Kaplan MCAT General Chemistry Review

Key Concept

When dealing with Lewis dot structures, we only deal with the eight valence electrons (s- and p-orbitals of the outer shell) on each atom. Remember that some atoms can expand their octets by utilizing the d-orbitals in this outer shell, but this will only take place with atoms in period 3 or greater.

Lewis Structures

A Lewis structure, or Lewis dot symbol, is the chemical symbol of an element surrounded by dots, each representing one of the s and/or p valence electrons of the atom. The Lewis symbols of the elements in the second period of the periodic table are shown in Table 3.1.

Table 3.1

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All of the Kaplan strategies are based on methodical approaches to problems and concepts, which save you time on Test Day.

Just as a Lewis symbol is used to represent the distribution of valence electrons in an atom, it can also be used to represent the distribution of valence electrons in a molecule. For example, the Lewis symbol of a fluoride ion, the Lewis structure of the diatomic molecule Certain steps must be followed in assigning a Lewis structure to a molecule. The steps are outlined here, using HCN as an example. (Hydrogen cyanide, an extremely poisonous compound, is found in small amounts in the pits of fruits such as cherries and apples.)

• Write the skeletal structure of the compound (i.e., the arrangement of atoms). In general, the least electronegative atom is the central atom. Hydrogen (always) and the halogens F, Cl, Br, and I (usually) occupy the end position.

In HCN, H must occupy an end position. Of the remaining two atoms, C is the least electronegative and, therefore, occupies the central position. The skeletal structure is as follows:

H-C-N

• Count all the valence electrons of the atoms. The number of valence electrons of the molecule is the sum of the valence electrons of all atoms present:

H has 1 valence electron;

C has 4 valence electrons;

N has 5 valence electrons; therefore,

HCN has a total of 10 valence electrons.

• Draw single bonds between the central atom and the atoms surrounding it. Place an electron pair in each bond (bonding electron pair).

H : C : N

Each bond has two electrons, so 10 - 4 = 6 valence electrons remain.

• Complete the octets (eight valence electrons) of all atoms bonded to the central atom, using the remaining valence electrons still to be assigned. (Recall that H is an exception to the octet rule because it can have only two valence electrons.) In this example, H already has two valence electrons in its bond with C.

• Place any extra electrons on the central atom. If the central atom has less than an octet, try to write double or triple bonds between the central and surrounding atoms using the nonbonding, unshared lone electron pairs.

The HCN structure above does not satisfy the octet rule for C because C possesses only four valence electrons. Therefore, two lone electron pairs from the N atom must be moved to form two more bonds with C, creating a triple bond between C and N. Finally, bonds are drawn as lines rather than pairs of dots.

H-CN :

Now, the octet rule is satisfied for all three atoms, because C and N have eight valence electrons and H has two valence electrons.

Formal Charge

In evaluating a Lewis structure to determine whether or not it may likely represent the actual arrangement of atoms in a compound, you will calculate the formal charge on each atom in the proposed Lewis structure. In doing so, you must be aware that you are assuming a perfectly equal sharing of all bonded electron pairs, regardless of actual differences in electronegativity, such that each electron pair is split evenly between the two atomic nuclei that share it. When you compare the number of electrons assigned to an atom in a Lewis structure (assigning one electron of each bonded pair to each of the atoms involved in the bond) to the number of electrons normally found in that atom’s valence shell, the difference between the two numbers is the formal charge. A fairly simple equation you can use to calculate formal charge is

Formal Charge = V - N_nonbonding - ½N_bonding

where V is the normal number of electrons in the atom’s valence shell, N_nonbonding is the number of nonbonding electrons, and N_bonding is the number of bonding

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Practice with many molecules and remembering the number of bonds that common central atoms usually form will make this formula easier to use on Test Day. For example, the nitrogen atom below normally has three bonds and one lone pair. Here, it is feeling generous and sharing more than usual; therefore, it will have a positive charge. If a molecule is selfish and shares less than usual, it will be negative (as we often see with oxygen atoms).

electrons. The charge of an ion or compound is equal to the sum of the formal charges of the individual atoms comprising the ion or compound.

Example: Calculate the formal charge on the central N atom of [NH₄]⁺.

Solution: The Lewis structure of [NH₄]⁺ is

Nitrogen is in group VA; thus it has five valence electrons. In [NH₄]⁺,

N has 4 bonds (i.e., eight bonding electrons and no nonbonding electrons).

So V = 5; N_bonding = 8; N_nonbonding = 0.

Formal charge = 5 - (8) - 0 = + 1

Thus, the formal charge on the N atom in [NH₄]⁺ is +1.

Let us offer a brief note of explanation on the difference between formal charge and oxidation number, as we are sure that you lay awake at night pondering such questions. It’s quite simple, really: Formal charge underestimates (ignores, actually) the effect of electronegativity differences, while oxidation numbers overestimate the effect of electronegativity, assuming that the more electronegative atom will in fact have a 100 percent share of the bonding electron pair. For example, in a molecule of CO₂ (carbon dioxide), the formal charge on each of the atoms would be 0, but the oxidation number of each of the oxygen atoms would be -2 and the carbon would have an oxidation number of +4. In reality, the distribution of electron density between the carbon and oxygen atoms lies somewhere between the extremes predicted by the formal charges and the oxidation states.

Resonance

As we’ve suggested, you may be able to draw two or more nearly identical Lewis structures that demonstrate the same arrangement between the atoms but differ in the specific placement of some pairs of electrons. These are called resonance structures. The actual electronic distribution in the real compound is a hybrid, or composite, of all the possible resonances. For example, SO₂ has three resonance structures, two of which are minor: O=S–O and O–S=O. The third is the major structure: O=S=O. The nature of the bonds within the actual compound is a hybrid of these three structures; indeed, spectral data indicate that the two S–O bonds are identical and equivalent. This phenomenon is known as resonance, and the actual structure of the compound is called the resonance hybrid. Resonance structures are expressed with a double-headed arrow between them.