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Chapter 8: Problem 36

(a) Construct a Lewis structure for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), in which each atom achieves an octet of electrons. (b) 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 \(\mathrm{O}_{2} ?\)

Short Answer

Expert verified
The Lewis structure for hydrogen peroxide (H2O2) is H - O - O - H, with each oxygen atom having 2 lone pairs. The O-O bond in H2O2 is a single bond, while in O2, it is a double bond. As bond order increases, bond length decreases, so the O-O bond in H2O2 will be longer than the O-O bond in O2.

Step by step solution

01

Determine the total number of valence electrons in H2O2

The key to constructing a Lewis structure is determining the number of valence electrons for each element. Hydrogen has 1 valence electron, and Oxygen has 6 valence electrons. For H2O2, we have: - Two hydrogen atoms, each with 1 valence electron, giving us a total of 2 valence electrons. - Two oxygen atoms, each with 6 valence electrons, giving us a total of 12 valence electrons. Adding these together, we have 14 valence electrons in H2O2.
02

Arrange the atoms and form bonds

We will arrange the atoms in such a way that each atom satisfies the octet rule (2 electrons for hydrogen, and 8 electrons for oxygen). Since there is only one type of bond (O-O) in the molecule besides the hydrogen atoms, we will start by placing the oxygen atoms in the center. The structure will look like this: H-O-O-H. Now, we will form single bonds between: 1) Each hydrogen atom and the adjoining oxygen atom. 2) The two central oxygen atoms. Each single bond consists of 2 electrons, and since we have three single bonds, 6 out of the 14 valence electrons would be used up in bonding.
03

Place the remaining electrons around the oxygen atoms

We have 8 valence electrons remaining. We will place them around the central oxygen atoms to complete their octet. Both of the oxygen atoms already have 4 electrons (from the single bonds with the other oxygen atom and the hydrogen atom). Each oxygen atom will need 4 more electrons. We will place these additional electrons as lone pairs on each of the oxygen atoms: two lone pairs per oxygen atom.
04

Construct the final Lewis structure for H2O2

With all the valence electrons assigned, the final Lewis structure for hydrogen peroxide (H2O2) is as follows: H - O - O - H : : Each oxygen atom has 2 lone pairs, and each hydrogen atom has a single bond to an oxygen atom, while both oxygen atoms have a single bond with the other.
05

Compare the O-O bond length in H2O2 with that in O2

In H2O2, the O-O bond is a single bond. In an O2 molecule, the O-O bond is a double bond. In general, as bond order increases, bond length decreases. Therefore, the O-O bond in H2O2 (single bond) will be longer than the O-O bond in O2 (double bond).

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Most popular questions from this chapter

Average bond enthalpies are generally defined for gas-phase molecules. Many substances are liquids in their standard state. coo (Section 5.7) By using appropriate thermochemical data from Appendix C, calculate average bond enthalpies in the liquid state for the following bonds, and compare these values to the gas-phase values given in Table 8.4: (a) \(\mathrm{Br}-\mathrm{Br}\), from \(\mathrm{Br}_{2}(l) ;\) (b) \(\mathrm{C}-\mathrm{Cl},\) from \(\mathrm{CCl}_{4}(l) ;\) (c) \(\mathrm{O}-\mathrm{O},\) from \(\mathrm{H}_{2} \mathrm{O}_{2}(I)\) (assume that the \(\mathrm{O}-\mathrm{H}\) bond enthalpy is the same as in the gas phase). (d) What can you conclude about the process of breaking bonds in the liquid as compared to the gas phase? Explain the difference in the \(\Delta H\) values between the two phases.

A major challenge in implementing the "hydrogen economy" is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, \(\mathrm{NaAlH}_{4}\) can release \(5.6 \%\) of its mass as \(\mathrm{H}_{2}\) upon decomposing to \(\mathrm{NaH}(s), \mathrm{Al}(s),\) and \(\mathrm{H}_{2}(\mathrm{~g}),\) NaAlH \(_{4}\) possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (a) Write a balanced equation for the decomposition of \(\mathrm{NaAlH}_{4}\) (b) Which element in \(\mathrm{NaAlH}_{4}\) is the most electronegative? Which one is the least electronegative? (c) Based on electronegativity differences, what do you think is the identity of the polyatomic anion? Draw a Lewis structure for this ion.

Which of these elements are unlikely to form covalent bonds: \(\mathrm{S}, \mathrm{H}, \mathrm{K}, \mathrm{Ar},\) Si? Explain your choices.

The substance chlorine monoxide, \(\mathrm{ClO}(g)\), is important in atmospheric processes that lead to depletion of the ozone layer. The ClO molecule has a dipole moment of \(1.24 \mathrm{D}\) and the (a) Determine the magnitude of \(\mathrm{Cl}-\mathrm{O}\) bond length is \(1.60 \mathrm{~A}\). 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 negative charge in the ClO molecule? (c) By using formal charges as a guide, propose the dominant Lewis structure for the molecule. Are the formal charges consistent with your answers to parts (a) and (b)? Can you reconcile any differences you find?

(a) Using average bond enthalpies, predict which of the following reactions will be most exothermic: (i) \(\mathrm{C}(g)+2 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g)\) (ii) \(\mathrm{CO}(g)+3 \mathrm{~F}_{2} \longrightarrow \mathrm{CF}_{4}(g)+\mathrm{OF}_{2}(g)\) (iii) \(\mathrm{CO}_{2}(g)+4 \mathrm{~F}_{2} \longrightarrow \mathrm{CF}_{4}(g)+2 \mathrm{OF}_{2}(g)\) (b) Explain the trend, if any, that exists between reaction exothermicity and the extent to which the carbon atom is bonded to oxygen.
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