Concepts of Chemical Bonding

Lewis Structures

We know that the formation of covalent bonds is achieved by an overlap of orbitals between two atoms that share a pair of valence electrons. The bonds are most often shown with either two dots or a line, and these representations are called Lewis symbols or Lewis structures.

For example:

So the bonds can be represented either with a line or a pair of electrons (two dots) and therefore, we can say that Lewis structures are electron-dot representations for molecules. Once we know this concept, it is relatively easy to interpret the structure. However, coming up with a reasonable Lewis structure from a formula is not as obvious. There are a few things to keep in mind when drawing a Lewis structure. These are the correct placement of the atoms in the molecule, determining the number of electrons involved in the bonding and non-bonding interactions, and keeping track of these electrons.

While hydrogen and the following elements have few electrons, the elements below the second row of the periodic table, have quite complex electron configurations and it may get challenging to find the right way of involving them in chemical bonding. To make this easier and avoid any possible confusion, Lewis suggested focusing only on the valence electrons since the covalent bonds are formed by the valence electrons.

As a reminder, valence electrons are the ones in the outermost energy level (principal quantum number – n). For example, the electron configuration of carbon is 1s²2s²2p². Out of these, the ones in the second energy level, n=2, are the 2s²2p²four electrons. So, carbon, being in group 4, has four valance electrons and that is true for all the other elements:

The number of valence electrons corresponds to the group number of the element.

For example, the following are the valence electrons for each element.

Na: 1, Mg: 2, B: 3, C: 4, N: 5.

Drawing Lewis Structures

In short, these are the steps you need to follow for drawing a Lewis structure:

1. Write the correct skeletal structure for the molecule.

* Hydrogen atoms are always terminal (only one bond)

* Put more electronegative elements in terminal positions

2. Sum the valence electrons from all the atoms.

3. Use a pair of electrons to form a bond between each pair of bound atoms.

4. Add the remaining electrons to satisfy the octet for a more electronegative atom first.

5. If any atoms lack an octet, make a double or triple bond to give them an octet.

Now, let’s go over these in more detail by drawing the Lewis structure of HCl.

The first thing in drawing a Lewis structure is to place the atoms in the correct location and since we only have two atoms here, we can only place them next to each other:

H Cl

And in general, as well, remember to always put the hydrogen on the periphery because it has one electron and can only make one bond.

You can see the bonding patterns of the most common elements in organic chemistry in the valency and formal charges post:

Next, sum the valence electrons. One from the H (group 1) and 7 from the Cl (group 7). Therefore, we have 8 valance electrons:

Out of these eight, two electrons must be used to make a bond between the atoms:

This leaves us with six electrons:

Remember to put the remaining electrons on the more electronegative atom first and then on the other(s) if more electrons are still available.

So, six electrons go to the Cl:

These electrons are called non-bonding electrons or lone pairs of electrons. And instead of saying the Cl has six electrons, even though it is not wrong to say so, we normally count them in pairs and say that the Cl has three lone pairs of electrons. The lone pairs are important because they define the geometry of the molecule and participate in chemical reactions.

Finally, check to make sure that the following elements obey the octet rule which applies to C, N, O, F, and all the other halogens. The octet rule states that some elements tend to have eight electrons around them, whether by bonding pattern or in their ionic form.

The bonding (shared) electrons are included in the electron count of each element:

Hydrogen does not follow the octet rule since it has one electron and can only form one bond which is referred to as the Duet rule.

In Lewis structures, the bonds can be shown either by dots or lines. For example, the previous structures can also be shown as follows:

Double and Triple Bonds in Lewis Structures

Let’s now draw the Lewis structure of CO₂. As always, the first step is to place the atoms correctly. Oxygen being more electronegative goes on the periphery:

For the valence electrons, we have 4 from the carbon and 2 x 6 = 12 from two oxygens, giving 16 valence electrons in total:

One bond between the carbon and each oxygen takes 4 electrons and we have 12 electrons left:

Place 6 on each oxygen and check for the octets:

The oxygens have 8 electrons. However, the carbon has only four and in these cases, you need to move one of the lone pairs, from the element that has an octet, to the element lacking an octet to make another bond:

The same is done for the other carbon since it still lacks an octet:

At this point, all the atoms have 8 electrons. Notice that there are four electrons between the carbon and each oxygen which means they make two double bonds:

Lewis Structure of Organic molecules

For the third part, let’s see how to work around a problem with a larger organic molecule where the placement of atoms is given and you need to add the bonds and any possible lone pairs:

For example, draw an acceptable Lewis structure for the following compound:

Count the total number of valence electrons:

Make a bond between each neighboring atom:

Count the remaining electrons and place them on the atoms in the order of their electronegativity:

Identify atoms lacking octet:

Move one lone pair from the oxygen to make a double band:

And this gives us the Lewis structure for this organic molecule.

