Basic Chemistry (5th Edition)
5th Edition
ISBN: 9780134138046
Author: Karen C. Timberlake
Publisher: PEARSON
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Chapter 10.3, Problem 10.20QAP
Interpretation Introduction
(a)
Interpretation:
Interpret the shape of CF4 according to VSEPR theory.
Concept Introduction:
The shape of a molecule can be determined with the help of VSEPR model in which Lewis structure of the molecule is drawn which shows the arrangement of surrounding atoms around the central atom.
The shape of molecule is decided by counting the lone pair of electrons along with bond pair and thus the shape of molecule is decided. Shape of molecule can be decided as:
| Bond pair | Lone pair | Geometry of molecule |
| 2 | 0 | Linear |
| 3 | 0 | Trigonal planar |
| 2 | 1 | Bent |
| 4 | 0 | Tetrahedral |
| 3 | 1 | Trigonal pyramidal |
| 2 | 2 | Bent |
| 5 | 0 | Trigonal bipyramidal |
| 4 | 1 | See saw |
| 3 | 2 | T shaped |
| 2 | 3 | Linear |
| 6 | 0 | Octahedral |
| 4 | 2 | Square planar |
Interpretation Introduction
(b)
Interpretation:
Interpret the shape of NCl3 according to VSEPR theory.
Concept Introduction:
The shape of a molecule can be determined with the help of VSEPR model in which Lewis structure of the molecule is drawn which shows the arrangement of surrounding atoms around the central atom.
The shape of molecule is decided by counting the lone pair of electrons along with bond pair and thus the shape of molecule is decided. Shape of molecule can be decided as:
| Bond pair | Lone pair | Geometry of molecule |
| 2 | 0 | Linear |
| 3 | 0 | Trigonal planar |
| 2 | 1 | Bent |
| 4 | 0 | Tetrahedral |
| 3 | 1 | Trigonal pyramidal |
| 2 | 2 | Bent |
| 5 | 0 | Trigonal bipyramidal |
| 4 | 1 | See saw |
| 3 | 2 | T shaped |
| 2 | 3 | Linear |
| 6 | 0 | Octahedral |
| 4 | 2 | Square planar |
Interpretation Introduction
(c)
Interpretation:
Interpret the shape of SeBr2according to VSEPR theory.
Concept Introduction:
The shape of a molecule can be determined with the help of VSEPR model in which Lewis structure of the molecule is drawn which shows the arrangement of surrounding atoms around the central atom.
The shape of molecule is decided by counting the lone pair of electrons along with bond pair and thus the shape of molecule is decided. Shape of molecule can be decided as:
| Bond pair | Lone pair | Geometry of molecule |
| 2 | 0 | Linear |
| 3 | 0 | Trigonal planar |
| 2 | 1 | Bent |
| 4 | 0 | Tetrahedral |
| 3 | 1 | Trigonal pyramidal |
| 2 | 2 | Bent |
| 5 | 0 | Trigonal bipyramidal |
| 4 | 1 | See saw |
| 3 | 2 | T shaped |
| 2 | 3 | Linear |
| 6 | 0 | Octahedral |
| 4 | 2 | Square planar |
Interpretation Introduction
(d)
Interpretation:
Interpret the shape of CS2 according to VSEPR theory.
Concept Introduction:
The shape of a molecule can be determined with the help of VSEPR model in which Lewis structure of the molecule is drawn which shows the arrangement of surrounding atoms around the central atom.
The shape of molecule is decided by counting the lone pair of electrons along with bond pair and thus the shape of molecule is decided. Shape of molecule can be decided as:
| Bond pair | Lone pair | Geometry of molecule |
| 2 | 0 | Linear |
| 3 | 0 | Trigonal planar |
| 2 | 1 | Bent |
| 4 | 0 | Tetrahedral |
| 3 | 1 | Trigonal pyramidal |
| 2 | 2 | Bent |
| 5 | 0 | Trigonal bipyramidal |
| 4 | 1 | See saw |
| 3 | 2 | T shaped |
| 2 | 3 | Linear |
| 6 | 0 | Octahedral |
| 4 | 2 | Square planar |
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Students have asked these similar questions
1.
Part 1: Naming Organic Compounds
он
H₁C-C-CH3
CH3
Br
CI CI
2. Br-CH-CH-CH₂
H₂C-CH-C= -CH-CH2-CH3
3.
HC-CH-CH-C-OH
5. H₂C-CH-CH₂-OH
7.
OH
4.
CH
CH₂-CH₂
6.
сно
CH-CH-CH-CH₂-CH₂
H₁₂C-CH-CH-CH-CH₁₂-CH₁₂
8.
OH
11
Organic Chemistry
Organic Nomenclature Practice
Name/Functional Group
n-butane
Formula
Structural Formula
(1) C4tt10
H3C
C-
(2) CH3CH2CH2 CH 3
H₂
-CH3
Н2
name & functional group
(1) and (2)
OH
H₁₂C
Н2
name only
(1) and (2)
name only
(1) and (2)
H₁C - = - CH₂
Н2
HC=C-C
CH3
Under aqueous basic conditions, nitriles will react to form a neutral organic intermediate 1 that has an N atom in it first, and then they will continue to react to
form the final product 2:
NC
H₂O
он-
H₂O
1
2
OH
Draw the missing intermediate 1 and the final product 2 in the box below. You can draw the two structures in any arrangement you like.
Click and drag to start drawing a
structure.
