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Polyatomic

Transformational properties of atomic orbitals

• When bonds are formed, atomic orbitals combine according to their symmetry.

• Symmetry properties and degeneracy of orbitals and bonds can be learned from corresponding character tables by their inspection holding in mind the following transformational properties:

Atomic orbital Transforms as S s x2+y 2+z 2

px x

py y py

pz z 2 2 2 2 dz2 z , 2z -x -y 2 2 dx2-y2 x -y

dxy xy dz2

dxz xz

dyz yz

1 Examples of atomic orbitals symmetry analysis

C2v Mulliken labels 2 2 2 A1 z x , y , z C2v D3h D4h Td Oh A2 Rz xy s a1 a1' a1g a1 a1g B1 x, R y xz b e t t px 1 e' u 2 1u B y, R yz t 2 x py b2 e' eu 2 t1u D3h pz a1 a2" a2u t2 t1u 2 2 2 A1’ x +y , z dz2 a1 a1' a1g e eg A2’Rz dx2-y2 a1 e' b1g e eg 2 2 t E’ (x,y) (x -y , xy) dxy a2 e' b2g t2 2g A ” b e t t2g 1 dxz 1 e" g 2 A ” z t 2 dyz b2 e" eg t2 2g E” (R x,R y) (xz, yz)

D4h Td Oh 2 2 2 2 2 2 2 2 2 A1g x +y , z A1 x +y +z A1g x +y +z 2 2 2 2 2 2 2 B1g x -y A2 Eg (2z -x -y , x -y ) 2 2 2 2 2 B2g xy E (2z -x -y , x -y ) T1g (R x,R y,R z)

Eg (R x,R y) (xz, yz) T1 (R x,R y,R z) T2g (xz, yz, xy)

A2u z T2 (x,y,z) (xz, yz, xy) T1u (x,y,z)

Eu (x, y) …

MO diagram of homonuclear diatomic

• Filling the resulting MO’s of homonuclear diatomic molecules with leads to the A B following results: 6σu

2πg AB # of e’s Bond # Bond 2p 2p order unpair. energy, 5σg e’s eV 1πu Li 2 61 0 1.1

Be 2 80 0 - 4σu 2s 2s B2 101 2 3.0

C2 122 0 6.4 3σg N2 143 0 9.9 2σu O2 162 2 5.2 1s 1s F2 181 0 1.4 1σg Ne 2 200 0 -

Bond order = ½ (#Bonding e’s - #Antibonding e’s)

2 MO-energy levels in N 2 • Photoelectron spectroscopy of simple molecules is an invaluable source of the information about their .

• The He-I photoelectron spectrum of gaseous N 2 below proves that there is the σ-π level inversion in this molecule. It also allows identify bonding (peaks with fine vibronic structure) and non-bonding MO (simple peaks) in it.

N N 6σu

2πg 2p 2p 15.6 5σg 16.7 1πu 18.8 4σu 2s 2s

3σg

E, eV 2σu 1s 1s 1σg

Let’s Play with some MO’s

water…water…water….

3 To begin….How to Deal with more than two ? You use some Acronyms! Sounds Good!

• SALCs – Symmetry Adapted Linear Combinations

• LGOs – Ligand Group Orbitals

Lets Separate the O for the H H

Oxygen

Hydrogen

To the Character Tables Ha Hb

4 z

O x H H y

O (p y) B2

O (p x) B1

O (p z)

A1 O (s)

z

O x H H y

Ha -Hb (s)

B1

Ha + H b (s)

A1

5 NOW Let’s Make Some MOs

LGOs

B1

A1

A1

Oxygen

NOW Let’s Make Some MOs

LGOs

B2

4 A 1

A1

3 A 1

A1

2 A 1 Oxygen Hydrogens

6 NOW Let’s Make Some MOs

LGOs

B2

4 A 1

A1

3 A 1

A1

2 A 1 Oxygen Hydrogens

NOW Let’s Make Some MOs

LGOs 2 B1

B1

A1 1B 2

1B 1

A1

Oxygen Hydrogens

7 NOW Let’s Make Some MOs

LGOs 2 B1

B1

A1 1B 2

1B 1

A1

Oxygen Hydrogens

NOW Let’s Make Some MOs

LGOs 2 B1

B1

4 A 1

A1

1B 2

3 A 1

1B 1

A1

2 A 1 Oxygen Hydrogens

8 NOW Let’s Make Some MOs

LGOs 2 B1

B1

4 A 1

A1 1B 2

3 A 1

1B 1

A1

2 A 1 Oxygen Hydrogens

Now for the Million Dollar Question

Why is Water Bent???

