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 \(\mathrm{NCS}^{-}\) 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 sub- stitution 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 \(\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}\) exhibit geometric isomerism? Does \(\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}\) exhibit optical isomerism?

Step by step solution

01

1. Calculate the energy of the photon in Joules

To convert the energy of the photon from cm⁻¹ to Joules, we will multiply the given value by the conversion factor: Energy in Joule = \(1.75 \times 10^{4}~\text{cm}^{-1}\) * \(1.986 \times 10^{-23}~\text{J}\) Energy in Joule = \(3.475 \times 10^{-19}~\text{J}\)

02

2. Calculate the wavelength of the photon

Using Planck's equation, \(E = h \cdot \frac{c}{\lambda}\), we can find the wavelength (λ) of the photon: \(3.475 \times 10^{-19}~\text{J} = (6.626 \times 10^{-34}~\text{J}\cdot\text{s}) \cdot \frac{3.00 \times 10^{8} \text{m s}^{-1}}{\lambda}\) Solving for λ, we get: \(\lambda = 5.73 \times 10^{-7}\) meters or 573 nm

03

3. Determine the hybridization of the N atom in NCS⁻

For the N atom in the NCS⁻ ligand, there are three electron domains surrounding it (a triple bond with the C atom, a single bond with the S atom, and one lone pair). According to VSEPR theory, this would be sp² hybridization (1 s orbital and 2 p orbitals). Thus, the hybridization of the nitrogen atom in the NCS⁻ ligand is sp².

04

4. Analyze the structure of Cr(NCS)₂(en)₂⁺ to determine isomerism

The Cr(NCS)₂(en)₂⁺ complex has two NCS⁻ ligands acting as monodentates and two ethylenediamine (en) ligands acting as bidentates. The coordination number of the complex is 6. Therefore, it exists as an octahedral complex. For geometric isomerism, we need to determine if there are geometrically different arrangements of the ligands around the central metal ion. Considering the arrangement of the ligands, there are two possible arrangements: 1. trans-Cr(NCS)₂(en)₂⁺: NCS⁻ ligands are opposite to each other. 2. cis-Cr(NCS)₂(en)₂⁺: NCS⁻ ligands are adjacent to each other. Therefore, the complex exhibits geometric isomerism. For optical isomerism, we need to determine if there is a non-superposable mirror image (chiral) of the complex. Only the cis-Cr(NCS)₂(en)₂⁺ form could possibly exhibit optical isomerism, as it has a chiral carbon in the ethylenediamine ligands. However, the mirror image of the complex is in fact superposable, with the ethylenediamine ligands changing positions but remaining geometrically similar. Thus, there is only one optically active form, and the complex does not exhibit optical isomerism.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!