Question

Which of the following molecules is not hypovalent?
A. ${\text{AlC}}{{\text{l}}_{\text{3}}}$
B. ${\text{AlB}}{{\text{r}}_{\text{3}}}$
C. ${\text{Al}}{{\text{F}}_{\text{3}}}$
D. All are hypovalent Because ${\text{Al}}{{\text{F}}_{\text{3}}}$ is an ionic compound

Answer 1

The molecules in which the central atom has less than 8 valence electrons are known as hypovalent molecules. Central atom hypovalent molecules do not obey the octet rule. The number of electrons in the valence shell of the central atom also depends on the type of bonding.

Complete step by step answer:
Determine the type of bond present in given compounds as follows:
In given compounds, central atom (${\text{Al}}$) is common only there is a change in bonding atom. Atoms ${\text{F}}$,${\text{Cl}}$and ${\text{Br}}$ are all halogens and present in the same group of the periodic table.

Atomic size of elements increases as we move from top to bottom in the periodic table. So the increase in order of the atomic size of ${\text{F}}$,${\text{Cl}}$and ${\text{Br}}$ is as follows :
${\text{F < Cl < Br}}$
Similarly, the order of the ionic size of ${{\text{F}}^{\text{ - }}}$,${\text{C}}{{\text{l}}^{\text{ - }}}$and ${\text{B}}{{\text{r}}^{\text{ - }}}$ is as follows :
${{\text{F}}^{\text{ - }}}{\text{ < C}}{{\text{l}}^{\text{ - }}}{\text{ < B}}{{\text{r}}^{\text{ - }}}$
The electronegativity of elements decreases as we move from top to bottom in the periodic table. So the decrease in order of electronegativity of ${\text{F}}$,${\text{Cl}}$and ${\text{Br}}$ is as follows:
${\text{F > Cl > Br}}$
In the case of compounds ${\text{Al}}{{\text{F}}_{\text{3}}}$,${\text{AlC}}{{\text{l}}_{\text{3}}}$and ${\text{AlB}}{{\text{r}}_{\text{3}}}$ cation (${\text{A}}{{\text{l}}^{{\text{3 + }}}}$) is common so an increase in the size of anion increases the covalent character of compounds. Due to the greater size of anions ${\text{C}}{{\text{l}}^{\text{ - }}}$and ${\text{B}}{{\text{r}}^{\text{ - }}}$ compounds ${\text{AlC}}{{\text{l}}_{\text{3}}}$and ${\text{AlB}}{{\text{r}}_{\text{3}}}$ are covalent compounds while due to greater electronegativity and smaller size of ${{\text{F}}^{\text{ - }}}$ ion compound${\text{Al}}{{\text{F}}_{\text{3}}}$ is an ionic compound.
Using the type of bond present determines the number of electrons in the valence shell of the central atom for all given compounds.
Atomic number of ${\text{Al}}$ is 13
So, the electronic configuration of a neutral ${\text{Al}}$ atom is 2, 8, 3

${\text{AlC}}{{\text{l}}_{\text{3}}}$ is a covalent compound so during covalent bond formation central ${\text{Al}}$ atom shares its 3 valence electrons with three ${\text{Cl}}$ atoms. As the central atom has less than 8 electrons so ${\text{AlC}}{{\text{l}}_{\text{3}}}$ is hypovalent compound.
So, option (A) ${\text{AlC}}{{\text{l}}_{\text{3}}}$ is not the correct option.

${\text{AlB}}{{\text{r}}_{\text{3}}}$ is a covalent compound so during covalent bond formation central ${\text{Al}}$ atom shares its 3 valence electrons with three ${\text{Br}}$ atoms. As the central atom has less than 8 electrons so ${\text{AlB}}{{\text{r}}_{\text{3}}}$ is hypovalent compound.

So, option (B) ${\text{AlB}}{{\text{r}}_{\text{3}}}$ is not the correct option.

${\text{Al}}{{\text{F}}_{\text{3}}}$ is an ionic compound so during ionic bond formation central ${\text{Al}}$ atom donates its outermost 3 electrons. So, ${\text{A}}{{\text{l}}^{{\text{3 + }}}}$ ion in ${\text{Al}}{{\text{F}}_{\text{3}}}$ has electronic configuration 2, 8. As the octet of the central atom is fulfilled so ${\text{Al}}{{\text{F}}_{\text{3}}}$ is not hypovalent compound.

Thus, the correct option is C.

Note: Though in all compounds${\text{Al}}{{\text{F}}_{\text{3}}}$,${\text{AlC}}{{\text{l}}_{\text{3}}}$and ${\text{AlB}}{{\text{r}}_{\text{3}}}$there is a bond between the metal and non-metal only ${\text{Al}}{{\text{F}}_{\text{3}}}$ is ionic compound while ${\text{AlC}}{{\text{l}}_{\text{3}}}$and ${\text{AlB}}{{\text{r}}_{\text{3}}}$ are covalent compounds