Ch01 testbank 문제은행 PDF

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testbank exam 시험 문제은행 중간고사 범위 테스트뱅크 챕터1...

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P1: PBU/OVY JWCL338-01

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JWCL338-Solomons-v1

1

THE BASICS: BONDING AND MOLECULAR STRUCTURE

SOLUTIONS TO PROBLEMS

Another Approach to Writing Lewis Structures When we write Lewis structures using this method, we assemble the molecule or ion from the constituent atoms showing only the valence electrons (i.e., the electrons of the outermost shell). By having the atoms share electrons, we try to give each atom the electronic structure of a noble gas. For example, we give hydrogen atoms two electrons because this gives them the structure of helium. We give carbon, nitrogen, oxygen, and fluorine atoms eight electrons because this gives them the electronic structure of neon. The number of valence electrons of an atom can be obtained from the periodic table because it is equal to the group number of the atom. Carbon, for example, is in Group IVA and has four valence electrons; fluorine, in Group VIIA, has seven; hydrogen, in Group 1A, has one. As an illustration, let us write the Lewis structure for CH3 F. In the example below, we will at first show a hydrogen’s electron as x, carbon’s electrons as o’s, and fluorine’s electrons as dots.

Example A 3 H , C , and F are assembled as H H H C F or H C F H H If the structure is an ion, we add or subtract electrons to give it the proper charge. As an example, consider the chlorate ion, ClO3 − .

Example B Cl , and O and an extra electron × are assembled as −

−

Ο Ο Cl Ο

or

Ο Ο Cl Ο

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THE BASICS: BONDING AND MOLECULAR STRUCTURE

1.1

14

N, 7 protons and 7 neutrons; 15 N, 7 protons and 8 neutrons

1.2 (a) one

(b) seven

1.3 (a) covalent

(c) four

(b) ionic

(d) three

(c) covalent

(e) eight

(f ) five

(d) covalent

1.4

−

O

••

O

O −

P O

••

••

2

••

JWCL338-01

−

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O

••

(d) H

F ••

O ••

N

O

(g) H

•

••

••

••

•

1.5 (a) H

••

••

••

(b) •• F ••

••

F ••

(e) H

••

O

O

••

F ••

C

••

H

(f )

H

B

H

(h) H

• •

O ••

H

••

O

••

H

C

O ••

••

H

O ••

–

H

H (c) H

S

O ••

••

O

••

P

O ••

H

H

O H 1.6 (a) H

O −

C

(c) − C

C N

(e) H

O

(f ) H

C

O−

H O (b) H

N H

−

(d) H

C O

−

C

−

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JWCL338-Solomons-v1

THE BASICS: BONDING AND MOLECULAR STRUCTURE

H 1.7 (a) H

C

O

(d) H

C+

+

(e) H

H

O C H

O

H

H

C

N

H

H

H

O

(f ) H

C

−

C

C

N

H

+

H +

(h) H

H

C

+

N

N

H

H H

H O−

O 1.8 (a) H

(g) H

H

H

(c)

H H

C

H

H

(b) H

O−

H

H

C

C O

O −

(b) and (c). Since the two resonance structures are equivalent, each should make an equa contribution to the overall hybrid. The C—O bonds should therefore be of equal length (the should be of bond order 1.5), and each oxygen atom should bear a 0.5 negative charge.

Ο

O 1.9 (a)

H

C

H

H

O (b)

H

−

C

C

C

+

−

H O

H C

H

−

C

H

H

H H (c) H

+

C

N

H H

H

C

H

N

H

H H (d) − C

+

H

H C

−

C

N H

C

N

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4

THE BASICS: BONDING AND MOLECULAR STRUCTURE

+

1.10 (a) CH3CH

CH

CH

+

OH

CH3CH

CH

CH

OH

CH

CH

OH

CH

CH

CH

+

CH3CH δ+ CH3CH (b) CH2

CH

CH +

CH

δ+ CH

CH

δ+ OH +

CH2

CH2

CH2 +

CH2 δ+

CH

δ+ CH

CH2

CH

CH

CH2

δ+ CH

CH

CH2

+

(c) +

+

δ+ δ+ (d) CH2

CH

δ+ −

Br δ− CH2

Br

CH

+

δ+ Br

CH

CH2+

CH2

+

CH2+

CH2

CH

CH2

(e) +

δ+ CH2

δ+ δ+

O (f )

