Question

In addition to $\mathit{C}{\mathit{O}}_{\mathbf{2}}$, two additional stable oxides of carbon form. The space filling models for $\mathit{C}{\mathit{O}}_{\mathbf{2}}$and the other two stable oxides are:

What are the formulas for the two additional stable oxides of carbon? Explain the bonding in each of these forms using the localized electron model.

Short answer

The molecular formulas are $CO$, $C{O}_2$and $C_3{O}_2$and $\sigma$ and$\pi$ bonds will be formed by overlapping orbitals.

Step-by-step solution

Description of the given question

Bonds are formulated by a pair of electrons shared by two atoms in the localised electron bonding concept. Because an orbital can carry two electrons, we commonly refer to them as bonding pairs or lone pairs and the electron pair that is shared is referred to as a bonding pair.

The molecular formulas and formation of bonds

The formula for the molecules mentioned in the question are $CO$, $C{O}_2$and $C_3{O}_2$.

Because it generates $2\sigma$bonds, carbon in $C{O}_2$will be hybridised (one with each oxygen atom ). With each oxygen atom, two $\pi$ bonds will be created. Above or below the Molecular plane and $\sigma$ bond, the initial $\pi$ bond will be placed. Because the molecule will lessen electron repulsion if these two $\pi$ bonds do not sit in the same plane, the second $\pi$ bond will be perpendicular to the preceding $\pi$bond.

$C{O}_2$is the second molecule. One $\sigma$and one $\pi$bond will form in this molecule. A $\sigma$bond will be formed directly between two atoms, with a $\pi$molecule positioned above or below it.

The final molecule is $C_3{O}_2$. The oxygen atoms are on the molecule's borders, and the carbon atoms are in the centre. Between the atoms in the molecule, four $\sigma$bonds will be formed (two between carbons and two between carbon and oxygen). To lessen electron repulsion, overlapping p orbitals will produce four $\pi$bonds. The first and third $\pi$bonds will be positioned above or below the molecular plane, while the second and last $\pi$bonds will be perpendicular to them. Each carbon atom will undergo a $sp$ hybridization process.