Oxidation number of Sulphur in $H_{2}SO_{4}$ is:

Question

Accepted Answer

The correct option is (4):

To determine the oxidation number of Sulfur (S) in $\mathrm{H_2SO_4}$, we need to apply the rules for assigning oxidation states. In a neutral compound, the sum of the oxidation numbers of all atoms must be zero. We know the standard oxidation states for Hydrogen (H) and Oxygen (O) in most compounds. Hydrogen typically has an oxidation state of $+1$ and Oxygen typically has an oxidation state of $-2$.

Let the oxidation number of Sulfur be $x$.
For $\mathrm{H\_2SO\_4}$:
There are 2 Hydrogen atoms, each with an oxidation state of $+1$. So, the contribution from Hydrogen is $2 	imes (+1) = +2$.
There are 4 Oxygen atoms, each with an oxidation state of $-2$. So, the contribution from Oxygen is $4 	imes (-2) = -8$.

Now, we set up the equation for the sum of oxidation numbers:
$(2 	imes (+1)) + x + (4 	imes (-2)) = 0$
$2 + x - 8 = 0$
$x - 6 = 0$
$x = +6$

Therefore, the oxidation number of Sulfur in $\mathrm{H_2SO_4}$ is $+6$.

The incorrect option is (1):
An oxidation state of $+4$ for sulfur in $\mathrm{H\_2SO\_4}$ would mean the sum of oxidation states is $2(+1) + (+4) + 4(-2) = 2 + 4 - 8 = -2$, which is not zero for a neutral molecule. This value is characteristic of sulfur dioxide or sulfurous acid.

The incorrect option is (2):
An oxidation state of $+3$ for sulfur in $\mathrm{H\_2SO\_4}$ would result in a total charge of $2(+1) + (+3) + 4(-2) = 2 + 3 - 8 = -3$, which is incorrect for a neutral compound. Sulfur does not commonly exhibit a $+3$ oxidation state in stable oxyacids.

The incorrect option is (3):
An oxidation state of $+5$ for sulfur in $\mathrm{H\_2SO\_4}$ would lead to a total charge of $2(+1) + (+5) + 4(-2) = 2 + 5 - 8 = -1$, which is incorrect for a neutral molecule. A $+5$ oxidation state is rarely observed for sulfur and is not found in sulfuric acid.