Solutions for Hybridization Exercises

Exercises


Exercise 1. Which is a stronger bond, a σ or π bond?
The sigma bond, σ, is stronger.

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Exercise 2. What hybridization is expected for the underlined atoms in the following?

a) CH3

First, we draw the Lewis structure.

Lewis structure of the CH<sub>3</sub> anion

The carbon is sp3 hybridized.

b) CH3SH

First, we draw the Lewis structure.

Lewis structure of CH<sub>3</sub>SH

Both the carbon and sulfur atoms are sp3 hybridized.

c) H2C=O

First, we draw the Lewis structure.

Lewis structure of H<sub>2</sub>CO

The carbon atom is sp2 hybridized.

d) BF3

First, we draw the Lewis structure.

Lewis structure of boron trifluoride

The boron atom is sp2 hybridized.

e) HCP

First, we draw the Lewis structure.

Lewis structure of HCP

The carbon atom is sp hybridized.

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Exercise 3. What is the hybridization of Br in BrO2?

First, we draw the Lewis structure.

Lewis structure of the bromite ion

The hybridization of Br is sp3

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Exercise 4. What is the hybridization of both N atoms in H2N2?

First, we draw the Lewis structure.

Lewis structure of H<sub>2</sub>N<sub>2</sub>

Both nitrogens are sp2 hybridized.

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Exercise 5. Describe the overlap of orbitals for H3C-CN as well as the hybridization of the two carbon atoms.

Draw the Lewis structure.

Lewis structure for H<sub>3</sub>CCN

The first carbon is sp3 hybridized. The second carbon is sp hybridized. Carbon and hydrogen bond via overlap of a hybridized sp3 orbital on carbon with a 1s orbital of hydrogen. The C-C bond is due to the overlap of a hybridized sp3 orbital on the first carbon with an sp hybridized orbital of the second carbon. The carbon and nitrogen bond is due to the overlap of an sp hybridized orbital on the carbon with an sp hybridized orbital on nitrogen.

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Exercise 6. Describe the overlap of orbitals in HCCCl. Use hybridization schemes.

Draw the Lewis structure.

Lewis structure for HCCCl

Both carbons are sp hybridized. The carbon and hydrogen are bonded due to the overlap of a 1s oribtal of hydrogen and a hybridized sp orbital of carbon. The carbon and chlorine bond is due to the overlap of a 3p orbital in chlorine and a hybridized sp orbital of carbon.

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Exercise 7. The following shows how the atoms are connected in the amino acid leucine.

Lewis structure of the amino acid Leucine

Draw a Lewis structure and then determine the hybridization of all non-terminal atoms.

Lewis structure of the amino acid leucine with all non-terminal atoms numbered.

Carbons 1, 2, 3, and 4 are sp3 hybridized. Carbon 5 is sp2 hybridized. The oxygens are sp3 hybridized, and the nitrogen is sp3 hybridized.

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