Vsepr Theory Homework (Page 2 of 3) in pdf

Geometry of Molecules

Rules for Drawing Lewis Structures

1. Count the total number of valence electrons in the

species. Remember to add electrons for anions and subtract

them for cations. (The number of valence electrons equals

the column number)

HCN

2. Express the total number as pairs by dividing the total by 2

2 Linear

2 0 Linear

180

N sp

3. Arrange the atoms in the correct molecular skeleton.

[Many times the first element is the central. Try the element

that bonds the most (the number of bonds are indicated by

3 trigonal

3 0 trigonal

120

N sp

-1

the valence of each column)]

4. Place the pairs around the atoms so that each atom has

an octet, except H. Start by placing 1 pair between each

Angular

3 trigonal

2 1

<120

symbol.

-1

(bent)

5. If an atom lacks an octet, consider double or triple bonds

between that atom and the adjacent atom.(Remember that

Halogens almost never form double bonds and H never

N sp

4 tetrahedral 4 0 tetrahedral

109

does)

6. The central atom may have an expanded octet (5 or 6

pairs), but only if it is from the 3rd period or higher.

4 tetrahedral 3 1 pryamidal

<109

Some Hints for Complicated Ones

4 tetrahedral 2 2 angular

<109

Skeleton: Not much more than above. The lowest

electronegative element is usually in the middle. H is never

in the middle. Connect each atom to the central with a single

trigonal

5 0

120-90 N

PCl

bond(a line or a pair of electrons)

bipyramid

Number of electrons: Use the column number of each

trigonal

teeter-

4 1

<120<90 Y

element in the molecule to total the number of electrons in

bipyramid

totter

the molecule. For a Positive ion subtract the positive number

trigonal

from the total. For a negative ion add the negative value to

3 2 T shaped

<90

ClF

bipyramid

the total. NH

= (5 + 1+1+1+1)-1

-1

trigonal

ClO

= (7+6+6+6)+1

2 3 linear

180

XeF

bipyramid

Arranging Electrons: Deduct two valence electrons that are

used in each of the bonds in the skeleton. Distribute the

remaining electrons as lone pairs so that each element has

6 octahedral 6 0 octahedral

an octet. If there are too few electrons convert single bonds

to multiple bonds.(Double bonds are usually limited to bonds

square

between C,N,O,S and maybe Se

6 octahedral 5 1

<90

BrF

pyramid

Some atoms can't get 8 (B,Al) without extraordinary work.

Expanded Octets: If there are too many electrons (each

element has an octet and there are some left) the central

square

6 octahedral 4 2

BrF

atom may have 10 or 12.(Will not happen to elements in the

planar

first or second row)

Vsepr Theory Homework Page 2

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