# How do pi bonds affect bond angle?

Feb 21, 2016

You may be referring to a situation where we are comparing the molecular geometry of two molecules where one has a higher bond order on certain bonds and the other doesn't.

For example, let's compare the bond angles of ${\text{ClFO}}_{3}$, ${\text{CH}}_{4}$, and ${\text{NH}}_{3}$ (http://cccbdb.nist.gov/).

• $\text{O"-"Cl"-"O}$ bond angle: ${115.30}^{\circ}$ (${\text{ClFO}}_{3}$)
• $\text{H"-"C"-"H}$ bond angle: ${109.47}^{\circ}$ (${\text{CH}}_{4}$)
• $\text{H"-"N"-"H}$ bond angle: ${106.67}^{\circ}$ (${\text{NH}}_{3}$)

Now, why the decrease going from ${\text{ClFO}}_{3}$ through ${\text{NH}}_{3}$? Let's look at their structures.

METHANE BOND ANGLE

Let's start from ${\text{CH}}_{4}$. It's the standard tetrahedral geometry, with a standard bond angle. There is no deviation away from the ideal bond angle of ${109.47}^{\circ}$.

AMMONIA BOND ANGLE

When we move from ${\text{CH}}_{4}$ to ${\text{NH}}_{3}$, let's ignore the central atom radius, because it's not significant. "109.47^@ -> 106.67^@.

Notice how there is a lone pair of electrons on ${\text{NH}}_{3}$, which takes up more space than a bonding pair. That lone pair repels the bonding pairs downwards and inwards, decreasing the bond angle. Hence, ${\text{NH}}_{3}$ has a smaller bond angle than ${\text{CH}}_{4}$, like we see above.

PERCHLORYL FLUORIDE BOND ANGLE

When we move from ${\text{CH}}_{4}$ to ${\text{ClFO}}_{3}$, again, ignore the radius difference, which is not significant. "109.47^@ -> 115.30^@.

Notice how we went from having four single bonds to one single bond and three double bonds. The bonding pair in each double bond has greater electron density than a single bond. The bonding pairs can repel each other to a greater extent because it is easier for electrons to instantaneously repel each other when they often appear near each other. Hence, the additional $\setminus m a t h b f \left(\pi\right)$ bonds contribute to a greater bond angle.

Additionally, you have the fluorine, which is the most electronegative atom on the molecule, and that means it draws electron density towards it.

That polarizes the molecule upwards through the $\text{Cl"-"F}$ bond, shifting the electron density of the $\setminus m a t h b f \left(\text{Cl"="O}\right)$ bonding pairs inwards, which means the electron density is more cramped.

If we suppose the electron density on each $\text{Cl"="O}$ bond is a dot, then try imagining a circle; the arc length is smaller for a smaller radius, so when those dots move inwards, the distance between electrons on average is smaller, thus increasing bonding pair repulsions. Thus, those bonding pairs repel each other more than before.

That, along with the presence of the additional $\pi$ bonds, increases the bond angle significantly with respect to ${\text{CH}}_{4}$.