(a)
Interpretation:
Two cations and two anions having the same electronic configuration as neon should be stated.
Concept Introduction:
Electron configuration of neon (Ne) is as follows,
Noble gases have the most stable electronic configurations; the valence shell is full. When forming ions, loss or gain of electron/s by an atom of a main group element takes place to attain the electronic configuration of the nearest noble gas in the periodic table.
There are two types of ions, cationic and anionic.
Cations are positively charged ions with less electrons than protons.
Anions are negatively charged ions and more electrons than protons.
(b)
Interpretation:
Two cations and two anions having the same electronic configuration as argon should be stated.
Concept Introduction:
Noble gases have the most stable electronic configurations; the valence shell is full. When forming ions, loss or gain of electron/s by an atom of a main group element takes place to attain the electronic configuration of the nearest noble gas in the periodic table.
When electrons are lost, cations are formed. They are positively charged ions and have fewer electrons than protons.
When electrons are gained, anions are formed. They are negatively charged ions and have more electrons than protons.
Electron configuration of argon (Ar) is

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Chapter 3 Solutions
CONNECT IA GENERAL ORGANIC&BIO CHEMISTRY
- An electrode process takes place at a metal-solution interface. Indicate the current condition that must be met for Faradaic rectification to occur.arrow_forwardAt a metal-solution interface, an electron is exchanged, and the symmetry factor beta < 0.5 is found in the Butler-Volmer equation. What does this indicate?arrow_forwardTopic: Photochemistry and Photophysics of Supramoleculesarrow_forward
- When two solutions, one of 0.1 M KCl (I) and the other of 0.1 M MCl (II), are brought into contact by a membrane. The cation M cannot cross the membrane. At equilibrium, x moles of K+ will have passed from solution (I) to (II). To maintain the neutrality of the two solutions, x moles of Cl- will also have to pass from I to II. Explain this equality: (0.1 - x)/x = (0.1 + x)/(0.1 - x)arrow_forwardCalculate the variation in the potential of the Pt/MnO4-, Mn2+ pair with pH, indicating the value of the standard potential. Data: E0 = 1.12.arrow_forwardGiven the cell: Pt l H2(g) l dis X:KCl (sat) l Hg2Cl2(s) l Hg l Pt. Calculate the emf of the cell as a function of pH.arrow_forward
- Introduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage Learning
