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 identify compounds where Sulphur has an oxidation state of +4, the oxidation state of Sulphur in each compound will be analyzed.
 

• \(SO_3\):
  Oxygen's oxidation state is -2. The equation for \(SO_3\) is \(x + 3(-2) = 0\), yielding \(x = +6\). Sulphur's oxidation state is +6.
• \(H_2SO_3\):
  Hydrogen is +1 and oxygen is -2. The equation is \(2(+1) + x + 3(-2) = 0\), resulting in \(x = +4\).
• \(SOCl_2\):
  Chlorine is -1 and oxygen is -2. The equation is \(x + (-2) + 2(-1) = 0\), giving \(x = +4\).
• \(SF_4\):
  Fluorine's oxidation state is -1. For \(SF_4\), \(x + 4(-1) = 0\) gives \(x = +4\).
• \(BaSO_4\):
  Barium is +2 and oxygen is -2. Thus, \(+2 + x + 4(-2) = 0\), so \(x = +6\).
• \(H_2S_2O_7\):
  Hydrogen is +1 and oxygen is -2. For two Sulphur atoms: \(2(+1) + 2x + 7(-2) = 0\), leading to \(x = +6\) for each Sulphur.

Sulphur is in the +4 oxidation state in the compounds \(H_2SO_3\), \(SOCl_2\), and \(SF_4\). There are 3 such compounds.