Drawing Lewis structures F F C F F

Lewis structures help us to see precisely how the atoms in a molecule are bonded together. They also help us predict the shapes and the properties of the molecules. As it turns out, Lewis structures are relatively easy to draw, as long as you follow the five steps...

Step 1: Write down the central atom. Usually the central atom will simply be the atom that there’s just one of. For example in CH4, the C is the central atom and in NF3, the N is the central atom. In BrF5, the central atom is the Br, and in SO2, the central atom is the S. Sometimes, there is no central atom, for instance in HBr, the H and the Br are just side by side. Likewise for the molecule F2.

H H C H H Step 2: Arrange the other atoms around the central atom. In CH4, after the central C is written down… the four H’s can simply be placed around it… H H C H H

F F N F Step 2: Arrange the other atoms around the central atom. In NF3, after the central N is written down… the three F’s can simply be placed around it… F F N F By the way, it really doesn’t matter whether these surrounding atoms go on top, bottom, left or right.

F F F Br F F Step 2: Arrange the other atoms around the central atom. In BrF5, after the central Br is written down… the five F’s can simply be placed around it… F F F Br F F Note that when there are more than four atoms surrounding the central one, you just scoot them around a bit to make room for them all.

O S O Step 2: Arrange the other atoms around the central atom. And in SO2, after the central S is written down… the two O’s can simply be placed around it… O S O

Step 3: Make bonds between the central atom and each of the surrounding ones. A bond is simply a pair of electrons being shared between two atoms. These are represented by a pair of dots. So for CH4… H We simply add electron pairs between the C and each H. H C H H

Step 3: Make bonds between the central atom and each of the surrounding ones. For NF3. We simply add electron pairs between the N and each F. F F N F

Step 3: Make bonds between the central atom and each of the surrounding ones. For BrF5. F We simply add electron pairs between the Br and each F. F F Br F F

Step 3: Make bonds between the central atom and each of the surrounding ones. And for SO2. We simply add electron pairs between the S and each O. O S O

X X Step 4: Complete the octets on every atom that needs it. An octet is simply a group of eight electrons – that’s what it takes to fill an atom’s valence level. With two above, two below, two to the left and two to the right, it’s easy to tell in an instant whether an atom’s octet is complete or not: X X Not complete Complete

Step 4: Complete the octets on every atom that needs it. Which atoms need an octet? All of the ones in group IV on the periodic table and up. (in other words, all those from C’s column to the right) I VIII II III IV V VI VII C

H H C H H Step 4: Complete the octets on every atom that needs it. So for CH4… H C H H is in group I; it does not need an octet, so leave all four H’s the way they are. H is in group I H C is in group IV C is in group IV; it does need an octet. But since it already has one, just leave it alone too! I II III IV V VI VII VIII H C

F F N F Step 4: Complete the octets on every atom that needs it. For NF3… F is in group VII F is in group VII; so it needs an octet. Just add electron pairs until all the F’s are complete: F N F N is in group V; so it also needs an octet. Just add one more pair right here. N is in group V I II III IV V VI VII VIII N F

F Br Step 4: Complete the octets on every atom that needs it. For BrF5… F is in group VII; so it needs an octet. Just add electron pairs until all the F’s are complete: Br is also in group VII but since it already has 10 drawn around it… I II III IV V VI VII VIII there’s no need to add any more. Br

O S O Step 4: Complete the octets on every atom that needs it. And for SO2… O is in group VI; so it needs an octet. Just add electron pairs until both the O’s are complete: O is in group VI O S O S is also in group VI. It also needs an octet. Just add two more pairs – top and bottom. I II III IV V VI VII VIII O S

Let’s review what we’ve done so far. Step 1: We wrote down the central atom. Step 2: We arranged the other atoms around the central atom. Step 3: We made bonds between the central atom and each surrounding atom. Step 4: We completed the octets on each atom that needed it (group IV and up). These first four steps give us a “rough draft” of our Lewis structure. In step 5, we need to edit that rough draft to make sure it contains the appropriate number of dots (electrons).

