Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 10 - Chemical Bonding II - Questions & Problems - Page 457: 10.6

Answer

Bond angle in a square planar structure is $90^{\circ}$ and bond angle in a tetrahedral structure is $109.5^{\circ}$. Therefore, repulsions in a square planar structure is higher than repulsions in a tetrahedral structure. Hence, the tetrahedral $CH_{4}$ molecule is more stable than the square planar $CH_{4}$ molecule.

Work Step by Step

Bond angle in a square planar structure is $90^{\circ}$ and bond angle in a tetrahedral structure is $109.5^{\circ}$. Therefore, repulsions in a square planar structure is higher than repulsions in a tetrahedral structure. Hence, the tetrahedral $CH_{4}$ molecule is more stable than the square planar $CH_{4}$ molecule.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.