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Chapter 8: Problem 89

One type of exception to the octet rule are compounds with central atoms having fewer than eight electrons around them. BeH_ and \(\mathrm{BH}_{3}\) are examples of this type of exception. Draw the Lewis structures for \(\mathrm{BeH}_{2}\) and \(\mathrm{BH}_{3} .\)

Short Answer

Expert verified
The Lewis structures for BeH2 and BH3 are as follows: BeH2: H - Be - H BH3: H | H - B - H In both structures, the central atom (Be in BeH2, and B in BH3) has fewer than 8 electrons around them, thus representing an exception to the octet rule.

Step by step solution

01

Determining Valence Electrons

Determine the valence electrons for the elements involved in the molecules BeH2 and BH3: Be, B, and H. Beryllium (Be): Group 2, 2 valence electrons Boron (B): Group 13, 3 valence electrons Hydrogen (H): Group 1, 1 valence electron
02

Drawing BeH2 Lewis Structure

To draw the Lewis structure for BeH2, first put Be atom (the central atom) in the center and draw two H atoms on either side of Be. 1. Since Be has 2 valence electrons and each H has 1 valence electron, the total number of valence electrons in BeH2 is 4. 2. Use 2 of these 4 valence electrons to form a single bond with each of the two H atoms. 3. Now, all the valence electrons are used up, and we have the Lewis structure for BeH2: H - Be - H
03

Drawing BH3 Lewis Structure

To draw the Lewis structure for BH3, first put B (the central atom) in the center and draw three H atoms around it. 1. B has 3 valence electrons, and each H has 1 valence electron, so the total number of valence electrons in BH3 is 6. 2. Use these 6 valence electrons to form single bonds with each of the three H atoms. 3. Now, all the valence electrons are used up, and we have the Lewis structure for BH3: H | H - B - H In both BeH2 and BH3, the central atom does not obey the octet rule, having only 4 and 6 electrons around them, respectively. These are examples of exceptions to the octet rule.

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