Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

$$ \begin{aligned} &\text { Match the following }\\\ &\begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \text { (a) }\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-} & \text { (p) Octahedral } \\ \text { (b) }\left[\mathrm{MnF}_{6}\right]^{4-} & \text { (q) Paramagnetic } \\\ \text { (c) }\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} & \text { (r) Square planar } \\ \text { (d) }\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} & \text { (s) } \mathrm{d}^{2} \mathrm{sp}^{3} \text { hybridization } \\ & \text { (t) Weak field ligand. } \\ \hline \end{array} \end{aligned} $$

Short Answer

Expert verified
(a) r, (b) t/q, (c) s, (d) p/q

Step by step solution

01

Analyze Complex (a)

Identify the geometry of \([\mathrm{Ni}(\mathrm{CN})_4]^{2-}\). Nickel is in a +2 oxidation state, and CN is a strong field ligand, leading to dsp\(^2\) hybridization, forming a square planar geometry. Hence, \([\mathrm{Ni}(\mathrm{CN})_4]^{2-}\) matches with (r) Square planar.
02

Analyze Complex (b)

Examine \([\mathrm{MnF}_{6}]^{4-}\). Manganese is in a +2 oxidation state with fluoride, a weak field ligand, leading to high spin and maintaining paramagnetic properties. The complex is likely to be octahedral. Thus, \([\mathrm{MnF}_{6}]^{4-}\) matches with (t) Weak field ligand and can also fit into (q) Paramagnetic.
03

Analyze Complex (c)

Consider \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\). Iron in a +3 oxidation state with CN, a strong field ligand, causes low-spin d\(^2\)sp\(^3\) hybridization, forming an octahedral complex, and is diamagnetic due to full pairing. Thus, \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\) matches with (s) d\(^2\)sp\(^3\) hybridization.
04

Analyze Complex (d)

Evaluate \([\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{6}]^{3+}\). Chromium in a +3 oxidation state results in d\(^3\)sp\(^3\) hybridization with water, a weak field ligand. The complex is octahedral and paramagnetic. Therefore, \([\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{6}]^{3+}\) matches with (p) Octahedral and (q) Paramagnetic.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Hybridization
Hybridization is a fundamental concept in coordination chemistry that helps us predict the geometry of coordination compounds. It involves the mixing of atomic orbitals to form new, hybrid orbitals that are involved in bonding. For example, in the case of \([\mathrm{Ni}(\mathrm{CN})_4]^{2-}\), hybridization involves dsp\(^2\) orbitals, resulting in square planar geometry. This is because the CN ligands are strong field ligands that cause pairing of electrons, allowing a low-energy hybridization pathway.

Another complex, \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\), demonstrates d\(^2\)sp\(^3\) hybridization, which is a characteristic of octahedral structures. Strong field ligands like the cyanide ion lead to electron pairing, making such a hybridization favorable. Understanding these hybridization types allows predictions of complex structures and their properties.
Ligand Field Theory
Ligand Field Theory (LFT) is an explanation of how ligands influence the energy levels of a central metal atom's d-orbitals. It explains the color, magnetism, and stability of coordination compounds by considering the impact of ligands on these d-orbitals.

In our examples, \([\mathrm{MnF}_{6}]^{4-}\), involves fluoride ligands, which are considered weak field ligands. This indicates that the d-orbitals remain as unpaired electrons due to minimal splitting, maintaining the paramagnetic property. In contrast, the \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\) complex uses strong field ligands that cause a significant splitting of the d-orbitals, filling all available lower-energy orbitals first, leading to diamagnetic properties.
Paramagnetic and Diamagnetic Properties
The magnetic properties of coordination compounds depend heavily on the electron configuration influenced by the ligands. When we discuss whether a compound is paramagnetic or diamagnetic, we look at the presence or absence of unpaired electrons.
  • Paramagnetic: Compounds with unpaired electrons. These compounds are attracted to magnetic fields. For example, \([\mathrm{MnF}_{6}]^{4-}\) is paramagnetic due to the weak field ligands that allow unpaired electrons.
  • Diamagnetic: Compounds where all electrons are paired. As a result, they are slightly repelled by a magnetic field. The \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\) complex exhibits diamagnetism because all electrons are paired, a result of the strong field cyanide ligands.
Understanding these properties can elucidate the behavior of metal complexes in external magnetic fields and aid in identifying the strength of the ligand fields present in coordination compounds.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The magnetic moment (spin only) of \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) is: \(\quad[\mathbf{2 0 1 1}]\) (a) \(1.41 \mathrm{BM}\) (b) \(5.64 \mathrm{BM}\) (c) \(1.28 \mathrm{BM}\) (d) \(2.82 \mathrm{BM}\)

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).

Which one of the following complexes is diamagnetic in nature? (1) \(\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]\) (2) \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}\) (3) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) (4) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}\) Select the correct answer (a) 1 and 2 (b) 2 and 3 (c) 1,2 and 4 (d) 1,3 and 4

The coordination number of \(\mathrm{Ni}^{2+}\) is 4 . \(\mathrm{NiCl}_{2}+\mathrm{KCN}\) (excess) \(\longrightarrow \mathrm{A}\) (Cyano complex) \(\mathrm{NiCl}_{2}+\) conc. \(\mathrm{HCl}\) (excess) \(\longrightarrow \mathrm{B}\) (chloro complex) The IUPAC name of \(\mathrm{A}\) and \(\mathrm{B}\) are (a) potassiumtetracyanonickelate(II), potassiumtetrachloronickelate (II) (b) tetracyanopotassiumnickelate (II), tetrachloropota-ssiumnickelate(II) (c) tetracyanonickel(II), tetrachloronickel(II) (d) potassium tetracyanonickel(II), potassium tetra-chloronickel(II)

A similarity between optical and geometrical isomerism is that (a) if in a compound one is present then the other will also be present (b) each gives equal number of isomers for a given compound (c) both are included in stereo isomerism (d) they have no similarity

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free