9 – linear XH 2 molecules

Molecular Orbitals of BeH 2

• BeH (D ) 2 ∞h Be 2 H

+ H Be H σσσu z 5σσσu s - s y 4σσσg π Group orbitals of 2H's: ππu 2.2 eV -13.5 eV

2px 2py πππu s + s + ψ - ψ σσσ + s s 2pz u σσσg

+ σσσg ψs + ψs 2s 3σσσu

2 2 2 -8.2 eV 1σg 2σg 3σu 2σσσg

D∞h ∞σv i 1s Σ + 1 1 x2+y 2, z 2 g 1σσσg + -128 eV Σu 1 -1 z

Πu 0 -2 (x,y)

10 Molecular Orbital Theory – Walsh diagram

The Walsh diagram shows what happens to the molecular orbitals for a set of molecules which are related in structure. In this case, the difference is the H-X-H bond angle which decreases from 180 o to 90 o

Molecular Orbital Theory – Walsh diagram

H

X

H O Water 104.5 ° H H

11 4) MO theory and (Walsh diagrams)

• Correlate changes in energy of MO’s between species AB x of high and lower symmetry, such as BeH 2 and H 2O.

z 5σσσu O H H 2b2

y 4a x 1 4σσσg

πππu b1

2 2 3a BeH 2 2σg 3σu linear 1 2 2 1 o BH 2 2a 1 1b 2 3a 1 , 131 2 2 2 o CH 2 2a 1 1b 2 3a 1 , 102 NH 2a 21b 23a 21b 1 , 103 o 1b2 2 1 2 1 1 3σσσu 2 2 2 2 o OH 2 2a 1 1b 2 3a 1 1b 1 , 105 + 2 2 2 2 o FH 2 2a 1 1b 2 3a 1 1b 1 , 113 2σσσg 2a1

Molecular Orbital Theory – BH 3

BH 3 has a C 3 principal axis of symmetry, 3 C 2 axes ( C3), 3 σv, and σh – it is in a D 3h point group

12 Molecular Orbital Theory – BH 3

The BH 3 molecule exists in the gas phase, but dimerizes to B 2H6 (which we will look at a bit later) z

2 BH 3 B2H6

The BH molecule is trigonal planar and we H 3 BH y will make the C 3 principal axis of symmetry the z axis, with the x and y axes in the plane H of the molecule. The y axis (arbitrary) will be along one of the B-H bonds. x

