Question
Explain the formation of ionic bonds in calcium chloride $(CaCl_2).$

Answer

Formation of calcium chloride $(CaCl_2):$
$i.$ The electronic configurations of calcium and chlorine are:
$Na (Z = 11): 1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^2 \ or  \ (2, 8, 8, 2)$
$Cl (Z = 17): 1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^5 \ or \ (2, 8, 7)$
$ii.$ Calcium has two electrons in its valence shell. It has tendency to lose two electrons to acquire the electronic configuration of the nearest inert gas, argon $(2, 8, 8).$
$iii.$ Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon $(2, 8, 8).$
$iv.$ During the combination of calcium and chlorine atoms, the calcium atom transfers its valence electrons to two chlorine atoms.
$v.$ Calcium atom changes into $Ca^{2+}$ ion while the two chlorine atoms change into two $Cl^–$ ions. These ions are held together by strong electrostatic force of attraction.
$vi.$ The formation of ionic bond$(s)$ between $Ca$ and $Cl$ can be shown as follows:
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