Question

name the following formula
1. Iron (ll) bromide = FeBr2
2. Iron (ll) sulfide
3. Copper (ll) chloride
4. Copper (l) nitride
5. Zinc oxide
6. Silver sulfide

Answer 1

Answer:

1)Iron (II) bromide:

FeBr₂

Fe(ll) is given in the question that means it's oxidation state is +2.

Fe oxidation state is the +2 and Br oxidation state is -1. These oxidation states can be exchanged during the formula writting.

2) Iron(II) Sulfide:

FeS

Fe oxidation state is +2 and the S oxidation state is -2. Both the oxidation numbers are same, Hence these are cancelled during the formula wtitting.

3) Copper(II) chloride:

CuCl₂

Cu oxidation state is +2

Cl oxidation state is - 1

Oxidation states exchanged during the formula wtritting.

4) Copper(I)nitride:

Cu₃N

Cu oxidation state +1 given according to the question

N oxidation state is - 3

5) Zinc oxide:

ZnO

Zn oxidation state is +2

Oxygen is having the - 2.

6) Silver sulfide:

Ag2S

Ag oxidation state is +1 and S oxidation state is - 2.

Explanation :

Generally During the formula writting the oxidation states of the atoms can be exchanged.

If we assume A oxidation state is +2 and the B oxidation state is - 3.

Then the formula becomes A₃B₂.