Problem 113 - A Complete Resource Book in Chemistry for JEE Main

Question

Arrange the following in order of decreasing number of unpaired electrons: 1\. \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) 2\. \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) 3\. \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4^{-}}\) 4\. \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (a) \(4,1,2,3\) (b) \(1,2,3,4\) (c) \(4,2,1,3\) (d) \(2,3,1,4\)

Step-by-Step Solution

Answer

The correct answer is (a) 4, 1, 2, 3.

Step 1: Determine Iron's Electron Configuration

Iron (Fe) has an atomic number of 26, meaning its electron configuration is \[ \text{[Ar]} 3d^6 4s^2 \]. For each compound, determine the oxidation state of Fe and accordingly adjust the electron configuration by removing the corresponding electrons from '4s' first, and then '3d'.

Step 2: Analyze \\([\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}]^{3+}\\)

The oxidation state of Fe is +3. Remove three electrons, resulting in the configuration \[ 3d^5 \] with 5 unpaired electrons. Water is a weak field ligand and does not cause pairing.

Step 3: Analyze \\([\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\\)

The oxidation state of Fe is +2. Remove two electrons to get \[ 3d^6 \]. With water as a weak field ligand, pairing doesn't occur, so there are 4 unpaired electrons.

Step 4: Analyze \\([\mathrm{Fe}(\mathrm{CN})_{6}]^{4-}\\)

The oxidation state of Fe is +2. The electron configuration is \[ 3d^6 \]. Cyanide is a strong field ligand, causing electrons to pair, resulting in no unpaired electrons.

Step 5: Analyze \\([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\\)

The oxidation state of Fe is +3. The electron configuration is \[ 3d^5 \]. With cyanide as a strong field ligand, pairing occurs, resulting in 1 unpaired electron.

Step 6: Order Based on Unpaired Electrons

Arrange the compounds by the decreasing number of unpaired electrons from steps 2-5: - \([\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}]^{3+}\): 5 unpaired electrons- \([\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\): 4 unpaired electrons- \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\): 1 unpaired electron- \([\mathrm{Fe}(\mathrm{CN})_{6}]^{4-}\): 0 unpaired electronsThe order is 4, 1, 2, 3.

Key Concepts

Unpaired Electrons

Unpaired electrons are electrons that occupy an atomic or molecular orbital by themselves rather than being paired up with another electron. In the chemistry of coordination compounds, these unpaired electrons play a crucial role in determining the magnetic properties of the compound.
When electrons are unpaired, the compound often exhibits paramagnetism, as these electrons generate a magnetic field that can interact with external magnetic fields.

• For example, in \( \left[\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}\right]^{3+} \), iron has an electron configuration of \(3d^5\) in its high-spin state due to the presence of weak field ligands like \(\mathrm{H}_2\mathrm{O}\), leading to five unpaired electrons.
• In comparison, \( \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \) has no unpaired electrons due to the pairing effect of the strong field cyanide ligands, resulting in diamagnetic behavior.

Understanding the number and effect of unpaired electrons is essential for predicting the magnetic properties of coordination complexes.

Electron Configuration

The electron configuration helps us understand how electrons are distributed in an atom. For transition metals like iron, the electron configuration determines how the electrons are removed or added when forming ions.
Iron (Fe) naturally has the electron configuration: \( \text{[Ar]} 3d^6 4s^2 \).
When it forms ions in coordination compounds, we alter this configuration based on its oxidation state.

• If iron has an oxidation state of +3, it means three electrons are lost. Typically, electrons are removed first from the outermost shell, the 4s orbital, and then from the 3d orbital, resulting in \(3d^5\).
• For an oxidation state of +2, only two electrons are removed from the 4s orbital, leading to \(3d^6\).

These changes in electron configurations are crucial for determining how ligands will interact with the central metal, affecting properties like magnetic behavior.

Oxidation States

The oxidation state, or oxidation number, of an element in a compound signifies how many electrons are lost or gained compared to its elemental form. This is particularly important in transition metals like iron, where multiple oxidation states are possible.
The oxidation state affects how many electrons are left around the central metal, directly impacting the electron configuration and subsequently the chemical behavior of the compound.

• For example, in \( \left[\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}\right]^{3+} \), the oxidation state of +3 indicates the loss of three electrons, leaving the iron in a \(3d^5\) configuration.
• Conversely, in \( \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \), an oxidation state of +2 suggests the loss of two electrons, allowing for the \(3d^6\) arrangement.

Recognizing and calculating oxidation states are vital for predicting compound reactivity, stability, and sentence structure.