Step 5: Compare how many electrons we’ve drawn to how many we should have drawn. So for CH4… We have drawn 8 electrons. And we now go back and see what groups the atoms are in: C is in group IV, so it contributes 4 valence electrons. H is in group I, so each of the four H’s contributes 1 valence electron. 4 + 4(1) = 4 + 4 = 8 C H This means our Lewis structure is supposed to have 8 electrons.

Step 5: Compare how many electrons we’ve drawn to how many we should have drawn. Since we have drawn 8 electrons and we were supposed to draw 8 electrons, there is no need to edit the rough draft at all. It is correct the way it is: C H

Step 5: Compare how many electrons we’ve drawn to how many we should have drawn. For NF3… 8 + 8 + 8 + 2 = 26 We have drawn 26 electrons. And we now go back and see what groups the atoms are in: N is in group V, so it contributes 5 valence electrons. F is in group VII, so each of the three F’s contributes 7 valence electrons. 5 + 3(7) = 5 + 21 = 26 N F This means our Lewis structure is supposed to have 26 electrons.

Step 5: Compare how many electrons we’ve drawn to how many we should have drawn. Since we have drawn 26 electrons and we were supposed to draw 26 electrons, there is no need to edit the rough draft at all. It is correct the way it is: N F

Step 5: Compare how many electrons we’ve drawn to how many we should have drawn. For BrF5… 8 + 8 + 8 + 8 + 8 = 40 We have drawn 40 electrons. And we now go back and see what groups the atoms are in: Br is in group VII, so it has 7 valence electrons. F is in group VII, so each of the five F’s has 7 valence electrons. 7 + 5(7) = 7 + 35 = 42 Br F This means our Lewis structure is supposed to have 42 electrons.

Step 5: Compare how many electrons we’ve drawn to how many we should have drawn. Since we have drawn 40 electrons but we were supposed to draw 42 electrons, we must add one more pair. F Whenever electron pairs must be added, simply draw them in on the central atom: Since it’s a little crowded down there, a little loop helps to show that this pair of electrons belongs on the Br. F F Br F F

Step 5: Compare how many electrons we’ve drawn to how many we should have drawn. Since we now have 42 electrons – the exact number we are supposed to have – the Lewis structure is finished. F Notice that the Br now has a total of 12 electrons: F F Br That’s way more than an octet, but sometimes it’s necessary to go beyond a stable octet when it comes to drawing Lewis structures. F F

Y X Y What if you need to add two pairs of electrons? Again just add them both to the central atom: Y X Y 24 22 Total electrons drawn: 20 Since we

And what if your rough draft has too many dots and you need to delete a pair or two? Let’s try SO2… 8 + 8 + 2 + 2 = 20 We have drawn 20 electrons. And we now go back and see what groups the atoms are in: S is in group VI, so it has 6 valence electrons. O is in group VI, so each of the two O’s has 6 valence electrons. 6 + 2(6) = 6 + 12 = 18 S O This means our Lewis structure is supposed to have 18 electrons.

And what if your rough draft has too many dots and you need to delete a pair or two? Since we have drawn 20 electrons but we were supposed to draw only 18 electrons, we must delete one pair. First, we need to define some terms: “Bonding pairs” are the electrons drawn between two atoms, like these two… “Nonbonding pairs” are all the others, like these… Whenever electron pairs must be deleted, erase a nonbonding pair on the central atom O S O Then erase a nonbonding pair on one of the surrounding atoms And then add in a pair of bonding electrons between those same two atoms

And what if your rough draft has too many dots and you need to delete a pair or two? Since we now have 18 electrons – the exact number we are supposed to have – the Lewis structure is finished. This group of four electrons is known as a “double bond.” O S O So whenever you are instructed to “make a double bond,” remember to erase one nonbonding pair on the central atom and one nonbonding pair on a surrounding atom, then add in a bonding pair between those same two atoms.

