Important Questions for Chemical Science Students - Cages and Metal Clusters

Most Important Questions on Cages and Metal Clusters (Download PDF)

1. Calculate the no. of skeletal electron pairs in [Pd5(CN)12]2-?

1. 6
2. 7
3. 8
4. 9

1. Determine the geometry of the complex [Rh6N(CO)17]3+.

1. Trigonal Prism
2. Trigonal Bipyramidal
3. Square pyramidal 
4. Pentagonal bipyramidal

1. Determine the geometry of the complex [Os5C(CO)15].

1. Trigonal Prism
2. Trigonal Bipyramidal
3. Square pyramidal 
4. Pentagonal bipyramidal

1. Os6 (CO)18 has a monocapped TBP structure. With the addition of 2 electrons, the structure changes to:

1. Trigonal Prism
2. Octahedron 
3. Square pyramidal 
4. Pentagonal bipyramidal

1. Calculate the Bonding molecular orbital of the following cage structure [Fe4(CO)12C]2-.

1. 35
2. 32
3. 44
4. 31

1. Determine the type of cluster of compound [Os6C(CO)17]. 

1. Arachno
2. Nido
3. hypo
4. Closo

1. Determine type of cluster of compound [Os6C(CO)17 {P(OMe)3}3].

1. Arachno
2. Nido
3. hypo
4. Closo

1. Determine the type of cluster of compound [Ru5N(CO)14]-.

1. Arachno
2. Nido
3. hypo
4. Closo

1. Determine the type of cluster of compound [Ni5(CO)12]2-.

1. Arachno
2. Nido
3. hypo
4. Closo

1. Calculate the M-M- Bond present in [(Fe3 (CO)12]. 

1. 3
2. 0
3. 2
4. 1

Answer Key:

1. C
2. A
3. C
4. B
5. D
6. D
7. C
8. D
9. A
10. A

Solutions:

Solution 1. Determination of structure for Skeletal Electron Pair:

Step I → Calculate TVE (Total Valence e- count.)

Step II → Add the value of the valence electron of interstitial atoms.

E.g., H → 1 e-

H2 → 2 e-

Carbon family → 4 e-

Nitrogen family → 5 e-

Step III →Calculate skeletal electron Pair (S) as:

S = (TVE - n x 12)/2

Electronic configuration of Pt = 4d104s0

Pd is a 10 e- contributor whereas CN is a 2e- contributor in nature method. 

TVE = 5 ×10 + 12 × 2 +2 = 76

S = (76 – 12 × 5)/2 

= (76 – 60)/2 

Skeletal electron pair = 8

Solution 2. Determination of structure for Geometry: -

Step I → Calculate TVE (Total Valence e- count.)

Step II → Add the value of the valence electron of interstitial atoms.

E.g., H → 1 e-

H2 → 2 e-

Carbon family → 4 e-

Nitrogen family → 5 e-

Step II → Compare with the table of geometries.

 Electronic configuration of Rh = 4d85s1 

       Hence electronic configuration by Ru = 9 e-

   TVE = (9 × 6) + 5 + (17 × 2) -3

  = 54 + 5 + 34 -3

              = 90

Solution 3. Determination of structure for Geometry: -

Step I → Calculate TVE (Total Valence e- count.)

Step II → Add the value of the valence electron of interstitial atoms.

E.g., H → 1 e-

H2 → 2 e-

Carbon family → 4 e-

Nitrogen family → 5 e-

Step II → Compare with the table of geometries.

[Os5C(CO)15]

Electronic configuration of Os = 5d66s2 

 Hence, electronic configuration by Os= 8 e-

   TVE = (8 × 5) + 4 + (15 × 2) 

  = 40 + 4 + 30 

            = 74

Compare TVE with table as:

It is square Pyramidal.

Solution 4. Determination of structure for Geometry: -

Step I → Calculate TVE (Total Valence e- count.)

Step II → Add the value of the valence electron of interstitial atoms.

E.g., H → 1 e-

H2 → 2 e-

Carbon family → 4 e-

Nitrogen family → 5 e-

Step II → Compare with the table of geometries.

 We know that Os6 (CO)18 has a monocapped TBP structure.

Calculate total valence electrons as:

Electronic configuration of Os = 5d66s2 

       Hence, electronic configuration by Os= 8 e-

   TVE = (8 ×6) +(18 × 2)

  = 48 + 36  

            = 84

On adding 2 electrons, the total valence electron becomes 86.

Compare TVE with the table as:

It is an octahedron in shape.

Solution 5. Determination of Bonding molecular orbital: -

Step I → Calculate TVE (Total Valence e- count.)

Step II → Add the value of the valence electron of interstitial atoms.

E.g., H → 1 e-

H2 → 2 e-

Carbon family → 4 e-

Nitrogen family → 5 e-

Step III →Calculate Bonding molecular orbital

BMO = (TVE)/2 

Calculate T.V.E as: 

C = 4 e- Contribution because it is encapsulated atom

Os = 3d64s2    = 8 e- contribution

             = 4+(8 x 4) + (12 x 2) +2

               = 62 e-

BMO = Total valence electron /2

            =62/2

           = 31

Solution 6. Determination of structure for HNCC: -

Step I → Calculate TVE (Total Valence e- count.)

