#### Question

Calculate the oxidation number of sulphur, chromium and nitrogen in H_{2}SO_{5}, `Cr_2O_7^(2-)` and `NO_3^-`. Suggest structure of these compounds. Count for the fallacy.

#### Solution 1

O.N. of S in H_{2}SO_{5}. By conventional method, the O.N. of S in H_{2}SO_{5} is 2 (+1) + x + 5 (-2) = 0 or x = +8 This is impossible because the maximum O.N. of S cannot be more than six since it has only six electrons in the valence shell. This fallacy is overcome if we calculate the O.N. of S by chemical bonding method. The structure of H_{2}SO_{5 }is

2 x (+1) + x + 2(-1) + 3 x (-2) = 0 or x = +6

(for H) (for S) for (O-O) (for other O atoms

`Cr in Cr_2O_7^(2-)`

2(x) + (-2xx7) = -2

2x - 14 = -2

2x = -2 + 14 x = +6

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

(for `O^-`) (for = O) for ->O

Thus, there is no fallacy about the O.N. of N in N0_{3}^{–}whether one calculates by conventional method or by chemical bonding method.

#### Solution 2

i)

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 H_{2}SO_{5} 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)

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

=> 2x - 14 = -2

=> x = +6

Here, there is no fallacy about the O.N. of Cr in `Cr_2O_7^(2-)`.

The structure of `Cr_2O_7^(2-)` is shown as follows:

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

iii)

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.