Transition Metals 4 These are the metals on the “bridge” or short part of the periodic table, running along the center. 4 These metals can have more than one charge. Sometimes, for example, iron (Fe) gives up 2 electrons, and sometimes 3.
If it helps, 4 Think about some brave Roman soldiers, fighting to the death on a bridge - each man wielding a metal sword and a metal shield.
Roman Numerals 1 = I 2 = II 3 = III 4 = IV
Transition Metals, con. 4 So iron can form both FeO and Fe 2 O 3, depending on the number of electrons the iron atoms gave up. In order to identify which “iron oxide” you’re talking about, you have to use Roman numerals to indicate the charge of the cation in the compound.
“Iron Oxide” 4 In FeO, the oxygen has a valence of 6, and so it takes two electrons, giving it a 2- charge. 4 It gets the two electrons from the one iron atom - so the iron atom must have a 2+ charge. 4 This makes the name iron (II) oxide, 4 where the Roman numeral II indicates that the charge is 2+
Iron Oxide 4 In Fe 2 O 3, the 3 different oxygen atoms would each take 2 electrons, for a total of 6- 4 Where did they get them? 4 From the 2 different iron atoms. If the total number of electrons taken is 6, then the total number of electrons given must be 6. 4 So gave up 3 electrons each. 4 each of the two iron atoms
FeI 2 4 Fe is a transition metal, so it can have more than one possible number of valence electrons. 4 But Iodine is a halogen, so it definitely has a valence of 7, so it will take 1 electron. 4 There are two iodines in the formula, and each of them must have taken 1 electron, for a total charge of 2- 4 So it must be iron (II) iodide
MnF 3 4 Manganese (not magnesium!) is a transition metal. Flourine is a halogen, (1-) 4 If each of the 3 fluorines took 1 electron, the manganese atom must have given up 3. 4 So it’s 4 manganese (III) fluoride
Another Example 4 What is the formula for aluminum fluoride? 4 Aluminum has 3 valence electrons, for a charge of 3+ 4 Fluorine, a halogen, has a valence of 7, so it wants one more electron, for a charge of 1- 4 It will take 3 fluorines to take all 3 of the aluminum valence electrons, so the formula will be AlF 3