Question

Elements X, Y, and Z have atomic numbers 6, 9, and 12, respectively. Which one forms:

(a) an anion? (b) a cation?

Answer 1

(a)

• Anions are negatively charged species that are formed by the gain of electrons by any element or molecule. Example, Chloride ion (${\mathrm{Cl}}^{-}$) is the anion formed by the gain of one electron by the element Chlorine ($\mathrm{Cl}$).
• $\mathrm{Cl}+{\mathrm{e}}^{-}\to {\mathrm{Cl}}^{-}$
• The atomic number of element $\mathrm{X}$ is 6 and its electronic configuration is 2,4.
• The element $\mathrm{X}$ has four electrons in its outermost or valence shell.
• Since, it is difficult to lose or gain four electrons, $\mathrm{X}$ will share its four electrons to attain the stable noble gas configuration. Hence, it neither forms an anion nor cation.
• The atomic number of element $\mathrm{Y}$ is 9 and its electronic configuration is 2,7.
• The element $\mathrm{Y}$ has seven electrons in its outermost or valence shell, so it needs only one more electron to attain a stable noble gas configuration (eight electrons in the outermost shell).
• Thus, the element $\mathrm{Y}$ gains one electron and form a ${\mathrm{Y}}^{-}$ ion, and hence, the element $\mathrm{Y}$will form an anion.
• $\mathrm{Y}+{\mathrm{e}}^{-}\to {\mathrm{Y}}^{-}$
• The atomic number of element $\mathrm{Z}$ is 12 and its electronic configuration is 2,8,2.
• Cations are positively charged species that are formed by the loss of electrons by any element or molecule. Example, Sodium ion (${\mathrm{Na}}^{+}$) is the cation formed by the loss of one electron from the element Sodium ($\mathrm{Na}$).
• $\mathrm{Na}\to {\mathrm{Na}}^{+}+{\mathrm{e}}^{-}$
• The element $\mathrm{Z}$ has two electrons in its outermost or valence shell, so it can easily donate these two electrons and attain a stable noble gas configuration (eight electrons in the outermost shell).
• Thus, the element $\mathrm{Z}$ loses two electrons and forms a $Z^{2+}$ ion, and hence, the element $\mathrm{Z}$ will form a cation.
• $\mathrm{Z}\to {\mathrm{Z}}^{2+}+2{\mathrm{e}}^{-}$

(b)

• Since, the element $\mathrm{Z}$ loses two electrons and forms a $Z^{2+}$ ion, and hence, the element $\mathrm{Z}$ will form a cation.
• $\mathrm{Z}\to {\mathrm{Z}}^{2+}+2{\mathrm{e}}^{-}$