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Class 11 Chemistry - Chapter Redox Reactions NCERT Solutions | Calculate the oxidation number of sulphu

Welcome to the NCERT Solutions for Class 11th Chemistry - Chapter Redox Reactions. This page offers a step-by-step solution to the specific question from Excercise 1 , Question 5: calculate the oxidation number of sulphur chromiu....
Question 5

Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O2- 7 and NO– 3. Suggest structure of these compounds. Count for the fallacy.

Answer

(a) H2SO5

Let assume oxidation number of S is x.

We know that,

Oxidation number of H = +1

Oxidation number of O = -2

Then we have

2(+1) + (x) + 5 (-2) = 0

⇒ 2 + x - 10 = 0

⇒ x  =  +8

However, the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S cannot be more than +6. The structure of H2SO5 is shown as follows:

Now , 2(+1) + 1(x) + 3(-2)  + 2 (-1) = 0

⇒ 2 + x - 6 -2 = 0

⇒ x  =  +6

Therefore, the O.N. of S is +6.

 

(b) Cr2O2-7

Let assume oxidation number of Cr is x.

We know that,

Oxidation number of O = -2

Then we have

 2(x) + 7 (-2) =  -2

⇒ 2x  -14  = -2

⇒ 2x   =  +12

x  = +6

Here, there is no fallacy about the O.N. of Cr in Cr2O72-

The structure of Cr2O72- is shown as follows: 

Here, each of the two Cr atoms exhibits the O.N. of +6. 

 

(c) NO3-

Let assume oxidation number of N is x.

We know that,

Oxidation number of O = -2

Then we have

1(x) + 3 (-2) =  -1

⇒ x  - 6 = -1

⇒ x   =  +5

Here, there is no fallacy about the O.N. of N in NO3-

The structure of  NO3-  is shown as follows:

The N atom exhibits the O.N. of +5. 

 

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