Question

free radicals are molecules or ions that have an unpaired electron. they tend to be very reactive and are thought to contribute to the aging process.
draw (on paper) a lewis structure for each of the substances below and indicate which are free radicals. for a free radical the unpaired electron goes on the least electronegative element, which is usually the central atom.
key: y = yes n = no

1. clo3
2. bro3
3. clo2-
4. o3

Explanation:

To determine if a species is a free radical, we check for an unpaired electron by calculating the total number of valence electrons and analyzing the Lewis structure. A free radical has an odd number of total valence electrons (since electrons are paired except for one unpaired electron).

1. $\boldsymbol{\ce{ClO3}}$

• Valence electrons: $\ce{Cl}$ (7) + 3×$\ce{O}$ (6) = $7 + 18 = 25$ (odd number).
• An odd number of electrons means one electron is unpaired. Thus, $\ce{ClO3}$ is a free radical.

2. $\boldsymbol{\ce{BrO3}}$

• Valence electrons: $\ce{Br}$ (7) + 3×$\ce{O}$ (6) = $7 + 18 = 25$ (odd number).
• An odd number of electrons means one electron is unpaired. Thus, $\ce{BrO3}$ is a free radical.

3. $\boldsymbol{\ce{ClO2^-}}$

• Valence electrons: $\ce{Cl}$ (7) + 2×$\ce{O}$ (6) + 1 (negative charge) = $7 + 12 + 1 = 20$ (even number).
• Even number of electrons means all electrons are paired. Thus, $\ce{ClO2^-}$ is not a free radical.

4. $\boldsymbol{\ce{O3}}$ (Ozone)

• Valence electrons: 3×$\ce{O}$ (6) = $18$ (even number).
• Even number of electrons means all electrons are paired (ozone has resonance structures with paired electrons). Thus, $\ce{O3}$ is not a free radical.

Answer:

s (v = Yes, n = No):

1. $\ce{ClO3}$: $\boldsymbol{v}$
2. $\ce{BrO3}$: $\boldsymbol{v}$
3. $\ce{ClO2^-}$: $\boldsymbol{n}$
4. $\ce{O3}$: $\boldsymbol{n}$