Please help with this!

Draw Lewis structures of all of the important resonance states for the following molecules or ions. Be sure to indicate all formal charges and all unshared electron pairs.

H3CCO+

C2H4Br+

C7H7+ (This is a ring with one H on each C)

C6H5NO2 (nitrobenzene)

The first has a double bonded O on one end C.

CH3- CHBr CH2Br- CH

Think on those for the second.

The last two are ring structures. The C7 apparently has several double bonds

The benzene ring with a nitro attached is straightforward.

The question is rather straightforward, I am uncertain what you are having difficulty with.

C2H4Br+

damn thiis is old I

i think this is wrong

This is the... second or third question EVER posted on this website, WOW!!!

Actually, no.

The first one was at 9:41 PM, Monday, June 13 of 2005. Weird stuff happened and Ms. Sue answered the question a decade later.

To draw the Lewis structure for C2H4Br+, we need to follow the following steps:

1. Determine the total number of valence electrons: Carbon (C) has 4 valence electrons, Hydrogen (H) has 1 valence electron, and Bromine (Br) has 7 valence electrons. In this case, there are 2 carbon atoms, 4 hydrogen atoms, and 1 bromine atom, so the total number of valence electrons is:

2(4) + 4(1) + 7 = 19 valence electrons

2. Decide which atom in the molecule will be the central atom: In this case, the carbon atom (C) should be the central atom as it's usually one of the least electronegative elements.

3. Connect the outer atoms to the central atom with single bonds: Start by connecting each hydrogen atom (H) to one of the carbon atoms (C) with a single bond.

H - C - H

4. Draw a lone pair on each outer atom (if applicable): Since hydrogen (H) is an exception and can't have any unshared electron pairs, we don't draw any lone pairs on the hydrogen atoms.

5. Distribute the remaining electrons around the atoms to satisfy the octet rule: Since we have already used 2 electrons for the single bonds, we are left with 19 - 2 = 17 electrons.

a) Distribute electrons around the central carbon atom (C): Carbon (C) wants to have 4 electron pairs, so place 4 electrons around it as lone pairs.

H - C - H
|
C (lone pair)

b) Distribute electrons around the bromine atom (Br): Bromine (Br) should have 7 electron pairs, so place 7 electrons around it as lone pairs.

H - C - H
|
C (lone pair)
|
Br (lone pairs)

6. Check if all atoms have a complete octet: In this case, both carbon (C) and bromine (Br) have complete octets, but the carbon on the right (C) still has a positive charge. This means that we need to rearrange the electrons to create a double bond between the carbon atoms.

H - C = C - H
|
Br (lone pairs)

7. Check the formal charges: To determine the formal charges, subtract the number of assigned electrons and half the number of shared electrons from the total valence electrons of each atom.

- Carbon on the left (C): 4 assigned electrons - 0 shared electrons - 4 unassigned electrons = 0 formal charge
- Carbon on the right (C): 4 assigned electrons - 2 shared electrons - 4 unassigned electrons = -2 formal charge
- Bromine (Br): 7 assigned electrons - 2 shared electrons - 4 unassigned electrons = +1 formal charge

So, the final Lewis structure for C2H4Br+ is:

H - C = C - H
|
Br (lone pairs)