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.

(i) H₂SO₅

Let the oxidation number(O.N.) of S be ‘x’

We know that,

O.N. of H = + 1

O.N. of O = -2

Therefore, we have,

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

2 + x-10 = 0

X = + 8

The O.N. of S is + 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.

(ii) Cr₂O₇^2-

Let the oxidation number (O.N.) of Cr be ‘x’

O.N. of O = -2

Therefore, we have,

2(x) + 7(-2) = -2

2x-14 = -2

X = + 6

The O.N. of Cr is + 6.

Here, there is no fallacy about the O.N. of Cr in Cr₂O₇^2-

The structure of Cr₂O₇^2- is shown as follows

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

(iii) NO₃^-

Let the oxidation number (O.N.) of N be ‘x’

O.N. of O = -2

Therefore, we have,

1(x) + 3(-2) = -1

1x-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.