Question

Which of the following is incorrect statement regarding \(\mathrm{A}=\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\)and \(\mathrm{B}=\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]+\) (1) A is more thermodinamically stable than B. (2) A shows optical isomerism. (3) \(A\) and \(B\) both are diamagnetic. (4) \(\mathrm{A}\) is outer orbital complex while \(\mathrm{B}\) is inner orbital complex.

Short answer

(3) A and B both are diamagnetic.

Step-by-step solution

Analyze the Complexes

Start by examining the given complexes \(\text{A} = [\text{Co}(\text{en})_2 \text{Cl}_2]^+ \) and \(\text{B} = [\text{Co}(\text{NH}_3)_4 \text{Cl}_2]^+ \). Note the ligands and the coordination numbers.

Review Stability Factors

A complex's stability is influenced by factors such as the nature of ligands. Ethylenediamine (en) in \(\text{A}\) is a bidentate ligand and generally forms more stable complexes due to the chelate effect, while \(\text{NH}_3\) in \(\text{B}\) is a monodentate ligand.

Identify Optical Isomerism

Determine if each complex shows optical isomerism. Since \(\text{A}\) has bidentate ligands (ethylenediamine), it can form optical isomers. Complex \(\text{B}\) has only monodentate ligands and does not show optical isomerism.

Check Magnetic Properties

To determine if they are diamagnetic, examine the electronic configuration. For \(\text{A}\) and \(\text{B}\), Co is in the +3 oxidation state (Co^3+ has 3d^6 configuration). \(\text{Co}^{3+}\) with a low-spin configuration (due to strong field ligands like \(\text{en}\) and \(\text{NH}_3}\)) will be diamagnetic.

Determine the Type of Complex

Classify whether each complex is an inner or outer orbital complex. Inner orbital complexes utilize inner d orbitals (3d) for hybridization, while outer orbital complexes utilize outer d orbitals (4d, 5d). Complex \(\text{A}\) generally forms an inner orbital complex, while complex \(\text{B}\)—because of its ligand properties—forms an outer orbital complex.

Identify the Incorrect Statement

Using the analysis from the previous steps, match each statement to its corresponding observation. Identify the statement that does not fit with the established properties of the complexes.

Key concepts

Complex Stability

When discussing the stability of coordination compounds, we consider factors such as the nature and number of ligands. In the case of complex \( A = [\text{Co}(\text{en})_2 \text{Cl}_2]^+ \), the ligand ethylenediamine (en) is bidentate, meaning it can coordinate to the metal center through two donor atoms. This creates a more stable complex due to the chelate effect, which increases stability by forming a ring structure. In contrast, complex \( B = [\text{Co}(\text{NH}_3)_4 \text{Cl}_2]^+ \) contains ammonium ligands (\( \text{NH}_3 \)), which are monodentate and only form one bond with the central metal ion. This results in a less stable complex compared to a chelate. Thus, complex A is more thermodynamically stable than complex B.

Optical Isomerism

Optical isomerism occurs in coordination compounds when they can form non-superimposable mirror images, often referred to as enantiomers. For a complex to exhibit optical isomerism, it usually requires asymmetric arrangement of the ligands. In complex \( A = [\text{Co}(\text{en})_2 \text{Cl}_2]^+ \), the presence of bidentate ligands like ethylenediamine helps create this asymmetry, enabling the formation of optical isomers. In contrast, complex \( B = [\text{Co}(\text{NH}_3)_4 \text{Cl}_2]^+ \), with only monodentate ligands, typically does not exhibit optical isomerism since its arrangement of ligands is usually symmetrical and does not allow for non-superimposable mirror images.

Magnetic Properties

The magnetic properties of coordination compounds are primarily determined by the electronic configuration of the central metal ion and the field strength of the ligands. For both complexes \( A \) and \( B \), the central metal ion is \( \text{Co}^{3+} \), which has a 3d6 electronic configuration. In a low-spin configuration, strong field ligands like ethylenediamine and ammonium can cause the electrons to pair up. This paired electron arrangement results in no unpaired electrons, leading to diamagnetism. Thus, both complexes \( A \) and \( B \) are diamagnetic because they possess low-spin configurations.

Inner and Outer Orbital Complexes

Coordination compounds can be classified based on the type of orbitals used in hybridization. Inner orbital complexes use inner d orbitals (3d) for hybridization, while outer orbital complexes utilize outer d orbitals (4d or 5d). For complex \( A = [\text{Co}(\text{en})_2 \text{Cl}_2]^+ \), it typically forms inner orbital complexes because it involves low-spin Co(III) with strong field ligands like en, leading to 3d orbital participation. In contrast, complex \( B = [\text{Co}(\text{NH}_3)_4 \text{Cl}_2]^+ \) is usually an outer orbital complex since it features monodentate ligands like ammonia, resulting in the use of 4d orbitals due to the weaker field strength of \( \text{NH}_3 \). Hence, statement (4) in the exercise is incorrect as \( A \) is typically an inner orbital complex, while \( B \) is an outer orbital complex.