Answer:
Here's what I get
Explanation:
A Lewis structure shows the valence electrons surrounding the atoms.
Your structure has two problems:
- It shows too many valence electrons
- It violates the octet rule for O — there are 10 electrons around the O atom.
Here's one way to draw a Lewis structure.
1. Draw a trial structure
Make F and O terminal atoms and give each one an octet (Fig. 1).
2. Count the valence electrons in the trial structure
5 BP + 15 LP = 10 + 30 = 40 electrons
3. Check the number of valence electrons available
1 S = 1 × 6 = 6 electrons
1 O = 1 × 6 = 6
4 F = 4 × 7 = 28
TOTAL = 40 electrons
The trial structure has the correct number of electrons.
4. Determine the formal charge on each atom.
To get the formal charges, we cut the covalent bonds in half.
Each atom gets the electrons on its side of the cut.
Formal charge = valence electrons in isolated atom - electrons on bonded atom
FC = VE - BE
(a) On S
VE = 6
BE = 5 bonding electrons = 5
FC = 6 - 5 = +1
(b) On O:
VE = 6
BE = 3 LP(six electrons) + 1 bonding electron = 7
FC = 6 - 7 = -1
(c) On F:
VE = 6
BE = 3 lone pairs(6 electrons) + 1 bonding electron = 6 + 1 =7
FC = 7 - 7 = 0
5. Minimize the formal charges
We must rearrange the valence electrons so that S gets one more and O gets one fewer.
Move a lone pair from the O to make an S=O double bond (Fig. 2).
6. Recalculate the formal charges
(a) On S
VE = 6
BE = (3 bonding electrons) = 6
FC = 6 - 6 = 0
(b) On O:
VE = 6
BE = 2 LP(four electrons) + 2 bonding electrons = 6
FC = 6 - 6 = 0
Fig. 2 shows the Lewis structure in which all atoms have a formal charge of zero.
