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

The compound cisplatin, \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\), has been studied extensively as an antitumor agent. The reaction for the synthesis of cisplatin is: $$\mathrm{K}_{2} \mathrm{PtCl}_{4}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{KCl}(a q)$$ Write the electron configuration for platinum ion in cisplatin. Most \(d^{8}\) transition metal ions exhibit square planar geometry. With this and the name in mind, draw the structure of cisplatin.

Short Answer

Expert verified
The electron configuration of the platinum ion in cisplatin is [Xe] 4f14 5d8. Cisplatin has a square planar geometry with the central Pt atom surrounded by two Cl- ions positioned together and two NH3 ligands positioned together on the other side of the square, approximately 90 degrees apart.

Step by step solution

01

Writing the electron configuration of the platinum ion in cisplatin

In the standard state, Platinum (Pt) has an electron configuration of [Xe] 4f14 5d9 6s1. In the reaction given, platinum in \(K_2 PtCl_4\) loses two electrons to form a Pt(II) ion. So the electron configuration of Pt(II) ends up as [Xe] 4f14 5d8 because two of the electrons in the 5d and 6s orbitals are removed forming a d8 configuration.
02

Drawing the structure of cisplatin

Because Pt(II) has a d8 configuration and most d8 transition metals have square planar geometry, cisplatin also exhibits similar geometry. This means that the central atom, Pt in this case, lies in the center with four ligands (NH3 and Cl in this case) attached at 90-degree angles from each other in a plane creating a square. The cis- in the name cisplatin represents that the same types of ligands (uptake Cl-) are beside each other. Therefore, the Cl- ions will be positioned next to each other(approximately 90 degrees apart) and the NH3 ligands will also be positioned together on the other side of the square.

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.

Transition Metal Complex
A transition metal complex is formed when central metal ions, like platinum in cisplatin, are bonded with molecules or ions known as ligands. These ligands donate electron pairs to the central metal, creating a coordination complex. Cisplatin, \( \text{Pt(NH}_3\text{)}_2\text{Cl}_2 \), is an example of a transition metal complex where platinum acts as the central metal ion. It is classified as a coordination compound, because it consists of the platinum ion surrounded by other molecules or ions.

Transition metals are unique because they can form a variety of complex ions with different oxidation states. This versatility comes from their ability to lose different numbers of electrons. In the case of cisplatin, platinum loses electrons to form a +2 charge, denoted as Pt(II). This loss of electrons allows it to bond with ligands such as ammonia (\( \text{NH}_3 \)) and chloride ions \( (\text{Cl}^-) \).

These complexes often have significant chemical properties and peculiar geometries, leading to specific shapes, which are essential for applications in different fields, including medicine, catalysis, and material science.
Electron Configuration
Electron configuration describes the arrangement of electrons in a metal atom or ion. Understanding this arrangement is key to predicting its chemical behavior. For platinum in its standard state, the electron configuration is \([\text{Xe}] \, 4f^{14} \, 5d^9 \, 6s^1\). When platinum loses two electrons to become Pt(II) in cisplatin, the electrons are removed first from the 5d and 6s orbitals.

This results in the electron configuration \([\text{Xe}] \, 4f^{14} \, 5d^8\) for Pt(II). This configuration is significant because it is classified as a \(d^8\) configuration, a common trait in transition metals that are prone to adopting specific geometries. Such information helps in understanding properties such as magnetism and chemical reactivity, integral for designing pharmaceuticals like cisplatin.

It is essential to recognize the influence of electron configuration on molecular geometry. In transition metal complexes, how the d-orbitals are filled can determine things like color, stability, and the shape of the formed compounds.
Square Planar Geometry
The square planar geometry is characteristic of many transition metal complexes with a \(d^8\) electron configuration, including cisplatin. It presents a specific arrangement where the central metal lies at the center of a square plane, and four ligands are positioned at 90-degree angles relative to each other.

In the case of cisplatin, platinum is situated at the center, with two \( \text{NH}_3 \) molecules and two \( \text{Cl}^- \) ions creating a planar square. Notably, the term 'cis' denotes that similar ligands are adjacent. Thus, the chloride ions are positioned next to each other, as are the ammonia molecules. This conformation is not merely structural but is crucial to the biological activity of cisplatin, where spatial arrangement influences its ability to interact and bind with DNA in cancer cells.

The prevalence of square planar geometry in \(d^8\) complexes like cisplatin reflects the underlying electronic structure. The distribution of electrons tends to favor this shape due to lower energy arrangements, leading to stable configurations suited to various chemical reactions and physical properties.

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

How many unpaired electrons are in the following complex ions? a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) (low-spin case) b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) c. \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\)

Write electron configurations for each of the following. a. \(\mathrm{Ti}, \mathrm{Ti}^{2+}, \mathrm{Ti}^{\mathrm{i}+}\) b. \(\operatorname{Re}, \mathrm{Re}^{2+}, \mathrm{Re}^{3+}\) c. \(\mathrm{Ir}, \mathrm{Ir}^{2+}, \mathrm{Ir}^{3+}\)

Consider the following data: $$\begin{aligned} \mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+} & & \mathscr{E}^{\circ}=1.82 \mathrm{~V} \\ \mathrm{Co}^{2+}+3 \mathrm{en} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{2+} & K &=1.5 \times 10^{12} \\ \mathrm{Co}^{3+}+3 \mathrm{en} \longrightarrow \mathrm{Co}(\mathrm{en}){ }^{3+} & K &=2.0 \times 10^{47} \end{aligned}$$ where en \(=\) ethylenediamine. a. Calculate \(\mathscr{E}^{\circ}\) for the half-reaction $$\mathrm{Co}(\mathrm{en})_{3}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{2+}$$ b. Based on your answer to part a, which is the stronger oxidizing agent, \(\mathrm{Co}^{3+}\) or \(\mathrm{Co}(\mathrm{en})_{3}{ }^{3+}\) ? c. Use the crystal field model to rationalize the result in part b.

A certain first-row transition metal ion forms many different colored solutions. When four coordination compounds of this metal, each having the same coordination number, are dissolved in wa ter, the colors of the solutions are red, yellow, green, and blue Further experiments reveal that two of the complex ions are para magnetic with four unpaired electrons and the other two are dia magnetic. What can be deduced from this information about th four coordination compounds?

. Which is more likely to be paramagnetic, \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) or \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) ? Explain.

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