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Chapter 20: Problem 88

The \(\mathrm{HO}_{3}\) radical was once thought of as a temporary reservoir of atmospheric OH radicals. Draw a Lewis structure of the species.

Short Answer

Expert verified
The \(\mathrm{HO}_{3}\) Lewis structure consists of one central Oxygen atom bonded to two other Oxygen atoms and the Hydrogen atom and also has an unpaired electron. This is representing the radical nature of the \(\mathrm{HO}_{3}\) molecule.

Step by step solution

01

Count the total number of valence electrons

In \(\mathrm{HO}_{3}\), there are one Hydrogen atom and three Oxygen atoms. Oxygen is in group 16 (or VI in older notation) on the periodic table and has 6 valence electrons. Hydrogen, a Group 1 (or I) element, has 1 valence electron. So, the total number of valence electrons is \(1*(1) + 3*(6) = 19\).
02

Begin drawing the Lewis structure

Start with the least electronegative atom in the center, which is Oxygen in this case (because Hydrogen can't be in the center). Put the others atoms (1 Hydrogen and 2 more Oxygens) around it and connect them with a pair of electrons (bond). Then complete the octets (8 valence electrons) for the Oxygen atoms (excluding central Oxygen for now).
03

Complete central atom's octet

Now, complete the octet for the central Oxygen. We will notice that we are short of 1 electron. That's due to the radical nature of \(\mathrm{HO}_{3}\) that it has an unpaired electron on the central Oxygen atom.Thus, the central atom has only 7 valence electrons. Remember that this is one of the special cases when constructing Lewis structures. The central atom here does not follow the octet rule.

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