Now, let’s pay attention to something interesting here. You may wonder why we did not move the lone pair from the nitrogen instead of the oxygen in the following way:

That is a good question and the answer is that it can be done, either way, i.e. we could have moved either of these lone pairs:

A and B are both correct Lewis structures of this molecule and are called resonance structures. Resonance structures can be interconverted to one another by moving some of the electrons:

The arrows that show the movement of electrons in resonance forms are called curved arrows which are used more often in organic chemistry.

Notice also that the oxygen and the nitrogen are charged in the resonance form B. These are called formal charges.

Check the following posts with practice problems for these important concepts:

Check Also

Practice

Determine the Lewis structures of the following molecules:

(a) BF₃	(b) CH₂O	(c) HCN	(d) BeCl₂	(e) CH₂Cl₂
(f) SOCl₂	(g) SO₂	(h) PCl₅	(i) XeO₄	(j) NCl₃
(k) SiCl₄	(l) SF₂	(m) H₂S	(n) SO₃	(o) COCl₂
(p) PCl₃	(q) OF₂	(r) BrF₅	(s) N₂O	(t) SF₆
(u) POCl₃	(x) XeF₂	(y) XeF₄	(z) C₂H₂

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(a) B is the central atom, so we can draw the skeletal structure:

There is a total of 3 + 3 x 7 = 24 electrons, and halogens on terminal positions are always going to have 3 lone pairs of electrons. Therefore, 3 x 6 = 18 electrons on the fluorenes, and the remaining 6 are used to make the three covalent bonds:

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(b) The carbon is the central atom, so we can draw the skeletal structure:

There is a total of 4 + 2×1 + 6 = 12 electrons, and 6 are already used for making the bond. The remaining 6 go to oxygen as it is the most electronegative atom.

Because the carbon lacks an octet, we use one lone pair from the oxygen to make a π bond with the carbon, and thus, a double bond is formed.

In general, remember that oxygens on terminal positions are always going to have 2 bonds and 2 lone pairs of electrons unless there is a formal charge on them.

Knowing these patterns for bonding, lone pairs, and formal charges will make these types of exercises a lot easier.

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(c) Carbon is the central atom, so we can draw the skeletal structure:

There is a total of 1 + 4 + 5 = 10 valence electrons, and we use four of them to make the bonds.The remaining 6 go on the nitrogen as lone pairs:

Because the carbon lacks an octet, we use two lone pairs from the nitrogen to make two additional bonds:

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(d) Be is the central atom, so we can draw the skeletal structure:

There is a total of 2 + 2×7 = 16 valence electrons, and we use four of them to make the bonds. The remaining 12 go on the Cl atoms as lone pairs:

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(e) Carbon is the central atom, so we can draw the skeletal structure:

There are 4 + 2 + 2×7 = 20 electrons, and 8 have been used to make four bonds. The remaining 12 go on the two chlorines as lone pairs:

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(f) Sulfur is the central atom, so we can draw the skeletal structure:

There are 6 + 6 + 2×7 = 26 electrons, and 6 are used to make 3 bonds. Two chlorines and the oxygen take 3 lone pairs, and the remaining one goes to the sulfur:

All the atoms have an octet, however, there are three lone pairs on the oxygen which gives it a -1 formal charge. Therefore, one of the lone pairs on the oxygen is used to make another bond with the sulfur, and even if it exceeds the octet, that is acceptable for elements in the third row and below because they have d orbitals.

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(g) Sulfur is the central atom, so we can draw the skeletal structure:

There are 3×6 = 18 electrons, and 4 of them are used to make 2 bonds. The two oxygens take 6 lone pairs, and the remaining one goes to the sulfur:

As it is drawn, the problems with this structure are that the sulfur lacks an octet and the oxygens have only one bond and three lone pairs. Remember, the normal valency of oxygens is having two bonds and two lone pairs otherwise a formal charge needs to be assigned.

Therefore, one lone pair from each oxygen is used to make an additional bond with the sulfur:

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(h) Phosphorous is the central atom, so we can draw the skeletal structure:

There are 5×7 + 5 = 40 electrons, out of which, 10 are used to make 5 covalent bonds. All the remaining 30 are divided between the five chlorine atoms, each taking 6 electrons as 3 lone pairs:

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(i) Xe is the central atom, so we can draw the skeletal structure:

There are 4×6 + 8 = 32 electrons and this time, instead of putting three lone pairs on the oxygen, we are going to directly add double bonds to leave two lone pairs for each oxygen:

All the electrons are used: 4 double bonds = 16, 8 lone pairs = 16.

