Question

An element M has the atomic number 9 and atomic mass 17. Its ion will be represented by:
A) M
B) $M^{2+}$
$\mathrm{C}) \mathrm{M}^{-}$
$\mathrm{D}) \mathrm{M}^{2-}$

Answer 1

When a neutral atom converts into ion, it does so in order to achieve stable configuration. Thus, it either loses or gains electrons so that it can achieve the noble gas configuration.

Complete step by step answer:
-An ion is formed when an electron gets added or removed from an element. When an electron gets removed from the element, it results in the formation of a positive ion known as cation and when electron gets added to an element, it results in the formation of a negative ion known as anion.
-Atomic number is defined as the number of electrons or number of protons present in an element.
-Now, let us discuss the atomic number of the element M.
Number of electrons in the given element M = 9
Electronic configuration for this element = $1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}^{5}$
-Now, we know that the most stable noble gas configuration corresponds to the atomic configuration as having 2 or 8 electrons in the last shell of electrons.
-Thus, the atom either loses or gains electrons to achieve this configuration.
Thus, since the electronic configuration of M is 2,7.
-It gains one electron to become an anion so that the noble gas configuration is achieved.
$\mathrm{M}+\mathrm{e}^{-} \rightarrow \mathrm{M}^{-}$

So, the correct answer is “Option C”.

Note: Note that an atom loses/gains electrons to achieve this noble gas configuration. But , this also depends on the energy of the atom. Suppose, for element M if it loses 7 electrons and becomes the noble gas configuration it requires a huge amount of energy to lose these electrons. But, rather it will gain one electron because the energy absorption is less. So, if an atom wants to become stable, it always tries to release/absorb the minimum amount of energy to achieve stability.