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Predict the ordering of the \(\mathrm{C}-\mathrm{O}\) bond lengths in \(\mathrm{CO}\), \(\mathrm{CO}_{2}\), and \(\mathrm{CO}_{3}^{2-}\)

Short Answer

Expert verified
The ordering of the C-O bond lengths in CO, CO2, and CO3虏鈦 is CO < CO2 < CO3虏鈦, based on the bond orders calculated using molecular orbital theory (3 for CO, 2 for CO2) and resonance structures (4/3 for CO3虏鈦), where higher bond orders correspond to shorter bond lengths.

Step by step solution

01

Use Molecular Orbital Theory for CO and CO2

Determine the bond orders of CO and CO2 using molecular orbital theory. In CO, there are 10 valence electrons forming one 蟽 bond and one 蟺 bond, making the bond order 3. In CO2, there are 16 valence electrons forming two 蟽 bonds and four 蟺 bonds, making the bond order 2 for an oxygen atom to a carbon atom in a CO2 molecule.
02

Use Resonance Structures for CO3虏鈦

Find the bond order of CO3虏鈦 using resonance structures. Carbonate has three resonance structures, where in each structure, one of the oxygen atoms forms a double bond with the carbon atom, and the other two oxygens form a single bond. These three resonance structures equally contribute to the real structure of the carbonate ion. Thus, the bond order for a C-O bond in CO3虏鈦 will be the average of the bond orders for each resonance structure, which is equal to (1 + 1 + 2)/3 = 4/3.
03

Compare Bond Orders and Predict Bond Lengths

Compare the bond orders obtained from steps 1 and 2 to predict the ordering of the C-O bond lengths in CO, CO2, and CO3虏鈦. Recall that a higher bond order indicates a stronger bond and a shorter bond length. The C-O bond orders are 3, 2, and 4/3 for CO, CO2, and CO3虏鈦, respectively. Consequently, the ordering of bond lengths will be CO < CO2 < CO3虏鈦.

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Most popular questions from this chapter

(a) What is meant by the term covalent bond? (b) Give three examples of covalent bonding. (c) A substance \(X Y\), formed from two different elements, boils at \(-33^{\circ} \mathrm{C}\). Is XY likely to be a covalent or an ionic substance? Explain.

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