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Step-by-Step Guide: Drawing Covalent Compounds & Polyatomic Ions, Lecture notes of Chemistry

The steps to draw lewis dot structures for covalent compounds and polyatomic ions using the lewis dot theory of bonding. It covers counting valence electrons, determining connectivity, connecting central atoms to outer atoms, completing octets on outer atoms, and adding lone pairs or forming double/triple bonds if needed. Three examples are provided for better understanding.

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LEWIS DOT STRUCTURE
The steps below outline general guidelines for drawing the Lewis dot structures for covalent compounds and polyatomic
ions with a central atom. This theory of bonding will become a powerful tool for understanding how chemical
substances behave physically and chemically. Three representative examples are provided as reference.
1. Count Valence Electrons: Using the molecular formula, find the total number of valence electrons:
Add up valence electrons for all atoms. We will most typically work with main groups elements, with valence
electrons in the outer most s & p orbitals.
For each negative (-) charge, add one electron to the total.
For each positive (+) charge, subtract one electron from the total.
HINT: For the vast majority of stable molecules or ions we encounter, the total number of valence electrons will be EVEN.
While molecules with odd numbers of electrons are found, they often are highly reactive radicals with unpaired electrons.
CH3CH2OH
1 Carbon: 2 x (4 valence electrons)
6 Hydrogens: 6 x (1 valence electron)
1 Oxygens: 1 x (6 valence electrons)
Total: 8 + 6 + 6 = 20 electrons
NO3-
1 Nitrogen: 1 x (5 valence electrons)
3 Oxygens: 3 x (6 valence electrons)
Monoanion: 1 extra electron
Total: 5 + 18 + 1 = 24 electrons
NO2-
1 Nitrogen: 1 x (5 valence electrons)
2 Oxygens: 2 x (6 valence electrons)
Monoanion: 1 extra electron
Total: 5 + 12 + 1 = 18 electrons
2. Determine Connectivity: Decide how the atoms are connected
together. In some cases this will be given, as in CH3CH2OH. Here the formula
suggests the three hydrogens are associated with one carbon, which is
attached to another carbon with two bonded hydrogens, which is then
connected to an oxygen which has an associated hydrogen.
When the connections are not provided based on the formula, use the general rules below:
Place the LEAST electronegative element in the center, surrounded
by the other atoms. With a weaker pull on electrons, the less
electronegative element can afford to share more electrons than
elements with higher electronegativities (which would be placed on
the outside of the molecule or ion).
Hydrogen will never be in the center. It can only form a duet of
electrons maximum (1s2), and therefore can only form 1 bond with
another element.
3. Connect the central atoms to the outer atoms with single bonds.
Each single line represents the sharing of 2 electrons; without sharing electrons, the atoms will not be associated with
one another in a covalent molecule.
Each single line corresponds to a bond order of 1. All else being equal, the higher the bond order, the strong the
covalent bond and the short the bond length. A bond order to 2 would represent the sharing of 2 electron pairs (or 4
electrons total). A bond order of 3 would represent the sharing of 3 electron pairs (or 6 electrons total). Each of the
bonds above currently have a bond order of 1. Proceed to the next steps to learn how higher bond orders are possible in
Lewis dot structures.
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LEWIS DOT STRUCTURE

The steps below outline general guidelines for drawing the Lewis dot structures for covalent compounds and polyatomic ions with a central atom. This theory of bonding will become a powerful tool for understanding how chemical substances behave physically and chemically. Three representative examples are provided as reference.

1. Count Valence Electrons: Using the molecular formula, find the total number of valence electrons:

 Add up valence electrons for all atoms. We will most typically work with main groups elements, with valence electrons in the outer most s & p orbitals.  For each negative (-) charge, add one electron to the total.  For each positive (+) charge, subtract one electron from the total.

HINT: For the vast majority of stable molecules or ions we encounter, the total number of valence electrons will be EVEN. While molecules with odd numbers of electrons are found, they often are highly reactive radicals with unpaired electrons.

CH 3 CH 2 OH

1 Carbon: 2 x (4 valence electrons) 6 Hydrogens: 6 x (1 valence electron) 1 Oxygens: 1 x (6 valence electrons) Total: 8 + 6 + 6 = 20 electrons

NO 3 -

1 Nitrogen: 1 x (5 valence electrons) 3 Oxygens: 3 x (6 valence electrons) Monoanion: 1 extra electron Total: 5 + 18 + 1 = 24 electrons

NO 2 -

1 Nitrogen: 1 x (5 valence electrons) 2 Oxygens: 2 x (6 valence electrons) Monoanion: 1 extra electron Total: 5 + 12 + 1 = 18 electrons

2. Determine Connectivity: Decide how the atoms are connected

together. In some cases this will be given, as in CH 3 CH 2 OH. Here the formula suggests the three hydrogens are associated with one carbon, which is attached to another carbon with two bonded hydrogens, which is then connected to an oxygen which has an associated hydrogen.

When the connections are not provided based on the formula, use the general rules below:  Place the LEAST electronegative element in the center , surrounded by the other atoms. With a weaker pull on electrons, the less electronegative element can afford to share more electrons than elements with higher electronegativities (which would be placed on the outside of the molecule or ion).  Hydrogen will never be in the center. It can only form a duet of electrons maximum (1s^2 ), and therefore can only form 1 bond with another element.

3. Connect the central atoms to the outer atoms with single bonds.

Each single line represents the sharing of 2 electrons; without sharing electrons, the atoms will not be associated with one another in a covalent molecule.

Each single line corresponds to a bond order of 1. All else being equal, the higher the bond order, the strong the covalent bond and the short the bond length. A bond order to 2 would represent the sharing of 2 electron pairs (or 4 electrons total). A bond order of 3 would represent the sharing of 3 electron pairs (or 6 electrons total). Each of the bonds above currently have a bond order of 1. Proceed to the next steps to learn how higher bond orders are possible in Lewis dot structures.

4. Complete octets on outer atoms by adding lone pair electrons.

Remember single bonds represent 2 electrons, and an octet represents 8 electrons total around an atom. Hydrogen will only have one bond to another atom to fulfill its duet (2 electrons maximum).

5. Adding any remaining valence electrons to the central atom as lone pair electrons.

Do this by comparing the total number of electrons in the structure at step 4 above, to the total number of electrons needed, as calculated in step 1. In the first two examples (CH 3 CH 2 OH) and NO 3 - , the total number of valence electrons has already been reached (20 & 24 electrons, respectively); no more electrons can be added to these structures.

In the case of nitrite (NO 2 - ), only 16 electrons are accounted for at step 4. Two additional electrons are needed to get to the total of 18 valence electrons; these are added as a lone pair to the central nitrogen atom (see structure at left). At this point, all of the valence electrons should be accounted for. Do not add any additional electrons.

6. If needed, form double or triple bonds to complete the octet of the central atom.

Remember additional electrons cannot be added at this point. Instead, pull lone pair electrons in from the outer atoms to form the double or triple bonds. In the case of CH 3 CH 2 OH, every atom has a complete octet. In step 4 above, the central nitrogen of NO 3 - only has 6 electrons. To reach a complete octet, use one the electron pairs on an outer oxygen to form an N=O double bond. Notice the modified structure to the right still has 24 electrons, but the nitrogen now has 8 total electrons around it.

Elements in groups 2A & 3A (columns with beryllium and boron) do not have large enough electronegativities to attain an octet. These elements can have less than 8 electrons.