O

−

C

C

−

H2C

δ+

H2C

CH3

CH3

− Oδ

δ− H2C (g) CH3

S

C CH3 +

CH2+ CH3

CH3 δ+ δ+ S CH2

S

CH2

+

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JWCL338-Solomons-v1

THE BASICS: BONDING AND MOLECULAR STRUCTURE

(h) CH3

O −

O

+

CH3 O

O −

+

N

2+

CH3

N

−

O O

N

(minor)

O −

δ−

+

CH3

N −

Oδ 1.11 (a) CH2 N(CH3)2 because all atoms have a complete octet (rule 3), and there are mor covalent bonds (rule 1). +

••

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(b) CH3

••

O

because it has no charge separation (rule 2).

C O

(c)

NH2

C

H N because it has no charge separation (rule 2).

1.12 (a) In its ground state, the valence electrons of carbon might be disposed as shown in th following figure. The electronic configuration of a ground state carbon atom The p orbitals are designated 2 px , 2 py , and 2 pz to indicat their respective orientations along the x , y, and z axes. Th 2s 2py assignment of the unpaired electrons to the 2 py and 2 px or bitals is arbitrary. They could also have been placed in the 2 p C 2p and 2 pz or 2 py and 2 pz orbitals. (To have placed them bot x 2pz in the same orbital would not have been correct, however, fo this would have violated Hund’s rule.) (Section 1.10A) The formation of the covalent bonds of methane from ind vidual atoms requires that the carbon atom overlap its orbitals containing single electro with 1s orbitals of hydrogen atoms (which also contain a single electron). If a groun state carbon atom were to combine with hydrogen atoms in this way, the result woul be that depicted below. Only two carbon-hydrogen bonds would be formed, and the would be at right angles to each other. The hypothetical formation of CH2 from a carbon atom in its ground state:

C + 2 H

H

C H

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6

THE BASICS: BONDING AND MOLECULAR STRUCTURE

(b) An excited-state carbon atom might combine with four hydrogen atoms as shown in th figure above. The promotion of an electron from the 2s orbital to the 2 pz orbital requires energy The amount of energy required has been determined and is equal to 400 kJ mol−1 . Thi expenditure of energy can be rationalized by arguing that the energy released when tw additional covalent bonds form would more than compensate for that required to excit the electron. No doubt this is true, but it solves only one problem. The problems tha cannot be solved by using an excited-state carbon as a basis for a model of methan are the problems of the carbon-hydrogen bond angles and the apparent equivalence o all four carbon-hydrogen bonds. Three of the hydrogens—those overlapping their 1 orbitals with the three p orbitals—would, in this model, be at angles of 90◦ with respec to each other; the fourth hydrogen, the one overlapping its 1s orbital with the 2s orbita of carbon, would be at some other angle, probably as far from the other bonds as th confines of the molecule would allow. Basing our model of methane on this excited state of carbon gives us a carbon that is tetravalent but one that is not tetrahedral, a it predicts a structure for methane in which one carbon-hydrogen bond differs from th other three. The hypothetical formation of CH4 from an excited-state carbon atom:

C + 4 H H H C H H

1.13 (a) Cis-trans isomers are not possible.

CH3

(b) CH3 C

and

C

H

H

CH3 C

C CH3

H

H

(c) Cis-trans isomers are not possible.

Cl

(d) CH3CH2 C H

and

C H

H

CH3CH2 C H

C Cl

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JWCL338-Solomons-v1

THE BASICS: BONDING AND MOLECULAR STRUCTURE

1.14 sp3 1.15 sp3 1.16 sp2 1.17 sp 1.18 (a)

−

H H

H (b)

There are four bonding pairs. The geometry is tetrahedral.

