Question

Which of the following is a covalent compound?
A.\[{\text{A}}{{\text{l}}_2}{{\text{O}}_3}\]
B.\[{\text{A}}{{\text{l}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3}\]
C.\[{\text{Al}}{{\text{F}}_3}\]
D.\[{\text{AlC}}{{\text{l}}_3}\]

Answer 1

Hint:In case of ionic compounds, order of covalent character can be determined using Fajan’s rule. On the other hand, in case of covalent compounds, order of ionic or polar character can be determined using dipole moment.

Complete answer:
Among \[{\text{A}}{{\text{l}}_2}{{\text{O}}_3}\] , \[{\text{A}}{{\text{l}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3}\] , \[{\text{Al}}{{\text{F}}_3}\] and \[{\text{AlC}}{{\text{l}}_3}\] ; we can clearly see compound \[{\text{A}}{{\text{l}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3}\] is a salt and salt are completely ionic as they gets completely dissociate into their respective ions. So the compound \[{\text{A}}{{\text{l}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3}\] is ionic and in its aqueous solution it will dissociate per follow: \[{\text{A}}{{\text{l}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3} \to 2{\text{A}}{{\text{l}}^{ + 3}} + 3{\text{SO}}_4^{ - 2}\] .
Now the order of ionic character or covalent character among rest can be determined using Fajan’s rule.
When a cation approaches an anion closely the positive charge of a cation attracts the electron cloud of the anion toward itself and this property of cation is known as polarizing power. Due to this polarizing power of cation, it also repels the positively charged nucleus of anion resulting in bulging of anion cloud toward the cation. This property of anion is known as polarization. The compounds having more polarizing power of cation and more polarization of anion are more covalent in nature than the compounds with low polarizing power of cation and low polarization of anion. A cation with smaller size and more effective nuclear charge has more polarizing power and corresponds to more covalent nature. An anion with bigger size and more charge have more polarization and correspond to more covalent nature.
Now among \[{\text{A}}{{\text{l}}_2}{{\text{O}}_3}\] , \[{\text{Al}}{{\text{F}}_3}\] and \[{\text{AlC}}{{\text{l}}_3}\] ; cation \[{\text{A}}{{\text{l}}^{ + 3}}\] is mutual for all the compounds but the differ in anion. The anion present in \[{\text{A}}{{\text{l}}_2}{{\text{O}}_3}\] , \[{\text{Al}}{{\text{F}}_3}\] and \[{\text{AlC}}{{\text{l}}_3}\] are \[{{\text{O}}^{ - 2}},{{\text{F}}^{ - 1}},{\text{C}}{{\text{l}}^{ - 1}}\] respectively. Now the anion \[{{\text{O}}^{ - 2}}\] is having greater charge than rest anion, so it will have more polarization and correspond to more covalent nature than the rest of the compound. In between \[{{\text{F}}^ - },{\text{C}}{{\text{l}}^ - }\] ; the charge on both the anion is same and we already know the anion with bigger size have more polarization. Therefore chloride anion will have more polarizing power than fluoride anion.
To summarize, the order of polarization of anion is: \[{{\text{O}}^{ - 2}} > {\text{C}}{{\text{l}}^ - } > {{\text{F}}^ - }\] .
Therefore order of covalent character is: \[{\text{A}}{{\text{l}}_2}{{\text{O}}_3} > {\text{AlC}}{{\text{l}}_3} > {\text{Al}}{{\text{F}}_3} > {\text{A}}{{\text{l}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3}\]
Thus, the correct option is A.

Note:
Ionic bonds are formed between two atoms or more atoms when there is complete transfer of one or more electrons from one atom to another. A covalent bond is formed by the mutual sharing of electrons between two atoms.