
What types of bonds are present in Aluminum Chloride?
$(A)$ Covalent
$(B)$ Ionic
$(C)$ Both $(A)$ and $(B)$
$(D)$ Neither $(A)$ nor $(B)$
Answer
498.3k+ views
Hint: Electronegativity determines whether a compound is ionic or covalent, as if the bonding atoms have a large difference in electronegativity then this causes transfer of electrons making an ionic bond. As the difference in electronegativity decreases the bond develops covalent character.
Complete answer:
Aluminum chloride forms a covalent bond because the electronegativity difference between \[Al\] and \[Cl\] is small. Tendency of aluminum (less electropositive) to lose an electron to form cation and chlorine (less electronegative) to accept an electron to form anion is less in this case. The electronegativity difference between aluminum and chlorine is \[1.55\]. And to form an ionic bond the electronegativity difference should be above $1.7$.
Aluminum also has very high ionization energy and for chlorine it is not easy to withdraw three electrons from the aluminum outermost shell because $Cl$ is less electronegative. Therefore, aluminum rather than losing $3$electrons shares its electron with three chlorine ions.
It can also be explained through the Fajan’s Rule: according to Fajan’s rule, the larger the anion, less is the effective nuclear charge that holds the valence electron of the ion in place. Since the last electron is loosely bound in large anions, it can easily be polarized by a cation, thereby making the compound more covalent.
So if we take example of $AlC{l_3}$ and $Al{F_3}$, Here fluorine is smaller than chlorine, also fluorine is enough electronegative to withdraw electrons from the outermost shell of aluminum therefore, $Al{F_3}$ is an ionic bond whether $AlC{l_3}$ is a covalent compound.
Thus the correct option for the given question is option (A).
Note:
Aluminum chloride is actually an ionic bond with a higher degree of covalency. This is due to polarization. Cations are very polarizing, while anions are very polarizable. The higher the charge of the cation will increase the degree of distortion of the electron cloud surrounding the anion.
Complete answer:
Aluminum chloride forms a covalent bond because the electronegativity difference between \[Al\] and \[Cl\] is small. Tendency of aluminum (less electropositive) to lose an electron to form cation and chlorine (less electronegative) to accept an electron to form anion is less in this case. The electronegativity difference between aluminum and chlorine is \[1.55\]. And to form an ionic bond the electronegativity difference should be above $1.7$.
Aluminum also has very high ionization energy and for chlorine it is not easy to withdraw three electrons from the aluminum outermost shell because $Cl$ is less electronegative. Therefore, aluminum rather than losing $3$electrons shares its electron with three chlorine ions.
It can also be explained through the Fajan’s Rule: according to Fajan’s rule, the larger the anion, less is the effective nuclear charge that holds the valence electron of the ion in place. Since the last electron is loosely bound in large anions, it can easily be polarized by a cation, thereby making the compound more covalent.
So if we take example of $AlC{l_3}$ and $Al{F_3}$, Here fluorine is smaller than chlorine, also fluorine is enough electronegative to withdraw electrons from the outermost shell of aluminum therefore, $Al{F_3}$ is an ionic bond whether $AlC{l_3}$ is a covalent compound.
Thus the correct option for the given question is option (A).
Note:
Aluminum chloride is actually an ionic bond with a higher degree of covalency. This is due to polarization. Cations are very polarizing, while anions are very polarizable. The higher the charge of the cation will increase the degree of distortion of the electron cloud surrounding the anion.
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