Lewis ("electron dot") Structures

  1. Consult the molecular formula and sum up all the valence electrons from the separate atoms. The Group number in the Periodic Table = the number of valence electrons for an atom.

    1. Add one for each (-) charge (extra electron);

    2. Subtract one for each (+) charge (missing electron).

    3. For example:

      CH3NH2 = 5(1) + 4 + 5 = 14 e-

      POCl3 = 5 + 6 + 3(7) = 32 e-

      NO3- = 5 + 3(6) + 1 = 24 e-

  2. Choose the central atom(s).

    • Almost always the least electronegative atom is the central atom.

    • For example, in ClO2, the Cl is the central atom; in SF5 the S is the central atom.

    • Occasionally, you will need to choose the unique atom, even when it is the most electronegative: e.g., the O in Cl2O.

    • A wrong choice usually will be signaled by your being unable to write a valid structure.

    Arrange the other atoms around the central atom, in accord with the normal valences of the atoms. That is, do not place more atoms around a central one than it normally can bond to.

    • For first and second row elements, the maximum valence = the Group number through Group IV; after that, it is 8 - (the Group number).

    • The difference is that up through IV, the atoms tend to donate electrons to get an octet, whereas beyond IV, they accept electrons.

    • Recognition of exceptions will come with experience.

    • You may find that you will have to place fewer atoms than normal around a central one; this is taken care of later.

    Hydrogen never is the central atom. It forms only one bond, so it must generally be in the outer layer of atoms. Therefore, place hydrogen atoms last.

  3. Insert pairs of electrons between all pairs of atoms that are to be bonded together. If this uses up all available electrons, go to Rule 6.

  4. Place any remaining electrons on peripheral atoms as unshared pairs, starting with the most electronegative such atom.

    • Fill this atom up to an octet.

    • Then proceed to the next most electronegative, and so on.

    • Remember that hydrogens can only have two electrons, and so cannot have any unshared pairs.

  5. If electrons still remain unused, place them on the central atoms as unshared pairs, again beginning with the most electronegative atom. Fill that atom to an octet. Then proceed to the next most electronegative, and so on.

    By application of Rules 4 and 5, the structures of our examples become:

  6. Examine the resulting structure. You will observe one of five situations:

    1. All of the atoms in the structure will have octets, except the hydrogens, which will have two electrons each. Go to Rule 7.

    2. The central atom is a Be with 4 electrons, or a B or Al with 6. These elements do not obey the octet rule. Go to Rule 7.

    3. The molecule has an odd number of electrons, which results in one of the central atoms having only 7 electrons. Go to Rule 7.

    4. The central atom has Z > 11, and has other than an octet. Go to Rule 7.

    5. The central atom is C, N, or O and

      1. the number of electrons in the molecule is even and the central atom lacks an octet; or

      2. the number of electrons in the molecule is odd and the central atom has fewer than 7 electrons.

        In either case, move an unshared pair from a peripheral atom to make a double bond to the central atom. If the central atom still has too few electrons, move another pair from the same atom to make a triple bond, or a pair from another atom to make a second double bond.

        When this action can be taken in more than one way, write all possible ways as separate structures. You have discovered resonance; the actual structure is a hybrid of all of the individual structures.

        Our two remaining example structures thus become:

  7. Examine every atom in the structure and assign it a formal charge as follows:

    formal charge = (number of valence electrons on the neutral, uncombined atom) - (number of covalent bonds to the atom in the current structure) - (the number of unshared electrons [not pairs] on the atom in the current structure)

    1. When counting covalent bonds, count double bonds (two pairs of electrons) as 2 and triple bonds (three pairs of electrons) as 3.

    2. The sum of all formal charges of all atoms must equal the given charge on the molecule or ion.

    3. No unreasonable charges should result from this process. That is, electronegative atoms should not get two (+) charges, metals should not get (-) charges, and so on. If this happens, you surely have done something wrong. Go back to Rule 1 and start over.

    4. If you get a lot of non-zero formal charges in a molecule that is neutral overall, you probably have misplaced the hydrogens. Go back to Rule 2.

      For example, for the O in POCl3, we have:

      6 - (1 covalent bond) - (6 unshared electrons) = -1

      whereas for the P:

      5 - (4 covalent bonds) - (0 unshared electrons) = +1

      The full set of formal charges for the example molecules is shown on the structures above.

  8. Rules 1-7 may allow some collections of atoms to form several valid structures with differing arrangements of atoms. This is OK; the alternative arrangements of atoms are called isomers. Isomers are particularly common among compounds of C, N, and O. For example, two isomeric Lewis structures can be written for C2H6O and four can be written for HCNO.

  9. Although four of the five cases in Rule 6 require no alteration of the structure, each case represents a normal bonding condition for the atoms involved. The Octet Rule actually applies only to C, N, O, and F. Bonding conditions unusual with respect to the Octet Rule will be found to be perfectly satisfactory if the orbital interactions are considered instead of forcing the molecule to use electron pair bonds.

  10. Application of Rule 1-7 also may lead to multiple structures having the same arrangement of atoms but different placement of electrons. This is OK too. These are resonance structures: the molecule actually resembles an average of all of the structures rather than any single one. Like the cases in Rule 9, these also will go away when we consider orbital interactions.

    Thanks to Professor Emeritus John Gordon, Kent State University, and Dr. Joyce Brockwell, Northwestern University, friends and former colleagues, for the original version of these Rules.

    This page last modified 1:23 PM on Monday June 11th, 2012.
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