Coordination Chemistry

(I) The complex exhibits ionization isomerism. Ionization isomerism arises when the exchange of ligands between the cationic and anionic parts results in the formation of different ions in solution.

(II) The complex exhibits linkage isomerism, as the ethylenediamine (en) ligand can bind through different sites.

(III) The complex exhibits geometric isomerism, as the arrangement of ligands around the metal ion can vary.

To solve the problem, we need to identify the relationship between the complex ions and .

1. Analyzing the Complex Ions:
Both complexes have the same formula: Co, 5 NH ligands, and a nitro group, with a 2+ charge. The difference lies in the ligand:
- In , the NO ligand is coordinated via the nitrogen atom (nitro, -NO ).
- In , the NO ligand is coordinated via an oxygen atom (nitrito, -ONO).

2. Identifying the Type of Isomerism:
The two complexes have the same molecular formula but differ in how the NO ligand is attached to the metal center. This type of isomerism, where the same ligand binds through different atoms, is called linkage isomerism. The NO /ONO ligand is ambidentate, meaning it can coordinate via either the N or O atom, leading to these two forms.

3. Considering Other Types of Isomerism:
- Coordination isomerism involves swapping ligands between cation and anion, but here both are cations with the same ligands.
- Geometric isomerism (e.g., cis-trans) is not applicable, as (with 5 identical ligands) lacks the necessary symmetry for such isomerism.
Linkage isomerism is the correct classification.

Final Answer:
The complex ions and are called linkage isomers.

Homoleptic Complexes

Step 1: Understanding homoleptic complexes. Homoleptic complexes are those in which all the ligands attached to the central metal atom are the same. Both and are examples of homoleptic complexes, as all the ligands in each complex are water molecules.

Step 2: Conclusion. Since Assertion (A) is true, and Reason (R) correctly explains it, the correct answer is option (A).

For Fe³⁺ (d⁵ configuration), the number of unpaired electrons depends on the ligand field strength.

(I) With a strong field ligand, the d-orbitals split, causing pairing of electrons. Thus, Fe³⁺ in the presence of a strong field ligand will have 0 unpaired electrons.

(II) With a weak field ligand, the d-orbitals do not split enough to cause electron pairing. Therefore, Fe³⁺ in the presence of a weak field ligand will have 5 unpaired electrons.

Coordination Chemistry – Question and Answer

(i) What happens to the colour of the complex when heated gradually?

Answer: The colour of the complex deepens when it is heated gradually. This occurs because thermal energy promotes electronic transitions between the metal ion and the ligands. As temperature increases, the ligand field interaction may change slightly, shifting the absorption spectrum and resulting in a more intensely coloured complex.

(ii) Write the electronic configuration for a ion if .

Answer: When , the complex forms a high-spin configuration. Thus, the electronic configuration is:

This means three electrons occupy the lower-energy orbitals and two occupy the higher-energy orbitals, following Hund’s rule of maximum multiplicity.

(iii) Write the hybridization and magnetic behaviour of the complex .
(Atomic number: Ni = 28)

Answer: In , nickel is in the zero oxidation state. The electron configuration of Ni is:

CO is a strong field ligand, so it causes pairing of the electrons. All electrons pair up, and the and orbitals hybridize to form four equivalent orbitals, giving:

• Hybridization:
• Geometry: Tetrahedral
• Magnetic behaviour: Diamagnetic (no unpaired electrons)

(i) Double Salts vs Complex Compounds

Double salts are ionic compounds formed by the combination of two simple salts, but they retain their individual identities in solution. These compounds do not involve coordinate bonds.

For example: is a double salt.

In contrast, complex compounds consist of a central metal ion bonded to ligands through coordinate covalent bonds. These ligands donate electron pairs to the metal.

Example of a complex compound: where "en" stands for ethylenediamine, a ligand.

(ii) Didentate vs Ambidentate Ligands

- Didentate ligands are ligands that can bind to a metal atom or ion through two donor atoms simultaneously. An example is ethylenediamine (en), which uses both nitrogen atoms to bind to the metal.

- Ambidentate ligands have two potential donor atoms, but can coordinate through only one atom at a time. A classic example is the thiocyanate ion: It can bind to the metal via the sulfur atom (S-bound) or the nitrogen atom (N-bound), but not both at once.

To solve the problem, we need to identify which of the given coordination compounds is diamagnetic.

1. Understanding Diamagnetism:
A species is diamagnetic if all its electrons are paired, resulting in no unpaired electrons and a magnetic moment of zero. We need to examine each coordination compound to determine the number of unpaired electrons in the central metal ion, considering the oxidation state, electronic configuration, and ligand field effects.

