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Chapter 20: Problem 124

Identify the correct statement (a) \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) is tetrahedral and paramagnetic (b) \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) is square planar and paramagnetic (c) \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) is square planar and paramagnetic (d) \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}\) is tetrahedral and diamagnetic

Short Answer

Expert verified
Option (d) is the correct statement.

Step by step solution

01

Analyze option (a)

The complex \( [\mathrm{Ni}(\mathrm{CN})_{4}]^{2-} \) contains a Ni ion in a \(d^{8}\) configuration. \(\mathrm{CN}^{-}\) is a strong field ligand which causes pairing of electrons. As a result, it will form a square planar geometry and will be diamagnetic, not tetrahedral nor paramagnetic. Thus, option (a) is incorrect.
02

Analyze option (b)

The complex \( [\mathrm{NiCl}_{4}]^{2-} \) has Ni in the \(d^{8}\) configuration. \(\mathrm{Cl}^{-}\) is a weak field ligand hence it does not cause pairing of electrons, leading to a tetrahedral geometry rather than square planar. The complex should be paramagnetic due to unpaired electrons. Thus, option (b) is incorrect.
03

Analyze option (c)

In \( [\mathrm{Ni}(\mathrm{CO})_{4}] \), CO is a strong field ligand and the Ni center becomes \(d^{10}\) due to the back-bonding property of CO, leading to square planar geometry and being diamagnetic (no unpaired electrons). However, the option states paramagnetic, so option (c) is incorrect.
04

Analyze option (d)

The complex \( [\mathrm{Cu}(\mathrm{CN})_{4}]^{3-} \) has Cu in the +1 oxidation state resulting in a \(d^{10}\) configuration. \(\mathrm{CN}^{-}\) is a strong field ligand, but with Cu in \(d^{10}\), it remains filled, resulting in a tetrahedral geometry and being diamagnetic (all electrons are paired). Thus, option (d) is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ligand Field Theory
Ligand Field Theory is a crucial concept in understanding the behavior and properties of coordination compounds. This theory explains how the five degenerate d-orbitals of a metal ion become non-degenerate when surrounded by a set of ligands. This splitting of d-orbitals is due to the electrostatic interactions between the metal cation and the ligands.

  • Strong field ligands, such as \(\text{CN}^-\), cause a large splitting of the d-orbitals. This often results in electron pairing within the lower energy orbitals, which can lead to diamagnetic properties.
  • Weak field ligands, like \(\text{Cl}^-\), cause a smaller splitting of the d-orbitals, often leading to unpaired electrons and resulting in paramagnetic compounds.
Ligand Field Theory is essentially an extension of Crystal Field Theory, taking into account the overlap of metal and ligand orbitals (molecular orbital considerations) to describe these interactions in a more detailed manner. Understanding this theory helps in predicting the magnetic and spectroscopic properties of complexes.
Crystal Field Theory
Crystal Field Theory originates from the concept that when ligands approach a metal ion, they create a crystal field. This field impacts the metal d-orbitals, causing them to split into different energy levels. The degree of this splitting depends on the nature of the ligands:

  • In octahedral complexes, the \(e_g\) set of d-orbitals experiences a higher increase in energy compared to the \(t_{2g}\) set, leading to a splitting pattern known as the crystal field splitting, denoted as \(\Delta \).
  • In tetrahedral complexes, the pattern is reversed with the \(t_2\) level being at a higher energy than the \(e\) level, leading to a smaller \(\Delta_t\) than \(\Delta_o\)
The magnitude of this splitting, \(\Delta\), determines many properties including the magnetic behavior of the compound. Complexes with large \(\Delta\) are usually diamagnetic as the lower-energy orbitals fill completely before any electron occupies the higher-energy ones.
Geometry of Complexes
The geometry of complexes is a key aspect that defines their bonding and properties. Depending on the configuration and type of ligands, a complex can adopt different geometries.

  • Square planar geometry is typically found in \(d^8\) metal centers with strong field ligands. In such configurations, the occupied orbitals form a square planar shape around the central atom. This is often seen in complexes like \([\mathrm{Ni}(\mathrm{CN})_{4}]^{2-}\).
  • Tetrahedral geometry is more common in systems with weak field ligands, such as \(\mathrm{Cl}^-\), where the ligands position themselves to minimize repulsion, creating a tetrahedral shape.
Understanding the geometry is important as it influences not just the physical shape, but also the magnetic and electronic properties of the complex.
Magnetic Properties of Complexes
Magnetic properties of complexes provide insights into the arrangement of electrons in d-orbitals and the overall electronic configuration. Complexes can be either paramagnetic or diamagnetic.

  • Paramagnetic complexes have unpaired electrons, which align with external magnetic fields. This is a common trait of complexes having weak field ligands like \(\mathrm{Cl}^-\), which do not pair up the electrons in higher energy orbitals.
  • Diamagnetic complexes have all electrons paired, resulting in no net magnetic moment. These are typically complexes with strong field ligands like \(\mathrm{CN}^-\), causing complete pairing of electrons in lower energy levels.
Studying the magnetic properties is crucial as it confirms the electron configuration of the metal ion and the extent of electron pairing attributed to the field strength of ligands.

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Most popular questions from this chapter

While \(\mathrm{Ti}^{3^{+}}, \mathrm{V}^{3^{+}}, \mathrm{Fe}^{3+}\) and \(\mathrm{Co}^{2^{+}}\)can afford a large num- ber of tetrahedral complexes, \(\mathrm{Cr}^{3+}\) never does this, the reason being (a) crystal field stabilisation energy in octahedral vis-脿-vis tetrahedral \(\mathrm{Cr}^{3+}\) system plays the deciding role (b) \(\mathrm{Cr}^{3^{+}}\)forces high crystal field splitting with a varieties of ligands (c) electronegativity of \(\mathrm{Cr}^{3+}\) is the largest among these trivalent 3 d-metals and so chromium prefers to be associated with as many ligands as its radius permits (d) both (b) and (c)

A \(0.01 \mathrm{M}\) complex of \(\mathrm{CoCl}_{2}\) and \(\mathrm{NH}_{3}\) (molar ratio \(1: 4\) ) is found to have effective molarity of \(0.02 \mathrm{M}\) (evaluated from colligative property). What is the formula of the complex? (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{3}\) (d) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}\)

The proposed complex with the nomenclature chlorodiaquatriammine cobalt (III) chloride can be represented wrongly as (a) \(\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right] \mathrm{Cl}_{2}\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{3}\right]\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (d) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}\right] \mathrm{Cl}_{3}\)

Which of the following complex ions will not show optical activity? (a) \(\left[\mathrm{Co}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]^{+}\) (b) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+}\) (c) \(\left[\mathrm{Pt}(\mathrm{Br})(\mathrm{Cl})\right.\) (I) \(\left(\mathrm{NO}_{2}\right)\) (Py) \(\left.\mathrm{NH}_{3}\right]\) (d) \(\operatorname{cis}=\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\)

Tetrahedral complexes of the types of \(\left[\mathrm{Ma}_{4}\right]\) and \(\left[\mathrm{Ma}_{3} \mathrm{~b}\right]\) (here \(\mathrm{M}=\) Metal, \(\mathrm{a}, \mathrm{b}=\) Achiral ligands) are not able to show optical isomerism because (a) these molecules/ions have non super imposable mirror images (b) these molecules possess a centre of symmetry (c) these molecules/ions possess a plane of symmetry and hence are achiral (d) these molecules/ions possess \(\mathrm{C}_{\mathrm{n}}\) axis of symmetry
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