>> View the other Key Equations and Concepts in this chapter

Lewis Structures

>> Parts of this equation/concept include:

A -	Valence Electrons
B -	Atoms and Monotomic Ions
C -	Multiatom Structures >> Covalent Molecules >> Polyatomic Ions >> Resonance Structures >> Formal Charges

Since Lewis structures show where all the electrons are, it is important that the number of valence electrons in the structure be exactly correct. The number of valence electrons is the number of s and p electrons in the outermost shell (highest n). The number of valence electrons can also be determined by counting across the period, skipping the transition metals, until the element is reached. It is also the group number.

If there is more than one atom, the number of valence electrons is the sum of all the atoms in the molecule or polyatomic ion.

For ions, add the value of a negative charge and subtract the value of a positive charge.

>> Example 1

How many valence electrons are in the following?

1. N
2. H₂S
3. CO₃^2–
4. NH₄⁺

Solution:

1. Nitrogen is in group 5A. It has five valence electrons.
2. Hydrogen has one valence electron, and sulfur has six. The total for the molecule is 2(1) + 6 = 8.
3. Carbon has four valence electrons; oxygen has six; then two for the charge. 4 + 3(6) + 2 = 24.
4. Nitrogen has five valence electrons; hydrogen has one, minus one for the charge. 5 + 4(1) – 1 = 8.

  >> back to the Top of the Page

A dot is used to represent each electron. The electrons go on one of four sides (up, down, left, right) of the element symbol. The first two electrons are paired up (it doesn't matter which side). The remaining electrons (or dots) are distributed on each side, then double up if necessary.

>> Example 2

1. N
2. Mg
3. Mg²⁺
4. F^–
5. Ne

Solution:

1. N has five valence electrons, so the Lewis structure is .
2. Mg has two valence electrons.
3. Mg²⁺ has no valence electrons, so its Lewis structure is Mg²⁺.
4. F^– has eight valence electrons, so its Lewis structure is .
5. Ne has eight valence electrons, so its Lewis structure is .

  >> back to the Top of the Page

>> Covalent Molecules

After counting valence electrons, the next step is to construct a skeletal structure. Normally, the first atom of the formula is also the central atom. The other atoms are arranged around this atom. If hydrogen is the first atom, it cannot be the central atom. In this case, the next atom would be the central atom. Because hydrogens will only bond once, they are on the outermost layer of the molecule.

For oxyacids (acids containing oxygen), the hydrogen is always attached to an oxygen rather than the central atom.

In organic molecules—those based on carbon—the carbons tend to form a chain, with each carbon surrounded by hydrogens. The way the formula is written can also be a clue to its skeletal structure. For example, there are two equally appropriate ways to arrange the atoms of C₂H₆O. Consequently, when the formula is written CH₃CH₂OH, the Lewis structure is , and when the formula is written CH₃OCH₃, the Lewis structure is .

After determining the skeletal structure, the electrons must be distributed. To determine how many electrons are involved in bonding and how many act as lone pairs, the S = N – A rule can be used. N is the number of electrons needed. Hydrogen needs two electrons; all other atoms need eight. A is the number of electrons available; it is the number of valence electrons counted in the first step. S is the number of shared electrons, or electrons involved in bonding. Since each line represents two electrons, S/2 are the number of bonds.

The electrons involved in bonding can be distributed first, using the bonds to connect the atoms and the extra ones as double bonds. In choosing where to put a double bond, recall that hydrogens and halogens (unless the halogen is the central atom) will not have more than one bond.

The electrons not involved in bonding will be distributed so that each atom is surrounded by eight electrons, unless the atom is hydrogen, which is surrounded only by two electrons.

The number of electrons in the Lewis structure must exactly match the number of electrons available.

>> Example 3

What is the Lewis structure of the following molecules?

1. PCl₃
2. H₂CO₃
3. CH₃NH₂

Solution:

1. 1. Count the number of valence electrons (A): P = 5, Cl = 7, so total ve = 5 + 3(7) = 26
   2. Determine the number of electrons needed (N): eight for each 4(8) = 32
   3. Determine the number of bonds (S): 32 – 26 = 6, so three bonds.
   4. Determine the skeletal structure: Phosphorus is the central atom. It is surrounded by chlorines:
      
   5. The chlorines are connected to the phosphorus, using all three bonds.
   6. The rest of the electrons are distributed around the atoms so that each has an octet.

      Hint 1: Distribute in pairs working from the outside in.

      Hint 2: Two electrons on each side (up, down, left, right) is an octet.

2. 1. Count the number of valence electrons: H = 1, C = 4, O = 6, so 2(1) + 4 + 3(6) = 24
   2. Number of electrons needed: H need 2; the others need 8. 2(2) + 4(8) = 36
   3. Number of bonds needed: S = N – A = 36 – 24 = 12, so six bonds.
   4. Skeletal structure. This is an oxyacid, so carbon is the central atom. It is surrounded by oxygen and the hydrogen is connected to one of the oxygens. Each oxygen is the same, so it doesn't matter which one.
      
