Question

A, B, C, D, and E are the elements whose atomic numbers are 4, 9, 11, 15, and 18, respectively.

Which two elements can form ions with an electronic configuration of an inert gas neon?

Answer 1

• An ion is termed as an atom or molecule that has lost or gained one or more valence electrons, resulting in a net positive or negative electrical charge.
• Example of ions: Magnesium ion (${\mathrm{Mg}}^{2+}$) and Chloride ion (${\mathrm{Cl}}^{-}$).
• Neon ($\mathrm{Ne}$) has an atomic number of 10. Its electronic configuration is 2, 8.
• According to the periodic table, Beryllium ($\mathrm{Be}$) has an atomic number of 4. Hence, element A is $\mathrm{Be}$. Its electronic configuration is 2, 2.
• Thus, $\mathrm{Be}$ has two electrons in the valence shell and it can easily lose or donate these two electrons to attain a stable noble gas configuration of Helium (He) and forms a positively charged Beryllium ion (${\mathrm{Be}}^{2+}$).
• $$\begin{gathered} \mathrm{Be}\to {\mathrm{Be}}^{2+}+2{\mathrm{e}}^{-} \\ (2,2)\left(2\right) \end{gathered}$$
• According to the periodic table, Fluorine ($\mathrm{F}$) has an atomic number of 9. Hence, element B is $\mathrm{F}$. Its electronic configuration is 2, 7.
• Thus, $\mathrm{F}$ has seven electrons in the valence shell and needs only one electron to attain a stable noble gas configuration of Neon and form a negatively charged Fluoride ion (${\mathrm{F}}^{-}$).
• $$\begin{gathered} \mathrm{F}+{\mathrm{e}}^{-}\to {\mathrm{F}}^{-} \\ (2,7)(2,8) \end{gathered}$$
• According to the periodic table, Sodium ($\mathrm{Na}$) has an atomic number of 11. Hence, element C is $\mathrm{Na}$. Its electronic configuration is 2, 8, 1.
• Thus, $\mathrm{Na}$ has only one electron in the valence shell and it can easily lose or donate its one electron to attain a stable noble gas configuration of Neon and forms a positively charged Sodium ion (${\mathrm{Na}}^{+}$).
• $$\begin{gathered} \mathrm{Na}\to {\mathrm{Na}}^{+}+{\mathrm{e}}^{-} \\ (2,8,1)(2,8) \end{gathered}$$
• According to the periodic table, Phosphorus ($\mathrm{P}$) has an atomic number of 15. Hence, element D is $\mathrm{P}$. Its electronic configuration is 2, 8, 5.
• Thus, $\mathrm{P}$ has five electrons in the valence shell and needs three electrons to attain a stable noble gas configuration of Argon ($\mathrm{Ar}$) and forms a negatively charged Phosphide ion (${\mathrm{P}}^{3-}$).
• $$\begin{gathered} \mathrm{P}+3{\mathrm{e}}^{-}\to {\mathrm{P}}^{3-} \\ (2,8,5)(2,8,8) \end{gathered}$$
• Argon ($\mathrm{Ar}$) has an atomic number of 18. Hence, element E is $\mathrm{Ar}$. Its electronic configuration is 2,8,8.
• Argon has a completely filled outermost or valence shell (complete octet), and therefore, its electronic configuration is highly stable.
• Due to the absence of free electrons in its outermost shell, Argon has zero tendency to accept or lose any electron.
• Hence, the two elements that can form ions with an electronic configuration of an inert gas Neon are B and C.