Solutions of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}(\) both octahedral \()\) and \(\left[\mathrm{CoCl}_{4}\right]^{2-}\) (tetrahedral) are colored. One is pink, one is blue, and one is yellow. Based on the spectrochemical series and remembering that the energy splitting in tetrahedral complexes is normally much less that that in octahedral ones, assign a color to each complex.

Short Answer

Expert verified
The colors of the complexes are: - \([Co(NH_3)_6]^{2+}\): Yellow (largest energy splitting, absorbs lower wavelength, complementary color) - \([Co(H_2O)_6]^{2+}\): Pink (intermediate energy splitting, absorbs mid-wavelength, complementary color) - \([CoCl_4]^{2-}\): Blue (smallest energy splitting, absorbs higher wavelength, complementary color)

Step by step solution

01

Spectrochemical Series

The spectrochemical series ranks the ligands in order of their ability to cause the energy splitting in the metal orbitals. The series can be given as: I^− < Br^− < Cl^− < F^− < OH^− < H_2O < NH_3 < en < NO_2^− < CN^− Ammonia (NH_3) and water (H_2O) are stronger ligands than chloride (Cl^−) in this series. Stronger ligands produce higher energy splitting.
02

Energy Splitting in Octahedral and Tetrahedral Complexes

Typically, the energy splitting in octahedral complexes (Δ_oct) is much larger than the energy splitting in tetrahedral complexes (Δ_tet). The tetrahedral complexes usually have 4/9 Δ_oct.
03

Assigning Colors Based on Absorbed Energy

The color of a complex is determined by the color of light it absorbs. The complementary color of the absorbed light is the color we see. The energy and wavelength of absorbed light are inversely related, which means that a higher energy transition requires a shorter wavelength. As we know, the energy of light increases in the order: red < orange < yellow < green < blue < violet. Based on the spectrochemical series, the complex with NH_3 ligands should experience the largest energy splitting, followed by the one with H_2O ligands. The complex with Cl^− ligands will have the lowest energy splitting as it is a tetrahedral complex. 1. \([Co(NH_3)_6]^{2+}\): The largest energy splitting - absorbs lower wavelength (blue/violet) - appears yellow/pale orange (complementary color). 2. \([Co(H_2O)_6]^{2+}\): The intermediate energy splitting - absorbs mid-wavelength (green/blue) - appears pink/red (complementary color). 3. \([CoCl_4]^{2-}\): The smallest energy splitting - absorbs higher wavelength (red/orange) - appears blue/green (complementary color). So, the colors of the complexes are: - \([Co(NH_3)_6]^{2+}\): Yellow - \([Co(H_2O)_6]^{2+}\): Pink - \([CoCl_4]^{2-}\): Blue

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Most popular questions from this chapter

A four-coordinate complex \(\mathrm{MA}_{2} \mathrm{B}_{2}\) is prepared and found to have two different isomers. Is it possible to determine from this information whether the complex is square planar or tetrahedral? If so, which is it?

If the lobes of a given \(d\)-orbital point directly at the ligands, will an electron in that orbital have a higher or lower energy than an electron in a \(d\)-orbital whose lobes do not point directly at the ligands?

Give the number of (valence) \(d\) electrons associated with the central metal ion in each of the following complexes: (a) \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right (\mathbf{b})\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2},(\mathbf{c}) \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]\) (d) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{ClO}_{4},(\mathbf{e})[\operatorname{Sr}(\mathrm{EDTA})]^{2-}\)

Given the colors observed for \(\mathrm{VO}_{4}^{3-}\) (orthovanadate ion), \(\mathrm{CrO}_{4}^{2-}\) (chromate ion), and \(\mathrm{MnO}_{4}^{-}\) (permanganate ion (see Exercise \(23.84 ),\) what can you say about how the energy separation between the ligand orbitals and the empty \(d\) orbitals changes as a function of the oxidation state of the transition metal at the center of the tetrahedral anion?

(a) In early studies it was observed that when the complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Br}\) was placed in water, the electrical conductivity of a 0.05\(M\) solution changed from an initial value of 191 \(\mathrm{ohm}^{-1}\) to a final value of 374 \(\mathrm{ohm}^{-1}\) over a period of an hour or so. Suggest an explanation for the observed results.(See Exercise 23.69 for relevant comparison data.) (b) Write a balanced chemical equation to describe the reaction. (c) \(A 500\)-mL solution is made up by dissolving 3.87g of the complex. As soon as the solution is formed, and before any change in conductivity has occurred, a 25.00-mL portion of the solution is titrated with 0.0100 \(\mathrm{M} \mathrm{AgNO}_{3}\) solution. What volume of AgNO \(_{3}\) solution do you expect to be required to precipitate the free \(\operatorname{Br}^{-}(a q) ?(\mathbf{d})\) Based on the response you gave to part (b), what volume of \(\mathrm{AgNO}_{3}\) solution would be required to titrate a fresh 25.00 -mL sample of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Br}\) after all conductivity changes have occurred?

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