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Chapter 8: Problem 90

Calculate the formal charge on the indicated atom in each of the following molecules or ions: (a) the central oxygen atom in \(\mathrm{O}_{3}\), (b) phosphorus in \(\mathrm{PF}_{6}^{-}\), (c) nitrogen in \(\mathrm{NO}_{2}\), (d) iodine in \(\mathrm{ICl}_{3}\), (e) chlorine in \(\mathrm{HClO}_{4}\) (hydrogen is bonded to \(\mathrm{O}\) ).

Short Answer

Expert verified
The formal charges on the indicated atoms in the molecules or ions are: (a) central oxygen atom in O3 = +2, (b) phosphorus in PF6^- = -1, (c) nitrogen in NO2 = +2, (d) iodine in ICl3 = 0, (e) chlorine in HClO4 = +3.

Step by step solution

01

Calculate Formal Charge on the central Oxygen in O3

1) Draw the Lewis structure of O3 Oxygen has 6 valence electrons, so in O3, we have 18 valence electrons in total. The Lewis structure will be: O=O-O with a lone pair on the terminal oxygen, and three lone pairs on each of the other oxygens. 2) Calculate Formal Charge on the central oxygen atom Valence Electrons of Central Oxygen: 6 Non-bonding electrons: 2 (from the lone pair) Bonding electrons: 4 (2 single bonds) Formal Charge = 6 - 2 - (1/2 x 4) = 6 - 2 - 2 = +2
02

Calculate Formal Charge on Phosphorus in PF6^-

1) Draw the Lewis structure of PF6^- Phosphorus has 5 valence electrons, and fluorine has 7 valence electrons. In PF6^-, we have a total of 48 valence electrons [(5 + 6 * 7) +1 = 48]. The Lewis structure will be a central phosphorus atom bonded to six fluorine atoms, with three lone pairs on each fluorine. 2) Calculate Formal Charge on Phosphorus Valence Electrons of Phosphorus: 5 Non-bonding electrons: 0 Bonding electrons: 12 (6 single bonds) Formal Charge = 5 - 0 - (1/2 x 12) = 5 - 6 = -1
03

Calculate Formal Charge on Nitrogen in NO2

1) Draw the Lewis structure of NO2 Nitrogen has 5 valence electrons and oxygen has 6 valence electrons, so in NO2 we have a total of 17 valence electrons (5 + 2 * 6 = 17). The Lewis structure will be: O=N-O with one lone pair on the singly bonded oxygen and two lone pairs on the other oxygen. 2) Calculate Formal Charge on Nitrogen Valence Electrons of Nitrogen: 5 Non-bonding electrons: 0 Bonding electrons: 6 (1 double bond and 1 single bond) Formal Charge = 5 - 0 - (1/2 x 6) = 5 - 3 = +2
04

Calculate Formal Charge on Iodine in ICl3

1) Draw the Lewis structure of ICl3 Iodine has 7 valence electrons and chlorine has 7 valence electrons, so in ICl3 we have a total of 28 valence electrons (7 + 3 * 7 = 28). The Lewis structure will be a central iodine atom bonded to three chlorine atoms and two lone pairs on the iodine. Each chlorine atom has three lone pairs. 2) Calculate Formal Charge on Iodine Valence Electrons of Iodine: 7 Non-bonding electrons: 4 (from the two lone pairs) Bonding electrons: 6 (3 single bonds) Formal Charge = 7 - 4 - (1/2 x 6) = 3 - 3 = 0
05

Calculate Formal Charge on Chlorine in HClO4

1) Draw the Lewis structure of HClO4 Chlorine has 7 valence electrons, oxygen has 6 valence electrons, and hydrogen has 1 valence electron, so in HClO4 we have a total of 32 valence electrons (7 + 4 * 6 + 1 = 32). The Lewis structure will consist of a central chlorine atom bonded to four oxygen atoms, with one of the oxygen atoms bonded to the hydrogen. Each oxygen will have two lone pairs and the hydrogen will have no lone pairs. 2) Calculate Formal Charge on Chlorine Valence Electrons of Chlorine: 7 Non-bonding electrons: 0 Bonding electrons: 8 (4 single bonds) Formal Charge = 7 - 0 - (1/2 x 8) = 7 - 4 = +3

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Most popular questions from this chapter

In the following pairs of binary compounds determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: (a) \(\mathrm{TiCl}_{4}\) and \(\mathrm{CaF}_{2}\), (b) \(\mathrm{ClF}_{3}\) and \(\mathrm{VF}_{3}\), (c) \(\mathrm{SbCl}_{5}\) and \(\mathrm{AlF}_{3}\).

Fill in the blank with the appropriate numbers for both electrons and bonds (considering that single bonds are counted as one, double bonds as two, and triple bonds as three). (a) Fluorine has valence electrons and makes bond(s) in compounds. (b) Oxygen has valence electrons and makes bond(s) in compounds. (c) Nitrogen has valence electrons and makes bond(s) in compounds. (d) Carbon has valence electrons and makes bond \((s)\) in compounds.

The substance chlorine monoxide, \(\mathrm{ClO}(\mathrm{g})\), is important in atmospheric processes that lead to depletion of the ozone layer. The ClO molecule has an experimental dipole moment of \(1.24 \mathrm{D}\), and the \(\mathrm{Cl}\) - \(\mathrm{O}\) bond length is \(1.60 \hat{A}\). (a) Determine the magnitude of the charges on the \(\mathrm{Cl}\) and \(\mathrm{O}\) atoms in units of the electronic charge, \(e\). (b) Based on the electronegativities of the elements, which atom would you expect to have a partial negative charge in the \(\mathrm{ClO}\) molecule? (c) Using formal charges as a guide, propose the dominant Lewis structure for the molecule. (d) The anion \(\mathrm{ClO}^{-}\)exists. What is the formal charge on the \(\mathrm{Cl}\) for the best Lewis structure for \(\mathrm{ClO}^{-}\)?

In the vapor phase, \(\mathrm{BeCl}_{2}\) exists as a discrete molecule. (a) Draw the Lewis structure of this molecule, using only single bonds. Does this Lewis structure satisfy the octet rule? (b) What other resonance structures are possible that satisfy the octet rule? (c) On the basis of the formal charges, which Lewis structure is expected to be dominant for \(\mathrm{BeCl}_{2}\) ?

(a) Construct a Lewis structure for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), in which each atom achieves an octet of electrons. (b) How many bonding electrons are between the two oxygen atoms? (c) Do you expect the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{H}_{2} \mathrm{O}_{2}\) to be longer or shorter than the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{O}_{2}\) ? Explain.
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