   5. Since the skeletal structure only uses five of the six bonds, another bond must be somewhere. The hydrogens already have their maximum number of bonds. The Lewis structure of the atom oxygen has two unpaired electrons, which implies that oxygen likes to bond twice. The oxygens bonded to the hydrogens already have two bonds, so a second (double) bond is most likely on the other oxygen. The rest of the electrons can then be distributed.
      
3. 1. Count valence electrons: C = 4, H = 1, N = 5, so 4 + 3(1) + 5 + 2(1) = 14.
   2. Count electrons needed: Two for hydrogen, eight for others, so 2(8) + 5(2) = 26.
   3. Determine number of bonds: 26 – 14 = 12, or six bonds.
   4. Skeletal structure. Hydrogen cannot be a connecting atom, so the carbon must be connected to the nitrogen. The way the formula is written it implies three hydrogens on the carbon and two on the nitrogen.
   5. Distribute the rest of the electrons. When all the atoms are connected, there are no more bonds, and only two more electrons. The only atom without an octet is nitrogen.
      

>> Polyatomic Ions

The only difference between Lewis structures of polyatomic ions and molecules is that you must consider the charge when counting valence electrons. Since electrons are negative, ions with a negative charge have extra electrons; ions with a positive charge are short electrons.

>> Example 4

What is the Lewis structure of the following ions?

1. SO₃^2–
2. NH₄⁺
3. CN^–

Solution:

1. 1. The number of valence electrons is 6 + 3(6) + 2 = 26.
   2. The number of electrons needed is 4(8) = 32.
   3. The number of bonds is 32 – 26 = 6, so three bonds.
   4. Three bonds is what is needed to connect the oxygens to the sulfur in the skeletal structure, so the rest of the electrons must be lone pairs.
      
2. 1. The number of valence electrons is 5 + 4(1) – 1 = 8.
   2. The number of needed electrons is 8 + 4(2) = 16.
   3. The number of bonds is 16 – 8 = 8, so four bonds.
   4. To connect the hydrogens to the nitrogen. Four bonds are needed. That also uses up all the electrons.
      
3. 1. The number of valence electrons is 4 + 5 + 1 = 10.
   2. The number of needed electrons is 2(8) = 16.
   3. The number of bonds is 16 – 10 = 6, so three bonds.
   4. With only two atoms all the bonds must be between these two atoms. One lone pair each atom will complete each atom's octet.
      

>> Resonance Structures

Lewis structures where a double bond can have two or more exactly equivalent positions. Each resonance structure shows a different equivalent position of the double bond. The true structure is an average of all the possible resonance structure.

>> Example 5

Draw all the resonance structures of the following.

1. SO₂
2. NO₃^–

Solution:

1. 
   
   
   

   Each sulfur-oxygen bond is actually 1 1/2 bond rather than either a double or a single bond.

2. 
   
   
   
   
   

   Each nitrogen-oxygen bond is actually a 1 1/3 bond rather than either a single or double bond.

>> Formal Charges

Formal charges are a method for determining the most appropriate Lewis structure. There is a formal charge for each atom in a structure. The sum of the formal charges will be the same as the charge on the compound.

A formal charge is calculated by subtracting the number of electrons assigned to an atom from the number of valence electrons contributed by the atom. Both electrons of a lone pair and half the electrons of a bond are "assigned."

Formal charge = valence electrons – assigned electrons

The best Lewis structure will have the lowest formal charges. Any structure with an atom having a formal charge of more than +1 or –1 is unlikely to be correct.

>> Example 6

What is the formal charge of each atom in the following structures of CH₄O? Which structure is correct?

Solution:

In all three structures, hydrogen has one bond. Therefore it is assigned one electron. It also has one valence electron, so its formal charge is zero in all the structures.

Carbon has four valence electrons. In the first structure, carbon has one lone pair (2) and three bonds (3), so five assigned electrons. Ve – ae = –1 = formal charge. In the second structure the carbon has no lone pairs and four bonds, so four assigned electrons. Formal charge = 4 – 4 = 0. In the third structure carbon has two lone pairs and two bonds, so six assigned electrons. Formal charge = 4 – 6 = –2.

Oxygen has six valence electrons. In the first structure the oxygen has one lone pair and three bonds, so five assigned electrons. Formal charge = 6 – 5 = +1. In the second structure the oxygen has two lone pairs and two bonds, so six assigned electrons. Formal charge = 6 – 6 = 0. In the third structure the oxygen has no lone pairs and four bonds, thus four assigned electrons. Formal charge = 6 – 4 = +2.

The correct structure is the middle one. The formal charge of each atom is zero.

>> View the other Key Equations and Concepts in this chapter