11: Lewis structures, covalent, and polar covalent bonding

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Covalent bonding and Lewis structures

\includegraphics[scale=0.5]{pic2.eps} — Figure 1:

\includegraphics[scale=0.5]{pic3.eps} — Figure 1:

$\begin{figure}\begin{center} \leavevmode \epsfbox{lec11_fig1.ps} {\small} \end{center}\end{figure}$

Clearly, these diagrams do not represent actual molecular geometries. In each diagram, each hydrogen has a stable doublet, while all other atoms have octets.

What does a lone pair look like? Again, we show an electron density plot for a water molecule below:

Consider, next, the molecule $O_2$. Since $O$ has 6 valence electrons, if we try allowing for one shared pair, we would end up with the following Lewis structure:

\[\stackrel{..}{\stackrel{\cdot O\cdot}{\stackrel{..}{}}}+\stackrel{..}{\stackrel{\cdot O\cdot}{\stackrel{..}{}}}\rightarrow \stackrel{..}{\stackrel{:O:}{\stackrel{..}{}}} \stackrel{..}{\stackrel{:O}{\stackrel{..}{}}}\]

Now the oxygen on the right does not have an octet. The way to rectify this is to allow a second shared pair:

\[\stackrel{..}{\stackrel{O:}{\stackrel{..}{}}}\stackrel{..}{\stackrel{:O}{\stackrel{..}{}}}\]

Considering all shared electrons, each oxygen has an octet. The sharing of two pairs of electrons gives rise to a double bond, of which this is a particular example. Another way to represent it is:

\[\stackrel{..}{\stackrel{O-}{\stackrel{..}{}}} \stackrel{..}{\stackrel{O}{\stackrel{..}{}}}\]

What about the molecule $N_2$. $N$ has 5 valence electrons, so if we try to include only 1 shared pair, the Lewis structure looks like:

\[:\ddot{N}:\ddot{N}:\]

which only gives each atom 6 electrons. A double bond gives rise to:

\[\cdot \ddot{N}::\ddot{N}\cdot\]

which is only $7$ electrons to each. However, sharing three pairs of electrons leads to the structure:

\[:N:::N:\]

This is an example of a triple bond, which can be represented as

\[:N\equiv N:\]

Formal charges

Up to now, we have assumed that electrons are equally shared between the two atoms in a covalent bond. While it is true for homonuclear diatomic (in which the two atoms are the same species), there is no experimental basis for the assumption that this holds for aheteronuclear diatomic (in which the two atoms are different). To see what the consequence of this assumption is, consider the molecule $CO$, whose Lewis structure is easily shown to be

\[:C:::O:\]

If the shared electrons are divided equally between the C and O atoms, then C has a total of 5 electrons (including its lone pair) and O has a total of 5 electrons (including its lone pair). Normally, C has 4 valence electrons and O has 6. Thus, formally, C has an extra electron, giving it a formal charge of -1. Oxygen formally has one electron too few, giving it a formal charge of +1. Formal charges are included in Lewis structures near the appropriate atoms:

\[\stackrel{-1}{:C:}:\stackrel{+1}{:O:}\]

Generally, the formal charge is given by:

$\begin{displaymath} {\rm Formal\ charge} = {\rm Group\ number} - {\rm number\ of... ...\ pairs} - {1 \over 2}({\rm number\ of\ electrons\ in\ bonds}) \end{displaymath}$

Formal charges can be used to distinguish between several possible Lewis structures. Generally, the Lewis structure with the smallest formal charges on individual atoms will be the correct one.

General rules for drawing Lewis structure

The following is a list of rules that can be used to determine the Lewis structure of a molecule:

Count up the total number of valence electrons. First add up the group numbers of all atoms in the molecule. If the molecule is an anion, add one electron for each unit of charge on the anion. If it is a cation, subtract one electron for each unit of charge on the cation.
Calculate the total number of electrons that would be needed for each atom to have an octet (or doublet for H).
Subtract the result of step 1 from the result of step 2. This is the total number of shared or bonding electrons.
Assign two bonding electrons to each bond.
If bonding electrons remain, assign them in pairs making some of the bonds double or triple bonds. (Usually, only C,N,O, and S can form double bonds, and only C and N can form triple bonds). There may be more than one way to do this. Keep all possible structures that result.
Assign remaining electrons as lone pairs, giving octets to all atoms except H.
Determine the formal charges and put them next to the appropriate atoms. (A formal charge of 0 need not be written explicitly). Check that the formal charges add up to the total charge on the molecule/ion. Do this for all structures obtained in step 5. The structure with the smallest formal charges should be considered as the preferred structure.

