Question

What are the oxidation numbers for the metals in each of the following coordination compounds?

(a)K[Au(Cl)4]..........K_______, Au_________

(b)Na4[Fe(Br)6]......Na_______, Fe________

Answer 1

The oxidation number of halogens will always be -1 unless oxygen is combined with them .

If Hydrogen is bonded to metal it will have +1 charge else for non metal bonding it will have -1 charge on it.

The given compounds are neutral, hence the sum of oxidation numbers will be zero and considering halogens have -1 oxidation state from the above reference:

a) K[Au(Cl)4] ->

K⁺¹ and [Au(Cl)4]^-1 , hence oxidation number for K = +1

To calculate for Au , the anion part has a total charge of -1 . Hence O(Au) + 4(-1) = -1 , so solving this equation, Au has an oxidation number O(Au) = -1 +4 = +3 .

b) Na4[Fe(Br)6] ->

Na⁺¹ and [Fe(Br)6]^-4 , hence oxidation number for Na = +1

To calculate for Fe , the anion part has a total charge of -4 . Hence O(Fe) +6(-1) = -4 , so solving the equation, Fe has an oxidation number O(Fe) = -4 +6 = +2 .