
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 8, Problem 8.46QP
Interpretation Introduction
Interpretation:
In the given set of anions which will be larger has to be identified and explained.
Concept Introduction:
- The distance between the nucleus and the valence shell of a cation or an anion is known as ionic radius. An ion is formed by either loss or gain of electrons from its valence shell.
- As we move down the group in periodic table the ionic radius increases as the electrons are added to new subshell. As we move across the period the ionic radius increases as the electrons are added to the same subshell.
- An anion is formed when an electron is added to the valence shell of an atom. The anion has a net negative charge in it. In anion the extra electron added occupies more space and maximizes the shielding.
- Anions will have larger size compared to cations.
- The reduction in the effective nuclear charge on the electron cloud, due to a difference in the attraction forces of the electrons in the nucleus is known as shielding effect
- When the proton number is greater than the electron, the size of the ion will be smaller due to less shielding. When the proton number is lesser than the electron, the size of the ion will be larger due to more shielding.
- A cation is formed when an electron is lost by an atom from its valence shell. The cation exhibit a net positive charge in it. In cation the shielding decreases as the electron is removed from the valence shell.
- If the total number of electrons is less than the total number of protons in the ion, then the protons present can effectively attract the valence shell decreasing the size of the ion and vice-versa.
- The trend in periodic table can be described as well. As we move down the group the ionic radius decreases as the electrons are added to a new shell. But as we move across a period in periodic table the ionic radius increases as the electrons are added to the same sub-shell.
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Module 4: Homework (Ch 9-10)
Question 28 of 30 (1 point) | Question Attempt: 1 of Unlimited
H₂SO
heat
OH
The mechanism of this reaction involves two carbocation intermediates, A and B.
Part 1 of 2
KHSO
4
rearrangement
A
heat
B
H₂O
2
OH
Draw the structure of A.
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Chapter 8 Solutions
General Chemistry
Ch. 8.2 - Practice Exercise An atom of a certain element has...Ch. 8.2 - Review of concepts
Identify the elements whose...Ch. 8.3 - Practice Exercise Arrange the following atoms in...Ch. 8.3 - Prob. 1RCCh. 8.3 - Prob. 2PECh. 8.3 - Prob. 2RCCh. 8.4 - Prob. 1PECh. 8.4 - Prob. 1RCCh. 8.5 - Prob. 1PECh. 8.5 - Prob. 1RC
Ch. 8.6 - Prob. 1PECh. 8.6 - Prob. 1RCCh. 8 - Prob. 8.1QPCh. 8 - Prob. 8.2QPCh. 8 - Prob. 8.3QPCh. 8 - Prob. 8.4QPCh. 8 - Prob. 8.5QPCh. 8 - Prob. 8.6QPCh. 8 - Prob. 8.7QPCh. 8 - Prob. 8.8QPCh. 8 - Prob. 8.9QPCh. 8 - Prob. 8.10QPCh. 8 - Prob. 8.11QPCh. 8 - Prob. 8.12QPCh. 8 - Prob. 8.13QPCh. 8 - 8.14 Use die first-row transition metals (Sc to...Ch. 8 - 8.15 In the periodic table, the element hydrogen...Ch. 8 - 8.16 A neutral atom of a certain element has 17...Ch. 8 - Prob. 8.17QPCh. 8 - Prob. 8.18QPCh. 8 - Prob. 8.19QPCh. 8 - Prob. 8.20QPCh. 8 - Prob. 8.21QPCh. 8 - 8.21 An ion M2+ derived from a metal in the first...Ch. 8 - Prob. 8.23QPCh. 8 - Prob. 8.24QPCh. 8 - Prob. 8.25QPCh. 8 - Prob. 8.26QPCh. 8 - Prob. 8.27QPCh. 8 - Prob. 8.28QPCh. 8 - Prob. 8.29QPCh. 8 - Prob. 8.30QPCh. 8 - 8.31 Which of these species are isoelectronic with...Ch. 8 - 8.32 Group the species that are isoelectronic:...Ch. 8 - Prob. 8.33QPCh. 8 - 8.34 How does atomic radius change as we move (a)...Ch. 8 - Prob. 8.35QPCh. 8 - Prob. 8.36QPCh. 8 - 8.37 On the basis of their positions in the...Ch. 8 - Prob. 8.38QPCh. 8 - 8.39 Which is the largest atom in Group 4A?
Ch. 8 - Prob. 8.40QPCh. 8 - Prob. 8.41QPCh. 8 - Prob. 8.42QPCh. 8 - Prob. 8.43QPCh. 8 - Prob. 8.44QPCh. 8 - Prob. 8.45QPCh. 8 - Prob. 8.46QPCh. 8 - Prob. 8.47QPCh. 8 - Prob. 8.48QPCh. 8 - 8.49 Define ionization energy. Ionization energy...Ch. 8 - Prob. 8.50QPCh. 8 - Prob. 8.51QPCh. 8 - Prob. 8.52QPCh. 8 - Prob. 8.53QPCh. 8 - Prob. 8.54QPCh. 8 - Prob. 8.55QPCh. 8 - Prob. 8.56QPCh. 8 - Prob. 8.57QPCh. 8 - Prob. 8.58QPCh. 8 - Prob. 8.59QPCh. 8 - Prob. 8.60QPCh. 8 - Prob. 8.61QPCh. 8 - Prob. 8.62QPCh. 8 - Prob. 8.63QPCh. 8 - Prob. 8.64QPCh. 8 - Prob. 8.65QPCh. 8 - Prob. 8.66QPCh. 8 - Prob. 8.67QPCh. 8 - 8.68 Why are the Group 1B elements more stable...Ch. 8 - Prob. 8.69QPCh. 8 - Prob. 8.70QPCh. 8 - Prob. 8.71QPCh. 8 - Prob. 8.72QPCh. 8 - Prob. 8.73QPCh. 8 - Prob. 8.74QPCh. 8 - Prob. 8.75QPCh. 8 - Prob. 8.76QPCh. 8 - Prob. 8.77QPCh. 8 - Prob. 8.78QPCh. 8 - Prob. 8.79QPCh. 8 - Prob. 8.80QPCh. 8 - Prob. 8.81QPCh. 8 - Prob. 8.82QPCh. 8 - Prob. 8.83QPCh. 8 - Prob. 8.84QPCh. 8 - Prob. 8.85QPCh. 8 - Prob. 8.86QPCh. 8 - Prob. 8.87QPCh. 8 - Prob. 8.88QPCh. 8 - Prob. 8.89QPCh. 8 - 8.90 Write the formulas and names of the oxides of...Ch. 8 - Prob. 8.91QPCh. 8 - Prob. 8.92QPCh. 8 - Prob. 8.93QPCh. 8 - Prob. 8.94QPCh. 8 - Prob. 8.95QPCh. 8 - Prob. 8.96QPCh. 8 - Prob. 8.97QPCh. 8 - Prob. 8.98QPCh. 8 - 8.99 Explain why the electron affinity of nitrogen...Ch. 8 - Prob. 8.100QPCh. 8 - Prob. 8.101QPCh. 8 - Prob. 8.102QPCh. 8 - Prob. 8.103QPCh. 8 - Prob. 8.104QPCh. 8 - Prob. 8.105QPCh. 8 - Prob. 8.106QPCh. 8 - Prob. 8.107SPCh. 8 - 8.108 In the late 1800s the British physicist Lord...Ch. 8 - Prob. 8.109SPCh. 8 - Prob. 8.110SPCh. 8 - Prob. 8.111SPCh. 8 - Prob. 8.112SPCh. 8 - Prob. 8.113SPCh. 8 - Prob. 8.114SP
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