/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 114 Match the list I and II, pick th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 114

Match the list I and II, pick the correct matching from the codes given below List I (a) \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]\) (b) \(\left[\mathrm{Cu}\left(\mathrm{CN}_{4}\right)\right]^{3}\) (c) \(\left[\mathrm{Cu}(\mathrm{CN})_{6}\right]^{3-}\) (d) \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) (e) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{-}\) List II 1\. square planar and \(1.73 \mathrm{BM}\) 2\. linear and zero 3\. octahedral and zero 4\. tetrahedral and zero 5\. octahedral and \(1.73\) BM (a) a-4, b-2, c-5, d-3, e-1 (b) a-4, b-5, c-2, d-1, e-3 (c) a-2, b-4, c-5, d-1, e-3 (d) a-5, b-4, c-1, d-3, e-2

Short Answer

Expert verified
Match (c): a-2, b-4, c-5, d-1, e-3.

Step by step solution

01

Analyzing the Structure - List I(a)

Analyze \([\text{Ag(CN)}_2]\)\. Silver (Ag) with two cyanide ions (CN\(-\)) forms a linear complex. Linear complexes typically suggest no geometric shift that would allow for unpaired electrons, thus zero magnetic moment.
02

Identifying the Structure - List I(b)

Analyze \([\text{Cu(CN)}_4]^{3}\). Copper (Cu) can form both square planar or tetrahedral structures. With four ligands, the more stable configuration is often square planar, typically having a magnetic moment of around 1.73 BM due to partial unpaired electron pairing.
03

Determining Geometry - List I(c)

Analyze \([\text{Cu(CN)}_6]^{3-}\). With six ligands, this complex tends to form an octahedral structure. The absence of unpaired electrons indicates a zero magnetic moment.
04

Examining Structure - List I(d)

Analyze \([\text{Cu(NH}_3)_4]^{2+}\). With four NH\(\_3\) ligands, this is likely to be a square planar complex with 1.73 BM due to the potential for some unpaired electron activity.
05

Analyzing Geometry - List I(e)

Analyze \([\text{Fe(CN)}_6]^{-}\). This typically forms an octahedral structure. Being a low-spin complex often leads to zero magnetic moment due to electron pairing.
06

Matching with List II

Using the previous evaluations: (a) Linear and zero - matches 2. (b) Square planar and 1.73 BM - matches 1. (c) Octahedral and zero - matches 3. (d) Square planar and 1.73 BM - matches 1. (e) Octahedral and zero - matches 3.

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 91影视!

