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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. 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
None of the statements are completely correct.

Step by step solution

01

Analyzing Option (a)

The compound \([\mathrm{Ni}( ext{CN})_4]^{2-}\) has Ni in +2 oxidation state, implying 3d\(^8\) electron configuration. The CN\(^-\) ligand is strong field, causing pairing of electrons. This results in a dsp\(^2\) hybridization and thus a square planar geometry. Since all electrons pair, it is diamagnetic, not paramagnetic. So, statement (a) is incorrect.
02

Analyzing Option (b)

The compound \([\mathrm{NiCl}_4]^{2-}\) has Ni in +2 oxidation state, thus in a 3d\(^8\) electron configuration. The Cl\(^-\) ligand is weak field, leading to sp\(^3\) hybridization and a tetrahedral geometry. Because there are unpaired electrons, the complex is paramagnetic, not square planar. Thus, statement (b) is incorrect.
03

Analyzing Option (c)

The compound \([\mathrm{Ni}( ext{CO})_4]\) is zero oxidation state for Ni, indicating a 3d\(^10\) electron configuration. CO is a strong-field ligand, causing pairing of electrons, leading to dsp\(^2\) hybridization, which is square planar. All electrons are paired, making it diamagnetic, not paramagnetic. Therefore, statement (c) is incorrect.
04

Analyzing Option (d)

The compound \([\mathrm{Cu}( ext{CN})_4]^{3-}\) involves Cu in the +1 oxidation state, resulting in a 3d\(^10\) electron configuration. CN\(^-\) is a strong-field ligand, causing dsp\(^2\) hybridization if significant, potentially leading it to be square planar rather than strictly tetrahedral. Regardless, with d\(^10\) configuration, it is indeed diamagnetic. However, tetrahedral geometry is unlikely due to CN\(^-\)'s nature, making the geometry part likely inaccurate. Thus, statement (d) is partly incorrect.

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Key Concepts

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

Hybridisation Theory
Hybridisation theory is a key concept in coordination chemistry that helps us understand the geometry of complexes. In this context, hybridisation refers to the mixing of atomic orbitals in a central metal atom to form new hybrid orbitals with different shapes and energy levels. These hybrid orbitals are what allow the central metal to bond with surrounding ligands.

For coordination complexes:
  • dsp\(^2\) hybridisation is typically associated with square planar geometry. This type of hybridisation occurs when inner shell d orbitals mix with an s and two p orbitals.
  • sp\(^3\) hybridisation generally results in tetrahedral geometry. It involves the mixing of one s orbital with three p orbitals.
The specific hybridisation depends on the nature of the ligands (whether strong or weak field) and the oxidation state/electronic configuration of the central metal.

For example, in the \([\text{Ni}^0(\text{CO})_4]\), carbon monoxide is a strong-field ligand, leading the nickel to undergo dsp\(^2\) hybridisation, achieving a square planar geometry with paired electrons, making it diamagnetic.
Crystal Field Theory
Crystal Field Theory (CFT) provides insights into the color and magnetic properties of coordination complexes. It explains how the metal ion and the surrounding ligands affect each other's energy by altering the degenerate nature of the d orbitals of the metal ion.

In CFT:
  • Strong-field ligands cause a large splitting of d orbials, often leading to electron pairing. This can result in low-spin complexes that are often diamagnetic.
  • Weak-field ligands cause a small splitting, allowing d-orbitals to remain unpaired, leading to high-spin configurations that are paramagnetic.
The color of coordination compounds arises because electrons can absorb specific wavelengths of light to move between these split d orbitals. For example, \([\text{NiCl}_4]^{2-}\) contains chloride, a weak-field ligand, resulting in tetrahedral geometry. The d orbitals are less split, causing the complex to be paramagnetic with unpaired electrons. That's why it's not square planar, contrary to what one might assume if thinking from a hybridisation-only lens.
Stereochemistry in Complexes
Stereochemistry deals with the spatial arrangement of ligands around the central metal atom in a complex. Understanding this arrangement is crucial since it influences the complex's physical and chemical properties. Most coordination compounds exhibit two common geometries: tetrahedral and square planar.

- Tetrahedral geometry: Involves four ligands symmetrically arranged around the central atom. This geometry is common for metals that form sp\(^3\)-hybridized bonds with ligands like Cl\(^-\).

- Square planar geometry: Seen in complexes where dsp\(^2\) hybridisation occurs. Such arrangements are stable particularly in complexes containing strong-field ligands, making planar complexes override tetrahedral configurations, as seen in \([\text{Ni}(\text{CN})_4]^{2-}\).
Paramagnetism and Diamagnetism
Paramagnetism and diamagnetism relate to the magnetic properties of a coordination complex, determined by the presence or absence of unpaired electrons.

- Paramagnetic substances have one or more unpaired electrons, which makes them attracted to magnetic fields. These are typical of complexes with weak-field ligands, as evaluated in the case of \([\text{NiCl}_4]^{2-}\).

- Diamagnetic substances have all paired electrons and are weakly repelled by a magnetic field. This occurs in low-spin complexes or when all the electrons are paired due to strong-field ligands, such as \([\text{Ni}(\text{CO})_4]\).

The magnetic property of a complex can be a practical indicator of whether its geometry follows square planar or tetrahedral coordination.
Oxidation States in Coordination Chemistry
An oxidation state in a coordination complex represents the charge left on the central metal ion when all ligands are removed. It is crucial for predicting the chemical reactivity and electronic configuration of the metal.

The electron configuration of the metal ion, as influenced by its oxidation state, dictates key properties of the complex:
  • Electrons are removed or added to fill the valence d-orbitals accordingly.
  • The oxidation state guides the selection of potential hybridisation and geometry, as seen in the plus two ( text{Ni}^{2+}) or zero ( text{Ni}^{0}) oxidation states affecting \([\text{Ni}(...)_x]\).
For example, the oxidation state of nickel notably influences its hybridisation in \([\text{Ni}(\text{CO})_4]\), allowing it to undergo dsp\(^2\) hybridisation, leading to a square planar complex rather than a different geometry.

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Most popular questions from this chapter

In which of the following pairs, the EAN of the central metal atom is not the same? (a) \(\left[\mathrm{FeF}_{6}\right]^{3+}\) and \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) (b) \(\left[\mathrm{Fe}\left(\mathrm{CN}_{6}\right)\right]^{3}\) and \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}\) (c) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) and \(\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-}\) (d) \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2}\)

The complex salt having the molecular composition \(\left[\mathrm{Co}\left(\mathrm{NO}_{2}\right)(\mathrm{SCN})(\mathrm{en})_{2}\right] \mathrm{Br}\) exhibits (a) linkage isomerism only (b) ionization isomerism only (c) cis-trans isomerism only (d) all of these

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 \(\mathrm{t}_{2 \mathrm{~g}} \rightleftharpoons \mathrm{e}_{\mathrm{g}}\) are responsible for colours of the co-ordination compounds Which of the following colour is not due to d-d transition of (a) Yellow colour of CdS. (b) Red colour of blood (c) Orange colour of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) in acidic medium. (d) Both (a) and (c).

The coordination number of a central metal atom in a complex is determined by (a) the number of only anionic ligands bonded to the metal ion (b) the number of ligands around a metal ion bonded pi-bonds (c) the number of ligands around a metal ion bonded by sigma and pi-bonds (d) the number of ligands around a metal ion bonded by sigma bonds

The pair having same magnetic movement is [2016] (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{2+}\) and \(\left.\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (c) \(\left[\mathrm{CoCl}_{4}\right]^{2}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (d) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{CoCl}_{4}\right]^{2}\)

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