Test: Organometallic Chemistry- 2 - Chemistry MCQ

Optical isomers are molecules that are mirror images of each other and cannot be superimposed, much like left and right hands. This phenomenon occurs due to the presence of a chiral centre in a molecule, which is typically a carbon atom bonded to four different groups.

Among the options provided, the following complex contains optical isomers:

• [Co(en)₃]³⁺ – This complex has three bidentate ethylenediamine (en) ligands, leading to chirality and the possibility of optical isomers.

The other complexes do not exhibit chirality because:

• [Co(H₂O)₄(en)]³⁺ – Contains one bidentate ligand and does not create a chiral environment.
• [CoCl(NH₃)₅]Cl₂ – This complex does not have a chiral centre due to the symmetric arrangement of ligands.
• [Zn(en)(NH₃)₂]²⁺ – Zinc is not chiral in this arrangement because it is a d¹⁰ metal ion and lacks a chiral centre.

In conclusion, the only complex with potential optical isomers is [Co(en)₃]³⁺.

The existence of two different coloured complexes with the composition of [Co(NH₃)₄Cl₂]⁺ can be attributed to a phenomenon known as geometrical isomerism.

• In geometrical isomerism, the arrangement of ligands around the central metal atom varies.
• This results in distinct spatial configurations, leading to different properties, such as colour.
• In this case, the positioning of the ammonia (NH₃) and chloride (Cl₋) ligands affects how light interacts with the complex.
• As a result, the two isomers exhibit noticeable differences in colour.

Understanding geometrical isomerism is crucial in coordination chemistry, as it explains why some complexes can exist in multiple forms with unique characteristics.

No of metal - metal bond= (18n - TVE)/2
Here, TVE = 2(6) + 6(3)

= 30
n = 2 (n denotes the number of metals present)

Hence, No of metal - metal bonds= (36-30)/2

= 3 metal - metal bonds.

The isolobal analogy relates molecular fragments based on their frontier orbitals and valence electron counts. According to this principle:

• CH (with 5 valence electrons) is isolobal to Co(CO)₃ (which also has 5 valence electrons).

• Co(CO)₃ achieves this electron count through Co's d-electrons and CO ligand contributions, matching CH's orbital and electronic structure.

Key Points:

• Option A aligns with the established isolobal relationship: CH ↔ Co(CO)₃.

• Other options mismatch electron counts or orbital configurations (e.g., CH₂ ↔ Ni(CO)₂ or CH ↔ Fe(CO)₄ are incorrect).

The correct statement is A: CH is isolobal to Co(CO)₃.

The classification of Organometallic compound  Rh6(CO)16 as a closo cluster is based on electron counting and Wade’s rules.

An arachno type structure refers to a type of cluster structure in metal carbonyl complexes, where the metal atoms form a "cage-like" structure with one fewer metal-metal bond than an "nido" structure. This structure is typically associated with a 3D cluster where the number of metal atoms is less than the total number of bonds, forming a "spider-like" arrangement.

• [Os₃(CO)₁₂]: This complex is an example of a cluster with an arachno structure. It consists of three osmium atoms, with twelve carbon monoxide ligands, and the structure forms a "cage" with arachno bonding characteristics.

In contrast:

• [Os₅(CO)₁₆]: This complex has a nido structure, where the metal atoms form a 3D polyhedron with one more metal-metal bond than in an arachno structure.
• [Ir₄(CO)₁₂]: This complex has a closo structure, where the metal atoms form a fully connected polyhedron.
• [Ph₆(CO)₁₆]: This complex is not typical of a metal carbonyl cluster with an arachno structure.

Therefore, the correct answer is B: [Os₃(CO)₁₂].

The Zintl ion B₅³⁺ is a type of cluster that forms a closo structure. In the context of polyhedral boron clusters, closo structures refer to completely closed polyhedra where the boron atoms are connected in a fully bonded arrangement, with each boron atom contributing to the overall structure.

• Closo clusters typically involve a structure where the cluster forms a fully connected 3D polyhedron. In this case, the B₅³⁺ ion represents a cluster with a closed bonding arrangement typical of closo structures.

Other options:

• Nido structures are those with one missing bond, leading to a "cage-like" structure.
• Arachnid or arachno structures have even fewer bonds, resulting in a "spider-like" configuration.
• Hypo structures have additional bonding that results in an open polyhedron.

Thus, B₅³⁺ corresponds to a closo structure, making option A the correct answer.

The thermal stability of metal carbonyl hydrides generally increases down a group in the periodic table due to enhanced metal-ligand bonding from better back-donation capabilities in larger metals. For the given compounds:

• HMn(CO)₅ (P): Manganese (Mn, Group 7, period 4).

• HTc(CO)₅ (Q): Technetium (Tc, Group 7, period 5).

