Question: Indicate the nature of binding in \[CC{l_4}\] and \[Ca{H_2}\]. A.Covalent in \[CC{l_4}\] and elect...

Question

Indicate the nature of binding in $CC{l_4}$ and $Ca{H_2}$ .
A.Covalent in $CC{l_4}$ and electro covalent in $Ca{H_2}$
B.Electro covalent in both $CC{l_4}$ and $Ca{H_2}$
C.Covalent in both $CC{l_4}$ and $Ca{H_2}$
D.Electro covalent in $CC{l_4}$ and covalent in $Ca{H_2}$

Answer 1

Solution

We have to know that the carbon tetrachloride is a chemical formula having the molecular formula, $CC{l_4}$ . And it is a liquid compound which does not have any colour and it has a sweet smell. The carbon tetrachloride contains one carbon atom and four chlorine atoms. And the calcium hydride is a chemical compound having the formula, $Ca{H_2}$ . An ionic bond is present between the C and H atom with a linear structure.

Complete answer:
We have to know that the carbon tetrachloride is a covalent compound. And there are four non – polar covalent bonds are present between the chlorine and carbon. The number of valence electrons present in carbon is equal to four and these four electrons participate in the formation of bonds. And a single electron of each Cl atom participates in the formation of bonds. Hence, the total number of electrons is equal to 8 and it makes the bonds. But others are non – bonding pairs of electrons. Therefore, total number of non – bonding electron is equal to $24$ or number of line pairs of electrons present in carbon tetrachloride is equal to $12$ . Here the carbon only creates covalent bond. Hence, the nature of binding in $CC{l_4}$ is covalent bond.
In calcium hydride, there is one calcium atom and two hydrogen atoms. And there is a formation of two ionic bonds between $C{a^{2 + }}$ and hydrogen ions. Since, one atom will remove their electron and other will gain that electrons and there is a formation of ionic bond between calcium and hydrogen. The calcium hydride is dissociated into $C{a^{2 + }}$ and ${H^ - }$ ions in the solution. And the bonding nature of binding in calcium hydride is electron covalent bond. Hence, option (A) is correct.
The nature of binding in $CC{l_4}$ and $Ca{H_2}$ is not electro covalent. Hence, option (B) is incorrect.
The nature of binding in $CC{l_4}$ and $Ca{H_2}$ is not covalent. Hence, option (C) is incorrect.
The nature of binding in $CC{l_4}$ is not electro covalent and the bond present between $Ca{H_2}$ is not covalent. Hence, option (D) is incorrect.

Hence, option (A) is correct.

Note:
We have to know that the nature of binding in carbon tetrachloride is covalent and in calcium hydride is equal to electro covalent bond. The chlorine atom in $CC{l_4}$ is symmetrically attached in corners and it is joined with a carbon atom by tetrahedral configuration. In calcium hydride, there will be loss of the calcium electron and that electrons are accepted by the hydrogen atoms. And the calcium ion is formed by the donation of electrons between the calcium and hydrogen.