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(j) N is the central atom and there are 5 + 3×7 = 26 electrons. Each chlorine taking three lone pairs leaves the nitrogen with one lone pair:

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(k) Silicon is the central atom, so we can draw the skeletal structure:

There are 4 + 4×7 = 32 electrons, and 8 have been used to make four bonds. The remaining 24 go on the two chlorines as lone pairs:

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(l) Sulfur is the central atom. There are 6 + 2×7 = 20 electrons, and 4 of them are used to make 2 bonds. The two fluorines take 6 lone pairs, and the remaining four electrons go to the sulfur:

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(m) Sulfur is the central atom, so we can draw the skeletal structure:

There are 6 + 2 = 8 electrons, and 4 of them are used to make 2 bonds. The remaining four electrons go to the sulfur:

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(n) Sulfur is the central atom, so we can draw the skeletal structure:

There are 3×6 + 6 = 24 electrons, and 6 of them are used to make 3 bonds. Three oxygens take 6 lone pairs and make an additional bond with the sulfur.

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(o) The carbon is the central atom, so we can draw the skeletal structure:

There is a total of 4 + 2×7 + 6 = 24 electrons, and 6 are already used for making the bond. The remaining 18 go to oxygen and the chlorine atoms as lone pairs. Because the carbon lacks an octet, we use one lone pair from the oxygen to make a π bond with the carbon, and thus, a double bond is formed.

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(p) P is the central atom and there are 5 + 3×7 = 26 electrons. Each chlorine taking three lone pairs leaves the nitrogen with one lone pair:

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(q) Oxygen is the central atom:

There are 6 + 2×7 = 20 electrons, and 4 of them are used to make 2 bonds. The two fluorines take 6 lone pairs, and the remaining four electrons go to the sulfur:

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(r) Br is the central atom, so we can draw the skeletal structure:

There are 5×7 + 7 = 42 electrons, out of which, 10 are used to make 5 covalent bonds. The remaining 30 are divided between the five fluorine atoms, each taking 6 electrons as 3 lone pairs, and Br takes the last pair of electrons:

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(s) Oxygen is more electronegative, therefore, it goes in a terminal position:

There are 2×5 + 6 = 16 valence electrons and four are already taken to make two bonds. The remaining 12 are distributed between the oxygen and nitrogen:

Because the middle nitrogen lacks an octet, two lone pairs from the nitrogen are used to make two pi bonds with the middle nitrogen and the Lewis structure of this compound can be given as follows:

There are formal charges here because the middle nitrogen and the oxygen are not in their common bonding pattern:

This is one resonance structure and the other would be:

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(t) Sulfur is the central atom, so we can draw the skeletal structure:

There are 6 + 6×7 = 48 valence electrons, and 12 are taken for making 6 covalent bonds. Each fluorine takes three lone pairs, so there are no electrons left: 48 – (12 + 6×6) = 0.

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(u) P is the central atom, so we can draw the skeletal structure:

There are 5 + 3×7 + 6 = 32 valence electrons out of which, 8 are already taken to make four covalent bonds. From the remaining 24, 18 go to the Cl atoms as three lone pairs per each, three 6 to the oxygen as three lone pairs.

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(v) Sulfur is the central atom, so we can draw the skeletal structure:

There are 6 + 4×7 = 34 valence electrons, and 8 are taken for making 4 covalent bonds. Each fluorine takes three lone pairs, so there are 34 – (8 + 4×6) = 2 electrons left which go the sulfur as a lone pair:

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(w) Br is the central atom, so we can draw the skeletal structure:

There are 7 + 3×7 = 28 electrons and 6 are taken to make three covalent bonds. Each fluorine takes 6 electrons, therefore there are 28 – (6 + 3×6) = 4 electrons left, which go on the Br as two lone pairs:

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(x) Xe is the central atom, so we can draw the skeletal structure:

There are 2×7 + 8 = 22 electrons and 4 are taken to make 2 covalent bonds. Each fluorine takes 3 lone pairs, so there are 22 – (4+2×6) = 6 electrons left which go to Xe as 3 lone pairs:

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(y)Xe is the central atom, so we can draw the skeletal structure:

There are 4×7 + 8 = 36 electrons and 8 are taken to make 4 covalent bonds. Each fluorine takes 3 lone pairs, so there are 36 – (8+4×6) = 4 electrons left which go to Xe as 2 lone pairs:

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(z) Because the hydrogens are always in terminal positions, the carbons must be connected, and therefore, we can draw a preliminary skeletal structure to start with:

There are 2×4 + 2×1 = 10 valence, and 6 have been used to make 3 covalent bonds. Therefore, the remaining 4 go to the carbon atoms as lone pairs on each:

The carbons atoms are lacking octet, so we use the two lone pairs for making 2 additional bonds between them:

Determine the Lewis structure of each ion: (a) NH₄⁺, (b) H₃O⁺, (c) CN^– (d) SCN^–, (e) CO₃^2-, (f) ClO₃^–, (g) SO₄^2- (h) PO₄^3-(i) SO₃²^– (j) NO₂^– (k) BF₄^– (l) NO₃^–.

