Bonding in Polyatomic Polyatomic Molecules Molecules Bonding in - - PowerPoint PPT Presentation

Bonding in Bonding in Polyatomic Polyatomic Molecules Molecules

Basically two ways to approach Basically two ways to approach polyatomics polyatomics. . First is to use First is to use delocalized delocalized M.O.’s M.O.’s where where e e -

• are not confined to a

are not confined to a single bond (region between 2 atoms) but can wander over 3 or single bond (region between 2 atoms) but can wander over 3 or more atoms. We will use this approach later for C bonding. more atoms. We will use this approach later for C bonding. Second is to use Second is to use hybridization of atomic hybridization of atomic orbitals

• rbitals and then use these

and then use these to form localized (usually) bonds. to form localized (usually) bonds. Hybridization combines Hybridization combines orbitals

• rbitals on the
• n the SAME

SAME nucleus to form

nucleus to form new new orbitals

• rbitals called hybrids. Hybrids have characteristics of both

called hybrids. Hybrids have characteristics of both the atomic the atomic orbitals

• rbitals from which they are formed.

from which they are formed. Example: Example: sp sp hybrid is formed by combining (adding) a 2s and a 2p hybrid is formed by combining (adding) a 2s and a 2p wave function or orbital on a single atom. wave function or orbital on a single atom.

Localized BeH Localized BeH2

2 orbitals

• rbitals:

: Begin by Begin by hybridizing hybridizing Be 2s, 2p Be 2s, 2p orbitals

• rbitals to give two

to give two sp sp hybrid hybrid

• rbitals
• rbitals --
• - each pointing in opposite direction:

each pointing in opposite direction: + + Be Be + + + + + + sp sp+

+ ≡

≡ ≡ ≡ ≡ ≡ ≡ ≡ sp spa

a

sp sp-

• ≡

≡ ≡ ≡ ≡ ≡ ≡ ≡ sp spb

b

Add sp Add spa

a + 1s

+ 1s to get localized to get localized Be Be-

• H bond

H bond

}

Add Add sp spb

b + 1s

+ 1s to get localized to get localized Be Be-

• H bond

H bond

}

H H H H sp sp+

+ = 2s+2p

= 2s+2p ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ sp spa

a and

and sp sp-

• = 2s

= 2s-

• 2p

2p ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ sp spb

b

. . . . . . . .

– – – –

+ + + +

sp sp hybrids hybrids on Be give localized bonds and a linear structure

• n Be give localized bonds and a linear structure

H1s + ( H1s + (sp sp-

• ≡

≡ ≡ ≡ ≡ ≡ ≡ ≡ sp spb

b )

) H1s + ( H1s + (sp sp+

+ ≡

≡ ≡ ≡ ≡ ≡ ≡ ≡ sp spa

a )

) Be Be H H H H Be Be and and

+ + .

. . .

+ +

. . . .

– – – –

BH BH3

3 Fragment:

Fragment: B B sp sp2

2: Combine

: Combine 2s + 2p 2s + 2px

x + 2p

+ 2py

y

sp sp2

2

hybrid hybrid

• Can explain by

Can explain by sp sp2

2 hybrid

hybrid which provides which provides 3 3 localized bonds. localized bonds. Note: (B 1s Note: (B 1s2

22s

2s2

22p

2p -

• 3 valence

3 valence e e-

• )

)

+ +

120˚ 120˚ Boron Nucleus Boron Nucleus Second sp Second sp2

2 points

points along this direction along this direction Third sp Third sp2

2

points along points along this direction this direction

Three of these Three of these in a plane in a plane pointing at 120˚ with respect pointing at 120˚ with respect to each other to each other 3 3 sp sp2

2

hybrids hybrids 120˚ 120˚

Overlap with H 1s to give 3 B Overlap with H 1s to give 3 B-

• H bonds

H bonds in a plane in a plane pointing pointing at at 120º 120º with respect to each other: BH with respect to each other: BH3

3

Predicts Predicts planar planar trigonal trigonal structure structure with equal with equal 120º H 120º H-

• B

B-

• H angles.

H angles. ( (sp sp2

2 hybrids

hybrids) ) (1/3 s, 2/3 p) (1/3 s, 2/3 p)

B B

Note B has 3 Note B has 3 valence electrons valence electrons 1s 1s2

22s

2s2

22p

2p + + + + + +

+ +H H + +H H + +H H

120˚ Put 1 valence e- from B and 1 e- from H in each of 3 bonds.