And what if your rough draft has too many dots and you need to delete a pair or two? Because you have erased four electrons and drawn in two, you have effectively deleted two electrons from your rough draft. Notice how each atom still has a stable octet. 8 8 8 O S O

Because this last concept is kind of tricky, let’s try one more like it: Let’s say we wanted to draw a Lewis structure for SeO3. Step 1: Write down the central atom: 8 + 8 + 8 + 2 = 26 Step 2: Arrange the three O’s around it: Step 3: Make bonds between the Se and each of the three O’s: O Step 4: Complete the octects on Se (group VI) and each O (group VI) Se O O Step 5: Compare the number of electrons we have drawn… Since we with the number we should have drawn: SeO3 = 6 + 3(6) = 6 + 18 = 24 Since we have 26 dots, but are only supposed to have 24, we erase two nonbonding pairs and draw in a bonding pair between the two.

Y X Y X Y What if you need to delete two pairs? 8 8 8 Sometimes this requires making two double bonds: Y X Y 16 18 Total electrons drawn: 20 Notice how each atom still has a stable octet. 8 8 Other times it requires making a triple bond: X Y Since we 10 12 Total electrons drawn: 14 Again note how each atom still has a stable octet.

The final question to answer: what about drawing Lewis structures for ions? Answer: you draw them the exact same way as you do molecules – following the same 5 steps described above. But in step 5, when it comes to determining how many electrons are supposed to be in the Lewis structure, you must incorporate the charge of the ion into your total. If it’s a positive ion, then electrons must have been lost, so the charge tells us how many to subtract from the total. And if it’s a negative ion, then electrons must have been gained, so the charge tells us how many to add to the total. Since we

For example, without the charge, NF41+ should have 5 + 4(7) = 5 + 28 = 33 electrons. But the charge of 1+ tells us that one electron has been lost, so that would bring the total down to 33 – 1 = 32. Or consider the SO3 2- ion. Without the charge it should have 6 + 3(6) = 6 + 18 = 24 electrons. But the charge of 2- tells us that two electrons have been gained, so that would bring the total up to 24 + 2 = 26. Since we

[ ] O O C O So let’s see what this looks like: 8 + 8 + 8 + 2 = 26 2- Let’s say we wanted to draw a Lewis structure for CO32-. Step 1: Write down the central atom: 8 + 8 + 8 + 2 = 26 Step 2: Arrange the three O’s around it: Step 3: Make bonds between the C and each of the three O’s: O Step 4: Complete the octects on C (group IV) and each O (group VI) [ ] 2- O C O Step 5: Compare the number of electrons we have drawn… with the number we should have drawn: CO32- = 4 + 3(6) = 4 + 18 = 22 Since we + 2 = 24 (because of the 2- charge) Since we have 26 dots, but are only supposed to have 24, we erase two nonbonding pairs and draw in a bonding pair between the two.

Now try problems 1-10 on the Lewis structures worksheet, using the left-hand boxes for practice, then copying your final answers into the right-hand boxes. Use the next few slides if you get stuck, or to check your final answers. Since we

H H N H 1. NH3 Step 1: Write down the central atom: Step 2: Arrange the three H’s around it: Step 3: Make bonds between the N and each of the three H’s: H Step 4: Complete the octect on N (group V) but NOT on the H’s (group I) H N H Step 5: Compare the number of electrons we have drawn… Since we 8 with the number we should have drawn: NH3 = 5 + 3(1) = 8 Since we have 8 dots, and we are supposed to have 8, we are all finished!

F F B F 2. BF3 8 + 8 + 8 = 24 Step 1: Write down the central atom: Step 2: Arrange the three F’s around it: 8 + 8 + 8 = 24 Step 3: Make bonds between the B and each of the three F’s: F Step 4: Complete the octects on the F’s (group VII) but NOT on the B (group III) Step 5: Compare the number of electrons we have drawn… F B F Since we with the number we should have drawn: BeF3 = 3 + 3(7) = 3 + 21 = 24 Since we have 24 dots, and we are supposed to have 24, we are finished.

H O H 3. H2O Step 1: Write down the central atom: Step 2: Arrange the two H’s around it: Step 3: Make bonds between the O and each of the two H’s: Step 4: Complete the octect on O (group VI) but NOT on the H’s (group I) H O H Step 5: Compare the number of electrons we have drawn… Since we 8 with the number we should have drawn: H2O = 2(1) + 6 = 8 Since we have 8 dots, and we are supposed to have 8, we are all finished.