• HRe(CO)₅ (R): Rhenium (Re, Group 7, period 6).

As we descend Group 7, the atomic size increases, leading to stronger back-donation into CO ligands and greater stabilization of the complex. Thus, the thermal stability follows:
HRe(CO)₅ > HTc(CO)₅ > HMn(CO)₅ → R > Q > P.

In this question, we're comparing the basicity of three metal carbonyl complexes:

• [Mn(CO)₅]⁻ (Manganese carbonyl anion)
• [Tc(CO)₅]⁻ (Technetium carbonyl anion)
• [Re(CO)₅]⁻ (Rhenium carbonyl anion)

The basicity of these complexes is influenced by the oxidation state of the metal and the ability of the metal center to accept electron density. Here are the key points:

1. Rhenium (Re) has a lower effective nuclear charge in its +1 oxidation state compared to Technetium (Tc) and Manganese (Mn). This means [Re(CO)₅]⁻ will have the greatest ability to donate electron density (and thus act as a stronger base), as the metal's orbitals are more easily available for donation.

2. Technetium (Tc), with its +1 oxidation state, lies between Manganese and Rhenium in terms of basicity. It has a slightly less available electron density than Rhenium but more than Manganese.

3. Manganese (Mn), being in a lower oxidation state (+1), has a higher effective nuclear charge, making it less basic. The electron density on the Mn center is relatively less available for donation compared to Tc and Re.

Thus, the basicity order is: [Re(CO)₅]⁻ > [Tc(CO)₅]⁻ > [Mn(CO)₅]⁻.

Therefore, the correct order is B: R > Q > P.

The number of IR bands in M(CO)₅ can be determined by considering the symmetry and number of distinct vibrations in the molecule. Here are the key points:

• M(CO)₅ is a coordination complex where M represents a metal atom.
• The complex has a trigonal bipyramidal geometry, which influences its vibrational modes.
• Each CO ligand can contribute to the vibrational spectrum.
• Due to the symmetry of the molecule, not all vibrations will be infrared (IR) active.
• Typically, a trigonal bipyramidal complex can exhibit a limited number of IR bands due to this symmetry.

In summary, M(CO)₅ will show a total of two IR bands based on its structure and the behaviour of its ligands.

Nitrosyl ligands can bind to transition metal atoms in two main geometries: linear and bent. This behaviour is determined by the charge of the nitrosyl ligand.

• The nitrosyl ligand can act as NO⁺ or NO in a linear arrangement.
• In a bent arrangement, it can behave as NO^– or NO⁺.
• Understanding these binding modes is crucial for studying the chemical properties and reactivity of metal complexes.

The stretching frequency νNO (in cm–1) for the NO ligand depends on the electron-donating or electron-withdrawing effects on the NO group, which influences the bonding interaction between NO and the metal center. Here's how each compound behaves in relation to its νNO:

NO+ (Nitrosonium ion): This species has a positive charge on the NO ligand, which pulls electron density away from the NO group, making the bond stronger. This results in a higher stretching frequency νNO.

NO (Neutral nitric oxide): The neutral NO molecule has less electron withdrawal compared to NO+, leading to a lower stretching frequency.

[NiCp(NO)]: In this complex, NO is bound to a metal (Ni), and the electron donation from the metal to the NO ligand slightly weakens the NO bond. This results in a lower νNO compared to NO and NO+.

[Cr(Cp)2(NO)4]: In this complex, the NO ligands interact with the metal center, but the metal is in a lower oxidation state compared to Ni, and the back-donation from the metal to the NO ligand weakens the bond even further, resulting in the lowest νNO.

Thus, the order of νNO (cm–1) is:

NO+ > NO > [NiCp(NO)] > [Cr(Cp)2(NO)4]

Therefore, the correct option is A: NO+ > NO > [NiCp(NO)] > [Cr(Cp)2(NO)4].

• In Fe(CO)₅​ there exists a synergic bond between Fe and CO. Due to synergic bond formation between metal and CO, the bond order of CO decreases.
• As bond order decreases the Bond length increases.Here the only option having greater bond length than 1.128A⁰ is option A- 1.158A⁰.

Hydroformylation is a chemical reaction that adds a formyl group to an alkene, producing aldehydes. A catalyst is crucial for facilitating this reaction, and certain metal complexes are commonly used. However, not all metal complexes are suitable catalysts for hydroformylation.

Among the options provided, H₂Rh(PPh₃)₂Cl does not function effectively as a catalyst for this process. Here’s a brief overview of the other options:

• HCo(CO)₄: This complex is a well-known catalyst for hydroformylation, effectively promoting the addition of the formyl group to alkenes.
• HCo(CO)₄PBu₃: Similar to the first, this compound is also suitable and can enhance catalytic activity.
• HRh(CO)(PPh₃)₃: This rhodium complex is another effective catalyst for hydroformylation, providing good yields of aldehydes.

