Question

The nature of bonding in $\text{ CC}{{\text{l}}_{\text{4}}}\text{ }$ and $\text{ Ca}{{\text{H}}_{\text{2}}}\text{ }$ is respectively
A) Ionic, Covalent
B) Covalent in both
C) Covalent, Ionic
D) Polar covalent in both

Answer 1

Hint: the atoms which have the low ionization enthalpy, readily form the ionic bond. The atoms which cannot easily donate or accept the electron to complete the octet prefer to form a covalent bond. The alkaline earth metal can easily donate its electrons, but the carbon cannot donate its 4 electrons to complete its octet.

Complete step by step answer:
The covalent bond is formed by the sharing of electrons between the participating atoms.
The atoms which have high ionization energy cannot transfer the electrons. The elements which have low electronegativity cannot accept the electron. Thus atoms prefer to share a pair of electrons so that both the atoms attain the nearest Noble gas configuration and achieve stability.
The bond which is produced due to the sharing of electrons is called the covalent bond.
Let's consider a carbon tetrachloride$\text{ CC}{{\text{l}}_{\text{4}}}\text{ }$. The electronic configuration of carbon is:
$\text{ C = 1}{{\text{s}}^{\text{2}}}\text{ 2}{{\text{s}}^{\text{2}}}\text{ 2}{{\text{p}}^{\text{2}}}$
And the electronic configuration of chlorine is as follows:
$\text{ Cl = }\left[ \text{Ne} \right]\text{ 3}{{\text{s}}^{\text{2}}}\text{ 3}{{\text{p}}^{5}}$
The carbon requires 4 more electrons to attain the nearest Noble gas configuration of$\text{ Ne }$. The carbon cannot donate the 4 electrons and it cannot accept the 4 electrons as it requires the higher energy to accommodate the electrons. The carbon can share its 4 electrons with the 4 chlorine atoms.
The chlorine atom requires one electron to complete its octet, thus it can accept the electron or share its electron with the atoms.
Since , carbon only forms the covalent bond, thus the four chlorine atoms and carbon atom share their pair of electrons to form a $\text{ CC}{{\text{l}}_{\text{4}}}\text{ }$.
Here, the $\text{ CC}{{\text{l}}_{\text{4}}}\text{ }$form the covalent bond.
The atoms which have a great electronegativity difference can complete their octet by losing and accepting an electron to complete the octet configuration. The ionic bond is defined as the electrostatic force of attraction which holds the oppositely charged ions together.
Let's have a look at the calcium hydride$\text{ Ca}{{\text{H}}_{\text{2}}}\text{ }$. The electronic configuration of calcium is as follows:
$\text{ Ca = }\left[ \text{Ar} \right]\text{ 4}{{\text{s}}^{\text{2}}}$
Calcium is an alkaline earth metal. The calcium can donate its two electrons to attain the nearest Noble gas configuration of$\text{Ar}$.
The electronic configuration of hydrogen is as follows,
$\text{ H = 1}{{\text{s}}^{\text{1}}}$
To attain the nearest noble gas configuration of $\text{ He }$ , the H requires an electron. Each calcium loses two electrons to form a $\text{ C}{{\text{a}}^{\text{2+}}}\text{ }$which can be can be captured by the two hydrogen atom to form hydride $\text{ }{{\text{H}}^{-}}\text{ }$.Such that each atom completes its octet.
Since one atom loses their electrons and the other accepts the electrons, the bond formed between the calcium and hydrogen is ionic. In solution, the calcium hydride $\text{ Ca}{{\text{H}}_{\text{2}}}\text{ }$dissociates into the $\text{ C}{{\text{a}}^{\text{2+}}}\text{ }$ and hydride$\text{ }{{\text{H}}^{-}}\text{ }$.
$\text{ Ca}{{\text{H}}_{\text{2}}}\text{ }\to \text{ C}{{\text{a}}^{\text{2+}}}\text{ + 2 }{{\text{H}}^{-}}\text{ }$
Here, the $\text{ Ca}{{\text{H}}_{\text{2}}}\text{ }$has an ionic bond.

Hence, (C) is the correct option.

Note: the atoms which have the lower ionization enthalpy readily form the ionic bond. However, the atoms which have higher ionisation enthalpy like carbon cannot form an ionic bond, instead of that, the atoms prefer to form a bond by the sharing of electrons. The atoms which have small electronegativity difference form a covalent bond. For example electronegativity of carbon $\text{ 2}\text{.55 }$ and chlorine is $\text{ 3}\text{.16 }$ and the difference is$\text{ 0}\text{.61 }$. If large electronegativity difference forms ionic bond for example, Calcium $\text{ 1}\text{.0 }$ and electronegativity of hydrogen are $\text{ 2}\text{.2 }$ and the difference is $\text{ 1}\text{.2}$ .