Problem 66 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

The magnetic moment of the divalent ion is approximately 5.9 B.M.

Step 1: Determine the Electron Configuration

The element with atomic number 25 is manganese (Mn). In its neutral state, the electron configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5\]A divalent ion means it loses two electrons. Typically, these electrons are removed from the outermost shell, which belongs to the 4s orbital in the case of manganese, resulting in the following configuration for \( \text{Mn}^{2+} \): \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^5\]

Step 2: Count the Number of Unpaired Electrons

In the electron configuration \(3d^5\), there are five electrons in the 3d orbital. According to Hund's rule, each of these electrons occupies its own orbital, resulting in five unpaired electrons.

Step 3: Calculate the Magnetic Moment

The magnetic moment \(\mu\) in Bohr Magneton (B.M.) can be calculated using the formula:\[\mu = \sqrt{n(n+2)}\]where \(n\) is the number of unpaired electrons. Substituting \(n = 5\) into the formula gives:\[\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \text{ B.M.}\]

Key Concepts

Electron Configuration

Understanding electron configuration is like looking at the blueprint of an atom. Electrons reside in orbitals around the nucleus, and each orbital can hold a set number of electrons, following the rules of quantum mechanics. For manganese ( Mn ), with an atomic number of 25, the neutral electron configuration is:

• 1s ² 2s ² 2p ⁶ 3s ² 3p ⁶ 4s ² 3d ⁵

Electrons fill orbitals from the lowest energy level upward, which means the 4s orbital fills before the 3d orbital, even though the number 4 seems higher than 3. When manganese forms a divalent ion ( Mn ²⁺), it loses two electrons. These electrons are typically removed from the outermost shell or highest energy level.
For Mn ²⁺, this means the electrons are removed from the 4s orbital, leaving the configuration as:

• 1s ² 2s ² 2p ⁶ 3s ² 3p ⁶ 3d ⁵

This arrangement is vital for understanding how magnetic properties arise.

Unpaired Electrons

Unpaired electrons play a crucial role in determining the magnetic properties of atoms and ions. When electrons fill orbitals, they follow Hund's rule: each orbital in a subshell gets one electron before any pairing occurs. This minimizes electron repulsion and stabilizes the atom.
For manganese ( Mn ²⁺), the 3d ⁵ configuration is key. It has five electrons, one in each of the 3d orbitals:

• This results in all five electrons being unpaired.
• Each unpaired electron acts like a tiny magnet.
• The more unpaired electrons present, the bigger the magnetic moment.

In general, the presence of unpaired electrons leads to paramagnetism, which means the atom or ion is attracted to magnetic fields. The number of unpaired electrons directly relates to the strength of this effect.

Bohr Magneton

The Bohr Magneton (B.M.) is a fundamental constant used to express the magnetic moment of an electron due to its angular momentum and spin. It is the natural unit of the magnetic moment and is represented as:

• \(\mu _B = \dfrac{e\hbar}{2m_e}\)
  • where \(e\) is the charge of the electron,\(\hbar \) is the reduced Planck's constant, and \(m_e\) is the mass of the electron.

However, calculating the magnetic moment of atoms or ions, we use a simplified formula involving unpaired electrons:

• \(\mu = \sqrt{n(n+2)}\)
  • where \(n\) is the number of unpaired electrons.

In the case of Mn²⁺, with five unpaired electrons, this results in a magnetic moment of approximately 5.9 B.M. This signifies that manganese ions possess considerable magnetic strength. Understanding the Bohr Magneton helps us connect quantum physics concepts with observable phenomena like magnetism in ions and atoms.