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1: (a) - 0, +6 and -2

In S₈, sulfur atoms share electrons with each other resulting in an oxidation state of 0. In SO₃, sulfur has shared its electrons with oxygen atoms to achieve a +6 oxidation state. In H₂S, sulfur has lost electrons to hydrogen, resulting in a -2 oxidation state.

2: (a) - 2/3

In Na₂SO₄, there are two Na⁺ cations and one SO₄²⁻ anion. To maintain electrical neutrality, the oxidation state of sulfur in the sulfate ion (SO₄²⁻) is +6. The average oxidation state of sulfur in Na₂SO₄ is then calculated by dividing the total charge of sulfur (+6) by the number of sulfur atoms (1), resulting in +6. Since there are 2 Na⁺ cations with a charge of +1 each, we need a combined charge of +2 to balance the sulfate ion. This is achieved by dividing +6 (oxidation state of sulfur) by 3 (number of Na atoms), giving an average oxidation state of sulfur in Na₂SO₄ as 2/3.

3: (c) - act as oxidising as well as reducing agents

Sulphur oxidation state can vary from -2 to +6. In the given compound, the oxidation state of Sulphur is +4, It can be reduced to -2 or get oxidized to +6. So, Sulphur can act as oxidizing as well as reducing agents.

4: (d) - No d-orbitals in its valence shell

Oxygen's electronic configuration limits its ability to achieve +4 and +6 oxidation states. While the physical state (gas vs. solid) doesn't play a role, oxygen's high electron affinity and lack of d-orbitals in its valence shell contribute to this limitation.

5: (a) - H₂O

In H₂O, oxygen shares its electrons with hydrogen atoms, resulting in an oxidation state of -2.