Chapter 4: Unit 11. Lewis Structures

Lewis Structures

Lewis Structures are electron dot representations of molecules. Two-electron bonds are drawn with a solid line and non-bonded electrons are drawn with dots.

Lewis structures contain only valence electrons. H gets two electrons and main group elements generally get eight electrons.

After placing all electrons in bonds and lone-pairs, it may be necessary to use lone pairs from terminal to form multiple bonds if an atom does not have an octet.

Stepwise directions for writing Lewis Dot structures:

1. Determine the total number of valence electrons
2. Determine which atom is the central atom (usually less electronegative atom is central atom. Hydrogen and fluorine can never be central atom)
3. Connect all other atoms to the central by a single covalent bond
4. Satisfy all the surrounding atoms with eight electrons by adding lone pair
5. Verify the total number of valence electrons of the structure. Extra electron should go to the central atom as lone pair.
6. If the central atom doesn’t have octet then share the lone pair from terminal atoms with central by making multiple bonds.
7. In case of ions, add or subtract valence electrons based on the charge of the ion.

Below are some examples of drawing Lewis dot structures.

Example#1: CCl₄

1. Total number of valence electrons: C → 4 + Cl → 7*4=28+ 4= total 32 electrons.

2. Since C is less electronegative, it should be the central atom.

3. Then after connecting all four Chlorine atoms by a single bond, the number of electrons remaining= 32-(4*2)= 32-8= 24 electrons.

4. Since each chlorine has only two electrons ( a single bond) surrounding it, they require 6 more electrons as lone pair i.e. 3 pairs of lone pairs should be placed on each chlorine. Four chlorine atoms should have 6*4= total 24 remaining electrons.

5. After satisfying octet for the surrounding atom, there is no extra electron to put on central atom.

6. Central atom C has achieved octet through 4 bonds ( 4*2= 8)

7. There is no need to form multiple bond.

Example #2: CO₂

1. Total number of valence electrons: C → 4 + ClO→ 6*2=12+ 4= total 16 electrons.

2. Since C is less electronegative, it should be the central atom.

3. Then after connecting two Oxygen atoms by a single bond, the number of electrons remaining= 16-(2*2)= 16-4= 12 electrons.

4. Since each oxygen has only two electrons ( a single bond) surrounding it, they require 6 more electrons as lone pair i.e. 3 pairs of lone pairs should be placed on each chlorine. Two oxygen atoms should have 6*2= total 12 remaining electrons.

5. After satisfying octet for the surrounding atom, there is no extra electron to put on central atom.

6. Central atom C has not achieved since there are only four electrons surrounding carbon by 2 bonds. ( 2*2= 4)

7. There is a need to form multiple bond. Each oxygen’s one lone pair is shared with carbon by making a bond between carbon and oxygen. This way Carbon gets extra 4 electrons to achieve octet.

Can you try other two examples: COCl₂ and NCl₃?

In the following example: CO₃^2- ion has total= 4*1 + 6*3 +2= 24 electrons.

Carbon is the central atom, after connecting all other atoms and satisfy the terminal atoms with 8 electrons, there is no more extra electron left to put on central. But Carbon still doesn’t have octet. So to satisfy the carbon with 8 electrons any terminal oxygen atom’s lone pairs are shared to form a double bond.

Normally ions are written with bracket with the charge mentioned outside the species.

In some cases  exception to octet rule is found. For example, Hydrogen follow duet rule, B oron requires six electrons. Any period three nonmetal (P, S, Cl) and higher elements can hold more than eight electrons by expanding their partially filled or empty “d” orbitals. Example of molecules with violated octet rule: PCl_{5, SF6}

Questions:

1. Draw the Lewis dot structures of following molecules
   a) SO₂
   b) O_{3
   c)} SO_{3
   d)} PCl₃
   e) H₂CO
   f) C₂H₄
   g) CH₃OH