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Magazine Chapter 14: Problem 25
What is the magnetic moment of \(\mathrm{K}_{3}\left[\mathrm{FeF}_{6}\right] ?\) (a) 5.91 B.M. (b) 4.89 B.M. (c) 3.87B.M. (d) 6.92 B.M.
Short Answer
Expert verified
The magnetic moment of \( \mathrm{K}_{3}\left[\mathrm{FeF}_{6}\right] \) is 5.91 B.M. (option a).
Step by step solution
01
Identify the Metal and its Oxidation State
In the complex \( \mathrm{K}_{3}[\mathrm{FeF}_{6}] \), the metal ion is \( \mathrm{Fe} \). The complex has three potassium ions \( \mathrm{K}^+ \) with a charge of +1 each. Therefore, the charge of the complex anion \( [\mathrm{FeF}_{6}]^{3-} \) must be -3. Thus, the oxidation state of \( \mathrm{Fe} \) is +3.
02
Determine the Electron Configuration
As \( \mathrm{Fe} \) is in oxidation state +3, it loses 3 electrons. The electron configuration of \( \mathrm{Fe}^{3+} \) is \([\mathrm{Ar}] \, 3d^5 \).
03
Determine the Number of Unpaired Electrons
\( \mathrm{Fe}^{3+} \) has a \( 3d^5 \) configuration, which means there are 5 electrons in the d-orbitals. According to Hund's rule, these electrons will fill each of the five \( d \) orbitals singly with parallel spins, resulting in 5 unpaired electrons.
04
Use the Formula to Calculate Magnetic Moment
The magnetic moment \( \mu \) can be calculated using the formula \( \mu = \sqrt{n(n+2)} \) Bohr Magneton (B.M.), where \( n \) is the number of unpaired electrons. Substituting \( n = 5 \), we get: \[ \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.91 \, \text{B.M.} \]
05
Verify the Options
The calculated magnetic moment is approximately 5.91 B.M., which matches option (a).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Oxidation State Determination
The oxidation state of an element in a compound indicates the theoretical charge it would have if all bonds were completely ionic. To determine the oxidation state of an element, follow these steps:
- Identify all ions in a compound and their known charges. For example, in the compound \( \mathrm{K}_{3}[\mathrm{FeF}_{6}] \), potassium \((\mathrm{K}^+)\) has a +1 charge.
- Find the total charge of the ions. Here, three potassium ions total a +3 charge.
- Deduce the charge of the anion. The charge on \([\mathrm{FeF}_{6}]^{3-}\) is -3, balancing the +3 from the potassium ions.
- The oxidation state of the central metal, iron (\(\mathrm{Fe}\)), is determined by solving the equation: charge of \(\mathrm{Fe} + \) charge of fluorides \(\mathrm{(usually, -1 \, each)} = -3\).
Electron Configuration
Electron configuration refers to the arrangement of electrons in an atom's orbitals. It dictates how electrons are distributed among the different atomic orbits and subshells.
- Start with the ground-state configuration of the neutral atom鈥攊n this case, iron (\(\mathrm{Fe}\)), which is \([\mathrm{Ar}]\,4s^2\,3d^6\).
- For \(\mathrm{Fe}^{3+}\), three electrons are removed, typically starting with the highest energy level or outermost shell: resulting in \([\mathrm{Ar}]\,3d^5\).
- Electron configurations are essential as they indicate what subshells a metal ion's electrons will occupy.
Unpaired Electrons
Unpaired electrons are electrons in an atom that are not paired with another electron in an orbital.
- According to the electron configuration \(3d^5\) for \(\mathrm{Fe}^{3+}\), there are 5 electrons in the d subshell.
- Using Hund's rule, distribute one electron to each of the five d orbitals before any pairing takes place.
- This results in each electron occupying its own orbital with parallel spins, resulting in 5 unpaired electrons.
Hund's Rule
Hund's Rule is a principle used to determine the electron arrangement in orbitals.
- It states that electrons will fill degenerate orbitals (orbitals with the same energy level) singly as much as possible with parallel spins before pairing starts.
- This minimizes electron repulsions in an atom, contributing to more stable electron arrangements.
- In the case of \(\mathrm{Fe}^{3+}\), with \(3d^5\), Hund's Rule explains why each of the five 3d orbitals contains a singly-occupied electron.
