Understanding Stability & Distortions in Square Planar Complexes: Jahn-Teller & Coordinati, Lecture notes of Geometry

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Coordination Chemistry II:Jahn-Teller, Square Planar

Complexes, Orbital Overlap

Method, and Electron Counting

Chapter 10 and Section 13.3Monday, November 30, 2015

Jahn-Teller Distortions

Jahn-Teller Theorem: electron configurations with

unequal occupancy

of degenerate orbitals

are not stable.

d

4

HS

A complex with such a configuration will undergo a

Jahn-Teller distortion

to

lower its energy.

no net changein energy

stabilization isdriving forcefor distortion

tetragonalelongation

Jahn-Teller Distortions

Jahn-Teller Theorem: electron configurations with

unequal occupancy

of degenerate orbitals

are not stable.

d

5

HS

Only some complexes can lower their energy by distorting:

no net changein energy

tetragonalelongation

no net changein energy

No stabilization!

No J-T distortion

Occurrence of Jahn-Teller Effects

Strong Jahn-Teller Effect

Weak Jahn-Teller Effect

t

2

e g

g

hs d

(^4)

t

2

e g

g

ls d

7

t

2

e g

g

d

9

t

2

e g

g

d

1

t

2

e g

g

d

2

t

2

e g

g

ls d

4

t

2

e g

g

hs d

6

t

2

e g

g

hs d

(^7)

t

2

e g

g

ls d

5

Partially occupied

e

g

orbitals (M–L σ*) lead to more pronounced distortions

than partially-occupied

t

2g

orbitals (non-bonding).

CFT of Square Planar Complexes

dz

dx

‐

y d

xz

dxy

dyz

Occurs

mostly

for

d

8

i.e.,

Pd

2+

Pt

2+

Ir

+

Au

3+

(majority

low

spin)

Strong

field

ligands

d

xz

dxy

dyz

d

z

dx

‐

y

d

xz

dyz

dz

dx

‐

y

dxy

Square Planar Complexes

σ & π LFT diagram

Be sure you know how

to derive this by

considering

║

, and

┴

separately

ML

6

Octahedral MO Diagram

e

g

e

g

1a

1

g

a

1

g

t

1

u

t

1

u

t

2

g

A

1

g

E

g

T

1

u

(n+1)

s

(n+1)

p

n

d

z

2

x

2

y

2

E

x g

y

z

T

1

u s

A

1

g^ xy xz yz

T

2

g

o

3e

σ

Using the AOM

The usefulness of the AOM is that it gives a good approximation of theenergies of the metal

d

orbitals in different coordination geometries.

the metal

d

orbitals are the frontier orbitals in most coordination complexes

the AOM can be used to predict changes to the metal

d

orbitals if the coordination

geometry is changed.

18 Electron Rule (Section 13.3)

The 18 electron rule is a loose formalism for describing stableelectron configurations for some transition metal coordinationcomplexes.

18 electrons is the maximum number of electrons that can be accommodated by themetal

nd

n

s

, and (

n

p

valence orbitals.

really the 18 electron rule is an extension of the octet rule to include

d

orbitals

the octet rule and the 18 electron rule are alternately know as the Effective AtomicNumber (EAN) rule.

σ-Only ML

6

Octahedral MO Diagram

e

g

e

g

a

1

g

a

1

g

t

1

u

t

1

u

t

2

g

A

1

g

E

g

T

1

u

(n+1)

s

(n+1)

p

n

d

z

2

x

2

y

2

E

x g

y

z

T

1

u s

A

1

g^ xy xz yz

T

2

g

anti-bonding

M–L σ*

bonding

M–L σ

M non-bonding

For an octahedral complex,

placing 6 electrons in the

metal

t

2

g

orbitals will give an

18 electron complex.

Donor-Pair Method - Example 1

tetramethylbis(trimethylphosphane)manganese

1. Remove all ligands as Lewis bases (closed octet ondonor atom)

PMe

3

CH

3

• 1. Determine the charge left on the metal after removing the Lewis Base ligands.

IV

in order to balance

charge, the manganese

must be 4+, this is the

metal

oxidation state

1. Using the metal oxidation state, determine the number of metal

d

electrons

d

electrons = #valence electrons in neutral metal – metal oxidation state

#d electrons



7





4







3

1. Add the number of metal

d

electrons to the number of electrons donated by the ligands to get

the total valence electron count.

2 :PMe

3

4 :CH

3

Mn

IV

4 e

8 e

3 e

15 e

So we’d say that

MnMe

4

(PMe

3

2

is a

d

3

Mn

IV

15-electron

complex

Donor Pair Method - Example 2

pentaamminechlorocobalt (2+)

1. Remove all ligands as Lewis bases (closed octet on donoratom)

NH

3

Cl

• 1. Determine the charge left on the metal after removing the Lewis Base ligands.

in this case the charge

from removing the Cl

-

must be added to theoverall charge on the

complex, giving us Co

III

1. Using the metal oxidation state, determine the number of metal

d

electrons

d

electrons = #valence electrons in neutral metal – metal oxidation state

#d electrons



9





3







6

1. Add the number of metal

d

electrons to the number of electrons donated by the ligands to get

the total valence electron count.

5 :NH

3

1 :Cl

Co

III

10 e

2 e

6 e

18 e

So [Co(NH

3

5

Cl]

2+

is a

d

6

Co

III

18-electron complex

Co

III

Neutral Ligand Method - Example 1

tetramethylbis(trimethylphosphane)manganese

1. Remove all ligands as neutral fragments, whichleaves a neutral metal
   1. Add up the electrons donated by the neutral ligands and on the neutral metal to get the totalvalence electron count

PMe

3

CH

3

Mn

0

2 :PMe

3

4 ·CH

3

Mn

0

4 e

4 e

7 e

15 e

Notice we again

determine that

MnMe

4

(PMe

3

2

is a 15

electron complex

1. We still need to determine the metal oxidation state.

metal OS = #one-electron donor ligands + charge on the complex

1. Now, using the metal oxidation state, determine the number of metal

d

electrons

d

electrons = #valence electrons in neutral metal – metal oxidation state

#d electrons



7





4







3

metal OS



4





0





 

4

Mn

IV

Neutral Ligand Method - Example 2

pentaamminechlorocobalt (2+)

1. Remove all ligands as neutral fragments, which leaves aneutral metal
   1. Using the metal oxidation state, determine the number of metal

d

electrons

d

electrons = #valence electrons in neutral metal – metal oxidation state

#d electrons



9





3







6

1. Add up the electrons donated by the neutral ligands and on the neutral metal to get the totalvalence electron count

5 :NH

3

1 ·Cl

Co

0

10 e

1 e

9 e

20 e

We forgot to account for

the charge on the

complex, which is 2+

+2 charge

–2 e

18 e

• 1. Determine the metal oxidation state.

metal OS = #one-electron donor ligands + charge on the complex

metal OS



1





2







3



NH

3

Cl

Co

0

Co

III