Problem 155 - A Complete Resource Book in Chemistry for JEE Main

Question

The correct order of magnetic moments (spin only values in B.M.) among the following is \(\quad\) [2004] (Atomic number of \(\mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27\) ) (a) \(\left[\mathrm{MnCl}_{4}\right]^{2-}>\left[\mathrm{CoCl}_{4}\right]^{2-}>\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}\) (b) \(\left[\mathrm{MnCl}_{4}\right]^{2->}[\mathrm{Fe}(\mathrm{CN})]^{4}>\left[\mathrm{CoCl}_{4}\right]^{2-}\) (c) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}>\left[\mathrm{MnCl}_{4}\right]^{2->}\left[\mathrm{CoCl}_{4}\right]^{2-}\) (d) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}>\left[\mathrm{CoCl}_{4}\right]^{2-}>\left[\mathrm{MnCl}_{4}\right]^{2-}\)

Step-by-Step Solution

Answer

The correct order is (a): \([\mathrm{MnCl}_4]^{2-} > [\mathrm{CoCl}_4]^{2-} > [\mathrm{Fe}(\mathrm{CN})_6]^{4-}\).

Step 1: Determine the Electronic Configuration

Find the electronic configuration of the central metal ions in each complex considering the oxidation state. For \([\mathrm{MnCl}_4]^{2-}\), \(\text{Mn}^{2+}\), electronic configuration is \([\text{Ar}] 3d^5\). For \([\mathrm{Fe}(\mathrm{CN})_6]^{4-}\), \(\text{Fe}^{2+}\), electronic configuration is \([\text{Ar}] 3d^6\). For \([\mathrm{CoCl}_4]^{2-}\), \(\text{Co}^{2+}\), electronic configuration is \([\text{Ar}] 3d^7\).

Step 2: Identify the Geometry and Field Strength

Determine the geometry and field ligand strength for each complex. Both \([\mathrm{MnCl}_4]^{2-}\) and \([\mathrm{CoCl}_4]^{2-}\) are tetrahedral (weak field), while \([\mathrm{Fe}(\mathrm{CN})_6]^{4-}\) is octahedral (strong field due to CN ligands).

Step 3: Calculate Unpaired Electrons

Calculate the number of unpaired electrons for each complex. In \([\mathrm{MnCl}_4]^{2-}\), all 5 \(3d\) electrons of \(\text{Mn}^{2+}\) remain unpaired. In \([\mathrm{CoCl}_4]^{2-}\), 3 electrons of \(\text{Co}^{2+}\) remain unpaired. In \([\mathrm{Fe}(\mathrm{CN})_6]^{4-}\), the strong field CN ligands cause pairing of electrons, resulting in 0 unpaired electrons.

Step 4: Calculate Magnetic Moment

Use the formula \(\mu = \sqrt{n(n+2)}\), where \(n\) is the number of unpaired electrons, to calculate the magnetic moment in Bohr Magnetons (B.M.). \([\mathrm{MnCl}_4]^{2-}\) has \(\mu = \sqrt{5(5+2)} = \sqrt{35}\). \([\mathrm{CoCl}_4]^{2-}\) has \(\mu = \sqrt{3(3+2)} = \sqrt{15}\). \([\mathrm{Fe}(\mathrm{CN})_6]^{4-}\) has \(\mu = \sqrt{0(0+2)} = 0\).

Step 5: Compare Magnetic Moments

Now, compare the magnetic moments: \([\mathrm{MnCl}_4]^{2-}\) (high spin - \(\sqrt{35}\) B.M.) > \([\mathrm{CoCl}_4]^{2-}\) (\(\sqrt{15}\) B.M.) > \([\mathrm{Fe}(\mathrm{CN})_6]^{4-}\) (0 B.M.). Therefore, the correct order is option (a) \([\mathrm{MnCl}_4]^{2-} > [\mathrm{CoCl}_4]^{2-} > [\mathrm{Fe}(\mathrm{CN})_6]^{4-}\).

Key Concepts

Electron Configuration

To understand magnetic properties, we must first examine an atom's electron configuration. Electrons fill orbitals following the Aufbau principle, which states that electrons occupy the lowest energy orbitals first. Knowing the number of electrons a metal ion has is essential, especially when considering coordination complexes.

