The least stable nucleus is:
A) $_2^4He$
B) $_6^{12}C$
C) $_8^{16}O$
D) $_{26}^{56}Fe$

Hint: From the given options, calculate the number of protons and the neutrons. If any one of both these numbers are not magic, then calculate the value of $\left( {\dfrac{n}{p}} \right)$ ratio , if it is equal to $1$ , then it is stable. If not, then it is unstable in nature.

Complete step by step solution:
The atom is said to be stable if it has maximum binding energy to hold the nucleus of the atom. And so the nucleus remains stable permanently. The binding energy is the amount of the energy that is required to separate a particle from the group of particles or the atoms. In order to find whether the atom is stable or not, the magic number is considered.

If the number of the protons or the electrons or the neutrons is equal to the magic number, then the atom is believed to be stable. If the $\left( {\dfrac{n}{p}} \right)$ ratio is equal to $1$, then the atom is considered as the radioactive substance but is stable. The magic numbers are $2,\,8,\,20,\,28,\,50,\,82,\,126$ .

(A) $_2^4He$ has the $p = 2$ and hence it is stable in nature.
(B) $_6^{12}C$ has the $p = 6$ , It is not a magic number. So let us consider $n = 6$ , it is also not a magic number. So it must be radioactive. $\left( {\dfrac{n}{p}} \right) = \dfrac{6}{6} = 1$ ratio is equal to $1$ . Hence it is stable.
(C) $_8^{16}O$ has the $p = 8$ , it is a magic number. Hence it is also stable.
(D) $_{26}^{56}Fe$ has the $p = 26$, It is not a magic number. So let us consider $n = 56$ , it is also not a magic number. So it must be radioactive. $\left( {\dfrac{n}{p}} \right) = \dfrac{{56}}{6} \ne 1$ . Hence it is least stable.

Thus the option (D) is correct.

Note: The unstable atom is also called the radioactive atoms. The unstable nucleus readily deteriorates spontaneously and produces certain radioactive substances since nuclear stability is the ability to stay constant without any spontaneous deterioration to produce any radioactivity.