General Chemistry: Principles and Modern Applications (11th Edition)
General Chemistry: Principles and Modern Applications (11th Edition)
11th Edition
ISBN: 9780132931281
Author: Ralph H. Petrucci, F. Geoffrey Herring, Jeffry D. Madura, Carey Bissonnette
Publisher: PEARSON
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Chapter 23, Problem 71E
Interpretation Introduction

(a)

Interpretation:

The Lewis structure of Hg22+ should be drawn.

Concept introduction:

Lewis structure is a representation of how valance shell electrons are arranged around individual atoms in a molecule. Steps of drawing Lewis structures are

  • Determine the total number of valence electrons in a molecule.
  • Draw a sketch of the molecule connecting all atoms to the central atom using only single bond. Central atom is the one which can form highest number of bonds. One bond indicates two electrons
  • The remaining electrons are distributed among the atoms as their octets are being completed.
  • If the octet of central atom is no completed, using lone pairs of terminal atoms, create multiple bonds.
  • Then determine the formal charge of all atoms

Formal charge = valence electrons − Nonbonding electrons − (1/2) x Bonding electrons

Interpretation Introduction

(b)

Interpretation:

The Lewis structure of Mn2O7 should be drawn.

Concept introduction:

Lewis structure is a representation of how valance shell electrons are arranged around individual atoms in a molecule. Steps of drawing Lewis structures are

  • Determine the total number of valence electrons in a molecule.
  • Draw a sketch of the molecule connecting all atoms to the central atom using only single bond. Central atom is the one which can form highest number of bonds. One bond indicates two electrons
  • The remaining electrons are distributed among the atoms as their octets are being completed.
  • If the octet of central atom is no completed, using lone pairs of terminal atoms, create multiple bonds.
  • Then determine the formal charge of all atoms

Formal charge = valence electrons − Nonbonding electrons − (1/2) x Bonding electrons

Interpretation Introduction

(c)

Interpretation:

The Lewis structure of OsO4 should be drawn.

Concept introduction:

Lewis structure is a representation of how valance shell electrons are arranged around individual atoms in a molecule. Steps of drawing Lewis structures are

  • Determine the total number of valence electrons in a molecule.
  • Draw a sketch of the molecule connecting all atoms to the central atom using only single bond. Central atom is the one which can form highest number of bonds. One bond indicates two electrons
  • The remaining electrons are distributed among the atoms as their octets are being completed.
  • If the octet of central atom is no completed, using lone pairs of terminal atoms, create multiple bonds.
  • Then determine the formal charge of all atoms

Formal charge = valence electrons − Nonbonding electrons − (1/2) x Bonding electrons

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Chapter 23 Solutions

General Chemistry: Principles and Modern Applications (11th Edition)

Ch. 23 - Complete and balance the following equations. If...Ch. 23 - By means of e chemical equation, give an example...Ch. 23 - Prob. 13ECh. 23 - Prob. 14ECh. 23 - Prob. 15ECh. 23 - Prob. 16ECh. 23 - Prob. 17ECh. 23 - According to Figure 23-8, G decreases with...Ch. 23 - Prob. 19ECh. 23 - Prob. 20ECh. 23 - Write plausible half-equations to represent each...Ch. 23 - Prob. 22ECh. 23 - Use electrode potential data from this chapter or...Ch. 23 - You are given these three reducing agents: Zn(s);...Ch. 23 - Prob. 25ECh. 23 - Prob. 26ECh. 23 - Prob. 27ECh. 23 - Use data from the text to construct a standard...Ch. 23 - When a soluble lead compound is added to a...Ch. 23 - Prob. 30ECh. 23 - Prob. 31ECh. 23 - If CO2(g) under pressure is passed into...Ch. 23 - Use equation (23.19) to determine [Cr2O72] in a...Ch. 23 - If a solution is prepared by dissolving 1.505 g...Ch. 23 - Prob. 35ECh. 23 - Prob. 36ECh. 23 - Prob. 37ECh. 23 - Prob. 38ECh. 23 - Will reaction (23.25) still be spontaneous in the...Ch. 23 - Prob. 40ECh. 23 - Prob. 41ECh. 23 - Prob. 42ECh. 23 - Prob. 43ECh. 23 - Prob. 44ECh. 23 - Prob. 45ECh. 23 - Prob. 46ECh. 23 - Prob. 47ECh. 23 - At 400C , 2Hg(I)+O2(g)2HgO(s) for the reaction...Ch. 23 - Prob. 49ECh. 23 - Prob. 50ECh. 23 - Prob. 51ECh. 23 - Prob. 52ECh. 23 - Prob. 53ECh. 23 - Prob. 54ECh. 23 - The text notes that in small quantities, zinc is...Ch. 23 - Prob. 56ECh. 23 - What formulas would you expect for the metal...Ch. 23 - For the straight-line graphs in Figure 23-8...Ch. 23 - Prob. 59ECh. 23 - Prob. 60ECh. 23 - Prob. 61ECh. 23 - Prob. 62ECh. 23 - Prob. 63ECh. 23 - Prob. 64ECh. 23 - Prob. 65ECh. 23 - Prob. 66ECh. 23 - Prob. 67ECh. 23 - Prob. 68ECh. 23 - Prob. 69ECh. 23 - Prob. 70ECh. 23 - Prob. 71ECh. 23 - For a coordination number of four, the radius of...Ch. 23 - Prob. 73ECh. 23 - Prob. 74FPCh. 23 - Several transition metal ions are found in cation...Ch. 23 - Prob. 76SAECh. 23 - Briefly describe each of the following ideas. phe...Ch. 23 - Prob. 78SAECh. 23 - Prob. 79SAECh. 23 - Prob. 80SAECh. 23 - Prob. 81SAECh. 23 - Prob. 82SAECh. 23 - Prob. 83SAECh. 23 - Prob. 84SAECh. 23 - Prob. 85SAECh. 23 - Provide the missing name or formula for the...Ch. 23 - Prob. 87SAECh. 23 - Prob. 88SAECh. 23 - Prob. 89SAECh. 23 - Prob. 90SAE
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