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Among \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right],\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) and \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) (a) \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{NiCl}_{4}\right]^{2}\) are diamagnetic and \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2}\) is paramagnetic (b) \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) and \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) are diamagnetic and \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) is paramagnetic (c) \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{NiCN}_{4}\right]^{2-}\) are diamagnetic and \(\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]^{2-}\) is paramagnetic (d) \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) is diamagnetic and \(\left[\mathrm{NiCl}_{4}\right]^{2}\) and \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) are paramagnetic

Short Answer

Expert verified
Option (c): \([\mathrm{Ni}( ext{CO})_{4}]\) and \([\mathrm{NiCN}_{4}]^{2-}\) are diamagnetic, \([\mathrm{NiCl}_{4}]^{2-}\) is paramagnetic.

Step by step solution

01

Introduction to the Problem

We need to determine which of the given nickel complexes are diamagnetic or paramagnetic, which involves examining the electronic structure and ligand field.
02

Determine the Electron Configuration of Ni

Nickel (Ni) has an atomic number of 28. Its ground state electron configuration is \([\text{Ar}] \, 3d^8 \, 4s^2\). When forming complexes, the 4s electrons are lost, leading to \3d^8\. We now need to consider how the ligands affect this configuration.
03

Analyze \([\mathrm{Ni}( ext{CO})_{4}]\)

CO is a strong field ligand and causes pairing of the electrons in Ni. This results in a \(3d^{10}\) configuration with no unpaired electrons, making \([\mathrm{Ni}( ext{CO})_{4}]\) diamagnetic.
04

Analyze \([\mathrm{Ni}( ext{CN})_{4}]^{2-}\)

CN is also a strong field ligand. When CN ligands form the complex \([\mathrm{Ni}( ext{CN})_{4}]^{2-}\), the electrons pair up leading to all electrons being paired in \(3d^{10}\). Thus, \([\mathrm{Ni}( ext{CN})_{4}]^{2-}\) is diamagnetic.
05

Analyze \([\mathrm{NiCl}_{4}]^{2-}\)

Cl is a weak field ligand and does not cause electron pairing. Therefore, \([\mathrm{NiCl}_{4}]^{2-}\) maintains the \(3d^8\) configuration with two unpaired electrons, making it paramagnetic.
06

Conclusion and Selection of Option

Given the analyses, we see that \([\mathrm{Ni}( ext{CO})_{4}]\) and \([\mathrm{Ni}( ext{CN})_{4}]^{2-}\) are diamagnetic, while \([\mathrm{NiCl}_{4}]^{2-}\) is paramagnetic. This matches option (c).

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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 coordination chemistry. It helps us understand how ligands affect the electronic structure of metals in complexes.
This theory arises as an extension of Crystal Field Theory and takes into account the overlap of ligand orbitals with metal orbitals. This interaction can lead to a splitting of the metal's d orbitals, which affects both the stability and properties of the complex formed.
Ligands can be categorized based on their field strength:
  • **Strong field ligands**, like CN− and CO, cause significant splitting of d orbitals. This often results in electron pairing within the lower energy orbitals. Complexes with strong field ligands tend to be low spin.
  • **Weak field ligands**, like Cl−, cause less splitting. This can result in higher spin complexes with unpaired electrons.
Understanding the influence of ligand field can help predict if a complex is diamagnetic or paramagnetic, which aligns with the behavior of the Ni complexes analyzed in the problem.
Electron Configuration
Electron configuration is key when analyzing the magnetic properties of metal complexes. For nickel, the initial electron configuration is \([ ext{Ar}] \, 3d^8 \, 4s^2 \).
Upon complex formation, nickel typically loses its 4s electrons.
This results in \(3d^8\) configuration.
The distribution of electrons among d orbitals in complexes is heavily influenced by the ligands attached.
In the provided exercise:
  • **In** \([\text{Ni}(\text{CO})_4]\), the strong field ligand CO causes electron pairing, yielding a \(3d^{10}\) configuration with all electrons paired.
  • **In** \([\text{Ni}(\text{CN})_4]^{2-}\), with CN also as a strong field ligand, electron pairing again results in a \(3d^{10}\) configuration.
  • **In** \([\text{NiCl}_4]^{2-}\), the weak field ligand Cl maintains a \(3d^8\) configuration with unpaired electrons remaining.
This understanding of electron configurations provides insights into the magnetic properties of these complexes, determining if they are diamagnetic or paramagnetic.
Magnetic Properties of Complexes
The magnetic properties of complexes are directly related to their electron configurations. An important distinction is whether a complex exhibits diamagnetic or paramagnetic behavior.
**Diamagnetic complexes** have all their electrons paired. They are not attracted to magnetic fields and may even experience slight repulsion.
**Paramagnetic complexes** have one or more unpaired electrons. These unpaired electrons create a net magnetic moment, making the complexes attracted to magnetic fields.
Analyzing the Ni complexes:
  • **\([\text{Ni}(\text{CO})_4]\)** has a \(3d^{10}\) configuration due to CO, making it diamagnetic.
  • **\([\text{Ni}(\text{CN})_4]^{2-}\)** also has a \(3d^{10}\) configuration, hence it is diamagnetic.
  • **\([\text{NiCl}_4]^{2-}\)** remains with a \(3d^8\) configuration and two unpaired electrons, rendering it paramagnetic.
Understanding these properties helps predict how a complex will interact with magnetic fields and why they exhibit such behaviors. This knowledge is crucial in fields like materials science and molecular magnetism.

