WEB Indicate whether each of the following is true or false. If the statement is false, correct it. (a) The coordination number of iron(III) in \(\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{4}(\mathrm{en})^{3+}\) is 5 . (b) \(\mathrm{Ni}(\mathrm{CN})_{6}^{4-}\) is expected to absorb at a longer wavelength than \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}{ }^{2+}\)

Short Answer

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Question: Determine the coordination number of iron(III) in the complex \(\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{4}(\mathrm{en})^{3+}\) and compare the absorption wavelengths of \(\mathrm{Ni}(\mathrm{CN})_{6}^{4-}\) and \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}{ }^{2+}\). Answer: The coordination number of iron(III) in the complex \(\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{4}(\mathrm{en})^{3+}\) is 6. The complex \(\mathrm{Ni}(\mathrm{CN})_{6}^{4-}\) is expected to absorb at a shorter wavelength than \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}{ }^{2+}\).

Step by step solution

01

(a) Determining the coordination number of iron(III)

The coordination number of a metal ion in a complex is determined by the number of ligand donor atoms directly bonded to it. In the complex \(\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{4}(\mathrm{en})^{3+}\), there are 4 ammonia ligands (\(\mathrm{NH}_{3}\)) and 1 ethylenediamine ligand (en). Each ammonia ligand provides 1 donor atom (N), while the ethylenediamine ligand (en) provides 2 donor atoms (2 N). To find the coordination number, sum up the number of donor atoms provided by all ligands: \(\text{Coordination number} = 4(\text{from } \mathrm{NH}_{3}s) + 2(\text{from } \mathrm{en}) = 6\) Thus, the statement is false, since the actual coordination number of iron(III) ion in the complex is 6.
02

(b) Comparing the absorption wavelengths of given complexes

To determine which complex has a longer absorption wavelength, we need to understand crystal field theory. The magnitude of the crystal field splitting parameter, \(\Delta_{o}\), affects the wavelength of light absorbed in a d-d transition. Complexes with stronger ligands (ligands that cause a larger splitting) will have a higher \(\Delta_{o}\) and will, therefore, absorb light at a lower wavelength according to the approximate relation \(\Delta_{o} = h\frac{c}{\lambda}\). Hence, the complex with a larger \(\Delta_{o}\) will have a shorter wavelength of absorption. The spectrochemical series orders ligands depending on their ability to split the d orbitals of a transition metal. Based on the spectrochemical series, the cyanide ligand (CN-) is a stronger field ligand compared to ammonia (NH3). Thus, the \(\Delta_{o}\) for \(\mathrm{Ni}(\mathrm{CN})_{6}^{4-}\) will be larger than that of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}{ }^{2+}\), resulting in the \(\mathrm{Ni}(\mathrm{CN})_{6}^{4-}\) complex absorbing light at a shorter wavelength than \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}{ }^{2+}\). The statement is false. The correct statement is: \(\mathrm{Ni}(\mathrm{CN})_{6}^{4-}\) is expected to absorb at a shorter wavelength than \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}{ }^{2+}\).

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Most popular questions from this chapter

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Why is \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) diamagnetic while \(\left[\mathrm{CoF}_{6}\right]^{3-}\) is paramagnetic?

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