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Chapter 9: Problem 66

In the vapor phase, beryllium chloride consists of discrete \(\mathrm{BeCl}_{2}\) molecules. Is the octet rule satisfied for Be in this compound? If not, can you form an octet around Be by drawing another resonance structure? How plausible is this structure?

Short Answer

Expert verified
No, the octet rule is not satisfied for Be in \(\mathrm{BeCl}_{2}\) compound. Also, no plausible resonance structure can be drawn that forms an octet around Be. Therefore, it remains associated with 4 electrons in this compound.

Step by step solution

01

Identify Valence Electrons

The element Beryllium (Be) is in group 2 of the periodic table, hence it has 2 valence electrons. Chlorine (Cl) is in group 7, hence it has 7 valence electrons. Therefore, in the molecule \(\mathrm{BeCl}_{2}\), Be is surrounded by only four electrons in total, which are shared with two Cl atoms.
02

Check Octet Rule

The octet rule states that atoms are more stable when their valence shell is full, typically with 8 electrons. As we've identified, Be is surrounded by four electrons, not eight, in \(\mathrm{BeCl}_{2}\). Thus, the octet rule is not satisfied for Be in this compound.
03

Suggest Resonance Structure

At this point, one may think of forming an octet around Be by suggesting a possible resonance structure. However, Be can't form an octet given that it only has four electrons to share or donate as it only has 2s orbital and thus can't expand its octet.
04

Determine Plausibility of the Resonance Structure

Considering the fact that Be can't expand its octet, no plausible resonance structure can be drawn that forms an octet around Be. Therefore, the compound \(\mathrm{BeCl}_{2}\) does not satisfy the octet rule in its molecular structure.

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Most popular questions from this chapter

In the Chemistry in Action essay "Just Say NO," nitric oxide is said to be one of about 10 of the smallest stable molecules known. Based on what you have learned in the course so far, write all the diatomic molecules you know, give their names, and show their Lewis structures.

The triiodide ion \(\left(\mathrm{I}_{3}^{-}\right)\) in which the I atoms are arranged in a straight line is stable, but the corresponding \(\mathrm{F}_{3}^{-}\) ion does not exist. Explain.

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