Problem 36 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Consider the following complex ions, \(P, Q\) and \(R\). \(P=\left[\mathrm{FeF}_{6}\right]^{3-}, Q=\left[\mathrm{V}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(R=\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is (a) \(R

Step-by-Step Solution

Answer

Therefore, the correct order is (b) \(Q < R < P\).

Step 1: Determine Oxidation States

First, we need to find the oxidation state of the metal ions in the complexes. In complex ion \( P = [\mathrm{FeF}_6]^{3-} \), iron typically has an oxidation state of +3. For complex ion \( Q = [\mathrm{V(H}_{2}\mathrm{O})_6]^{2+} \), vanadium has an oxidation state of +2. And for complex ion \( R = [\mathrm{Fe(H}_{2}\mathrm{O})_6]^{2+} \), iron has an oxidation state of +2.

Step 2: Calculate Number of Unpaired Electrons

Next, calculate the number of unpaired electrons. In \( P \) with \( \text{Fe}^{3+} \) (3d\(^5\)) there are 5 unpaired electrons. For \( Q \) with \( \text{V}^{2+} \) (3d\(^3\)) there are 3 unpaired electrons. In \( R \) with \( \text{Fe}^{2+} \) (3d\(^6\)), there are 4 unpaired electrons as water is a weak field ligand causing a high-spin configuration.

Step 3: Calculate Spin-Only Magnetic Moment

The spin-only magnetic moment \( \mu \) can be calculated using the formula \( \mu = \sqrt{n(n+2)} \), where \( n \) is the number of unpaired electrons. \( P \) with 5 unpaired electrons has \( \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \) B.M. \( Q \) with 3 unpaired electrons has \( \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \) B.M. \( R \) with 4 unpaired electrons has \( \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \) B.M.

Step 4: Arrange in Order of Magnetic Moment

Arrange the complexes based on the calculated magnetic moments: \( ?\) with 3.87 B.M., \( R \) with 4.90 B.M., and \( P \) with 5.92 B.M. Thus, the order from lowest to highest magnetic moment is \( Q < R < P \).

Key Concepts

Complex Ions

When it comes to understanding complex ions, it's important to focus on an atom surrounded by molecules or ions, called ligands. These ligands are attached to the metal atom or ion in the center.
In our example, we have three distinct complex ions:

• \( \left[\mathrm{FeF}_{6}\right]^{3-} \) - Here, the central iron atom is surrounded by six fluoride ions.
• \( \left[\mathrm{V}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+} \) - In this case, the central vanadium ion is surrounded by six water molecules.
• \( \left[\mathrm{Fe}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+} \) - Similar to the vanadium complex, but with iron at its center.

Understanding what ligands do is crucial. Fluoride in \( [\mathrm{FeF}_{6}] \) and water in the other two complexes are ligands impacting the properties like color, reactivity, and magnetism of complexes. These interactions are pivotal in determining the arrangement and behavior of electrons within the metal ion.

Oxidation States

Oxidation states are a way to keep track of electron transfers in chemical reactions. It's an imaginary charge that an atom would have if all bonds to atoms of different elements were 100% ionic.
In the context of our complex ions, determining the oxidation state of the central metal is key:

• For \( P \) \([\mathrm{FeF}_{6}]^{3-}\), the oxidation state of iron is +3. This means that the iron atom has "lost" three electrons compared to its elemental state.
• In \( Q \) \([\mathrm{V}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\), vanadium is in the +2 oxidation state.
• In \( R \) \([\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\), iron holds a +2 oxidation state, meaning it's less oxidized than in \( P \).

Knowing the oxidation state helps anticipate the number of unpaired electrons and therefore predict magnetic characteristics of the complex ions.

Number of Unpaired Electrons

The number of unpaired electrons in a complex ion is crucial in determining its magnetic properties. These are electrons in the outer shells of atoms or ions that are not paired with another electron having an opposite spin. The unpaired electrons significantly contribute to the magnetic moment.
For example:

• In complex ion \( P \), safety \,\((\mathrm{FeF}_{6})^{3-}\), the \( \mathrm{Fe}^{3+} \) ion has a 3d\(^5\) configuration. With all five electrons unpaired, the ion is strongly paramagnetic.
• \( Q \) has 3d\(^3\) configuration due to \( \mathrm{V}^{2+} \), presenting three unpaired electrons, which leads to a moderate magnetism.
• For \( R \) with \( \mathrm{Fe}^{2+} \), the configuration is 3d\(^6\). Here, water being a weak field ligand does not pair up some electrons causing four unpaired.

The pattern of unpaired electrons impacts both the calculation of spin-only magnetic moments and the order of increasing magnetic strengths among these ions.