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Chapter 8: Problem 64

Draw the Lewis structures for each of the following molecules or ions. Identify instances where the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state how many electrons surround these atoms: $(\mathbf{a}) \mathrm{PF}_{6}^{-},(\mathbf{b}) \mathrm{BeCl}_{2},(\mathbf{c}) \mathrm{NH}_{3},(\mathbf{d}) \mathrm{XeF}_{2} \mathrm{O}\( (the Xe is the central atom), (e) \)\mathrm{SO}_{4}^{2-}$.

Short Answer

Expert verified
In summary, the Lewis structures for the given molecules or ions are as follows: a) PF6-: All atoms obey the octet rule. b) BeCl2: The Be atom does not obey the octet rule and has only 4 electrons around it. c) NH3: All atoms obey the octet rule. d) XeF2O: The Xe atom does not fully obey the octet rule and has 12 electrons around it. e) SO4^2-: All atoms obey the octet rule.

Step by step solution

01

Drawing Lewis Structure for PF6-

PF6- has a total of 5 valence electrons on the central P atom, 6 valence electrons on each of the 6 F atoms, and an extra electron due to its -1 charge. The total number of valence electrons is thus 5 + 6*6 + 1 = 42. Place the P atom in the center, surrounded by the 6 F atoms. Then, put a single bond between each P-F pair with 2 shared electrons. Lastly, complete the octet rule by assigning the remaining 24 electrons (6 unshared pairs) to each F atom. The Lewis structure for PF6- is complete. The octet rule is obeyed in PF6-.
02

Drawing Lewis Structure for BeCl2

BeCl2 has a total of 2 valence electrons on the central Be atom and 7 valence electrons on each of the two Cl atoms. The total number of valence electrons is thus 2 + 7*2 = 16. Place the Be atom in the center, surrounded by the 2 Cl atoms. Put a single bond between Be-Cl pairs to share 2 electrons and assign the remaining 12 electrons (6 unshared pairs) to each Cl atom. The Lewis structure for BeCl2 is complete. The octet rule is not obeyed in BeCl2 since Be has only 4 electrons around it.
03

Drawing Lewis Structure for NH3

NH3 has a total of 5 valence electrons on the central N atom and 1 valence electron on each of the three H atoms. The total number of valence electrons is thus 5 + 1*3 = 8. Place the N atom in the center, surrounded by the 3 H atoms. Put a single bond between each N-H pair and share 2 electrons. Assign the remaining 2 electrons (1 unshared pair) to the N atom. The Lewis structure for NH3 is complete. The octet rule is obeyed in NH3.
04

Drawing Lewis Structure for XeF2O

XeF2O has a total of 8 valence electrons on the central Xe atom, 7 valence electrons on each of the two F atoms, and 6 valence electrons on the O atom. The total number of valence electrons is thus 8 + 7*2 + 6 = 28. Place the Xe atom in the center, surrounded by the 2 F atoms and the O atom. Put single bonds between each Xe-F and Xe-O pair to share 6 electrons. Complete the octet rule for each Xe-F and Xe-O atom. Also add 3 lone pairs (6 unshared electrons) to the Xe atom. The Lewis structure for XeF2O is complete. The octet rule is not fully obeyed in XeF2O because the Xe atom has 12 electrons around it.
05

Drawing Lewis Structure for SO4^2-

In the SO4^2- ion, there are 6 valence electrons on the central S atom, 6 valence electrons on each of the four O atoms, and two extra electrons due to the -2 charge. The total number of valence electrons is 6 + 6*4 + 2 = 32. Place the S atom in the center, surrounded by the four O atoms. Put a double bond between the S-O1 and S-O2 pair, and single bonds between the S-O3 and S-O4 pair for a total of 10 shared electrons. Assign the remaining 16 electrons (2 unpaired pairs) around each O atom except those with the double bond. The Lewis structure for SO4^2- is complete. The octet rule is obeyed in SO4^2-.

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Most popular questions from this chapter

Under special conditions, sulfur reacts with anhydrous liquid ammonia to form a binary compound of sulfur and nitrogen. The compound is found to consist of \(69.6 \% \mathrm{~S}\) and \(30.4 \% \mathrm{~N}\). Measurements of its molecular mass yield a value of \(184.3 \mathrm{~g} / \mathrm{mol}\). The compound occasionally detonates on being struck or when heated rapidly. The sulfur and nitrogen atoms of the molecule are joined in a ring. All the bonds in the ring are of the same length. (a) Calculate the empirical and molecular formulas for the substance. (b) Write Lewis structures for the molecule, based on the information you are given. (Hint: You should find a relatively small number of dominant Lewis structures.) (c) Predict the bond distances between the atoms in the ring. (Note: The \(\mathrm{S}-\mathrm{S}\) distance in the \(\mathrm{S}_{8}\) ring is \(205 \mathrm{pm} .\) ) \((\mathbf{d})\) The enthalpy of formation of the compound is estimated to be $480 \mathrm{~kJ} / \mathrm{mol}^{-1} . \Delta H_{f}^{\circ}\( of \)\mathrm{S}(g)\( is \)222.8 \mathrm{~kJ} / \mathrm{mol}$. Estimate the average bond enthalpy in the compound.

Predict the chemical formula of the ionic compound formed between the following pairs of elements: (a) Al and Cl, (b) \(\mathrm{Mg}\) and \(\mathrm{O},(\mathbf{c}) \mathrm{Zn}\) and \(\mathrm{Cl}\), (d) \(\mathrm{Li}\) and \(\mathrm{O}\).

You and a partner are asked to complete a lab entitled "Carbonates of Group 2 metal" that is scheduled to extend over two lab periods. The first lab, which is to be completed by your partner, is devoted to carrying out compositional analysis and determine the identity of the Group 2 metal (M). In the second lab, you are to determine the melting point of this compound. Upon going to lab you find two unlabeled vials containing white powder. You also find the following notes in your partner's notebook-Compound 1: \(40.04 \% \mathrm{M}\) and \(12.00 \%\) C, \(47.96 \%\) O (by mass), Compound \(2: 69.59 \% \mathrm{M}\), \(6.09 \% \mathrm{C},\) and \(24.32 \% \mathrm{O}\) (by mass). (a) What is the empirical formula for Compound 1 and the identity of $\mathrm{M} ?(\mathbf{b})$ What is the empirical formula for Compound 2 and the identity of \(\mathrm{M}\) ? Upon determining the melting points of these two compounds, you find that both compounds do not melt up to the maximum temperature of your apparatus, instead, the compounds decompose and liberate colorless gas. (c) What is the identity of the colorless gas? (d) Write the chemical equation for the decomposition reactions of compound 1 and 2. (e) Are compounds 1 and 2 ionic or molecular?

(a) Does the lattice energy of an ionic solid increase or decrease (i) as the charges of the ions increase, (ii) as the sizes of the ions increase? (b) Arrange the following substances not listed in Table 8.1 according to their expected lattice energies, listing them from lowest lattice energy to the highest: MgS, KI, GaN, LiBr.

The substances \(\mathrm{NaF}\) and \(\mathrm{CaO}\) are isoelectronic (have the same number of valence electrons). (a) What are the charges on each of the cations in each compound? (b) What are the charges of each of the anions in each compound? (c) Without looking up lattice energies, which compound is predicted to have the larger lattice energy? (d) Using the lattice energies in Table 8.1 , predict the lattice energy of ScN.
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