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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) 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.

Short Answer

Expert verified
The Lewis structure of hydrogen peroxide (H₂O₂) is as follows: H-O-O-H : : There are 2 bonding electrons between the two oxygen atoms in H₂O₂. The O-O bond in H₂O₂ is expected to be longer than the O-O bond in O₂, as the bond order in H₂O₂ is 1, while in O₂, it is 2. Higher bond orders usually lead to shorter bond lengths.

Step by step solution

01

1. Calculate the total number of valence electrons

To draw the Lewis structure for H_2O_2, we must first determine the total number of valence electrons. Hydrogen has 1 valence electron, and oxygen has 6 valence electrons. There are 2 hydrogen atoms and 2 oxygen atoms in H_2O_2, so we have: Total valence electrons = 2*(1) + 2*(6) = 2 + 12 = 14
02

2. Connect the atoms and form bonds

Place the two oxygen atoms in the center and connect them to each other with a single bond (2 electrons). Then, connect each hydrogen atom to an oxygen atom (2 electrons each): H-O-O-H Now we've used 6 of our 14 valence electrons.
03

3. Distribute the remaining valence electrons

We have 8 more valence electrons to distribute. First, complete the octet of both oxygen atoms by adding lone pairs of electrons: H-O-O-H : : Now both oxygen atoms have an octet of electrons, and all 14 valence electrons are used. So the correct Lewis structure for H_2O_2 is: H-O-O-H : : #b) Number of bonding electrons between the two oxygen atoms#
04

Count the bonding electrons between the two oxygen atoms

Looking at the Lewis structure of H_2O_2, there is a single bond between the two oxygen atoms: H-O-O-H : : A single bond consists of 2 bonding electrons, so there are 2 bonding electrons between the two oxygen atoms in H_2O_2. #c) Comparing the O-O bond length in H2O2 and O2#
05

1. Determine the bond order for both molecules

To compare the O-O bond length in H_2O_2 and O_2, we need to look at the bond order. The bond order is the number of shared electron pairs between two atoms, i.e., the number of bonding pairs. In H_2O_2, there is a single bond between the two oxygen atoms, so the bond order is 1. In O_2, there is a double bond between the two oxygen atoms (O=O), so the bond order is 2.
06

2. Compare the bond lengths and give an explanation

A higher bond order usually means a shorter bond length, as more electron pairs shared between the atoms lead to stronger and shorter bonds. Therefore, we expect the O-O bond in H_2O_2 (with a bond order of 1) to be longer than the O-O bond in O_2 (with a bond order of 2).

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Most popular questions from this chapter

Based on Lewis structures, predict the ordering, from shortest to longest, of \(\mathrm{N}-\mathrm{O}\) bond lengths in $\mathrm{NO}^{+}, \mathrm{NO}_{2}^{-},\( and \)\mathrm{NO}_{3}^{-}$.

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