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 species has the greatest number of unpaired electrons (a) \(\mathrm{Ge} ;\) (b) \(\mathrm{Cl} ;\) (c) \(\mathrm{Cr}^{3+}\) (d) Br -?

Short Answer

Expert verified
The species with the greatest number of unpaired electrons is \(Cr^{3+}\) as it has 3 unpaired electrons.

Step by step solution

01

Identify Atomic Numbers

The atomic numbers for each species are as follows: Ge (32), Cl (17), Cr (24), Br (35). These atomic numbers are found on the periodic table.
02

Determine the electron configuration

Write down the electron configurations according to the Aufbau principle, Hund's rule, and Pauli's exclusion principle.\n For Ge: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^2\]\n For Cl: \[1s^2 2s^2 2p^6 3s^2 3p^5\]\n For Cr3+: Chromium has a special electron configuration. Its ground-state electron configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\] however in case of Cr3+, three electrons are lost, two from 4s orbital and one from 3d, hence the configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\]\n For Br-: When Bromine gains an electron, become Br-. Its electron configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\]
03

Identify Number of Unpaired Electrons

Now, count the number of unpaired electrons in the last shell for each species. Ge has 2, Cl has 1, Cr3+ has 3 and Br- has 0 unpaired electron.

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.

Unpaired Electrons
Unpaired electrons are electrons that occupy an atomic or molecular orbital on their own, without a partner with opposite spin. These electrons play a significant role in chemical reactivity and magnetism. When examining the electron configurations of elements or ions, look for partially filled orbitals to identify unpaired electrons.
In the given exercise, each species: Ge, Cl, Cr^{3+}, and Br^- has specific unpaired electrons. Ge has two unpaired electrons in its 4p orbital, Cl possesses one unpaired electron in the 3p orbital, Cr^{3+} has three unpaired electrons residing within the 3d orbital, and Br^- has no unpaired electrons as its orbitals are fully paired. Understanding the counting of unpaired electrons helps in predicting an element's magnetic properties, with more unpaired electrons leading to stronger magnetic effects, a concept known as paramagnetism.
Aufbau Principle
The Aufbau Principle is a fundamental guide for constructing the electron configuration of atoms in ground state. Its name is derived from the German word 'Aufbau,' which means 'building up.' According to this principle, electrons occupy orbitals from lowest to highest energy.
Here's how it works:
  • Begin filling orbitals from 1s, then 2s, 2p, 3s, and continue onwards, following the increasing order of energy levels.
  • Each orbital can hold a maximum of two electrons with opposite spins.
For instance, in the complete electron configuration of Ge mentioned in the exercise, each orbital fills according to energy hierarchy starting from 1s up to the 4p orbital. It ensures that at each step, the lowest energy arrangement for the electrons is achieved, which is crucial for describing an atom's ground state.
Hund's Rule
Hund's Rule states that for electrons filling degenerate orbitals, which are orbitals of equal energy, electrons will fill the orbitals singly as much as possible, pairing up only when necessary. This minimizes electron-electron repulsion and results in greater stability.
To apply Hund's Rule:
  • Place one electron in each orbital of the same energy before starting to pair.
  • Ensure that unpaired electrons in degenerate orbitals have parallel spins to reduce repulsive interactions.
An example from the exercise is Cr^{3+}. After removing three electrons, we end up with three unpaired electrons in the 3d orbitals. According to Hund's Rule, these electrons will occupy different d orbitals with their spins aligned, ensuring maximum stability.
Pauli Exclusion Principle
The Pauli Exclusion Principle is a principle stating that no two electrons within an atom can have the same set of four quantum numbers, ensuring that each electron in an atom has its unique "address." The principle plays a critical role in determining electron configurations and understanding atomic structure.
Key aspects of Pauli Exclusion Principle include:
  • Each atomic orbital can hold a maximum of two electrons.
  • The two electrons occupying the same orbital must have opposite spins, usually denoted as \( +\frac{1}{2} \) and \( -\frac{1}{2} \).
In the electron configurations mentioned in the exercise, this principle ensures that within each orbital, especially in multi-electron atoms, electrons avoid having the same set of quantum numbers, leading to the unique distribution seen in the configurations of Ge, Cl, Cr^{3+}, and Br^-.

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

In multielectron atoms many of the periodic trends can be explained in terms of \(Z_{\text {eff }}\) Consider the following statements and discuss whether or not the statement is true or false. (a) Electrons in a \(p\) orbital are more effective than electrons in the \(s\) orbitals in shielding other electrons from the nuclear charge. (b) \(\mathrm{Z}_{\text {eff }}\) for an electron in an \(s\) orbital is lower than that for an electron in a \(p\) orbital in the same shell. (c) \(Z_{\text {eff }}\) is usually less than \(Z.\) (d) Electrons in orbitals having \(\ell=1\) penetrate better than those with \(\ell=2.\) (e) \(\mathrm{Z}_{\text {eff }}\) for the orbitals of the elements \(\mathrm{Na}(3 s)\) \(\mathrm{Mg}(3 s), \mathrm{Al}(3 p), \mathrm{P}(3 p),\) and \(\mathrm{S}(3 p)\) are in the order \(Z_{\text {eff }}(\mathrm{Na})<\mathrm{Z}_{\text {eff }}(\mathrm{Mg})>\mathrm{Z}_{\text {eff }}(\mathrm{Al})<\mathrm{Z}_{\text {eff }}(\mathrm{P})>\mathrm{Z}_{\text {eff }}(\mathrm{S}).\)

For the atom \(^{119} \mathrm{Sn}\), indicate the number of (a) protons in the nucleus; (b) neutrons in the nucleus; (c) \(4 d\) electrons; (d) 3s electrons; (e) 5 \(p\) electrons; (f) electrons in the valence shell.

Explain why the radii of atoms do not simply increase uniformly with increasing atomic number.

Describe how the ionization energies of the ions \(\mathrm{He}^{-}, \mathrm{Li}^{-}, \mathrm{Be}^{-}, \mathrm{B}^{-}, \mathrm{C}^{-}, \mathrm{N}^{-}, \mathrm{O}^{-},\) and \(\mathrm{F}^{-}\) vary with atomic number.

How would you expect the sizes of the hydrogen ion, \(\mathrm{H}^{+},\) and the hydride ion, \(\mathrm{H}^{-},\) to compare with that of the H atom and the He atom? 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