Q11.22P - Chemistry: Molecular Nature Of Matter And Change

Question

Describe the hybrid orbitals used by the central atom and the type(s) of bonds formed in

(a) ${\mathit{O}}_{\mathbf{3}}$

(b) ${{\mathit{I}}_{\mathbf{3}}}^{\mathbf{-}}$

(c) $\mathit{C}\mathit{O}\mathit{C}{\mathit{I}}_{\mathbf{2}}$ (C is central)

Step-by-Step Solution

Answer

Answer

The hybridization and the types of bonds are:

1.  ${\mathit{O}}_{\mathbf{3}}\mathbf{=}\mathit{s}{\mathit{p}}^{\mathbf{3}}$ hybridized, the bonds formed are two sigma and one pi-bond.

2.  ${{\mathit{I}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{=}\mathit{s}{\mathit{p}}^{\mathbf{3}}\mathit{d}$ hybridized, the bonds formed are two sigma and three pi-bonds.

3.  $\mathit{C}\mathit{O}\mathit{C}{\mathit{I}}_{\mathbf{2}}\mathbf{=}\mathit{s}{\mathit{p}}^{\mathbf{2}}$ hybridized and the bonds formed are three sigma and one pi-bond

Step 1: Hybridization

Hybridization may be defined as the bonding of a hybrid orbital in which the hybrid orbital is formed after the mixing of the orbital of the atom of different energy.

Step 2: Subpart (a) Ozone Geometry.

The ozone has a hybridization of $\mathit{s}{\mathit{p}}^{\mathbf{3}}$. The valence shell of all the O-atom has a p-orbital having an electronic configuration forming a single and double bond.

$\mathit{O}\mathbf{=}\mathbf{1}{\mathit{s}}^{\mathbf{2}}\mathbf{2}{\mathit{s}}^{\mathbf{2}}\mathbf{2}{\mathit{p}}^{\mathbf{4}}$

     

Here, the bonds formed are two sigma and one pi-bond.

Step 3: Subpart (b) Iodide Ion Geometry.

The iodide ion has hybridization of $\mathit{s}{\mathit{p}}^{\mathbf{3}}\mathit{d}$. The valence shell of all the I-atom p-orbital has an electronic configuration:

$\mathit{I}\mathbf{=}\mathit{K}\mathit{r}\mathbf{4}{\mathit{d}}^{\mathbf{10}}\mathbf{5}{\mathit{s}}^{\mathbf{2}}\mathbf{5}{\mathit{p}}^{\mathbf{5}}$

The geometry of the ${{\mathit{I}}_{\mathbf{3}}}^{\mathbf{-}}$ is trigonal by-pyramidal and the shape is T-shape as shown below.

Here, the bonds formed are two sigma and three pi-bonds.

The three pi-bonds are because of the presence of three lone pairs of electrons.

Step 4: Subpart (3) Phosgene Geometry.

The ozone has hybridization of $\mathit{s}{\mathit{p}}^{\mathbf{2}}$. The valence shell of all the O-atom has a p-orbital having an electronic configuration:

$$\begin{gathered} \mathit{C}\mathbf{=}\mathbf{1}{\mathit{s}}^{\mathbf{2}}\mathbf{2}{\mathit{s}}^{\mathbf{2}}\mathbf{2}{\mathit{p}}^{\mathbf{2}} \\ \mathit{O}\mathbf{=}\mathbf{1}{\mathit{s}}^{\mathbf{2}}\mathbf{2}{\mathit{s}}^{\mathbf{2}}\mathbf{2}{\mathit{p}}^{\mathbf{4}} \\ \mathit{C}\mathit{I}\mathbf{=}\mathbf{1}{\mathit{s}}^{\mathbf{2}}\mathbf{2}{\mathit{s}}^{\mathbf{2}}\mathbf{2}{\mathit{p}}^{\mathbf{6}}\mathbf{3}{\mathit{s}}^{\mathbf{2}}\mathbf{3}{\mathit{p}}^{\mathbf{6}} \end{gathered}$$

The geometry of the Phosgene is Trigonal Planar.

Here, the bonds formed are three sigma and one pi-bond between carbon and oxygen.