1. Chapter 10 Molecular Structure: Solids and Liquids
10.1
Electron Configuration of Ionic Compounds
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2. 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+
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3. Formation of Mg2+
Magnesium achieves an octet by losing its two valence electrons.
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4. Formation of Negative Ions
In ionic compounds, nonmetals
Achieve an octet arrangement.
Gain electrons.
Form negatively charged ions with 3-, 2-, or 1- charges.

5. Formation of a Chloride, Cl-
Chlorine achieves an octet by adding an electron to its valence electrons.
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6. 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.

7. Ions of Transition Metals
Example: Fe forms Fe3+ and Fe2+
Fe s22s22p63s23p64s23d6
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.

8. 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

9. 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 ─