In the chemical reaction, the equation of which is 2KI + SO3 = K2SO3 + I2, the oxidizing agent is

Analysis:

The given reaction is:

\[ 2KI + SO_3 = K_2SO_3 + I_2 \]

To identify the oxidizing agent, we need to determine the oxidation states of the elements involved before and after the reaction.

Before the reaction:

• In KI: K is +1, I is -1.
• In SO3: O is -2, so S is +6 (S + 3*(-2) = 0 => S = +6).

After the reaction:

• In K2SO3: K is +1, O is -2, so S is +4 (2*(+1) + S + 3*(-2) = 0 => S = +4).
• In I2: The oxidation state of an element in its elemental form is 0.

Now let's see which element's oxidation state decreased (indicating reduction, and thus it is part of the oxidizing agent):

• Potassium (K): +1 to +1 (no change).
• Sulfur (S): +6 to +4 (decreased, so sulfur is reduced).
• Iodine (I): -1 to 0 (increased, so iodine is oxidized).
• Oxygen (O): -2 to -2 (no change).

Since sulfur in SO3 is reduced (from +6 to +4), SO3 acts as the oxidizing agent. The question asks for the oxidizing agent. Among the options, we need to find the species or element that is reduced or contains the element that is reduced.

Let's check the options:

• 1) I- in potassium iodide: Iodine has an oxidation state of -1. It is oxidized to I2 (0). So, I- is the reducing agent, not the oxidizing agent.
• 2) O-2 in sulfur(VI) oxide: Oxygen has an oxidation state of -2. Its oxidation state does not change.
• 3) K+1 in potassium iodide: Potassium has an oxidation state of +1. Its oxidation state does not change.
• 4) S+6 in sulfur(VI) oxide: Sulfur has an oxidation state of +6. It is reduced to +4 in K2SO3. Therefore, sulfur in SO3 is the element that gets reduced, making SO3 the oxidizing agent.

The question asks for the oxidizing agent. In the reaction, SO3 accepts electrons from KI, and the sulfur atom within SO3 is reduced. Thus, SO3 is the oxidizing agent.

Answer: 4