Problem 70 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following high spin complexes has the largest CFSE (Crystal Field Stabilization Energy)? (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2^{+}}\) (b) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (c) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{2+}\) (d) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3^{+}}\)

Step-by-Step Solution

Answer

\([\mathrm{Cr(H}_2\mathrm{O})_6]^{3+}\) has the largest CFSE with \(-1.2\Delta\).

Step 1: Understand High Spin Complexes

High spin complexes have unpaired electrons and typically form with weak field ligands such as water (\(\mathrm{H}_2\mathrm{O}\)). For these complexes, the electrons fill the \(e_g\) orbitals only after the \(t_{2g}\) orbitals are fully half-filled.

Step 2: Calculate Electron Configuration

Determine the electron configurations for the metal ions in each complex assuming a high spin state. - For \([\mathrm{Cr(H}_2\mathrm{O})_6]^{2+}\), chromium in \(+2\) oxidation state has an electron configuration of \([\mathrm{Ar}]\ 3d^4\).- For \([\mathrm{Cr(H}_2\mathrm{O})_6]^{3+}\), chromium in \(+3\) oxidation state has an electron configuration of \([\mathrm{Ar}]\ 3d^3\).- For \([\mathrm{Mn(H}_2\mathrm{O})_6]^{2+}\), manganese in \(+2\) oxidation state has an electron configuration of \([\mathrm{Ar}]\ 3d^5\).- For \([\mathrm{Mn(H}_2\mathrm{O})_6]^{3+}\), manganese in \(+3\) oxidation state has an electron configuration of \([\mathrm{Ar}]\ 3d^4\).

Step 3: Assign Electrons in Oribitals

Assign electrons to \(t_{2g}\) and \(e_g\) orbitals for each complex using high spin configurations.- \(\mathrm{Cr}^{2+}\): \(t_{2g}^3 e_g^1\)- \(\mathrm{Cr}^{3+}\): \(t_{2g}^3\)- \(\mathrm{Mn}^{2+}\): \(t_{2g}^3 e_g^2\)- \(\mathrm{Mn}^{3+}\): \(t_{2g}^3 e_g^1\)

Step 4: Calculate CFSE for Each Complex

CFSE (Crystal Field Stabilization Energy) is calculated as: \[\text{CFSE} = (\text{Number of } t_{2g} \text{ electrons} \times -0.4 \Delta ) + (\text{Number of } e_g \text{ electrons} \times +0.6 \Delta )\]- For \(\mathrm{Cr}^{2+}\): \((-0.4\Delta \times 3) + (+0.6\Delta \times 1) = -0.6\Delta\)- For \(\mathrm{Cr}^{3+}\): \((-0.4\Delta \times 3) + (+0.6\Delta \times 0) = -1.2\Delta\)- For \(\mathrm{Mn}^{2+}\): \((-0.4\Delta \times 3) + (+0.6\Delta \times 2) = 0\Delta\)- For \(\mathrm{Mn}^{3+}\): \((-0.4\Delta \times 3) + (+0.6\Delta \times 1) = -0.6\Delta\)

Step 5: Determine the Largest CFSE

Compare the CFSE values: - \([\mathrm{Cr(H}_2\mathrm{O})_6]^{2+}\) has \(-0.6\Delta\).- \([\mathrm{Cr(H}_2\mathrm{O})_6]^{3+}\) has \(-1.2\Delta\).- \([\mathrm{Mn(H}_2\mathrm{O})_6]^{2+}\) has \(0\Delta\).- \([\mathrm{Mn(H}_2\mathrm{O})_6]^{3+}\) has \(-0.6\Delta\).Thus, \([\mathrm{Cr(H}_2\mathrm{O})_6]^{3+}\) has the largest CFSE.

Key Concepts

Understanding High Spin Complexes

In the world of chemistry, high spin complexes are intriguing formations where the arrangement of electrons is key. These complexes are characterized by unpaired electrons, which is quite distinct when compared to low spin complexes where electrons are more paired.
High spin complexes typically form in the presence of weak field ligands, such as water (\(\mathrm{H}_2\mathrm{O}\)). These ligands do not exert strong splitting of the d-orbitals in the transition metals, allowing more electrons to remain unpaired. - A high spin complex is often favored by ligands that do not bring significantly large splitting of the d-orbitals.- Electron filling priority in high spin complexes: first, each available orbital is singly occupied before pairing begins. Understanding these foundational aspects illuminates why high spin complexes tend to have unpaired electrons and play into the calculation of their electron configurations and crystal field stabilization energies (CFSE).

Calculating Electron Configuration

The electron configuration of a metal ion in a complex is a blueprint to its properties and behaviors. For high spin complexes, this involves considering the oxidation state and how electrons fill the available d-orbitals.- **Oxidation State**: Determine the oxidation state of the metal ion within the complex. For example, in \([\mathrm{Cr(H}_2\mathrm{O})_6]^{3+}\), chromium is in the +3 state, whereas in \([\mathrm{Mn(H}_2\mathrm{O})_6]^{3+}\), manganese is also in a +3 state.- **Electron Configuration in High Spin State**: Knowing the oxidation state allows for the determination of how many d-orbitals are occupied. For instance, - \(\mathrm{Cr}^{2+}\) yields an electron configuration of \([\mathrm{Ar}]\ 3d^4\), meaning there are 4 electrons to arrange in the d orbitals.
- \(\mathrm{Cr}^{3+}\) has \([\mathrm{Ar}]\ 3d^3\), with 3 d-orbital electrons to place.
- Detailing this configuration allows us to assign the exact distribution of electrons, critical for determining the CFSE and understanding the physical and chemical properties of the complex.

Exploring t2g and eg Orbitals

In transition metal complexes, the splitting of the d-orbitals into \(t_{2g}\) and \(e_g\) orbitals plays a crucial role in their behavior, especially in high spin configurations.When ligands approach, the degeneracy of the d-orbitals is lifted, as ligands exert forces unevenly across the field.
- **\(t_{2g}\) Orbitals** - Lower energy set of orbitals, consisting of the \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\). - Electrons preferentially fill these orbitals first in high spin complexes, performing a crucial role in stability.- **\(e_g\) Orbitals** - Higher in energy and consist of the \(d_{x^2-y^2}\) and \(d_{z^2}\) orbitals. - In high spin scenarios, these orbitals are filled with remaining electrons only after the \(t_{2g}\) orbitals are half-occupied.
Evaluating how electrons populate these orbitals explains how the crystal field stabilization energy is calculated. It sheds light on why some complexes possess higher stabilities based solely on how they manage their d-block electrons across these differentiated orbitals.