Write Lewis structures that obey the octet rule for the following species. Assign the formal charge for each central atom. a. \(\mathrm{POCl}_{3}\) e. \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) b. \(\mathrm{SO}_{4}^{2-}\) f. \(\mathrm{XeO}_{4}\) c. \(\mathrm{ClO}_{4}^{-}\) g. \(\mathrm{ClO}_{3}^{-}\) d. \(\mathrm{PO}_{4}^{3-}\) h. \(\mathrm{NO}_{4}^{3-}\)

Short Answer

Expert verified
The short version of the answer to the question is: a) \(\mathrm{POCl}_{3}\): Lewis structure: [P](=O)(-Cl)3, Formal charge of \(P\) = 0. b) \(\mathrm{SO}_{4}^{2-}\): Lewis structure: [S](=O)4 with two resonance structures, Formal charge of \(S\) = 0. c) \(\mathrm{ClO}_{4}^{-}\): Lewis structure: [Cl](=O)4, Formal charge of \(Cl\) = 0. d) \(\mathrm{PO}_{4}^{3-}\): Lewis structure: [P](=O)4 with three resonance structures, Formal charge of \(P\) = 0. e) \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\): Lewis structure: [S](=O)2(-Cl)2, Formal charge of \(S\) = 0. f) \(\mathrm{XeO}_{4}\): Lewis structure: [Xe](=O)4, Formal charge of \(Xe\) = 0. g) \(\mathrm{ClO}_{3}^{-}\): Lewis structure: [Cl](=O)3(-O), Formal charge of \(Cl\) = 0. h) \(\mathrm{NO}_{4}^{3-}\): Lewis structure: [N](=O)4, Formal charge of \(N\) = +1.

Step by step solution

01

a) \(\mathrm{POCl}_{3}\)

1. First, count the total number of valence electrons in the molecule: \(P\) has 5 valence electrons, and each \(Cl\) has 7 valence electrons. Thus, there are a total of (5 + 3*7) = 26 valence electrons. 2. Next, place the central atom (\(P\) in this case) and connect it to the surrounding atoms (\(Cl\)) with single bonds. This will use 6 of the valence electrons (2 each bond). Remaining valence electrons = 20. 3. Now, fill the octets of surrounding atoms (\(Cl\)) before filling the central atom. Each \(Cl\) needs 6 more electrons to complete its octet. All the remaining 20 electrons are used for this. 4. Finally, find the formal charge of the central atom \(P\). Formal charge = (valence electrons - non-bonding electrons - 1/2 bonding electrons). Formal charge of \(P\) is (5 - 0 - (1/2)* 6) = 0.
02

b) \(\mathrm{SO}_{4}^{2-}\)

1. Count the total number of valence electrons: \(S\) has 6, each \(O\) has 6, and there are 2 extra from the charge. Total valence electrons = (6 + 4*6 + 2) = 32. 2. Place the central atom (\(S\)) and connect it to the surrounding atoms (\(O\)) with single bonds, using 8 valence electrons. Remaining valence electrons = 24. 3. Fill the octets of surrounding atoms (\(O\)). Each \(O\) needs 6 more electrons to complete its octet. All of the remaining 24 electrons are used for this. 4. Calculate the formal charge of the central atom \(S\). Formal charge = (6 - 0 - (1/2)* 8) = 0.
03

c) \(\mathrm{ClO}_{4}^{-}\)

1. Count the total number of valence electrons: \(Cl\) has 7, and each \(O\) has 6 valence electrons, plus 1 extra from the charge. Thus, the total is (7 + 4*6 + 1) = 32 valence electrons. 2. Place the central atom (\(Cl\)) and connect it to the surrounding atoms (\(O\)) with single bonds, using 8 valence electrons. The remaining valence electrons = 24. 3. Fill the octets of surround... -*-

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which compound in each of the following pairs of ionic substances has the most exothermic lattice energy? Justify your answers. a. \(\mathrm{NaCl}, \mathrm{KCl}\) b. \(\mathrm{LiF}, \mathrm{LiCl}\) c. \(\mathrm{Mg}(\mathrm{OH})_{2}, \mathrm{MgO}\) d. \(\mathrm{Fe}(\mathrm{OH})_{2}, \mathrm{Fe}(\mathrm{OH})_{3}\) e. \(\mathrm{NaCl}, \mathrm{Na}_{2} \mathrm{O}\) f. \(\mathrm{MgO}, \mathrm{BaS}\)

Use the formal charge arguments to rationalize why \(\mathrm{BF}_{3}\) would not follow the octet rule.

Given the following information: Heat of sublimation of \(\operatorname{Li}(s)=166 \mathrm{~kJ} / \mathrm{mol}\) Bond energy of \(\mathrm{HCl}=427 \mathrm{~kJ} / \mathrm{mol}\) Ionization energy of \(\operatorname{Li}(g)=520 . \mathrm{kJ} / \mathrm{mol}\) Electron affinity of \(\mathrm{Cl}(g)=-349 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy of \(\operatorname{LiCl}(s)=-829 \mathrm{~kJ} / \mathrm{mol}\) Bond energy of \(\mathrm{H}_{2}=432 \mathrm{~kJ} / \mathrm{mol}\) Calculate the net change in energy for the following reaction: $$ 2 \mathrm{Li}(s)+2 \mathrm{HCl}(g) \longrightarrow 2 \mathrm{LiCl}(s)+\mathrm{H}_{2}(g) $$

Use Coulomb's law, $$ V=\frac{Q_{1} Q_{2}}{4 \pi \varepsilon_{0} r}=2.31 \times 10^{-19} \mathrm{~J} \cdot \mathrm{nm}\left(\frac{Q_{1} Q_{2}}{r}\right) $$ to calculate the energy of interaction for the following two arrangements of charges, each having a magnitude equal to the electron charge.

Hydrogen has an electronegativity value between boron and carbon and identical to phosphorus. With this in mind, rank the following bonds in order of decreasing polarity: \(\mathrm{P}-\mathrm{H}, \mathrm{O}-\mathrm{H}\), \(\mathrm{N}-\mathrm{H}, \mathrm{F}-\mathrm{H}, \mathrm{C}-\mathrm{H}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free