Which of the following complex is considered to be tetrahedral?

[FeCl₄]^-

The correct answer is option 1. \([FeCl_4]^-\).

In coordination chemistry, the geometry of a complex depends on several factors, including the electronic configuration of the central metal ion, the oxidation state, and the nature of the ligands (whether they are strong or weak field ligands).

Tetrahedral Geometry: Typically favored by \(d^0\), \(d^5\) (high-spin), and \(d^{10}\) metal ions with weak field ligands.

Square Planar Geometry: Typically favored by d8 metal ions with strong field ligands or in a low oxidation state.

Analysis of Each Complex:

1. \([FeCl_4]^-\)

Central metal ion: \( \text{Fe}^{3+} \) (\(d^5\) configuration).

Oxidation state: +3.

Ligands: Chloride (\( \text{Cl}^- \)), which are weak field ligands.

Electronic Configuration: \(3d^5\).

With weak field ligands like chloride, \(Fe^{3+}\) often adopts a high-spin state which favors a tetrahedral geometry to minimize electron-electron repulsion in a high-spin \(d^5\) configuration. The shape is tetrahedral.

2. \([Ni(CN)_4]^{2-}\)

Central metal ion: \( \text{Ni}^{2+} \) (\(d^8\) configuration).

Oxidation state: +2.

Ligands: Cyanide (\( \text{CN}^- \)), which is a strong field ligand.

Electronic Configuration: \(3d^8\).

Strong field ligands like cyanide lead to pairing of electrons in the d orbitals, favoring a square planar geometry for \(d^8\) metal ions. The shape is square Planar.

3. \([PtCl_4]^{2-}\)

Central metal ion: \( \text{Pt}^{2+} \) (\(d^8\) configuration).

Oxidation state: +2.

Ligands: Chloride (\( \text{Cl}^- \)), weak field ligands.

Electronic Configuration: \(5d^8\).

Platinum(II) often adopts a square planar geometry due to the relativistic effects and the tendency of \(d^8\) metal ions to form square planar complexes. The shape is square Planar.

4. \([PdCl_2(NH_3)_2]\)

Central metal ion: \( \text{Pd}^{2+} \) (\(d^8\) configuration).

Oxidation state: +2.

Ligands: Chloride (\( \text{Cl}^- \)) and ammonia (\( \text{NH3} \)), moderate field ligands.

Electronic Configuration: \(4d^8\).

Palladium(II) often adopts a square planar geometry due to its \(d^8\) configuration. The shape is square Planar.

Summary

\([FeCl_4]^-\): \(Fe^{3+}\) with weak field ligands (chloride) favors a tetrahedral geometry due to its high-spin \(d^5\) configuration.

\([Ni(CN)_4]^{2-}\): \(Ni^{2+}\) with strong field ligands (cyanide) favors a square planar geometry due to its \(d^8\) configuration.

\([PtCl_4]^{2-}\): \(Pt^{2+}\) with weak field ligands (chloride) favors a square planar geometry due to its \(d^8\) configuration and relativistic effects.

\([PdCl_2(NH_3)_2]\): \(Pd^{2+}\) with weak/moderate field ligands (chloride and ammonia) favors a square planar geometry due to its \(d^8\) configuration.

Thus, the complex that is considered to be tetrahedral is:\([FeCl_4]^-\)