Question

One mole of hydrazine $(N_2H_4)$ loses 10 moles of electrons in a reaction to form a new compound $X$. Assuming that all the nitrogen atoms in hydrazine appear in the new compound, what is the oxidation state of nitrogen in $X$? (Note: There is no change in the oxidation state of hydrogen in the reaction)

• A. $-1$
• B. $-3$
• C. $+3$
• D. $+5$
• E. $+1$

Hint:

In oxidation problems:
• Always calculate initial oxidation state first
• Total increase in oxidation state = electrons lost
• Divide change among identical atoms carefully
• Hydrogen often remains +1 unless specified otherwise

Answer 1

The Correct Option is C

Concept: This is a redox-based oxidation state problem. The key idea is:
• Loss of electrons $\Rightarrow$ Oxidation
• Increase in oxidation number $\Rightarrow$ Equal to number of electrons lost
• Total change in oxidation number must equal total electrons lost Also, hydrogen oxidation state remains unchanged, so only nitrogen contributes to the electron loss.

Step 1: Find initial oxidation state of nitrogen in $N_2H_4$ Let oxidation state of each nitrogen atom be $x$. We know: \[ \text{Oxidation state of H} = +1 \] Total molecule is neutral: \[ 2x + 4(+1) = 0 \] \[ 2x + 4 = 0 \] \[ 2x = -4 \] \[ x = -2 \] Thus, initial oxidation state of each nitrogen = $-2$

Step 2: Understand electron loss Given: \[ \text{1 mole of } N_2H_4 \text{ loses 10 moles of electrons} \] This means: \[ \text{Total increase in oxidation number} = +10 \]

Step 3: Distribute oxidation change among nitrogen atoms Each molecule contains: \[ 2 \text{ nitrogen atoms} \] So increase per nitrogen atom: \[ \frac{10}{2} = 5 \]

Step 4: Calculate final oxidation state Initial oxidation state = $-2$ Increase = $+5$ \[ \text{Final oxidation state} = -2 + 5 = +3 \]

Step 5: Verify logic physically
• Nitrogen is oxidized (loses electrons)
• Oxidation state increases from negative to positive
• Final value $+3$ is chemically reasonable (seen in compounds like $HNO_2$) Final Answer: \[ \boxed{+3} \]