In the given exercise, we have complexes with Mn, Fe, and Co. We know their atomic numbers are 25, 26, and 27, respectively. But in each complex, these metals are in different oxidation states, meaning they have lost some electrons.

- For \([\text{MnCl}_4]^{2-}\), \(\text{Mn}^{2+}\) has lost two electrons, resulting in the electron configuration \([\text{Ar}] 3d^5\).- For \([\text{Fe(CN}_6]^{4-}\), \(\text{Fe}^{2+}\) also loses two electrons, resulting in the electron configuration \([\text{Ar}] 3d^6\).- Finally, \([\text{CoCl}_4]^{2-}\) with \(\text{Co}^{2+}\) results in \([\text{Ar}] 3d^7\).

Recognizing these configurations helps predict magnetic properties. The number of unpaired electrons significantly affects a compound34s magnetic moment.

Crystal Field Theory

Crystal Field Theory (CFT) explains how the electron distribution in transition metal complexes affects their properties, including magnetic moments. CFT focuses on the interaction between electrons in d orbitals and ligands surrounding a metal ion.

In CFT, we consider the splitting of d orbitals into different energy levels due to the presence of surrounding ligands. The energy gap between these levels is influenced by the geometry and nature of the ligands.

- **Tetrahedral Geometry:** In \([\text{MnCl}_4]^{2-}\) and \([\text{CoCl}_4]^{2-}\), the coordination is tetrahedral. Chloride, a weak field ligand, causes minimal splitting, allowing for more unpaired electrons within these 3d orbitals.- **Octahedral Geometry:** For \([\text{Fe(CN}_6]^{4-}\), cyanide is a strong field ligand, causing significant splitting of the 3d orbitals and leading to electron pairing.

This difference in field strength explains why \([\text{Fe(CN}_6]^{4-}\) displays a lower magnetic moment despite having a similar number of d electrons as \([\text{MnCl}_4]^{2-}\) and \([\text{CoCl}_4]^{2-}\). The ability of CN to induce pairing emphasizes how ligand strength impacts electron arrangement and, consequently, magnetic properties.

Coordination Complexes

A coordination complex forms when metal ions bond with ligands. The nature of these ligands and the structure of the complex profoundly influence its properties.

In our exercise, we examined three complexes:

• \([\text{MnCl}_4]^{2-}\)
• \([\text{Fe(CN}_6]^{4-}\)
• \([\text{CoCl}_4]^{2-}\)

The metal-ligand bond's strength, whether it's strong or weak, depends on the ligand's identity and the coordination geometry:

- **Tetrahedral vs. Octahedral:** Tetrahedral complexes, like \([\text{MnCl}_4]^{2-}\) and \([\text{CoCl}_4]^{2-}\), typically involve weaker field ligands and accommodate unpaired electrons easily.- **Ligand Field Strength:** Cyanide ligands in \([\text{Fe(CN}_6]^{4-}\) are a good example of strong field ligands, which most often lead to electron pairing in octahedral complexes.

Understanding these components and their interactions is the key to predicting magnetic behavior in coordination complexes.

Magnetic Properties Calculation

Calculating the magnetic moment gives insight into a compound's magnetic properties. The number of unpaired electrons in a coordination complex primarily decides its magnetic moment.

To find the magnetic moment, you can use the formula:\[\mu = \sqrt{n(n+2)}\]where \( n \) is the number of unpaired electrons.

Applying this:

• For \([\text{MnCl}_4]^{2-}\), all 5 d electrons remain unpaired, giving us \( \mu = \sqrt{5(5+2)} = \sqrt{35} \).
• For \([\text{CoCl}_4]^{2-}\), 3 unpaired electrons result in \( \mu = \sqrt{3(3+2)} = \sqrt{15} \).
• For \([\text{Fe(CN}_6]^{4-}\), strong field ligands lead to no unpaired electrons, resulting in \( \mu = 0 \).

Differences in \( n \) contribute to the distinct magnetic moments observed, highlighting the influence of both electron configuration and ligand strength. These calculations elucidate why \([\text{MnCl}_4]^{2-}\) exhibits the highest magnetic moment, followed by \([\text{CoCl}_4]^{2-}\), and finally \([\text{Fe(CN}_6]^{4-}\).