CH CH4

4 Structure. If all C

• Structure. If all C-
• H bonds same, need

H bonds same, need hybridization hybridization that that provides 4 equal localized bonds. provides 4 equal localized bonds. sp sp3

3 does the job.

does the job. 2s + 2p 2s + 2px

x + 2p

+ 2py

y + 2p

+ 2pz

z ≡

≡ ≡ ≡ ≡ ≡ ≡ ≡ sp sp3

3 gives

gives 4 hybrid 4 hybrid orbitals

• rbitals which point to

which point to the corners of a tetrahedron. Angle between is the corners of a tetrahedron. Angle between is 109 109º º28 28’ ’ sp sp3

3 [1/4 s,

[1/4 s, 3/4 p]. Tetrahedral hybrids. 3/4 p]. Tetrahedral hybrids. Since C has Z = 6, 1s Since C has Z = 6, 1s2

22s

2s2

22p

2p2

2 →

→ → → → → → → 4 valence 4 valence e e -

• 4 H atoms have 4 x 1s

4 H atoms have 4 x 1s → → → → → → → → 4 valence 4 valence e e -

• 8 valence e

8 valence e-

• go into 4 bonds constructed of H 1s and each sp

go into 4 bonds constructed of H 1s and each sp3

3 gives

gives 4 equivalent localized electron pair bonds sp 4 equivalent localized electron pair bonds sp3

3 + H (1s).

+ H (1s).

Experimentally, find CH Experimentally, find CH4

4 has

has 4 bonds 4 bonds of equal length (1.043Å,

• f equal length (1.043Å,

arranged at corners of tetrahedron (109º28’ apart). arranged at corners of tetrahedron (109º28’ apart).

↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ E E        

2s 2s2

2

2p 2p 1s 1s2

2

sp sp3

3

sp sp3

3

sp sp3

3

sp sp3

3

1s 1s2

2

Isolated Carbon atom Isolated Carbon atom C atom in CH C atom in CH4

4

Geometry of carbon sp Geometry of carbon sp3

3/H 1s Bonds in methane (CH

/H 1s Bonds in methane (CH4

4):

): sp sp3

3 hybridization on C leads to 4 bonds. CH

hybridization on C leads to 4 bonds. CH4

4 is a

is a good example. good example.

C C H H H H H H H H

4 C sp3/H 1s bonds All H All H-

• C

C-

• H

H angles are angles are 109º 28 109º 28′ ′ ′ ′ ′ ′ ′ ′

Summary of Hybridization Results Summary of Hybridization Results BeH BeH2

2

2 2 sp sp linear H linear H-

• Be

Be-

• H

H BH BH3

3

3 3 sp sp2

2

• trig. plane (120º)
• trig. plane (120º)

CH CH4

4

4 4 sp sp3

3

tetrahedral tetrahedral (109º28’ (109º28’ H H-

• C

C-

• H angles)

H angles) Example Example Groups Attached Groups Attached Hybrid Hybrid Geometry Geometry to Center Atom to Center Atom

2pz 2py 2px 1s 1s 1s

Localized Bonds and Lone Pair Electrons Localized Bonds and Lone Pair Electrons Lone pair electrons are essentially valence electrons which Lone pair electrons are essentially valence electrons which do not become involved in bonding. do not become involved in bonding. These are handled These are handled beautifully by hybrid orbital picture. beautifully by hybrid orbital picture. NH3 Could make Could make 3 3 bonds bonds From 3, 2p From 3, 2p orbitals

• rbitals

and 3, 1s H orbital: and 3, 1s H orbital: This predicts 90º geometry 2e - remaining in 2s2 of N. 3H, 1s (no choice) N: 1s22s22p3 Nitrogen Nitrogen nucleus nucleus H atom 1s

• rbital

N

107˚

H H H Geometry of NH Geometry of NH3

3 found to be

found to be Trigonal Trigonal Pyramidal Pyramidal 3 H project down. 3 H project down. 107º angle is suspiciously close to 109º28” 107º angle is suspiciously close to 109º28” predicted by sp predicted by sp3

3 hybridization!

hybridization! Try hybridizing Try hybridizing sp sp orbitals

• rbitals on N to form
• n N to form 4 sp

4 sp3

3 combinations.

combinations. Remember, must account for Remember, must account for 5 5 N valence N valence e e -

• !

!

4N sp 4N sp3

3 orbitals

• rbitals combine with 3 1s H

combine with 3 1s H orbitals

• rbitals to give 3 sp

to give 3 sp3

3, 1s M.O.’s

, 1s M.O.’s leaving one sp leaving one sp3

3 hybrid left

hybrid left Put 1 Put 1e e -

• from

from N N into each M.O. 3 into each M.O. 3e e -

• Put 1

Put 1e e -

• from each H into M.O.’s 3

from each H into M.O.’s 3e e -

• Put 2

Put 2e e -

• from

from N N into free sp into free sp3

3 orbital.

