Which having higher bond angle and why "ClO"_2 or "ClO"_2^-?

1 Answer
Apr 5, 2018

Well, draw out the Lewis structure.


"Cl": 7 valence electrons
"O": 6 valence electrons

Therefore,

"ClO"_2: 7 + 2 xx 6 = 21 valence electrons

"ClO"_2^(-): 7 + 2 xx 6 + 1 = 22 valence electrons

So now, put the largest atom in the middle ("Cl") and surround it with the other atoms. Put two bonds on "Cl" by default, since it is the central atom. Add three lone pairs on each oxygen and begin to form the appropriate pi bonds using those electrons.

Do "ClO"_2^- first, then take one electron off of "Cl" to get "ClO"_2. For "ClO"_2 you will have 11 valence electrons on "Cl":

"ClO"_2^- really has resonance going on, so the hybrid structure has 1.5-bonds. On the other hand, "ClO"_2 has two double bonds instead.

FACTOR 1: BONDING ELECTRONS

The bonding electrons in "ClO"_2 repel each other more, since there are more electrons in a double bond than a 1.5-bond. That would increase the bond angle of "ClO"_2 compared to "ClO"_2^-.

FACTOR 2: NONBONDING ELECTRONS

There is one more nonbonding electron in "ClO"_2^- to repel the bonding electrons, which would decrease the bond angle of "ClO"_2^- compared to "ClO"_2.

Therefore, bb("ClO"_2) would have the larger bond angle. In fact, "ClO"_2 has /_OClO = 117.4033^@, whereas "ClO"_2^- has an approximate bond angle of /_OClO ~~ 110^@ (pm 2^@)).