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.
(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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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 5: Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O2- 7 and NO– 3.....
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