Question

The oxidation number of nitrogen in \(\left(\mathrm{N}_{2} \mathrm{H}_{5}\right)^{+}\)is (a) \(-2\) (b) \(+2\) (c) \(+3\) (d) \(-3\)

Short answer

The oxidation number of nitrogen in \((\mathrm{N}_{2}\mathrm{H}_{5})^{+}\)is -2.

Step-by-step solution

Understand the concept of oxidation numbers

An oxidation number of an atom is the charge that atom would have if the compound was composed of ions. It's a bookkeeping tool used to keep track of electrons in chemical reactions. In a neutral molecule, the sum of the oxidation numbers must equal zero, while in an ion, it must equal the ion's charge.

Assign known oxidation numbers

Hydrogen normally has an oxidation number of +1, unless it's bonded to metals where it can be -1. In \(\left(\mathrm{N}_{2}\mathrm{H}_{5}\right)^{+}\), hydrogen is not bonded to metal and therefore each hydrogen will have an oxidation number of +1.

Determine the charge of the ion

The entire ion \(\left(\mathrm{N}_{2}\mathrm{H}_{5}\right)^{+}\) has a charge of +1. This will be used to balance the oxidation numbers.

Set up the equation to solve for the oxidation number of nitrogen

Let the oxidation number of nitrogen be x. Since there are two nitrogen atoms, their combined oxidation number will be 2x. The molecule contains five hydrogen atoms, each with an oxidation number of +1, contributing a total of +5 to the oxidation number. The sum of the oxidation numbers must equal the charge of the ion: \[2x + 5(1) = +1\].

Calculate the oxidation number of nitrogen

Solve the equation from step 4 for x: \[2x + 5 = 1\] \[2x = 1 - 5\] \[2x = -4\] \[x = -4/2\] \[x = -2\]. Therefore, the oxidation number of nitrogen is -2.

Key concepts

Balancing Oxidation Numbers

When dealing with redox reactions, you'll find that balancing oxidation numbers is a critical skill. It's akin to balancing your checkbook — just as you ensure that your income and expenses are equal, in redox reactions, you need to make sure that the loss and gain of electrons are balanced. This is because the number of electrons lost by one substance must equal the number of electrons gained by another.

To balance oxidation numbers, identify the changes in oxidation states of the elements involved, then use coefficients to balance the reaction. Remember, the overall charge should remain consistent on both sides of the equation.

For example, in the reaction between hydrogen and oxygen to form water, the oxidation number of oxygen changes from 0 in O₂ to -2 in H₂O, while the oxidation number of hydrogen changes from 0 in H₂ to +1 in H₂O. To balance the oxidation numbers, you would multiply each molecule of oxygen by 2 and each molecule of hydrogen by 1, successfully balancing the equation and ensuring an equal exchange of electrons.

Electron Bookkeeping

Understanding electron bookkeeping, or the accounting of electron transfers, is essential in chemistry, especially in redox reactions. Oxidation numbers serve as a useful tool for tracking these electrons.

Electrons are either lost or gained during chemical reactions, and each loss (oxidation) or gain (reduction) can be recorded using oxidation numbers. Just as bookkeepers must ensure that debits equal credits, chemists use oxidation numbers to verify that all electrons are accounted for.

An important takeaway is that the oxidation number does not necessarily represent a real charge on a molecule but rather a hypothetical one used for tracking purposes. For a molecule to be stable, its oxidation numbers must be balanced. When they aren't, a redox reaction is likely occurring, indicating a transfer of electrons between species.

Oxidation State of Nitrogen

The oxidation state of nitrogen can vary widely, from -3 in ammonia (NH₃) to +5 in nitrates (NO₃^-). This is due to nitrogen's ability to form multiple bonds and its position in the periodic table.

In the given exercise, determining the oxidation state of nitrogen in \(\left(\mathrm{N}_{2}\mathrm{H}_{5}\right)^{+}\) requires that we count all valence electrons and take into account the overall charge of the ion. Following the steps provided, we found out that nitrogen in \(\left(\mathrm{N}_{2}\mathrm{H}_{5}\right)^{+}\) has an oxidation number of -2. It's noteworthy that this isn't the most common oxidation state of nitrogen, but in complex ions or unusual structures, it's not rare to encounter such varying oxidation states.

When dealing with compounds that contain elements like nitrogen which can have multiple oxidation states, it's crucial to follow the steps of assigning and calculating oxidation numbers methodically to ensure the right oxidation state is determined.