CHEMISTRY 115 HOMEWORK 8 NAME _______________

DUE MAY 30, 01 COMP HMWRK:none

  1. The complex [NiF4]2- is paramagnetic but [Ni(CN)4]2- is diamagnetic. Explain difference.

The fluoro ligand is a weak field ligand so that the electrons are expected to be unpaired and the four coordinate structure expected for it is the tetrahedron while the cyano ligand is a strong field ligand that would lead to paired electrons and a square planar arrangement.

Using Valence Bond Theory
             

F-

F-

F-

F-
 

[Ar]

_||_

_||_

_||_

_|__

_|__

xx

xx

xx

xx
       

3d

    

4s

  

4p

  
           

CN-

CN-

CN-

CN-

  
 

[Ar]

_||_

_||_

_||_

_||_

xx

xx

xx

xx

___
       

3d

    

4s

  

4p

  

Thus the tetrafluoronickelate(II) ion is sp3 hybridized to tetrahedral while the tetracyanonickelate(II) ion is dsp2 hybridized to square planar

 

2. Predict the number of unpaired electrons in each of the following complex ions.

A. [V(H2O)6]3+

C. [Cr(CO)6]

V+3 would have a [Ar] 3d2 configuration

Cr0 would have an [Ar] 4s1 3d5 configuration

H2O is a weaker field ligand

CO is a strong field ligand

Even with the +3 V, probably expect weak field

Even with Cro, probably expect strong field

Expect 2 d electrons to be unpaired

Expect pairing to d6 so 0 unpaired electrons

B. [CoF4]2- (tetrahedral; why?)

D. [PdCl4]2- (square planar; why?)

Co2+ would have a [Ar] 3d7 configuration

Pd2+ would have a [Kr] 4d8 configuration

F is a weak field ligand so expect a high spin complex which leads to tetrahedral structure

Pd2+ is a 2nd series transition metal so that all complexes are expected to be low spin

d7 configuration has minimum crystal field stabilization so often find tetrahedral structures for this configuration

d8 configurations are the most likely to have square planar structure because of large split in energy between dxy and dx2-y2 orbitals
   

3. Draw the valence bond (localized electron pairs) and crystal field diagrams for each of the following.

A. [V(H2O)6]3+ C. [Cr(CO)6]

V3+

      

O

O

O

O

O

O

          

[Ar]

_|_

_|_

__

xx

xx

xx

xx

xx

xx

__

__

__

__

__
     

3d

    

4s

  

4p

      

4d

    

Cr0

      

CO

CO

CO

CO

CO

CO

          

[Ar]

_||_

_||_

_||_

xx

xx

xx

xx

xx

xx

__

__

__

__

__
     

3d

    

4s

  

4p

      

4d

    

Octahdedral arrangement is d2sp3 so need to choose orbitals accordingly.

B. [CoF4]2- D. [PdCl4]2-
 

Co2+

          

Cl-

Cl-

Cl-

Cl-
 

[Ar]

_||_

_||_

_|_

_|__

_|__

xx

xx

xx

xx
       

3d

    

4s

  

4p

  
 

Pd2+

        

Cl-

Cl-

Cl-

Cl-

  
 

[Kr]

_||_

_||_

_||_

_||_

xx

xx

xx

xx

___
       

4d

    

5s

  

5p

  

A. [V(H2O)6]3+

C. [Cr(CO)6]

 
           

__

__

Eg

 

  
 

__

__

Eg

            
               

Do

  
     

Do

            

_|_

_|_

__

T2g

  

_||_

_||_

_||_

T2g

  
                   

H2O is normally a weak field ligand while CO is a strong field ligandà size of Do

B. [CoF4]2-

D. [PdCl4]2-
         

__

dx2-y2

  
               
           

D

  
         

_||_

dxy

  
               

_|_

_|_

_|_

T2g

  

_||_

dz2

  
   

Dt

          

_||_

_||_

Eg

  

_||_

_||_

dxz,dyz

  
               

Different arrangement of d orbitals in tetrahedral and square planar D.