Key Concepts

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

Complex Geometry
In coordination chemistry, the geometry of a complex plays a crucial role in determining its properties, such as magnetism and color. Different metal ions and ligands interact to form specific shapes that minimize energy. For instance:
  • Linear geometry: Occurs when two ligands are present, as seen in \([\text{Ag(CN)}_2]\). Here, the silver and cyanide ions form a straight line, leading to no geometric spatial effect that would generally allow unpaired electrons. As a result, these complexes usually have a zero magnetic moment.
  • Square planar geometry: This is common for complexes with four ligands, such as \([\text{Cu(NH}_3)_4]^{2+}\), where the ligands are located at the corners of a square around the metal. The electron arrangements in this configuration can lead to partial unpaired electrons, contributing to magnetic moments like 1.73 BM.
  • Tetrahedral geometry: Like square planar, tetrahedral complexes contain four ligands but differ as they form a triangular pyramid. This can sometimes result with zero magnetic moment when electrons pair thoroughly.
  • Octahedral geometry: With six ligands, these complexes, such as \([\text{Fe(CN)}_6]^{-}\), offer room for various arrangements that generally promote electron pairing, leading to no magnetic moment.
Understanding the geometry helps explain how electrons distribute, impacting the magnetic properties and color of complexes.
Magnetic Moments
A magnetic moment signifies the extent to which a complex might interact with a magnetic field. It is affected by the electronic structure, the geometry of the complex, and the nature of the ligands. A complex's magnetic moment is often expressed in bohr magnetons (BM). A value of 1.73 BM suggests that some unpaired electrons exist.
  • Zero magnetic moment: Common in environments supporting electron pairing, such as octahedral complexes, which accommodate six ligands around the central metal, as seen in \([\text{Cu(CN)}_6]^{3-}\). Efficient pairing leads to no unpaired electrons, hence zero magnetic moment.
  • Non-zero magnetic moment: Complexes like the square planar \([\text{Cu(NH}_3)_4]^{2+}\) exhibit magnetic moments, evidenced by 1.73 BM. Here, unpaired electrons remain due to how electronic configuration and ligand-field splitting create inequivalent d-orbitals.
Understanding magnetic moments helps in predicting the behavior of complexes in magnetic fields and explaining phenomena such as color and reactivity.
Ligands
Ligands are ions or molecules that donates at least one pair of electrons to a metal center to form a coordination complex. Common ligands include atmospheric molecules like ammonia (NH\(_3\)) and cyanide (CN\(^-\)). Ligands influence the geometry and electronic distribution within a complex.
  • Cyanide (CN\(^-\)): A strong field ligand producing considerable field splitting, often leading to low-spin complexes, which results in electron pairing. This is illustrated in complexes like \([\text{Fe(CN)}_6]^{-}\) that have an octahedral shape and zero magnetic moment.
  • Ammonia (NH\(_3\)): A moderate field ligand, as seen in \([\text{Cu(NH}_3)_4]^{2+}\) which forms square planar geometry and may show non-zero magnetic moments due to less field splitting allowing some unpaired electrons.
Ligands dictate much of the coordination chemistry by determining the nature of electronic transitions and splitting patterns, crucial to the behavior of the complex.
Electronic Structure
Electronic structure, crucial in coordination chemistry, highlights how electrons arrange in an atom or complex's orbitals. Configuration depends greatly on factors like the central metal's properties and the presence of specific ligands, influencing the geometry and magnetic characteristics.
  • d-Orbital splitting: In complexes, ligands affect the energy levels of the d-orbitals. The extent of this splitting is influenced by the ligand's field strength, altering the electrons' distribution. For instance, strong field ligands like CN\(^-\) produce substantial splitting, leading to paired electrons often forming low-spin complexes such as \([\text{Fe(CN)}_6]^{-}\) with zero magnetic moment.
  • High-spin vs Low-spin: When ligands cause less d-orbital splitting, resulting complexes may be high-spin, having a larger number of unpaired electrons. Conversely, significant splitting leads to low-spin states, favoring paired electrons. Whether a complex is high- or low-spin vastly influences its color, reactivity, and magnetism.
Comprehending electronic structures helps predict the magnetic and optical properties of the complex, enhancing our understanding of their potential applications in catalytic and material sciences.

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 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

While \(\mathrm{Ti}^{3+}, \mathrm{V}^{3+}, \mathrm{Fe}^{3+}\) and \(\mathrm{Co}^{2+}\) can afford a large number 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) \(\mathrm{Cr}^{3^{3}}\) 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)

Which one of the following has an optical isomer? [2010] (a) \(\left[\mathrm{Zn}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2}\right]^{2+}\) (b) \(\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\) (c) \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}(\mathrm{en})\right]^{3+}\) (d) \(\left[\mathrm{Zn}(\mathrm{en})_{2}\right]^{2+}\) \((\mathrm{en}=\) ethylenediamine \()\)

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 concentrated HCl is added to a solution of \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2^{+}}\)ion, an intense blue colour develops due to the formation of which one of the following? (a) \(\left[\mathrm{CoCl}_{4}\right]^{2}\) (b) \(\left[\mathrm{CoCl}_{6}\right]^{-}\) (c) \(\left[\mathrm{CoCl}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}\right]^{+}\) (d) \(\left[\mathrm{CoCl}_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]\)
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation
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.