O.N. of S in $SO_4^{2-}$ is:

Question

Accepted Answer

The correct option is (2):

To determine the oxidation number (O.N.) of sulfur (S) in the sulfate ion ($SO_4^{2-}$), we need to apply the rules for assigning oxidation numbers. First, we know that the oxidation number of oxygen is typically -2, except in peroxides or superoxides. In the sulfate ion, there are four oxygen atoms. Therefore, the total contribution from the oxygen atoms is $4 	imes (-2) = -8$. The overall charge of the sulfate ion is -2. The sum of the oxidation numbers of all atoms in a polyatomic ion must equal the charge of the ion. Let the oxidation number of sulfur be $x$. So, we can set up the equation: $x + (-8) = -2$. Solving for $x$, we get $x = -2 + 8$, which means $x = +6$. Thus, the oxidation number of sulfur in the sulfate ion is +6. This high positive oxidation state for sulfur indicates that it has lost many of its valence electrons to the more electronegative oxygen atoms.

The incorrect option is (1):
An oxidation number of 4 would mean that the sulfur atom has a lower positive charge than it actually possesses in the sulfate ion. If the O.N. of S were +4, the total charge would be $4 + 4 \times (-2) = 4 - 8 = -4$, which does not match the actual ion charge of -2.

The incorrect option is (3):
An oxidation number of 8 for sulfur is not typically observed as sulfur's maximum possible oxidation state, being in Group 16, is +6, corresponding to the loss of all its valence electrons (3s2 3p4). An O.N. of +8 would imply sulfur is exceeding its maximum valence.

The incorrect option is (4):
An oxidation number of zero usually indicates a pure elemental substance or a neutral compound where the elements have not gained or lost electrons. In the sulfate ion, sulfur is clearly bonded to more electronegative oxygen atoms and the ion itself carries a charge, so its oxidation number cannot be zero.