World of Chemistry, 3rd edition
3rd Edition
ISBN: 9781133109655
Author: Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher: Brooks / Cole / Cengage Learning
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Question
Chapter 4.2, Problem 2RQ
Interpretation Introduction
Interpretation: The reason for name of polyatomic ion to be different from name of a binary compound needs to be explained.
Concept Introduction :
The term poly is used for more than two, polyatomic ion contains more than one atom.
Binary compounds are the common chemical compounds, where they are made by the two distinct elements. There are three types of binary compounds are present as ionic compounds, covalent compounds, binary acids. There are ionic, covalent, and metallic bonds are present between the binary compounds.
Expert Solution & Answer
Answer to Problem 2RQ
For the binary compounds, the cation is written first according to its oxidation state and the anion is placed second with the suffix −ide.
In polyatomic ion, the atoms are covalently bonded and the cation position is first than anions. The charge of the polyatomic ion is unbalanced.
Explanation of Solution
Polyatomic ions are bonded with a covalent bond. The atoms bonded in a polyatomic ion having an unbalanced charge. Compounds formed from such a combination of ions are polyatomic ionic compounds.
In binary compounds, two different elements are combined and form binary compounds.
Conclusion
There are ionic, covalent, and metallic bonds are present between the binary compounds. Binary compound the naming is based on the oxidation state and suffix for the anions. In polyatomic ion, the position of the cation is just before the anion. Covalent bonding is present in polyatomic ion.
Chapter 4 Solutions
World of Chemistry, 3rd edition
Ch. 4.1 - Prob. 1RQCh. 4.1 - Prob. 2RQCh. 4.1 - Prob. 3RQCh. 4.1 - Prob. 4RQCh. 4.1 - Prob. 5RQCh. 4.1 - Prob. 6RQCh. 4.2 - Prob. 1RQCh. 4.2 - Prob. 2RQCh. 4.2 - Prob. 3RQCh. 4.2 - Prob. 4RQ
Ch. 4.2 - Prob. 5RQCh. 4.2 - Prob. 6RQCh. 4 - Prob. 1ACh. 4 - Prob. 2ACh. 4 - Prob. 3ACh. 4 - Prob. 4ACh. 4 - Prob. 5ACh. 4 - Prob. 6ACh. 4 - Prob. 7ACh. 4 - Prob. 8ACh. 4 - Prob. 9ACh. 4 - Prob. 10ACh. 4 - Prob. 11ACh. 4 - Prob. 12ACh. 4 - Prob. 13ACh. 4 - Prob. 14ACh. 4 - Prob. 15ACh. 4 - Prob. 16ACh. 4 - Prob. 17ACh. 4 - Prob. 18ACh. 4 - Prob. 19ACh. 4 - Prob. 20ACh. 4 - Prob. 21ACh. 4 - Prob. 22ACh. 4 - Prob. 23ACh. 4 - Prob. 24ACh. 4 - Prob. 25ACh. 4 - Prob. 26ACh. 4 - Prob. 27ACh. 4 - Prob. 28ACh. 4 - Prob. 29ACh. 4 - Prob. 30ACh. 4 - Prob. 31ACh. 4 - Prob. 32ACh. 4 - Prob. 33ACh. 4 - Prob. 34ACh. 4 - Prob. 35ACh. 4 - Prob. 36ACh. 4 - Prob. 37ACh. 4 - Prob. 38ACh. 4 - Prob. 39ACh. 4 - Prob. 40ACh. 4 - Prob. 41ACh. 4 - Prob. 42ACh. 4 - Prob. 43ACh. 4 - Prob. 44ACh. 4 - Prob. 45ACh. 4 - Prob. 46ACh. 4 - Prob. 47ACh. 4 - Prob. 48ACh. 4 - Prob. 49ACh. 4 - Prob. 50ACh. 4 - Prob. 51ACh. 4 - Prob. 52ACh. 4 - Prob. 53ACh. 4 - Prob. 1STPCh. 4 - Prob. 2STPCh. 4 - Prob. 3STPCh. 4 - Prob. 4STPCh. 4 - Prob. 5STPCh. 4 - Prob. 6STPCh. 4 - Prob. 7STPCh. 4 - Prob. 8STPCh. 4 - Prob. 9STPCh. 4 - Prob. 10STPCh. 4 - Prob. 11STP
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