4.E: Exercise
- Page ID
- 83073
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Additional Exercises for Chapter 4:
1. For each of the following molecules, draw the Lewis structure (using lines for shared electron pairs and dots for unshared electrons), then determine the geometry and bond angles. (A) PH3; (B) CCl4; (C) H2Se; (D) OCS; (E) SO3
2. For each of the following molecules or polyatomic ions, draw the Lewis structure (using lines for shared electron pairs and dots for unshared electrons), then determine the geometry and bond angles. (A) NH4(+); (B) ClO2(-); (C) HCO2(-); (D) CO3(2-)
3. Cyanogen has the formula C2N2, is connected with the two carbons in the middle, NCCN. Draw the Lewis structure for cyanogen, using lines for shared electron pairs and dots for unshared electrons. What are the bond angles around each carbon in this molecule?
4. Hydrogen peroxide, H2O2, is connected with the two oxygens in the middle, HOOH. Draw the Lewis structure for cyanogen, using lines for shared electron pairs and dots for unshared electrons. What are the bond angles around each carbon in this molecule?
5. Each of the following compounds contains both ionic and covalent bonds. Draw the Lewis structures, using lines for shared electron pairs and dots for unshared electrons. (A) KOH; (B) CsCN
6. Name these compounds. (A) NH3; (B) CCl4; (C) P5O7; (D) KOH; (E) CsCN
7. Write the formula for these compounds. (A) trinitrogen hexafluoride; (B) carbon disulfide; (C) barium chloride
8. Rank the following bonds from longest to shortest: N≡N, F-F, O=O, Br-Br.
9. Using Figure 4.4.2, classify the bonds in each molecule as non-polar, polar, or ionic. (A) NH3; (B) CH4; (C) CaCl2; (C) CCl4
10. Explain why neither CH4 nor CCl4 molecules are polar.
11. Explain why the bond lenght of Br2 was longer than F2 in #8.
12. What is the difference between ionic and covalent bonds? Use LiCl to explain why an ionic bond keeps lithium and chlorine together. Use HCl to explain why a covalent bond keeps hydrogen and chlorine together.
Answers:
6. (A) ammonia (common name); (B) carbon tetrachloride; (C) pentaphosphorous heptoxide; (D) potassium hydroxide; (E) cesium cyanide (D & E are ionic compounds, not covalent molecules.)
7. (A) N3F6; (B) CS2; (C) BaCl2 (Barium chloride ionic and the formula can be predicted based on the charges, Ba(2+) & Cl(-), so no numeric prefixes are used.)
8. (longest) Br-Br > F-F > O=O > N≡N (shortest)
9. (A) NH3 - polar; (B) CH4 - non-polar (C) CaCl2 - ionic; (C) CCl4 - polar
10. CH4 cannot be polar because it does not have polar bonds. While CCl4 has polar bonds, the tetrahedral shape spreads those bonds out evenly (symmetrical), so no one side of the molecule is more negative or positive, and thus the molecule is not polar overall.
11. Atoms further down the periodic table have more shells of electrons stacked up on the outside, thus taking up more space and extending further from the nucleus. Bromine (4 shells of electrons) has larger radius than fluorine (2 shells). The larger bromine atoms cannot get as close when they bond so have longer bond length.
12. In an ionic bond, the atoms completely exchange electrons, creating charged ions. In a covalent bond, the atoms share electrons. In LiCl the Li(+) cation is “bonded” to the Cl(-) ion due to attraction of opposite charges. In HCl the two atoms must stay close to both possess (share) the same pair of electrons.