Writing Lewis Structures: More Violating The Octet Rule Answers


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You should be able to figure out my diagrams by now.

Example #1 - PF6¯

Determine the total number of valence electrons available:

One phosphorous has 5 valence electrons
Six fluorine, each with 7 valence electron, totals 42
The negative charge means one electron has been added
This means there are 5 + 42 + one = 48 valence electrons, making 24 pairs, available.

Determine a provisional electron distribution by arranging the electron pairs (E.P.) until all available pairs have been distributed:


Example #2 - XeF2

Determine the total number of valence electrons available:

One xenon has 8 valence electrons
Two fluorine, each with 7 valence electron, totals 14
This means there are 22 valence electrons, making 11 pairs, available.

Determine a provisional electron distribution by arranging the electron pairs (E.P.) until all available pairs have been distributed:


Example #3 - ICl4¯

Determine the total number of valence electrons available:

One iodine has 7 valence electrons
Four chlorine, each with 7 valence electron, totals 28
One negative charge means one electron
This means there are 36 valence electrons, making 18 pairs, available.

Determine a provisional electron distribution by arranging the electron pairs (E.P.) until all available pairs have been distributed:


Example #4 - SO3¯

1. This compound is covalent.

2. Determine the total number of valence electrons available:

One sulfur has 6 valence electrons
Three oxygen, each with 6 valence electrons, totals 18
This means there are 3 + 28 + 1 = 32 valence electrons, making 16 pairs, available.

There is a double bond between sulfur and each oxygen. This makes 12 electrons in the valence shell of sulfur. Each oxygen has two non-bonding pairs. You'll see the structure in the multiple bonds tutorial.


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