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Carbon monoxide acts as a donor and reacts with certain metals to give metal carbonyls. This is due to (a) presence of one sigma and two pi bonds between \(\mathrm{C}\) and \(\mathrm{O}(\cdot \mathrm{C} \equiv \mathrm{O}:)\) (b) presence of a lone pair on carbon atom in CO molecule (c) presence of lone pair on oxygen atom in CO molecule (d) poisonous nature of CO.

Short Answer

Expert verified
Carbon monoxide acts as a donor in the formation of metal carbonyls due to the presence of a lone pair on the carbon atom in the CO molecule (option b).

Step by step solution

01

Understand the bonding in CO

Carbon monoxide (CO) has a triple bond between the carbon (C) and oxygen (O) atoms. This includes one sigma bond (σ) and two pi bonds (π). Furthermore, the carbon atom has a lone pair of electrons, while the oxygen atom generally has two lone pairs in this molecule.
02

Identify the donor sites

For CO to act as a ligand and form metal carbonyls, it needs to have a site that can donate electron density to the metal. The lone pair of electrons on carbon is the electron-donating site.
03

Rule out incorrect options

While the presence of sigma and pi bonds is a characteristic of the CO molecule, it is the lone pair on the carbon that is directly involved in the formation of the metal carbonyl complexes. Thus, options (a), (c), and (d) are not the correct reasons for CO's ability to form metal carbonyls.
04

Choose the correct option

The ability of carbon monoxide to form metal carbonyls is due to the presence of a lone pair on the carbon atom. The lone pair can be donated to a metal to form a coordinate covalent bond, resulting in a metal carbonyl complex.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carbon Monoxide as Ligand
Carbon monoxide (CO) may be more widely recognized for its toxic effects on living organisms, but in the realm of chemistry, CO is a vital ligand in many metal complexes. A ligand is an ion or molecule that binds to central metal atoms to form coordination compounds. What makes CO so special as a ligand is its unique molecular structure.

CO is composed of a carbon atom triple-bonded to an oxygen atom. Yet, it's not the triple bond that endows CO with its electron-donating capability; it is the presence of a lone pair of electrons on the carbon atom. This allows CO to form a bond with metal atoms where it can 'donate' these lone electrons. Metal carbonyls form when carbon monoxide acts as a ligand, utilizing this lone pair to bind to the metal, creating fascinating structures utilized in various industrial and chemical processes.
Coordinate Covalent Bond
When we talk about carbon monoxide bonding to a metal atom to form a metal carbonyl, we're referring to the formation of a coordinate covalent bond. Unlike a regular covalent bond, where each atom provides one of the electrons in the shared pair, in a coordinate bond one atom provides both electrons for the bond.

This is exactly what happens with CO; the carbon atom donates its lone pair to the empty orbital of a metal, forming a stable complex. This is an essential distinction that drives the unique properties and behaviors of metal carbonyls. These types of bonds are pivotal in the chemistry of coordination compounds and enable the formation of complex structures that have significant roles in catalysis and materials science.
Lone Pair Electron Donation
Lone pair electron donation is at the heart of the formation of metal carbonyls. The 'lone pair' refers to the pair of valence electrons on an atom that are not used in bonding. In carbon monoxide, the carbon atom has such a pair of electrons that are pivotal in its interaction with metals.

The ability of these electrons to be donated to another atom—like those in transition metals—allows for the creation of the coordinate covalent bond, which marks the complexity and functionality of the metal-ligand interaction. This electron donation is fundamental not only in the formation of metal carbonyls but also in a much broader set of chemical reactions and bonding scenarios across inorganic and organic chemistry.

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Most popular questions from this chapter

A metal \(X\) reacts with aqueous \(\mathrm{NaOH}\) solution to form \(Y\) and a highly inflammable gas. Solution \(Y\) is heated and \(\mathrm{CO}_{2}\) is poured through it. \(Z\) precipitates out and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is formed. \(Z\) on heating gives \(\mathrm{Al}_{2} \mathrm{O}_{3}\). Identify \(X, Y\) and \(Z\). $$\begin{array}{lll} {\boldsymbol{X}} & {\boldsymbol{Y}} & {\boldsymbol{Z}} \\ (a) \mathrm{Al} &\mathrm{NaAlO}_{2} & \mathrm{Al}(\mathrm{OH})_{3} \\ (b) \mathrm{Al}_{2} \mathrm{O}_{3} & \mathrm{NaAlO}_{2} & \mathrm{Al}_{2} \mathrm{CO}_{3} \\ (c) \mathrm{Al}_{2} \mathrm{O}_{3} & {\left[\mathrm{Na}_{2} \mathrm{AlO}_{2}\right]^{+} \mathrm{OH}^{-}} & \mathrm{Al}(\mathrm{OH})_{3} \\\ (d) \mathrm{Al} & \mathrm{Al}(\mathrm{OH})_{3} & \mathrm{Al}_{2} \mathrm{O}_{3} \end{array}$$

\(\mathrm{SiCl}_{4} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} X \stackrel{\text { Heat }}{\longrightarrow} Y \stackrel{\mathrm{NaOH}}{\longrightarrow} Z\) \(X, Y\) and \(Z\) in the above reaction are $$ \begin{array}{lll} X & Y & {Z} \\ (A)\mathrm{SiO}_{2} & \mathrm{Si} & \mathrm{NaSi} \\ (B)\mathrm{Si}(\mathrm{OH})_{4} & \mathrm{SiO}_{2} & \mathrm{Na}_{2} \mathrm{SiO}_{3} \\ (C)\mathrm{Si}(\mathrm{OH})_{4} & \mathrm{Si} & \mathrm{SiO}_{2} \\ (D)\mathrm{SiO}_{2} & \mathrm{SiCl}_{4} & \mathrm{Na}_{2} \mathrm{SiO}_{3} \end{array} $$

Which of the following properties correctly explain \(\mathrm{SiO}_{2} ?\) (a) Linear, basic (b) Tetrahedral, acidic (c) Tetrahedral, basic (d) Linear, acidic

Aluminium exhibits \(+3\) oxidation state. As we move down the group, \(+1\) oxidation state gets more stable, This is a consequence of (a) increasing size of the atom (b) inert pair effect (c) electron deficient nature (d) \(p \pi-p \pi\) bonding.

Which of the following compounds are formed when \(\mathrm{BCl}_{3}\) is treated with water? (a) \(\mathrm{H}_{3} \mathrm{BO}_{3}\) (b) \(\mathrm{B}_{2} \mathrm{H}_{6}\) (c) \(\mathrm{B}_{2} \mathrm{O}_{3}\) (d) \(\mathrm{HBO}_{2}\)

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