Question

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

• A. \(SO_3\)
• B. \(H_2SO_3\)
• C. \(H_2S_2O_7\)
• D. \(BaSO_4\)

Answer 1

The Correct Option is B

To determine which compound has the central atom with a +4 oxidation state, we need to calculate the oxidation state of sulfur in each compound:

1. \(SO_3\):
   • Oxygen has a common oxidation state of -2.
   • Let's assume the oxidation state of sulfur (S) is \(x\).
   • The equation based on the oxidation state is: \(x + 3(-2) = 0\).
   • Solving, \(x - 6 = 0 \rightarrow x = +6\).
   • Hence, in \(SO_3\), sulfur has an oxidation state of +6.
2. \(H_2SO_3\):
   • Hydrogen has an oxidation state of +1.
   • Let's assume the oxidation state of sulfur (S) is \(x\).
   • The equation is \(2(+1) + x + 3(-2) = 0\).
   • Solving, \(2 + x - 6 = 0 \rightarrow x = +4\).
   • Hence, in \(H_2SO_3\), sulfur has an oxidation state of +4.
3. \(H_2S_2O_7\):
   • Oxygen: -2, Hydrogen: +1.
   • Assuming oxidation state of sulfur is \(x\).
   • The equation is \(2(+1) + 2x + 7(-2) = 0\).
   • Solving, \(2 + 2x - 14 = 0 \rightarrow 2x = +12 \rightarrow x = +6\).
   • Thus, in \(H_2S_2O_7\), sulfur has an oxidation state of +6.
4. \(BaSO_4\):
   • Barium (Ba) has a +2 oxidation state.
   • Assuming sulfur has an oxidation state of \(x\).
   • The equation is: \(+2 + x + 4(-2) = 0\).
   • Solving, \(2 + x - 8 = 0 \rightarrow x = +6\).
   • Thus, in \(BaSO_4\), sulfur has an oxidation state of +6.

After evaluating oxidation states, we conclude that the compound where the sulfur has an oxidation state of +4 is \(H_2SO_3\). This makes option \(H_2SO_3\) the correct answer.