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Question
a lewis structure for chlorate (clo₃⁻) is shown, however, its formal charges are not minimized. starting from this structure, complete the correct structure with minimized formal charges.
Step1: Calculate formal charges
The formula for formal charge is \(FC = V - N - \frac{B}{2}\), where \(V\) is the number of valence electrons, \(N\) is the number of non - bonding electrons, and \(B\) is the number of bonding electrons.
For \(Cl\) in the given structure: \(V = 7\), \(N = 0\), \(B = 6\). \(FC=7 - 0-\frac{6}{2}=4\)
For single - bonded \(O\): \(V = 6\), \(N = 6\), \(B = 2\). \(FC = 6-6-\frac{2}{2}=- 1\)
Step2: Minimize formal charges
To minimize formal charges, we need to convert a single bond to a double bond.
Move a lone pair from one of the \(O\) atoms to form a \(Cl = O\) double bond.
For \(Cl\) after double - bond formation: \(V = 7\), \(N = 0\), \(B = 8\). \(FC=7 - 0-\frac{8}{2}=3\)
For double - bonded \(O\): \(V = 6\), \(N = 4\), \(B = 4\). \(FC = 6 - 4-\frac{4}{2}=0\)
For single - bonded \(O\): \(V = 6\), \(N = 6\), \(B = 2\). \(FC=-1\) (there are two single - bonded \(O\) atoms now)
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The correct Lewis structure has one \(Cl = O\) double bond and two \(Cl - O\) single bonds. The \(Cl\) atom has no lone pairs, the double - bonded \(O\) has two lone pairs, and each single - bonded \(O\) has three lone pairs.