3. Polarity

L2CHEM

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First ‘electronegativity’ must be covered at least briefly.

Basically, electronegativity is how strongly an atom attracts shared electrons in a bond.

Molecules have different electronegativity. Fluorine, has the strongest electronegativity out of all the elements. In the periodic table, it goes like:

B	C	N	O	F
Al	Si	P	S	Cl

As the elements get ‘closer’ to fluorine on the periodic table, the stronger the electronegativity is. In the table above, aluminium $Al$ would have the weakest electronegativity, since it’s the furtherest away from fluorine.

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When we say ‘polarity’, we are asking if a molecule is polar or not. Polar molecules, are molecules that have a difference in electronegativity between the ions/molecules in a bond, whereas non-polar molecules, are molecules that DON’T have an electronegativity difference in a bond.

An ion/molecule that has a higher electronegativity will ‘take the ${\delta }^{-}$ sign’, to indicate that it’s more electronegative, and likewise, a less electronegative ion/molecule will ‘take the ${\delta }^{+}$’ sign to indicate that it’s less electronegative.

For instance, $H_{2}O$ has a overall shape of bent, where $O$ is more electronegative, as it’s closer to $F$, and $H$ is less electronegative. $O$ takes the ${\delta }^{-}$ sign, since it’s more electronegative, and $H$ takes the ${\delta }^{+}$ sign because it’s less electronegative.

When one takes ${\delta }^{-}$, and the other takes ${\delta }^{+}$, we say that the molecules have ‘created dipoles’. Dipoles will be created where there is a electronegativity difference. For example, $C{H}_4$, will have a tetrahedral structure with 4 dipoles, as $C$ is more electronegative than $H$. Each $H$ takes the ${\delta }^{+}$ sign, and creates dipoles. We can perhaps say that the molecule has 4 dipoles…

After writing down the ${\delta }^{\pm }$, then we determine whether the dipoles cancel out. This works exactly like adding vectors in physics. If they cancel out, just like how vectors cancel out, we can say that the dipoles cancel out, whereas if they don’t cancel out, we can say that the dipoles don’t cancel out.

What this ‘dipole cancellation’ tells us is whether if the molecule is polar or not. If the dipoles DO cancel out, then it’s non-polar, but if they DO NOT cancel out, then it’s polar.

This is related to shapes, too. If shape is symmetrical, it’s non-polar. If non-symmetrical, it’s polar.

Symmetry of the shape, and dipoles cancel out are closely related, since shape often determines the creation of dipoles, and whether they cancel out or not.

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$C{H}_4$’s dipoles cancel out, since it’s tetrahedral(symmetrical shape), and the ${\delta }^{+}$ cancel out, since $H({\delta }^{+})$ is in all 4 directions. Just like how vectors cancel out, the dipoles cancel out here as well. Therefore, $C{H}_4$ is a non-polar molecule, because dipoles cancel out due to the symmetrical tetrahedral shape.

$H_{2}O$’s dipoles don’t cancel out, since it’s bent(non-symmetrical shape), and the ${\delta }^{+}$ don’t cancel out, since the dipoles aren’t cancelled out due to the non-symmetrical shape. Therefore, $H_{2}O$ is a polar molecule, due to its non-symmetrical shape, and the fact that dipoles don’t cancel out.