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Chapter 23: Problem 56

Give the number of (valence) \(d\) electrons associated with the central metal ion in each of the following complexes: (a) \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\), (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}\) (c) \(\mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]\), (d) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{ClO}_{4}\), (c) \([\mathrm{Sr}(\mathrm{EDTA})]^{2-}\) -

Short Answer

Expert verified
The number of valence d-electrons in each complex are: (a) 5 d-electrons (b) 5 d-electrons (c) 10 d-electrons (d) 3 d-electrons (e) 0 d-electrons

Step by step solution

01

(Step 1: Identification of Central Metal Ion)

In each complex, identify the central metal ion. In this case, they are all transition metal ions. (a) Fe (b) Mn (c) Ag (d) Cr (e) Sr
02

(Step 2: Determining Oxidation States)

Determine the oxidation state of the central metal ion in each complex. (a) In \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\), the total charge of the complex is 0, and K is +1, so [Fe(CN)6] is -3. The charge on CN is -1, and there are 6 CN ligands, so the charge on Fe is +3. Thus, Fe has an oxidation state of +3. (b) In $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}$, the total charge of the complex is 0, and NO3 is -1, so [Mn(H2O)6] is +2. H2O has 0 charge, so Mn has an oxidation state of +2. (c) In \(\mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]\), the total charge of the complex is 0, and Na is +1, so [Ag(CN)2] is -1. The charge on CN is -1, and there are 2 CN ligands, so the charge on Ag is +1. Thus, Ag has an oxidation state of +1. (d) In $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{ClO}_{4}$, the total charge of the complex is 0, and ClO4 is -1, so [Cr(NH3)4Br2] is +1. NH3 has 0 charge, and Br is -1 with 2 Br ligands, so Cr has an oxidation state of +3. (e) In \([\mathrm{Sr}(\mathrm{EDTA})]^{2-}\), the total charge of the complex is -2, and EDTA is 4-, so Sr has an oxidation state of +2.
03

(Step 3: Finding d-electron count)

Use the electronic configuration of each central metal cation to find the number of d-electrons. (a) Fe+3: Fe has an atomic number of 26, so its electronic configuration is [Ar] 3d6 4s2. Fe+3 loses 3 electrons, so its configuration is [Ar] 3d5. Fe+3 has 5 d-electrons. (b) Mn+2: Mn has an atomic number of 25, so its electronic configuration is [Ar] 3d5 4s2. Mn+2 loses 2 electrons, so its configuration is [Ar] 3d5. Mn+2 has 5 d-electrons. (c) Ag+1: Ag has an atomic number of 47, so its electronic configuration is [Kr] 4d10 5s1. Ag+1 loses 1 electron, so its configuration is [Kr] 4d10. Ag+1 has 10 d-electrons. (d) Cr+3: Cr has an atomic number of 24, so its electronic configuration is [Ar] 3d5 4s1. Cr+3 loses 3 electrons, so its configuration is [Ar] 3d3. Cr+3 has 3 d-electrons. (e) Sr+2: Sr has an atomic number of 38, so its electronic configuration is [Kr] 4d0 5s2. Sr+2 loses 2 electrons, so its configuration is [Kr] 4d0. Sr+2 has 0 d-electrons. In conclusion: (a) \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) has 5 d-electrons. (b) $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}$ has 5 d-electrons. (c) \(\mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]\) has 10 d-electrons. (d) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{ClO}_{4}$ has 3 d-electrons. (e) \([\mathrm{Sr}(\mathrm{EDTA})]^{2-}\) has 0 d-electrons.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation State
The oxidation state of a central metal ion plays a pivotal role in determining the properties of transition metal complexes. It directly influences the complex's reactivity, color, and magnetic behavior. Understanding the oxidation state begins with a simple rule: the sum of oxidation states of all atoms in a neutral molecule must be zero, and in an ion, it must equal the ion's charge. This is exemplified in the complex \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\), where potassium has a +1 oxidation state and since there are 3 potassium ions contributing a total of +3 charge, the iron must balance this with a -3 charge, after accounting for the 6 cyanide ligands each with a -1 charge.

