Chapter U2.10, Problem 6E

Interpretation Introduction

Interpretation: The three molecules which have bent shape needs to be listed.

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 arrangement of atoms in molecule.

Hybridization of any atom indicates the molecular geometry of 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 VSEPR theory.

Answer to Problem 6E

Explanation of Solution

For Lewis structure of H₂S molecule, calculate total number of valence electrons:

H₂S molecule = 6 electrons in S + 1 x 2 valence electrons in H = 8 electrons

Hybridization = Number of sigma bonds + Number of lone pair

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

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

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

Hybridization = Number of sigma bonds + Number of lone pair

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

For Lewis structure of H₂Se molecule, calculate total number of valence electrons:

H₂Se molecule = 6 electrons in O+ 1 x 2 valence electrons in H = 8 electrons

Hybridization = Number of sigma bonds + Number of lone pair

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

Thus, with sp³ hybridization, the molecular shape must be tetrahedral but due to presence of two lone pair on central atom, the geometry must be bent shape with bond angle less than 109°.

Conclusion

The presence of lone pairs on central atom changes the molecular geometry of molecules.