Step II → Add the value of the valence electron of interstitial atoms.

E.g., H → 1 e-

H2 → 2 e-

Carbon family → 4 e-

Nitrogen family → 5 e-

Step III →Calculate Skeletal electron Pair (S)

S = (TVE - n x 12)/2 

n = number of metals in the given cluster.

Step IV 

The structure is arranged according to the wade rule, if skeletal electron pair is,

S=n -1 Super closo

S=n Hyper closo

S=n +1 Closo

S=n+2 Nido

S=n +3 Arach no

S=n +4 Hypho

Calculate T.V.E as:

C = 4 e- Contribution because it is encapsulated atom

Os = 3d64s2    = 8 e- contribution

               = 4+(8 x 6) + (17 x 2) 

               = 86 e- 

 Skeletal Electron pair = (TVE – 12xN)/2

                                       = (86 – 12×6)/2

                                         = 7

                                     S = n+1 hence, it is an example of Closo

Solution 7. Determination of structure for HNCC: -

Step I → Calculate TVE (Total Valence e- count.)

Step II → Add the value of the valence electron of interstitial atoms.

E.g., H → 1 e-

H2 → 2 e-

Carbon family → 4 e-

Nitrogen family → 5 e-

Step III →Calculate Skeletal electron Pair (S)

S = (TVE - n x 12)/2 

n = number of metals in the given cluster.

Step IV 

The structure is arranged according to the wade rule, If skeletal electron pair is,

S=n -1 Super closo

S=n Hyper closo

S=n +1 Closo

S=n+2 Nido

S=n +3 Arachno

S=n +4 Hypho

           Ligand Electronic configuration

              Co           2

P(OMe)3           2

Calculate T.V.E of [Os6C(CO)17 {P(OMe)3}3] as:

                  C = 4 e- Contribution because it is encapsulated atom

               Os = 5d66s2    = 8 e- contribution

         = 4+(8 x 6) + (17 x 2) + (2 x 3)

           = 92

 Skeletal Electron pair = (TVE – 12xN)/2

                                       = (92 – 12×6)/2

                                       = 10

                                   10 = 6+4

              S = n+4 hence, it is an example of Hypho

Solution 8. Determination of structure for HNCC: -

Step I → Calculate TVE (Total Valence e- count.)

Step II → Add the value of the valence electron of interstitial atoms.

E.g., H → 1 e-

H2 → 2 e-

Carbon family → 4 e-

Nitrogen family → 5 e-

Step III →Calculate Skeletal electron Pair (S)

S = (TVE - n x 12)/2 

n = number of metals in the given cluster.

Step IV 

The structure is arranged according to the wade rule, if skeletal electron pair is,

S=n -1 Super closo

S=n Hyper closo

S=n +1 Closo

S=n+2 Nido

S=n +3 Arach no

S=n +4 Hypho

           Ligand Electronic configuration

              Co           2

Calculate T.V.E of [Ru5N(CO)14]- as:

N = 5 e- Contribution because it is encapsulated atom

 Ru = 4d75s1    = 8 e- contribution

               = 5+(8 x 5) + (14 x 2) + 1

               = 74

 Skeletal Electron pair = (TVE – 12xN)/2

                          = (74 – 12× 5)/2

                          S = 7

                        7 = 6+1

              S = n+1 hence, it is an example of Closo

Solution 9. Determination of structure for HNCC: -

Step I → Calculate TVE (Total Valence e- count.)

Step II → Add the value of the valence electron of interstitial atoms.

E.g., H → 1 e-

H2 → 2 e-

Carbon family → 4 e-

Nitrogen family → 5 e-

Step III →Calculate Skeletal electron Pair (S)

S = (TVE - n x 12)/2 

n = number of metals in the given cluster.

Step IV 

The structure is arranged according to the wade rule, if skeletal electron pair is,

S=n -1 Super closo

S=n Hyper closo

S=n +1 Closo

S=n+2 Nido

S=n +3 Arach no

S=n +4 Hypho

           Ligand Electronic configuration

              Co           2

Calculating T.V.E of [Ni5(CO)12]2-

      Ni= 3d85s2    = 10 e- contribution

               = (10 x 5) + (12 x 2) + 2

               = 76

 Skeletal Electron pair = (TVE – 12xN)/2

                             = (76 – 12×5)/2

                          S = 8

                           8 = 5+3

              S = n+3 hence, it is an example of arachno

Solution 10. Ligand Electronic configuration

                         CO             2

Fe = 3d65s2 → 8 e- contributor

T.V.E = 12 X 2 + 3 X 8 

          = 24 + 24

         = 48

Total M-M Bond = (18 X n – T.V.E)/2

= (54 – 48)/2 = 3

No. M-M Bond present in compound = 3

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