(a) NH₄⁺	(b) H₃O⁺	(c) CN^–	(d) SCN^–	(e) CO₃^2-
(f) ClO₃^–	(g) SO₄^2-	(h) PO₄^3-	(i) SO₃²^–	(j) NO₂^–
(k) BF₄^–	(l) NO₃^–

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(a) Nitrogen is the central atom, so we can draw the skeletal structure:

There is a positive charge, so to determine the number of electrons, we subtract 1 from the total number of valence electrons:

5 + 4 – 1 = 8 electrons

All the electrons are used to make the 4 covalent bonds, and the positive charge is on the nitrogen:

Remember, the formula for calculating the formal charge:

FC= V – (N + B)

Where:

V – number of valence electrons
N – number of nonbonding electrons
B – number of bonds

So, for the nitrogen it would be: 5 – (0 + 4) = +1

Try memorizing the bonding and formal charge patterns to make this process easier:

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The central atom is the oxygen, so we can draw the skeletal structure:

The total number of electrons is 6 + 3 – 1 = 8, out of which, 6 are used to make three covalent bonds, and one goes to the oxygen as a lone pair. The oxygen is positively charged since it only has one lone pair of electrons:

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Connect the atoms together and count the number of electrons by adding 1 since there is a negative charge:

#e^– = 4 + 5 + 1 = 10

Two electrons are used to make the covalent bond, therefore, the remaining 8 go to nitrogen as 3 lone pairs, and two to the carbon as a lone pair:

Use two lone pairs from the nitrogen to make a triple bond with the carbon which is left with a lone pair and therefore, a negative charge:

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Carbon is the central atom, so we can draw the skeletal structure:

There are 6 + 4 + 5 + 1 = 16 electrons, and 4 are already taken to make two covalent bonds. The remaining 12 are divided between the S and N as 3 lone pairs:

One possibility is to use two lone pairs from the nitrogen, and the other is to once form the nitrogen and one form the sulfur:

These two are resonance forms of the same compound so using either one would be correct.

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There are 4 + 3×6 + 2 = 24 electrons. The carbon goes in the middle, and the oxygens take 6 electron each as three lone pairs:

The carbon lacks an octet, so we use a lone pair from one oxygen to make a double with it. The other two oxygen are then negatively charged:

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There are 7 + 3×6 + 1 = 26 electrons. The chlorine goes in the middle with one lone pair, and the oxygens take 6 electrons each as three lone pairs:

As drawn, the oxygens are going to be negatively charged, which would mean a -3 ionic charge. Therefore, two oxygen will use one of their lone pairs to make a double with the chlorine. The other oxygen is then negatively charged:

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Sulfur is the central atom, so we can draw the skeletal structure:

There are 4×6 + 6 + 2 = 32 electrons, and 8 of them are used to make 4 bonds. The four oxygens take 3 lone pairs each and no electrons are left:

All the electrons have been used; however, two oxygens need to share a pair of electrons with the sulfur otherwise the ionic would be -4:

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Phosphorous is the central atom and there are 5 + 4×6 + 3 = 32 electrons. Give each oxygen three lone pairs:

All the electrons are used and the only thing to fix is the ionic charge. As drawn, it would be a -4, therefore, we give one lone pair from an oxygen to make a double bond with the P which can exceed the octet:

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Sulfur is the central atom, so we can draw the skeletal structure:

There are 3×6 + 6 + 2 = 26 electrons, and 6 of them are used to make 3 bonds. Three oxygens take 3 lone pairs each and one goes to the sulfur:

All the electrons have been used; however, one oxygen needs to share a pair of electrons with the sulfur otherwise the ionic would be -3:

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Nitrogen is the central atom, so we can draw the skeletal structure:

There are 5 + 2×6 + 1 = 18 electrons, and 4 are used to make the two covalent bonds. Both oxygens get 6 electrons as three lone pairs, and nitrogen gets one lone pair:

One lone pair from oxygen is used to make a π bond with the nitrogen and thus making the ionic charge -1:

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B is the central atom, so we can draw the skeletal structure:

There is a total of 3 + 4×7 + 1 = 32 electrons, and 8 are used to make the covalent bonds. Halogens on terminal positions are always going to have 3 lone pairs of electrons, so 4×6 = 24 electrons go on the fluorenes. The boron has a -1 charge since it has four bonds and it is in group 3: