Problem 116 - A Complete Resource Book in Chemistry for JEE Main

Question

The number of unpaired electrons present in \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) are, respectively (a) 0,0 (b) 0,4 (c) 1,2 (d) 2,4Among the following four sets, the one having the same geometry and same magnetic property for both complexes is (a) \(\left[\mathrm{Co}\left(\mathrm{CN}_{6}\right)\right]^{3-}\) and \(\left[\mathrm{CoF}_{6}\right]^{3-}\) (b) \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\) and \(\left[\mathrm{MnF}_{6}\right]^{3-}\) (c) \([\mathrm{Cr}(\mathrm{CN})]^{3-}\) and \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (d) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\)

Step-by-Step Solution

Answer

(b) 0, 4 and (a) have similar geometry and magnetic properties.

Step 1: Determine Iron's Oxidation State

For \( \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \), we need to find the oxidation state of Fe. Cyanide (CN) is a strong field ligand and gives a \(-1\) charge per ligand. The complex charge is \(-4\), so:\[-4 = x + 6(-1)\] \[x = +2\]. For \( \left[\mathrm{Fe}(\mathrm{H}_{2} \mathrm{O})_{6}\right]^{2+} \), water is a neutral ligand. The complex's total charge is \(+2\), so the oxidation state of iron is also \(+2\).

Step 2: Determine d-electron Count

For both complexes, iron is in the \(+2\) state, meaning it loses 2 electrons from its normal 3d\(^6\)4s\(^2\) configuration, i.e., 3d\(^6\).

Step 3: Determine Electron Configuration with Ligands

In \( \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \), Cyanide is a strong field ligand, so it pairs up electrons within the 3d orbitals following the low spin state. In \( \left[\mathrm{Fe}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+} \), water is a weak field ligand, leading to a high spin state.

Step 4: Calculate Unpaired Electrons

In \( \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \), all 3d electrons pair up due to strong field ligands, resulting in 0 unpaired electrons. For \( \left[\mathrm{Fe}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+} \), the weak field does not pair the electrons completely, resulting in 4 unpaired electrons.

Step 5: Compare Geometry and Magnetic Property

Both \( \left[\mathrm{Co(CN)}_{6}\right]^{3-} \) and \( \left[\mathrm{CoF}_6\right]^{3-} \) have similar geometry (octahedral) but differ in magnetic properties; one is diamagnetic and the other is paramagnetic, which does not match. \( \left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-} \) and \( \left[\mathrm{MnF}_6\right]^{3-} \) are similarly disparate. \([\mathrm{Cr}(\mathrm{CN})]^{3-}\) and \(\left[\mathrm{Cr(\mathrm{H}_2O)}_6\right]^{3+}\) respectively have different properties. The pairs \( \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \) and \( \left[\mathrm{Fe(\mathrm{H}_2O)}_{6}\right]^{2+} \) differ in both geometry and magnetism.

Key Concepts

Crystal Field Theory

Crystal Field Theory (CFT) is a model used to understand the electronic structure of transition metal complexes. It explains how metal ions and ligands interact, particularly focusing on the effect of the ligands' electric fields on the d-orbitals of the metal atoms. This theory is essential in predicting the geometry and magnetic properties of complex ions.

In CFT, the spatial arrangement of ligands around a central metal ion causes a splitting of the degenerate d-orbitals into higher and lower energy levels. The extent of this splitting depends on the nature of the ligands. Strong field ligands, like cyanide ( ext{CN}^-), cause a large splitting and typically result in a low-spin configuration. This means the electrons prefer to pair up in the lower energy orbitals before occupying higher energy ones. The coordination complex \([ ext{Fe}( ext{CN})_6]^{4-}\) is an example where such splitting leads to a low-spin state with all paired electrons.

On the other hand, weak field ligands such as water ( ext{H}_2 ext{O}) induce a smaller splitting, resulting in a high-spin state where more unpaired electrons are present. For example, in \([ ext{Fe(H}_2 ext{O)}_6]^{2+}\), the small splitting does not suffice to pair up the electrons, resulting in unpaired electrons.

Unpaired Electrons

Unpaired electrons greatly influence the magnetic properties of transition metal complexes. According to the spin-only formula, the magnetic moment of a complex can be calculated using the number of unpaired electrons. Paramagnetic substances have unpaired electrons, making them attracted to magnetic fields. Diamagnetic compounds have all their electrons paired and are therefore weakly repelled by a magnetic field.

In the coordination sphere of \([ ext{Fe}( ext{CN})_6]^{4-}\), the strong field of cyanide ligands results in no unpaired electrons. Hence, the complex \([ ext{Fe}( ext{CN})_6]^{4-}\) is diamagnetic. However, within \([ ext{Fe}( ext{H}_2 ext{O})_6]^{2+}\), the hexaaquairon(II) complex, the water molecules, being weak field ligands, result in a configuration with four unpaired electrons, making this complex paramagnetic.

The number of unpaired electrons is crucial when determining the magnetic properties and behavior of coordination compounds. This knowledge helps in applications across chemistry such as material science and in developing compounds for MRI technologies.

Oxidation States

The oxidation state of a central metal atom in a complex is vital for understanding its chemical behavior and reactivity. It is inferred by considering the charges contributed by the ligands and the overall charge of the complex.

In the complex \([ ext{Fe}( ext{CN})_6]^{4-}\), each cyanide ligand contributes a \-1 charge. Given the overall charge of \-4 for the complex, the oxidation state of iron ( ext{Fe}) can be calculated: \[-4 = x + 6(-1)\], solving gives \(+2\). Similarly, for \([ ext{Fe(H}_2 ext{O)}_6]^{2+}\), each water molecule is neutral, and the complex bears a \(+2\) charge, making the oxidation state of iron again \(+2\).

Determining the oxidation state is essential not only for understanding the electron configuration but also for predicting the reactivity and types of reactions that such complexes might undergo. It provides fundamental insights into the electron-sharing process between metals and ligands, shaping the electronic characteristics of the compound.