In contrast, H₂Rh(PPh₃)₂Cl lacks the necessary reactivity and coordination properties to effectively catalyse this reaction, making it unsuitable in this context.

The catalyst used for the oxidation of ethylene to acetaldehyde plays a crucial role in this chemical reaction.

• Ru(PPh₃)₃Cl: A well-known catalyst used in various organic transformations.
• Co₂(CO)₈ and H₂: This combination can facilitate the oxidation process.
• TiCl₄ and AlEt₃: Often used in chemical reactions requiring strong Lewis acids.
• PdCl₂ and CuCl: A common pairing in catalytic processes.

In this context, the preferred catalyst for the oxidation of ethylene is PdCl₂ and CuCl.

The catalytic cycle of hydrogenation of alkanes using RhCl(PPh₃)₃ as a catalyst involves different electron counts of the Rh complexes. Understanding these complexes is crucial for grasping the mechanism of the reaction.

• 14-electron complexes are typically less stable and may participate in the reaction.
• 16-electron complexes can also play a role and are often seen as intermediates.
• 18-electron complexes are generally the most stable and are frequently involved in the catalytic process.

In this specific case, the reaction involves:

• Both 16- and 18-electron Rh complexes.
• The presence of 14- and 16-electron Rh complexes is possible but less relevant.

Therefore, the most accurate representation of the catalytic cycle includes:

• The involvement of 16- and 18-electron Rh complexes during the hydrogenation process.

The Wacker process is used for the oxidation of ethylene (C₂H₄) to acetaldehyde (CH₃CHO). The catalyst for this reaction is typically a palladium chloride complex. Specifically, [PdCl₄]²⁻ (palladium(II) chloride complex) is used as the catalyst, along with a co-catalyst, usually copper chloride (CuCl₂).

In the Wacker process:

• Ethylene is oxidized by molecular oxygen (O₂) in the presence of the palladium catalyst to form acetaldehyde.
• Palladium acts as the central metal, facilitating the reaction by coordinating to the ethylene and enabling the addition of oxygen.

Thus, the correct catalyst used in the Wacker process is [PdCl₄]²⁻, making Option B the correct answer.

Homogeneous catalysts play a crucial role in hydroformylation, a chemical process that converts alkenes into aldehydes. The choice of catalyst can significantly affect the efficiency and selectivity of the reaction.

• Cobalt (Co) is the most commonly used catalyst in hydroformylation. It offers several advantages:
• High activity and selectivity for desired products.
• Ability to operate under mild conditions.
• Chromium (Cr) and Titanium (Ti) are less frequently used but can be effective in specific scenarios. They may provide:
• Different reaction pathways.
• Unique catalytic properties that might be beneficial for certain substrates.
• Vanadium (V) is typically not used in hydroformylation, as it does not perform as well as the other metals.

In summary, cobalt is the preferred homogeneous catalyst for hydroformylation due to its efficiency and adaptability. Other metals like chromium and titanium can be used in specific cases, but cobalt remains the standard choice.

Metals used in automobile catalytic converters include:

• Platinum (Pt)

• Palladium (Pd)

• Rhodium (Rh)

These metals play a crucial role in the catalytic process, helping to convert harmful emissions into less harmful substances. Each metal contributes uniquely:

• Platinum is highly effective for oxidation reactions.
• Palladium is often used for both oxidation and reduction reactions.
• Rhodium is particularly effective in reducing nitrogen oxides.

In summary, all three metals are essential for efficient functioning of catalytic converters, contributing to improved air quality.

Active catalytic species for hydroformylation involves the use of specific metal complexes to facilitate the reaction. The following are notable catalysts:

• RuCl₂(PPh₃)₃: A ruthenium-based catalyst known for its efficiency in hydroformylation processes.

• HCo(CO)₃: This cobalt complex is also effective in promoting hydroformylation reactions.

• RuCl(PPh₃)₃: Another ruthenium compound, this variant shows utility in hydroformylation.

• K₂PtCl₆: A platinum-based catalyst that can be used in hydroformylation, albeit less commonly than others.

Among these, HCo(CO)₃ is highlighted as a prominent active catalytic species.

Wilkinson's catalyst is a well-known compound in chemistry, particularly in the field of catalysis. Here are some key points about it:

• It is coordinatively saturated, meaning that it has a complete set of ligands around the metal centre.

• It does not obey the 18-electron rule, which is a guideline used to predict the stability of transition metal complexes.

• Wilkinson's catalyst is commonly used for the oxidation of alcohols, making it valuable in organic synthesis.

• It is an iridium complex, which plays a significant role in the preparation of important pharmaceutical products.

