The molecular formula of the given compound is to be written. The hybridization of each carbon and nitrogen atom is to be determined. The geometry of each carbon and nitrogen atom is to be described. Concept introduction: Mixing of atomic orbitals to form degenerate molecular orbitals is called hybridization. Hybridization state of each atom is to be calculated from the total number of sigma bond pairs and lone pairs of that atom. If there are two electron domains on an atom, then the atom is s p hybridized. s p orbitals are formed by mixing of one s and one p . If there are three electron domains on an atom, then the atom is s p 2 hybridized. s p 2 orbitals are formed by mixing of one s and two p . If there are four electron domains on an atom, then the atom is s p 3 hybridized. s p 3 orbitals are formed by mixing of one s and three p . A single bond has a sigma bond. A double bond contains a sigma bond and a pi bond. A triple bond contains a sigma bond and two pi bonds.

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Chapter 9, Problem 106AP

Interpretation Introduction

Interpretation: The molecular formula of the given compound is to be written. The hybridization of each carbon and nitrogen atom is to be determined. The geometry of each carbon and nitrogen atom is to be described.

Concept introduction:

Mixing of atomic orbitals to form degenerate molecular orbitals is called hybridization.

Hybridization state of each atom is to be calculated from the total number of sigma bond pairs and lone pairs of that atom.

If there are two electron domains on an atom, then the atom is sp hybridized. sp orbitals are formed by mixing of one s and one p.

If there are three electron domains on an atom, then the atom is sp2 hybridized. sp2 orbitals are formed by mixing of one s and two p.

If there are four electron domains on an atom, then the atom is sp3 hybridized. sp3 orbitals are formed by mixing of one s and three p.

A single bond has a sigma bond.

A double bond contains a sigma bond and a pi bond.

A triple bond contains a sigma bond and two pi bonds.

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