Molecular Orbital Theory – D3h Character Table

D3h E 2C 3 3C 2 σh 2S 3 3σv

A1’ 1 1 1 1 1 1

A2’ 1 1 -1 1 1 -1 E’ 2 -1 0 2 -1 0

A1” 1 1 1 -1 -1 -1

A2” 1 1 -1 -1 -1 1 E” 2 -1 0 -2 1 0

13 Molecular Orbital Theory – D3h Character Table

D3h E 2C 3 3C 2 σh 2S 3 3σv

A1’ 1 1 1 1 1 1

A2’ 1 1 -1 1 1 -1

E’ 2 -1 0 2 -1 0

A1” 1 1 1 -1 -1 -1

A2” 1 1 -1 -1 -1 1

E” 2 -1 0 -2 1 0

Molecular Orbital Theory – D3h Character Table

D3h E 2C 3 3C 2 σh 2S 3 3σv

2s A1’ 1 1 1 1 1 1

A2’ 1 1 -1 1 1 -1

E’ 2 -1 0 2 -1 0

A1” 1 1 1 -1 -1 -1

A2” 1 1 -1 -1 -1 1

E” 2 -1 0 -2 1 0

14 Molecular Orbital Theory – D3h Character Table

D3h E 2C 3 3C 2 σh 2S 3 3σv

A1’ 1 1 1 1 1 1

A2’ 1 1 -1 1 1 -1

E’ 2 -1 0 2 -1 0

A1” 1 1 1 -1 -1 -1

A2” 1 1 -1 -1 -1 1 2p z E” 2 -1 0 -2 1 0

Molecular Orbital Theory – D3h Character Table

D3h E 2C 3 3C 2 σh 2S 3 3σv

A1’ 1 1 1 1 1 1

A2’ 1 1 -1 1 1 -1

E’ 2 -1 0 2 -1 0

A1” 1 1 1 -1 -1 -1

A2” 1 1 -1 -1 -1 1 2p x 2p y E” 2 -1 0 -2 1 0 doubly degenerate

15 Molecular Orbital Theory – D3h Character Table

D3h E 2C 3 3C 2 σh 2S 3 3σv

A1’ 1 1 1 1 1 1

A2’ 1 1 -1 1 1 -1

E’ 2 -1 0 2 -1 0

A1” 1 1 1 -1 -1 -1

A2” 1 1 -1 -1 -1 1

E” 2 -1 0 -2 1 0

Molecular Orbital Theory – LGOs on H atoms

In BH 3 we need three LGOs, formed from linear combinations of the H 1s orbitals H

H H

What happens if we carry out the D 3h symmetry operations on this group of H 1s orbitals? How many remain unchanged?

EC3 C2 σh S3 σv

3 0 1 3 0 1

16 Molecular Orbital Theory – LGOs on H atoms

EC3 C2 σh S3 σv

3 0 1 3 0 1

The resulting row of characters is also obtained by adding the

characters of the A1’ and E’ representations

D3h E 2C 3C σh 2S 3σ 3 2 3 v

A1’ 1 1 1 1 1 1 E’ 2 -1 0 2 -1 0 LGOs 3 0 1 3 0 1

Molecular Orbital Theory – LGOs for BH 3

Ψ(a 1’) = (1/ √3)( Ψ1 + Ψ2 + Ψ3)

Ψ(e’) 1 = (1/ √6)(2 Ψ1 – Ψ2 – Ψ3)

Ψ(e’) 2 = (1/ √2)( Ψ2 – Ψ3)

17 Molecular Orbital Theory – LGOs for BH 3

nodal planes

e’

a1’

Molecular Orbital Theory

18 Molecular Orbital Theory – BH 3

Empty anti-bonding MOs

Empty non-bonding MO

Three bonding MOs are filled, accounting for the three B-H σ bonds

Molecular Orbital Theory

19 Molecular Orbital Theory – NH 3

The ammonia molecule,

NH 3, has C 3v symmetry, with a C 3 principal axis of symmetry and 3 vertical planes of symmetry

z

N y H H H x

Molecular Orbital Theory – NH 3

Part of the character table of C 3v

C3v E 2C 3 3σv

A1 1 1 1

A2 1 1 -1 E 2 -1 0

20 Molecular Orbital Theory – NH 3

Part of the character table of C 3v

C3v E 2C 3 3σv

2s and 2p z A1 1 1 1 orbitals on N

A2 1 1 -1

E 2 -1 0

Molecular Orbital Theory – NH 3

Part of the character table of C 3v

C3v E 2C 3 3σv

A1 1 1 1

A2 1 1 -1

2p x and 2p y E 2 -1 0 orbitals on N

21 Molecular Orbital Theory – NH 3

Part of the character table of C 3v

C3v E 2C 3 3σv

A1 1 1 1

A2 1 1 -1

E 2 -1 0

LGOs 3 0 1 a1 + e

2) Molecular Orbitals of NH 3 (C 3v )

• NH 3 (C 3v : E, 2C 3, 3 σv)

z N 3 H NH3 2e -13.5 eV H(3) H(1) H(2) 4a1 y e - s2 + s3 x -15.5 eV

e (2p 2p ) The symmetry of 3H’s group orbitals: x, y a1 2s1 - s2 - s3

Γr = 3E+0C 3+σv = A 1 + E a (2p ) 3a 1 z 1 s1 + s2 + s3 C E 2C 3σ 3v 3 v -17.0 eV symmetry adapted linear 2 2 2 combinations A1 1 1 1 z x +y , z 1e (SALC) of three 1s a1 (2s) orbitals can be A2 1 1 -1 found with help of -25.6 eV the "projection E 2 -1 0 (x,y) operator" technique (F.A. -31.0 eV Cotton, p. 114) 2a1

