Question

With three named examples explain how hybridization of atomic orbitals May influence the shape and bond angles of chemical entities

Answer 1

Question #83953, Chemistry / Inorganic Chemistry | for completion

With three named examples explain how hybridization of atomic orbitals May influence the shape and bond angles of chemical entities

Answer:

We can use Lewis dot structures to determine bonding patterns in molecules. We can then use VSEPR to predict molecular shapes, based on the valence electron pairs of the Lewis structures. Once we know a molecular shape, we can start to look at the physical properties of compounds. For example, we should now be able to predict which molecules will be polar. Polarity exists when there is a separation of charge within a molecule. This will arise from polar bonds within the molecule, due to differences in electronegativity values between bonded atoms. For example, HF is a polar compound. Fluorine is much more electronegative than hydrogen and the shared pair of bonding electrons will spend more time near the F nucleus, than near the H nucleus.

The direction of a dipole moment (charge imbalance) is usually indicated by the presence of an arrow, as shown below for HF.

This indicates that the H will carry a partial positive charge $(\delta +)$ and the F will carry a partial negative charge $(\delta -)$ . All diatomic molecules containing atoms of different electronegativities will be polar molecules. This will affect their physical properties (melting and boiling points, solubilities, etc.).

In larger molecules (more than two atoms), the polarity of the overall compound will be determined by the presence of polar bonds and the molecular shape.

For example, we can compare carbon dioxide, $\mathrm{CO}_{2}$ to sulfur dioxide, $\mathrm{SO}_{2}$ . Their Lewis structures are shown below.

$$\begin{array}{c c} \overline {{\mathrm {O}}} = \mathrm {C} = \overline {{\mathrm {O}}} \\ \overline {{\mathrm {O}}} \end{array} \qquad \begin{array}{c c} \overline {{\mathrm {O}}} = \overline {{\mathrm {S}}} = \overline {{\mathrm {O}}} \\ \overline {{\mathrm {O}}} \end{array}$$

Carbon and sulfur have the same electronegativity, much less than that of oxygen. So, in both compounds the bonds will be equally polar. However, they have very different physical properties, $\mathrm{CO}_{2}$ boils at $-78^{\circ}\mathrm{C}$ , and $\mathrm{SO}_{2}$ boils

at $+22.8^{\circ}\mathrm{C}$, a $100^{\circ}$ difference. This must depend on more than just the presence of the two polar bond in each molecule. What makes the difference is the **molecular shape**.

$\mathrm{CO}_{2}$ will be a **linear** molecule, because there are only two electron pairs on the central carbon atom. It will have the shape shown below.

As indicated by the arrows, there are two very polar bonds in this molecule. However, because of the molecular shape of $\mathrm{CO}_{2}$, they are pointing in opposite directions, and will **cancel out**. $\mathrm{CO}_{2}$ is a **non-polar** compound, due to its molecular shape.

Shown below is the shape of an $\mathrm{SO}_2$ molecule. Its **molecular geometry** will be **trigonal** because of the **three valence electron pairs** on sulfur, two bonding pairs and a lone pair. This will give it a **bent**molecular shape. In this molecule, the dipoles are **not** pointing in opposite directions, and will not cancel out. They will, in fact add, and give a **net dipole moment**. $\mathrm{SO}_2$ is a **polar** compound, which explains (as we will soon see) its elevated boiling point.

(Quiz For a similar example, draw the Lewis structures of $\mathrm{BF}_3$ (-99) and $\mathrm{NH}_3$ (-33)

Another indication of the importance of molecular shape can be seen by comparing the physical properties of $\mathrm{CHCl}_3$ and $\mathrm{CCl}_4$. $\mathrm{CHCl}_3$ dissolves in water, and $\mathrm{CCl}_4$ does not. Why?

They will both have **tetrahedral** geometries, with 4 valence pairs of electrons on each C. The C-Cl bonds will all be **polar**. Their shapes are shown below.

In the $\mathrm{CHCl}_3$ molecule, the three polar C-Cl bonds add (vector addition) to give a net dipole moment to the molecule. In $\mathrm{CCl}_4$, the four polar C-Cl bonds will cancel out, making this a non-polar molecule. Water is a polar solvent, which only interacts with other polar species, "likes dissolve likes".

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