PREVIEW Covalent Bonding and Shapes of Molecules 1

PREVIEW Covalent Bonding and Shapes of Molecules1

CHAPTER 1

Solutions to the Problems Problem 1.1 Write and compare the ground-state electron configurations for each pair of elements: (a) Carbon and silicon C (6 electrons) 1s 22s 22p 2 Si (14 electrons) 1s 2 2s 2 2p 6 3s 2 3p 2 Both carbon and silicon have four electrons in their outermost (valence) shells.(b) Oxygen and sulfur O (8 electrons) 1s 2 2s 2 2p 4 S (16 electrons) 1s 2 2s 2 2p 6 3s 2 3p 4 Both oxygen and sulfur have six electrons in their outermost (valence) shells.(c) Nitrogen and phosphorus N (7 electrons) 1s 2 2s 2 2p 3 P (15 electrons) 1s 2 2s 2 2p 6 3s 2 3p 3 Both nitrogen and phosphorus have five electrons in their outermost (valence) shells.Problem 1.2 Show how each chemical change leads to a stable octet.(a) Sulfur forms S 2- .(b) Magnesium forms Mg 2+ .S 2-

(18 electrons): 1s

2 2s 2 2p 6 3s 2 3p 6

S (16 electrons): 1s

2 2s 2 2p 6 3s 2 3p 4

Mg (12 electrons): 1s

2 2s 2 2p 6 3s 2 Mg 2+

(10 electrons): 1s

2 2s 2 2p 6 Problem 1.3 Judging from their relative positions in the Periodic Table, which element in each set is more electronegative?(a) Lithium or potassium In general, electronegativity increases from left to right across a row and from bottom to top of a column in the Periodic Table. This is because electronegativity increases with increasing positive charge on the nucleus and with decreasing distance of the valence electrons from the nucleus. Lithium is closer to the top of the Periodic Table and thus more electronegative than potassium.(b) Nitrogen or phosphorus Nitrogen is closer to the top of the Periodic Table and thus more electronegative than phosphorus.(c) Carbon or silicon Carbon is closer to the top of the Periodic Table and thus more electronegative than silicon.Problem 1.4 Classify each bond as nonpolar covalent, or polar covalent, or state that ions are formed.(a) S-H (b) P-H (c) C-F (d) C-Cl Recall that bonds formed from atoms with an electronegativity difference of less than 0.5 are considered nonpolar covalent and with an electronegativity difference of 0.5 or above are considered a polar covalent bond.S-H2.5 - 2.1 = 0.4Nonpolar covalent P-H2.1 - 2.1 = 0Nonpolar covalent C-F4.0 - 2.5 = 1.5Polar covalent C-Cl3.0 - 2.5 = 0.5Polar covalent Problem 1.5 Using the symbols δ- and δ+, indicate the direction of polarity in each polar covalent bond.(a) C-N(b) N-O .

δ+δ-

C-N

δ-δ+

.N-O Nitrogen is more electronegative than carbon Oxygen is more electronegative than nitrogen Bond Difference in electronegativity Type of bond 1 / 4

PREVIEW 2Chapter 1 (c) C-Cl .

δ+δ-

C-Cl Chlorine is more electronegative than carbon Problem 1.6 Draw Lewis structures, showing all valence electrons, for these molecules.(a) C2H6 (b) CS 2 (c) HCN

HC HC HH HH

SCSCH N

Problem 1.7 Draw Lewis structures for these ions, and show which atom in each bears the formal charge.(a) CH3NH3

• (b) CO 3

2- (c) HO -

Methylammonium ion Carbonate ion Hydroxide ion

H OCO O

OHHCNH

H H H + Problem 1.8 Draw Lewis structures and condensed structural formulas for the four alcohols with molecular formula C4H10O.Classify each alcohol as primary, secondary, or tertiary.

CCH

• CH

3 OH CH 3 CHCH 2 OHCH 3 CH 3 CH H H C H H C H H C H H OH C C C H H H H H H H C H H OH CH HH C H H C H O H C H H H C C C O H H H H H H C H H H H CH 3CH 2CH 2CH 2OH CHCH3 OH CH 3CH 2 PrimaryPrimarySecondaryTertiary Problem 1.9 Draw structural formulas for the three secondary amines with molecular formula C 4 H 11 N.N H C C H H C H H H H H C H H H CHN C H H C H H HH C H H H H CHN C H H H H C H H H CH H H Problem 1.10 Draw condensed structural formulas for the three ketones with molecular formula C5H10O.

CH 3CH 2CH 2 O CCH 3 CCH 3 O CH CH 3 CH 3CH 3CH 2 O CCH 2CH

3 2 / 4

PREVIEW Covalent Bonding and Shapes of Molecules3 Problem 1.11 Draw condensed structural formulas for the two carboxylic acids with molecular formula C

4H8O2.

CH 3CH 2CH 2CO 2H CHCH 3 CH 3 CO 2H Problem 1.12 Draw structural formulas for the four esters with molecular formula C4H8O2.C H H H C H H C O OC H H HHC H H C O OC H H C H H H HC O OC H H C H H C H H H HC O OC C H C H H H HH H Problem 1.13 Predict all bond angles for these molecules.(a) CH3OH CO H H H H 109.5 o 109.5 o (b) PF 3 P F F F 109.5 o

••

(c) H2CO3 (Carbonic Acid) O C O O HH 109.5 o 109.5 o 120 o 120 o 120 o

MCAT Practice: Questions

Fullerenes

• The geometry of carbon in diamond is tetrahedral, while carbon’s geometry in graphite is trigonal planar. What is the

geometry of the carbons in C 60 ?

• They are all tetrahedral.
• They are all trigonal planar.
• They are all pyramidal with bond angles near 109.5°.
• They are not perfectly trigonal planar but have an extent of pyramidalization. The curve of the

buckyball surface is curved requiring some extent of pyramidilization. 3 / 4

PREVIEW 4Chapter 1

• Because of their spherical shape, C

60 moleclues are used as nanoscale ball bearings in grease and lubricants. We can estimate the size of these ball bearings by examining C-C bond distances. Carbon-carbon bond distances vary between approximately 120pm (pm = picometers) and 155pm. Roughly, what is the diameter of C 60?

• 10 pm
• 100 pm
• 1,000 pm C

60 is approximately 8 bonds across so approximately 1,000 pm in diameter.

• 10,000 pm
• What best describes the C-C-C bond angles in C

60 ?

• They are exactly 120

o .

• They are a bit larger than 120

o .

• They are a bit smaller than 120

o . The five-membered rings in the C 60 structure reduce bond angles.

• They are near 109.5°.

Problem 1.14 Which molecules are polar? For each molecule that is polar, specify the direction of its dipole moment.(a) CH2Cl2 A molecular dipole moment is determined as the vector sum of the bond dipoles in three-dimensional space. Thus, by superimposing the bond dipoles on a three-dimensional drawing, the molecular dipole moment can be determined.Note that on the following diagrams, the dipole moments from the C-H bonds are ignored because they are small.C Cl Cl H H

µ = 1.60 D

(b) HCN CNH

µ = 2.98 D

(c) H2O2 The H2O2 molecule can rotate around the O-O single bond, so we must consider the molecular dipole moments in the various possible conformations. Conformations such as the one on the left have a net molecular dipole moment, while conformations such as the one the right below do not. The presence of at least some conformations (such as that on the left) that have a molecular dipole moment means that the entire molecule must have an overall dipole moment, in this case

µ = 2.2 D.H

O O HOO HH µ = 2.2 D

• / 4