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Chapter 4 Covalent Compounds - websites.rcc.edu

Chapter 4 1 Chapter 4 Covalent Compounds Solutions to In- Chapter Problems H is surrounded by two electrons, giving it the noble gas configuration of He. Cl is surrounded by eight electrons, giving it the noble gas configuration of Ar. H ClH Cl+ This arrangement gives each Cl an octet and the electronic configuration of Ar. Cl Cl Cl Cl+ Atoms with one, two, or three valence electrons form one, two, or three bonds, respectively. Atoms with four or more valence electrons form enough bonds to give an octet. a. F forms one bond. (8 7 valence e = 1 bond) c. Br forms one bond. (8 7 valence e = 1 bond) e. P forms three bonds. (8 5 valence e = 3 bonds) b.

Chapter 4–1 Chapter 4 Covalent Compounds Solutions to In-Chapter Problems 4.1 H is surrounded by two electrons, giving it the noble gas configuration of He. Cl is surrounded by eight electrons, giving it the noble gas configuration of Ar.

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Transcription of Chapter 4 Covalent Compounds - websites.rcc.edu

1 Chapter 4 1 Chapter 4 Covalent Compounds Solutions to In- Chapter Problems H is surrounded by two electrons, giving it the noble gas configuration of He. Cl is surrounded by eight electrons, giving it the noble gas configuration of Ar. H ClH Cl+ This arrangement gives each Cl an octet and the electronic configuration of Ar. Cl Cl Cl Cl+ Atoms with one, two, or three valence electrons form one, two, or three bonds, respectively. Atoms with four or more valence electrons form enough bonds to give an octet. a. F forms one bond. (8 7 valence e = 1 bond) c. Br forms one bond. (8 7 valence e = 1 bond) e. P forms three bonds. (8 5 valence e = 3 bonds) b.

2 Si forms four bonds. (8 4 valence e = 4 bonds) d. O forms two bonds. (8 6 valence e = 2 bonds) f. S forms two bonds. (8 6 valence e = 2 bonds) Ionic bonding is observed in CaO since Ca is a metal and readily transfers electrons to the nonmetal oxygen. Covalent bonding is observed in CO2 since carbon is a nonmetal and does not readily transfer electrons. Arrange the the the bonds and lone bonds 2 e used1 Br x 7 e = 7 e 1 H x 1 e = 1 e 8 e lone [1]Step [2]Step [3]8 e usedBrHBrHH Bra. Arrange the the the bonds and lone bonds 8 e usedStep [1]Step [2]Step [3]14 e CHHF four atoms around CH CHHFH CHHFno octet1 C x 4 e = 4 e 1 F x 7 e = 7 e 3 H x 1 e = 3 e 14 e lone pairs.

3 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or Compounds 4 2 Use the same steps as in Answer to draw the Lewis structure. HCOHHCHHH After placing all electrons in bonds and lone pairs, use a lone pair to form a multiple bond if an atom does not have an octet. Follow the stepwise procedure in Example A valid Lewis structure for formic acid contains a carbon oxygen double bond. The C atom has four bonds and each O atom has two bonds and two lone pairs.

4 HCOHO H NHCHHCOOHb. Lewis structure:a. molecular formula: C2H5NO2 Boron is surrounded by only six electrons, so the Lewis structure for BBr3 does not follow the octet rule. Br BBrBrsix electrons O O atom needs two lone pairs and the N atom needs one. b. There are 10 electrons around phosphorus. c. Phosphorus and hydrogen do not follow the octet rule. Resonance structures are two or more Lewis structures having the same arrangement of atoms but a different arrangement of electrons. Draw the second resonance structure as in Sample Problem in the text, or Example The two resonance structures in each example differ in the location of a double bond and a lone pair.

5 HCONHCCOOHHH 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. Chapter 4 3 ONO OCOH OCOH b. O SOO SOThe location of a lone pair and double bond is different. N NON NON NOThree resonance structures for nitrous oxide: Name the Compounds using the following two-step method: Step [1]: Name the first nonmetal by its element name and the second using the suffix -ide. Step [2]: Add prefixes to show the number of atoms of each element.

6 A. CS2 = carbon disulfide c. PCl5 = phosphorus pentachloride b. SO2 = sulfur dioxide d. BF3 = boron trifluoride a. silicon dioxide = SiO2 c. sulfur trioxide = SO3 b. phosphorus trichloride = PCl3 d. dinitrogen trioxide = N2O3 The shape around each carbon atom in ethylene, H2C=CH2, is trigonal planar, since each C is surrounded by three groups (2 H s and 1 C). has two lone has four has one lone pyramidalB has three planar H NH The ion has a bent geometry (b), because the N is surrounded by four groups. Electronegativity increases across a row of the periodic table because the nuclear charge increases (excluding the noble gases).

7 Electronegativity decreases down a column of the periodic table because the atomic radius increases, pushing the valence electrons farther from the nucleus. a. Na, Li, H b. Be, C, O c. I, Cl, F d. B, N, O 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or Compounds 4 4 Calculate the electronegativity difference between the two atoms and use the following rules: less than (nonpolar); (polar Covalent ); greater than (ionic).

8 A. HF = ( ) = = polar d. ClF = ( ) = = polar b. MgO = ( ) = = ionic e. H2O = ( ) = = polar c. F2 = ( ) = 0 = nonpolar f. NH3 = ( ) = = polar The head of the bond dipole arrow points towards the more electronegative atom. !+!+!+!+!"!"!"!" To determine the overall polarity of a molecule, identify the polar bonds, determine the shape around individual atoms, and then decide if the individual bond dipoles cancel or reinforce. H ClH CHHCHHHH CN!+ !+polarc.!"!+polard.!" (The four C Cl bond dipoles cancel.)HCHFFClCClClCl!"!"!"!"!"!"nonpol ar(all nonpolar bonds)!+ All C H and C C bonds are nonpolar. H CCOHHHHHH CCHHOH!

9 "!+!+!"polar C O, O HBoth C's are tetrahedral(four atoms around each C). !+tetrahedral(four atoms)trigonal planar(three groups) Solutions to End-of- Chapter Problems In Covalent bonding, atoms share electrons to attain the electronic configuration of the noble gas closest to them in the periodic table. In ionic bonding, one atom donates electrons to the other atom. a. LiCl: ionic; the metal Li donates electrons to the nonmetal chlorine. HCl: Covalent ; H and Cl share electrons since both atoms are nonmetals and the electronegativity difference is not large enough for electron transfer to occur. b. KBr: ionic; the metal K donates electrons to the nonmetal bromine.

10 HBr: Covalent ; H and Br share electrons since both atoms are nonmetals and the electronegativity difference is not large enough for electron transfer to occur. 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. Chapter 4 5 In Covalent bonding, atoms share electrons to attain the electronic configuration of the noble gas closest to them in the periodic table. In ionic bonding, one atom donates electrons to the other atom.


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