Additionally, predicting the oxidation state in multi-atom ligands such as EDTA, a strong chelating agent that binds through multiple atoms, can be more complex. Here, recognizing the ligand's overall charge is critical. For instance, in \([\mathrm{Sr}(\mathrm{EDTA})]^{2-}\), where EDTA is 4-, the strontium must have a +2 oxidation state to account for the charge.
Transition Metal Complexes
Transition metal complexes contain a central transition metal ion surrounded by molecules or anions, called ligands. These complexes exhibit unique chemical and physical properties due to the varied coordination numbers, geometries, and electronic configurations of the metal ions. Central to understanding these properties is the d-electron count, which gives insight into the metal's behavior in bonding and reactions. For instance, in the complex \(\left[\mathrm{Mn}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}\), the manganese ion is surrounded by six water molecules, which are neutral ligands. The d-electron count for Mn+2 is 5, which tells us about the complex's magnetic properties and potential reactivity.
Ligand Field Theory
Ligand field theory (LFT) is an advanced explanation of how ligands affect the distribution of the d-orbital electrons in transition metal complexes. These interactions can split the d-orbitals into different energy levels, leading to characteristic absorption of light and the colorful nature of many complexes. LFT helps predict electronic transitions and the resultant color of the complex, contributing to the field of spectroscopy. For example, in the complex \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right]\mathrm{ClO}_{4}\), the presence of ligands such as \(\mathrm{NH}_{3}\) and \(\mathrm{Br}^{-}\) creates an octahedral field, splitting the d-orbitals of chromium and affecting the complex's absorbance spectrum and its observable color.
Coordination Chemistry
Coordination chemistry is the study of compounds that feature an array of coordinate covalent bonds between a central metal atom and one or more ligands. These bonds are formed when ligands donate a pair of electrons to an empty orbital on the metal ion, creating a coordination complex. Understanding coordination chemistry requires knowledge of coordination numbers, the types of ligands involved, and the geometry of the complex. The number and arrangement of ligands around a central atom can dramatically affect a complex's stability, solubility, and reactivity. For instance, in \(\mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]\), the silver ion forms a linear complex with two cyanide ions, which is indicative of a coordination number of two. This affects not just the shape, but also the potential biological activity of the compound, as in the case of many silver-based antimicrobials.

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Most popular questions from this chapter

The complex \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) contains five unpaired electrons. Sketch the energy-level diagram for the \(d\) arbitals, and indicate the placement of electrons for this complex ion. Is the ion a high-spin or a low-spin complex?

(a) Sketch a diagram that shows the definition of the crystal-field splitting energy \((\Delta)\) for an octahedral crystal field. (b) What is the relationship between the magnitude of \(\Delta\) and the energy of the \(d-d\) transition for a \(d^{2}\) complex? (c) Calculate \(\Delta\) in \(\mathrm{kJ} / \mathrm{mol}\) if a \(d^{1}\) complex has an absorption maximum at \(545 \mathrm{~nm}\).

Write names for the following coordination compounds: (a) \(\left[\mathrm{Cd}(\mathrm{en}) \mathrm{Cl}_{2}\right]\) (b) \(\mathrm{K}_{4}\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]\) (c) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{s}\left(\mathrm{CO}_{3}\right)\right] \mathrm{Cl}\) (d) \(\left[\operatorname{lr}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\left(\mathrm{NO}_{3}\right)_{3}\)

Pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right),\) abbreviated py, is the molecule (a) Would you expect pyridine to act as a monodentate or bidentate ligand? (b) For the equilibrium reaction $$\left[\mathrm{Ru}(\mathrm{py})_{4}(\mathrm{bipy})\right]^{2+}+2 \mathrm{py} \Longrightarrow\left[\mathrm{Ru}(\mathrm{py})_{6}\right]^{2+}+\mathrm{bipy}$$ would you predict the equilibrium constant to be larger or smaller than one?

A classmate says, "A weak-field ligand usually means the complex is high spin." Is your clasmmate correct? Explain.
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