Problem 100
Question
Ferrous ion change to \(\mathrm{X}\) ion, on reacting with acidified hydrogen peroxide. The number of d-electrons present in \(X\) and its magnetic moment (in \(\mathrm{BM})\) are respectively (a) 5 and \(4.9\) (b) 4 and \(5.92\) (c) 6 and \(6.95\) (d) 5 and \(5.92\)
Step-by-Step Solution
Verified Answer
The answer is (d) 5 and 5.92.
1Step 1: Analyze Original Ion
Identify the electron configuration of the initial ion. Ferrous ion is Fe\(^{2+}\) which originates from iron (Fe) with atomic number 26. Fe\(^{2+}\) has lost 2 electrons, so its electron configuration is \([Ar] 3d^6\), having 6 d-electrons.
2Step 2: Determine Reaction Behavior
Understand how ferrous ion Fe\(^{2+}\) changes on reacting with acidified hydrogen peroxide. In such reactions, Fe\(^{2+}\) is oxidized to ferric ion Fe\(^{3+}\). The electron configuration for Fe\(^{3+}\) is \([Ar] 3d^5\). Therefore, ion \(X\) is Fe\(^{3+}\) and has 5 d-electrons.
3Step 3: Calculate Spin-Only Magnetic Moment
The spin-only magnetic moment \( \mu \) is given by the formula \( \mu = \sqrt{n(n+2)} \), where \(n\) is the number of unpaired electrons. For Fe\(^{3+}\), there are 5 unpaired d-electrons (3d half-filled). Thus, \( \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \) BM.
4Step 4: Match Answers with Calculations
Compare the number of d-electrons and the calculated magnetic moment to the provided options. We have 5 d-electrons and a magnetic moment of approximately 5.92 BM. The correct choice matching these values is option (d) 5 and 5.92.
Key Concepts
Oxidation StatesElectron ConfigurationMagnetic Moment
Oxidation States
Oxidation states are essential for understanding how atoms interact during chemical reactions. The oxidation state indicates the degree of oxidation, or electron loss, of an atom in a chemical compound. In coordination chemistry, we often look at the change in oxidation state to determine how a metal ion behaves under different conditions.
The process through which the ferrous ion (\( \text{Fe}^{2+} \)) changes to ferric ion (\( \text{Fe}^{3+} \) ) when reacting with acidified hydrogen peroxide is a classic example of oxidation.
In this reaction, iron loses one additional electron, moving from an oxidation state of +2 in ferrous ion to +3 in ferric ion. This electron loss is crucial because it affects not only the charge but also the electron configuration of the ion, further influencing its reactivity and properties.
The process through which the ferrous ion (\( \text{Fe}^{2+} \)) changes to ferric ion (\( \text{Fe}^{3+} \) ) when reacting with acidified hydrogen peroxide is a classic example of oxidation.
In this reaction, iron loses one additional electron, moving from an oxidation state of +2 in ferrous ion to +3 in ferric ion. This electron loss is crucial because it affects not only the charge but also the electron configuration of the ion, further influencing its reactivity and properties.
Electron Configuration
Understanding electron configuration is the key to predicting and explaining the chemical behavior of elements, especially transition metals.
For iron, which has an atomic number of 26, the ground state electron configuration is \([ \text{Ar} ] 3d^6 4s^2\). When it loses two electrons to form ferrous (\( \text{Fe}^{2+} \)), the configuration changes to \([ \text{Ar} ] 3d^6 \), showing that the two electrons are removed from the 4s orbital and one from the 3d orbital.
Upon further oxidation to form ferric (\( \text{Fe}^{3+} \)), another electron is removed from the 3d orbital, changing the configuration to \([ \text{Ar}] 3d^5\). This change results in a half-filled d orbital, which is notably stable due to the symmetrical arrangement of electrons.
The electron configuration not only dictates the chemical reactivity but also impacts the magnetic properties of the substance.
For iron, which has an atomic number of 26, the ground state electron configuration is \([ \text{Ar} ] 3d^6 4s^2\). When it loses two electrons to form ferrous (\( \text{Fe}^{2+} \)), the configuration changes to \([ \text{Ar} ] 3d^6 \), showing that the two electrons are removed from the 4s orbital and one from the 3d orbital.
Upon further oxidation to form ferric (\( \text{Fe}^{3+} \)), another electron is removed from the 3d orbital, changing the configuration to \([ \text{Ar}] 3d^5\). This change results in a half-filled d orbital, which is notably stable due to the symmetrical arrangement of electrons.
The electron configuration not only dictates the chemical reactivity but also impacts the magnetic properties of the substance.
Magnetic Moment
The magnetic moment of a transition metal ion is a measure of its magnetic properties, mostly arising from the spin of unpaired electrons. Generally, ions with more unpaired electrons exhibit stronger magnetic properties.
In our example, ferric ion (\( \text{Fe}^{3+} \)) has 5 unpaired electrons in the \( 3d \) orbital. The formula to determine the spin-only magnetic moment is given by \( \mu = \sqrt{n(n+2)} \), where \(n\) represents the number of unpaired electrons.
For the ferric ion, this calculation results in a magnetic moment of approximately \(5.92 \text{ BM}\) (Bohr Magnetons). The high value reflects the presence of several unpaired electrons, making \( \text{Fe}^{3+} \) paramagnetic. This property is significant in coordination chemistry and plays a vital role in understanding the behavior of coordination compounds in magnetic fields.
In our example, ferric ion (\( \text{Fe}^{3+} \)) has 5 unpaired electrons in the \( 3d \) orbital. The formula to determine the spin-only magnetic moment is given by \( \mu = \sqrt{n(n+2)} \), where \(n\) represents the number of unpaired electrons.
For the ferric ion, this calculation results in a magnetic moment of approximately \(5.92 \text{ BM}\) (Bohr Magnetons). The high value reflects the presence of several unpaired electrons, making \( \text{Fe}^{3+} \) paramagnetic. This property is significant in coordination chemistry and plays a vital role in understanding the behavior of coordination compounds in magnetic fields.
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