4. BrF3 Step 1: Write down the central atom: Step 2: Arrange the three F’s around it: 8 + 8 + 8 + 2 = 26 Step 3: Make bonds between the Br and each of the three F’s: F Step 4: Complete the octects on Br (group VII) and each F (group VII) F Br F Step 5: Compare the number of electrons we have drawn… Since we with the number we should have drawn: BrF3 = 7 + 3(7) = 7 + 21 = 28 Since we have 26 dots, but are supposed to have 28, we simply add an extra pair on the central atom.

5. SF6 8 + 8 + 8 + 8 + 8 + 8 = 48 Step 1: Write down the central atom: Step 2: Arrange the six F’s around it: F Step 3: Make bonds between the S and each of the six F’s: F F S Step 4: Complete the octects on S (group VI) and each F (group VII) F F Step 5: Compare the number of electrons we have drawn… F Since we with the number we should have drawn: SF6 = 6 + 6(7) = 6 + 42 = 48 Since we have 48 dots, and we are supposed to have 48, we are all finished.

O C O 6. CO2 8 + 2 + 8 + 2 = 20 Step 1: Write down the central atom: Step 2: Arrange the two O’s around it: Step 3: Make bonds between the C and each of the two O’s: Step 4: Complete the octects on C (group IV) and each O (group VI) O C O Step 5: Compare the number of electrons we have drawn… Since we with the number we should have drawn: CO2 = 4 + 2(6) = 4 + 12 = 16 Since we have 20 dots, but are only supposed to have 16, we must delete two pairs. Erase these nonbonding pairs… Then make a double bond… Then erase these pairs… And make a second double bond. That’s it.

7. HF Step 1: Write down the central atom: (actually, since there are only two atoms, just write them down side-by-side). Step 2: (With no central atom, this step doesn’t apply.) Step 3: Make a bond between the H and the F: Step 4: Complete the octect on F (group VII) but NOT on H (group I) H F Step 5: Compare the number of electrons we have drawn… Since we 8 with the number we should have drawn: HF = 1 + 7 = 8 Since we have 8 dots, and are supposed to have 8, we are finished.

8. Br2 Step 1: Write down the central atom: (actually, since there are only two atoms, just write them down side-by-side). Step 2: (With no central atom, this step doesn’t apply.) Step 3: Make a bond between the two Br’s: 8 + 6 =14 Step 4: Complete the octects on both Br’s (group VII) Br Br Step 5: Compare the number of electrons we have drawn… Since we with the number we should have drawn: Br2 = 2(7) = 14 Since we have 14 dots, and are supposed to have 14, we are finished.

9. CO Step 1: Write down the central atom: (actually, since there are only two atoms, just write them down side-by-side). Step 2: (With no central atom, this step doesn’t apply.) Step 3: Make a bond between the C and the O: 8 + 6 =14 Step 4: Complete the octects on both C (group IV) and O (group VI) C O Step 5: Compare the number of electrons we have drawn… Since we with the number we should have drawn: CO = 4 + 6 = 10 Since we have 14 dots, but are only supposed to have 10, we must delete two pairs. Erase these nonbonding pairs… Then make a double bond… Then erase these pairs… And make a triple bond. That’s it.

10. SO32- Step 1: Write down the central atom: 8 + 8 + 8 + 2 = 26 Step 2: Arrange the three O’s around it: O Step 3: Make bonds between the S and each of the three O’s: Step 4: Complete the octects on S (group VI) and each O (group VI) [ ] 2- O S O Step 5: Compare the number of electrons we have drawn… Since we with the number we should have drawn: SO32- = 6 + 3(6) = 6 + 18 = 24 + 2 = 26 (because of the 2- charge) Since we have 26 dots, and are supposed to have 26, we are finished… But don’t forget to add the brackets and the charge

Now that you have learned how to draw Lewis structures, try the rest of the worksheet. Since we