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Chapter 6: Problem 37

Draw Lewis structures for the following polyatomic ions. a. \(\mathrm{CO}_{3}^{2-}\) b. \(\mathrm{O}_{2}^{2-}\) c. \(\mathrm{PO}_{4}^{3-}\)

Short Answer

Expert verified
The Lewis structures for the three polyatomic ions have been drawn. For \(CO^3_{2-}\), Carbon (C) forms a double bond with one Oxygen and single bonds with other two, and each Oxygen has 6 lone pair of electrons and -1 charge each. For \(O^2_{2-}\), two Oxygen atoms form a double bond, and each have 6 lone pair of electrons. For \(PO^4_{3-}\), Phosphorus is centralized and forms single bonds with four Oxygen atoms. Three oxygens have 6 lone pair of electrons and -1 charge each, and one oxygen atom remains neutral.

Step by step solution

01

Draw the Lewis Structure for \(CO^3_{2-}\)

Start by drawing the central atom, Carbon (C), and then draw a bond (a line) from Carbon to each Oxygen (O). Carbon has 4 valence electrons and each Oxygen has 6, so this molecule has a total of 24 electrons. Arrange these 24 electrons around the atoms (after considering 6 electrons for 3 bonding lines, as each bonding line represents 2 electrons) in such a way that every Oxygen has 8 (an octet), and distribute the remaining electrons. Any extra electrons can be placed on the central atom. Add charges in the end to atoms as required. Final structure will have Carbon (C) in the center double bonded with one Oxygen and single bonded with other two, each having 6 lone pair of electrons and -1 charge each.
02

Draw the Lewis Structure for \(O^2_{2-}\)

This molecule is made up of two Oxygen atoms and overall charge is -2. Each Oxygen atom has 6 valence electrons, bringing total up to 12. Add the two extra required for the -2 charge, resulting in a total of 14 electrons that need to be distributed. Draw a double bond (two lines) between the Oxygen atoms. Allocate the rest as lone pairs on each Oxygen (each oxygen atom takes 6 lone pair of electrons). No charges required on individual atoms as overall molecule is electrically neutral.
03

Draw the Lewis Structure for \(PO^4_{3-}\)

The central atom is Phosphorus (P), surrounded by four Oxygen (O) atoms. Phosphorus has 5 valence electrons, each oxygen atom has 6, so total valence electrons are 5+4*6 = 29. As the ion has a -3 charge, three extra electrons must be added, bringing the total to 32. Draw the Phosphorus in the center, surrounded by four singly bonded Oxygens. After considering 8 electrons for the 4 bonds, distribute the remaining 24 electrons around the Oxygens, in such a way that every Oxygen gets an octet. Add charges to individual atoms in the end, resulting in 3 oxygen atoms having -1 charge each and one being electrically neutral.

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Most popular questions from this chapter

Why does \(F\) generally form covalent bonds with great polarity?

Ionic compounds tend to have higher boiling points than covalent substances do. Both ammonia, \(\mathrm{NH}_{3},\) and methane \(\mathrm{CH}_{4},\) are covalent compunds, yet the boiling point of ammonia is \(130^{\circ} \mathrm{C}\) higher than that of methane. What might account for this large difference?

Describe a weakness of using Lewis structures to model covalent compounds.

Describe the energy changes that take place when two atoms form a covalent bond.

Predict whether the bonds between the following pairs of elements are ionic, polar covalent, or nonpolar covalent. a. \(\mathrm{Na}-\mathrm{F}\) b. \(\mathrm{H}-\mathrm{I}\) c. \(\mathrm{N}-\mathrm{O}\) d. \(\mathrm{Al}-\mathrm{O}\) e. \(\mathrm{S}-\mathrm{O}\) f. \(\mathrm{H}-\mathrm{H}\)
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