Metal Carbonyls and Metal Nitrosyls MCQ Quiz - Objective Question with Answer for Metal Carbonyls and Metal Nitrosyls - Download Free PDF

CONCEPT:

Reactivity of Metal Carbonyl Complexes with Hydride Sources

• Transition metal carbonyls like Cr(CO)₆ can undergo substitution and reduction reactions with nucleophiles like NaBH₄.
• NaBH₄ acts as a hydride donor, replacing a CO ligand and forming a metal–hydride complex like [Cr(CO)₅H]^-.
• This hydride complex (A) is electron-rich and can act as a nucleophile or Lewis base toward other metal carbonyls.
• Oxidative addition of H₂ can also occur on 16-electron hydride complexes to form bridging or terminal dihydrogen species.

EXPLANATION:

• Step 1: Reaction of Cr(CO)₆ with NaBH₄
  • One CO is replaced by a hydride from BH₄^-, giving: A = [Cr(CO)₅H]^-
• Step 2: Compound A reacts with another Cr(CO)₆ molecule
  • The hydride bridges both Cr centers with CO loss, forming: B = [(CO)₅Cr–H–Cr(CO)₅]^-
• Step 3: Compound A reacts with BH₃
  • BH₃ accepts the hydride from Cr–H to form a Cr–BH₄ bond: C = [Cr(CO)₄BH₄]^-

Step 1: Formation of Compound A

Cr(CO)₆ reacts with NaBH₄. NaBH₄ is a hydride donor and replaces one CO ligand with a hydride (H^-), producing:

Cr(CO)₆ + NaBH₄ → [Cr(CO)₅H]^- + CO + Na⁺

Compound A = [Cr(CO)₅H]^-

Step 2: Formation of Compound B

Compound A reacts with another molecule of Cr(CO)₆. The hydride bridges the two chromium centers, and one CO ligand is released:

[Cr(CO)₅H]^- + Cr(CO)₆ → [(CO)₅Cr–H–Cr(CO)₅]^- + CO

Compound B = [(CO)₅Cr–H–Cr(CO)₅]^-

Step 3: Formation of Compound C

Compound A also reacts with BH₃, in which the hydride shifts to BH₃, forming a Cr–BH₄ unit with loss of one CO ligand:

[Cr(CO)₅H]^- + BH₃ → [Cr(CO)₄BH₄]^- + CO

Compound C = [Cr(CO)₄BH₄]^-

• This matches the species given in Option 2 exactly.

Therefore, the correct answer is: Option 2

CONCEPT:

CO Displacement and Dihydrogen Complex Formation in Organometallic Chemistry

• Metal carbonyls like Mo(CO)₆ are known to undergo substitution reactions with neutral ligands such as trialkylphosphines (PR₃).
• This reaction leads to the stepwise displacement of CO ligands and formation of mixed-ligand complexes like [Mo(CO)_x(PR₃)_y].
• These types of complexes can further react with H₂ to form dihydrogen complexes, in which H₂ binds in an η² (side-on) fashion to the metal center.

EXPLANATION:

​Mo(CO)6 + 2 PPr3 → [Mo(CO)3(PPr3)2]
[Mo(CO)3(PPr3)2] + H2 → [Mo(CO)3(PPr3)2(η2-H2)]

• Starting complex: Mo(CO)₆
• Upon reaction with P(i-Pr)₃, some CO ligands are displaced. A stable configuration is:
  • Compound A = [Mo(CO)₃(P(i-Pr)₃)₂]
  • 3 CO + 2 bulky phosphines = 18-electron complex
• Upon addition of H₂, the complex can bind H₂ in an η²-mode (side-on), without oxidative addition:
  • Compound B = [Mo(CO)₃(P(i-Pr)₃)₂(η²-H₂)]
• This matches the pair given in Option 2.

Correct answer [Mo(CO)₃(P(i-Pr)₃)₂] and [Mo(CO)₃(P(i-Pr)₃)₂(η²-H₂)]

CONCEPT:

M-C Bond Length in Metal Carbonyl Complexes

• The M-C bond length in metal carbonyl complexes is primarily influenced by the degree of π-backbonding.
• π-backbonding occurs when electrons are transferred from the filled metal d-orbitals to the π*-orbitals of the CO ligand.
• The extent of π-backbonding depends on the electron density of the metal, which is influenced by its oxidation state.

