Question

The pair(s) of complexes wherein both exhibit tetrahedral geometry is(are)
(Note: py = pyridine, Given: Atomic numbers of Fe, Co, Ni and Cu are 26, 27, 28 and 29, respectively)

• A. \([FeCl_4]^{-} \quad \text{and} \quad [Fe(CO)_4]^{2-}\)
• B. \([Co(CO)_4]^{-} \quad \text{and} \quad [CoCl_4]^{2-}\)
• C. \([Ni(CO)_4] \quad \text{and} \quad [Ni(CN)_4]^{2-}\)
• D. \([Cu(py)_4]^{+} \quad \text{and} \quad [Cu(CN)_4]^{3-}\)

Answer 1

The Correct Option is A, B, D

Step 1: Understanding the Given Complexes
The given reactions represent the dissociation of different metal-ligand complexes. The metal ions undergo various changes in oxidation states, and we are interested in the hybridization and geometry of the resulting complexes.
Step 2: Understanding the Hybridization
For each metal complex, the central metal ion undergoes hybridization depending on its electronic configuration and the nature of the ligand.
- [FeCl₄]^– → Fe³⁺, 3d⁵ (weak field ligand)
The iron ion, in the +3 oxidation state, has a 3d⁵ configuration. This is typical for a weak field ligand, leading to sp³ hybridization, resulting in a tetrahedral geometry.
- [Fe(CO)₄]^–2 → Fe^2–, 3d¹⁰ → sp³
In this case, the iron ion has a 3d¹⁰ configuration, indicating a fully filled d-orbital. CO is a strong field ligand, which promotes sp³ hybridization, leading to tetrahedral geometry.
- [Co(CO)₄]^– → Co^–, 3d¹⁰ → sp³
Similarly, for the cobalt ion with a 3d¹⁰ configuration, sp³ hybridization occurs, resulting in a tetrahedral geometry due to the strong field nature of CO.
- [CoCl₄]^2– → Co²⁺, 3d⁷ (weak field ligand) → sp³
In the case of Co²⁺ with a 3d⁷ configuration, the weak field ligand (Cl^–) leads to sp³ hybridization, which results in tetrahedral geometry.
- [Ni(CO)₄] → Ni, 3d¹⁰ → sp³
For nickel with a 3d¹⁰ configuration, CO acts as a strong field ligand, leading to sp³ hybridization and tetrahedral geometry.
- [Ni(CN)₄]^2– → Ni²⁺, 3d⁸ (strong field ligand) → dsp²
In this case, the nickel ion has a 3d⁸ configuration, and the strong field ligand CN^– promotes dsp² hybridization, leading to square planar geometry.
- [Cu(py)₄]⁺ → Cu⁺, 3d¹⁰ → sp³
The copper ion with a 3d¹⁰ configuration undergoes sp³ hybridization, resulting in tetrahedral geometry.
- [Cu(CN)₄]^3– → Cu⁺, 3d¹⁰ → sp³
Again, for copper with a 3d¹⁰ configuration, CN^– is a strong field ligand, and sp³ hybridization leads to tetrahedral geometry.
Step 3: Conclusion
In all the complexes with a 3d¹⁰ electronic configuration, the only possible hybridization and geometry is sp³ and tetrahedral. However, for complexes with a 3d⁸ configuration, such as [Ni(CN)₄]^2–, dsp² hybridization leads to square planar geometry.