Question

From the given list, the number of compounds with +4 oxidation state of Sulphur_____.
\(SO_3​, H_2​SO_3​, SOCl_2​, SF_4​, BaSO_4​, H_2​S_2​O_7\)​.

Answer 1

Correct Answer: 3

To determine the number of compounds where Sulphur exhibits a +4 oxidation state, let's examine the oxidation state of Sulphur in each compound. 
 

• \(SO_3\):
  The oxidation state of O is -2. For \(SO_3\), the equation is \(x + 3(-2) = 0\), giving \(x = +6\). Therefore, Sulphur is in the +6 state.
• \(H_2SO_3\):
  H has an oxidation state of +1 and O is -2. The total is \(2(+1) + x + 3(-2) = 0\), so \(x = +4\).
• \(SOCl_2\):
  Cl has an oxidation state of -1 and O is -2. The equation is \(x + (-2) + 2(-1) = 0\), giving \(x = +4\).
• \(SF_4\):
  F has an oxidation state of -1. For \(SF_4\), \(x + 4(-1) = 0\) results in \(x = +4\).
• \(BaSO_4\):
  Ba is +2 and O is -2. Thus, \(+2 + x + 4(-2) = 0\), so \(x = +6\).
• \(H_2S_2O_7\):
  H is +1 and O is -2. Split into two Sulphur atoms: \(2(+1) + 2x + 7(-2) = 0\), leading to \(x = +6\) for each Sulphur.

The compounds with Sulphur in the +4 oxidation state are \(H_2SO_3\), \(SOCl_2\), and \(SF_4\). Hence, the number of such compounds is 3.

Answer 2

The compounds with +4 oxidation state of Sulfur are H₂SO₃, SOCl₂, and SF₄, giving a total of 3.