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Class 11
Chemistry
Redox Reactions
calculate the oxidation number of sulphur chromiu...

Question:
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O^2- 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) Cr₂O^2-₇

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 Cr₂O₇^2-

The structure of Cr₂O₇^2-is shown as follows:

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

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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Question:
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O^2- ₇ and NO^– ₃. Suggest structure of these compounds. Count for the fallacy.

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Question: 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.

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Question:
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O^2- ₇ and NO^– ₃. Suggest structure of these compounds. Count for the fallacy.