Question

One mole of \(\mathrm{N}_{2} \mathrm{H}_{4}\) loses ten moles of electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in \(\mathrm{Y}\) ? (There is no change in the oxidation state of hydrogen). a. \(+2\) b. \(-2\) c. \(+3\) d. \(+4\)

Short answer

The oxidation state of nitrogen in compound \(\mathrm{Y}\) is +3.

Step-by-step solution

Calculate Initial Oxidation State

The compound \(\mathrm{N}_2\mathrm{H}_4\) (hydrazine) contains nitrogen and hydrogen. Assuming the oxidation state of hydrogen is +1, each molecule of nitrogen must balance this: \(2x + 4(+1) = 0\) for charge neutrality, where \(x\) is the oxidation state of nitrogen. Solving gives \(2x + 4 = 0\), so \(2x = -4\) and \(x = -2\). Each nitrogen initially has an oxidation state of \(-2\).

Electrons Lost Per Nitrogen Atom Calculation

The problem states that 10 moles of electrons are lost. For one mole of \(\mathrm{N}_2\mathrm{H}_4\), this means each nitrogen atom loses 5 electrons, as 10 electrons are divided between 2 nitrogen atoms.

Determine New Oxidation State

Starting from an oxidation state of \(-2\), each nitrogen loses 5 electrons: \(-2 + 5 = +3\). Thus, the new oxidation state for each nitrogen in compound \(\mathrm{Y}\) is +3.

Verify Result Against New Compound's Neutrality

If each nitrogen in \(\mathrm{Y}\) is +3, this results must hold for the total nitrogen count without affecting hydrogen oxidation states (unchanged at +1). Our derived oxidation states align with the reaction conditions, conferring correctness upon our process.

Key concepts

Hydrazine

Hydrazine is a chemical compound with the formula \( \mathrm{N}_2\mathrm{H}_4 \). It plays a central role in many chemical reactions due to its composition of nitrogen and hydrogen atoms. Often used as a reducing agent, it can participate in processes that involve the transfer of electrons. In this exercise, hydrazine is used to find the oxidation state of nitrogen in a new compound, \( \mathrm{Y} \). The compound is made up of two nitrogen atoms and four hydrogen atoms.

• The oxidation state of a hydrogen atom is typically \(+1\).
• To maintain charge neutrality, the total oxidation state of hydrazine must equal zero.

By calculating the initial oxidation state of nitrogen in hydrazine, we are able to understand its role and transformation in different chemical reactions.

Electron transfer

During reactions, the concept of electron transfer is pivotal to understanding how compounds change. Electrons are negatively charged subatomic particles. In oxidation-reduction reactions, compounds gain or lose electrons, which changes the oxidation state of the elements involved. In the given problem, one mole of hydrazine loses ten moles of electrons:

• Each nitrogen atom in \( \mathrm{N}_2\mathrm{H}_4 \) starts with an oxidation state of \(-2\).
• The loss of 10 moles of electrons is attributed to both nitrogen atoms, meaning each atom individually loses five electrons.

By losing electrons, nitrogen's charge and thus its oxidation state become more positive, helping to calculate the new state of compound \( \mathrm{Y} \).

Charge neutrality calculations

Charge neutrality calculations ensure that the charge of molecules remains balanced, which is crucial in determining oxidation states. The principle behind charge neutrality is that the sum of the oxidation states in a compound must equal its overall charge.In the exercise with hydrazine:

• Initially, the oxidation state calculation is set by the equation: \(2x + 4(+1) = 0\) where \(x\) is the oxidation state of nitrogen.
• This simplifies to \(x = -2\), showing the initial charge neutrality in hydrazine.

When transforming to a new compound \( \mathrm{Y} \), although hydrogens' oxidation states do not change, the charge neutrality must hold, confirming the calculated oxidation state of nitrogen as \(+3\) post-electron transfer. This ascertains that the charges are balanced correctly considering the partner elements involved.