EXPLANATION:

• [V(CO)₆]⁻:
  • The vanadium metal has a negative charge, which increases its electron density and enhances π-backbonding with the CO ligands.
  • Enhanced π-backbonding strengthens the metal-carbon π-bond but weakens the C-O bond, leading to a shorter M-C bond.
• [Cr(CO)₆]:
  • Chromium is in a neutral oxidation state, so its electron density and π-backbonding ability are moderate.
  • The M-C bond length in this case is intermediate compared to [V(CO)₆]⁻ and [Mn(CO)₆]⁺.
• [Mn(CO)₆]⁺:
  • Manganese has a positive charge, reducing its electron density and weakening π-backbonding with the CO ligands.
  • Weaker π-backbonding leads to a longer M-C bond.

BOND LENGTH ORDER:

• Shortest M-C Bond: [V(CO)₆]⁻ (strongest π-backbonding)
• Intermediate M-C Bond: [Cr(CO)₆] (moderate π-backbonding)
• Longest M-C Bond: [Mn(CO)₆]⁺ (weakest π-backbonding)

Therefore, the correct order of M-C bond length is: [V(CO)₆]⁻ < [Cr(CO)₆] < [Mn(CO)₆]⁺.

Concept:

The metal-metal (M-M) bond order in transition metal complexes can be analyzed by calculating the total electron count and the number of electrons involved in bonding. A higher bond order indicates a shorter bond length. The bond order is determined based on the filling of bonding molecular orbitals: σ, π, and δ orbitals.

Explanation:

• [Mo₂(SO₄)₄]⁴⁻:

  • Charge on complex = -4.

  • Applying the formula: -4 = 2M + (-8) (since each SO₄ group is -2).

  • Solving: 2M = 4,

  • Each Mo in +2 oxidation state contributes 6 electrons, so total electrons from two Mo atoms = 6 + 6 = 12 electrons.

  • Electrons in the complex for bond formation = Total electrons - Non-bonding electrons.

  • Electrons in complex = 12 - 4 (non-bonding) = 8 bonding electrons.

  • These 8 bonding electrons are filled as: σ²π⁴δ², giving a bond order of 4.

• [Mo₂(SO₄)₄]³⁻:

  • Charge on complex = -3.

  • Applying the formula: -3 = 2M + (-8) (since each SO₄ group is -2).

  • Solving: 2M = 5,

  • Each Mo in +2 oxidation state contributes 6 electrons, so total electrons from two Mo atoms = 6 + 6 = 12 electrons.

  • Electrons in the complex for bond formation = Total electrons - Non-bonding electrons.

  • Electrons in complex = 12 - 5 (non-bonding) = 7 bonding electrons.

  • These 8 bonding electrons are filled as: σ²π⁴δ¹, giving a bond order of 3.5.

• [Mo₂(HPO₄)₄]²⁻:

  • Charge on complex = -2.

  • Applying the formula: -2 = 2M + (-8) (since each HPO₄ group is -2).

  • Solving: 2M = 6,

  • Each Mo in +2 oxidation state contributes 6 electrons, so total electrons from two Mo atoms = 6 + 6 = 12 electrons.

  • Electrons in the complex for bond formation = Total electrons - Non-bonding electrons.

  • Electrons in complex = 12 - 6 (non-bonding) = 6 bonding electrons.

  • These 8 bonding electrons are filled as: σ²π⁴, giving a bond order of 3.

• ​Lesser the bond order, longer will be the bond length:
  • 
     

Conclusion:

The correct order of M-M bond length is: Option 1: [Mo₂(SO₄)₄]⁴⁻ < [Mo₂(SO₄)₄]³⁻ < [Mo₂(HPO₄)₄]²⁻.

Concept:

In organometallic chemistry, the reactivity and structure of complexes can be studied using IR spectroscopy, as different types of carbonyl (CO) bonds exhibit distinct IR frequencies. For example, terminal CO ligands display higher stretching frequencies, while bridging or acyl CO ligands show lower frequencies. Additionally, ligand substitution and rearrangement reactions, as seen in this problem, can alter these IR frequencies due to changes in the bonding environment around the metal center.

• Nucleophilic Displacement: This reaction involves a nucleophilic displacement, where the sulfur-containing ligand (ClCH₂CH₂SCH₃) displaces chloride, forming a new Fe-S bond. This is typical of nucleophilic substitution reactions in organometallic complexes.
• IR Frequency Interpretation: The IR bands at higher frequencies (1980 and 1940 cm^-1) correspond to terminal CO ligands, while the lower frequencies (1920 and 1630 cm^-1) suggest the presence of a bridging or acyl CO bond.

Explanation:

The reaction of [(η⁵-C₅H₅)Fe(CO)₂]^- with ClCH₂CH₂SCH₃ proceeds as follows:

[(η⁵-C₅H₅)Fe(CO)₂]^- + ClCH₂CH₂SCH₃ → A (C₁₀H₁₂FeO₂S)

A + heat → B

Complex A has IR bands at 1980 and 1940 cm^-1, indicating the presence of terminal CO ligands. Upon heating, A rearranges to form complex B, which shows IR bands at 1920 and 1630 cm^-1, consistent with the formation of a bridging or acyl CO bond. The reaction mechanism involves:

• Formation of Complex A: The sulfur reagent loses Cl^- and bonds to Fe as an alkyl ligand, forming A with two terminal CO ligands.
• Rearrangement to Complex B: Upon heating, the S atom bonds to Fe, and the alkyl group migrates to a carbonyl carbon, forming an acyl C=O bond in B. This results in two different C=O stretching frequencies due to the presence of an ordinary carbonyl and an acyl CO bond.

The reaction and structures of A and B are shown below:

Conclusion:

The correct answer is Option 2, as it accurately depicts the structures of A and B based on the IR data and reaction mechanism.

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.​

​

• Every ligand is a σ donor at first. 
• CO donates its electron to metal at first and forms a 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 *. 

​

• 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 a 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 [Mn(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 Mn(CO)6+

• In the complex [Os(CO)6]2+, there is two positive charge on the metal atom and six CO ligands.
• Due to the presence of two positive charges, the extent of metal-ligand back donation is low.
• Thus, the CO stretching frequency in [Os(CO)6]2+ will be higher than the complex Mn(CO)6+

• In the free CO, the CO stretching frequency will be more than [Mn(CO)6]+

• Hence, the order of decreasing carbonyl stretching frequencies is C > B > D > A

Concept:

NO ligand

• These are the derivatives of NO⁺
• Two types of NO ligands are there, one who donate1 electron (bent) and the other who donates three electrons (linear)

CO ligand

• CO ligand is σ-donor as well as Π-acceptor

Explanation:

The IR stretching frequency for nitrosyls is as follow:

ν_NO (bent NO) = 1525-1690 cm^-1

ν_NO (linear NO) = 1650-1900 cm^-1 

The molecular orbital diagram of NO is given below:

Here in this energy level diagram you can see that the HOMO of the NO ligand is Π* orbital.

Now you see the energy level diagram of CO ligand:

Here for CO ligand the HOMO is σ orbital.

Therefore from the above explanation:

• The NO ligand in bent form donates one electron, therefore statement A is incorrect
• The IR stretching frequency of the nitrosyl ligand also lie in between the given range of statement, therefore this statement B is correct.
• The HOMO of NO ligand and CO ligands are π* and σ respectively, therefore statement C is correct. 

The correct statement are B and C.

Conclusion:

Hence the correct answer is B and C.

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

Explanation:-

The reaction pathway is shown below

Conclusion:-

• Hence, the compound D formed in the following reaction sequence is​ C6H6

The correct answer is \(\eta^5\)-Cp2Ni

Concept:-

• Oxidation states: Understanding oxidation states is vital to solving this problem. The oxidation state, also called oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound.
• Stability of oxidation states: Typically, metals in lower oxidation states are more susceptible to oxidation because they can lose more electrons readily. Conversely, those in higher oxidation states are usually less favorable for further oxidation (although there may be exceptions according to the specific chemistry of the element).
• Metallocenes: The compounds given in the question are metallocenes, where a metal atom is sandwiched between two cyclopentadienyl anions. These are highly conjugated systems making them relatively stable.
• Transition metals: Iron (Fe), Cobalt (Co), and Ruthenium (Ru) are transition metals. They can show various oxidation states. The chemistry of transition metals centers around the variability of their oxidation states, the formation of complex ions, and the colors their compounds can produce.
•  
• EAN rule:   
• The EAN rule is given by Sidgwick.
• According to the EAN rule, the EAN of metal is equal to the sum of electrons on metal and the electrons donated by the ligands and the EAN is equal to the atomic number of the next noble gas.
• The EAN rule is applicable to organometallic complexes. The rule is similar to the 18 electron rule.
• According to this rule, the sum of valence electrons of transition metals or metal ions is equal to 18.
• The complexes in which the EAN rule is followed are considered stable. The first row of transition metal carbonyls mostly obeys the 18-electron rule. There are many complexes that do not follow the 18 electron rule 

the EAN of \(\eta^5 \)-Cp2Fe is 5+5+8=18 : stable ,so it will not react

for,\(\eta^5 \) -Cp2CoCl is 5+5+9+1=20, unstable can react.

for,\(\eta^5\) Cp2Ni is 5+5+10=20, unstable can react.

for, \(\eta^5 \) Cp2Co is 5+5+9=19, unstable can react.

replacing \(\eta^5\)Cp with (will reduce EAN with 5) nitric oxide NO-(will increase EAN by 3 ) in liner form.

So, options A and D are wrong, B and C can react but Nickel in η⁵ -Cp₂Ni is indeed in a state where it's susceptible to ligand substitution. Ni(II) complexes are often high-spin and thus often more reactive, which can make them more prone to exchange one ligand for another.

Nitric oxide is both a strong π-acceptor and a strong σ-donor ligand. Its ability to interact strongly with the metal center and to facilitate ligand substitution can be enhanced in situations where the metal is electron-rich and has an available coordination site, as is true with the Ni(II) in η⁵ -Cp₂Ni.

If you were given that η⁵ -Cp₂Ni is the easiest for the substitution of the η⁵-Cp group with nitric oxide, it's possible that specific experimental evidence or high-level computational chemistry results have demonstrated this reactivity preference of η⁵ -Cp₂Ni in the context of the other metallocene compounds listed.

Conclusion:-

so, The substitution of η5 -Cp  group with nitric oxide is the easiest for η5 -Cp2Ni

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.​

​

• Every ligand is a σ donor at first. 
• CO donates its electron to metal at first and forms a 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 *. 

​

• 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 a 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 [Mn(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 Mn(CO)6+

• In the complex [Os(CO)6]2+, there is two positive charge on the metal atom and six CO ligands.
• Due to the presence of two positive charges, the extent of metal-ligand back donation is low.
• Thus, the CO stretching frequency in [Os(CO)6]2+ will be higher than the complex Mn(CO)6+

• In the free CO, the CO stretching frequency will be more than [Mn(CO)6]+

• Hence, the order of decreasing carbonyl stretching frequencies is C > B > D > A

Concept:-

• Oxidative Addition involves the increase in both the oxidation state and coordination number of the metal center by 2 units by two new metal-ligand bonds.
• An example of an oxidative reaction is shown below:

Explanation:-

The reaction pathway for the Oxidative Addition reaction is shown below:

• In the above reaction, it is shown that Me-I undergoes oxidative addition reaction to Ir metal center through anti-addition.

Mechanism:

Conclusion:-

Hence,  the product in the reaction will be

Correct option is (a) 

Concept:-

• Absorptions in IR spectra are usually quoted in wavenumber as:

\(\nu _{CO} (cm^{-1}) =\frac{1}{2\pi }\sqrt{\frac{k}{\mu }}\)

• For a mode of vibration to be Infrared (IR) active, it must give rise to a change in the molecular electric dipole moment.
• The carbonyl stretching modes (\(\nu _{CO}\)) for some mononuclear metal carbonyl complexes are given below; where X is a general group other than CO.

