notes 6 – ionic compounds (part 2) sci 10chemistry

Notes 6 – Ionic compounds (Part 2)

Sci 10 Chemistry

Ionic compounds with a transition metal:

Almost all transition metals (in the centre of the periodic table) are able to form more than one cation.

i.e.: Copper can form the cations

Cu 1+ , and Cu 2+ Iron can be Fe 2+ or Fe 3+

Roman Numerals

I = 1

II = 2

III = 3

IV = 4

V = 5

VI = 6

Ionic compounds with transition metals –Nomenclature (naming)

Name to formula: i.e. : copper (I) chloride

1. Write the elements’ symbols with their charges (the charge of the cation (the transition metal) is given in the brackets)

Cu 1+ Cl -

2. If the charges cancel (add to zero), you are done Rewrite the formula CuCl.


If the charges do not add to zero: Use the crossover method to determine the

number of each atom required to have a neutral compound

i.e. : iron (II) chloride

Fe 2+ Cl -1

FeCl2

See if you can reduce the subscripts. If not, you are done

Give it a try!

What is the formula for:

osmium (II) sulfide

Os 2+ S 2-

The charges add to zero so the answer is:

OsS

Give it a try!

What is the formula for:

nickel (III) fluoride Don’t forget that the III (3) in the brackets

tells us the charge of nickel

Ni 3+ F 1-

The charges do not add to zero, so crossover method:

NiF3


Formula to name: The charge of the transition metal is written in

brackets using Roman numerals

1. Write the name of the cation (the transition metal)

2. Use the inverse crossover method to find the charge of the transition metal

3. Write the charge next to the name of the cation in brackets (use Roman numerals)

4. Write the name of the anion with the ending –ide.

Ionic compounds with transition metals –Nomenclature (naming) i.e.: Fe2O3

iron… (II or III?) Inverse crossover to find the charge of iron (2+ or 3+)

• This time the subscripts go up to become the charges

Fe2O3

So we have Fe 3+ = iron (III) and O2-

Make sure the charge found for the anion (the non-metal) is its actual charge according to the periodic table. If it is, you are done.

The complete name is: iron (III) oxide

Give it a try!

NiF3

nickel …. (II or III?)

Inverse crossover Ni1F3 Ni 3+ F 1-

1- is the actual charge of fluoride so this works

We have nickel (III) hereThe name is: nickel (III) fluoride


PbO2 lead… (II or IV?) Inverse crossover:

Pb1O2 Pb 2+ O 1-

Is that the actual charge of the non-metal (oxygen)? No, so we have to multiply the charges of both atoms to

have the actual (correct) charge on OMultiply both charges by 2:

Pb 2+ x 2 O 1- x 2

Pb 4+ O 2- so it’s lead (IV) Complete name: lead (IV) oxide

Give it a try!

FeOiron…(II) or (III)?Inverse crossover

Fe1O1Fe 1+O 1- *not the charge of O

Multiply both charges by 2 to have the actual (correct) charge on O:

Fe 1+x 2 O1+ x 2 = Fe 2+ O 2-

It is iron (II)Complete name: iron (II) oxide

Careful!

We NEVER use Roman numerals for metals with one charge!

To do:

Notes 6 Practice– Ionic Compounds with a Transition Metal

notes 6 – ionic compounds (part 2) sci 10chemistry

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