Example: Ethylene \((C_2 H_4)\).
Step 1: Total valence electrons = 24 + 4 = 12 Step 2: Total electrons needed for octets/doublets = 28 + 42 = 24 Step 3: Total shared/bonding electrons = 24 - 12 = 12. Total electrons in lone pairs = 12 - 12 = 0. Step 4: Figure 2: $\begin{figure}\begin{center} \leavevmode \epsfbox{lec11_fig3.ps} {\small} \end{center}\end{figure}$* Note that, in order to use up all of the bonding electrons, the \(C=C\) double bond is necessary. For this case, all of the formal charges work out to be 0, which is shown straightforwardly.

Example: Ethylene $(C_2 H_4)$.

Step 1: Total valence electrons = 2*4 + 4 = 12

Step 2: Total electrons needed for octets/doublets = 2*8 + 4*2 = 24

Step 3: Total shared/bonding electrons = 24 - 12 = 12. Total electrons in lone pairs = 12 - 12 = 0.

Step 4:

Figure 2:

$\begin{figure}\begin{center} \leavevmode \epsfbox{lec11_fig3.ps} {\small} \end{center}\end{figure}$

Note that, in order to use up all of the bonding electrons, the $C=C$ double bond is necessary. For this case, all of the formal charges work out to be 0, which is shown straightforwardly.

Resonant structures

Sometimes more than one viable Lewis structure is possible. Consider the ozone $O_3$ molecule, whose possible Lewis structures are:

\[\stackrel{..}{\stackrel{O:}{\stackrel{..}{}}}\stackrel{\stackrel{..}{:O:}}{\stackrel{}{_{_{+1}}}}\stackrel{\stackrel{_{_{-1}}}{..}}{\stackrel{O:}{\stackrel{..}{}}}\]

and

\[\stackrel{\stackrel{_{_{-1}}}{..}}{\stackrel{:O:}{\stackrel{..}{}}}\stackrel{\stackrel{..}{O}}{\stackrel{}{_{_{+1}}}}\stackrel{..}{\stackrel{:O}{\stackrel{..}{}}}\]

From these, one would predict that the two bond $OO$ bond lengths are different. However, experimentally, they are known to be the same, so it seems that the octet rule is breaking down. A "bogus" fix to this problem is to define something called resonantstructures. In a resonant structure, the molecule is said to exist in a hybrid state between the two. In a sense, the molecule is both structures at once. The resonant structure is represented as:

\[\left \{ \stackrel{..}{\stackrel{O:}{\stackrel{..}{}}}\stackrel{\stackrel{..}{:O:}}{\stackrel{}{_{_{+1}}}}\stackrel{\stackrel{_{_{-1}}}{..}}{\stackrel{O:}{\stackrel{..}{}}} \longleftrightarrow \stackrel{\stackrel{_{_{-1}}}{..}}{\stackrel{:O:}{\stackrel{..}{}}}\stackrel{\stackrel{..}{O}}{\stackrel{}{_{_{+1}}}}\stackrel{..}{\stackrel{:O}{\stackrel{..}{}}} \right \}\]

Breakdown of the octet rule

Given the crudeness of the octet rule, it should not be surprising that there are cases for which it does not work. Below, some of examples of octet rule breakdown are described:

Molecules with an odd number of electrons

The octet can only work for molecules with an even number of valence electrons. While this covers most cases, there are a few molecules that have an odd number of electrons. As an example, consider the molecule NO. N has 5 valence electrons and O has 6, giving a total of 11. Thus, one of the atoms in the Lewis structure for NO cannot have an octet. The correct Lewis structure for NO is:

\[ \ ^{:\dot{N}:} \stackrel{:\ddot{O}}{\stackrel{..}{}}\]

which gives a formal charge of 0 to each atom. Giving an octet to N and 7 electrons to the oxygen would give higher formal charges.

Octet deficient molecules

Some molecules are stable, even though they possess too few electrons to achieve octets for all atoms. As an example of this, consider the $BF_3$ molecule. Applying the standard rules would lead to a Lewis structure of the form

Figure 3:
$\begin{figure}\begin{center} \leavevmode \epsfbox{lec11_fig4.ps} {\small} \end{center}\end{figure}$

which gives a double bond to Fluorine as well as a formal charge of +1. Fluorine can never have a positive formal charge, and formation of a double bond between F and B is incorrect. The way out of this mess is to deny the boron atom a full octet. Rather, we let B have a sextet, which gives a Lewis structure:

Figure 4:
$\begin{figure}\begin{center} \leavevmode \epsfbox{lec11_fig5.ps} {\small} \end{center}\end{figure}$

Although boron only has 6 electrons around it, all formal charges are 0, and the molecule is stable.

Valence shell expansion

For molecules involving atoms in periods greater than the second, Lewis structures can become more complicated. Consider the example of the molecule $SF_6$.

