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Find the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25 . a. \(6.9 \mathrm{BM}\) b. \(5.9 \mathrm{BM}\) c. \(4.9 \mathrm{BM}\) d. \(3.0 \mathrm{BM}\)

Short Answer

Expert verified
The magnetic moment is approximately 5.9 BM, matching option b.

Step by step solution

01

Determine the Element

The atomic number given is 25, which corresponds to the element manganese (Mn) on the periodic table.
02

Identify the Charge and Electronic Configuration

The ion is divalent, meaning it has a +2 charge. For Mn, the ground state electronic configuration is \[ \text{Mn: [Ar] 4s}^2 3d^5 \]. When it loses two electrons to form a +2 ion, they are removed from the 4s and 3d subshells, resulting in \[ \text{Mn}^{2+}: [Ar] 3d^5 \].
03

Determine the Number of Unpaired Electrons

The 3d subshell in \[ \text{Mn}^{2+} \] has 5 electrons. These are configured as \[ \uparrow \downarrow, \uparrow, \uparrow, \uparrow, \uparrow \], resulting in 5 unpaired electrons.
04

Apply the Magnetic Moment Formula

The magnetic moment \(\mu\) is calculated using the formula: \[\mu = \sqrt{n(n+2)}\]\text{ BM, where \( n \) is the number of unpaired electrons.} Substitute \( n = 5 \): \[\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \text{ BM}\]
05

Choose the Correct Answer

Compare the calculated magnetic moment (approximately 5.9 BM) with the given options. The closest match is option (b) \(5.9 \mathrm{BM}\).

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Key Concepts

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

Divalent Ion
A divalent ion is an ion that carries a 2+ charge. This results when an atom loses two electrons. This can occur with many elements, especially transition metals. The loss of electrons leads to changes in the chemical properties of the element. When an element such as manganese (Mn) forms a divalent ion, it typically loses these electrons from the outermost s or d orbitals.

Manganese in its neutral state normally has an electronic configuration of \([ \text{Ar} ] 4s^2 3d^5\). When Mn becomes a divalent ion \( \text{Mn}^{2+} \), it loses two electrons, resulting in a new configuration of \([ \text{Ar} ] 3d^5\). This process greatly impacts its electronic structure and magnetic properties.
Electronic Configuration
Electronic configuration is the arrangement of electrons in a chemical element’s atomic or molecular orbitals. Each element has a unique electronic configuration that describes the distribution of electrons in different atomic orbitals. For example, the electronic configuration of manganese (Mn) in the ground state is \([ \text{Ar} ] 4s^2 3d^5\).

In the case of ions, the electronic configuration changes as electrons are added or removed. For \( \text{Mn}^{2+} \), electrons are removed from the outermost shells, the 4s and then 3d. This leads to the configuration \([ \text{Ar} ] 3d^5\). This change is important because it influences the atom’s chemical reactivity and magnetic moment.
Unpaired Electrons
Unpaired electrons are electrons that occupy an atomic orbital alone, without a pairing electron. These are crucial for understanding an atom's magnetic properties. In the case of \( \text{Mn}^{2+} \), the electronic configuration \([ \text{Ar} ] 3d^5\) reveals the presence of 5 unpaired electrons. These electrons are organized in such a way that they occupy separate d-orbitals, resulting in unpaired states.

The number of unpaired electrons directly affects the magnetic moment of the atom or ion. More unpaired electrons generally mean a stronger magnetic moment. In manganese's divalent ion form, the 5 unpaired electrons contribute substantially to its magnetic characteristics.
Magnetic Moment Formula
The magnetic moment is a measure of the object's tendency to interact with magnetic fields. For an atom or ion, it depends significantly on the presence of unpaired electrons.

The formula used to calculate the magnetic moment is given by: \(\mu = \sqrt{n(n+2)}\\) \text{Bohr Magnetons (BM)}, where \( n \) is the number of unpaired electrons.

  • For \( \text{Mn}^{2+} \), with 5 unpaired electrons, the calculation is \( \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \text{ BM} \).

This calculation tells us how magnetically active the ion is due to its unpaired electrons, providing key insights into its magnetic properties.

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