Hybridization 1

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Copyright©2000 by HoughtonMifflin Company. All rights reserved.

1

Covalent Bonding: OrbitalsChapter 09




2

The four bonds around C are of equal length and Energy




3

Can you explain this based on your knowledge

of electron energy levels?

6C

1s2 2s2 2p2

Bonding from s is

Different from bondingFrom p. In addition the

Angles should be within

900!




4

How to generate four equal orbitals?




5

Hint:

A key to wave mechanics is

superposition

Which is creating new waves from interference of old ones




6

An energy-level diagram showing the

formation of four sp

3

orbitals.




7

Hybridize at n=2 sp3 sp3 sp3 sp3

s px py pz

s px py pz

Hybridization

sp3 sp3 sp3 sp3

Four sp3 orbitals of equal length, energy and in tetrahedral shape

Ground state of C 1s2 2s2 sp2

Promote electron at n=2 2s1

2p3




8

Hybridization

The mixing of atomic orbitals to formspecial orbitals for bonding.

The atoms are responding as needed to givethe minimum energy for the molecule.




9

Valence Bond Theory and NH3

N – 1s22s22p3

3 H – 1s1

If use the

three 2p orbitals

predict 900

Actual H-N-H bond angle is 107.30

How do you explain this?

2s 2px 2py 2pz




10

2s 2px 2py 2pzOriginal

Mix 1s and 3pAnd generate four

Equivalent sp3

Hybridized orbitals

sp3 sp3 sp3 sp3

3 bonding orbitals

Which can accommodate

The 1s1 electron from

hydrogen

1 sp3

lone

pair

Consider the n=2 for N




11

The nitrogen atom in ammonia is sp3

hybridized.




12




13

An orbital energy-level diagram for sp2

hybridization. Note that one p orbital

remains unchanged.




14




15

When an s and two p orbitals are mixed to form a set

of three sp2 orbitals, one p orbital remains unchanged

and is perpendicular to the plane of the hybrid orbitals.




16




17

Figure 9.13: (a) The orbitals used to form

the bonds in ethylene. (b) The Lewis

structure for ethylene.




18




19 Sigma bond (s) –

electron density between the 2 atomsPi bond (p) – electron density above and below plane of nuclei

of the bonding atoms




20

The s bonds in ethylene. Note that for each bond the

shared electron pair occupies the region directly between

the atoms.




21

A sigma (s

) bond centers along theinternuclear axis.

A pi (p) bond occupies the space above and

below the internuclear axis.

CC

H H

HH

s

p




22

(a)The orbitals used

to form the bondsin ethylene.

(b) The Lewis

structure forethylene.




23

The orbital energy-level diagram for the

formation of sp hybrid orbitals on carbon.




24




25

When one s orbital and one p orbital are

hybridized, a set of two sp orbitals oriented at

180 degrees results.




26

The orbitals of an sp hybridized

carbon atom.




27




28

The orbital arrangement for an sp2hybridized

oxygen atom to form CO2.

8O 1s22s22p4



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The hybrid orbitals in the CO2 molecule.



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(a) The orbitals used to form the bonds in carbon dioxide. Note

that the carbon-oxygen double bonds each consist of one s bond

and one p bond. (b) The Lewis structure for carbon dioxide.

2sp orbitals from C to form two double bonds

( ) A h b idi d i (b) h b d i h



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(a) An sp hybridized nitrogen atom. (b) The s bond in the N2

molecule. (c) The two p bonds in N2 are formed when electron

pairs are shared between two sets of parallel p orbitals. (d) The

total bonding picture for N2.




Sigma (s) and Pi Bonds (p)

Single bond1 sigma bond

Double bond 1 sigma bond and 1 pi bond

Triple bond 1 sigma bond and 2 pi bonds

How many s and p bonds are in the acetic acid

(vinegar) molecule CH3COOH?

C

H

H

CH

O

O Hs bonds = 6 + 1 = 7

p bonds = 1

10.5




How to generate more than 4

bonds?

Remember the PCl5, SF6, etc…

The s and 3p can generate 4 orbitals

Include d orbitals to generate more!




s p p p d d d d d

Hybridize 1s and 3p and 1d

Result in 5 dsp3 orbitals

dsp3 dsp3 dsp3 dsp3 dsp3

1 2 3 4 5

A f d 3 h b id bi l h h




A set of dsp3 hybrid orbitals on a phosphorus atom.

Note that the set of five dsp3 orbitals has a trigonal

bipyramidal arrangement. (Each dsp3 orbital also has a

small lobe that is not shown in this diagram.)

(a) The PCl5 molecule (b) The orbitals used to form the bonds in




(a) The PCl5 molecule. (b) The orbitals used to form the bonds in

PCl5. The phosphorus uses a set of five dsp3 orbitals to share

electron pairs with sp3 orbitals on the five chlorine atoms. The

other sp3 orbitals on each chlorine atom hold lone pairs.

f 6 bi l ?



Copyright©2000 by Houghton 37

s p p p d d d d d

Hybridize 1s and 3p and 2d

Result in 6 d2sp3 orbitals

d2sp3 ds2 p3 d2sp3 d2sp3 d2sp3 d2sp3

How to form 6 orbitals?

1 2 3 4 5 6

Hybridization 1

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