An intermediate formed during hydroformylation of olefins using Co₂(CO)₈ as a catalyst is a significant concept in organometallic chemistry. The process involves the following key points:

• The catalyst, Co₂(CO)₈, plays a crucial role in the reaction.
• During the hydroformylation, an intermediate complex is generated.
• Among the potential intermediates, the most relevant one is HCo(CO)₄.

Understanding this intermediate is essential as it impacts the efficiency and selectivity of the hydroformylation process, which is vital for producing aldehydes from olefins.

Correct Answer :- B

Explanation : The Monsanto acetic acid process is the major commercial production method for acetic acid. Methanol, which can be generated from synthesis gas ("syn gas", a CO/H2 mixture), is reacted with carbon monoxide in the presence of a catalyst to afford acetic acid. In essence, the reaction can be thought of as the insertion of carbon monoxide into the C-O bond of methanol, i.e. the carbonylation of methanol.

Ferrocene is an organometallic compound consisting of a sandwich structure with an iron (Fe) atom between two cyclopentadienyl anions (C₅H₅⁻). When acetyl chloride reacts with ferrocene in the presence of AlCl₃ (which acts as a Lewis acid), the acetyl group (CH₃CO) is introduced onto the cyclopentadienyl ring via an electrophilic aromatic substitution reaction.

The reaction can be represented as:

The result is the formation of acetylferrocene, where one of the cyclopentadienyl rings is substituted with an acetyl group (-COCH₃).

So, the product of the reaction is acetylferrocene.

Infrared Stretching Frequency (νCO) and Bond Strength

The infrared stretching frequency (νCO) depends on the bond strength of the C≡O bond. Stronger bonds result in higher frequencies.

H₃B←CO (R):

CO acts as a σ-donor to BH₃, a Lewis acid. This σ-donation withdraws electron density from CO, strengthening the C≡O bond.

νCO increases compared to free CO, making R the highest.

Free CO (Q):

Standard νCO ≈ 2143 cm⁻¹.

Metal Carbonyls:

[Mn(CO)₆]⁺ (P):

Mn⁺ has a higher oxidation state (+1), reducing back-donation into CO's π* orbitals. This results in a weaker C≡O bond compared to R and Q, but stronger than [V(CO)₆] (S).

[V(CO)₆] (S):

Vanadium (likely in a lower oxidation state, e.g., 0 or -1) allows stronger back-donation, significantly weakening the C≡O bond.

Order of νCO:

R (H₃B←CO) > Q (Free CO) > P (Mn(CO)₆⁺) > S ([V(CO)₆])

The correct sequence is: R > Q > P > S, corresponding to Option D.

The Vilsmeier reaction is a method used to introduce an aldehyde group (–CHO) onto an aromatic ring or a compound with similar reactivity. This reaction is generally performed using DMF (dimethylformamide) and phosphorus oxychloride (POCl₃), resulting in the formylation of the compound.

When ferrocene undergoes a Vilsmeier reaction, the reaction introduces a formyl group (–CHO) onto one of the cyclopentadienyl rings of the ferrocene molecule. This results in the formation of ferrocenecarbaldehyde.

In the image you provided, Option A shows the correct structure with a CHO (formyl) group attached to the cyclopentadienyl ring of ferrocene, which is the product of the Vilsmeier reaction.

Thus, the correct answer is A.

Hydroformylation, also known as oxo synthesis or oxo process, is an industrial process for the production of aldehydes from alkenes. It is important because aldehydes are easily converted into many secondary products. For example, the resulting aldehydes are hydrogenated to alcohols that are converted to detergents.

When propene reacts with CO and H2 then butanal is formed.

H2 + CO + CH3CH=CH2 → CH3CH2CH2CHO ("normal")

H2 + CO + CH3CH=CH2 → (CH3)2CHCHO ("iso")

Hence B is correct

The given complex is written as:

This indicates a metal complex with cyclopentadienyl  and cyclohexadienyl  ligands, along with phosphine (PMe₃​) and hydride (H) groups. The central metal is likely a second-row transition metal from the periodic table, and the coordination environment suggests it is from group 6.

• Zr (Zirconium) and Nb (Niobium) are in the 4th and 5th rows, respectively.
• Mo (Molybdenum) is a second-row transition metal in group 6, which fits the given compound, and it is known to form complexes with the described ligands.
• R₄ is not a valid metal, so it is not an option.

Thus, the correct second-row transition metal in the complex is Mo (Molybdenum), making Option C the correct answer.

The reactivity order of alkene follows:   Norbornene > cyclohexane > 1-butene > ethylene.

Norbornene compound is more reactive due to more substituted than Cyclohexane. We know that More substituted alkene is highly reactive compared to less substituted alkyl. That is why cyclohexane is less reactive compared to norbornene.