The correct order of increasing bond angle in the following species is?

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

##stackrel(117.40^@)overbrace("ClO"_2) > stackrel(~~111^@)overbrace("ClO"_2^(-)) > stackrel(110.88^@)overbrace("Cl"_2"O"),##

if the bond angle in ##"ClO"_2^(-)## is about ##111^@##.

##stackrel(117.40^@)overbrace("ClO"_2) > stackrel(110.88^@)overbrace("Cl"_2"O") > stackrel(~~110^@)overbrace("ClO"_2^(-)),##

if the bond angle in ##"ClO"_2^(-)## is about ##110^@##.

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 of ##"Cl"## is ##3.16##, and that of ##"O"## is ##3.44##.

• ##"Cl"_2"O"## has oxygen at the center (with chlorine having its typical valency as a halogen), and has ##7+7+6 = 20## total to distribute. The major structure is:

Since oxygen is more electronegative, the negative electron is mostly concentrated onto oxygen, which is closer to each -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 ##"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^@##, because...

Its actual bond angle is about ##color(blue)(110.88^@)##.

• ##"ClO"_2## has ##7+6+6 = 19## total 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^@##.

Since ##"O"## atom is larger than ##"Cl"## atom, that also contributes to the substantially larger bond angle than in ##"Cl"_2"O"##.

We also have the one less valence electron on ##"Cl"## than in ##"ClO"_2^(-)##, giving less "lone-pair" repulsion, and thus less contraction of the ##"O"="Cl"="O"## bond angle by the ##"Cl"## valence electrons, relative to one more valence electron on ##"Cl"##. This further increases the bond angle.

Its actual bond angle is about ##color(blue)(117.40^@)##.

• ##"ClO"_2^(-)## has ##7+6+6+1 = 20## total valence electrons to distribute. The major resonance structure is:

The bond order of each ##"Cl"stackrel(--" ")(_)"O"## bond in the resonance hybrid structure (roughly ##1.5##) is lower than the bond order in each ##"Cl"="O"## bond in ##"ClO"_2## (pretty much ##2##).

Therefore, there is less electron density in each ##"Cl"stackrel(--" ")(_)"O"## bond, allowing the ion's ##"O"-"Cl"-"O"## bond angle to contract a little, relative to the same bond angle in ##"ClO"_2##.

However, the fourth valence electron on ##"Cl"## contracts the bond angle relative to ##"ClO"_2## even more than if there were only three nonbonding valence electrons on ##"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^@##.

However, this source unfortunately lists a poor-precision angle of ##110^@ pm 2^@##, which adds some confusion.

Therefore, the bond angle order is questionable.

##color(blue)(stackrel(117.40^@)overbrace("ClO"_2) > stackrel(~~111^@)overbrace("ClO"_2^(-)) > stackrel(110.88^@)overbrace("Cl"_2"O")),##

if the bond angle in ##"ClO"_2^(-)## is about ##111^@##.

##color(blue)(stackrel(117.40^@)overbrace("ClO"_2) > stackrel(110.88^@)overbrace("Cl"_2"O") > stackrel(~~110^@)overbrace("ClO"_2^(-))),##

if the bond angle in ##"ClO"_2^(-)## is about ##110^@##.

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