From the given list, the number of compounds with +4 oxidation state of Sulphur_____.
\(SO_3, H_2SO_3, SOCl_2, SF_4, BaSO_4, H_2S_2O_7\).
\(SO_3, H_2SO_3, SOCl_2, SF_4, BaSO_4, H_2S_2O_7\).
Correct Answer: 3
Approach Solution - 1
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.
Approach Solution -2
The compounds with +4 oxidation state of Sulfur are H2SO3, SOCl2, and SF4, giving a total of 3.
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