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Chapter 21: Problem 55

The CrF_ \(4-\) ion is known to have four unpaired electrons. Does the \(\mathrm{F}^{-}\) ligand produce a strong or weak field?

Short Answer

Expert verified
The CrF₄⁻ ion has a tetrahedral geometry with a Chromium (Cr) atom at the center in the 0 oxidation state. The electronic configuration of Cr in this complex leads to the presence of four unpaired electrons distributed across the higher and lower energy d orbitals after crystal field splitting occurs. Since more unpaired electrons occupy the higher energy orbitals, this indicates that the F⁻ ligand produces a weak field.

Step by step solution

01

Determine the central metal ion and its oxidation state

We need to first identify the central metal ion in the given complex ion. In this case, it's Chromium (Cr). Then, we should determine the oxidation state of the central metal ion in the complex. Since the fluorine ligands have a charge of -1 each, and the overall charge of the complex ion is -1, the oxidation state of Cr is: \[Cr^{(4-4)} = Cr^{0}\]
02

Identify the electronic configuration of the central metal ion

Now, we look at the electronic configuration of Cr. The atomic number of Cr is 24, and its electronic configuration is: \[1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^5\ 4s^1\] Since the oxidation state of Cr in the complex is 0, the electronic configuration remains the same. The electrons in the 3d and 4s orbitals are the most relevant to the problem at hand.
03

Determine the coordination number and geometry of the complex

The CrF₄⁻ ion is a tetrahedral complex, as it has four F⁻ ligands surrounding the central Cr atom. Therefore, its coordination number is 4.
04

Analyze the crystal field splitting in the complex

For tetrahedral complexes, the crystal field splitting separates the d orbitals into a higher energy set of three orbitals (d(xy), d(xz), and d(yz)) and a lower energy set of two orbitals (d(z^2) and d(x^2-y^2)). In CrF₄⁻, the four unpaired electrons can be arranged in the d orbitals as follows: - Three electrons in the higher energy set (d(xy), d(xz), d(yz)) - One electron in the lower energy set (d(z^2) or d(x^2-y^2)) As there are more unpaired electrons in the higher energy orbitals than in the lower energy orbitals, this indicates that the crystal field splitting is relatively small and that the F⁻ ligand results in a weak field.
05

Conclude whether the F⁻ ligand produces a strong or weak field

The presence of four unpaired electrons in the CrF₄⁻ ion, which mainly occupy the higher energy 3d orbitals after crystal field splitting, indicates a weak field created by the F⁻ ligands.

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Most popular questions from this chapter

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

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

Use standard reduction potentials to calculate $\mathscr{C}^{\circ}, \Delta G^{\circ},\( and \)K$ (at 298 K) for the reaction that is used in production of gold: $$2 \mathrm{Au}(\mathrm{CN})_{2}-(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q)$$ The relevant half-reactions are $$\begin{aligned} \operatorname{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-0.60 \mathrm{V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-1.26 \mathrm{V} \end{aligned}$$

When 6\(M\) ammonia is added gradually to aqueous copper(II) nitrate, a white precipitate forms. The precipitate dissolves as more 6\(M\) ammonia is added. Write balanced equations to explain these observations. [Hint: \(\mathrm{Cu}^{2+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} . ]\)

Draw all geometrical and linkage isomers of \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{NO}_{2}\right)_{2}\)
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