2 mole of \(\mathrm{N}_{2} \mathrm{H}_{4}\) loses 16 mole of electron is being converted into a new compound (A). Assuming that all the N-appears in A. What is the oxidation state ' \(\mathrm{N}\) ' in \(\mathrm{A}\).

Redox reactions, a shorthand for reduction-oxidation reactions, involve the transfer of electrons between two substances. These reactions are fundamental to chemistry, as they are involved in energy production, corrosion, combustion, and many more processes. In a redox reaction, one element or compound loses electrons (oxidation) while another gains electrons (reduction). The substance that loses electrons is known as the reducing agent, while the one that gains electrons is called the oxidizing agent.

For example, in our exercise, \( \mathrm{N}_{2} \mathrm{H}_{4} \) undergoes oxidation as it loses electrons to form a new compound. Understanding redox reactions is crucial for grasping the concept of electron transfer and its effect on the oxidation state of the elements involved.

The oxidation number, or oxidation state, is a concept that helps us describe the degree of oxidation of an atom in a compound. Determining oxidation numbers is essential for analyzing redox reactions, as it allows us to track the flow of electrons. The oxidation state is a hypothetical charge an atom would have if all bonds to atoms of different elements were 100% ionic.

An atom's oxidation state can change in a chemical reaction, indicating whether it has undergone oxidation or reduction. For instance, in the provided exercise, nitrogen has an initial oxidation state of -2 in \( \mathrm{N}_{2} \mathrm{H}_{4} \). To determine the change in oxidation state during the reaction, we look at the number of electrons lost or gained by the atom. Recognizing that each nitrogen atom loses 4 electrons in the process allows us to conclude an increase in the oxidation state by 4.

Balancing redox equations is a methodical approach to ensure the conservation of mass and charge in a chemical reaction. It's a crucial skill in chemistry as unbalanced equations do not represent the actual reaction. The balance between the total numbers of oxidized and reduced electrons is vital for a correct redox equation. To achieve this, we typically separate the redox reaction into two half-reactions – one for oxidation and one for reduction. Each half-reaction is balanced separately, considering both mass and charge.

In our exercise, we don't have the complete redox reaction but need to know the principles for balancing redox equations. Balancing requires maintaining the same number of atoms for each element and the same total charge on both sides of the equation. By focusing on the electron transfer, we ensure that the same number of electrons is lost in oxidation as is gained in reduction. This ensures the charge balance that is crucial for the reaction to occur physically.