Problem 17 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The correct order of the spin-only magnetic moment of metal ions in the following low-spin complexes, \(\left[\mathrm{V}(\mathrm{CN})_{6}\right]^{4-},\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-},\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\), and \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\), is (a) \(\mathrm{Cr}^{2+}>\mathrm{Ru}^{3+}>\mathrm{Fe}^{2+}>\mathrm{V}^{2+}\) (b) \(\mathrm{V}^{2+}>\mathrm{Cr}^{2+}>\mathrm{Ru}^{3+}>\mathrm{Fe}^{2+}\) (c) \(\mathrm{V}^{2+}>\mathrm{Ru}^{3+}>\mathrm{Cr}^{2+}>\mathrm{Fe}^{2+}\) (d) \(\mathrm{Cr}^{2+}>\mathrm{V}^{2+}>\mathrm{Ru}^{3+}>\mathrm{Fe}^{2+}\)

Step-by-Step Solution

Answer

(a) \( \text{Cr}^{2+} > \text{Ru}^{3+} > \text{Fe}^{2+} > \text{V}^{2+} \).

Step 1: Determine the d-electron count for each complex

For each complex, identify the oxidation state of the metal and determine the number of d-electrons it has. Start with \( [\mathrm{V}(\mathrm{CN})_{6}]^{4-} \). Vanadium in the complex has an oxidation state of +2 (from V to \( \text{V}^{2+} \)), which gives it \( 3d^3 \) configuration (5 total d-electrons minus 2 for the charge). Next, \( [\mathrm{Fe}(\mathrm{CN})_{6}]^{4-} \), where iron has an oxidation state of +2, thus with a \( 3d^6 \) configuration. In \( [\mathrm{Ru}(\mathrm{NH}_{3})_{6}]^{3+} \), ruthenium is in the +3 oxidation state, resulting in a \( 4d^5 \) configuration. Lastly, for \( [\mathrm{Cr}(\mathrm{NH}_{3})_{6}]^{2+} \), chromium in +2 state leads to a \( 3d^4 \) configuration.

Step 2: Calculate the spin-only magnetic moment for each complex

The spin-only magnetic moment \( \mu \) is calculated by the formula: \( \mu = \sqrt{n(n+2)} \), where \( n \) is the number of unpaired electrons. In low-spin complexes: 1. \( [\mathrm{V}(\mathrm{CN})_{6}]^{4-} \) has 0 unpaired electrons (\( n = 0 \)), \( \mu = 0 \). 2. \( [\mathrm{Fe}(\mathrm{CN})_{6}]^{4-} \) has 0 unpaired electrons (\( n = 0 \)), \( \mu = 0 \). 3. \( [\mathrm{Ru}(\mathrm{NH}_{3})_{6}]^{3+} \) has 1 unpaired electron (\( n = 1 \)), \( \mu = \sqrt{3} \approx 1.73 \). 4. \( [\mathrm{Cr}(\mathrm{NH}_{3})_{6}]^{2+} \) has 2 unpaired electrons (\( n = 2 \)), \( \mu = \sqrt{8} \approx 2.83 \).

Step 3: Arrange complexes in decreasing order of magnetic moment

Based on the spin-only magnetic moments calculated: \( [\mathrm{Cr}(\mathrm{NH}_{3})_{6}]^{2+} \) has the highest moment, followed by \( [\mathrm{Ru}(\mathrm{NH}_{3})_{6}]^{3+} \). Both \( [\mathrm{V}(\mathrm{CN})_{6}]^{4-} \) and \( [\mathrm{Fe}(\mathrm{CN})_{6}]^{4-} \) have no magnetic moment due to zero unpaired electrons. The correct order is \( \text{Cr}^{2+} > \text{Ru}^{3+} > \text{V}^{2+} = \text{Fe}^{2+} \).

Key Concepts

Low-spin complexes

Low-spin complexes are fascinating structures in coordination chemistry, where ligands cause paired electron arrangements in the metal's d-orbitals. These complexes occur when strong field ligands, like cyanide (CN) or ammonia (NH₃), produce a significant crystal field splitting energy. This energy pushes the electrons to pair up in the lower energy d-orbitals, rather than occupying higher energy levels even as unpaired electrons.

• Strong field ligands create larger separation between energy levels or orbitals.
• Electrons prefer to pair up in low energy orbitals whenever possible.
• This pairing minimizes the number of unpaired electrons, thus reducing magnetic properties.

Understanding low-spin complexes is crucial because these paired electron arrangements greatly influence the magnetic properties of the metals. For instance, in the exercise, both \( [\mathrm{V}(\mathrm{CN})_6]^{4-} \) and \( [\mathrm{Fe}(\mathrm{CN})_6]^{4-} \) are low-spin and have no unpaired electrons, making their magnetic moment zero.

Unpaired electrons

Unpaired electrons are electrons that occupy orbitals alone without another electron pairing up. The presence of these unpaired electrons is what directly contributes to a substance's magnetic properties. For a given metal ion, the number of unpaired electrons determines the spin-only magnetic moment, calculated using the formula:\[\mu = \sqrt{n(n+2)}\]where \( n \) is the number of unpaired electrons. The more unpaired electrons present, the higher the magnetic moment. In our exercise, the complexes are evaluated based on their number of unpaired electrons:

• \( [\mathrm{Cr}(\mathrm{NH}_3)_6]^{2+} \) has 2 unpaired electrons, leading to the highest magnetic moment.
• \( [\mathrm{Ru}(\mathrm{NH}_3)_6]^{3+} \) has 1 unpaired electron, resulting in a lower magnetic moment.
• Both \( [\mathrm{V}(\mathrm{CN})_6]^{4-} \) and \( [\mathrm{Fe}(\mathrm{CN})_6]^{4-} \) have no unpaired electrons, which means no magnetic moment.

Recognizing the role of unpaired electrons helps us understand why certain complexes exhibit magnetism while others do not.

d-electron configuration

The d-electron configuration of a metal ion is critical in determining many of its chemical properties, including its color and magnetism. The configuration involves the specific distribution of electrons in the d-orbitals. To identify it correctly, one must consider the metal's oxidation state and the electron count within its outer shell.

• The exercise provides the electron configuration for each metal ion within the complexes. For example, \( \text{V}^{2+} \) in \( [\mathrm{V}(\mathrm{CN})_6]^{4-} \) has a \( 3d^3 \) configuration, resulting from an initial 5 d-electrons less the oxidation state change (removing 2 electrons).
• Similarly, the \( \text{Fe}^{2+} \) ion in \( [\mathrm{Fe}(\mathrm{CN})_6]^{4-} \) holds a \( 3d^6 \) configuration.

Understanding the d-electron configuration allows us to assess which d-orbitals will be occupied and if there might be electron pairing. This information is crucial in identifying the number of unpaired electrons and consequently calculating the magnetic moment. By mastering d-electron configurations, one gains deeper insights into the subtle yet powerful interactions influencing complex structures.