Organometallic Chemistry Questions-Answer| Detailed Solutions| Inorganic| CSIR NET, GATE

Organometallic Chemistry-I

Organometallic Chemistry Questions-Answer detailed solutions 

Organometallic compound is called metal attach with an organic compound. Or metal bond with an organic molecule such as Methyl bond with Lithium like this Me-Li, Ethyl group directed bond with Lithium like this Et-Li. Where lithium is metal so, we can call it organolithium compound. It is used as a catalyzed in organic synthesis.

Which books are best for Organometallic chemistry?

First of all, talk to inorganic organometallic chemistry, we have to refereed Ajay Kumar Organometallics & Bioinorganic chemistry book, and another most popular book B.D Gupta Organometallic chemistry. Or you can use it as a textbook Inorganic chemistry James E. Huheey, Ellen A. Keiter, Richard L. Keiter, fourth edition.

Questions with a solution on Organometallic chemistry

Few Tricky Questions with a solution of Organometallic Chemistry or Organometallic Compound including bellow. You can easy to solve those questions If you have clear those concepts of the organometallic chemistry chapter.

Q.1. Among the given boranes and heteroboranes, the example which belongs to the 'closo' type is

(D) B₅H₈^-

(B) [C₂B₉H₁₁]^2- 

(C) GeC₂B₉H₁₁

(D) B₆H₁₀

This is the organometallic chemistry Q.1

Answer: (C) GeC₂B₉H₁₁

Hints:  Ge has 4 valence electrons are present like BH, where BH has 4 valence electrons are present. So, Ge and BH two are equivalents.

• Similarly, Carbon has 4 valence electrons like BH. So, C and BH are equivalent two each other.

Solution of Question one.

Q.2. [CpMoCl₂]₂ obeys the 18-electron rule. The correct structure of this compound is (atomic number of Mo = 42)

(A)

(B)

(C)

(D)

This is the organometallic chemistry Question Number 2

 

Answer: (C)

Hints: 

Electrons count of organometallic compound [CpMoCl₂]₂ from molecular structure 

• Compound (A) Electron’s count = +6 (Mo) +5 (Cp)+2 (2Cl) +3 (M-M) =16 e’s
• Compound (B)Electron’s count = +6 (Mo) +5 (Cp) +2 (2Cl) +1 (M-M) =14 e’s
• Incase Compound (C) Electron’s count = +6 (Mo) +5 (Cp) +2 (2µ2 -Cl) +4 (2µ2 -Cl) +1 (M-M) =18 e’s
• In Compound (C) Electron’s count = +6 (Mo)+5 (Cp) +1 (Cl) +1 (µ2 -Cl) +2 (µ2 -Cl) +1 (M-M) =16 e’s

Solution Q.2

• Bridging Ligand µ2 -Cl contributes 3 electrons Mo-Cl-Mo, one bond makes one electron donate by Cl, and other bond forms donate the lone pair to metal.   

Q.3. Among the given platinum (II) complexes, the one that is thermally the most unstable is

(A)

(B)

(C)

(D)

Organometallic chemistry question number three.

Answer: (C)

Hints:

Total valence electron count = 10 + 2×2 + 1 + 1 = 16 e’s. All organometallic compound has the same electrons count. But 

• In the case of Compound (A), which absents beta hydrogen, that is why Beta elimination is not possible. So, this compound is stable.
• Compound (B) has ring stain, beta elimination rate comparatively lower than compound (C). So, compound B is comparatively less unstable.
• Compound (C) has Beta hydrogens and the beta elimination rate is high compared to other compounds. So, compound C is more unstable.
• Compound (D) presents beta hydrogens but beta elimination is not possible, due to the anti position of Metal and Beta hydrogen. 

For any Beta, elimination required syn position of metal and beta hydrogens.       

Compounds (C) have higher numbers of Beta Hydrogens compared to others. 

The higher number of Beta hydrogens in this compound is more unstable.

Q.4.The hapticity of cycloheptatriene, (C₇H₈), in Mo(C₇H₈)(CO)3 is ……………………………

(A)

(B) 6

(C)

(D)

Organometallic chemistry question number four.

Answer: (B) 6

Hints: 

The electrons count in Mo(C₇H₈)(CO)₃ organometallic compound gives

6 (Mo valence electrons) + x + 3 × 2 =18 e’s

6 + x + 6 = 18 e’s

X = 18 -12 = 6

• Where x = number of electrons contribution of C₇H₈ Ligand = Hapticity of the C₇H₈.

Therefore, the hapticity of this ligand = 6.

Q.5. The structure type and Shape of the polyhedral (skeletal) framework of the carbonate, Me₂C₂B₁₀H₁₀. respectively are

A) nido and dodecahedron

B) closo and icosahedron

C) nido and icosahedron

D) closo and dodecahedron

Organometallic chemistry question number five.

Answer: (B) closo and icosahedron

Hints:

• Polyhedral electrons count (PEC) = (TVE - 2×n)/2

Where, TVE = Total valence electrons and n = Number of metals.

This is for the non-transitional metal

• In case transitional metal PEC or Skeletal electrons count (SEC) = (TVE - 12×n)/2

Me₂C₂B₁₀H₁₀ compound TVE = 2×1 + 4×2 + 3×10 + 1×10 = 50 e’s

PEC or SEC = [50 - 2× (Number of Carbon + Number of Boron)]/2 = [50 - 2×(2+10)]/2 =[50 - 2×12]/2 = [50 – 24] = 26/2 = 13

• Number of Skeletal Electrons is 13 called Icosahedrons.
• Number of total metal =12, PEC = 12 +1 = n + 1, closo 

Q.6. The rate of substitution for the following vary with L in the order

A) CH3- >Cl- >Ph- > H-

B) Cl- >Ph- > H- > CH₃-

C) Ph- >CH3- > H- >Cl-

D) H- > CH₃->Ph- > Cl-

Organometallic chemistry question number six.

Answer: (D) H- > CH3->Ph- > Cl-

Hints:

Q.7.  The product formed in the reaction of MeMn(CO)₅ with ¹³CO is

A) (Me¹³CO)Mn(CO)5

B) (MeCO)Mn(CO)₅

C) (MeCO)Mn(CO)₄(¹³CO)

D) (Me¹³CO)Mn(CO)₄(¹³CO)

Organometallic chemistry question number seven.

Answer: (C) (MeCO)Mn(CO)₄(¹³CO)

Hints: 

This is one type of Migratory insertion reaction Follow those steps. The reaction rate is high or this reaction is faster than the normal substitution reaction. That is why the M.I (Migratory Insertion) is more favorable compared to substitution reactions.

Migratory Insertion reaction in organometallic chemistry.

• In case the CH3 group or ligand migrates on CO.
• Formation of a vacant on metal.
• Then ¹³CO Insertion this vacant orbital. 

The final product is (MeCO)Mn(CO)₄(¹³CO). 

Q.8. Hydroboration of 2-butyne with (C₆H₁₁)₂BH yields the intermediate U, which on treatment with I2 and NaOMe at -78 C gives product V. The structure of U and V are

(A)

(B)

(C)

(D)

Organometallic chemistry question number eight.

Answer:  (D)

Hints: