Organometallic Compounds MCQ Quiz - Objective Question with Answer for Organometallic Compounds - 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)]

Explanation of Isomer Shifts in Sn Mössbauer Spectra:

The isomer shift in Mössbauer spectroscopy is related to the electron density around the nucleus. The general rule is:

• Electron Donor Ligands increase electron density at the central metal and cause a higher isomer shift.
• Electron Acceptor Ligands withdraw electron density from the central metal, resulting in a lower isomer shift.

Analysis of the Compounds:

• Compound W: Contains CO₅ and P⁻ ligands. Phosphine (P⁻) is an electron donor, increasing electron density at the tin (Sn) center. This results in a higher isomer shift.
• Compound Y: Contains Cl⁻ (chlorine), which is an electron-withdrawing ligand. This decreases electron density on the tin (Sn) center, leading to a lower isomer shift compared to W.
• Compound Z: Similar to Y, with a lower isomer shift than W.

Therefore, the correct option is W > Y.

CONCEPT:

IR Frequencies of Carbonyl Complexes and Electron Donation

• CO stretching frequencies (ν_CO) are highly sensitive to the electron density at the metal center.
• More electron-donating ligands (like Cp* = C₅Me₅) increase metal → CO backbonding, which lowers ν_CO.
• Less electron-donating ligands result in higher ν_CO.
• Trend of CO stretching frequency (from high to low): CpCp* > Cp₂ > Cp*₂

EXPLANATION:

1. a. Cp₂Ti(CO)₂:
   • Has two Cp (C₅H₅) ligands → moderate donor strength
   • → Corresponds to ii. 1956 and 1875 cm⁻¹
2. b. CpCp*Ti(CO)₂:
   • One Cp, one Cp* → stronger donation than Cp₂, but less than Cp*₂
   • → Corresponds to iii. 1930 and 1850 cm⁻¹
3. c. Cp*₂Ti(CO)₂:
   • Two Cp* ligands → strongest donation, highest backbonding → lowest ν_CO
   • → Corresponds to i. 1979 and 1897 cm⁻¹

• a – ii → Cp₂Ti(CO)₂ → 1956 and 1875 cm⁻¹
• b – iii → CpCp*Ti(CO)₂ → 1930 and 1850 cm⁻¹
• c – i → Cp*₂Ti(CO)₂ → 1979 and 1897 cm⁻¹

Therefore, the correct answer is: Option 4 ✅

CONCEPT:

Oxidative Addition in Organometallic Complexes

• Oxidative addition is a key step in many catalytic cycles (e.g. in cross-coupling reactions).
• It increases the oxidation state and coordination number of the metal center.
• Complexes that favor oxidative addition usually:
  • Contain low oxidation state transition metals
  • Have empty orbitals to accept electron pairs
  • Are electron-rich or stabilized by π-acceptor ligands like CO and phosphines

EXPLANATION:

• Complex X: [RhCl₃(H₂O)₃]
  • Rh is in +3 oxidation state
  • Octahedral geometry → prone to ligand substitution and aquation reactions
  • It can undergo oxidative addition under suitable conditions, especially with ligands like H₂, halogens
• Complex Y: [Ir(CO)(Cl)(PPh₃)₂]
  • Iridium is in +1 oxidation state — typical for oxidative addition
  • CO and PPh₃ are strong π-acceptor ligands, making the metal electron-rich
  • Hence, very prone to oxidative addition

Correct answer Both X and Y are prone to oxidative addition

Concept:

The given organometallic compound is type of carbene. There are two types of carbene:

1. Fischer carbene; (CO)₅M=C(OMe)R
2. Schrock carbene; L_nM=C

So the given organometallic compound is a type of Fischer carbene

Explanation:

• Fischer carbene is singlet and electrophilic in nature
• It generally obeys 18e^- rule
• It is two electron donor
• Metal is present in the low oxidation state as compare to the Schrock carbene
• Metal is electron rich and therefore carbene is electron deficient

• Carbene carbon should have at least one Z type group (OR, SH, SR, NH₂)
• Actual structure of Fischer carbene lies between the above two resonating structure and there is partial double bond
• Due to the this resonating structure, rational barrier across the metal and carbene carbon is low because there is no complete double bond present
• From the above explanation it is clear that Fischer carbene is two electron donor, therefore statement A is correct.
• However statement B is wrong as it is electrophilic and the rotational barrier across double bond is low so statement C is also correct.

Conclusion:

Hence, the correct statements are A and C and therefore the correct answer is option 3.

Concept:

→ The given clusters are higher carbonyl clusters.

Higher carbonyl clusters refer to compounds that contain multiple carbonyl (CO) functional groups bound to a central metal atom. 

→ To determine these structures we first need to calculate skeletal electrons.

S = 

where N = number of metals present in the cluster.

TVE = total valence electron

[N+1] is closo type

[N+2] is Nido type

[N+3] is arachno type

[N+4] is hypo type

Explanation:

→ [Fe5(CO)14N]− 

Fe have 8 valence electrons, CO contributes 2 electrons, N have 5 valence electrons + 1 for negative charge

TVE = 8 × 5 + 14 × 2 + 5 + 1 = 74 electrons

S = 

S = 

Number of Metals present in this structure is 5, but S is 7 thus, it is N+2 type i.e., Nido

→ [Co6(CO)13N]−

Co have 9 valence electrons, CO contributes 2 electrons, N have 5 valence electrons + 1 for negative charge

TVE = 9 × 6 + 13 × 2 + 5 + 1 = 74 electrons

S = 

S = 

Number of Metals present in this structure is 6, but S is 7 thus, it is N+1 type i.e., Closo.

Conclusion:
The correct answer is nido-, closo-.

Concept:

• Azulene is bicyclic organic system. It consists of cycloheptatriene and cyclopentadiene rings. 
• Presence of 10 pi-electrons with sp² hybridised carbons makes it aromatic and thus shows Friedel-Crafts substitution reactions.
• The organometallic compounds follows 18e^- rules. The metal center with18 valence e^-s is considered stable.
• Aromatic systems usually can change their hapticity depending on the electron requirement for stable system.

Explanation:

Lets say, the hapticity of cyclopentadiene and cycloheptatriene rings is x and y, respectively.

Contribution of electron by each CO = 2

Valence electrons of Fe = 8

Valence electrons of Mo = 6

electron shared by other metal = 1

(a) Calculating hapticity of Fe containing Azulene based system:

applying 18 e^- rule on Fe₁ :

           8 + 4 +1 + x = 18 

                            x = 5 

applying 18 e- rule on Fe₂ :

           8 + 6 +1 + y = 18

                             y = 3

hapticity of rings should be 5 and 3 for cyclopentadiene and cycloheptatriene respectively.

(a) Calculating hapticity of Mo containing Azulene based system:          

applying 18 e- rule on Mo₁ :

​           6 + 6 + 1 + x = 18

                              x = 5

applying 18 e- rule on Mo₂ :

          6 + 6 + 1 + y =18 

                             y = 5

Hence, the hapticity of rings should be 5 for both cyclopentadiene and cycloheptatriene.

Conclusion:

The set of structures showing the correct hapticity of azulene is:

​

The correct answer is 1, 2, 1  

Concept:-

• Crystal Field Theory (CFT): CFT is a model that describes the breaking of degeneracies of electron orbital states, usually d or f orbitals, due to a static electric field produced by a surrounding charge distribution (ligands). This can help you understand the electronic structure of the complex.
• Molecular Orbital Theory (MOT): This theory provides a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule. It gives a detailed insight into bond formation and could potentially be used to predict the types of bonding in the complex.
• Metal-Metal Multiple Bonds: This concept can help you understand when and why metal-metal multiple bonds form. It's especially useful when working with transition metals, like Mo. For example, if you know that the metal atom has d-orbitals available for bonding, and conditions are favorable, you can predict the possibility of multiple bonds (σ, π, δ).

Explanation:-

• Like several other transition metal carboxylate complexes, [Mo2(CH3CO2)4]  adopts a Chinese lantern structure. 
• Each Mo(II) center in Mo₂(CH₃CO₂)₄ has four d valence electrons.
• These eight d-electrons form one σ, two π bonds, and one δ bond, creating a bonding electron configuration of σ2π4δ2.
• Each of these bonds are formed by the overlapping of pairs of d orbitals.[4] The four acetate groups bridge the two metal centers.
• The Mo-O bond between each Mo(II) center and O atom from acetate has a distance of 2.119 Å, and the Mo-Mo distance between the two metal centers is 2.0934 Å.

Number of M-M bond = (18 x n - TVE) / 2
Number of M-M bond = ( 18 x 2 - (2X6 + 4X4)) / 2
Number of M-M bond = ( 36 - 28 ) / 2 = 4

So, the only option 1 is to have a 4 M-M bond.

Conclusion:-

So, The number of metal-metal bond(s), with 𝜎, 𝜋, and 𝛿 character, present in [Mo2(CH3CO2)4] complex is(are), 1,2,1 i.e option 1

Concept:-

To find the number of valence electrons of a metal, you can use the following steps:

• Determine the electron configuration of the metal. This can be done by using the periodic table and filling the orbitals in order of increasing energy, starting with the lowest energy level.

• Identify the valence electrons. The valence electrons are the outermost electrons in the atom, and they are the ones involved in chemical bonding. For transition metals, the valence electrons are the electrons in the outermost d orbitals and the s orbital.

• Count the number of valence electrons. The number of valence electrons for transition metals is usually equal to the group number of the metal, with the exception of the first-row transition metals, which have two fewer valence electrons than their group number. For example, the group number of iron (Fe) is 8, so it has 8 valence electrons. However, since it is a first-row transition metal, it actually has 6 valence electrons.

• Take into account any charge on the metal. If the metal has a positive charge, subtract that number from the total number of valence electrons. If it has a negative charge, add the absolute value of that number to the total number of valence electrons.

Explanation:-

• ​Electron contribution of each Ru center for

= 8 + 4 × 2 + 2

= 18

• ​Electron contribution of the central Ru metal for

= 8 + 2 × 2 + 4

= 16

Thus, it is not a stable complex.

• ​Electron contribution of the central Ru metal for

= 8 + 2 × 2 + 2

= 14

Thus, it is not a stable complex.

• Electron contribution of the central Ru metal for

= 8 + 2 × 2 + 6

= 18

Thus, the central Ru metal follows the 18-electron rule.

• But the electron contribution of the other two Ru metal for

= 8 + 3 × 2 + 3

= 17

Thus, it is not a stable complex.

Conclusion:-

• Hence, in the solid state, the stable structure of the metal cluster [Ru3(CO)10(PPh3)2] is

Concept:

→ Metal carbon distance for metallocenes depends upon the presence of unpaired electrons.

→ As the number of unpiared electrons for a system increases, metal carbon bond length also increases.

Explanation:

→ Ni(η5 − Cp)2 have 20 electrons (TVE = 10 × 1+ 2 × 5)

→ Co(η5 − Cp)2 have 19 electrons (TVE =  9 × 1+ 2 × 5)

→ Fe(η5 − Cp)2 have 18 electrons (TVE =  8 × 1+ 2 × 5)

Out of these electrons 12 electrons are involved in ligand group orbitals means that 12 of the valence electrons in the ligands are involved in bonding to the metal ion through their group orbitals because metals presentt in above complexes are in +2 oxidation state.

The remaining electrons would be metal-centered electrons, which are typically involved in d-orbital hybridization and contribute to the metal-ligand bonding.. Thus,

Ni(η5 − Cp)2 have 8 electrons  ( 20 - 12 electrons = 8 electrons)

Co(η5 − Cp)2 have 7 electrons ( 19 - 12 electrons = 7 electrons)

Fe(η5 − Cp)2 have 6 electrons  (18 - 12 electrons = 6 electrons)

Arrangements of these electrons is as:

Ni(η5 − Cp)2 have two unpaired electrons.

Co(η5 − Cp)2 have one unpaired electron.

Fe(η5 − Cp)2 ​have one unpaired electron.

Thus, the order of metal c arbon bond is Ni(η5 − Cp)2 > Co(η5 − Cp)2 > Fe(η5 − Cp)2.

Conclusion:
The correct answer is option 4.

Explanation:

• The 1-norbornyl ligand is seen to form a stable complex with transition metal Cobalt through a Co-C bond.
• This can be attributed to the fact that it is bulky and it is less prone to β-Hydrogen elimination reaction.
• The oxidation state of metal cobalt in the complex  is +IV.
• The electronic configuration in the state is 3d⁵.
• It has been experimentally found that the complex is tetrahedral and it norbornyl induces pairing in the system.
• The CFSE of tetrahedral is not that strong to induce electron pairing and no low spin complexes of the first transition series have been identified before this.
• So, the CFSE looks like this:

​

• So, the number of unpaired electrons is one, and the geometry is tetrahedral.

Concept:

• Complexes following 18e^-  rule are considered to be thermodynamically stable
• Nitrosyl ligand combines with metal in two forms : 1) Bent Nitrosyl and 2) Linear Nitrosyl.
• Neutral Nitrosyl contributes 3e^- in linear form and 1e  in bent form

Explanation:

Electron contribution by

• n⁵ cyclopentadiene = 5
• Fe = 8
• CO=2 (in bridging as well as non-bridging form)
• bent NO = 1
• Linear NO =3

(a) 

Electron count at metal center = 8 (Fe)+ 5(cyclopentadienyl) + 3(Nitrosyl) + 2 (µ2-CO) +1 (Fe-Fe) = 19e^-

Given structure is not following 18e^- rule, therefore, it is not thermodynamically stable.

(b)

Electron count at metal center = 8 (Fe)+ 5(cyclopentadienyl) + 3(Nitrosyl) + 2 (µ2-CO) = 18

It follows the 18e^- rule and is thermodynamically stable.

(c) 

Electron count at each metal center = 8 (Fe)+ 5(cyclopentadienyl) + 1(Nitrosyl) + 2 (µ2-CO) + 1 (Fe-Fe) = 17

Given structure is not following 18e- rule, therefore, it is not thermodynamically stable.

(d)

Electron count at each metal center = 8 (Fe)+ 5(cyclopentadienyl) + 1(Nitrosyl) + 2 (µ2-CO) +  = 16

Given structure of organometallic compound is not following 18e- rule, therefore, it is not thermodynamically stable.

Conclusion:

Hence, the thermodynamically stable structure of

[(η5-C5H5)Fe(µ2-CO)(NO)]2

Concept:

M-M Bond in Complex:

• A σ bond between two metal atoms forms through the overlap of orbitals from each atom.

• π bonds arise when d_xz or d_xy orbitals overlap.

• δ bonds are created by face-to-face overlap of d_xy or orbitals.

Re₂Cl₈:

• The Re compound, featuring two d⁴ Re(III) centers, forms a quadruple Re-Re bond without bridging ligand interactions.

• Each Re atom contributes four d-electrons, resulting in a bonding configuration.

• The orbital is thought to participate in bonding with Cl⁻ ligands.

• Evidence for quadruple bonding is derived from the structure of [Re₂Cl₈]^2-, which exhibits an eclipsed arrangement of Cl^- ligands, considered sterically unfavored.

• The δ bond, formed when d_xy orbitals are aligned, locks the complex in an eclipsed conformation.

Key Points for Calculating Bond Order:

• Electron Count: The total number of bonding electrons between two metal centers contributes to the bond order. More electrons shared between two metals increase the bond order.
• Ligands: Ligands like Cl^- and PMe₂Ph can donate or withdraw electron density, affecting the overall bonding between two metal centers.
• Oxidation State: The oxidation state of the rhenium centers impacts the electron count, with higher oxidation states often resulting in fewer electrons available for bonding.

Explanation: 

• K₂Re₂Cl₈:

  • Oxidation state of Re in given complex:

    • ​(+2) +2x + (-8) = 0 

    • 2x = 6, x = 3

    • Re in +3 oxidation state have 4 valence electrons and two Re have total 8 valence electron which on filling will result in .

    • Bond Order = 

• Re₂Cl₄(PMe₂Ph)₄

  • Oxidation state of Re in given complex:

    • ​2x + (-4) + 0 = 0

    • 2x = 4, x = +2

    • Re in +2 oxidation state have 5 valence electrons and two Re have total two electrons which on filling will result in 

    • Bond Order = 

• Re₂Cl₄(PMe₂Ph)₄Cl

  • ​Oxidation state of Re in given complex:

    • ​2x + (-4) + 0 + (-1) = 0

    • 2x = +5

    • x = +2 and +3.

    • Re in +2 oxidation state have 5 valence electrons and Re in +3 oxidation state have 4 valence electrons which on filling will result in 

    • Bond Order = 

Conclusion:

The correct Re-Re bond order follows the sequence K₂Re₂Cl₈ > Re₂Cl₄(PMe₂Ph)₄Cl > Re₂Cl₄(PMe₂Ph)₄

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    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