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More Lewis Structures

>> Parts of this equation/concept include:

A -	Oxoacids
B -	Expanded Octet
C -	Deficient Octets
D -	Molecules with Odd Number of Electrons

In the skeletal structure of oxoacids, the oxygen atoms are bonded to the central atom and the hydrogen atoms are bonded to oxygen atoms (one hydrogen to each oxygen). Otherwise, the procedure is the same as that for other Lewis structures.

>> Example 1

What is the Lewis structure of the following?

1. HNO₃
2. H₂CO₃

Solution:

1. The hydrogen is bonded to the oxygens, not the nitrogen. To complete the octet on the nitrogen there must a double bond. One of the oxygens that is not bonded to the hydrogen will double-bond. (Because there are two equal choices there is a resonance structure.)

   

2. The structure is similar to that of Example a. To minimize the formal charge on oxygen, each hydrogen is bonded to a different oxygen atom.

   

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Many atoms expand their octet. Only atoms with d orbitals can expand their octet. This requires that the atom have a principal quantum number of 3 or more. Therefore these atoms will be in the third or higher period of the periodic table and have an atomic number of 12 or more.

Note: Although these atoms can expand their octet, they do not always do so. Only the central atom will expand its octet. After drawing a structure in the normal way, if the formal charges on the molecule are decreased by creating a double bond, the double bond will form.

>> Example 2

Will the following have an expanded octet?

1. ClO₃^–
2. SCl₂

Solution:

1. Using the S = N – H rule, the molecule needs 32 electrons and has 7 + 3(6) + 1 = 26, so there are three bonds. This leads to a Lewis structure of

   

   However, the formal charge on chlorine is +2 and that on each oxygen is –1. If the chlorine double-bonds to two of the oxygens, the formal charge on chlorine and those two oxygen atoms is zero. Therefore the more likely structure is

   

2. Using the S = N – H rule, this molecule needs 24 electrons and has 20 electrons. This predicts two bonds. So the structure is

   

   The formal charge on each atom is zero. There is no point in changing the structure.

Many of the expanded octet structures do not lend themselves to the S = N – H rule because it assumes that the atoms need eight electrons, and if an atom expands its octet, it doesn't have (need) eight electrons. It is usually fairly obvious when it is not going to work. However, the Lewis structure must still have exactly the right number of electrons. One method is to complete the octet on each terminal atom and any leftover electrons will go on the central atom.

>> Example 3

What is the Lewis structure for the following atoms?

1. XeF₄
2. PCl₅
3. I₃^–

Solution:

1. The number of valence electrons "needed" is 8(5) = 40 and they have 8 + 4(7) = 36, so S = N – H; 36 = 4, so two bonds. With a formula of XeF₄, having two bonds is obviously impossible. This is a clue that this molecule probably expands its octet. The best way to approach this is to connect everything to the central atom, fill the octets of the terminal (outside) atoms, and add any leftover electrons to the central atom.

   

2. Unless they are the central atom, halogens only bond once. Consequently, it seems likely that all five chlorines bond to the phosphorus. The molecule needs 6(8) = 48 electrons and has 5 + 5(7) = 40 electrons, so S = N – H = 48 – 40 = 8, or four bonds. That isn't going to work. Since phosphorus is in the third period of the periodic table, it can expand its octet. So the Lewis structure is

   

3. If you use S = N – H, it needs 24 electrons and has only 22, which predicts one bond. This won't work, so use the alternate method of filling the octects of the outside atoms and putting the extra electrons on the central atom.

   

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A few atoms may have deficient octets. The most likely atom to have a deficient octet is boron. Boron may be surrounded by only six electrons or it may have a complete octet. The boron atom has three valence electrons. The number 6 comes from the pairing of those three electrons. It is the most common atom with a deficient octet, because it is the only nonmetal with fewer than four valence electrons. (Lewis structures are used for covalent molecules that are generally composed of nonmetals.)

The other atoms that may have a deficient octet are aluminum and beryllium. These atoms are metalloids, so they may covalently bond, even if the bonds are normally considered ionic.

Aluminum is stable when surrounded by six electrons. (Like boron, it has three valence electrons.) Beryllium is stable when surrounded by four electrons. (The atom has two valence electrons.)

The S = N – H rule will not work for deficient octets.

>> Example 4

What is the Lewis structure for AlCl₃?

Solution:

This compound has 3 + 3(7) = 24 electrons. The Lewis structure of AlCl₃ is

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Some molecules do not have an even number of electrons. For this situation it is impossible for all atoms to have an octet and even for all electrons to be paired. Electronegativity is used to determine which atom will have seven instead of eight electrons. Since electronegative atoms attract electrons, it will be the least electronegative atom that is shorted an electron.

>> Example 5

What is the Lewis structure of ClO₂?

Solution:

The total number of electrons is 7 + 2(6) = 19 electrons. The atom that is the least electronegative in this case is chlorine.

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