/*! 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 42 Consider an octahedral complex, ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider an octahedral complex, \(\mathrm{MA}_{2} \mathrm{~B}_{4}\). How many geometric isomers are expected for this compound? Will any of the isomers be optically active? If so, which ones?

Short Answer

Expert verified
The octahedral complex \(\mathrm{MA}_2\mathrm{B}_4\) has 2 geometric isomers: cis and trans. None of the isomers are optically active.

Step by step solution

01

Understanding the Octahedral Complex

An octahedral complex has six coordination sites around the central metal ion. In this question, the complex has the formula \(\mathrm{MA}_2\mathrm{B}_4\), meaning there are 2 identical ligands of type \(\mathrm{A}\) and 4 identical ligands of type \(\mathrm{B}\).
02

Counting Geometric Isomers

For the \(\mathrm{MA}_2\mathrm{B}_4\) complex, the \(\mathrm{A}\) ligands can either be adjacent (cis) or opposite (trans) to each other. These arrangements lead to two possible geometric isomers: cis isomer and trans isomer.
03

Optically Active Isomer

Optical isomers (enantiomers) occur when a molecule is chiral, typically having no plane of symmetry. The trans isomer of \(\mathrm{MA}_2\mathrm{B}_4\) is not chiral because it has a plane of symmetry through the \(\mathrm{M}\) and \(\mathrm{A}\) ligands. However, the cis isomer can be chiral when each \(\mathrm{B}\) ligand is different from others, but since all \(\mathrm{B}\) ligands are the same here, neither isomer is optically active.

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 91Ó°ÊÓ!

Key Concepts

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

Geometric Isomers
Geometric isomers, also known as cis-trans isomers, are types of stereoisomers. They differ in the spatial arrangement of atoms around a central atom. In octahedral complexes like \(\mathrm{MA}_2\mathrm{B}_4\), the geometric isomers arise when the ligands are positioned differently around the metal center.
For \(\mathrm{MA}_2\mathrm{B}_4\), there are two possible geometric isomers. In the "cis" isomer, the two \(\mathrm{A}\) ligands are adjacent to each other. Conversely, in the "trans" isomer, the two \(\mathrm{A}\) ligands are positioned opposite each other.
This difference in arrangement leads to distinct properties and reactivity patterns in the isomers. Understanding which isomer is present can be crucial in chemical applications, particularly in catalysis and materials science. Geometric isomerism plays an essential role in the chemistry of coordination compounds.
Optical Activity
Optical activity refers to the ability of a compound to rotate the plane of polarized light. This characteristic is often a property of chiral molecules, which lack an internal plane of symmetry.
In the context of the \(\mathrm{MA}_2\mathrm{B}_4\) complex, we need to analyze its potential for optical activity. Generally, octahedral complexes can exhibit chirality if they have a distinct lack of symmetry. However, both the "cis" and "trans" isomers of this complex are not optically active.
  • The trans form has a plane of symmetry passing through the metal and the two identical ligands \(\mathrm{A}\).
  • The cis form lacks chirality because all \(\mathrm{B}\) ligands are identical.
Thus, the isomers of \(\mathrm{MA}_2\mathrm{B}_4\) do not affect polarized light and will not exhibit optical activity under these conditions.
Octahedral Coordination
Octahedral coordination refers to the arrangement of six ligands symmetrically around a central metal ion. It's one of the most common and significant coordination geometries found in complexes.
In octahedral geometry, the six positions are equivalent if identical ligands surround the central metal. The \(\mathrm{MA}_2\mathrm{B}_4\) complex highlights how subtle changes like ligand type and position impact the overall properties. The presence of two different ligand types (\(\mathrm{A}\) and \(\mathrm{B}\)) allows us to explore different isomers.
This coordination environment is crucial in many applications, including catalysis, medicine, and electronics. Recognizing the principles of octahedral coordination helps in predicting compound behavior and reactivity.
Ligand Arrangement
Ligand arrangement in a coordination compound determines the compound's geometry and impacts its chemical and physical properties. Understanding \(\mathrm{MA}_2\mathrm{B}_4\) involves observing how the ligands arrange themselves around the central metal ion.
Ligands can vary in size, charge, and bonding ability, influencing their arrangement. In the discussed complex, four identical ligands \(\mathrm{B}\) and two identical ligands \(\mathrm{A}\) can result in different arrangements like cis and trans isomers.
The correct placement of ligands affects the stability, color, and possible interactions in mixed-ligand systems. In chemistry, precisely identifying ligand arrangements aids in predicting compound behavior, synthesis outcomes, and potential applications in complex systems.

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

When Alfred Werner was developing the field of coordination chemistry, it was argued by some that the optical activity he observed in the chiral complexes he had prepared was due to the presence of carbon atoms in the molecule. To disprove this argument, Werner synthesized a chiral complex of cobalt that had no carbon atoms in it, and he was able to resolve it into its enantiomers. Design a cobalt(III) complex that would be chiral if it could be synthesized and that contains no carbon atoms. (It may not be possible to synthesize the complex you design, but we will not worry about that for now.)

Draw the structure for \(\mathrm{Pt}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\left(\mathrm{NH}_{3}\right)_{2}\) and use it to answer the following questions: (a) What is the coordination number for platinum in this complex? (b) What is the coordination geometry? (c) What is the oxidation state of the platinum? (d) How many unpaired electrons are there? [Sections 23.2 and 23.6\(]\)

The complexes \(\left[\mathrm{CrBr}_{6}\right]^{3-}\) and \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) are both known. (a) Draw the \(d\) -orbital energy-level diagram for octahedral Cr(III) complexes. (b) What gives rise to the colors of these complexes? (c) Which of the two complexes would you expect to absorb light of higher energy?

Consider the tetrahedral anions \(\mathrm{VO}_{4}^{3-}\) (orthovanadate ion), \(\mathrm{CrO}_{4}^{2-}\) (chromate ion), and \(\mathrm{MnO}_{4}^{-}\) (permanganate ion). (a) These anions are isoelectronic. What does this statement mean? (b) Would you expect these anions to exhibit \(d-d\) transitions? Explain. (c) As mentioned in "A Closer Look" on charge-transfer color, the violet color of \(\mathrm{MnO}_{4}^{-}\) is due to a ligand-to-metal charge transfer (LMCT) transition. What is meant by this term? (d) The LMCT transition in \(\mathrm{MnO}_{4}^{-}\) occurs at a wavelength of \(565 \mathrm{nm}\). The \(\mathrm{CrO}_{4}^{2-}\) ion is yellow. Is the wavelength of the LMCT transition for chromate larger or smaller than that for \(\mathrm{MnO}_{4}^{-} ?\) Explain. (e) The \(\mathrm{VO}_{4}^{3-}\) ion is colorless. Do you expect the light absorbed by the LMCT to fall in the UV or the IR region of the electromagnetic spectrum? Explain your reasoning.

Solutions of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (both octahedral), and \(\left[\mathrm{CoCl}_{4}\right]^{2-}\) (tetrahedral) are colored. One is pink, one is blue, and one is yellow. Based on the spectrochemical series and remembering that the energy splitting in tetrahedral complexes is normally much less than that in octahedral ones, assign a color to each complex.

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.