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.

Question

Want to see the full answer?

Answer 1

Question

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

1) a) Give the dominant Intermolecular Force (IMF) in a sample of each of the following compounds. Please show your work. (8) SF2, CH,OH, C₂H₂ b) Based on your answers given above, list the compounds in order of their Boiling Point from low to high. (8)

19.78 Write the products of the following sequences of reactions. Refer to your reaction road- maps to see how the combined reactions allow you to "navigate" between the different functional groups. Note that you will need your old Chapters 6-11 and Chapters 15-18 roadmaps along with your new Chapter 19 roadmap for these. (a) 1. BHS 2. H₂O₂ 3. H₂CrO4 4. SOCI₂ (b) 1. Cl₂/hv 2. KOLBU 3. H₂O, catalytic H₂SO4 4. H₂CrO4 Reaction Roadmap An alkene 5. EtOH 6.0.5 Equiv. NaOEt/EtOH 7. Mild H₂O An alkane 1.0 2. (CH3)₂S 3. H₂CrO (d) (c) 4. Excess EtOH, catalytic H₂SO OH 4. Mild H₂O* 5.0.5 Equiv. NaOEt/EtOH An alkene 6. Mild H₂O* A carboxylic acid 7. Mild H₂O* 1. SOC₁₂ 2. EtOH 3.0.5 Equiv. NaOEt/E:OH 5.1.0 Equiv. NaOEt 6. NH₂ (e) 1. 0.5 Equiv. NaOEt/EtOH 2. Mild H₂O* Br (f) i H An aldehyde 1. Catalytic NaOE/EtOH 2. H₂O*, heat 3. (CH,CH₂)₂Culi 4. Mild H₂O* 5.1.0 Equiv. LDA Br An ester 4. NaOH, H₂O 5. Mild H₂O* 6. Heat 7. MgBr 8. Mild H₂O* 7. Mild H₂O+

Li+ is a hard acid. With this in mind, which if the following compounds should be most soluble in water? Group of answer choices LiBr LiI LiF LiCl