B

H

Be

F

(c)

There are two bonding pairs about the central atom. The geometry is linear.

F

+

H

There are four bonding pairs. The geometry is tetrahedral.

N

H

H H

(d) There are two bonding pairs and two nonbonding pairs. The geometry is angular.

S

H

H H

(e)

There are three bonding pairs. The geometry is trigonal planar.

B H

H F

(f)

There are four bonding pairs around the central atom. The geometry is tetrahedral.

C

F

F

F

F

(g)

Si

F

F

F

There are four bonding pairs around the central atom. The geometry is tetrahedral.

−

(h)

Cl

C Cl

Cl

There are three bonding pairs and one nonbonding pair around the central atom. The geometry is trigonal pyramida

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8

THE BASICS: BONDING AND MOLECULAR STRUCTURE

120° F C C 120° 1.19 (a) F F F

180° (b) CH3 C C H 1.20 H

O

C H

trigonal planar at each carbon atom 180° (c) H C N

CH3 linear

H

H

C

C

H

H

linear

H

CH3 1.21 CH3CHCHCHCH3

CH3

(CH3)2CHCH(CH3)CH(CH3)2

or

CH3 CH3

1.22 (a)

CH

=

CH3

CH3

CH2 CH3

(b)

CH CH3

(c)

CH2

ΟΗ

OH

H C

C

CH3

CH3

(d) CH 3

=

CH2

CH3

=

CH2 CH2

CH2 CH2

CH2

=

CH3

=

CH2 CH

(e) CH3

CH3

ΟΗ OH

(f ) CH2

CH2

CH3

CH2

CH3

C

=

O (g) CH3

(h)

Ο

C

CH2

CH3

=

CH2

CH2 Cl

CH3

CH

CH

=

Cl

CH3

CH2

CH3

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JWCL338-Solomons-v1

THE BASICS: BONDING AND MOLECULAR STRUCTURE

1.23 (a) and (d) are constitutional isomers with the molecular formula C5 H12 . (b) and (e) are constitutional isomers with the molecular formula C5 H12 O. (c) and (f) are constitutional isomers with the molecular formula C6 H12 .

H

1.24 (a) H

H

Cl

H

H

O

H

C

C

C

C

C

C

H

H

H

H

H H

C

(c) H

H

H

O

H H (b) H H

C

C

C

C

C

C C

H

H

H H H

C C

H H

H

H

C

H H C H

H H

Cl 1.25 (a)

C H

H

(Note that the Cl atom and the three H atoms may be written at any of the four positions.)

H Cl

Cl (b)

or

C Cl

H

H

H

and so on

C

H Cl H

Cl (c)

and others

C

H

Br H

(d)

C

H

Cl C

H H

H

C

and others

H H

H

H

C

C

C C

H H

H

H

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10

THE BASICS: BONDING AND MOLECULAR STRUCTURE

Problems Electron Configuration 1.26 (a) no

(b) yes

(c) yes

(d) no

(e) yes

(f ) yes

(g) yes

(h) yes

Lewis Structures

O P 1.27 (a)

O

C

O

(b)

H

O

S

O

H (c)

Cl

Cl Cl

O (d)

H

C

N

O

1.28

CH3

O

CH3

CH 3

O

H 3C

H

Carbon FC = 0 Oxygen FC = 0

Carbon FC = 0 Oxygen FC = 0

Methyl Carbon Carbonyl Carbo Oxygen FC = 0 O

O

H 3C

OH

Methyl Carbon FC = 0 Carbonyl Carbon FC = 0 Carbonyl Oxygen FC = 0 Hydroxyl Oxygen FC = 0

H 3C

OCH3

Methyl Carbon FC = 0 Carbonyl Carbon FC = Carbonyl Oxygen FC = Alkoxy Oxygen FC = 0...