Chapter 10 Solutions
Basic Chemistry (5th Edition)
Ch. 10.1 - Prob. 10.1QAPCh. 10.1 - Determine the total number of valence electrons...Ch. 10.1 - Prob. 10.3QAPCh. 10.1 - Draw the Lewis structure for each of the following...Ch. 10.1 - Prob. 10.5QAPCh. 10.1 - If the available number of valence electrons for a...Ch. 10.1 - Prob. 10.7QAPCh. 10.1 - Draw the Lewis structure for each of the following...Ch. 10.2 - What is resonance?Ch. 10.2 - Prob. 10.10QAP
Ch. 10.2 - Draw resonance structures for each of the...Ch. 10.2 - Prob. 10.12QAPCh. 10.3 - Prob. 10.13QAPCh. 10.3 - 10.14 Choose the shape (1 to 6) that matches each...Ch. 10.3 - Prob. 10.15QAPCh. 10.3 - Prob. 10.16QAPCh. 10.3 - Prob. 10.17QAPCh. 10.3 - Prob. 10.18QAPCh. 10.3 - Use VSEPR theory to predict the shape of each of...Ch. 10.3 - Prob. 10.20QAPCh. 10.3 - Prob. 10.21QAPCh. 10.3 - Draw the Lewis structure and predict the shape for...Ch. 10.4 - Describe the trend in electronegativity as...Ch. 10.4 - Prob. 10.24QAPCh. 10.4 - Prob. 10.25QAPCh. 10.4 - Prob. 10.26QAPCh. 10.4 - Prob. 10.27QAPCh. 10.4 - Prob. 10.28QAPCh. 10.4 - Prob. 10.29QAPCh. 10.4 - Prob. 10.30QAPCh. 10.4 - Prob. 10.31QAPCh. 10.4 - Prob. 10.32QAPCh. 10.5 - Prob. 10.33QAPCh. 10.5 - Prob. 10.34QAPCh. 10.5 - Prob. 10.35QAPCh. 10.5 - Prob. 10.36QAPCh. 10.5 - Prob. 10.37QAPCh. 10.5 - Prob. 10.38QAPCh. 10.6 - Prob. 10.39QAPCh. 10.6 - Prob. 10.40QAPCh. 10.6 - Prob. 10.41QAPCh. 10.6 - Prob. 10.42QAPCh. 10.6 - Prob. 10.43QAPCh. 10.6 - Prob. 10.44QAPCh. 10.7 - 10.45 Using Figure 10.6, calculate the heat change...Ch. 10.7 - 10.46 Using Figure 10.6, calculate the heat change...Ch. 10.7 - 10.47 Using Figure 10.6. calculate the heat change...Ch. 10.7 - 10.48 Using Figure 10.6. calculate the heat change...Ch. 10.7 - 10.49 Using Figure 10.6 and the specific heat of...Ch. 10.7 - 10.50 Using Figure 10.6 and the specific heal of...Ch. 10.7 - 10.51 An ice bag containing 275 g of ice at 0°C...Ch. 10.7 - Prob. 10.52QAPCh. 10 - Prob. 10.53FUCh. 10 - Prob. 10.54FUCh. 10 - Prob. 10.55FUCh. 10 - Prob. 10.56FUCh. 10 - Prob. 10.57FUCh. 10 - Prob. 10.58FUCh. 10 - Prob. 10.59UTCCh. 10 - Prob. 10.60UTCCh. 10 - Prob. 10.61UTCCh. 10 - Prob. 10.62UTCCh. 10 - Prob. 10.63UTCCh. 10 - Prob. 10.64UTCCh. 10 - Prob. 10.65UTCCh. 10 - Prob. 10.66UTCCh. 10 - 10.67 Use your knowledge of changes of state to...Ch. 10 - Prob. 10.68UTCCh. 10 - Prob. 10.69UTCCh. 10 - Prob. 10.70UTCCh. 10 - Prob. 10.71UTCCh. 10 - Prob. 10.72UTCCh. 10 - Prob. 10.73AQAPCh. 10 - Prob. 10.74AQAPCh. 10 - Prob. 10.75AQAPCh. 10 - Prob. 10.76AQAPCh. 10 - Prob. 10.77AQAPCh. 10 - Prob. 10.78AQAPCh. 10 - Prob. 10.79AQAPCh. 10 - Prob. 10.80AQAPCh. 10 - Prob. 10.81AQAPCh. 10 - Prob. 10.82AQAPCh. 10 - Prob. 10.83AQAPCh. 10 - Prob. 10.84AQAPCh. 10 - Prob. 10.85AQAPCh. 10 - Prob. 10.86AQAPCh. 10 - Prob. 10.87AQAPCh. 10 - Prob. 10.88AQAPCh. 10 - Prob. 10.89AQAPCh. 10 - Prob. 10.90AQAPCh. 10 - Prob. 10.91AQAPCh. 10 - Prob. 10.92AQAPCh. 10 - Prob. 10.93AQAPCh. 10 - Prob. 10.94AQAPCh. 10 - Indicate the major type of intermolecular...Ch. 10 - Prob. 10.96AQAPCh. 10 - Prob. 10.97AQAPCh. 10 - Prob. 10.98AQAPCh. 10 - Prob. 10.99AQAPCh. 10 - Prob. 10.100AQAPCh. 10 - Prob. 10.101AQAPCh. 10 - Prob. 10.102AQAPCh. 10 - Prob. 10.103CQCh. 10 - Prob. 10.104CQCh. 10 - Prob. 10.105CQCh. 10 - Prob. 10.106CQCh. 10 - Prob. 10.107CQCh. 10 - Prob. 10.108CQCh. 10 - Prob. 10.109CQCh. 10 - Prob. 10.110CQCh. 10 - Prob. 10.111CQCh. 10 - Prob. 10.112CQCh. 10 - Prob. 13CICh. 10 - Prob. 14CICh. 10 - Prob. 15CICh. 10 - Prob. 16CI
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