Carbon monoxide (CO) forms bonds to a variety of metals and metal ions. Its ability to bond to iron in hemoglobin is the reason that \(\mathrm{CO}\) is so toxic. The bond carbon monoxide forms to metals is through the carbon atom: $$ \mathrm{M}-\mathrm{C} \equiv \mathrm{O} $$ a. On the basis of electronegativities, would you expect the carbon atom or the oxygen atom to form bonds to metals? b. Assign formal charges to the atoms in CO. Which atom would you expect to bond to a metal on this basis? c. In the MO model, bonding MOs place more electron density near the more electronegative atom. (See the HF molecule in Figs. 9.43 and 9.44.) Antibonding MOs place more electron density near the less electronegative atom in the diatomic molecule. Use the MO model to predict which atom of carbon monoxide should form bonds to metals.

Short Answer

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Based on electronegativities, we expect the carbon atom to form bonds to metals as it is less electronegative than oxygen. Formal charges of both carbon and oxygen in CO are 0, providing inconclusive results. However, using the MO model, we find that the carbon atom has more electron density available for bonding in antibonding molecular orbitals. Therefore, the carbon atom of carbon monoxide should form bonds to metals.

Step by step solution

01

Calculate electronegativities of carbon and oxygen

Use the Pauling electronegativity scale to find the electronegativities of carbon and oxygen. Carbon has an electronegativity of 2.5, while oxygen has an electronegativity of 3.5.
02

Compare electronegativities

Since the oxygen atom has a higher electronegativity than carbon, it attracts the electron density more strongly, making it more negative in nature. Typically, metals bond to more electropositive atoms. In this case, the carbon atom, being less electronegative, is more likely to bond to the metal. #b. Assign formal charges to the atoms in CO#
03

Determine formal charges

To calculate the formal charges of carbon and oxygen in CO, we need to compare the number of valence electrons in their neutral states to the electrons assigned to them in the molecule.
04

Formal charge of carbon

Carbon has 4 valence electrons in its neutral state. In CO, there is a triple bond between carbon and oxygen, and a lone pair on carbon. This means that 2 electrons are assigned to carbon as part of the 2 bonding pairs (1 electron from each pair) and 2 electrons from the lone pair. Therefore, the formal charge on carbon is 4 - 4 = 0.
05

Formal charge of oxygen

Oxygen has 6 valence electrons in its neutral state. In CO, two of these electrons are assigned to oxygen as part of the two bonding pairs (1 electron from each pair) and 4 electrons from the two lone pairs. Therefore, the formal charge on oxygen is 6 - 6 = 0.
06

Determine which atom bonds to metal

Based on formal charges, both carbon and oxygen atoms have a formal charge of 0, so it is not conclusive which atom would bond to the metal using this method alone. #c. Use the MO model to predict the bonding atom#
07

Analyze bonding molecular orbitals

In the MO model, bonding molecular orbitals place more electron density near the more electronegative atom. In CO, the more electronegative atom is oxygen. Thus, in bonding molecular orbitals, we would expect more electron density around oxygen.
08

Analyze antibonding molecular orbitals

Alternatively, antibonding molecular orbitals place more electron density near the less electronegative atom in the diatomic molecule. In CO, the less electronegative atom is carbon. Thus, we would expect more electron density around carbon in antibonding molecular orbitals.
09

Predict which atom bonds to metals

Since metals typically bond to atoms with more electron density available for bonding, the MO model suggests that the atom in CO with greater electron density available for bonding is carbon. Therefore, we predict that the carbon atom of carbon monoxide should form bonds to metals.

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Most popular questions from this chapter

Although nitrogen trifluoride \(\left(\mathrm{NF}_{3}\right)\) is a thermally stable compound, nitrogen triodide \(\left(\mathrm{N} \mathrm{I}_{3}\right)\) is known to be a highly explosive material. \(\mathrm{NI}_{3}\) can be synthesized according to the equation $$ \mathrm{BN}(s)+3 \mathrm{IF}(g) \longrightarrow \mathrm{BF}_{3}(g)+\mathrm{NI}_{3}(g) $$ a. What is the enthalpy of formation for \(\mathrm{NI}_{3}(s)\) given the enthalpy of reaction \((-307 \mathrm{kJ})\) and the enthalpies of formation for $\mathrm{BN}(s)(-254 \mathrm{kJ} / \mathrm{mol}), \operatorname{IF}(g)(-96 \mathrm{kJ} / \mathrm{mol})\( and \)\mathrm{BF}_{3}(g)(-1136 \mathrm{kJ} / \mathrm{mol}) ?$ b. It is reported that when the synthesis of \(\mathrm{NI}_{3}\) is conducted using 4 moles of IF for every 1 \(\mathrm{mole}\) of \(\mathrm{BN}\) , one of the by-products isolated is \(\left[\mathrm{IF}_{2}\right]^{+}\left[\mathrm{BF}_{4}\right]^{-} .\) What are the molecular geometries of the species in this by-product? What are the hybridizations of the central atoms in each species in the by-product?

A variety of chlorine oxide fluorides and related cations and anions are known. They tend to be powerful oxidizing and fluorinating agents. \(\mathrm{FClO}_{3}\) is the most stable of this group of compounds and has been studied as an oxidizing component in rocket propellants. Draw a Lewis structure for \(\mathrm{F}_{3} \mathrm{ClO}\) , $\mathrm{F}_{2} \mathrm{ClO}_{2}^{+},\( and \)\mathrm{F}_{3} \mathrm{ClO}_{2}$ . What is the molecular structure for each species, and what is the expected hybridization of the central chlorine atom in each compound or ion?

For each of the following molecules or ions that contain sulfur, write the Lewis structure(s), predict the molecular structure (including bond angles), and give the expected hybrid orbitals for sulfur. $$ \begin{array}{l}{\text { a. } \mathrm{SO}_{2}} \\ {\text { b. } \mathrm{SO}_{3}}\end{array} $$ $$ \text {c} \mathrm{s}_{2} \mathrm{O}_{3}^{2-}\left[\begin{array}{c}{\mathrm{o}} \\\ {\mathrm{s}-\mathrm{s}-\mathrm{o}} \\ {\mathrm{o}} \\\ {\mathrm{o}}\end{array}\right]^{2-} $$ e. \(\mathrm{SO}_{3}^{2-}\) f. \(\mathrm{SO}_{4}^{2-}\) g. \(\mathrm{SF}_{2}\) h. \(\mathrm{SF}_{4}\) i. \(\mathrm{SF}_{6}\) j. \(\mathrm{F}_{3} \mathrm{S}-\mathrm{SF}\) k. \(\mathrm{SF}_{5}+\)

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? Place the species in order of increasing bond length and bond energy. $$ \text {a} \mathrm{CN}^{+} \quad \text { b. CN } \quad \text { c. } \mathrm{CN}^{-} $$

Many important compounds in the chemical industry are derivatives of ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right) .\) Two of them are acrylonitrile and methyl methacrylate. Complete the Lewis structures, showing all lone pairs. Give approximate values for bond angles \(a\) through \(f\) . Give the hybridization of all carbon atoms. In acrylonitrile, how many of the atoms in the molecule must lie in the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in methyl methacrylate and acrylonitrile?

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