Problem 56 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the current order of spin only magnetic moment (in B.M.) of \(\mathrm{Mn}^{2+}, \mathrm{Cr}^{2+}\) and \(\mathrm{V}^{2+}\) ? (a) \(\mathrm{Mn}^{2+}>\mathrm{V}^{2+}>\mathrm{Cr}^{2+}\) (b) \(\mathrm{V}^{2+}>\mathrm{Cr}^{2+}>\mathrm{Mn}^{2+}\) (c) \(\mathrm{Mn}^{2+}>\mathrm{Cr}^{2+}>\mathrm{V}^{2+}\) (d) \(\mathrm{Cr}^{2+}>\mathrm{V}^{2+}>\mathrm{Mn}^{2+}\)

Step-by-Step Solution

Answer

The order is (c) \( \mathrm{Mn}^{2+} > \mathrm{Cr}^{2+} > \mathrm{V}^{2+} \).

Step 1: Determine the Number of Unpaired Electrons

To find the spin-only magnetic moment, first determine the number of unpaired electrons in each ion. For \[ \mathrm{Mn}^{2+} \], the electron configuration is \([\mathrm{Ar}]3d^5\), so it has 5 unpaired electrons. For \( \mathrm{Cr}^{2+} \), the electron configuration is \([\mathrm{Ar}]3d^4\), which means it has 4 unpaired electrons. Lastly, for \( \mathrm{V}^{2+} \), the electron configuration is \([\mathrm{Ar}]3d^3\), which results in 3 unpaired electrons.

Step 2: Use the Magnetic Moment Formula

Calculate the spin-only magnetic moment \( \mu \) using the formula: \[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \]where \( n \) is the number of unpaired electrons. Substitute the number of unpaired electrons for each ion to compute their magnetic moments.

Step 3: Calculate Magnetic Moment for Mn²⁺

For \( \mathrm{Mn}^{2+} \), with 5 unpaired electrons:\[ \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \text{B.M.} \]

Step 4: Calculate Magnetic Moment for Cr²⁺

For \( \mathrm{Cr}^{2+} \), with 4 unpaired electrons:\[ \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{B.M.} \]

Step 5: Calculate Magnetic Moment for V²⁺

For \( \mathrm{V}^{2+} \), with 3 unpaired electrons:\[ \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{B.M.} \]

Step 6: Compare Magnetic Moments

Now compare the calculated magnetic moments: \( \mathrm{Mn}^{2+}: 5.92 \, \text{B.M.} \), \( \mathrm{Cr}^{2+}: 4.90 \, \text{B.M.} \), \( \mathrm{V}^{2+}: 3.87 \, \text{B.M.} \). Thus, the order is \( \mathrm{Mn}^{2+} > \mathrm{Cr}^{2+} > \mathrm{V}^{2+} \).

Key Concepts

Unpaired Electrons

The concept of unpaired electrons is crucial for understanding magnetic properties in atoms. Electrons possess a property called spin, which can be thought of as a tiny magnetic field.
Atoms with unpaired electrons have net magnetic moments and are susceptible to external magnetic fields.
To determine unpaired electrons within an atom or ion, we first evaluate its electron configuration. This provides a map of where each electron resides and how it interacts with others.
For transition metals like manganese (Mn), chromium (Cr), and vanadium (V), the electrons populate the d-orbitals. Here's how unpaired electrons are determined:

• Check the electron configuration of the metal ion.
• Identify how many electrons are present in the d-orbitals.
• Determine which of these are not paired with others.
  
  For example, in ext{Mn}^{2+}, there are five electrons filling the 3d orbital, all unpaired. In contrast, ext{Cr}^{2+} and ext{V}^{2+} have four and three unpaired d-electrons respectively. The presence of these unpaired electrons directly affects the magnetic and chemical properties of the ions.

Transition Metals Chemistry

Transition metals, often found in the center of the periodic table, bring unique chemistry due to their d-orbitals. Unlike main-group elements, transition metals can exist in multiple oxidation states.
This variability arises because transition metals can lose different numbers of their d electrons.
These metals often have unpaired d-electrons leading to interesting magnetic properties.
Some key features of transition metal chemistry include:

• Variable oxidation states: Transition metals can form a variety of different cations.
• Complex formation: They interact with ligands to form complex ions.
• Color: Many transition metals exhibit vivid colors due to d-d electron transitions.
• Magnetism: Unpaired electrons in their d-orbitals often make transition metals paramagnetic or even ferromagnetic.
  For example, ext{Mn}^{2+}, ext{Cr}^{2+}, and ext{V}^{2+} all exhibit interesting magnetic moments essential in various industrial applications, including catalysts and materials for electronic devices.

Electron Configuration

Electron configuration is a systematic way of describing the arrangement of electrons in atoms or ions. It outlines which orbitals are occupied and can give insight into an element's chemical behavior.
The format is based on the electron filling order or Aufbau principle and Hund's rule, which guide the placement of electrons in orbitals.
Transition metals like those in our example such as ext{Mn}^{2+}, ext{Cr}^{2+}, and ext{V}^{2+}, are especially notable because small changes in electron configurations can lead to different properties.

• For ext{Mn}, the neutral atom has an electron configuration of [ ext{Ar}] 3d^5 4s^2. When it becomes ext{Mn}^{2+}, two electrons are removed, typically from the 4s orbital, leading to [ ext{Ar}] 3d^5, which contains five unpaired electrons.
• Likewise, ext{Cr}^{2+} derives from a neutral ext{Cr} atom. Its electron configuration is [ ext{Ar}] 3d^5 4s^1, and as a 2+ ion, it results as [ ext{Ar}] 3d^4 with four unpaired electrons.
• Finally, once ext{V} loses two electrons, it transforms into ext{V}^{2+}, presenting as [ ext{Ar}]3d^3.

Understanding electron configurations allows for predicting magnetic properties and reactivity patterns of transition metals important in both academia and industry.