A nuclear decay is possible if the mass of the parent nucleus exceeds the total mass of the decay particles. If $M(A\mathit{,}Z\mathit{)}$ denotes the mass of a single neutral atom of an element with mass number $A$ and atomic number $Z$, then the minimal condition that the $\beta$ decay

${\mathrm{X}}_{\mathrm{Z}}^{\mathrm{A}}\to {\mathrm{Y}}_{\mathrm{Z}+1}^{\mathrm{A}}+{\beta }^{-}+{\overline{\mathrm{v}}}_{\mathrm{e}}$

will occur is $\left({\mathrm{m}}_{\mathrm{a}}\right.$, denotes the mass of the $\beta$ particle and the neutrino mass $m_{\mathrm{y}}$ can be neglected) :

Consider the following statements
$\left(\mathrm{i}\right)$ All isotopes of an element have the same number of neutrons
$\left(\mathrm{ii}\right)$ Only one isotope of an element can be stable and non-radioactive
$\left(\mathrm{iii}\right)$ All elements have isotopes
$\left(\mathrm{iv}\right)$ All isotopes of Carbon can form chemical compounds with Oxygen- $16$ 

The correct option regarding an isotope is

Consider the following statements ($\mathrm{X}$ and $\mathrm{Y}$ stand for two different elements)
I. ${}_{32} {\mathrm{X}}^{65}$ and ${}_{33}\mathrm{Y}^{65}$ are isotopes.
II. ${}_{42}\mathrm{X}^{96}$ and ${}_{42}\mathrm{Y}^{25 }$ are isotopes.
III. ${}_{85}\mathrm{X}^{174}$ and ${}_{88}\mathrm{Y}^{177}$ have the same number of neutrons.
IV. ${}_{92}\mathrm{X}^{235}$ and ${}_{94}\mathrm{Y}^{235}$ are isobars.

The correct statements are