Draw Lewis dot structures, including appropriate resonance forms, and assign formal charges

to each of those structures, for the molecule urea, with chemical formula NH2CONH2. You
should find at least 3 such structures, some of which will have formal charges which are
nonzero. Using VSEPR theory, predict the NCO bond angle.

These boards do not support Lewis dot structures. Here is a site that shows the structure; perhaps it will be of some help in setting up the dot structures.

To draw the Lewis dot structures for urea (NH2CONH2), we need to determine the total number of valence electrons in the molecule.

1. Start by identifying the atoms and their valence electron count:
- N (nitrogen) has 5 valence electrons
- H (hydrogen) has 1 valence electron
- C (carbon) has 4 valence electrons
- O (oxygen) has 6 valence electrons

2. Calculate the total number of valence electrons:
Nitrogen (N) = 1 atom x 5 valence electrons = 5 electrons
Hydrogen (H) = 4 atoms x 1 valence electron = 4 electrons
Carbon (C) = 1 atom x 4 valence electrons = 4 electrons
Oxygen (O) = 2 atoms x 6 valence electrons = 12 electrons

Total = 5 + 4 + 4 + 12 = 25 valence electrons

Now, let's draw the Lewis dot structures for urea:

Structure 1:
H - N = C = O
|
H

In this structure, nitrogen forms a double bond with one of the carbon atoms, and nitrogen and carbon each have a single bond with an oxygen atom. Each hydrogen atom is bonded to the nitrogen atom.

Structure 2:
H - N = C = O
||
H

In this structure, nitrogen forms a double bond with one of the carbon atoms, and nitrogen and carbon each have a single bond with an oxygen atom. Both hydrogen atoms are bonded to the nitrogen atom.

Structure 3:
H - N = C = O
||
H

In this structure, nitrogen forms a double bond with one of the carbon atoms, and nitrogen and carbon each have a single bond with an oxygen atom. Both hydrogen atoms are bonded to the carbon atom.

To assign formal charges to each structure, we need to consider the valence electrons and the bonding pairs.

For Structure 1:
Nitrogen (N) has 5 valence electrons - 4 bonding electrons - 1 lone pair = 0 formal charge
Hydrogen (H) has 1 valence electron - 1 bonding electron = 0 formal charge
Carbon (C) has 4 valence electrons - 2 bonding electrons = 0 formal charge
Oxygen (O) has 6 valence electrons - 2 bonding electrons = 0 formal charge

For Structure 2:
Nitrogen (N) has 5 valence electrons - 2 bonding electrons - 1 lone pair = +1 formal charge
Hydrogen (H) has 1 valence electron - 1 bonding electron = 0 formal charge
Carbon (C) has 4 valence electrons - 2 bonding electrons = 0 formal charge
Oxygen (O) has 6 valence electrons - 2 bonding electrons = 0 formal charge

For Structure 3:
Nitrogen (N) has 5 valence electrons - 2 bonding electrons = -1 formal charge
Hydrogen (H) has 1 valence electron - 1 bonding electron = 0 formal charge
Carbon (C) has 4 valence electrons - 3 bonding electrons = +1 formal charge
Oxygen (O) has 6 valence electrons - 2 bonding electrons = 0 formal charge

Using VSEPR theory, we can predict the NCO bond angle by looking at the electron pairs and atoms around the central nitrogen atom. In all three structures, the nitrogen atom is the central atom bonded to a carbon atom and two oxygen atoms.

The NCO bond angle in urea is approximately 180 degrees. This is because the nitrogen atom is surrounded by three regions of electron density (two single bonds and one double bond), adopting a linear molecular geometry.

Note: The formal charges help us identify which resonance structure(s) are more stable than others. In this case, the second structure (with the nitrogen having a +1 formal charge) is less stable compared to the other structures (with formal charges of 0), making the first and third structures the major contributors to the resonance hybrid.