### 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 - PF**_{6}¯

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 - XeF**_{2}

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 - ICl**_{4}¯

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 - SO**_{3}¯

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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