Question

Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, `"Cr"_2"O"_7^(2-)` and `"NO"_3^-`. Suggest structure of these compounds. Count for the fallacy.

Answer 1

1)
+1 x -2
\[\ce{H2SO5}\]

2(+1) + 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:

\[\begin{array}{cc}
\ce{^{-2}O}\phantom{...........}\\
||\phantom{.........}\\
\ce{^{+1}H - ^{-2}O - S^{\text{x}} - ^{-1}O - ^{-1}O - ^{+1}H}\\
||\phantom{.........}\\
\ce{^{-2}O}\phantom{............}\end{array}\]

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.

2)
 x     2- 
\[\ce{Cr2 O^{2-}_7}\]

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

⇒ 2x - 14 = -2

⇒ x = +6

Here, there is no fallacy about the O.N. of Cr in `"Cr"_2"O"_7^(2-)`.

The structure of `"Cr"_2"O"_7^(2-)` is shown as follows:

\[\begin{array}{cc}
\ce{^{2-}O}\phantom{...........}\ce{^{2-}O}\phantom{....}\\
||\phantom{..............}||\phantom{..}\\
\ce{^{2-}O = Cr^{+6} - ^{2-}O - Cr^{+6} - O^{2-}}\\
|\phantom{..............}|\phantom{...}\\
\ce{_{1-}O^-}\phantom{.........}\ce{_{1-}O^-}\phantom{....}\end{array}\]

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

3)
 x  2-
\[\ce{N O^-_3}\]

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

⇒ x - 6 = -1

⇒ x = +5

Here, there is no fallacy about the O.N. of N in `"NO"_3^(-)`

The structure of `"NO"_3^(-)` is shown as follows

The N atom exhibits the O.N. of +5.