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Transition Metals
Formulas & Nomenclature
Transition metals
with ONE charge
• Silver - always a “+1”
• Cadmium - always a “+2”
• Zinc - always a “+2”
Transition Metals with
MORE than ONE charge
• In order to write the formula of a
compound that contains a transition
metal (other than silver, cadmium or
zinc), you need to know the oxidation
number.
• One way the oxidation number is
expressed in the name is by the use of
a ROMAN NUMERAL
• The roman numeral indicates the
oxidation number of the transition metal
Name of metal + (Roman Numeral) + Name of
(I)
NONMETAL +
(II)
“-ide” ending
(III)
(IV)
Write the formula for
iron (III) oxide
• Step 1 - determine the oxidation
numbers for each element
•
Fe+3
O-2
• The roman numeral (III) shows that the
charge on the iron is “+3”. The charge of
the oxide ion is found on the periodic
table.
Write the formula for
iron (III) oxide
*Step 2 - “cross charges” if necessary
Fe+3 O-2
The iron ion will lose three electrons while
the oxide ion gains two. You must have 2
iron ions for every 3 oxide ions for the
charges to balance:
Fe2O3
Write the formula for
iron (III) oxide
• Step 3 - evaluate your answer
Check to be sure the overall charge on
the compound is ZERO
Fe2O3
2(+3) + 3(-2) = 0
Practice Problems
•
1.
2.
3.
4.
5.
6.
Write the correct formula for the
following ionic compounds:
Copper (I) nitride
Copper (II) nitride
Lead (II) bromide
Lead (IV) bromide
Chromium (II) sulfide
Chromium (III) sulfide
Naming Ionic Compounds with
Transition Metals
• Transition metals and metals on the
right side of the periodic table often
have more than one oxidation number.
To tell which oxidation number is used
the name of the formula must indicate
the charge. The charge is written as a
Roman Numeral in parentheses after
the metal cation’s name.
CuO
• Oxide’s oxidation number is always -2
• There is one oxide ion, therefore the
overall negative charge is -2
• There must be an overall +2 charge to
equal the negative charge.
• There is one copper ion, therefore its
charge must be +2
CuO
• 1 Copper (x) + 1 Oxygen (-2) = 0
x -2 = 0
x = +2
Name :
Copper (II) oxide
SnCl4
• Chlorine’s charge is always -1
• There are 4 chloride ions, so the overall
negative charge is -4
• There must be an overall charge of +4
to equal the negative charge.
• There is one tin ion, therefore its charge
must be +4
SnCl4
• 1(x) + 4(-1) = 0
x -4 = 0
x = +4
Name:
Tin (IV) chloride
Fe3N2
• Nitride’s charge is always -3
• There are 2 nitride ions, so the overall
negative charge is -6
• There must be an overall positive
charge of +6 to equal the negative
charge
• There are 3 iron ions, so the charge of
one iron ion must be +2
Fe3N2
• 3(x) + 2(-3) = 0
3x - 6 = 0
3x = 6
x = 2
Name:
Iron (II) nitride
Practice Problems
Write the correct names for the following
ionic compounds
1. PbI2
2. PbI4
3. SnF2
4. SnF4
5. CrI3
6. CuCl
7. Fe2O3
Stock vs Classical Names
ELEMENT
STOCK NAME
CLASSICAL NAME
Copper
Cu+1
Cu+2
Copper (I)
Copper (II)
Cuprous
Cupric
Iron
Fe+2
Fe+3
Iron (II)
Iron (III)
Ferrous
Ferric
Lead
Pb+2
Pb+4
Lead (II)
Lead (IV)
Plumbous
Plumbic
Mercury
Hg+1
Hg+2
Mercury (I)
Mercury (II)
Mercurous
Mercuric
Sn+2
Sn+4
Tin (II)
Tin (IV)
Stannous
stannic
Tin
CATION
Practice Problems
• Fe2O3
• PbCl2
• Sn3P4
• Hg3N