Top Questions-Answer on Organometallic Chemistry with detailed solution| Part-3| CSIR NET GATE

Organometallic chemistry questions and answer sessions for CSIR NET GATE IIT JAM, and equivalent entrance examinations like DRDO, NTPC, TIFR, BHU.

Organometallic Chemistry -3 Questions and Answer sessions

Hello, everyone today we will discuss organometallic chemistry -3 questions and answer sessions for CSIR NET GATE IIT JAM, and equivalent entrance examinations like DRDO, NTPC, TIFR, BHU, etc. We are already discussing a few topics under organometallic compound, hapticity, Metal-Metal bond count, 18 electrons rule of a stable compound, beta elimination reaction, etc. You can go through and check back click here.

Come back to the point after checking that session we hope to develop some ideas on the organometallic compound and catalyst chapter. Now today we will discuss some of the top 10 tricky and logical questions and solutions on organometallic compound chemistry which may help develop more thought knowledge and information about these topics.

The best books on organometallic chemistry

We have to earlier talk about the best books on organometallic chemistry. We will reply again that the best books are organometallic chemistry by Ajay Kumar or Inorganic chemistry Principles of Structure and Reactivity by James E. Huheey, Ellen A. Keiter Richard L. Keiter. Those books help you to build a concept.

Let’s start the Question-Answer session with a detailed solution that helps lots of practices and mindset on exam hole.

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That’s important everyday practice questions and answers topic-wise

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That is why we have to suggest that lots of question practice are important for quality build and good mark or good rank any national level examination. Before the exam, you have to do it.

Questions-Answer solution:

Now come back to our main topics, today we will discuss organometallic chemistry question-answer detailed solution. mainly we have to solve those questions but first, you try to understand and read the question and solve it. After that check the answer paper. If have any mistakes by you or need to improve then check.

Q.7. Binding modes of NO are 18 electron compounds [Co(CO)3 (NO)] and [Ni(πœ‚πŸ“ –Cp)(NO)], respectively, are

A. Linear and bent         B. bent and linear

C. Linear and linear      D. bent and bent

Organometallic Chemistry-3 Questions.7
Organometallic Chemistry-3 Questions.7

Answer: Correct option is (C) Linear and linear 


The binding modes of Nitrosyl (NO), It may be either linear (electron count =3) or bent (electron count =1).

Binding mode of NO (Linear or bent)
Binding mode of NO (Linear or bent)

Electron count is according to covalent electron count. 

The valence electron of Co = 9 and Ni =10 and πœ‚πŸ“ –Cp =5; each CO =2 and for linear NO =3 and bent NO =1.

The valence electron count of metal complex [Co(CO)3 (NO)] is 9 + 3 × 2 + x =  9 + 6 + x 

The metal complex follows the 18 electron stability rule,

9 + 6 + x =18 ; x =3; linear


The valence electron count of metal complex [Ni(πœ‚πŸ“ –Cp)(NO)] is 10 + 5 + x

The metal complex follows the 18 electron stability rule,

10 + 5 + x =18 ; x =3; linear

Q.8. The role of copper salt as co-catalyst in the Wacker process is

A. Oxidation of Pd(0) by Cu(II)

B. Oxidation of Pd(0) by Cu(I) 

C. Oxidation of Pd(II) by Cu(I)

D. Oxidation of Pd(II) by Cu(II)

Organometallic Chemistry-3 Questions.8
Organometallic Chemistry-3 Questions.8

Answer: Correct option is (A) Oxidation of Pd(0) by Cu(II)


The reaction of the Wacker process is

Pd  + 2CuCl2  → PdCl2  + 2CuCl

Clearly, we see that oxidation of Pd(0) to Pd(II)

As a co-catalyst, the role of copper salt (Cucl2) in the Wacker process is the oxidation of Pd(0) by Cu(II).

Q.9.  For typical Fischer and Schrock carbenes, consider the following statements

A. The oxidation state of the metal is low in Fischer carbene and high in Schrock carbene

B. Auxiliary ligands are 𝝅-acceptor in Fischer carbene and non- 𝝅-acceptor in Schrock carbene

C. Substituents on carbene carbon are non- 𝝅-donor in Fischer carbene and 𝝅-donor in Schrock carbene

D. Carbene carbon is electrophilic in Fischer carbene and nucleophilic in Schrock carbene

The correct statements are

1. A, B, and C              2. A, B, and D          3. B, C, and D             4. A, C, and D

Organometallic Chemistry-3 Questions.9
Organometallic Chemistry-3 Questions.9

Answer: The correct option is (2) A, B, and D


Before attempting, this question we have to understand what is the difference between the Fischer carbene complex and the Schrock carbene complex.

Fischer-type carbene complex Schrock-type carbene complex
Low oxidation state metal center high oxidation state metal center
Good Ο€-acceptor metal ligandsGood Ο€-donor metal ligands
Middle and late transition metals Fe(0), Mo(0), Cr(0) Early transition metals Ti(IV), Ta(V)
Ο€-donor substituents on the carbene atom such as alkoxy and alkylated amino groups. Hydrogen and alkyl substituents on carbenoid carbon.
Carbene carbon is Electrophilic in Fischer carbene. Carbene carbon is a Nucleophile in Fischer carbene.
An example of a Fischer-type carbene complex is (OC)5Cr=C(NR2)Ph.  An example of a Schrock carbene complex is  Ta(=C(H)But)(CH2But)3

Difference between Fischer-type carbene complex and Schrock-type carbene complex with molecular bonding wise.

