Polar bonds, Lewis structures, formal charge and resonance

The Lewis structure is like the “source code” of chemistry: you will read the shape, polarity, hybridization and even the reactions of a molecule from it over the next two weeks. A wrong Lewis structure means a wrong shape and wrong polarity. That is why this week we concentrate on a single skill: systematically writing the correct Lewis structure for any molecule or ion.

Golden rules:

1. The electron count is everything. First count the total valence electrons, and count again at the end.
2. The central atom is the least electronegative one (except H; H is always terminal).
3. Formal charge is the referee between several structures.
4. In resonance the atoms do not move, only the electron pairs move.

Electronegativity and bond polarity

Electronegativity (EN) is the tendency of an atom in a bond to attract the shared electrons. It increases to the right and upward in the periodic table; the most electronegative element is F (4.0).

When two different atoms bond, the electrons are not shared equally. The more electronegative atom carries a partial negative charge ($\delta^-$), and the other a partial positive charge ($\delta^+$). This is a polar covalent bond:

$\overset{\delta+}{\mathrm{H}}-\overset{\delta-}{\mathrm{Cl}}$

Bond type is a continuum:

These limits are not sharp, only a rough guide. If the instructor gives their own limits in the exam, use those.

The recipe for writing a Lewis structure

Step 1. Count the total valence electrons. For an anion add as many as the charge, for a cation subtract.

Step 2. Build the skeleton. The central atom is usually the atom that appears once in the formula and is the least electronegative. H and F are always terminal. In oxyacids, H is bonded to O (H–O–S in $\mathrm{H_2SO_4}$).

Step 3. Connect the center to each outer atom with a single bond (each bond is 2 electrons).

Step 4. Give the remaining electrons first to the outer atoms as lone pairs and complete their octets (nothing is added to H, 2 electrons complete it).

Step 5. If electrons are still left, put them on the central atom.

Step 6. If the central atom’s octet is incomplete, turn a lone pair on an outer atom into a double bond with the center. Repeat if necessary (triple bond).

Example 1: formaldehyde, $\mathrm{H_2CO}$.

1) e−: 2(1) + 4 + 6 = 12
2) skeleton: C central, two H and O attached
3) 3 single bonds: 6 e− used, 6 left
4) 3 pairs on O (6 e−): 0 left
5) C has only 6 e− (3 bonds) → no octet
6) turn one pair from O into a C=O double bond
          H
           \
            C = O:   (2 lone pairs on O)
           /
          H
check: C 8 e−, O 8 e−, H 2 e−, total 12 e−

Example 2: the chlorate ion, $\mathrm{ClO_3^-}$.

1) e−: 7 + 3(6) + 1 = 26
3) 3 Cl–O single bonds: 6 e−, 20 left
4) 3 pairs on each O: 18 e−, 2 left
5) 1 lone pair on Cl: 0 left
   every atom has an octet → square brackets and the charge: [ ... ]−

Formal charge

It compares the electrons an atom “owns” in the Lewis structure with the valence electrons of the free atom:

$\text{FC} = (\text{valence e}^-) - (\text{nonbonding e}^-) - (\text{number of bonds})$

The number of bonds is half the bonding electrons (single 1, double 2, triple 3). The sum of all the formal charges is 0 in a molecule and the charge of the ion in an ion.

If there are several valid structures, the best one is the one that:

1. has formal charges closest to zero,
2. carries the negative formal charge on the more electronegative atom,
3. does not have charges of the same sign on neighboring atoms.

Example 3: two candidates for $\mathrm{CO_2}$.

                 O = C = O          O ≡ C − O
formal charges:  0   0   0          +1  0  −1

The one on the left is preferred because all its charges are zero. The one on the right also obeys the octet rule but has charge separation.

PITFALL · formal charge ≠ oxidation number. Formal charge shares the electrons equally, while the oxidation number gives all the bonding electrons to the more electronegative atom. In $\mathrm{CO_2}$ the formal charge of C is 0 and its oxidation number is +4.

Resonance

For some species a single Lewis structure is not enough. Ozone $\mathrm{O_3}$ (18 e⁻):

  O = O − O:        ←→        :O − O = O
 (0) (+1) (−1)               (−1) (+1) (0)

The two structures differ only in the position of the double bond. In experiments the two O–O bonds have equal length (between a single and a double bond). The real molecule is the resonance hybrid, the average of these structures; the molecule does not flip back and forth between the two structures. That is why the double-headed arrow ($\leftrightarrow$) is not an equilibrium arrow.

The average bond order:

$\text{bond order} = \frac{\text{total number of bonds (for the atom pair)}}{\text{number of bond positions}} \qquad \mathrm{O_3}: \tfrac{3}{2} = 1.5$

In $\mathrm{NO_3^-}$ and $\mathrm{CO_3^{2-}}$ there are three equivalent structures, and the average bond order is $4/3$.

TIP · recognizing resonance. If a double bond can be placed in more than one position between atoms of the same kind, there is resonance. The positions of the atoms never change; if they did, it would be a different molecule (an isomer).

Exceptions to the octet rule

1. An odd number of electrons (odd-electron species): NO (11 e⁻), $\mathrm{NO_2}$ (17 e⁻). An unpaired electron remains on at least one atom; these are called free radicals.
2. Incomplete octet: compounds of B and Be. In $\mathrm{BF_3}$ there are 6 electrons around B. The structure with only single bonds is preferred because a B=F double bond is unfavorable in terms of formal charge.
3. Expanded octet: elements from period 3 onward can carry more than 8 electrons at the center. $\mathrm{PCl_5}$ (10 e⁻ on P), $\mathrm{SF_6}$ (12 e⁻ on S), $\mathrm{XeF_4}$ (12 e⁻ on Xe). Period 2 elements (C, N, O, F) can never exceed the octet.

Bond length and bond enthalpy

As the bond order increases there is more electron density between the two atoms: the bond gets shorter and stronger.

Bond enthalpy is the energy needed to break 1 mol of bonds in the gas phase; it is always positive. Breaking bonds takes energy, forming bonds releases energy. The enthalpy of a reaction can be estimated with the average values in the table:

$\Delta H \approx \sum \text{BE}(\text{bonds broken}) - \sum \text{BE}(\text{bonds formed})$

Example 4. Hydrogenation of ethylene: $\mathrm{C_2H_4 + H_2 \rightarrow C_2H_6}$.

broken: 1 C=C (611) + 1 H–H (436)            = 1047 kJ
formed:  1 C–C (347) + 2 new C–H (2 × 414)   = 1175 kJ
ΔH ≈ 1047 − 1175 = −128 kJ

We did not count the four unchanged C–H bonds; they are on both sides and cancel.

PITFALL · the direction of the formula. The ΔH°f formula was “products − reactants”. The bond enthalpy formula looks reversed: “broken (reactant) − formed (product)”. Remember the logic: breaking +, forming −.

Common mistakes