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Chapter 8: Problem 85

Write Lewis structures that obey the octet rule (duet rule for H) for each of the following molecules. Carbon is the central atom in \(\mathrm{CH}_{4}\) , nitrogen is the central atom in \(\mathrm{NH}_{3},\) and oxygen is the central atom in \(\mathrm{H}_{2} \mathrm{O}\) . a. \(\mathrm{F}_{2} \quad\) e. \(\mathrm{NH}_{3}\) b. \(\mathrm{O}_{2} \quad\) f. \(\mathrm{H}_{2} \mathrm{O}\) c. \(\mathrm{CO} \quad\) g. \(\mathrm{HF}\) d. \(\mathrm{CH}_{4}\)

Short Answer

Expert verified
The short version of the Lewis structures for each molecule: a. F2: F-F with 3 lone pairs on each F. b. O2: O=O with 2 lone pairs on each O. c. CO: C≡O with a lone pair on O. d. CH4: C(H)(H)(H)(H) e. NH3: N(H)(H)(H) with 1 lone pair on N. f. H2O: H-O-H with 2 lone pairs on O. g. HF: H-F with 3 lone pairs on F.

Step by step solution

01

Count Valence Electrons

For each of the given molecules, determine the total number of valence electrons. Valence electrons are the electrons in the outermost shell of each atom. a. F2: Fluorine has 7 valence electrons, so F2 has 14 electrons. b. O2: Oxygen has 6 valence electrons, so O2 has 12 electrons. c. CO: Carbon has 4 valence electrons and Oxygen has 6, so CO has 10 electrons. d. CH4: Carbon has 4 valence electrons and Hydrogen has 1, so CH4 has 8 electrons. e. NH3: Nitrogen has 5 valence electrons and Hydrogen has 1, so NH3 has 8 electrons. f. H2O: Oxygen has 6 valence electrons and Hydrogen has 1, so H2O has 8 electrons. g. HF: Hydrogen has 1 valence electron and Fluorine has 7, so HF has 8 electrons.
02

Distribute Electrons

Now, for each molecule, connect the atoms to the central atom using single bonds, and distribute the remaining valence electrons as lone pairs around terminal atoms to obey the octet rule (duet rule for Hydrogen). a. F2: Connect F-F with a single bond. b. O2: Connect O=O with a double bond. c. CO: Connect C≡O with a triple bond. d. CH4: Connect C-H with single bonds. e. NH3: Connect N-H with single bonds. f. H2O: Connect O-H with single bonds. g. HF: Connect H-F with a single bond.
03

Assign Lone Pairs

Finally, assign the remaining valence electrons as lone pairs around the terminal atoms to complete the octet (duet for Hydrogen). a. F2: F-F with 3 lone pairs on each F. b. O2: O=O with 2 lone pairs on each O. c. CO: C≡O with a lone pair on O. d. CH4: No lone pairs needed. e. NH3: 1 lone pair on N and no lone pairs needed for H atoms. f. H2O: 2 lone pairs on O and no lone pairs needed for H atoms. g. HF: No lone pairs needed for H and 3 lone pairs needed for F. The Lewis structures for each respective molecule are as follows: a. F2: F:F b. O2: O::O c. CO: C:::O d. CH4: H | C-H | H e. NH3: H | N-H | H f. H2O: H-O-H | g. HF: H-F

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Most popular questions from this chapter

One type of exception to the octet rule are compounds with central atoms having fewer than eight electrons around them. BeH_ and \(\mathrm{BH}_{3}\) are examples of this type of exception. Draw the Lewis structures for \(\mathrm{BeH}_{2}\) and \(\mathrm{BH}_{3} .\)

Which of the following incorrectly shows the bond polarity? Show the correct bond polarity for those that are incorrect. a. \(^{\delta+} \mathrm{H}-\mathrm{F}^{\delta-} \quad\) d. $\delta^{+} \mathrm{Br}-\mathrm{Br}^{\delta-}$ b. \(^{\delta+} \mathrm{Cl}-\mathrm{I}^{\delta-} \qquad\) e. \(\quad\) e. $\quad ^{\delta+}\mathrm{O}-\mathrm{P}^{\delta-}$ c. \(\quad \delta+\mathrm{Si}-\mathrm{S}^{\delta-}\)

Without using Fig. 8.3, predict which bond in each of the following groups will be the most polar. a. $\mathrm{C}-\mathrm{F}, \mathrm{Si}-\mathrm{F}, \mathrm{Ge}-\mathrm{F} \quad\( c. \)\mathrm{S}-\mathrm{F}, \mathrm{S}-\mathrm{Cl}, \mathrm{S}-\mathrm{Br}$ b. \(\mathrm{P}-\mathrm{Cl}\) or \(\mathrm{S}-\mathrm{Cl} \quad\) d. \(\mathrm{Ti}-\mathrm{Cl}, \mathrm{Si}-\mathrm{Cl}, \mathrm{Ge}-\mathrm{Cl}\)

An alternative definition of electronegativity is $$\text { Electronegativity } = \text { constant } (\mathrm{I.E.}-\mathrm{E.A.})$$ where I.E. is the ionization energy and E.A. is the electron affinity using the sign conventions of this book. Use data in Chapter 7 to calculate the \((\mathrm{I} . \mathrm{E} .-\mathrm{E} \cdot \mathrm{A} .)\) term for \(\mathrm{F}, \mathrm{Cl}, \mathrm{Br}\) and \(\mathrm{I}\). Do these values show the same trend as the electronegativity values given in this chapter? The first ionization energies of the halogens are 1678, 1255, 1138, and 1007 kJ/mol, respectively. (Hint: Choose a constant so that the electronegativity of fluorine equals 4.0. Using this constant, calculate relative electronegativities for the other halogens and compare to values given in the text.)

Use the following data to estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for magnesium fluoride. $$\mathrm{Mg}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{MgF}_{2}(s)$$ $\begin{array}{l}{\text { Lattice energy }} & {-22913 . \mathrm{kJ} / \mathrm{mol}} \\ {\text { First ionization energy of } \mathrm{Mg}} & \quad{735 \mathrm{kJ} / \mathrm{mol}} \\ {\text {Second ionization energy of } \mathrm{Mg}} & \quad {1445 \mathrm{kJ} / \mathrm{mol}}\\\\{\text { Electron affinity of } \mathrm{F}} & {-328 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Bond energy of } \mathrm{F}_{2}} & \quad {154 \mathrm{kJ} / \mathrm{mol}} \\\ {\text { Enthalpy of sublimation for } \mathrm{Mg}} & \quad {150 . \mathrm{kJ} / \mathrm{mol}} \end{array}$
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