/*! 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 60 The complex ion \(\mathrm{PdCl}_... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 60

The complex ion \(\mathrm{PdCl}_{4}{ }^{2-}\) is diamagnetic. Propose a structure for \(\mathrm{PdCl}_{4}^{2-}\)

Short Answer

Expert verified
The proposed structure for the \(\mathrm{PdCl}_{4}^{2-}\) complex ion is square-planar with the Pd(II) ion as the central atom and four Cl^- ions as ligands. This structure is diamagnetic with no unpaired electrons present.

Step by step solution

01

Identify the oxidation state

We have \(\mathrm{PdCl}_{4}^{2-}\) as the complex ion. Each chlorine atom has a -1 charge, and there are four chlorine atoms. Therefore, the central Pd ion has a charge of +2. So, Pd is in the oxidation state of +2.
02

Write the electron configuration of Pd and Pd(II)

Write the electron configuration of the central atom, Pd (atomic number 46): Pd: \([Kr] 4d^{10} 5s^0\) Now, consider the Pd(II) configuration after losing two electrons: Pd(II): \([Kr] 4d^8\)
03

Determine the coordination number and geometry

The coordination number is the number of ligands (Cl^- in this case) attached to the central atom (Pd). Since we have four Cl^- ions acting as ligands, the coordination number is 4. For a coordination number of 4, there are two possible geometries: tetrahedral and square planar.
04

Determine whether the complex is tetrahedral or square planar

To determine the geometry for the \(\mathrm{Pd^{2+}}\) ion, we need to check if the ion follows the 18-electron rule. For \(\mathrm{Pd^{2+}}\), we have: \([Kr] 4d^8\) By adding the four lone pairs from the Cl^- ions, we get the total number of valence electrons as follows: \(8 + 4(2) = 16\) Since the 18-electron rule is not satisfied, it is unlikely to form a tetrahedral complex. Therefore, the geometry is square-planar in which, the Pd(II) ion has a filled d-orbital, resulting in a diamagnetic complex. Conclusion:
05

Proposed structure

The proposed structure for the \(\mathrm{PdCl}_{4}^{2-}\) complex ion is square-planar with the Pd(II) ion as the central atom and four Cl^- ions as ligands. This structure is diamagnetic with no unpaired electrons present.

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!

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

What is the lanthanide contraction? How does the lanthanide contraction affect the properties of the \(4 d\) and \(5 d\) transition metals?

Use standard reduction potentials to calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) for the reaction that is used in production of gold: $$2 \mathrm{Au}(\mathrm{CN})_{2}^{-}(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q)$$ The relevant half-reactions are $$\begin{aligned}\mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ} &=-0.60 \mathrm{~V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{E}^{\circ} &=-1.26 \mathrm{~V} \end{aligned}$$

The carbonate ion \(\left(\mathrm{CO}_{3}^{2-}\right)\) can act as either a monodentate or a bidentate ligand. Draw a picture of \(\mathrm{CO}_{3}^{2-}\) coordinating to a metal ion as a monodentate and as a bidentate ligand. The carbonate ion can also act as a bridge between two metal ions. Draw a picture of a \(\mathrm{CO}_{3}^{2-}\) ion bridging between two metal ions.

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

Name the following complex ions. a. \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) c. \(\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{5}{ }^{3-}\) b. \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}^{+}\) d. \(\mathrm{Co}(\mathrm{SCN})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}^{+}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

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.