Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 20: Problem 48

Draw all geometrical isomers of \(\mathrm{Pt}(\mathrm{CN})_{2} \mathrm{Br}_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} .\) Which of these isomers has an optical isomer? Draw the various optical isomers.

Short Answer

Expert verified
The complex \(\mathrm{Pt}(\mathrm{CN})_{2}\mathrm{Br}_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\) has an octahedral geometry with three different ligands and can form two geometrical isomers: fac-isomer and mer-isomer. The fac-isomer has optical activity, as it has an enantiomeric pair of optical isomers with ligands in clockwise and counterclockwise sequences, whereas the mer-isomer does not possess optical isomers.

Step by step solution

01

Identifying the coordination geometry

As the platinum atom has a coordination number of 6, the coordination geometry is octahedral. In an octahedral geometry, the central atom is located at the center of an octahedron, with the six ligands surrounding it at equal distances and angles.
02

Arranging the ligands into geometrical isomers

For the given complex, we have three different ligands: CN, Br, and H2O. As there are six potential positions around the platinum atom for these ligands, we can draw different geometrical isomers with this arrangement. We look for fac (facial) and mer (meridian) isomers in the complex: 1. fac-isomer: Three different ligands (CN, Br, H2O) occupy adjacent positions on the same face of the octahedron, while the other face has similar ligands in the same arrangement. 2. mer-isomer: The three different ligands (CN, Br, H2O) occupy three alternate positions, creating a meridian on the octahedron.
03

Identifying optical isomers

For the given complex, the fac-isomer is the only one that displays optical activity, as it has a non-superimposable mirror image. The mer-isomer does not have a mirror image that is non-superimposable, so it does not possess optical isomers.
04

Drawing optical isomers of the fac-isomer

The fac-isomer has an enantiomeric pair as its optical isomers. These isomers are non-superimposable mirror images of each other: - In the first enantiomer, we can label the positions of the ligands as [CN]-[Br]-[H2O] in a clockwise arrangement on the front face and in the opposite sense on the back face. - In the second enantiomer (the mirror image of the first), the arrangement of ligands will be in a counterclockwise sequence on the front face and in a clockwise sequence on the back face. The resulting optical isomers are the enantiomeric pair of the fac-isomer, distinguished by their arrangement of ligands in a clockwise or counterclockwise sequence.

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.

Start your free trial

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

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

One of the classic methods for the determination of the manganese content in steel involves converting all the manganese to the deeply colored permanganate ion and then measuring the absorption of light. The steel is first dissolved in nitric acid, producing the manganese(II) ion and nitrogen dioxide gas. This solution is then reacted with an acidic solution containing periodate ion; the products are the permanganate and iodate ions. Write balanced chemical equations for both of these steps.

When concentrated hydrochloric acid is added to a red solution containing the \(\operatorname{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) complex ion, the solution turns blue as the tetrahedral \(\mathrm{CoCl}_{4}^{2-}\) complex ion forms. Explain this color change.

The complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) is paramagnetic with one unpaired electron. The complex ion \(\mathrm{Fe}(\mathrm{SCN})_{6}^{3-}\) has five unpaired electrons. Where does SCN \(^{-}\) lie in the spectrochemical series relative to \(\mathrm{CN}^{-} ?\)

Which of the following crystal field diagram(s) is(are) correct for the complex given? a. \(\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) (tetrahedral) b. \(\mathrm{Mn}(\mathrm{CN})_{6}^{3-}\) (strong field) c. \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) (square planar, diamagnetic)

A coordination compound of cobalt(III) contains four ammonia molecules, one sulfate ion, and one chloride ion. Addition of aqueous \(\mathrm{BaCl}_{2}\) solution to an aqueous solution of the compound gives no precipitate. Addition of aqueous \(\mathrm{AgNO}_{3}\) to an aqueous solution of the compound produces a white precipitate. Propose a structure for this coordination compound.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

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.

Sign-up for free