Ionic Compound Notes
Ions
Most __________ are formed when a metal _________ ____ at least one electron.
Most ______ are formed when a non-metal _________ at least one electron.
*Polyatomic Ions are formed when “_____ ________” form covalent
bonds, but have an __________ amount of protons and electrons.
Practice Questions
1) What cation is formed when the following metal atoms loose all
their valence electrons?
a. Na
e. Ag
i. *Fe
b. Mg
f. Li
j. *Fe
c. Al
g. In
k. *Cu
d. Be
h. Ca
l. *Cu
2) What anion is formed when the following non-metal atoms gain
enough electrons to fill their valence levels?
a. F
c. S
e. Cl
b. O
d. N
f. Br
3) List the elements that can form anions in a binary ionic
compound. Where are they found on the periodic Table?
Ionic Bonds
Ionic bonds are generated because of the ___________ between
cations and anions.
____________ charges attract each other because of the coulombic
forces.
The net charge of all ionic compounds must be __________, even
though they are made of individually ___________ particles.
For every electron that is missing on the ___________, it must be
relocated somewhere else on an _______________.
* Polyatomic ions follow this rule, too.
Formulas – Practice Questions
4) Write the formula for the compound formed between these
cations and anions.
a. Mg+2 Br
____________
e. Ba+2 O-2
____________
b. Ca+2 S-2
____________
f. Cu+1 O-2
____________
c. Al+3
I-1
____________
g. Fe+3 Cl-1
____________
d. K+1
N-3
____________
h. Ru+3
_________
-1
S-2
5) Write the formula for the compound formed between these
cations and anions. These have polyatomic ions in them.
Polyatomic are surrounded by parenthesis when there are more
than one of the ions in the compound.
Ex: Mg(NO3)2 is formed from Mg+2 & NO3-1 ions
a.
Be+2 IO3-1 ____________
b.
Ag+1 ClO3-1 ____________
c.
Na+1
d.
NH4+1 I-1
PO4
-3
____________
____________
e.
e. Sr+2
NO3-1 ___________
f.
f. Mo+5
SO4-2 ___________
g.
g. V+3
OH-1 ___________
Ionic Solids: Crystalline Lattices
A _________ _____________ is formed between the positive and
negative ions.
Cations are always ____________ than their original atoms because
electrons have been lost.
Anions are always __________ than their original atoms because
electrons have been gained.
The cations and anions to surround each other in the crystalline lattice.
Sodium chloride would look something like this:
- + - + + - + - +
-
Chloride anion (negative)
+
Sodium cation (positive)
- + - + -
Different ionic compounds form different shaped crystals because of
the ions sizes and arrangements.
Practice Questions:
6) Which is a bigger cation:
Na+1 or Mg+2 why?
#p+
#e-
electron configurations
7) Which is a bigger anion:
F-1 or O-2 why?
#p+
#e-
8) Which is a bigger ion:
or O-2 why?
Mg+2
#p+
#e-
Al+3 or Ga+3 why?
N-3 or P-3 why?
electron configurations
Naming Ionic Compounds
To go from the formula to the name of an ionic compound:
Name the _________ first, then the _________
Include ___________ ___________ for the charge of a metal that has
multiple oxidation states. (some transition metals and semi-metals)
Binary Ionic Compounds
Ternary Ionic Compounds
Examples:
*if the cation is ammonium,
and the anion is elemental,
add “-ide” to the anion.
KF =
NH4Cl =
CuI =
MgSO4 =
Practice Naming
9) Given the formula of these binary ionic compounds, determine
the proper name.
a. MgO
c. Na2S
e. AuBr
b. KCl
d. TiN
f. Fe2O3
10) Given the formula of these ternary ionic compounds, determine
the proper name.
a. Mg(NO3)2
b. NH4Cl
c. Na2CO3
d. LiC2H3O2
e. Fe(CN)3
f. K2SO4
g. Sn(OH)2
h. Cu2CrO4
To go from the name to the formula of an ionic compound:
1) Determine the _______ of the cation.
2) Determine the _________ of the anion.
3) Add the right number of cations and anions until the net charge of the
compound is ________. Use the ________ common multiple to
determine how many of each ion is needed.
4) Use ______________ to isolate any polyatomic ion if there needs to
be more than one of the ions in the compound.
magnesium iodide
1) Mg+2 magnesium is an alkaline earth metal, therefore +2 charge
2) I-1
iodine is a halogen, therefore -1 charge
3) add one Mg+2 to two I-1’s to make the net charge 0
copper (II) phosphate
1) Cu+2 copper’s charge is +2 because the roman numeral is (II)
2) PO4-3 phosphate is a polyatomic ion – memorize them
3) +2 and -3 the lowest common multiple is 6 so…
Cu+2 & Cu+2 & Cu+2 and PO4-3 & PO4-3
Practice Formulas
11) Determine the proper formula for these binary ionic compounds.
a. Lithium sulfide
e. Aluminum oxide
b. Sodium chloride
f. Iron (II) nitride
c. Calcium oxide
g. gallium chloride
d. Nickel (III) chloride
h. tin (II) selenide
12) Determine the proper formula for these ternary ionic compounds
a.
Lithium iodate
e.
Cobalt (II) carbonate
b.
Aluminum nitrate
f.
Osmium (VIII) hydroxide
c.
Calcium acetate
g.
Scandium cyanide
d.
Molybdenum (V) chromate
h.
Ammonium phosphide
Answers to the practice questions
1)a. Na+1 b. Mg+2 c. Al+3 d. Be+2 e. Ag+1 f. Li+1 g. In+3 h. Ca+2
2)a. F-1 b. O-2 c. S-2 d. N-3 e. Cl-1
3) C, N, O, F, P, S, Cl, As, Se, Br, Te, I, At – they are all non-metals
4) a. MgBr2 b. CaS c. AlI3 d. K3N e. BaO f. Cu2O g. FeCl3 h. Ru2S3
5) a. Be(IO3)2 b. AlClO3 c. Na3PO4 d. NH4I e. Sr(NO3)2 f. Mo2(SO4)5 g. V(OH)3
6) Na+1 has 11p+, Mg+2 has 12 p+, they both have 10 e- - the more protons in the nucleus,
the stronger the attraction to the nucleus, the smaller the particle (Na+1 is bigger)
Ga+3 has electrons in the 3rd energy level, Al+3 has electrons in the 2nd energy level
7) F-1 < O-2 why? More protons in F, same # e- (10 in each)
N-3 < P-3 why? P-3 has e- in the 3rd energy level, N-3 has e-in the 2nd energy level
8) Mg+2 < O-2 why? More protons in Mg+2 (both have 10 e-)
9) a. magnesium oxide b. potassium chloride c. sodium sulfide d. titanium (III) nitride
e. gold (I) bromide f. iron (III) oxide
10) a. magnesium nitrate b. ammonium chloride c. sodium carbonate d. lithium acetate
e. iron (III) cyanide f. potassium sulfate g. tin (II) hydroxide h. copper (I) chromate
11) a. Li2S b. NaCl c. CaO d. NiCl3 e. Al2O3 f. Fe3N2 g. GaCl3 h. SnSe
12) a. LiIO3 b. Al(NO3)3 c. Ca(C2H3O2)2 d. Mo2(CrO4)5 e. CoCO3 f. Os(OH)8
g. Sc(CN)3 h. (NH4)3P