Chapter 10 Molecular Structure: Solids and Liquids 10.1 Electron Configuration of Ionic Compounds Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Metals Form Positive Ions Octets by losing all of their valence electrons. Positive ions with the electron configuration of the nearest noble gas. Positive ions with fewer electrons than protons. Group 1A(1) metals  ion 1+ Group 2A(2) metals  ion 2+ Group 3A(13) metals  ion 3+ Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Formation of Mg2+ Magnesium achieves an octet by losing its two valence electrons. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Formation of Negative Ions In ionic compounds, nonmetals Achieve an octet arrangement. Gain electrons. Form negatively charged ions with 3-, 2-, or 1- charges.

Formation of a Chloride, Cl- Chlorine achieves an octet by adding an electron to its valence electrons. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Ions of Transition Metals Form positively charged ions. Lose s valence electrons to form 2+ ions. Lose d electrons to form ions with higher positive charges. Do not form octets like representative metals.

Ions of Transition Metals Example: Fe forms Fe3+ and Fe2+ Fe 1s22s22p63s23p64s23d6 Loss of two 4s electrons Fe2+ 1s22s22p63s23p64s03d6 Loss of two 4s electrons and one d electron Fe3+ 1s22s22p63s23p64s23d5 Note: The 3d subshell is half-filled and stable.

Learning Check Identify the ion with the following electron configurations: A. 1s22s22p63s23p64s23d104p65s0 and 37 protons B. 1s22s22p63s23p64s03d2 and 22 protons C. 1s22s22p63s23p64s23d104p6 and 35 protons

Solution Identify the ion with the following electron configurations: A. 1s22s22p63s23p64s23d104p65s0 and 37 protons = Rb+ B. 1s22s22p63s23p64s03d2 and 22 protons = Ti2+ C. 1s22s22p63s23p64s23d104p6 and 35 protons = Br ─