Which of the following complex is considered to be tetrahedral?

[FeCl₄]^-

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

To determine which complex is considered to be tetrahedral, we need to understand the geometry of each complex based on the ligand arrangement around the central metal ion.

Here are the options provided:

1. \([FeCl_4]^-\): This complex has an iron ion \((Fe)\) surrounded by four chloride ions \((Cl^-)\). Chloride ions are typically considered as weak-field ligands, favoring a high-spin configuration, and the geometry around Fe in this complex is tetrahedral.

2. \([Ni(CN)_4]^{2-}\): This complex consists of a nickel ion \((Ni)\) coordinated with four cyanide ions \((CN^-)\). Cyanide ions are strong-field ligands, leading to a low-spin configuration, and the geometry around Ni in this complex is square planar, not tetrahedral.

3. \([PtCl_4]^{2-}\): This complex has a platinum ion \((Pt)\) coordinated with four chloride ions \((Cl^-)\). Chloride ions are weak-field ligands, favoring a high-spin configuration, and the geometry around Pt in this complex is tetrahedral.

4. \([PdCl_2(NH_3)_2]\): This complex contains a palladium ion \((Pd)\) coordinated with two chloride ions \((Cl^-)\) and two ammonia molecules \((NH3_)\). Chloride ions are weak-field ligands, and the geometry around Pd in this complex is square planar due to the presence of the two ammonia ligands.

The complex that is considered tetrahedral is: \([FeCl_4]^-\). This complex features a tetrahedral geometry around the iron ion (Fe) due to the coordination with four chloride ions (Cl^-), which are weak-field ligands.