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

Which of the following high-spin complexes has the greatest number of unpaired electrons? (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) \((c)\left[\operatorname{re}\left(H_{2} O\right)_{6}\right]^{2+}\) \((b)\left[\operatorname{Mn}\left(H_{2} O\right)_{6}\right]^{2+}\) \((d)\left[N i\left(H_{2} O\right)_{6}\right]^{2+}\)

Short Answer

Expert verified
[Mn(H2O)6]^{2+} has the greatest number of unpaired electrons (5).

Step by step solution

01

Identify the Oxidation State of the Metal

For each complex, determine the oxidation state of the metal. - For \([\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{6}]^{3+}\), the oxidation state of Cr is +3.- For \([\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\), the oxidation state of Fe is +2.- For \([\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\), the oxidation state of Mn is +2.- For \([\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\), the oxidation state of Ni is +2.
02

Determine Electron Configuration of Metal Ions

Find the electron configuration of each metal ion in its oxidation state:- \(\text{Cr}^{3+}\) has an electronic configuration of \([\text{Ar}] 3d^3\).- \(\text{Fe}^{2+}\) has an electronic configuration of \([\text{Ar}] 3d^6\).- \(\text{Mn}^{2+}\) has an electronic configuration of \([\text{Ar}] 3d^5\).- \(\text{Ni}^{2+}\) has an electronic configuration of \([\text{Ar}] 3d^8\).
03

Identify High-Spin Configuration and Count Unpaired Electrons

Determine the high-spin electron configuration and count the unpaired electrons:- \(\text{Cr}^{3+}\): High spin \(3d^3\) means all are unpaired, so 3 unpaired electrons.- \(\text{Fe}^{2+}\): High spin \(3d^6\) means 4 unpaired electrons.- \(\text{Mn}^{2+}\): High spin \(3d^5\) means all are unpaired, so 5 unpaired electrons.- \(\text{Ni}^{2+}\): High spin \(3d^8\) means 2 unpaired electrons.
04

Conclusion

Compare the number of unpaired electrons in each complex: - [Cr(H2O)6]^{3+}: 3 unpaired electrons - [Fe(H2O)6]^{2+}: 4 unpaired electrons - [Mn(H2O)6]^{2+}: 5 unpaired electrons - [Ni(H2O)6]^{2+}: 2 unpaired electrons The complex with the greatest number of unpaired electrons is [Mn(H2O)6]^{2+}.

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.

High-Spin Complexes
High-spin complexes play an important role in coordination chemistry. These complexes occur when ligands surrounding a metal ion induce minimal splitting of its d-orbitals. The ligands' weak field strength means that the energy difference between the d-orbitals—known as crystal field splitting—is small.
This allows electrons to occupy higher energy orbitals rather than pairing up in lower ones, creating unpaired electrons. As a result, high-spin complexes tend to be paramagnetic.
These properties impact the magnetic behavior of the complexes and their reactivity. Common metal ions like \(\text{Fe}^{2+}\), \(\text{Mn}^{2+}\), and others surrounded by weak field ligands typically form high-spin complexes.
  • High-spin complexes exhibit weak field ligands.
  • They display smaller d-orbital splitting, resulting in higher energy orbital occupation.
  • Leading to more unpaired electrons, they exhibit paramagnetism.
Understanding high-spin complexes helps predict the electronic structures that are crucial in various areas of chemical research and application.
Unpaired Electrons
Unpaired electrons are a significant feature in defining the magnetic properties of a coordination complex. In high-spin complexes, the minimized splitting of d-orbitals results in electron configurations where more electrons remain unpaired.
These unpaired electrons contribute to a complex's paramagnetic nature, as they are attracted to magnetic fields. The number of unpaired electrons directly correlates with the level of magnetism—more unpaired electrons mean stronger paramagnetism.
In the selection problem, counting the unpaired electrons helps identify which complex is the most paramagnetic. \(\text{Mn}^{2+}\) in its high-spin state confirms five unpaired electrons, representing maximum paramagnetism among the options.
  • Unpaired electrons affect how compounds interact with magnetic fields.
  • They thus play a vital role in applications involving magnetic materials and devices.
  • More unpaired electrons equate to stronger paramagnetic properties.
Recognizing the importance of unpaired electrons aids in understanding and manipulating material properties.
Electron Configuration
Electron configuration is fundamental to predicting a coordination complex's behavior. It describes the arrangement of electrons within an atom or ion, indicating which d-orbitals the electrons occupy.
Transitional metal ions like \(\text{Cr}^{3+}, \ ext{Fe}^{2+}, \text{Mn}^{2+},\) and \(\text{Ni}^{2+}\) each have specific configurations based on their oxidation states. These configurations tell us the number of unpaired electrons and whether a complex will be high-spin or low-spin.
For example, \(\text{Mn}^{2+}\) exhibits a \(3d^5\) configuration, leading to five unpaired electrons in a high-spin state. Understanding these configurations gives insight into an ion's chemical and physical properties.
  • Electron configurations reflect an ion’s electronic structure in various chemical environments.
  • They determine the magnetic and optical properties of metal complexes.
  • Analyzing electron configurations guides the identification and prediction of high-spin and low-spin states.
Grasping these concepts is crucial for exploring and exploiting chemical systems.
Oxidation States
The oxidation state of a metal in a complex reflects its net charge after accounting for the electrons involved in bonding with ligands. It dictates the electron configuration and overall properties of the metal ion within its coordination environment.
In our exercise, recognizing the correct oxidation states of \(\text{Cr}^{3+}\), \(\text{Fe}^{2+}\), \(\text{Mn}^{2+}\), and \(\text{Ni}^{2+}\) is the initial step. This permits accurate determination of each metal's electron configuration.
Identifying oxidation states also reveals how the metal interacts with ligands, influencing the energy levels and photochemical behaviors of the complex.
  • An oxidation state is crucial for understanding coordination chemistry.
  • Determines electron removal or addition count for defining orbital occupation.
  • It affects the reactivity and stability of coordination complexes.
Mastery of the concept aids comprehensive comprehension of chemical bonding and coordination chemistry.

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

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