​

Explanation:-

• For the metal carbonyl complex M(CO)5X, the point group is C_4v.
• The symmetries of CO stretching modes for the C_4v point group is

A₁, A₁, B₁, and E.

• Among these vibrational modes only A1, A1, and E are IR active.
• Thus for the complex M(CO)5X, the number of carbonyl stretching modes observed in the IR spectra is 3.

Conclusion:-

• Hence, option 1 is correct.

CONCEPT:

M-C Bond Length in Metal Carbonyl Complexes

• The M-C bond length in metal carbonyl complexes is primarily influenced by the degree of π-backbonding.
• π-backbonding occurs when electrons are transferred from the filled metal d-orbitals to the π*-orbitals of the CO ligand.
• The extent of π-backbonding depends on the electron density of the metal, which is influenced by its oxidation state.

EXPLANATION:

• [V(CO)₆]⁻:
  • The vanadium metal has a negative charge, which increases its electron density and enhances π-backbonding with the CO ligands.
  • Enhanced π-backbonding strengthens the metal-carbon π-bond but weakens the C-O bond, leading to a shorter M-C bond.
• [Cr(CO)₆]:
  • Chromium is in a neutral oxidation state, so its electron density and π-backbonding ability are moderate.
  • The M-C bond length in this case is intermediate compared to [V(CO)₆]⁻ and [Mn(CO)₆]⁺.
• [Mn(CO)₆]⁺:
  • Manganese has a positive charge, reducing its electron density and weakening π-backbonding with the CO ligands.
  • Weaker π-backbonding leads to a longer M-C bond.

BOND LENGTH ORDER:

• Shortest M-C Bond: [V(CO)₆]⁻ (strongest π-backbonding)
• Intermediate M-C Bond: [Cr(CO)₆] (moderate π-backbonding)
• Longest M-C Bond: [Mn(CO)₆]⁺ (weakest π-backbonding)

Therefore, the correct order of M-C bond length is: [V(CO)₆]⁻ < [Cr(CO)₆] < [Mn(CO)₆]⁺.

Concept:-

13C NMR:

• The Carbon-13 nuclear magnetic resonance or 13C NMR spectroscopy is the application of nuclear magnetic resonance spectroscopy to carbon (C). 13C NMR spectroscopy helps to identify the nature or environment of C atoms in organic molecules just as 1H NMR spectroscopy helps to detect the H atoms.
• The nature or environment of C atoms in organic molecules can be identified by chemical shift (δ) values.
• The chemical shift values of some common functional groups are as follows:

• In multinuclear NMR all spin-active nuclei can couple with each other and that the multiplicity of the coupling is given by

Spin multiplicity = 2nI + 1

where n = the number of equivalent nuclei that are being coupled to

and I  = magnetic spin number

Explanation:-

• For the complex [(\(\eta^5\))CpRh(CO)3], if the CO is present as terminal CO, then each CO is attached to only one NMR active Rh metal.

• Thus, the multiplicity will be

​= (2 × 1 × \(\frac{1}{2}\)) + 1   [n =1, 103Rh, I=1/2 (100%)]

= 2 (doublet)

• This does not satisfy the condition that shows a triplet at -65°C.
• For the complex [(\(\eta^5\))CpRh(CO)3], if the CO is present as μ2-CO, then each CO is attached to two NMR active Rh metal.

• Thus, the multiplicity will be

​= (2 × 2 × \(\frac{1}{2}\)) + 1   [n =2, 103Rh, I=1/2 (100%)]

= 3 (triplet)

• This satisfies the condition that shows a triplet at -65°C.
• For the complex [(\(\eta^5\))CpRh(CO)3], if two CO is present as μ2-CO and one CO is present as terminal CO, then it does not satisfy the condition that shows a triplet at -65°C.

Conclusion:-

• Hence, the compound [(\(\eta^5\))CpRh(CO)3] shows a triplet at -65°C due to the presence of μ2-CO