Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 2 - Atoms, Molecules, and Ions - Additional Exercises: 96

Answer

a) Tellurium b) Rubidium c) Argon d) Astatine

Work Step by Step

a) Oxygen lies in group (family) 16. In group 16, lies Tellurium, whose most stable ion is $Te^{2-}$. Tellurium has atomic number = 52, and thus number of protons = 52. Since the ion has a 2- charge on it, number of electrons will be two more than the number of protons. Thus number of electrons of $Te^{2-}$ = 54 + 2 = 54. b) The alkali metal family has the element rubidium (Rb), whose most stable ion is $Rb^{+}$. Its atomic number = 37 , which means it has 37 protons, and 37 electrons. BUT, in its ion form, it loses one electron to form the $Rb^{+}$ ion. Thus number of electrons of $Rb^{+}$ = 36. c) 18 protons mean that the atomic number of that noble gas is 18. Noble gas of atomic number 18 is Argon ( Refer to periodic table ) d) 85 protons mean that the atomic number of that halogen is 85. Halogen of atomic number 85 is Astatine Refer to periodic table )
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