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Most popular questions from this chapter

Formula of ferrocene is (a) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}\) (b) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3+}\) (c) \(\left[\mathrm{Fe}(\mathrm{CO})_{3}\right]\) (d) \(\left[\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \mathrm{Fe}\right]\)

The coordination compound is a complex substance which contains a central metal atom or ion surrounded by oppositely charged ions or neutral molecules. These compounds exhibit structural as well as stereoisomerism. Hybridisation theory explains the geometry of the complex. Crystal field theory explains the colour of complexes and magnetic properties. Which one of the following does not show optical activity? (a) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+}\) (b) \(\left[\mathrm{Pt}(\mathrm{Br})(\mathrm{Cl})(\mathrm{I})\left(\mathrm{NO}_{2}\right)\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}\right)\left(\mathrm{NH}_{3}\right)\right]^{-}\) (c) \(\operatorname{Cis}\left[\mathrm{Co}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]^{+}\) (d) \(\operatorname{Cis}\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\)

Identify the coordination compounds (a) \(\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Fe}(\mathrm{CN})_{2} \cdot 4 \mathrm{KCN}\) (c) \(\mathrm{KCl} \cdot \mathrm{MgCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{CuSO}_{4} \cdot 4 \mathrm{NH}_{3}\)

When degenerate d-orbitals of an isolated atom/ion are brought under the impact of magnetic field of ligands, the degeneracy is lost. The two newly formed sets of d-orbitals, depending upon nature and magnetic field of ligands are either stabilized or destabilized. The energy difference between the two sets whenever lies in the visible region of the electromagnetic spectrum, then the electronic transition \(\mathrm{t}_{2 \mathrm{~g}} \rightleftharpoons \mathrm{e}_{8}\) are responsible for colours of the co- ordination compounds \(\mathrm{Ti}^{3+}(\mathrm{aq})\) is purple while \(\mathrm{Ti}^{4+}(\mathrm{aq})\) is colourless because (a) The difference between \(\mathrm{t}_{2 \mathrm{~g}}\) and \(\mathrm{e}_{\mathrm{g}}\) of \(\mathrm{T} \mathrm{i}^{4+}\) is quite high and does not fall in visible region. (b) There is no crystal field effect in \(\mathrm{Ti}^{4+}\) (c) \(\mathrm{Ti}^{4+}\) has \(\mathrm{d}^{0}\) configuration. (d) \(\mathrm{Ti}^{4+}\) is very small ion than \(\mathrm{Ti}^{-3+}\) and does not adsorb any radiation.

The stability constants of the complexes formed by a metal ions \(\left(\mathrm{M}^{2^{+}}\right)\)with \(\mathrm{NH}_{3}, \mathrm{CN}^{-}, \mathrm{H}_{2} \mathrm{O}\) and 'en' are of the order of \(10^{11}, 10^{27}, 10^{15}\) and \(10^{8}\) respectively. Then (en = ethylene diamine) (a) en is the strongest ligand (b) these values cannot predict the strength of the ligand (c) \(\mathrm{CN}^{-}\)is the strongest ligand (d) all ligands are equally strong

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