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Chapter 21: Problem 51

How many unpaired electrons are in the following complex ions? a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) (low-spin case) b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) c. \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\)

Short Answer

Expert verified
The number of unpaired electrons in the given complex ions are: a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) (low-spin case): 0 unpaired electrons b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): 2 unpaired electrons c. \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\): 1 unpaired electron

Step by step solution

01

Identify the oxidation state for each metal

In each complex, we first need to determine the oxidation state of the central metal atom: a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) : The total charge of the complex is 2+, and ammonia (NH3) is neutral, so the oxidation state of Ru is +2. b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) : The total charge of the complex is 2+, and water (H2O) is neutral, so the oxidation state of Ni is +2. c. \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\) : The total charge of the complex is 3+, and ethylenediamine (en) is neutral, so the oxidation state of V is +3.
02

Write the electron configurations

Next, we will write the electron configurations of the central metal atoms in their corresponding oxidation states: a. Ru+2: Ru has an atomic number of 44, and its electron configuration is [Kr] 4d7 5s1. After losing 2 electrons, we have [Kr] 4d6. b. Ni+2: Ni has an atomic number of 28, and its electron configuration is [Ar] 3d8 4s2. After losing 2 electrons, we have [Ar] 3d8. c. V+3: V has an atomic number of 23, and its electron configuration is [Ar] 3d3 4s2. After losing 3 electrons, we have [Ar] 3d2.
03

Determine the number of unpaired electrons

For each electron configuration, count the number of unpaired electrons in the d orbitals: a. [Kr] 4d6 (low-spin case): In the low-spin case, electrons preferentially occupy the lower energy orbitals first. Here, the d orbitals are completely filled before any electron jumps into a higher energy orbital. Thus, there are no unpaired electrons, and the answer is 0. b. [Ar] 3d8: The d orbitals fill in the following order: 鈫戔啌鈫戔啌鈫戔啌鈫戔啈, so there are 2 unpaired electrons. c. [Ar] 3d2: The d orbitals fill in the following order: 鈫戔啌鈫�, so there is 1 unpaired electron.
04

Final Answers

The number of unpaired electrons in the complex ions are: a. Ru(NH3)6^2+ (low-spin case): 0 unpaired electrons b. Ni(H2O)6^2+: 2 unpaired electrons c. V(en)3^3+: 1 unpaired electron

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Most popular questions from this chapter

Amino acids can act as ligands toward transition metal ions. The simplest amino acid is glycine \(\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)\). Draw a structure of the glycinate anion \(\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-}\right)\) acting as a bidentate ligand. Draw the structural isomers of the square planar complex \(\mathrm{Cu}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2}\right)_{2}\)

a. In the absorption spectrum of the complex ion [Cr(NCS) \(\left._{6}\right]^{3-}\), 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 \(\left[\mathrm{Cr}(\mathrm{NCS})_{6}\right]^{3-}\) 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? \(\left[\mathrm{Cr}(\mathrm{NCS})_{6}\right]^{3-}\) undergoes substitution by ethylenediammine (en) according to the equation \(\left[\mathrm{Cr}(\mathrm{NCS})_{6}\right]^{3-}+2 \mathrm{en} \longrightarrow\left[\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}\right]^{+}+4 \mathrm{NCS}^{-}\) Does \(\left[\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}\right]^{+}\) exhibit geometric isomerism? Does \(\left[\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}\right]^{+}\) exhibit optical isomerism?

Sketch and explain the most likely pattem for the crystal field diagram for the complex ion trans-diamminetetracyanonickelate(II), where \(\mathrm{CN}^{-}\) produces a much stronger crystal field than \(\mathrm{NH}_{3}\). Explain completely and label the \(d\) orbitals in your diagram. Assume the \(\mathrm{NH}_{3}\) ligands lie on the \(z\) axis.

Molybdenum is obtained as a by-product of copper mining or is mined directly (primary deposits are in the Rocky Mountains in Colorado). In both cases it is obtained as \(\mathrm{MoS}_{2}\), which is then converted to \(\mathrm{MoO}_{3}\). The \(\mathrm{MoO}_{3}\) can be used directly in the production of stainless steel for high-speed tools (which accounts for about \(85 \%\) of the molybdenum used). Molybdenum can be purified by dissolving \(\mathrm{MoO}_{3}\) in aqueous ammonia and crystallizing ammonium molybdate. Depending on conditions, either \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Mo}_{2} \mathrm{O}_{7}\) or \(\left(\mathrm{NH}_{4}\right)_{6} \mathrm{Mo}_{7} \mathrm{O}_{24} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) is obtained. a. Give names for \(\mathrm{MoS}_{2}\) and \(\mathrm{MoO}_{3}\). b. What is the oxidation state of Mo in each of the compounds mentioned above?

Carbon monoxide is toxic because it binds more strongly to iron in hemoglobin (Hb) than does \(\mathrm{O}_{2}\). Consider the following reactions and approximate standard free energy changes: $$\begin{array}{cl}\mathrm{Hb}+\mathrm{O}_{2} \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ}\end{array}$$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$\mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2}$$
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