Exceptional Cases of Evaluation of Oxidation Numbers

Exceptional Cases of Evaluation of Oxidation Numbers

The rules described earlier are usually helpful in determination of the oxidation number of a specific atom in simple molecules but these rules fail in following cases. In these cases, the oxidation numbers are evaluated using the concepts of chemical bonding involved.

Type I. In molecules containing peroxide linkage in addition to element-oxygen bonds. For example,

(i) Oxidation number of S in H2SO5

(Permonosulphuric acid or Caro's acid)

By usual method;  

or    

But this cannot be true as maximum oxidation number for S cannot exceed + 6. Since S has only 6 electrons in its valence shell. This exceptional value is due to the fact that two oxygen atoms in shows peroxide linkage as shown below,

 

Therefore the evaluation of o.n. of sulphur here should be made as follows,

2 × (+1)   +   x    +   3 × (–2)   +   2 × (–1)

(for H) (for S)        (for O)         (for O–O)    

or   2  +  x – 6 – 2 = 0    or          x = + 6.

(ii) Oxidation number of S in H2S2O8

(Peroxidisulphuric acid or Marshall's acid)

By usual method; 

1 × 2 + 2x + 8 (–2) = 0

2x = + 16 – 2 = 14   or    x = + 7

Similarly Caro's acid, Marshall's acid also has a peroxide linkage so that in which S shows +6 oxidation state.

 

Therefore the evaluation of oxidation state of sulphur should be made as follow,

2 × (+1)   +   2 × (x)    +   6 × (–2)   +   2 × (–1) = 0

(for H)      (for S)                 (for O)              (for O–O)         

or    2  +  2x – 12 – 2 = 0  or  x = + 6.

(iii)  Oxidation number of Cr in CrO5

(Blue perchromate)

By usual method ;   x – 10 = 0  or   x = + 10

This cannot be true as maximum O. N. of Cr cannot be more than + 6. Since Cr has only five electrons in 3d orbitals and one electron in 4s orbital. This exceptional value is due to the fact that four oxygen atoms in CrO5are in peroxide linkage.

The chemical structure of CrO5 is

                                                

Therefore, the evaluation of o.n. of Cr should be made as follows

x    +   1 × (– 2)   +   4  (–1) = 0

(for Cr)        (for O)      (for O–O)     

or  x – 2 – 4 = 0  or  x = + 6.

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