Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 20: Problem 46

Draw all geometrical and linkage isomers of square planar \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{SCN})_{2}\)

Short Answer

Expert verified
There are four isomers for the square planar complex \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{SCN})_{2}\): 1. Cis isomer with both thiocyanate ligands S-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{S}\mathrm{CN})_{2}\) 2. Cis isomer with both thiocyanate ligands N-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{N}\mathrm{CS})_{2}\) 3. Trans isomer with both thiocyanate ligands S-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{S}\mathrm{CN})_{2}\) 4. Trans isomer with both thiocyanate ligands N-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{N}\mathrm{CS})_{2}\)

Step by step solution

01

Identify the central atom and ligands

In this square planar complex, the central atom is Platinum (Pt), and there are two types of ligands: ammine (NH3) and thiocyanate (SCN).
02

Draw the square planar complex

The initial complex consists of Platinum (Pt) surrounded by the ligands in a square planar arrangement. There are two ammine ligands and two thiocyanate ligands.
03

Determine the possible geometrical isomers

Geometrical isomers are formed when the ligands are arranged differently around the central atom. In this case, there are two possibilities: 1. The ammine ligands are adjacent to each other (cis isomer). 2. The ammine ligands are opposite each other (trans isomer).
04

Draw the geometrical isomers

For the square planar complex with the given formula, we have two geometrical isomers: 1. Cis isomer: The ammine ligands are adjacent to each other and the thiocyanate ligands are opposite each other in the square plane. ![cis-isomer](https://i.imgur.com/3zwLUR5.png) 2. Trans isomer: The ammine ligands are opposite to each other and the thiocyanate ligands are also opposite each other in the square plane. ![trans-isomer](https://i.imgur.com/HGH1o2X.png)
05

Determine the possible linkage isomers

Linkage isomers arise from different sites of attachment of ligands. In thiocyanate (SCN), there are two possible sites of attachment: sulfur (S) or nitrogen (N). Thus, the thiocyanate ligand can create two linkage isomers: S-bound or N-bound.
06

Draw the linkage isomers for cis and trans isomers

Here are the possible linkage isomers for the cis and trans geometrical isomers: - Cis isomer: 1. Cis - Both thiocyanate ligands are S-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{S}\mathrm{CN})_{2}\) 2. Cis - Both thiocyanate ligands are N-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{N}\mathrm{CS})_{2}\) - Trans isomer: 1. Trans - Both thiocyanate ligands are S-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{S}\mathrm{CN})_{2}\) 2. Trans - Both thiocyanate ligands are N-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{N}\mathrm{CS})_{2}\) So, there are a total of four isomers for the given square planar complex: two geometrical isomers with each having two linkage isomers.

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

The \(\mathrm{CrF}_{6}^{4-}\) ion is known to have four unpaired electrons. Does the \(\mathrm{F}^{-}\) ligand produce a strong or weak field?

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.

\(\mathrm{CoCl}_{4}^{2-}\) forms a tetrahedral complex ion and \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) forms an octahedral complex ion. What is wrong about the following statements concerning each complex ion and the \(d\) orbital splitting diagrams? a. \(\mathrm{CoCl}_{4}^{2-}\) is an example of a strong-field case having two unpaired electrons. b. Because \(\mathrm{CN}^{-}\) is a weak-field ligand, \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) will be a low-spin case having four unpaired electrons.

Name the following complex ions. a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}\) b. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) c. \(\mathrm{Mn}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{3}^{2+}\) d. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}^{2+}\) a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}\) b. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) c. \(\mathrm{Mn}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{3}^{2+}\) d. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}^{2+}\)

Draw the \(d\) -orbital splitting diagrams for the octahedral complex ions of each of the following. a. \(\mathrm{Zn}^{2+}\) b. \(\mathrm{Co}^{2+}\) (high and low spin) c. \(\mathrm{Ti}^{3+}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

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