Most Expected Questions for Chemical Science - Organometallic Compounds! (Download PDF)

Most Important Questions on Organometallic Compounds (Download PDF)

1. Which of the following follows the 18 electron rule?

1. CpMn (CO)₂CPh₃
2. Fe (CO)₄
3. Cr (CO)₂ (NCMe)₂
4. IrCl₃(PPh₃)₂(AsPh₂)]^-

2. The structure of Rh6(CO)16 is:

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

3. Correct order of M-C bond order:

1. Cr (CO)₄ < [Ti (CO)₄ ]^-2 < [Mn(CO)₄]+ < [V(CO)₄]^-
2. [Mn (CO)₄]+ < [V(CO)₄]- < Cr(CO)₄ < [Ti(CO)₄ ]^-2
3. [Mn (CO)₄]+ < Cr(CO)₄ < [V(CO)₄]- < [Ti(CO)₄ ]^-2
4. [Cr(CO)₄] < [Mn(CO)₄]+ < [V(CO)₄]- < [Ti(CO)₄ ]^-2

4. In the oxidative addition of H₂ to [Ir (PPh₃)₂(CO)X], X can be Cl, Br, I , the rate order of rate with different X is:

1. [Ir(PPh₃)₂(CO)Cl] <[Ir(PPh₃)₂(CO)I] < [Ir(PPh₃)₂(CO)Br]
2. [Ir(PPh₃)₂(CO)I] < [Ir(PPh₃)₂(CO)Cl] < [Ir(PPh₃)₂(CO)Br]
3. [Ir(PPh₃)₂(CO)I] < [Ir(PPh₃)₂(CO)Br] < [Ir(PPh₃)₂(CO)Cl]
4. [Ir(PPh₃)₂(CO)Cl] < [Ir(PPh₃)₂(CO)Br] < [Ir(PPh₃)₂(CO)I]

5. [Fe(ƞ5cp)(CO)₂(C₂H₅)] cannot undergo beta hydride elimination because

1. It does not have beta hydrogen.
2. It cannot form a coplanar transition state.
3. It violates bredt’s rule.
4. It is a stable 18 e- complex.

6. Determine the Metal-Metal bond present in (n4-C4H4)2Fe2 (CO)3.

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

7. Determine the structure and type of centre respectively of compound [Fe₄(CO)₁₂C]^2- .

1. Octahedral, Arachno
2. Triangular Bipyramidal, Nido
3. butterfly, Arachno
4. Square Bipyramidal, Closo

8. On reducing Fe3(CO)12 with an excess of sodium, a carbonylated ion is formed. The ion is isoelectronic with:

1. [Mn(CO)₅]-
2. [Ni(CO)₄]
3. [Mn(CO)₅]+
4. [V(CO)₆]-

9. Which of the following sets is isolobal to BH?

1. Fe(CO)₄ , CH₂ , CpCo
2. Fe(CO)₃ , C , CpCo
3. Fe(CO)₃ , CH₂ , CpCo 
4. Fe(CO)₃ , CH₂ , CpMn

10. Determine the unknown quantity in [CpW (CO)X]₂. It is given that this complex contains a W-W single bond.

1. 6
2. 3
3. 4
4. 5

Solutions:

Solution 1:

In CpMn (CO)₃CPh₃ , contribution of CPh3 is 1 because it has 1 extra electron to donate.

Total valence electron count of CpMn(CO)3CPh3 = 5+7+(3×2)+1 = 19 e-

In Fe(CO)4 , Total valence electron count of Fe(CO)₄ = 8+ (4×2) = 16 e-

In Cr(CO)₂ (NCMe)₂ , contribution of (NCMe)2 is 2 because after forming 3 bonds with CMe , N has 1 lone pair left , so, total valence electron count of Cr(CO)2 (NCMe)2 : 6 +(2×2)+(2×2) = 14 e-

In IrCl3(PPh3)2(AsPh2)]-, contribution of AsPh2 is 1 because As belongs to the N family and they have a tendency to form 3 bonds. Here, As will form only 2 bonds, so 1 electron is left for bonding.

Total valence electron count of IrCl3(PPh3)2(AsPh2)]- = 9+(1×3) +(2×2)+1+1 = 18e-

Solution 2:

In [Rh6(CO)16], no. of metals >4 so, it is a high nuclearity metal carbonyl cluster in which 12 electrons are always present outside the Polyhedra.

Total electron count (TEC) of [Rh6(CO)16] = (6×9) +(16×2) =86

So, polyhedral electron pair = (TEC- 12×N)/2

= [86-(12×6)]/2 = 7

No. of electron pair = 6+1

Its structure is like [B6H6]-2.

So, the structure is closo.

Solution 3: 

Due to backbonding, there will be more e- density present in carbonyl antibonding orbital due to which bond strength of C-O bond decreases while the bond strength of M-C bond increases. More the electron density on the central metal atom, more will be the backbonding. Hence, the order of M-C bond order is:

[Mn (CO)4] + < Cr(CO)4 < [V(CO)4]- < [Ti(CO)4 ]-2

Solution 4:

In oxidative addition of any ligand, the increase in oxidation state by +2 takes place and along with this, increase in coordination no. of +2 takes place due to addition of two ligands. Oxidative addition occurs much more readily in case of electron rich metal. In all the given complexes, only the halide ligand differs. On changing X to Cl, Br, I, electron donation increases, metal becomes more electron rich, hence, rate of oxidative addition increases.

Solution 5:

[Fe(ƞ5cp)(CO)2(C2H5)] cannot undergo beta hydride elimination because it is a stable 18 electron complex due to which it directly cannot undergoes beta hydride elimination. It has beta hydrogen. It can also form coplanar transition state. Here, bredt’s rule is not applicable because this is applicable to cyclic bridged complexes.

Solution 6: 

Solution 7:

Solution 8:

Solution 9:

Solution 10:

Answer Key:

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

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