Problem 58 - Chemistry The Central Science

Question

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

Step-by-Step Solution

Answer

The number of valence d electrons of central metal ions in the given complexes are: (a) 5 (b) 5 (c) 10 (d) 3 (e) 0

Step 1: Determine the oxidation state of central metal ions

Use the charge on the complex and the charges on ligands to find the oxidation state of the central metal ion in each complex. (a) K3[Fe(CN)6] (b) [Mn(H2O)6](NO3)2 (c) Na[Ag(CN)2] (d) [Cr(NH3)4Br2]ClO4 (e) [Sr(EDTA)]^2−

Step 2: Find number of valence d electrons using the electronic configuration of the metal ion

(a) In K3[Fe(CN)6], the overall charge on the complex is 0. Therefore, the charge on Fe is -3. The atomic number of Fe is 26, and its ground state electronic configuration is [Ar]3d^64s^2. When Fe loses 3 electrons, it forms Fe(III) with the electronic configuration: [Ar]3d^5. The number of valence d electrons is 5. (b) In [Mn(H2O)6](NO3)2, the overall charge on the complex is +2. The charge on Mn is +2 since the complex is neutral. The atomic number of Mn is 25, and its ground state electronic configuration is [Ar]3d^54s^2. When Mn loses 2 electrons, it forms Mn(II) with the electronic configuration: [Ar]3d^5. The number of valence d electrons is 5. (c) In Na[Ag(CN)2], the overall charge on the complex is 0. Therefore, the charge on Ag is +1. The atomic number of Ag is 47, and its ground state electronic configuration is [Kr]4d^105s^1. When Ag loses 1 electron, it forms Ag(I) with the electronic configuration: [Kr]4d^10. The number of valence d electrons is 10. (d) In [Cr(NH3)4Br2]ClO4, the overall charge on the complex is +1. The charge on Cr is +3 since NH3 ligands are neutral, and Br has a charge of -1. The atomic number of Cr is 24, and its ground state electronic configuration is [Ar]3d^54s^1. When Cr loses 3 electrons, it forms Cr(III) with the electronic configuration: [Ar]3d^3. The number of valence d electrons is 3. (e) In [Sr(EDTA)]^2−, Sr has a charge of +2, and the charge on the EDTA ligand is -4. Therefore, the overall charge on the complex is -2. The atomic number of Sr is 38, and its ground state electronic configuration is [Kr]5s^2. Sr has no valence d electrons with the electronic configuration: [Kr]. The number of valence d electrons is 0. So, the number of valence d electrons of central metal ions in the given complexes are: (a) 5 (b) 5 (c) 10 (d) 3 (e) 0

Key Concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and are crucial in determining how an element will chemically interact with other elements. In coordination complexes, valence electrons particularly involve d electrons in transition metals.

To find the number of valence d electrons in a complex, we must first determine the oxidation state of the central metal ion, which in turn influences its electronic configuration. For instance, as seen in the original exercise, when iron (\(\text{Fe}\)) in \(\text{K}_3[\text{Fe}(\text{CN})_6]\) loses three electrons, it transitions from a neutral state to a +3 oxidation state, revealing 5 d valence electrons left in its configuration: \(\text{[Ar]}3d^5\).

Valence electrons play a fundamental role in bond formation and a metal’s chemical properties. Understanding this concept is key to analyzing and predicting chemical reactions and the stability of complexes.

Oxidation State

The oxidation state of an element in a compound describes the degree of oxidation (loss of electrons) of that element. It is denoted by a positive or negative integer. In coordination chemistry, determining the oxidation state of the central metal ion is crucial for understanding its electronic structure and bonding capacity.

In the given exercises, central metal ion oxidation states are determined by considering the charges of the entire complex and its ligands. For example, for \(\text{K}_3[\text{Fe}(\text{CN})_6]\), the overall neutral charge means the iron has a charge of +3 because each cyanide (\(\text{CN}^-\)) contributes -1, and the three potassium's contribute +1 each. Recognizing these charge interactions helps in calculating the central atom's valency and predicting its behavior in reactions.

Electronic Configuration

The electronic configuration of an atom describes the distribution of its electrons in the atomic orbitals. It speaks volumes about the chemical behavior of an element, particularly in transition metals.

Determining the electronic configuration involves first identifying the oxidation state. From this, you subtract the corresponding number of electrons from the neutral atom's configuration. For example, with manganese (\(\text{Mn}\)) in \([\text{Mn}(\text{H}_2\text{O})_6](\text{NO}_3)_2\), manganese in the +2 oxidation state has lost 2 electrons yielding the configuration \(\text{[Ar]}3d^5\), revealing 5 valence d electrons.

This configuration is essential to predict physical properties and reactivity. Metals often participate in d-d electron transitions, especially when forming colored complexes.

Transition Metals

Transition metals are elements known for their ability to form a variety of oxidation states and complex ions. They are found in groups 3 through 12 of the periodic table and are characterized by the presence of d electrons, which play a significant role in their chemical and physical properties.

The typical properties of transition metals include the formation of colored compounds, variable oxidation states, and the capacity to form stable complex ions. For example, transition metals such as iron and chromium have partially filled d subshells which enable them to bond with various ligands and exhibit different oxidation states, which is evident in the original step by step solution.

Their unique ability to form complexes with ligands makes them critical catalysts in both biological and industrial processes, adding to their prominence in chemical reactions.