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Chapter 8: Problem 34

(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 three lone pairs on each oxygen atom. The O-O bond in H2O2 is expected to be longer than the O-O bond in O2.

Step by step solution

01

Valence Electrons

In order to construct the Lewis structure of hydrogen peroxide, we first need to determine the number of valence electrons for each atom. In the periodic table, hydrogen (H) has 1 valence electron and oxygen (O) has 6 valence electrons. So, for two hydrogen atoms and two oxygen atoms, we have the following number of valence electrons: 2(H) 脳 1 + 2(O) 脳 6 = 14 valence electrons
02

Connect Atoms and Distribute Electrons

We know that hydrogen can only form 1 bond, so each hydrogen atom will be bonded to one of the oxygen atoms. Initially, connect the atoms with single bonds and distribute the remaining electrons: - Connecting 1 H to 1 O forms a single bond and uses 2 valence electrons, leaving 12 electrons. - Connecting the 2nd H to the other O also forms a single bond and uses another 2 electrons, leaving 10 electrons. - Now, we connect the two oxygen atoms forming a single bond, using up another 2 electrons, leaving 8 electrons. We distribute the remaining 8 electrons as lone pairs to complete the octet requirement, prioritizing oxygen atoms as hydrogen has achieved its duet rule already.
03

Verify Octet Rule

Let's verify if all atoms have achieved the octet rule with the current Lewis structure: - Both hydrogen atoms are connected with single bonds, so they have achieved their duet rule (2 electrons each). - Both oxygen atoms are connected to one hydrogen and the other oxygen atom with single bonds (2 electrons for each bond) and have 3 lone pairs each (2 electrons for each lone pair). So, they have achieved the octet rule with 8 electrons each.
04

Compare Bonds in H2O2 and O2

In the Lewis structure of H2O2, the oxygen-oxygen bond is a single bond. In O2, the oxygen-oxygen bond is a double bond. Double bonds are generally shorter and stronger than single bonds due to the added electron density between the atoms. So, we can expect the O-O bond in H2O2 (single bond) to be longer than the O-O bond in O2 (double bond). #Answer (a):# The Lewis structure of hydrogen peroxide (H2O2) can be represented as: H | O - O - H With three lone pairs on each oxygen atom. #Answer (b):# The O-O bond in H2O2 is expected to be longer than the O-O bond in O2.

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Most popular questions from this chapter

Barium azide is \(62.04 \%\) Ba and \(37.96 \%\) N. Each azide ion has a net charge of 1 -. (a) Determine the chemical formula of the azide ion. (b) Write three resonance structures for the azide ion. (c) Which structure is most important? (d) Predict the bond lengths in the ion.

(a) What is the octet rule? (b) How many electrons must a sulfur atom gain to achieve an octet in its valence shell? (c) If an atom has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{3}\), how many electrons must it gain to achieve an octet?

What is the Lewis symbol for each of the following atoms or ions: (a) \(\mathrm{Ca}\), (b) \(\mathrm{P}\), (c) \(\mathrm{Mg}^{2+}\), (d) \(S^{2-}\) ?

The reaction of indium, In, with sulfur leads to three binary compounds, which we will assume to be purely ionic. The three compounds have the following properties: $$ \begin{array}{lll} \hline \text { Compound } & \text { Mass \% In } & \text { Melting Point }\left({ }^{\circ} \mathrm{C}\right) \\ \hline \text { A } & 87.7 & 653 \\ \text { B } & 78.2 & 692 \\ \text { C } & 70.5 & 1050 \\ \hline \end{array} $$ (a) Determine the empirical formulas of compounds A, B, and C. (b) Give the oxidation state of In in each of the three compounds. (c) Write the electron configuration for the In ion in each compound. Do any of these configurations correspond to a noble-gas configuration? (d) In which compound is the ionic radius of In expected to be smallest? Explain. (e) The melting point of ionic compounds often correlates with the lattice energy. Explain the trends in the melting points of compounds A, B, and \(\mathrm{C}\) in these terms.

(a) What are valence electrons? (b) How many valence electrons does a nitrogen atom possess? (c) An atom has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2} .\) How many valence electrons does the atom have?
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