Let us apply the rules to this case:

Step 1: Total valence electrons = 6 + 6*7 = 48

Step 2: Total electrons needed for octets = 7*8 = 56

Step 3: Total shared/bonding electrons = 56 - 48 = 8.

Now we have a problem because we have 6 bonds but only 8 electrons to put in them. If each bond is to receive a pair of electrons, we would need 12 shared electrons rather than 8.

The solution to this problem is to allow for more bonding electrons. Extra bonding electrons are assigned to the central atom, S in this case, which can accommodate an expanded valence shell due to its relatively high atomic number. Now, step 2 becomes:

Step 2: Total electrons needed for 6 octets + one 12-tet = 60.

Step 3: Total shared/bonding electrons = 60 - 48 = 12.

Step 4:

Figure 5:
$\begin{figure}\begin{center} \leavevmode \epsfbox{lec11_fig6.ps} {\small} \end{center}\end{figure}$

This example suggests that step 6 of the usual Lewis structure rules should be replaced by

6'. Assign lone pairs to the terminal atoms to give them octets. If any electrons remain, assign them to the central atoms as lone pairs.

Rule 5 now becomes irrelevant.

Another example is the $I_{3}^{-}$ anion. According to the rules:

Step 1: Total valence electrons = 7*3 + 1 = 22

Step 2: Total electrons for octets = 8* 3 = 24

Step 3: Total shared/bonding electrons = 24 - 22 = 2.

Two electrons is not enough for the two bonds in the $I_{3}^{-}$ ion. Thus, one of the iodines will have its valence shell expanded. We need two extra electrons, so step 2 now becomes:

Step 2: Total electrons for 2 octets + one 10-tet = 26.

Step 3: Total shared/bonding electrons = 26 - 22 = 4.

The resulting Lewis structure is:

Figure 6:
$\begin{figure}\begin{center} \leavevmode \epsfbox{lec11_fig7.ps} {\small} \end{center}\end{figure}$

\includegraphics[scale=1.0]{ClO3+.eps} — Figure 7: VSEPR shapes

$\includegraphics[scale=1.0]{VSEPR_shapes.eps}$

The reason for these assignments will become clear through a study of several examples:

1. $CO_2$
The Lewis diagram for $CO_2$ is: \[\stackrel{..}{\stackrel{O}{\stackrel{..}{}}}\stackrel{::C::}{\stackrel{ \ }{ \ }}\stackrel{..}{\stackrel{O}{\stackrel{..}{}}}\]; Thus, the steric number of the central carbon is; \[SN =2+0=2\]
and, according to the assignment table, the shape is linear:; \[O=C=O\]
The linear shape can be understood on the basis of the electron repulsion principle: The central carbon is surrounded by two sets of shared electron pairs, which want to achieve a maximum separation between them. The maximum separation they can achieve is at a $180^\circ$ angle between the pairs, giving rise to the linear configuration.

2. $ClO_{3}^{+}$
The Lewis diagram for this cation is was shown earlier. The steric number of the central chlorine is: \[SN=3+0=3\]
and the geometry, according to the table, is trigonal planar:
Figure 8:; $\begin{figure}\begin{center} \leavevmode \epsfbox{lec11_fig9.ps} {\small} \end{center}\end{figure}$; Again, this geometry can be seen to achieve a maximum separation between the three pairs of shared electrons around the chlorine.
3. $NO_{2}^{-}$
The Lewis structure for this anion is a resonant form:: \[\left \{ \left [ \stackrel{..}{\stackrel{:O}{\stackrel{..}{}}}\stackrel{..}{\stackrel{:N:}{\stackrel{ \ }{ \ }}}\stackrel{..}{\stackrel{:O}{\stackrel{..}{}}}\right ]^{-1}\longleftrightarrow \left [ \stackrel{..}{\stackrel{O:}{\stackrel{..}{}}}\stackrel{..}{\stackrel{:N:}{\stackrel{ \ }{ \ }}}\stackrel{..}{\stackrel{O:}{\stackrel{..}{}}}\right ]^{-1}\right \}\]
The steric number of the nitrogen is; \[SN=2+1=3\]
Hence the basic shape should be trigonal planar. However, one of the atoms will be missing from this structure and replaced by the lone pair on the nitrogen. Hence its shape will actually be bent, with a bond angle less than $120^\circ$ owing to the fact that the lone pair is spatially more delocalized than the bonding pairs. This spatial delocalization of the lone pair causes the bonding pair to be repelled from the lone pair at greater distances than could be achieved by a 120 degree angle, giving rise to the smaller bend angle in the structure:; Figure 9:; $\begin{figure}\begin{center} \leavevmode \epsfbox{lec11_fig10.ps} {\small} \end{center}\end{figure}$; Because of the resonant structure, the bond lengths are the same.

Contributors and Attributions

Mark Tuckerman (New York University)