# How do the bond angles compare for "ClO"_2, "ClO"_2^(-), and "Cl"_2"O"? Explain why.

Jun 30, 2016

I got a questionable result, based on two sources giving me borderline bond angles.

$\stackrel{{117.40}^{\circ}}{\overbrace{\text{ClO"_2) > stackrel(~~111^@)overbrace("ClO"_2^(-)) > stackrel(110.88^@)overbrace("Cl"_2"O}}} ,$

if the bond angle in ${\text{ClO}}_{2}^{-}$ is about ${111}^{\circ}$.

$\stackrel{{117.40}^{\circ}}{\overbrace{{\text{ClO"_2) > stackrel(110.88^@)overbrace("Cl"_2"O") > stackrel(~~110^@)overbrace("ClO}}_{2}^{-}}} ,$

if the bond angle in ${\text{ClO}}_{2}^{-}$ is about ${110}^{\circ}$.

It's just too close to be sure either way unless your book has more precise bond angles in the answer key.

For reference, the electronegativity of $\text{Cl}$ is $3.16$, and that of $\text{O}$ is $3.44$.

• $\text{Cl"_2"O}$ has oxygen at the center (with chlorine having its typical valency as a halogen), and has $7 + 7 + 6 = 20$ total valence electrons to distribute. The major resonance structure is: Since oxygen is more electronegative, the negative electron density is mostly concentrated onto oxygen, which is closer to each bonding-electron pair (the electron density is closer together when you look near oxygen).

Therefore, oxygen's share of electron density repels the bonding-electron pairs more easily in each $\text{Cl"-"O}$ bond than if the electronegativity difference was smaller.

This competes with the lone-pair repulsion, which would have contracted the bond angle...

Overall, the competing effects stack to increase the bond angle to a bit more than the expected ${109.5}^{\circ}$, because...

Its actual bond angle is about $\textcolor{b l u e}{{110.88}^{\circ}}$.

• ${\text{ClO}}_{2}$ has $7 + 6 + 6 = 19$ total valence electrons to distribute (yes, it's paramagnetic). The major resonance structure is: With two double bonds, the bonding-electron pairs repel each other more so than with comparable single bonds, increasing the bond angle above the standard ${109.5}^{\circ}$.

Since $\text{O}$ atom is larger than $\text{Cl}$ atom, that also contributes to the substantially larger bond angle than in $\text{Cl"_2"O}$.

We also have the one less valence electron on $\text{Cl}$ than in ${\text{ClO}}_{2}^{-}$, giving less "lone-pair" repulsion, and thus less contraction of the $\text{O"="Cl"="O}$ bond angle by the $\text{Cl}$ valence electrons, relative to one more valence electron on $\text{Cl}$. This further increases the bond angle.

Its actual bond angle is about $\textcolor{b l u e}{{117.40}^{\circ}}$.

• ${\text{ClO}}_{2}^{-}$ has $7 + 6 + 6 + 1 = 20$ total valence electrons to distribute. The major resonance structure is: The bond order of each $\text{Cl"stackrel(--" ")(_)"O}$ bond in the resonance hybrid structure (roughly $1.5$) is lower than the bond order in each $\text{Cl"="O}$ bond in ${\text{ClO}}_{2}$ (pretty much $2$).

Therefore, there is less electron density in each $\text{Cl"stackrel(--" ")(_)"O}$ bond, allowing the ion's $\text{O"-"Cl"-"O}$ bond angle to contract a little, relative to the same bond angle in ${\text{ClO}}_{2}$.

However, the fourth valence electron on $\text{Cl}$ contracts the bond angle relative to ${\text{ClO}}_{2}$ even more than if there were only three nonbonding valence electrons on $\text{Cl}$. How much smaller of a bond angle we get, I'm not sure.

I cannot find a more precise actual bond angle than ${111}^{\circ}$.

However, this source unfortunately lists a poor-precision angle of ${110}^{\circ} \pm {2}^{\circ}$, which adds some confusion.

Therefore, the bond angle order is questionable.

$\textcolor{b l u e}{\stackrel{{117.40}^{\circ}}{\overbrace{\text{ClO"_2) > stackrel(~~111^@)overbrace("ClO"_2^(-)) > stackrel(110.88^@)overbrace("Cl"_2"O}}}} ,$

if the bond angle in ${\text{ClO}}_{2}^{-}$ is about ${111}^{\circ}$.

$\textcolor{b l u e}{\stackrel{{117.40}^{\circ}}{\overbrace{{\text{ClO"_2) > stackrel(110.88^@)overbrace("Cl"_2"O") > stackrel(~~110^@)overbrace("ClO}}_{2}^{-}}}} ,$

if the bond angle in ${\text{ClO}}_{2}^{-}$ is about ${110}^{\circ}$.

It's just too close to be sure either way unless your book has more precise bond angles in the answer key.