2. Option A: [Ni(CN) ] :
- Oxidation State: CN has a charge of -1, so 4 CN contribute -4. The overall charge is -2, so Ni’s oxidation state is +2 (4(-1) + x = -2, x = +2).
- Electron Configuration: Ni (atomic number 28) has a ground state configuration of [Ar] 4s 3d . Ni loses two electrons from the 4s orbital: [Ar] 3d .
- Geometry and Ligand Field: [Ni(CN) ] is a 4-coordinate complex. Ni (d ) with CN (a strong field ligand) typically forms a square planar geometry due to the strong field splitting. In a square planar field, the d orbitals split as: d , d , d , d , d . The 8 d-electrons fill the lower orbitals: (d ) (d ) (d ) (d ) , leaving d empty. All electrons are paired.
- Result: No unpaired electrons, so [Ni(CN) ] is diamagnetic.

3. Option B: [NiCl ] :
- Oxidation State: Cl has a charge of -1, so 4 Cl contribute -4. The overall charge is -2, so Ni’s oxidation state is +2.
- Electron Configuration: Ni is 3d .
- Geometry and Ligand Field: [NiCl ] is a 4-coordinate complex. Cl is a weak field ligand, and for Ni (d ), 4-coordinate complexes are typically tetrahedral (not square planar, as tetrahedral splitting is smaller). In a tetrahedral field, the d orbitals split as: e (d , d ) and t (d , d , d ), with e lower in energy. The 8 electrons fill as: (e) (t ) . The t set has 3 orbitals, so 4 electrons distribute as 2 paired, 2 unpaired (e.g., (d ) (d ) (d ) ).
- Result: Two unpaired electrons, so [NiCl ] is paramagnetic.

4. Option C: [Fe(CN) ] :
- Oxidation State: CN contributes -6, overall charge is -3, so Fe’s oxidation state is +3 (6(-1) + x = -3, x = +3).
- Electron Configuration: Fe (atomic number 26) is [Ar] 4s 3d . Fe is [Ar] 3d .
- Geometry and Ligand Field: [Fe(CN) ] is a 6-coordinate octahedral complex. CN is a strong field ligand. In an octahedral field, d orbitals split into t (lower) and e (higher). For d Fe with a strong field ligand, the splitting is large, so it’s low spin: (t ) (e ) , with 5 electrons in t (3 orbitals): (t ) has 1 unpaired electron (e.g., 2 paired, 1 unpaired).
- Result: One unpaired electron, so [Fe(CN) ] is paramagnetic.

5. Option D: [CoF ] :
- Oxidation State: F contributes -6, overall charge is -3, so Co’s oxidation state is +3.
- Electron Configuration: Co (atomic number 27) is [Ar] 4s 3d . Co is [Ar] 3d .
- Geometry and Ligand Field: [CoF ] is octahedral. F is a weak field ligand. For d Co with a weak field ligand, the splitting is small, so it’s high spin: (t ) (e ) . t has 4 electrons (2 paired, 1 unpaired per orbital), and e has 2 electrons (1 unpaired per orbital), totaling 4 unpaired electrons.
- Result: Four unpaired electrons, so [CoF ] is paramagnetic.

Final Answer:
The diamagnetic species is [Ni(CN) ] (option A).

The color of a coordination compound is largely influenced by the ligand field and the d-electron transitions of the metal ion.

In the case of , water is a weak field ligand that causes a small splitting of the d-orbitals in . This allows the absorption of light in the visible spectrum, giving the solution a green color.

In contrast, in , cyanide is a strong field ligand that causes a large splitting of the d-orbitals in , leading to no available electronic transitions in the visible region. As a result, the solution of is colorless because no visible light is absorbed.

The complex exhibits ionization isomerism

In ionization isomerism, two different compounds have the same molecular formula but differ in the way the ions are arranged. One of the isomers will have the ion as the anion, while the other will have the ion as the anion.

The difference in the arrangement of ions leads to the formation of different ions in solution, which are responsible for the isomerism.

Low spin tetrahedral complexes are not known because tetrahedral complexes typically have a weaker ligand field splitting energy compared to octahedral complexes. In a tetrahedral field, the splitting energy ( ) is smaller, which does not create a significant enough energy difference between the d-orbitals to cause pairing of electrons. Therefore, in a tetrahedral complex, the electrons tend to occupy all available orbitals singly (high-spin state), and low-spin tetrahedral complexes are not stable.