Chemistry Lecture *39-. Lewis Pot Structures

Lewis dot structures are diagrams that show how atoms in a

molecule are bonded together. Simple Lewis dot structures can

be made by -following the -following steps-

I. Praw a dot diagram o-P each atom in the molecule.

2-. Petermine how many unpaired electrons are on each atom -

this is the number o£ bonds it will -Porna

3. Arrange the atoms so that an unpaired electron can pair up

with another unpaired electron.

4-. I-P you wish, draw a line to substitute -for a pair o-P shared

electrons.

5". check your work. Is each atom making the correct number

o-P bonds? Poes hydrogen now have two electrons, and do the

other atoms have eight?

Praw the Lewis structure -Por Hz.

H- -H^ H:H

Notice that each hydrogen -Porms one bond.

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Draw the Lewis structure -Por

H- H ' / 0;

H - - b :I

i * H

Notice that each hydrogen -Porms one bond. Oxygen -Perms two

bonds since it starts with two unpaired electrons.

Draw NH3

, A/- H ' H

A / - -

H

Notice that nitrogen starts with 3 unshared electrons, and -Perms

3 bonds.

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Draw

, c • H - H - n*

(J /+H. , ' c . - H ' —^ t t ' C i « pr

Notice that carbon starts with 4- unshared o-P electrons, and

•Porms 4- bonds.

This procedure works with simple molecules. But it is help-Pul to

use another procedure when the molecules become more

iplicated.come

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The -following procedure works -Por both simple and complicated

molecules.

I. Praw a dot diagram o-P each atom +0 determine the number o£

unpaired electrons on each atom. This will be the number o-P

bonds it can -Porm.

Z. Estimate the location o-P the atoms. The atom with the most

unpaired electrons will probably be in the middle, and the other

atoms will surround and bond with it.

3. Count the total number o-P valence electrons -Prom all oP the

atoms.

4-. Place pairs o£ electrons between the central atom and the

surrounding or terminal atoms. Keep track o-P the number o£

pairs distributed.

5". Place the remaining electrons in pairs on the terminal atom

until each has an octet. But do not put extra electrons

on hydrogen (it only wants 2- electrons).

G. I-P there are any electrons still lePt over, place them on the

central atom.

T. I-P the central atom does not have an octet, move pairs oP

unshared electrons -Prom the terminal atoms so they are

between the central atom and the terminal atom

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Praw the Lewis structure Por

The dot structures -Por C and Cl are

We see that C can Porm 4- bonds and each Cl can Porm one

bond. C is likely to be the central atom.

The total number o£ valence electrons is

C- 1 x 4 = 4-

CI-. 4- X T = 2-0

Total * =32-

Prawing the basic structure

Cl

Ct C

c|Adding pairs oP electrons between the C and CI's

c|

CI

Cl

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Q electrons have been used up. l£ we distribute pairs o£

electrons to the CI's we get

i ..- C - c j

i- -

"

This uses up all the valence electrons. Each chlorine -Perms one

bond, carbon -forms 4- bonds, and everyone has an octet.

Everyone is now happy.

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Draw the Lewis structure Por PH3.

. r H -P can -Porm 3 bonds, H can Porm one bond.

P-- I x *5- = 5"

H. 3 x I =3

Total o-P 0 valence electrons.

With P in the center, we can distribute G electrons.

U.-P- M

We have two electrons lePt over. Do we put them on a

hydrogen? Nope! Hydrogen only wants 2- electrons. Put the

electrons on the P.

- P - HH

P is Porming 3 bonds, the H's are each Porming one bond. P has

an octet and each H has 2. electrons. Everyone is happy.

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Draw the Lewis structure -Por NBr?

N can -Porm 3 bonds, and each B»r can -Porm one bond.

N-. I X ̂ = 5"

Br. 3 x T = 2-1Total oP 2-G valenc-e elec-trons.

The basic, struc-ture is

Adding electrons between N and the B>r's uses G elec-trons.

- rl

Distributing the elec-trons to the Br's uses 18 electrons -Por a total

o-P 2-4-. The remaining 2- electrons are placed on the N.

6r - A/- 6r;l "

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Draw the Lewis structure -Por CO

Carbon can -Porm 4- bonds, and eac-h oxygen can -Porm 2- bonds.

G: I X 4- - 4

O. 2- x G = 12-

Total o£ IG valence electrons.

We put carbon in the middle and oxygen at each end, then

distribute electrons between the atoms.

This uses up 4- electrons with 12. le-Pt over. We put electrons

around the oxygens.« «• I r

- - 0 ;

This uses up all the electrons. But carbon only has 4- electrons.

So we move an unshared pair o-P electrons -Prom each oxygen and

put them in between the carbon and oxygen to make double bonds.

£-0;

Carbon -Porms 4- bonds, each oxygen -Porms 2- bonds, and

everyone has an octet.

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Prow HCN

H • - c'Hydrogen oan -Porm one bond, oarbon oan -Porm 4- bonds, and

nitrogen oan -Porm 3 bonds. Carbon will be in the middle.

H-- I x I = I

G-. I X 4- - 4-N. I x 5" = 5"

Total o-P IO valenoe electrons.

We put oarbon in the middle and hydrogen and nitrogen at eaoh

end. We put eleotrons in between the atoms.

This uses up 4- eleotrons. We put the remaining G around nitrogen.

H ~ '

Carbon only has 4- eleotrons. So we move two pairs o-P

eleotrons -Prom nitrogen and put them between the nitrogen and

the oarbon. This -Porms a triple bond.

C and N have an ootet, H has a pair o-P eleotrons, and everyone

makes the appropriate number o-P bonds.

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