How many unpaired electrons are present in each of the following? (a) \(\left[\mathrm{CoF}_{6}\right]^{3-}\) (high spin) (b) \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\) (low spin) (c) \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}(\text { low spin })\) (d) \(\left[\mathrm{MnCl}_{6}\right]^{4-}\) (high spin) (e) \(\left[\mathrm{RhCl}_{6}\right]^{3-}\) (low spin)

Short Answer

Expert verified
Unpaired electrons in each complex: (a) 4, (b) 1, (c) 1, (d) 5, (e) 0

Step by step solution

01

Determine Electron Configuration of Metal Ion

For the metal ions in each complex, we need to know the atomic number to determine the number of electrons and thus the electronic configuration. For Co in (a), the atomic number is 27; for Mn in (b), (c), and (d), it is 25; and for Rh in (e), it is 45.
02

Calculate the Oxidation States

Use the charges of the ligands and the overall charge of the complex to calculate the oxidation state of the metal ion. For (a) \(\mathrm{CoF}_{6}\)^{3-}\, each F is -1, contributing -6, so Co must be +3. For (b) \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\) and (c) \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}\), each CN is -1, contributing -6, so Mn is +3 in (b) and +2 in (c). For (d) \(\left[\mathrm{MnCl}_{6}\right]^{4-}\), each Cl is -1, contributing -6, so Mn is +2, and for (e) \(\left[\mathrm{RhCl}_{6}\right]^{3-}\), Rh is +3.
03

Assign Electrons According to Spin

High spin complexes have more unpaired electrons since electrons are added to the orbitals by following Hund's rule. Low spin complexes have paired electrons wherever possible. Use the electron count from the oxidation state in Step 2 to place electrons starting from the lowest available energy level.
04

Determine the Number of Unpaired Electrons

For (a) high spin \(\mathrm{Co}^{3+}\) has an electron configuration \(d^6\), which leads to 4 unpaired electrons. For (b) low spin \(\mathrm{Mn}^{3+}\) with \(d^4\) configuration has 1 unpaired electron. For (c) low spin \(\mathrm{Mn}^{2+}\) with \(d^5\) has 1 unpaired electron. For (d) high spin \(\mathrm{Mn}^{2+}\) with \(d^5\) also has 5 unpaired electrons, and for (e) low spin \(\mathrm{Rh}^{3+}\) with \(d^6\) has 0 unpaired electrons.

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Key Concepts

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

Electron Configuration
Electron configuration is the arrangement of electrons in an atom's or ion's orbitals, which are defined by quantum numbers. Imagine it as a map that shows where each electron lives within an atom. Understanding electron configuration is essential for predicting the chemical, electrical, and magnetic properties of an atom.

For transition metals, the configuration can vary based on oxidation states, as electrons are removed from the highest energy levels first. In the complexes from the exercise, we start with the basic electron configuration of the neutral atom (using the periodic table) and then adjust it based on the metal's oxidation state determined in the exercise. The 'd' orbital electrons are the ones involved in bonding and magnetism, hence they are the primary focus for these transition metal complexes.
Oxidation States
The oxidation state, often referred to as oxidation number, is a measure of the degree of oxidation of an atom in a chemical compound. Think of it as an imaginary charge that an atom would have if all bonds were purely ionic. To find the oxidation state of the metal in a coordination complex, like the examples in the exercise, you subtract the sum of the ligands' charges from the overall charge of the complex.

This is crucial because the oxidation state of the metal affects its electron configuration, which in turn dictates properties such as color, magnetism, and reactivity. In our exercise, different oxidation states of cobalt, manganese, and rhodium dictate the number of electrons in the 'd' orbitals and hence the number of unpaired electrons.
Spin Complexes
Spin complexes relate to the spin of electrons in the d-orbitals of transition metals. When we refer to 'high spin' and 'low spin' complexes, we're talking about the arrangement of electrons in these orbitals, which can have a significant impact on the properties of the complex, including color and magnetism.

High spin complexes occur when a metal ion is surrounded by weak-field ligands, which don't exert enough energy to pair up the electrons. As a result, electrons fill all the orbitals singly before any pairing occurs. Conversely, low spin complexes form when strong-field ligands are present, and electrons pair up in lower energy orbitals first, leading to fewer unpaired electrons. This has been applied in our exercise to predict the number of unpaired electrons in each complex.
Hund's Rule
Hund's Rule is like a guideline for filling subshells with electrons. It tells us that electrons will fill each degenerate orbital in a subshell singularly before pairing up. The rule is all about maximizing the number of unpaired electrons and minimizing repulsion between them.

Imagine a bus with seats in pairs (representing orbitals): people (electrons) would rather sit alone than next to someone. Only when all seats have one person, do people start sitting next to each other. This is what Hund's Rule suggests for electrons and it affects the properties of atoms, ions, and molecules significantly. In coordination chemistry, Hund's Rule helps predict the number of unpaired electrons in high spin complexes, leading to the exercise solutions given.

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Most popular questions from this chapter

Predict the products of each of the following reactions and then balance the chemical equations. (a) Fe is heated in an atmosphere of steam. (b) \(\mathrm{NaOH}\) is added to a solution of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) (c) FeSO \(_{4}\) is added to an acidic solution of \(\mathrm{KMnO}_{4}\) (d) Fe is added to a dilute solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (e) A solution of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{HNO}_{3}\) is allowed to stand in air. (f) \(\mathrm{FeCO}_{3}\) is added to a solution of \(\mathrm{HClO}_{4}\) (g) Fe is heated in air.

Draw diagrams for any cis, trans, and optical isomers that could exist for the following (en is ethylenediamine): (a) \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{NO}_{2}\right) \mathrm{Cl}\right]^{+}\) (b) \(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) (c) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{4}\right]\) (d) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+}\) (e) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\)

Why is the formation of slag useful during the smelting of iron?

Would you expect the \(\mathrm{Mg}_{3}\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]_{2}\) to be diamagnetic or paramagnetic? Explain your reasoning.

A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?

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