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QUESTION

# 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.