22 Molecular Orbital Theory

Molecular Orbital Theory – NH 3

on N

N-H bonding orbitals

23 Molecular Orbital Theory – T d

C3

Methane has T d symmetry, a cubic point group

The C 3 axes in CH 4 coincide with the C-H z bonds

The C 2 and S 4 axes coincide with the x, y, and z axes y x

Molecular Orbital Theory – Td character table

Part of the T d Character Table

Td E 8C 3 3C 2 6S 4 6s d

A1 1 1 1 1 1

A2 1 1 1 -1 -1 E 2 -1 2 0 0

T1 3 0 -1 1 -1

T2 3 0 -1 -1 1

24 Molecular Orbital Theory – Td character table

Part of the T d Character Table

Td E 8C 3 3C 2 6S 4 6s d 2s of C A1 1 1 1 1 1

A2 1 1 1 -1 -1 E 2 -1 2 0 0

T1 3 0 -1 1 -1

T2 3 0 -1 -1 1

Molecular Orbital Theory – Td character table

Part of the T d Character Table

Td E 8C 3 3C 2 6S 4 6s d

A1 1 1 1 1 1

A2 1 1 1 -1 -1 E 2 -1 2 0 0

T1 3 0 -1 1 -1 2p x, 2p y, and 2p are triply T2 3 0 -1 -1 1 z degenerate

25 Molecular Orbital Theory – Td character table

Part of the T d Character Table

Td E 8C 3 3C 2 6S 4 6s d

A1 1 1 1 1 1

A2 1 1 1 -1 -1 E 2 -1 2 0 0

T1 3 0 -1 1 -1

T2 3 0 -1 -1 1

LGO 4 1 0 0 2 = a 1 + t 2 s

Molecular Orbital Theory – Td character table

Wavefunctions for the LGOs for methane hydrogens

ψ(a 1) = ½ ( ψ1 + ψ2 + ψ3 + ψ4)

ψ(t 2)1 = ½ ( ψ1 - ψ2 + ψ3 - ψ4)

ψ(t 2)2 = ½ ( ψ1 + ψ2 - ψ3 - ψ4)

ψ(t 2)3 = ½ ( ψ1 - ψ2 - ψ3 + ψ4)

26 Molecular Orbital Theory – LGOs for methane

C atomic orbitals

H ligand group orbitals

Molecular Orbital Theory - methane

27 Molecular Orbital Theory – LGOs for F atoms

What if the atoms attached to the central are not H, and have p orbitals, such as ?

2) Molecular Orbitals of NH 3 (C 3v )

• NH 3 (C 3v : E, 2C 3, 3 σv)

z N 3 H NH3 2e -13.5 eV H(3) H(1) H(2) 4a1 y e - s2 + s3 x -15.5 eV

e (2p 2p ) The symmetry of 3H’s group orbitals: x, y a1 2s1 - s2 - s3

Γr = 3E+0C 3+σv = A 1 + E a (2p ) 3a 1 z 1 s1 + s2 + s3 C E 2C 3σ 3v 3 v -17.0 eV symmetry adapted linear 2 2 2 combinations A1 1 1 1 z x +y , z 1e (SALC) of three 1s a1 (2s) orbitals can be A2 1 1 -1 found with help of -25.6 eV the "projection E 2 -1 0 (x,y) operator" technique (F.A. -31.0 eV Cotton, p. 114) 2a1

28 3) Molecular Orbitals of CH 4 (T d)

4 H

The symmetry of 4H’s group orbitals: C

t2 Γr = 4E+1C 3+0C 2+0S 4+2 σd = A 1 + T 2 2t2 -13.5 eV s1+s2-s3-s4 z 3a1 t2 H(2) t2 H(1) -11.7 eV a1 y C x s1-s2+s3-s4 H(4) H(3) t t2 (2px, 2py, 2pz) -14.8 eV 2 (-14, PhES)

Td 1t2 A x2+y 2+z 2 s1-s2-s3+s4 1 -22.3 eV A2 E a1 a1 (2s) T 1 -25.7 eV T (x,y,z) 2 (-23, PhES) s +s +s +s 2a1 1 2 3 4

29