Problem 46
Question
The species having tetrahedral shape is (a) \(\left[\mathrm{PdCl}_{4}\right]^{2-}\) (b) \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) (c) \(\left[\mathrm{Pd}(\mathrm{CN})_{4}\right]^{2-}\) (d) \(\left[\mathrm{NiCl}_{4}\right]^{2-}\)
Step-by-Step Solution
Verified Answer
The species with a tetrahedral shape is \( \left[\mathrm{NiCl}_{4}\right]^{2-} \).
1Step 1: Identify Coordination Number
Determine the coordination number for each complex, which is the number of ligand donor atoms bonded to the central metal atom. For all the complexes given, the coordination number is 4 because each metal is surrounded by 4 ligands.
2Step 2: Understand the Coordinate Geometry
For coordination number 4, there are two possible geometries: tetrahedral and square planar. Tetrahedral complexes are more common for lighter metals or those with a d10 electronic configuration.
3Step 3: Analyze Electronic Configurations
Examine the electron configuration of the central metal atoms in each complex. - Pd is in d8 state in both \( \left[\mathrm{PdCl}_{4}\right]^{2-} \) and \( \left[\mathrm{Pd(CN)}_{4}\right]^{2-} \) , typically forming square planar complexes. - Ni is in d8 state in \( \left[\mathrm{Ni(CN)}_{4}\right]^{2-} \), forming square planar due to the strong field of \(CN^-\).- Ni in \( \left[\mathrm{NiCl}_{4}\right]^{2-} \) is a d10 configuration making it tetrahedral.
4Step 4: Identify Ligand Field Strength
Consider the field strength of the ligands: \(CN^-\) is a strong field ligand often leading to square planar configurations. \(Cl^-\) is a weak field ligand, which does not typically cause pairing of electrons, favoring tetrahedral geometry.
5Step 5: Determine the Shape
Based on coordination number, metal electron configuration, and ligand field strength, deduce the geometry: - \( \left[\mathrm{NiCl}_{4}\right]^{2-} \) is tetrahedral.- The others are square planar due to either high d-block electron count or strong field ligands.
Key Concepts
Coordination NumberTetrahedral GeometryLigand Field Strength
Coordination Number
In coordination chemistry, the coordination number is a fundamental concept that helps us understand how many atoms or ions are bonded to a central metal ion in a complex. It is determined by counting the number of ligand donor atoms that are directly attached to the metal center. In the context of the exercise, each complex has a coordination number of 4. This is because each metal center is bound to four ligands:
- Coordination number 4 allows complexes to exhibit distinct geometries. These geometries can significantly affect a complex's chemical properties.
- The common geometries for these numbers include tetrahedral and square planar arrangements, depending on various factors such as the metal's electronic configuration and the type of ligands.
Tetrahedral Geometry
Tetrahedral geometry is one of the possible structures for complexes with a coordination number of 4. The name tetrahedral comes from the geometry where four atoms are symmetrically arranged around a central atom, forming a shape that is reminiscent of a pyramid with a triangular base. This shape is very common in coordination complexes, especially those with certain electronic configurations:
- A tetrahedral shape is favored in cases where the metal center avoids electron-electron repulsion as opposed to planar arrangements.
- In a tetrahedral complex, the bond angles are approximately 109.5°, which helps minimize these repulsions.
Ligand Field Strength
Ligand field strength is a key concept for determining the geometry of a metal complex. It refers to the ability of a ligand to split the d-orbitals of the central metal ion. The field strength of a ligand can influence whether a complex will form a tetrahedral or square planar shape:
- Strong field ligands, like \(CN^{-}\), typically lead to large splitting of d-orbitals. This generally results in a preference for square planar complexes as seen in\(\left[\mathrm{Ni(CN)}_{4}\right]^{2-}\).
- Weak field ligands, such as \(Cl^{-}\), result in smaller splitting and thus often favor the formation of tetrahedral complexes, as they do not lead to efficient pairing of d-electrons.
Other exercises in this chapter
Problem 44
\(\mathrm{Ag}^{+}+\mathrm{NH}_{3} \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)\right]^{+} ; k_{1}=6.8 \times 10^{-3}\) \(\left[\mathrm{Ag}\le
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Statement-1 : The geometrical isomers of the complex \(\left[M\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]\) are optically inactive. Statement-2 : Bo
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Write the formulae of the following complexes: (i) Pentaamminechlorocobalt(III) (ii) Lithium tetrahydroaluminate(III).
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The geometry of \(\mathrm{Ni}(\mathrm{CO})_{4}\) and \(\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\) are (a) both square planar (b) tetrahedral
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