Balancing A Redox Reaction

We are given two reactions, and we are to determine the major products X and Y. The reaction types can be identified as follows:
Step 1: The reaction between (sodium acetate) and (bromine-substituted ethene) leads to the substitution of the bromine atom, resulting in an alkylated product: . Thus, product X is .
Step 2: The second reaction involves the addition of a strong base or another suitable agent to break the ester bond and form the final product Y, which is . Thus, the major products X and Y are and , corresponding to option (4).

Step 1: Oxidation with KMnO₄ - The given starting compound is a dicarboxylic acid. - Treatment with and under heat leads to oxidative cleavage, forming a cyclic ketone.
Step 2: Bromination with under UV light - The cyclohexanone undergoes bromination via radical substitution at the allylic position, leading to the formation of bromo-cyclohexene.

1. Oxidation state of Vanadium (x) in Rb₄Na[HV₁₀O₂₈]:
Let the oxidation state of V be .
Oxidation state of Rb is +1 and Na is +1.
Oxidation state of O is -2.
The compound is neutral, so the sum of oxidation states is zero.
However, vanadium in this polyoxometalate is usually in the +5 oxidation state. Let's verify: Thus, .

2. Oxidation state of Chlorine (y) in Ca(ClO₂)₂:
Let the oxidation state of Cl be .
Oxidation state of Ca is +2.
Oxidation state of O is -2.
The compound is neutral, so the sum of oxidation states is zero.
Thus, .

3. Sum of Oxidation States (x + y):
Therefore, the sum of and is 8.

Final Answer: 8.

Step 1: Determine the initial oxidation states. In , hydrogen has an oxidation state of +1, and oxygen has -2. In , fluorine is 0 because it is a diatomic molecule.
Step 2: Determine the final oxidation states. In , hydrogen remains at +1, and fluorine is -1. In , hydrogen is +1, oxygen is -2 (like in water), and fluorine is -1.
Step 3: Analyze the changes in oxidation states. Hydrogen's oxidation state does not change (+1 in all compounds). Fluorine's oxidation state changes from 0 in to -1 in and , indicating reduction. Oxygen's oxidation state does not change in this reaction.
Step 4: Conclusion. Since fluorine's oxidation state decreases, it is reduced. Oxygen and hydrogen do not undergo any redox changes in terms of oxidation states in this particular reaction.