Living by Chemistry
Living by Chemistry
2nd Edition
ISBN: 9781464142314
Author: Angelica M. Stacy
Publisher: W. H. Freeman
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Chapter U2.37, Problem 9E
Interpretation Introduction

Interpretation:

The tetrahedron shape of CH4 molecule and the similarities with NH3 and H2O needs to be explained.

Concept introduction:

The Lewis structure of an organic compound represents the bonding of atoms with lone pairs (if any). It indicates the bonds with atoms and also the arrangement of atoms in a molecule.

Hybridization of an atom indicates the molecular geometry of a molecule. The formula to check the hybridization can be written as:

Hybridization = Number of sigma bonds + Number of lone pair

The molecular geometry of molecules with lone pair on central atom is given by the VSEPR theory.

Expert Solution & Answer
Check Mark

Answer to Problem 9E

All these three molecules have the same type of hybridization and should have tetrahedral geometry but due to the presence of lone pairs on the central atom, the molecular geometry changes to a pyramidal and bent shape.

Living by Chemistry, Chapter U2.37, Problem 9E , additional homework tip  1Living by Chemistry, Chapter U2.37, Problem 9E , additional homework tip  2Living by Chemistry, Chapter U2.37, Problem 9E , additional homework tip  3

Explanation of Solution

For Lewis structure of CH4 molecule, calculate total number of valence electrons:

CH4 molecule = 4 electrons in C + 4 × 1 valence electrons in H = 8 electrons

Living by Chemistry, Chapter U2.37, Problem 9E , additional homework tip  4

Hybridization = Number of sigma bonds + Number of lone pair

Hybridization of C atom = 4 + 0 = 4 = sp3

Thus, with sp3 hybridization, the molecular shape must be tetrahedral and must be fit in a tetrahedron.

For Lewis structure of NH3 molecule, calculate total number of valence electrons:

NH3 molecule = 5 electrons in N + 3 × 1 valence electrons in H = 8 electrons

Living by Chemistry, Chapter U2.37, Problem 9E , additional homework tip  5

Hybridization = Number of sigma bonds + Number of lone pair

Hybridization of N atom = 3 + 1 = 4 = sp3

Thus, with sp3 hybridization, the molecular shape must be tetrahedral but due to the presence of lone pair on the central N atom, the geometry alters to trigonal pyramidal shape.

For Lewis structure of H2O molecule, calculate total number of valence electrons:

H2O molecule = 6 electrons in O + 1 × 2 valence electrons in H = 8 electrons

Living by Chemistry, Chapter U2.37, Problem 9E , additional homework tip  6

Hybridization = Number of sigma bonds + Number of lone pair

Hybridization of O atom = 2 + 2 = 4 = sp3

Thus, with sp3 hybridization, the molecular shape must be tetrahedral but due to the presence of two lone pairs on the central O atom, the geometry alters to a bent shape.

Conclusion

Although all these three molecules have same type of hybridization and should have tetrahedral geometry but due to presence of lone pairs on central atom, the molecular geometry changes to pyramidal and bent shape.

