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 central palladium atom surrounded by four chloride ions at the corners of a square. This structure allows all electrons to be paired, resulting in a diamagnetic complex.

Step by step solution

01

Electron Configuration of Palladium and Chlorine

Determine the electron configuration of palladium (Pd) and chlorine (Cl). The electron configuration of Pd is [Kr]4d\(^{10}\)5s\(^0\), and for Cl, it is [Ne]3s\(^2\)3p\(^5\). Since we have a 2- charge on the complex ion, that means palladium will donate two electrons to form bonds with the four chlorines.
02

Formation of Coordination Bonds

The Pd atom will form coordination bonds with four Cl atoms, with each chlorine donating one electron to form the bond. Since each Cl has one unpaired electron in its 3p orbital, the Pd atom can accommodate the electron from each chlorine, making a total of four coordination bonds within the complex.
03

Structure and Geometry of the Complex

The \(\mathrm{PdCl}_{4}^{2-}\) complex has a central palladium atom surrounded by four chloride ions. Considering that the square planar geometry is common for complexes with a coordination number of 4, it is most likely that the structure of the \(\mathrm{PdCl}_{4}^{2-}\) complex will also be square planar. In this geometry, all four chloride ions are positioned in the same plane as the central palladium atom, with each chloride ion at the corners of a square and the palladium atom at the center.
04

Confirmation of Diamagnetism

in the proposed square planar structure, all electrons are paired up in the \(\mathrm{PdCl}_{4}^{2-}\) complex. Palladium donates two electrons, one to each of two chlorines. Now all chlorines have a complete 3p orbital. Since there are no unpaired electrons, the complex is diamagnetic, which matches the given information. In conclusion, the proposed structure for the \(\mathrm{PdCl}_{4}^{2-}\) complex ion is square planar with all electrons paired, resulting in a diamagnetic complex.

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!

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

Oxalic acid is often used to remove rust stains. What properties of oxalic acid allow it to do this?

Draw all geometrical and linkage isomers of \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{NO}_{2}\right)_{2}\).

Which of the following statement(s) is(are) true? a. The coordination number of a metal ion in an octahedral complex ion is 8 . b. All tetrahedral complex ions are low-spin. c. The formula for triaquatriamminechromium(III) sulfate is \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}\left(\mathrm{NH}_{3}\right)_{3}\right]_{2}\left(\mathrm{SO}_{4}\right)_{3}\) d. The electron configuration of \(\mathrm{Hf}^{2+}\) is \([\mathrm{Xe}] 4 f^{12} 6 s^{2}\). e. Hemoglobin contains \(\mathrm{Fe}^{3+}\).

Carbon monoxide is toxic because it binds more strongly to iron in hemoglobin (Hb) than does \(\mathrm{O}_{2}\). Consider the following reactions and approximate standard free energy changes: $$ \begin{aligned} \mathrm{Hb}+\mathrm{O}_{2} & \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\circ} &=-70 \mathrm{~kJ} \\ \mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO} & \Delta G^{\circ} &=-80 \mathrm{~kJ} \end{aligned} $$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$ \mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2} $$

Rank the following complex ions in order of increasing wavelength of light absorbed. $$ \mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}, \mathrm{Co}(\mathrm{CN})_{6}^{3-}, \mathrm{CoI}_{6}^{3-}, \mathrm{Co}(\mathrm{en})_{3}^{3+} $$

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