During the conversion of \(\mathrm{NH}_{2} \mathrm{OH} \longrightarrow \mathrm{N}_{2} \mathrm{O}\), the equivalent weight of \(\mathrm{NH}_{2} \mathrm{OH}\) is (mol. wt. of \(\mathrm{NH}_{2} \mathrm{OH}\) is \(\mathrm{M}\) ) (1) \(M\) (2) \(\mathrm{M} / 2\) (3) \(\mathrm{M} / 4\) (4) M/5 (5) \(2 \mathrm{M}\)

Oxidation numbers are a way to keep track of electrons in chemical reactions. They help us understand how electrons are gained or lost during a reaction. Each element in a molecule is assigned an oxidation number based on certain rules. For example, in \(\text{NH}_{2} \text{OH}\), hydrogen generally has an oxidation number of +1, while oxygen has an oxidation number of -2. The sum of the oxidation numbers in a neutral molecule must always be zero. Therefore, for \(\text{NH}_{2} \text{OH}\), we get: \(\text{N} + 2(\text{H}(+1)) + \text{O}(-2) + \text{H}(+1) = 0\). Solving this equation, we find that the oxidation number of nitrogen (N) is -1.

Chemical reactions involve the rearrangement of atoms and the transfer of electrons between substances. To understand the conversion of \(\text{NH}_{2} \text{OH} \to \text{N}_{2} \text{O}\), we need to look at the changes in oxidation numbers. In \(\text{N}_{2} \text{O}\), the oxidation number of nitrogen is calculated considering that oxygen’s oxidation number is -2, and the sum of oxidation states in a neutral molecule should be zero. So we have: \(2n - 2 = 0\), which simplifies to \(n = +1\). Thus, each nitrogen in \(\text{N}_{2} \text{O}\) has an average oxidation number of +1. Understanding these changes provides insight into the number of electrons transferred during the reaction.

The molecular weight (also known as molar mass) is the sum of the atomic weights of all the atoms in a molecule. For \(\text{NH}_{2} \text{OH}\), suppose the molecular weight is denoted as \(\text{M}\). When calculating the equivalent weight, we use the molecular weight and the number of moles of electrons transferred. In the given reaction, nitrogen goes from an oxidation state of -1 to +1, implying a transfer of 2 electrons. So, the equivalent weight is the molecular weight divided by the number of moles of electrons transferred. Thus, the equivalent weight of \(\text{NH}_{2} \text{OH}\) in this reaction is \(\frac{\text{M}}{2}\).