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) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (d) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\)

Step-by-Step Solution

Answer

Cr(III) complex has the largest CFSE.

Step 1: Determine the oxidation state of metal ions

For each complex, identify the oxidation state of the metal ion. - In \( \left[\text{Cr}(\text{H}_2\text{O})_6\right]^{2+} \), Cr is in the +2 oxidation state. - In \( \left[\text{Cr}(\text{H}_2\text{O})_6\right]^{3+} \), Cr is in the +3 oxidation state. - In \( \left[\text{Mn}(\text{H}_2\text{O})_6\right]^{2+} \), Mn is in the +2 oxidation state. - In \( \left[\text{Mn}(\text{H}_2\text{O})_6\right]^{3+} \), Mn is in the +3 oxidation state.

Step 2: Determine d-electron configuration

Calculate the number of d-electrons for each metal in its oxidation state:- Cr(II): \(\text{d}^4\)- Cr(III): \(\text{d}^3\)- Mn(II): \(\text{d}^5\)- Mn(III): \(\text{d}^4\)

Step 3: Identify the High Spin configuration

For each d-electron configuration, determine the high spin electronic arrangement. In high spin complexes, the electrons occupy all five d-orbitals before pairing:- Cr(II) \(\text{d}^4\): \(t_{2g}^2e_g^2\)- Cr(III) \(\text{d}^3\): \(t_{2g}^3e_g^0\)- Mn(II) \(\text{d}^5\): \(t_{2g}^3e_g^2\)- Mn(III) \(\text{d}^4\): \(t_{2g}^3e_g^1\)

Step 4: Calculate CFSE for each complex

Use the formula \( ext{CFSE} = -0.4 imes n_{t_{2g}} imes igtriangleup + 0.6 imes n_{e_g} imes igtriangleup \) where \( igtriangleup \) is the crystal field splitting energy, assume \( P=0 \) for simplicity in high spin:- Cr(II): \(-0.4(2) + 0.6(2) = +0.4\bigtriangleup\)- Cr(III): \(-0.4(3) + 0.6(0) = -1.2\bigtriangleup\)- Mn(II): \(-0.4(3) + 0.6(2) = -0.6\bigtriangleup\)- Mn(III): \(-0.4(3) + 0.6(1) = -0.6\bigtriangleup\)

Step 5: Determine the largest CFSE

Compare the CFSE values calculated:- Cr(II) has a CFSE of \(+0.4\bigtriangleup\)- Cr(III) has a CFSE of \(-1.2\bigtriangleup\)- Mn(II) has a CFSE of \(-0.6\bigtriangleup\)- Mn(III) has a CFSE of \(-0.6\bigtriangleup\)The smallest negative CFSE value represents the largest stabilization. Hence, Cr(III) has the largest CFSE with \(-1.2\bigtriangleup\).

Key Concepts

High Spin Complexes

High spin complexes are a fascinating phenomenon found primarily in coordination chemistry. They involve metal ions that are surrounded by ligands, creating a specific arrangement of electrons in the d-orbitals. In these complexes, the electron arrangement fills all available d-orbitals with single electrons as much as possible before pairing them up.

This occurs because in high spin complexes, the energy required to pair up electrons in a single orbital is greater than the energy needed to distribute them across multiple orbitals. The higher energy level, known as the crystal field splitting energy ( ∆ ), dictates this behavior. As a result, complexes with small ∆ values often form high spin arrangements.

The typical configuration can be remembered as "spreading out before pairing." For example, in a d^5 configuration of a complex like [Mn(H_2O)_6]^{2+} , the electrons might be distributed as t_{2g}^3e_g^2 , with electrons occupying all available orbitals before doubling up.

d-electron Configuration

The d-electron configuration of a metal ion is a crucial concept that dictates the chemical behavior, stability, and magnetic properties of the ion within a complex. Transition metals are known for their partially filled d-orbitals, and their unique properties often arise from these d-electron configurations. Understanding how to determine the d-electron configuration involves looking at the oxidation state of the metal and its position in the periodic table.

To find the d-electron count, you first need to consider the atomic number of the element, subtract the electrons that have been removed to reach the specific oxidation state, and account for any additional configuration adjustments. For example, chromium in its +2 oxidation state, expressed as Cr^{2+} , has a d^4 configuration, while in the +3 state ( Cr^{3+} ), it has a d^3 configuration.

• Cr^{2+}: d^4
• Cr^{3+}: d^3
• Mn^{2+}: d^5
• Mn^{3+}: d^4

These configurations play a pivotal role in the overall stability and properties of the metal complexes.

Oxidation States of Transition Metals

Transition metals are known for having multiple oxidation states, which add to their versatility in forming various compounds and complexes. Each oxidation state of a metal is indicative of its overall electronic configuration, affecting how the metal will bond and interact with surrounding ligands.

For instance, the oxidation state can define how many electrons are withdrawn from the d-orbitals, altering both the charge and the d-electron count of the ion. In reality, the oxidation state reflects the "hypothetical charge" the metal ion would have if all bonds to constituents were purely ionic. The formal charge is calculated based on the assumption of an entirely ionic scenario, simplifying predictions on electronic behavior.

Consider [Cr(H_2O)_6]^{2+} with chromium in the +2 oxidation state, where two electrons have been removed from its total count. Contrastingly, in [Cr(H_2O)_6]^{3+} , three electrons are removed, confirming a +3 state. The possible oxidation states facilitate a wide range of chemical and physical properties for transition metal complexes, making their chemistry rich and engaging for study.

Crystal Field Splitting Energy

Understanding crystal field splitting energy ( ∆ ) is essential when discussing complexes, particularly in explaining their color and magnetism. This concept revolves around the distribution of electrons in the metal d-orbitals when surrounded by ligands. These ligands create an electric field, causing a split in the energy levels of the d-orbitals.

The splitting results in forming two sets of orbitals: the lower energy t_{2g} set and the higher energy e_g set. The difference in energy between these sets is referred to as ∆ . In high spin complexes, such as [Cr(H_2O)_6]^{3+} , this energy difference determines how electrons are distributed across these orbitals.

• The value of ∆ is influenced by the nature of the ligands and the metal ion.
• The arrangement of electrons in these split orbitals contributes to the complex's overall stability, represented by the Crystal Field Stabilization Energy (CFSE).
• High spin complexes are often seen with smaller ∆ values, as electrons occupy higher energy orbitals before pairing up, minimizing repulsion within the orbitals.

Exploring how ∆ and CFSE affects the properties of a complex deepens our understanding of chemical bonding and coordination chemistry.