Problem 168
Question
Which one of the following has a square planar geometry? [2007] (a) \(\left[\mathrm{FeCl}_{4}\right]^{2-}\) (b) \(\left[\mathrm{NiCl}_{4}\right]^{2}\) (c) \(\left[\mathrm{PtCl}_{4}\right]^{2-}\) (d) \(\left[\mathrm{CoCl}_{4}\right]^{2-}\)
Step-by-Step Solution
Verified Answer
\([\mathrm{PtCl}_4]^{2-}\) has a square planar geometry.
1Step 1: Determine the Oxidation State
For each complex, we first identify the oxidation state of the central metal atom.(a) For \(\left[ \mathrm{FeCl}_{4} \right]^{2-}\), let the oxidation state of iron be \(x\). We have:\[x + 4(-1) = -2 \x - 4 = -2 \x = +2\](b) For \(\left[ \mathrm{NiCl}_{4} \right]^{2-}\), considering \(x\) as the oxidation state of nickel:\[x + 4(-1) = -2 \x - 4 = -2 \x = +2\](c) For \(\left[ \mathrm{PtCl}_{4} \right]^{2-}\), let \(x\) be the oxidation state of platinum:\[x + 4(-1) = -2 \x - 4 = -2 \x = +2\](d) For \(\left[ \mathrm{CoCl}_{4} \right]^{2-}\), let \(x\) be the oxidation state of cobalt:\[x + 4(-1) = -2 \x - 4 = -2 \x = +2\]
2Step 2: Determine the Electron Configuration
Next, we determine the electron configuration of each metal ion related to their oxidation state. Focus on the 3d and 4s orbitals (or equivalent d orbitals for higher transition metals).(a) Iron in \(\left[ \mathrm{FeCl}_{4} \right]^{2-}\) is \(\mathrm{Fe}^{2+}\) with electron configuration 3d\(^{6}\).(b) Nickel in \(\left[ \mathrm{NiCl}_{4} \right]^{2-}\) is \(\mathrm{Ni}^{2+}\) with electron configuration 3d\(^{8}\).(c) Platinum in \(\left[ \mathrm{PtCl}_{4} \right]^{2-}\) is \(\mathrm{Pt}^{2+}\) with electron configuration 5d\(^{8}\).(d) Cobalt in \(\left[ \mathrm{CoCl}_{4} \right]^{2-}\) is \(\mathrm{Co}^{2+}\) with electron configuration 3d\(^{7}\).
3Step 3: Determine Possible Geometry
Identify the possible geometries for each ion based on d-orbital electron configuration and ligand characteristics.(a) \(\mathrm{Fe}^{2+}\) with 3d\(^{6}\) configuration can form tetrahedral complexes, not square planar.(b) \(\mathrm{Ni}^{2+}\) typically forms tetrahedral complexes due to its 3d\(^{8}\) electron configuration but can also form square planar.(c) \(\mathrm{Pt}^{2+}\) with 5d\(^{8}\) tends to form square planar complexes due to its larger d-orbital involvement and ligand field stabilization.(d) \(\mathrm{Co}^{2+}\) with 3d\(^{7}\) generally favors tetrahedral geometry.
4Step 4: Verify Preferred Geometry
Consider the electron count and field strengths:- \(\mathrm{Pt}^{2+}\) (\(\left[ \mathrm{PtCl}_{4} \right]^{2-}\)) typically organizes in a low-spin square planar geometry due to the strong ligand field stabilization in its dx\(^2\)-y\(^2\) orbital.- Other complexes are more likely tetrahedral due to their ligand field and electron configuration.- \(\left[ \mathrm{PtCl}_{4} \right]^{2-}\) is known for square planar geometry based on experimental evidence and ligand strength.
Key Concepts
Oxidation StateElectron ConfigurationLigand Field Theory
Oxidation State
Understanding oxidation state is crucial when analyzing the nature of compounds. The oxidation state, also known as oxidation number, refers to the hypothetical charge that an atom would have if all bonds were 100% ionic. For transition metals in coordination complexes, determining oxidation state helps us know the charge on the metal after accounting for the surrounding ligands.
To determine the oxidation state of a central metal in a complex, consider the overall charge of the complex and the charges contributed by the ligands:
To determine the oxidation state of a central metal in a complex, consider the overall charge of the complex and the charges contributed by the ligands:
- The ligands, such as chloride (Cl^{-} ), typically have known charges.
- Write a simple equation to solve for the unknown oxidation state of the metal, using the formula:
x + n( ext{charge of ligand}) = ext{overall charge of the complex}
Electron Configuration
Each element's electron configuration provides us with a map of its electrons, ordered based on increasing energy levels. This is essential for understanding the reactivity and geometry of molecules, especially complexes involving transition metals.
The focus lies on the d-block elements where the electron configuration can significantly influence geometry:
The focus lies on the d-block elements where the electron configuration can significantly influence geometry:
- The electron configuration is affected by the oxidation state of the metal. For example, removing two electrons from Fe, Ni, or Pt will result in 0_{3} m{d}0_{6}Fe^{2+} or m{d}0_{8}Ni^{2+}, and _{3} m{d}0_{8}Pt^{2+}.
- By knowing this configuration, we can predict how different electrons will pair up or remain unpaired, influencing whether a complex adopts a specific geometry.
Ligand Field Theory
Ligand Field Theory (LFT) is a powerful tool that explains the properties and behaviors of complexes, especially focusing on splitting of d-orbitals due to ligand interactions. This theory helps us understand why certain geometries are preferred.
The key concepts of Ligand Field Theory include:
The key concepts of Ligand Field Theory include:
- Ligands approaching a transition metal cause energy levels of d-orbitals to split into higher and lower energy states.
- The extent of this splitting depends on the nature of the ligands and their positions around the metal.
Square Planar vs Tetrahedral Geometry
The possible geometries are determined by different factors:- For low-spin d0_{8} complex like [PtCl_{4}]^{2-}, strong field ligands lead to a significant splitting where electrons prefer lower energy slots forming a square planar geometry.
- In complexes where d-orbital splitting is weaker, such as with CoCl_{4}^{2-} and FeCl_{4}^{2-}, a tetrahedral geometry is more common.
Other exercises in this chapter
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