a. In the absorption spectrum of the complex ion \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) there is a band corresponding to the absorption of a photon of light with an energy of \(1.75 \times 10^{4} \mathrm{cm}^{-1}\). Given 1 \(\mathrm{cm}^{-1}=1.986 \times 10^{-23} \mathrm{J},\) what is the wavelength of this photon? b. The \(\mathrm{Cr}-\mathrm{N}-\mathrm{C}\) bond angle in \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) is predicted to be \(180^{\circ} .\) What is the hybridization of the \(\mathrm{N}\) atom in the NCS - ligand when a Lewis acid-base reaction occurs between \(\mathrm{Cr}^{3+}\) and \(\mathrm{NCS}^{-}\) that would give a \(180^{\circ}\) \(\mathrm{Cr}-\mathrm{N}-\mathrm{C}\) bond angle? \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) undergoes substitution by ethylenediamine (en) according to the equation \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}+2 \mathrm{en} \longrightarrow \mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}+4 \mathrm{NCS}^{-}\) Does \(\operatorname{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}\) exhibit geometric isomerism? Does \(\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}\) exhibit optical isomerism?

Short Answer

Expert verified
The wavelength of the photon is approximately \(571\: \mathrm{nm}\). The hybridization of the N atom in the NCS- ligand is sp. The complex \(\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}\) exhibits geometric isomerism, with possible arrangements being trans and cis isomers. The complex can exhibit optical isomerism, depending on the arrangement of the ligands; while the trans isomer does not exhibit optical isomerism, the cis isomer does.

Step by step solution

01

Find the Wavelength of the Photon

The energy of a photon can be calculated using the formula: \(E_{photon} = \dfrac{hc}{\lambda}\), where h is the Planck's constant, c is the speed of light, and \(\lambda\) is the wavelength of the photon. Given the energy per cm and a conversion factor for Joules, we can find the wavelength. Step 1: Convert the given energy to Joules. Given energy: \(1.75 \times 10^4\: \mathrm{cm}^{-1}\), Conversion factor: \(1\:\mathrm{cm}^{-1} = 1.986 \times 10^{-23} \:\mathrm{J}\), Energy in Joules: \((1.75 \times 10^4 \:\mathrm{cm}^{-1}) \times (1.986 \times 10^{-23} \:\mathrm{J}\:\mathrm{cm}) = 3.475 \times 10^{-19} \mathrm{J}\) Step 2: Calculate the wavelength. Using the energy-wavelength relationship: \(\lambda = \dfrac{hc}{E_{photon}}\), where h = \(6.626 \times 10^{-34}\: \mathrm{Js}\) and c = \(3.0 \times 10^8\: \mathrm{m/s}\), \(\lambda = \dfrac{(6.626 \times 10^{-34}\: \mathrm{Js})(3.0 \times 10^8\: \mathrm{m/s})}{3.475 \times 10^{-19} \mathrm{J}} \approx 5.71 \times 10^{-7} \:\mathrm{m}\) or \(571 \:\mathrm{nm}\) The wavelength of the photon is approximately \(571\: \mathrm{nm}\).
02

Determine the Hybridization of N Atom

To obtain a \(180^{\circ}\) Cr-N-C bond angle, the hybridization of the N atom in the NCS- ligand should result in a linear geometry around the N atom. The required hybridization is sp.
03

Analyze Geometric Isomerism

Geometric isomerism occurs when there are different arrangements of ligands around a central metal atom. In the complex \(\rm{Cr(NCS)_2(en)_2^{+}}\), there are two NCS- ligands and two en ligands around the Cr atom. The possible arrangements for these ligands are trans and cis isomers. Hence, the complex exhibits geometric isomerism.
04

Analyze Optical Isomerism

Optical isomerism occurs when a compound has a non-superimposable mirror image, which means it cannot be laid on top of its mirror image and have all their atoms match up. In this case, the trans isomer of \(\rm{Cr(NCS)_2(en)_2^{+}}\) has a plane of symmetry and does not exhibit optical isomerism. However, the cis isomer does not have a plane of symmetry, thus it can exhibit optical isomerism. Therefore, the complex can exhibit optical isomerism, depending on the arrangement of the ligands.

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Most popular questions from this chapter

Which of the following ligands are capable of linkage isomerism? Explain your answer. $$\mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}, \mathrm{NO}_{2}^{-}, \mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}, \mathrm{OCN}^{-}, \mathrm{I}^{-}$$

Name the following coordination compounds. a. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}\) b. \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{I}_{3}\) c. \(\mathrm{K}_{2}\left[\mathrm{PtCl}_{4}\right]\) d. \(K_{4}\left[P_{t} C l_{6}\right]\) e. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}\) f. \(\left[\operatorname{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right]\)

\(\mathrm{CoCl}_{4}^{2-}\) forms a tetrahedral complex ion and \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) forms an octahedral complex ion. What is wrong about the following statements concerning each complex ion and the \(d\) orbital splitting diagrams? a. \(\mathrm{CoCl}_{4}^{2-}\) is an example of a strong-field case having two unpaired electrons. b. Because \(\mathrm{CN}^{-}\) is a weak-field ligand, \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) will be a low-spin case having four unpaired electrons.

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The complex ion \(\mathrm{PdCl}_{4}^{2-}\) is diamagnetic. Propose a structure for \(\mathrm{PdCl}_{4}^{2-}\)

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