From the given list, the number of compounds with +4 oxidation state of Sulphur ________.
SO₃, H₂ SO₃, SOCl₂, SF₄, BaSO₄, H₂ S₂ O₇

Solution

Chemistry ma’am · 3 steps

Find sulphur's oxidation number in each compound by making the charges add up to zero, then count the +4s.

1. Set up the rule. In a neutral compound all oxidation numbers add to zero. Take O as -2, H as +1, Cl and F as -1, Ba as +2, and let sulphur be x.

   Compound	Equation	Oxidation state of S
   SO₃	x + 3(-2) = 0	+6
   H₂SO₃	2(+1) + x + 3(-2) = 0	+4
   SOCl₂	x + (-2) + 2(-1) = 0	+4
   SF₄	x + 4(-1) = 0	+4
   BaSO₄	(+2) + x + 4(-2) = 0	+6
   H₂S₂O₇	2(+1) + 2x + 7(-2) = 0	+6

2. Two sulphurs. In H₂S₂O₇ there are two S atoms, so 2x = 12 and each sulphur is +6. There are no peroxide links here, so every oxygen is -2.
3. So. Sulphur is +4 in H₂SO₃, SOCl₂ and SF₄, so the answer is 3.

The common slip: 4. Treating SO₃ like the sulphite ion SO₃²⁻ puts S at +4 and gives 4. Neutral SO₃ has no 2- charge, so S is +6.

The idea to keep. Sulphur's common states are -2, +4 and +6. Sulphite-type species, SOCl₂ and SF₄ sit at +4, while sulphates, SO₃ and pyrosulphuric acid sit at +6.

Answer from the NTA final answer key. Solution by PrepWiser.