Chapter 3: Problem 83

One type of exception to the octet rule are compounds with central atoms having fewer than eight electrons around them. \(\mathrm{BeH}_{2}\) 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 the given compounds are as follows: For BeH2: Be -- H -- H For BH3: B -- H | H | H Both central atoms exhibit fewer than 8 electrons around them, making them exceptions to the octet rule.

Step by step solution

Count the total number of valence electrons

To determine the valence electrons, we'll count the valence electrons for each atom within the compound. For example, Beryllium (Be) and Boron (B) have 2 and 3 valence electrons respectively, while Hydrogen (H) has 1 valence electron. For BeH2: Be: 2 valence electrons H: 1 valence electron each (2 Hydrogens present) Total valence electrons: \(2 + 2(1) = 4\) For BH3: B: 3 valence electrons H: 1 valence electron each (3 Hydrogens present) Total valence electrons: \(3 + 3(1) = 6\)

Determine the central atom and arrange the surrounding atoms

The central atom in a compound is typically the one with the lowest electronegativity, excluding Hydrogen. For BeH2, the central atom is Be, and for BH3, the central atom is B. Arrange Hydrogen atoms around the central atom.

Distribute the electrons

Now, we will distribute the valence electrons to form bonds and fill the outer shells of the atoms. Hydrogen requires 2 electrons to be stable, while Be and B can accommodate fewer than 8 electrons. For BeH2, we need 4 electrons. Each electron pair will create a bond between Be and one H atom: Be -- H -- H For BH3, we need 6 electrons. Each electron pair will create a bond between B and one H atom: B -- H | H | H

Verify the structure and check for exceptions

We have now drawn the Lewis structures for both compounds. Let's verify whether they are exceptions to the octet rule. The central atom Be has 4 electrons around it, and B has 6 electrons. Both central atoms have fewer than 8 electrons around them, which confirms that these structures are indeed exceptions to the octet rule.

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One type of exception to the octet rule are compounds with central atoms having fewer than eight electrons around them. \(\mathrm{BeH}_{2}\) 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

Step by step solution

Count the total number of valence electrons

Determine the central atom and arrange the surrounding atoms

Distribute the electrons

Verify the structure and check for exceptions

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