Q11.24P - Chemistry: Molecular Nature Of Matter And Change

Question

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

(a) $\mathit{B}\mathit{r}{\mathit{F}}_{\mathbf{3}}$;     (b) $\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathbf{=}\mathit{C}\mathit{H}$;     (c)  $\mathit{S}{\mathit{O}}_{\mathbf{2}}$

Step-by-Step Solution

Answer

Answer

The hybridization and the type of bond formed are:

1. $\mathit{B}\mathit{r}{\mathit{F}}_{\mathbf{3}}\mathbf{=}\mathit{s}{\mathit{p}}^{\mathbf{2}}$ hybridized and all three bonds are sigma bonds.

2. ${\mathit{C}}_{\mathbf{3}}{\mathit{H}}_{\mathbf{4}}\mathbf{=}\mathit{s}\mathit{p}$ hybridized and here six-sigma and two-pi bonds can be observed

3. $\mathit{S}{\mathit{O}}_{\mathbf{2}}\mathbf{=}\mathit{s}{\mathit{p}}^{\mathbf{2}}$ hybridized and here two sigma and two pi bonds are present.

Step 1: (Subpart a) Bromine fluoride Geometry.

The $\mathit{B}{\mathit{F}}_{\mathbf{3}}$ has a hybridization of $\mathit{s}{\mathit{p}}^{\mathbf{2}}$. The valence shell of all the Br-atom and F-atom have p-orbitals involved having an electronic configuration as:

$$\begin{gathered} \mathit{F}\mathbf{=}\mathbf{1}{\mathit{s}}^{\mathbf{2}}\mathbf{2}{\mathit{s}}^{\mathbf{2}}\mathbf{2}{\mathit{p}}^{\mathbf{5}} \\ \mathit{B}\mathit{r}\mathbf{=}\mathit{A}\mathit{r}\mathbf{4}{\mathit{s}}^{\mathbf{2}}\mathbf{3}{\mathit{d}}^{\mathbf{10}}\mathbf{4}{\mathit{p}}^{\mathbf{5}} \end{gathered}$$

The geometry of the Boron Fluoride is Trigonal Planar.

                

Here, all three bonds are sigma bonds as the first bond formation between two atoms is sigma always.

Step 2: Subpart (b) Prop-1-yne Geometry.

The $\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathbf{=}\mathit{C}\mathit{H}$ has hybridization of sp. The valence shell of all the C-atom and H-atom has p-orbital and s-orbital with the electronic configuration of:

$$\begin{gathered} \mathit{F}\mathbf{=}\mathbf{1}{\mathit{s}}^{\mathbf{2}}\mathbf{2}{\mathit{s}}^{\mathbf{2}}\mathbf{2}{\mathit{p}}^{\mathbf{5}} \\ \mathit{C}\mathbf{=}\mathbf{1}{\mathit{s}}^{\mathbf{2}}\mathbf{2}{\mathit{s}}^{\mathbf{2}}\mathbf{2}{\mathit{p}}^{\mathbf{2}} \end{gathered}$$

The geometry of the Prop-1-yne is Trigonal Planar.

Here, six-sigma bonds can be observed that are formed by head-to-head overlapping of electron density.

The other two pi-bonds are weaker comparatively and are formed by sidewise overlapping.

Step 3: Subpart (c) Sulfur dioxide Geometry.

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

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

The geometry of the Sulfur dioxide is Trigonal planar and the shape is a bent shape.

Here, two sigma and two pi bonds are present.

The first bonding between S and O is sigma and the second bonding is pi.