Explanation:

To determine the oxidation number of nitrogen in compound Y, we need to analyze the given information and apply the rules for assigning oxidation numbers.

Given Information:
- 2 moles of N2H4 loses 20 moles of electrons to form compound Y.
- There is no change in the oxidation number of hydrogen.

Rule for assigning oxidation numbers:
- The oxidation number of an element in a neutral compound is 0.
- The sum of oxidation numbers of all atoms in a compound is equal to the charge of the compound.
- Hydrogen usually has an oxidation number of +1 when bonded to nonmetals.
- The sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion.

Analysis:
1. From the given information, we know that 2 moles of N2H4 lose 20 moles of electrons. This means that each mole of N2H4 loses 10 moles of electrons.
2. Since there is no change in the oxidation number of hydrogen, the oxidation number of hydrogen in N2H4 is +1.
3. Let's assume the oxidation number of nitrogen in compound Y is x.
4. In N2H4, there are 2 moles of nitrogen. Since each nitrogen atom loses 10 moles of electrons, the total electrons lost by nitrogen in N2H4 is 2 * 10 = 20 moles.
5. The total electrons lost by nitrogen must be equal to the total electrons gained by the other element(s) in compound Y.
6. Since there is no change in the oxidation number of hydrogen, the total electrons gained by hydrogen in compound Y is 2 * 10 = 20 moles.
7. Now, we can set up the equation based on the given information:
20 (moles of electrons gained by hydrogen) = 20 (moles of electrons lost by nitrogen)
20 = 2 * x (moles of electrons gained/lost per mole of nitrogen)
x = 10
8. Therefore, the oxidation number of nitrogen in compound Y is +10.

Conclusion:
The oxidation number of nitrogen in compound Y is +10. None of the options provided (a=3, b=-3, c=1, d=5) match with the calculated oxidation number.

Answer is 3