Calculate the Oxidation Number of Sulphur, Chromium and Nitrogen in H2SO5, Cr2O72- and NO3-

• H2SO5 by conventional method.

Consider x to be the oxidation number of Sulphur.

⇒2+1+x=5-2=0 

⇒x=+8 

Because S only has 6 valence electrons, it cannot be 8 or more than 8. Calculating S oxidation using the chemical bonding approach eliminates this misconception.

From its structure HO–SO2–O–O–H.

⇒-1+x+2+2+2-1=0 

⇒x=+6 

Therefore, the oxidation number od S is +6.

• Oxidation number for Chromium in Cr2O72-.

Consider x to be the oxidation number of Chromium in Dichromate ion.

⇒2x+7-2=-2 

⇒x=+6 

As a result, Cr’s oxidation number in dichromate ion is +6. This is correct, and no fallacy exists.

• Oxidation number for Nitrogen in NO3–.

Consider x to be the oxidation number of Nitrogen in Nitrate ion.

⇒x+3-2=-1 

From the structure O––N+O–O–

⇒x+1-1+1-2=0 

⇒x=+5 

As a result, Nitrogen’s oxidation number in Nitrate ion is +5. This is correct, and no fallacy exists.