Beryllium shows some covalent characteristics in some of its compounds, unlike the other alkaline earth halides. Give a possible explanation for this phenomenon.

Short Answer

Expert verified
The covalent characteristics of some beryllium compounds, unlike other alkaline earth halides, can be attributed to its small atomic radius. This results in a strong attraction between the nucleus and its valence electrons, making it difficult for beryllium to lose its outer electrons and form ionic compounds like other alkaline earth halides. Instead, beryllium tends to share electrons with other atoms, forming covalent bonds in its compounds.

Step by step solution

01

Understanding Beryllium's atomic structure

Beryllium is an element in Group 2 of the periodic table, with an atomic number of 4. It has a total of 4 electrons occupying its electron shells, with 2 electrons present in its outer shell. This means it needs to lose only two electrons to achieve a stable noble gas configuration. Its electronic configuration is 1s²2s².
02

Explaining the covalent character of Beryllium halides

In contrast to the other elements in Group 2, Beryllium has a relatively small atomic radius and its outer electrons are closer to the nucleus. This results in a strong attraction between the nucleus and the outer electrons, making it difficult for Beryllium to readily lose its valence electrons and form ionic compounds. As a result, Beryllium tends to share electrons with other atoms, thereby forming covalent bonds in its compounds, such as beryllium fluoride (BeF₂) and beryllium chloride (BeCl₂).
03

Comparing Beryllium with other alkaline earth halides

Other alkaline earth halides, such as magnesium chloride (MgCl₂) and calcium chloride (CaCl₂), have relatively larger atomic radii compared to beryllium. This leads to weaker attraction between the nucleus and the valence electrons, allowing these elements to easily lose their outer electrons and form ionic compounds. Thus, alkaline earth halides tend to exhibit ionic characteristics, unlike beryllium halides. In conclusion, the covalent characteristics observed in some beryllium compounds can be attributed to its small atomic radius, which leads to a strong attraction between its nucleus and valence electrons, making it challenging for beryllium to lose its outer electrons and form ionic compounds like other alkaline earth halides.

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

Lead forms compounds in the \(+2\) and \(+4\) oxidation states. All lead(II) halides are known (and are known to be ionic). Only \(\mathrm{PbF}_{4}\) and \(\mathrm{PbCl}_{4}\) are known among the possible lead(IV) halides. Presumably lead(IV) oxidizes bromide and iodide ions, producing the lead(II) halide and the free halogen: Suppose \(25.00 \mathrm{~g}\) of a lead(IV) halide reacts to form \(16.12 \mathrm{~g}\) of a lead(II) halide and the free halogen. Identify the halogen.

The Group 5 A elements can form molecules or ions that involve three, five, or six covalent bonds; \(\mathrm{NH}_{3}, \mathrm{AsCl}_{5}\), and \(\mathrm{PF}_{6}^{-}\) are examples. Draw the Lewis structure for each of these substances, and predict the molecular structure and hybridization for each. Why doesn't \(\mathrm{NF}_{5}\) or \(\mathrm{NCl}_{6}^{-}\) form?

Write the Lewis structure for \(\mathrm{O}_{2} \mathrm{~F}_{2}\). Predict the bond angles and hybridization of the two central oxygen atoms. Assign oxidation states and formal charges to the atoms in \(\mathrm{O}_{2} \mathrm{~F}_{2}\). The compound \(\mathrm{O}_{2} \mathrm{~F}_{2}\) is a vigorous and potent oxidizing and fluorinating agent. Are oxidation states or formal charges more useful in accounting for these properties of \(\mathrm{O}_{2} \mathrm{~F}_{2}\) ?

You travel to a distant, cold planet where the ammonia flows like water. In fact, the inhabitants of this planet use ammonia (an abundant liquid on their planet) much as earthlings use water. Ammonia is also similar to water in that it is amphoteric and undergoes autoionization. The \(K\) value for the autoionization of ammonia is \(1.8 \times 10^{-12}\) at the standard temperature of the planet. What is the \(\mathrm{pH}\) of ammonia at this temperature?

In large doses, selenium is toxic. However, in moderate intake, selenium is a physiologically important element. How is selenium physiologically important?

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