A student draws four resonance structures for nitromethane (CH3NO2). How many lewis structures should have been drawn to represent nitromethane?

To determine how many resonance structures should have been drawn to represent nitromethane (CH3NO2), we first need to understand what resonance structures are.

Resonance structures are different representations of a molecule where the placement of electrons can be depicted differently. These structures arise when there is more than one way to arrange the electrons in a molecule, while maintaining the same overall connectivity of atoms.

Now, for nitromethane (CH3NO2), the Lewis structure can be drawn considering the octet rule. Let's go step by step:

1. Determine the total number of valence electrons present in the molecule.
Carbon (C) contributes 4 valence electrons (VE).
Hydrogen (H) contributes 1 VE each.
Nitrogen (N) contributes 5 VE.
Oxygen (O) contributes 6 VE each.

Therefore, the total number of valence electrons in nitromethane (CH3NO2) is:
(4 carbon VE) + (3 hydrogen VE) + (1 nitrogen VE) + (2 oxygen VE) = 12 VE.

2. Identify the central atom. In nitromethane (CH3NO2), the carbon atom (C) acts as the central atom.

3. Connect the carbon atom (C) to the other atoms using single bonds. In this case, connect the carbon atom to three hydrogen atoms (H) and one nitrogen atom (N).

4. Distribute the remaining valence electrons to complete the octet of the atoms, starting with the outer atoms (hydrogen and nitrogen) and then the central carbon atom. For oxygen atoms, they can exceed the octet (due to their ability to accommodate extra electrons).

5. Calculate the formal charges on each atom to ensure the most stable arrangement of electrons. Remember that formal charge is calculated using the formula:
Formal Charge = Valence Electrons - Lone Pair Electrons - 0.5 * Bonding Electrons

Once you have drawn one Lewis structure for nitromethane (CH3NO2), you can then consider the resonance structures.

In the case of nitromethane (CH3NO2), the correct number of resonance structures should have been drawn to represent it is just one, not four.