Ch. 12 Worksheet 1 (12.1 & 12.2)
1. How are ionic bonds and covalent bonds different? Ionic – electrons not shared. Cation (metal) has
completely lost at least 1 electron. Anion (nonmetal) has gained at least 1 electron. Covalent –
electrons are shared (equally or not). Nonmetal & metal.
2. How does a polar covalent bond differ from a nonpolar covalent bond? polar – electrons are shared
unequally. Bonding atoms have different electronegativities.
3. If a very electronegative atom is bonded to a very non-electronegative atom, what type of bond is likely
to form? an ionic bond. The very electronegative atom pulls all of the electron to it.
4. If a non-electronegative atom is bonded to a slightly electronegative atom, what type of bond is likely to
form? polar covalent bond (electrons are shared, but unequally so there is a slight dipole.)
5. If 2 atoms with the same electronegativity form a bond, what type of bond will it be?
pure covalent (nonpolar covalent). equal sharing of electrons
6. For each of the following binary molecules, write the type of bond expected to form between them.
(ionic, polar covalent, covalent) Tell which atom is the most negative.
a. H – H nonpolar covalent f. C – C nonpolar covalent
b. H – F polar covalent g. NaBr ionic (metal/nonmetal)
+ - + -
c. F – F nonpolar covalent h. CO polar covalent
d. KF ionic i. OH polar covalent
+ - - +
e. NaF ionic j. NO polar covalent
+ - + -
12.3 & 12.4 Predicting Formulas for Ionic Compounds Using Electron Configurations
1. a. Why do metals lose electrons to form ions? So that their electron configuration is the same as the
previous noble gas. (More energetically favorable)
b. When does a metal stop losing electrons? When it has the same electron configuration as the previous
2. Use its electron configuration to show why Oxygen forms an O2- ion and not an O3- ion.
O (8 e) 1s22s22p4 O2- (10 e) 1s22s22p6- same as Neon. If it gained 3 electrons, the electron
configuration would be 1s22s22p63s1. (Not favorable)
3. Write the electron configurations for the pairs of atoms given below. Use them to predict the formula
for an ionic compound formed from these elements.
a. Mg Mg2+ S [Ne]3s23p4 6 valence MgS
2 2- 2 6
[Ne] 3s [Ne] S [Ne]3s 3p = [Ar]
2 valence electrons
b. K [Ar]4s1 1 valence Cl [Ne]3s23p5 7 valence e KCl
K+ [Ar] [Ne]3s23p6 or [Ar]
c. Cs [Xe]6s1 1 valence F [He]2s22p5 7 valence e CsF
Cs+ [Xe] F- [Ne]
d. Ba [Xe]6s2 2 valence Br [Ar]4s23d104p5 7 valence (only s & p) BaBr2
Ba2+ [Xe] Br- [Ar]4s23d104p6 = [Kr]
e. Li 1s22s1 1 valence F [He]2s22p5 7 valence e LiF
Li+ 1s2 F- [Ne]
f. Al [Ne]3s23p1 3 valence O 1s22s22p4 6 valence Al2O3
Al3+ [Ne] O2- 1s22s22p6
(Thinking Question) Why are cations smaller than their parent atoms? Why are anions larger?
Cations have lost electrons in the outer (larger) shell so are smaller after they have lost their electrons.
Anions have gained electrons and, even though the electrons do not go into a larger shell (they generally just fill
up a shell), more negative charge with the same amount of positive to attract and hold the negative charge,
makes for a larger particle.