Textbook Question
Look at the location of elements A, B, C, and D in the following periodic table:
Which hydride has the lowest boiling point?
Ch.22 - The Main Group Elements
McMurry8th EditionChemistryISBN: 9781292336145
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Table of contents
- Ch.1 - Chemical Tools: Experimentation & Measurement170
- Ch.2 - Atoms, Molecules & Ions160
- Ch.3 - Mass Relationships in Chemical Reactions169
- Ch.4 - Reactions in Aqueous Solution199
- Ch.5 - Periodicity & Electronic Structure of Atoms102
- Ch.6 - Ionic Compounds: Periodic Trends and Bonding Theory92
- Ch.7 - Covalent Bonding and Electron-Dot Structures61
- Ch.8 - Covalent Compounds: Bonding Theories and Molecular Structure59
- Ch.9 - Thermochemistry: Chemical Energy122
- Ch.10 - Gases: Their Properties & Behavior155
- Ch.11 - Liquids & Phase Changes54
- Ch.12 - Solids and Solid-State Materials73
- Ch.13 - Solutions & Their Properties120
- Ch.14 - Chemical Kinetics98
- Ch.15 - Chemical Equilibrium125
- Ch.16 - Aqueous Equilibria: Acids & Bases110
- Ch.17 - Applications of Aqueous Equilibria147
- Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium86
- Ch.19 - Electrochemistry154
- Ch.20 - Nuclear Chemistry96
- Ch.21 - Transition Elements and Coordination Chemistry156
- Ch.22 - The Main Group Elements187
- Ch.23 - Organic and Biological Chemistry51
Chapter 22, Problem 22.29b
Consider the six second- and third-row elements in groups 4A–6A of the periodic table:
Possible structures for the binary fluorides of each of these elements in its highest oxidation state are shown below.
(b) Explain why the fluorides of nitrogen and phosphorus have different molecular structures but the fluorides of carbon and silicon have the same molecular structure.
Verified step by step guidance1
Step 1: Identify the elements in groups 4A–6A of the periodic table. These include carbon (C), silicon (Si), nitrogen (N), and phosphorus (P).
Step 2: Understand the highest oxidation states for these elements when they form binary fluorides. For nitrogen and phosphorus, the highest oxidation states are +5, forming NF5 and PF5. For carbon and silicon, the highest oxidation states are +4, forming CF4 and SiF4.
Step 3: Consider the molecular geometry of these compounds. NF5 does not exist due to nitrogen's inability to expand its octet, while PF5 has a trigonal bipyramidal structure. CF4 and SiF4 both have a tetrahedral structure.
Step 4: Explain the difference in molecular structures for nitrogen and phosphorus fluorides. Nitrogen, being in the second period, cannot expand its octet, limiting it to forming NF3 with a trigonal pyramidal shape. Phosphorus, in the third period, can expand its octet, allowing PF5 to form with a trigonal bipyramidal shape.
Step 5: Explain the similarity in molecular structures for carbon and silicon fluorides. Both carbon and silicon form tetrahedral structures (CF4 and SiF4) because they both achieve a stable configuration by sharing four pairs of electrons with fluorine atoms, and neither can expand their octet.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It is determined by the number of bonding pairs and lone pairs of electrons around the central atom, which influences the shape of the molecule. For example, carbon and silicon both have tetrahedral geometries in their fluorides due to their similar bonding characteristics and the presence of four equivalent bonds.
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Molecular Geometry with Two Electron Groups
Hybridization
Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals that can accommodate bonding. The type of hybridization affects the molecular structure; for instance, nitrogen undergoes sp3 hybridization in NF3, leading to a trigonal pyramidal shape, while phosphorus can exhibit sp3d hybridization in PF5, resulting in a trigonal bipyramidal structure. This difference in hybridization explains the distinct molecular geometries of nitrogen and phosphorus fluorides.
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Oxidation States and Bonding
The oxidation state of an element in a compound indicates the degree of oxidation or reduction it has undergone. In the case of binary fluorides, the highest oxidation states of carbon and silicon lead to similar bonding characteristics, resulting in analogous molecular structures. Conversely, nitrogen and phosphorus, despite being in the same group, can exhibit different oxidation states and bonding behaviors, leading to variations in their molecular structures.
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Related Practice
Textbook Question
Select the group 4A element that best fits each of the following descriptions.
b. Is the least dense semimetal
Textbook Question
In the following pictures of binary hydrides, ivory spheres
represent H atoms or ions, and burgundy spheres represent
atoms or ions of the other element.
(1)
(2)
(3)
(4)
(b) What is the oxidation state of hydrogen in compounds (1), (2), and (3)? What is the oxidation state of the other
element?
Textbook Question
Complete and balance the equation for each of the following reactions.
b. Ca(s) + H2O(l) →
Textbook Question
Which compound in each of the following pairs is more ionic?
(b) P4O6 or Ga2O3
Textbook Question
Look at the location of elements A, B, C, and D in the following periodic table:
(b) Classify each oxide as basic, acidic, or amphoteric.