Q11.21P - 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) $\mathit{N}{{\mathit{O}}_{\mathbf{3}}}^{\mathbf{-}}$

(b) $\mathit{C}{\mathit{S}}_{\mathbf{2}}$

(c) $\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{O}$

Step-by-Step Solution

Answer

Answer

The hybrid orbital used by the central atom and the types of bonds formed is:

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

2.  $\mathit{C}{\mathit{S}}_{\mathbf{2}}\mathbf{=}\mathit{s}\mathit{p}$ hybridized and the bonds formed are two sigma and two pi-bond.

3.  $\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{=}\mathit{s}{\mathit{p}}^{\mathbf{2}}$ 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) Nitrate Ion

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

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

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

The first bond is always the sigma bond between any two atoms.

The second and third bonds are pi-bonds.

Step 3: Subpart (b) Carbon Sulphide

The Carbon Supplied has hybridization sp. The valence shell of C-atom and S-atom has a p-orbital having an electronic configuration as:

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

 

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

The first bond is always the sigma bond between any two atoms.

The second bonds are pi-bond formed between carbon and sulfur.

Step 4: Subpart (c) Formaldehyde

Carbon Supplied has hybridization $\mathit{s}{\mathit{p}}^{\mathbf{2}}$. The valence shell of all the atoms C-atom and S-atom has a p-orbital having an electronic configuration.

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

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

The first bond is the sigma bond between any two atoms.

The second between carbon and oxygen is pi-bond.