Metal Carbonyls and Metal Nitrosyls MCQ Quiz in தமிழ் - Objective Question with Answer for Metal Carbonyls and Metal Nitrosyls - இலவச PDF ஐப் பதிவிறக்கவும்

Na[Mn(CO)₅] + CH₂ = CH - CH₂ - Cl → 

Mechanism:

Correct option is (a)

Concept

• Back bonding occurs as electrons pass from one atom’s atomic orbital to another atom’s or ligand’s anti-bonding orbital.This form of bonding will occur between atoms in a compound when one atom has a lone pair of electrons and the other has a vacant orbital next to it.​

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• Every ligand is a σ donor at first. 
• CO donates its electron to metal at first and forms bond with it.
• The metal will donate electrons to carbon and shows back bonding.The electrons of metal will enter LUMO (lowest u occupied molecular orbital) of CO which is antibonding molecular orbital π ^*. 

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• As we know whenever an electron enters in an antibonding orbital then the bond order decreases and hence the bond strength also decreases. So due to back bonding between metal and carbons the CO bond strength decreases and there arises slight double bond character between metal and carbon, and the bond order between the carbon and oxygen in carbonyl (CO) decreases from three to two.
• Hence M-C bond strength increase and the C-O bond strength decreases.
• M-C bond strength ∝ 1 / C-O bond strength
• Stretching frequency is given by \(ν= {{1 \over 2 \pi c} \sqrt{k \over μ} } \)   where k represents bond strength. According to this ν (frequency) ∝ k (bond strength)
• Thus, the greater the negative charge on the metal, the more will be the extent of back bonding with the carbonyl (CO) group, and hence lower will be its CO stretching frequency.

Explanation: 

• In the complex V(CO)6+, there is a positive charge on the metal atom and six (CO) ligand. 
• As there is no negative charge on the metal for the delocalization with the carbonyl (CO) group, the CO stretching frequency will be higher in the complex V(CO)6+

• In the complex Co(CO)4-, there is a negative charge on the metal atom and four CO ligands.
• The mono negative charge will undergo delocalization with the four CO ligands, thus the CO stretching frequency will be lower in the complex Co(CO)4-

• In the complex Fe(CO)4-2, there is two negative charges on the metal atom and four CO ligands.
• The two negative charges will undergo delocalization with the four CO ligands, thus the CO stretching frequency will be least in the complex Fe(CO)4-2
• Due to two negative charges, the CO stretching frequency in the complex Fe(CO)4-2 will be lower than Co(CO)4-

• Hence, Fe(CO)4-2 will have the lowest CO stretching frequency.

Concept:

• The number of CO bands in the IR spectrum can be helpful in determining the isomers.
• Group theory is usually applied to predict the number of CO-stretching bands
• A popular application of it is the distinction of isomers of metal carbonyl complexes.

Explanation:

Set (i): trigonal bipyramidal isomers

(A) axial‐Fe(CO)4L belongs to the point group of C_3V, it shows 3 CO bands in IR spectroscopy.

(B) equitorial‐Fe(CO)4L belongs to C_2v point group. So, it gives 4 CO bands in IR.

Set (ii): Octahedral isomers

(C) fac‐Mo(CO)3L₃ gives 2 CO bands in IR.

(D) mer‐Mo(CO)3L₃ gives 3 CO bands in IR.

Conclusion:

number of CO bands in IR for given isomers is:

(A) 3

(B) 4

(C) 2

(D) 3