In which of the following compound central atom has +4 oxidation state?

To find the compound in which the central sulfur (S) atom has an oxidation state of +4, we can determine the oxidation state of sulfur in each of the given options by applying the rules for assigning oxidation numbers:

1. Hydrogen (H) generally has an oxidation state of +1 when bonded to non-metals.
2. Oxygen (O) generally has an oxidation state of -2 in its compounds.
3. Barium (Ba) is an alkaline earth metal (Group 2) and always has an oxidation state of +2 in its compounds.
4. The sum of the oxidation states of all atoms in a neutral molecule is equal to 0.

Let's calculate the oxidation state of the central sulfur atom (S), represented by $x$, in each compound:

1. ${\mathrm{SO}}_3$ (Sulfur trioxide):
The molecule is neutral, so the sum of oxidation numbers is 0:
$x+3(-2)=0$
$x-6=0$
$x=+6$
Here, S is in the +6 oxidation state.

2. $H_2{\mathrm{SO}}_3$ (Sulfurous acid):
The molecule is neutral, so the sum of oxidation numbers is 0:
$2(+1)+x+3(-2)=0$
$2+x-6=0$
$x-4=0$
$x=+4$
Here, S is in the +4 oxidation state. This matches our required condition.

3. $H_2{S}_2{O}_7$ (Pyrosulfuric acid / Oleum):
The molecule is neutral, so the sum of oxidation numbers is 0:
$2(+1)+2x+7(-2)=0$
$2+2x-14=0$
$2x-12=0$
$2x=12$
$x=+6$
Here, S is in the +6 oxidation state.

4. ${\mathrm{BaSO}}_4$ (Barium sulfate):
The molecule is neutral, so the sum of oxidation numbers is 0:
$(+2)+x+4(-2)=0$
$2+x-8=0$
$x-6=0$
$x=+6$
Here, S is in the +6 oxidation state.

Thus, the sulfur atom has a +4 oxidation state in $H_2{\mathrm{SO}}_3$.