Difference between Fischer-type carbene complex and Schrock-type carbene complex (MO)
Bonding model of Difference between Fischer-type carbene complex and Schrock-type carbene complex

We see the electron-deficient carbon (which acts as an electrophile) in the Fischer carbene complex and on the other hand electron-rich carbon ligand (Which acts as a nucleophile) in the Schrock carbene complex.

Q.10.  The refluxing of RhCl3.3H2O with an excess of PPh3 in ethanol gives a complex A. Complex A and the valence electron count on rhodium are, respectively,

A. [RhCl(PPh3) 3], 16        B. [RhCl(PPh3)5], 16

C. [RhCl(PPh3)3], 18        D. [RhCl(PPh3) 5], 18

Organometallic Chemistry-3 Questions.10
Organometallic Chemistry-3 Questions.10

Answer: The correct option is (A). [RhCl(PPh3)3], 16


The preparation of Wilkinson's catalyst

RhCl3.3H2O  +  excess of PPh3   → [RhCl(PPh3)5]

The electron count of Wilkinson's catalyst [RhCl(PPh3) 3] is 9 + 1 + 3  × 2 = 9 + 1 + 6 = 16 e complex.

Q.11. The 𝛽-hydrogen elimination will be facile in

A. 1             B. 2               C. 3                      D. 4

Organometallic Chemistry-3 Questions.11
Organometallic Chemistry-3 Questions.11

Answer: Correct option is (A) 1


The complex is stable when beta hydrogen is absent. If Beta hydrogen is present in the metal complex then the metal complex is unstable due to the formation of beta hydride elimination. And produced the stable olefine (condition it will be stable) and Metal hydrogen bond.

Beta hybrid elimination (sp3 hybridized)
Beta hybrid elimination (sp3 hybridized) facile

Another condition of beta elimination is beta position carbon (or any atom) will be sp3 hybridized. If facile elimination occurs the condition is sp3 hybridized carbon (attach beta hydrogen). 

Q.12.  The W-W bond order in [W(πœ‚πŸ“ –C5H5)(Β΅-Cl)(CO)2]2 is

A. Three           B. Two

C. One              D. Zero

Organometallic Chemistry-3 Questions.12
Organometallic Chemistry-3 Questions.12

Answer: correct option is (D) Zero


The number of Metal-Metal bond = (18n - V.E) / 2

Where n is the total number of metal present in the metal complex. And V.E = Velence Electrons count.

W-W bond order calculation.
W-W bond order calculation.

The valence electron count (V.E) of metal complex [W(πœ‚πŸ“ –C5H5)(Β΅-Cl)(CO) 2]2 is 6× 2 + 5 ×2 + 3 × 2 + 2 × 4 = 12 + 10 + 6 +8 = 36.

Where the number of metal atoms (n) = 2.

Valence electron of W = 6; πœ‚πŸ“ –C5H5 =5; CO =2; and Bridnig ligand Β΅-Cl = 3;

The W-W bond order = (18 × 2 - 36) /2 = (36 - 36)/2 = 0. The bond order of W-W is zero in the given metal complex.

Q.13.  For the reaction of [Fe(πœ‚5 -C5H5)(CH3)(CO)2] with PMe3, the main intermediate is

1. [Fe(πœ‚5 -C5H5)(CH3)(CO) 2(PMe3)]

2. [Fe(πœ‚5 - C5H5)(COCH3)(CO)]

3. [Fe(πœ‚3 - C5H3)(CH3)(CO) 2]

4. [Fe(πœ‚3 - C5H5)(COCH3)(CO)(PMe3)]

Organometallic Chemistry-3 Questions.13
Organometallic Chemistry-3 Questions.13

Answer: correct option is [Fe(πœ‚5 - C5H5)(COCH3)(CO)]


For the fast rate of the reaction formation of Migratory insertion after the addition of PMe3 ligand.

Another path is possible: First (i) Dissociation after that (ii) Addition. The whole reaction is the dissociative reaction. But this reaction does not favor kinetically. This path reaction is slow compared to the Migratory insertion path.

Migratory insertion reaction intermediate (According to the faster rate of the reaction).

The starting complex [Fe(πœ‚5 -C5H5)(CH3)(CO)2] is 18 electron species and stable. The reaction with PMe3.

And the formation of the final product [Fe(πœ‚3 - C5H5)(COCH3)(CO)(PMe3)] (the electron count18
stability) via the intermediate complex  [Fe(πœ‚5 - C5H5)(COCH3)(CO)] (the electron count 16 e ). It is PMe3 induced migratory carbonyl insertion reaction.

Q.14.  Among the following, species expected to show fluxional behavior are

A. [NiCl4]2- (tetrahedral),

B. IF7 (pentagonal bipyramidal),

C. [CoF6]3- (octahedral),

D. Fe(CO) 6 (trigonal bipyramidal)

1. B and C            2. B and D

3. C and D            4. A and D

Organometallic Chemistry-3 Questions.14
Organometallic Chemistry-3 Questions.14

Answer: Correct option is (B) IF7 (pentagonal bipyramidal)


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