Chapter U2 Solutions

Living by Chemistry

Ch. U2.29 - Prob. 3ECh. U2.29 - Prob. 4ECh. U2.29 - Prob. 5ECh. U2.29 - Prob. 6ECh. U2.29 - Prob. 7ECh. U2.30 - Prob. 1TAICh. U2.30 - Prob. 1ECh. U2.30 - Prob. 2ECh. U2.30 - Prob. 3ECh. U2.30 - Prob. 4ECh. U2.30 - Prob. 5ECh. U2.31 - Prob. 1TAICh. U2.31 - Prob. 1ECh. U2.31 - Prob. 2ECh. U2.31 - Prob. 3ECh. U2.31 - Prob. 4ECh. U2.31 - Prob. 5ECh. U2.31 - Prob. 6ECh. U2.31 - Prob. 7ECh. U2.31 - Prob. 8ECh. U2.32 - Prob. 1TAICh. U2.32 - Prob. 1ECh. U2.32 - Prob. 2ECh. U2.32 - Prob. 3ECh. U2.32 - Prob. 4ECh. U2.32 - Prob. 5ECh. U2.32 - Prob. 6ECh. U2.32 - Prob. 7ECh. U2.32 - Prob. 8ECh. U2.32 - Prob. 9ECh. U2.33 - Prob. 1TAICh. U2.33 - Prob. 1ECh. U2.33 - Prob. 2ECh. U2.33 - Prob. 3ECh. U2.33 - Prob. 4ECh. U2.33 - Prob. 5ECh. U2.33 - Prob. 6ECh. U2.33 - Prob. 7ECh. U2.33 - Prob. 8ECh. U2.33 - Prob. 9ECh. U2.33 - Prob. 10ECh. U2.33 - Prob. 11ECh. U2.34 - Prob. 1TAICh. U2.34 - Prob. 1ECh. U2.34 - Prob. 2ECh. U2.34 - Prob. 3ECh. U2.34 - Prob. 4ECh. U2.34 - Prob. 5ECh. U2.35 - Prob. 1TAICh. U2.35 - Prob. 1ECh. U2.35 - Prob. 2ECh. U2.35 - Prob. 3ECh. U2.35 - Prob. 4ECh. U2.35 - Prob. 5ECh. U2.35 - Prob. 6ECh. U2.36 - Prob. 1TAICh. U2.36 - Prob. 1ECh. U2.36 - Prob. 2ECh. U2.36 - Prob. 3ECh. U2.36 - Prob. 4ECh. U2.36 - Prob. 5ECh. U2.36 - Prob. 6ECh. U2.36 - Prob. 7ECh. U2.37 - Prob. 1TAICh. U2.37 - Prob. 1ECh. U2.37 - Prob. 2ECh. U2.37 - Prob. 3ECh. U2.37 - Prob. 4ECh. U2.37 - Prob. 5ECh. U2.37 - Prob. 6ECh. U2.37 - Prob. 7ECh. U2.37 - Prob. 8ECh. U2.37 - Prob. 9ECh. U2.38 - Prob. 1TAICh. U2.38 - Prob. 1ECh. U2.38 - Prob. 2ECh. U2.38 - Prob. 3ECh. U2.38 - Prob. 4ECh. U2.38 - Prob. 5ECh. U2.38 - Prob. 6ECh. U2.38 - Prob. 7ECh. U2.39 - Prob. 1TAICh. U2.39 - Prob. 1ECh. U2.39 - Prob. 2ECh. U2.39 - Prob. 3ECh. U2.39 - Prob. 4ECh. U2.39 - Prob. 5ECh. U2.39 - Prob. 6ECh. U2.39 - Prob. 7ECh. U2.40 - Prob. 1TAICh. U2.40 - Prob. 1ECh. U2.40 - Prob. 2ECh. U2.40 - Prob. 3ECh. U2.40 - Prob. 4ECh. U2.40 - Prob. 5ECh. U2.41 - Prob. 1TAICh. U2.41 - Prob. 1ECh. U2.41 - Prob. 2ECh. U2.41 - Prob. 3ECh. U2.41 - Prob. 4ECh. U2.41 - Prob. 5ECh. U2.41 - Prob. 6ECh. U2.41 - Prob. 7ECh. U2.41 - Prob. 8ECh. U2.42 - Prob. 1TAICh. U2.42 - Prob. 1ECh. U2.42 - Prob. 2ECh. U2.42 - Prob. 4ECh. U2.42 - Prob. 5ECh. U2.42 - Prob. 6ECh. U2.43 - Prob. 1TAICh. U2.43 - Prob. 1ECh. U2.43 - Prob. 2ECh. U2.43 - Prob. 3ECh. U2.43 - Prob. 4ECh. U2.43 - Prob. 6ECh. U2.43 - Prob. 7ECh. U2.44 - Prob. 1TAICh. U2.44 - Prob. 1ECh. U2.44 - Prob. 2ECh. U2.44 - Prob. 3ECh. U2.44 - Prob. 4ECh. U2.44 - Prob. 5ECh. U2.44 - Prob. 6ECh. U2.44 - Prob. 7ECh. U2.44 - Prob. 8ECh. U2.44 - Prob. 9ECh. U2.45 - Prob. 1TAICh. U2.45 - Prob. 1ECh. U2.45 - Prob. 2ECh. U2.45 - Prob. 3ECh. U2.45 - Prob. 4ECh. U2.45 - Prob. 5ECh. U2.45 - Prob. 6ECh. U2.45 - Prob. 7ECh. U2.46 - Prob. 1TAICh. U2.46 - Prob. 1ECh. U2.46 - Prob. 2ECh. U2.46 - Prob. 3ECh. U2.46 - Prob. 4ECh. U2.46 - Prob. 5ECh. U2.46 - Prob. 6ECh. U2.46 - Prob. 7ECh. U2.46 - Prob. 8ECh. U2.46 - Prob. 9ECh. U2.47 - Prob. 1TAICh. U2.47 - Prob. 1ECh. U2.47 - Prob. 2ECh. U2.47 - Prob. 3ECh. U2.47 - Prob. 4ECh. U2.47 - Prob. 5ECh. U2.47 - Prob. 6ECh. U2.47 - Prob. 7ECh. U2.47 - Prob. 8ECh. U2.47 - Prob. 9ECh. U2.48 - Prob. 1TAICh. U2.48 - Prob. 1ECh. U2.48 - Prob. 2ECh. U2.48 - Prob. 3ECh. U2.48 - Prob. 4ECh. U2.48 - Prob. 6ECh. U2 - Prob. C6.1RECh. U2 - Prob. C6.2RECh. U2 - Prob. C6.3RECh. U2 - Prob. C6.4RECh. U2 - Prob. C6.5RECh. U2 - Prob. C7.1RECh. U2 - Prob. C7.2RECh. U2 - Prob. C7.3RECh. U2 - Prob. C7.4RECh. U2 - Prob. C7.5RECh. U2 - Prob. C8.1RECh. U2 - Prob. C8.2RECh. U2 - Prob. C8.3RECh. U2 - Prob. C8.4RECh. U2 - Prob. C8.5RECh. U2 - Prob. C8.6RECh. U2 - Prob. C9.1ECh. U2 - Prob. C9.2ECh. U2 - Prob. C9.3ECh. U2 - Prob. C9.4ECh. U2 - Prob. C9.5ECh. U2 - Prob. 1RECh. U2 - Prob. 2RECh. U2 - Prob. 3RECh. U2 - Prob. 4RECh. U2 - Prob. 5RECh. U2 - Prob. 6RECh. U2 - Prob. 7RECh. U2 - Prob. 8RECh. U2 - Prob. 1STPCh. U2 - Prob. 2STPCh. U2 - Prob. 3STPCh. U2 - Prob. 4STPCh. U2 - Prob. 5STPCh. U2 - Prob. 6STPCh. U2 - Prob. 7STPCh. U2 - Prob. 8STPCh. U2 - Prob. 9STPCh. U2 - Prob. 10STPCh. U2 - Prob. 11STPCh. U2 - Prob. 12STPCh. U2 - Prob. 13STPCh. U2 - Prob. 14STPCh. U2 - Prob. 15STPCh. U2 - Prob. 16STPCh. U2 - Prob. 17STPCh. U2 - Prob. 18STPCh. U2 - Prob. 19STPCh. U2 - Prob. 20STP
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