Question

1. for the following molecules (a) give the number of π -bonds present in the shown structures, (b) draw all the resonance structures that illustrate the atoms that can support the shown charge, and (c) circle the most stable resonance structure.

Explanation:

Step1: Identify π - bonds

Double - bonds and triple - bonds contain π - bonds. In the given structure, the C = N double - bond has 1 π - bond, and each C = O double - bond has 1 π - bond. There are 2 C = O double - bonds and 1 C = N double - bond.
The total number of π - bonds is 3.

Step2: Draw resonance structures

1. For the C = N double - bond and the adjacent negatively - charged carbon: The lone pair on the negatively - charged carbon can delocalize towards the C = N double - bond, forming a new C = C double - bond and a negative charge on the nitrogen.
2. For the C = O double - bonds: The lone pairs on the oxygen atoms can participate in resonance. The lone pair on the oxygen of the C = O bond adjacent to the negatively - charged carbon can delocalize, forming a new C = C double - bond and a negative charge on the oxygen.
3. The resonance structures show the movement of π - electrons and lone pairs to distribute the negative charge over different atoms in the molecule.

Step3: Determine the most stable resonance structure

The most stable resonance structure is the one that has the negative charge on the more electronegative atom. Among nitrogen and oxygen, oxygen is more electronegative. So, the resonance structure where the negative charge is on the oxygen atom (from the C = O bond) is more stable.

Answer:

(a) 3
(b) Resonance structures are drawn by showing the movement of π - electrons and lone pairs to distribute the negative charge. (It is difficult to draw here in text - based format, but the general process is as described above).
(c) The resonance structure with the negative charge on the oxygen atom (from the C = O bond) is circled as the most stable one.