Question

How many cyanide ions do you need to balance an ammonium ion?

Answer 1

Hint: Ammonium ion is represented as $N{{H}_{4}}^{+}$and cyanide ion is represented as $C{{N}^{-}}$. Both have opposite charges, and to balance them we need to firstly understand the amount of charge each carries and then figure out how many ions or atoms are needed to make it neutral.

Complete step-by-step answer:Firstly, in this question it is not mentioned that what compound exactly is needed to be formed after the balancing of both the atoms.
So, we assume here the simplest compound that can be formed after balancing both the opposite charges. For example, let us assume that Ammonium cyanide, the simplest salt formed after the combination of given two ions, is to be formed.
Firstly, let us see the charges in them
Ammonium ion $\left (N{{H}{4}}^{+} \right)$ is formed when an ammonia molecule reacts with hydrogen ion (which has a charge of +1). As a result, the ammonium ion develops a unit positive charge $\left( +1 \right)$.
In Cyanide ion $\left( C{{N}^{-}} \right)$ , the carbon produces a complete negative charge (-1), as well as a lone pair of electrons.
Hence, now it is clear to us that ammonium ion has $\left( +1 \right)$charge and cyanide ion has $\left( -1 \right)$charge. Since they have equal and opposite charges, they cancel each other, and the compound Ammonium cyanide thus formed is balanced and neutral.

Therefore, we can say that one ammonium ion is required to balance one cyanide ion. Both when combined form a neutral and stable salt.

Note:In this question we assume that both ions react to form ammonium cyanide and hence one molecule of each was sufficient to balance each other. However, different reactions and products so formed would require different numbers of both the ions to balance each other.