A Lewis structure shows the valence electrons surrounding the atoms.
It shows too many valence electrons
It violates the octet rule for O — there are 10 electrons around the O atom.
Count the valence electrons in the trial structure
5 BP + 15 LP = 10 + 30 = 40 electrons
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.
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
We must rearrange the valence electrons so that S gets one more and O gets one fewer.
(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
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Question is Incomplete:
Given that S is the central atom, draw a Lewis structure of OSF4 in which the formal charges of all atoms are zero. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons.
