Which of the following is an oxidation-reduction reaction?

A. [tex]ZnS (s) + 2 O_2 (g) \longrightarrow ZnSO_4 (s)[/tex]

B. [tex]CaO (s) + H_2 O (l) \longrightarrow Ca(OH)_2 (s)[/tex]

C. [tex]6 Li_2 O (s) + P_4 O_{10} (g) \longrightarrow 4 Li_3 PO_4 (s)[/tex]

D. [tex]SO_2 (g) + H_2 O (l) \longrightarrow H_2 SO_3 (aq)[/tex]

To determine which of the given reactions is an oxidation-reduction (redox) reaction, we need to analyze the changes in the oxidation states of the elements involved in each reaction. Let’s look at each reaction individually and determine if there is any change in oxidation states, which is a hallmark of redox reactions.

### Reaction 1
[tex]\[ \text{ZnS (s) + 2 O}_2 \text{(g) } \rightarrow \text{ ZnSO}_4 \text{(s)} \][/tex]

- Reactants:
- $ \text{Zn} $ in $ \text{ZnS} $: Oxidation state = +2
- $ \text{S} $ in $ \text{ZnS} $: Oxidation state = -2
- $ \text{O}_2 $: Oxidation state = 0 (elemental form)

- Products:
- $ \text{Zn} $ in $ \text{ZnSO}_4 $: Oxidation state = +2
- $ \text{S} $ in $ \text{ZnSO}_4 $: Oxidation state = +6
- $ \text{O} $ in $ \text{ZnSO}_4 $: Oxidation state = -2

In this reaction, sulfur's oxidation state changes from -2 to +6, and oxygen's oxidation state changes from 0 to -2. This indicates that both oxidation (increase in oxidation state) and reduction (decrease in oxidation state) are occurring, so this is a redox reaction.

### Reaction 2
[tex]\[ \text{CaO (s) + H}_2\text{O (l) } \rightarrow \text{ Ca(OH)}_2 \text{(s)} \][/tex]

- Reactants:
- $ \text{Ca} $ in $ \text{CaO} $: Oxidation state = +2
- $ \text{O} $ in $ \text{CaO} $: Oxidation state = -2
- $ \text{H} $ in $ \text{H}_2\text{O} $: Oxidation state = +1
- $ \text{O} $ in $ \text{H}_2\text{O} $: Oxidation state = -2

- Products:
- $ \text{Ca} $ in $ \text{Ca(OH)}_2 $: Oxidation state = +2
- $ \text{O} $ in $ \text{Ca(OH)}_2 $: Oxidation state = -2
- $ \text{H} $ in $ \text{Ca(OH)}_2 $: Oxidation state = +1

The oxidation states of all elements remain the same in both reactants and products, indicating no redox reaction.

### Reaction 3
[tex]\[ 6 \text{Li}_2\text{O (s) + P}_4\text{O}_{10} \text{(g) } \rightarrow 4 \text{Li}_3\text{PO}_4 \text{(s)} \][/tex]

- Reactants:
- $ \text{Li} $ in $ \text{Li}_2\text{O} $: Oxidation state = +1
- $ \text{O} $ in $ \text{Li}_2\text{O} $: Oxidation state = -2
- $ \text{P} $ in $ \text{P}_4\text{O}_{10} $: Oxidation state = +5
- $ \text{O} $ in $ \text{P}_4\text{O}_{10} $: Oxidation state = -2

- Products:
- $ \text{Li} $ in $ \text{Li}_3\text{PO}_4 $: Oxidation state = +1
- $ \text{P} $ in $ \text{Li}_3\text{PO}_4 $: Oxidation state = +5
- $ \text{O} $ in $ \text{Li}_3\text{PO}_4 $: Oxidation state = -2

The oxidation states of all elements remain the same in both reactants and products, indicating no redox reaction.

### Reaction 4
[tex]\[ \text{SO}_2 \text{(g) + H}_2\text{O (l) } \rightarrow \text{ H}_2\text{SO}_3 \text{(aq)} \][/tex]

- Reactants:
- $ \text{S} $ in $ \text{SO}_2 $: Oxidation state = +4
- $ \text{O} $ in $ \text{SO}_2 $: Oxidation state = -2
- $ \text{H} $ in $ \text{H}_2\text{O} $: Oxidation state = +1
- $ \text{O} $ in $ \text{H}_2\text{O} $: Oxidation state = -2

- Products:
- $ \text{S} $ in $ \text{H}_2\text{SO}_3 $: Oxidation state = +4
- $ \text{O} $ in $ \text{H}_2\text{SO}_3 $: Oxidation state = -2
- $ \text{H} $ in $ \text{H}_2\text{SO}_3 $: Oxidation state = +1

The oxidation states of all elements remain the same in both reactants and products, indicating no redox reaction.

### Conclusion
After analyzing the oxidation states for each reaction, the only reaction with changes in oxidation states, indicating it is a redox reaction, is:

[tex]\[ \text{ZnS (s) + 2 O}_2 \text{(g) } \rightarrow \text{ ZnSO}_4 \text{(s)} \][/tex]

Thus, the redox reaction is the first one, with the index [tex]$ 0 $[/tex].