes can be described by crystal field theory, taking into account the configuration of d electrons and any potential geometric distortions [3, 4].

22.1.

Determine which complex is responsible for the color of the aqueous copper(II) solution, specify which color region of the visible spectrum it mainly absorbs, and estimate the wavelength of this absorption.

Model Answer

The complex responsible for the color of the solution is the hexaaquacopper(II) complex Cu(H2O)6 2+ [4]. This compound absorbs light mainly in the orange region, which is the complementary color of blue, i.e. λ1 = 620 nm [4].

22.2.

Using the experimental absorbance values, show how to calculate the corrected absorbance A' by correcting for the absorbance of the free copper(II) ions in the solution. [VISUAL]

Model Answer

The corrected absorbance is calculated as A' = A - A_Cu * (V_Cu / 20.00), where A_Cu = 0.080 is the absorbance of the copper solution without ligand (tube 12) [5]. For tube 5: A' = 0.329 - 0.080 * (4.00 / 20.00) = 0.313 [5].

22.3.

Plot the corrected absorbance A' with respect to x, where x represents the volume of copper solution added. [VISUAL]

Model Answer

The plot of corrected absorbance A' with respect to the volume of copper solution added x shows two intersecting straight lines that reach a maximum at x_max = 4.00 mL [5].

22.4.

Assuming that the copper ion is the limiting reagent, determine the corrected absorbance A' with respect to x.

Model Answer

If the copper ion is the limiting reagent, the concentration of the complex is equal to the total copper concentration added [5]: [Z] = [Cu2+]_total = x / V_total * C_Cu. Thus, the corrected absorbance is directly proportional to x [5].

22.5.

Assuming that the ligand is the limiting reagent, determine the corrected absorbance A' with respect to x.

Model Answer

Assuming that the ligand is the limiting reagent, the concentration of the complex is determined by the total ligand concentration: [Z] = [L]_total / n = (20.00 - x) / (V_total * n) * C_L [3]. Thus, the corrected absorbance is proportional to (20.00 - x) [3].

22.6.

Show that the intersection of the two straight lines occurs when xmax = 20/(1+n).

Model Answer

At the intersection, the complex concentration is equal under both limiting conditions: x / 20 * C_Cu = (20 - x) / (20 * n) * C_L [3]. Assuming C_Cu = C_L, we get x = (20 - x) / n, which simplifies to x_max = 20 / (1 + n) [3].

22.7.

Deduce the molecular formula of the amine aqua copper (II) complex Z.

Model Answer

From the intersection of the two straight lines on the graph, x_max = 4.00 mL [5]. Since x_max = 20 / (1 + n), 4.00 = 20 / (1 + n), yielding n = 4 [3]. The molecular formula of the complex is [Cu(H2O)2(NH3)4]2+ [6].

22.8.

Assuming a regular octahedral frame of the ligands around the copper center, draw and fill the electronic levels of the d orbitals along an energetic axis. [VISUAL]

Model Answer

Cu(II) possesses 9 d electrons and water is a low-field ligand [6]. In a regular octahedral field, the d orbitals split into a triply degenerate t2g level (d_xy, d_yz, d_xz) at lower energy and a doubly degenerate eg level (d_z2, d_x2-y2) at higher energy, filled as (t2g)6 (eg)3 [6].

22.9.

Determine the crystal field splitting of a low-field octahedral Mn(II) complex, draw its d-orbital energy level diagram, and fill the electrons consistent with its spin state. The involved ligands will be taken on the z axis. [VISUAL]

Model Answer

MnII possesses 5 d electrons and water is a low-field ligand, so the filled diagram in an octahedral field consists of 5 unpaired electrons, filled as (t2g)3 (eg)2 [6, 7].

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