• rbital.

Of 5 valence Of 5 valence e e -

• in N, 2 go into one sp

in N, 2 go into one sp3

3 orbital, 3 go into other 3 sp

• rbital, 3 go into other 3 sp3

3,s.

,s. (combined with H (1s)) (combined with H (1s)) One of the driving forces for the tetrahedral configuration is t One of the driving forces for the tetrahedral configuration is that hat it puts bonding and lone pair electron groups as far away from it puts bonding and lone pair electron groups as far away from each other as possible. each other as possible. This minimizes This minimizes repulsions

repulsions.

.

Essentially a tetrahedral arrangement of bonds with lone pair taking the 4th position of the tetrahedron. N H H H Geometry of NH3: Trigonal Pyramidal

• •
• Lone Pair

Lone Pair

N N Electron pair repulsion effect is largest for small central atom Electron pair repulsion effect is largest for small central atoms s like B, N, O. like B, N, O. As go to larger central atoms (e.g. S or metals) frequently As go to larger central atoms (e.g. S or metals) frequently find this effect not so large and start to get things closer find this effect not so large and start to get things closer to pure p orbital bonds (90º structures) to pure p orbital bonds (90º structures) For example, H For example, H2

2S has H

S has H-

• S

S-

• H angle of 92º.

H angle of 92º. Approximately Approximately Tetrahedral Tetrahedral

~ ~ 109° 28’

109° 28’

O O

H H H H

Lone pairs stick into and out of paper

H H2

2O:

O:

4 sp 4 sp3

3 hybrids. Use

• hybrids. Use 2

2 to make MO bonds with H 1s. to make MO bonds with H 1s. 2 2e e -

• (valence) from O and

(valence) from O and 2e- from 2H from 2H → → → → → → → → 4e- in in 2 2 M.O. M.O.’ ’s s 2 2e e -

• from O into one sp

from O into one sp3

3, 2 other

, 2 other e e -

• into last sp

into last sp3

3

Predicts H Predicts H-

• O

O-

• H bond

H bond

• f 109º28”
• f 109º28” -
• Find 105º

Find 105º Electron Arrangement: Electron Arrangement: 2 sp 2 sp3

3 lone pairs

lone pairs 2 sp 2 sp3

3, 1s

, 1s MO bonds MO bonds O 1s22s22p4 (6 valence e-) 2H 1s (2e-) Three Atoms Three Atoms (O, H, H) in (O, H, H) in The Plane of The Plane of The Paper The Paper

• •
• •

Single and Multiple Bonds in Carbon Compounds Single and Multiple Bonds in Carbon Compounds sp sp3

3 hybridization on C leads to 4 bonds. CH

hybridization on C leads to 4 bonds. CH4

4 is a

is a good example. CH good example. CH3

3(CH

(CH3

3) is another example:

) is another example:

C C CH CH3

3

H H H H H H

3 C sp3/H 1s bonds 1 C 1 C-

• C

C sp sp3

3/sp

/sp3

3

bond bond

Can keep replacing Can keep replacing C C-

• H bonds with C

H bonds with C-

• C

C

• bonds. When all 4 sp
• bonds. When all 4 sp3

3

• rbitals
• rbitals form bonds

form bonds around a given around a given carbon we call that carbon we call that center saturated. center saturated.

H H H H H H H H

sp sp2

2

sp sp2

2

sp sp2

2

sp sp2

2

sp sp2

2

sp sp2

2

Carbon can also exhibit sp Carbon can also exhibit sp2

2 hybridization: C

hybridization: C2

2H

H4

4

(ethylene) (ethylene)

C C C C

C C C C H H H H H H H H

120˚ 120˚ 120˚ 120˚

2 of sp2 orbitals on each C yield C C-

• H bonds

H bonds ( total of 4 ) Remaining sp2 on each c overlap each other → → → → gives C C-

• C bond

C bond This uses 3 valence e - from each C to form 6 sp2 bonds (σ σ σ σ bonds) (Each Each C contributes 1 valence e - to C-C bond.)

– + –

C C C

– +

Up to this point, need not be planar because Up to this point, need not be planar because σ σ σ σ σ σ σ σ bond symmetric bond symmetric under rotation. under rotation. This leaves us still with 1 This leaves us still with 1 e e-

• on each C and one unused 2p orbital.
• n each C and one unused 2p orbital.

Now Now overlap

• verlap 2p

2p

• rbitals
• rbitals to give

to give

π π π π π π π π bond

bond.

.

π π π π π π π π bond

bond 2p

+

C