QQuestionChemistry
QuestionChemistry
Draw the Lewis structure for the polyatomic hydroperoxy anion (HO^- 2). be sure to include all resonance structures that satisfy the octect rule.
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Answer
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Step 1:: Identify the total number of valence electrons in the molecule.
The hydroperoxy anion (HO$^-2$) consists of one hydrogen atom (1 valence electron), one oxygen atom (6 valence electrons), and two extra negative charges (2 $\times$ 6 = 12 valence electrons).
Therefore, the total number of valence electrons in the molecule is 1 + 6 + 12 = 19 valence electrons.
Step 2:: Determine the central atom and arrange the remaining atoms around it.
In this case, oxygen is the central atom, and hydrogen is attached to it. Place the hydrogen atom and the negative charges on opposite sides of the oxygen atom.
Step 3:: Place the remaining valence electrons as lone pairs on the atoms.
Distribute the remaining 19 - (2 $\times$ 6) = 7 valence electrons as lone pairs on the oxygen atom and hydrogen atom.
Since oxygen can form double bonds and hydrogen can only form one bond, place two lone pairs on the oxygen atom and one lone pair on the hydrogen atom.
Step 4:: Satisfy the octet rule for each atom.
The oxygen atom has 6 electrons from bonds and 2 lone pairs, totaling 8 electrons, so it satisfies the octet rule. The hydrogen atom has 2 electrons from the bond, also satisfying the octet rule.
Step 5:: Draw resonance structures.
Since the negative charges are delocalized in the molecule, we need to draw resonance structures to account for this delocalization. Draw two additional structures where the negative charges are distributed on the oxygen atoms.
Final Answer
The Lewis structure for the hydroperoxy anion (HO$^- 2$) includes the following resonance structures: \begin{array}{ccc} \text{Structure 1:} & \text{Structure 2:} & \text{Structure 3:} \ \chemfig{H-O=-(-[:90]O)^{-}} & \chemfig{H-O=O(-[:90]O)^{-}} & \chemfig{O=-(-[:90]O)^{-}-H} \end{array} These structures represent the delocalization of the negative charges in the molecule.
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