Q10.65CP - Chemistry: Molecular Nature Of Matter And Change

Question

Both aluminium and iodine form chlorides, ${\mathbf{Al}}_2{\mathbf{Cl}}_6$ and ${\mathbf{l}}_2{\mathbf{Cl}}_6$, with “bridging” Cl atoms. The Lewis structures are

1. What is the formal charge on each atom?
2. Which of these molecules has a planar shape? Explain.

 

Step-by-Step Solution

Answer

a) The Formal change on Al = -1

    The formal charge on bridging Cl=1 

    The formal charge on the end Cl = 0

    The Formal charge on I = -1

    The formal charge on bridging Cl = 1

     The formal charge on the end Cl = 0

b)  The molecule ${\mathrm{l}}_2{\mathrm{Cl}}_6$ has a planar shape.

Step 1: Calculation of Formal Charge of Al 2 Cl 6 and l 2 Cl 6 ,

The molecules ${\mathrm{Al}}_2{\mathrm{Cl}}_6$ and ${\mathrm{l}}_2{\mathrm{Cl}}_6$ both have 3C – 4e bridge bonds. They both have a somewhat similar structure. Let us calculate the formal charge on each atom for both molecules.

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{of} \mathrm{an} \mathrm{atom}=\mathrm{no}.\mathrm{of} {\mathrm{valencee}}^{-}–[\mathrm{no}. \mathrm{of} \mathrm{unshared} \text{\hspace{0.17em}}\mathrm{valence} {\mathrm{e}}^{-}\\ +\frac{1}{2}\mathrm{no}. \mathrm{of} \mathrm{shared} \mathrm{valence} {\mathrm{e}}^{-}]\end{array}$

 For the molecule ${\mathrm{Al}}_2{\mathrm{Cl}}_6$ the formal charge on each atom would be,

$\begin{array}{c}\mathrm{Formal} \mathrm{change} \mathrm{on} \mathrm{Al}=3{\mathrm{e}}^{-} -\left(0{\mathrm{e}}^{- }+\frac{1}{2}\mathrm{x}8{\mathrm{e}}^{-}\right)\\ =3{\mathrm{e}}^{-} -4{\mathrm{e}}^{-}\\ =-1\end{array}$

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{bridging} \mathrm{Cl}=7{\mathrm{e}}^{-} -\left(4{\mathrm{e}}^{-} +\frac{1}{2}\mathrm{x}4{\mathrm{e}}^{-}\right)\\ =7{\mathrm{e}}^{-} -6{\mathrm{e}}^{-}\\ =1\end{array}$

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{end} \mathrm{Cl}=7{\mathrm{e}}^{-} -\left(6{\mathrm{e}}^{-} +\frac{1}{2}\mathrm{x}2{\mathrm{e}}^{-}\right)\\ =7{\mathrm{e}}^{-} -7{\mathrm{e}}^{-}\\ =0\end{array}$

For the molecule, ${\mathbf{l}}_2{\mathbf{Cl}}_6$ the formal charge on each atom would be,

$\begin{array}{c}\mathrm{Formal} \mathrm{change} \mathrm{on} \mathrm{l}=7{\mathrm{e}}^{-} -\left(4{\mathrm{e}}^{- }+\frac{1}{2}\mathrm{x}8{\mathrm{e}}^{-}\right)\\ =7{\mathrm{e}}^{-} -8{\mathrm{e}}^{-}\\ =-1\end{array}$

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{bridging} \mathrm{Cl}=7{\mathrm{e}}^{-} -\left(4{\mathrm{e}}^{-} +\frac{1}{2}\mathrm{x}4{\mathrm{e}}^{-}\right)\\ =7{\mathrm{e}}^{-} -6{\mathrm{e}}^{-}\\ =1\end{array}$

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{end} \mathrm{Cl}=7{\mathrm{e}}^{-} -\left(6{\mathrm{e}}^{-} +\frac{1}{2}\mathrm{x}2{\mathrm{e}}^{-}\right)\\ =7{\mathrm{e}}^{-} -7{\mathrm{e}}^{-}\\ =0\end{array}$

Step 2: Molecule having Planar Shape

The molecules ${\mathrm{Al}}_2{\mathrm{Cl}}_6$ and ${\mathrm{l}}_2{\mathrm{Cl}}_6$ both are dimers of ${\mathrm{AlCl}}_3$ and ${\mathrm{lCl}}_3$, having 3C – 4e bridge bonds. They both have a somewhat similar structure. But still, molecule ${\mathrm{l}}_2{\mathrm{Cl}}_6$ has a planar shape.

In ${\mathrm{l}}_2{\mathrm{Cl}}_6$, each I atom is sp³d² hybridized. It has 2 bonds that are of  3C−4e⁻ and 4 bonds which are of  2C−2e⁻ (I atom is bonded to two bridging Cl atoms and two terminal Cl atoms as a square planar shape).

Due to two lone pairs of electrons on I, the angle between the two lone pairs is 180^o. 

Hence, the shape is rectangular planar as all the iodine and the chlorine atoms lie in the same plane. Hence, the molecule ${\mathrm{l}}_2{\mathrm{Cl}}_6$ has a planar shape.

While in the molecule ${\mathrm{Al}}_2{\mathrm{Cl}}_6$, each Al atom is sp³ hybridized. It has 2 bonds that are of 3C−4e⁻ and 4 bonds which are of  2C−2e⁻ (Al atom is bonded to two bridging Cl atoms and two terminal Cl atoms as tetrahedral shape). 

And it does not have lone pairs on Al. Hence, Aluminium and the chlorine atoms are not in the same plane and the molecule ${\mathrm{Al}}_2{\mathrm{Cl}}_6$ is not planar.

Therefore, in the molecule ${\mathrm{Al}}_2{\mathrm{Cl}}_6$, the formal change on Al = -1, bridging Cl = 1 and on end Cl = 0;

In the molecule ${\mathrm{l}}_2{\mathrm{Cl}}_6$, the formal change on I = -1, bridging Cl = 1 and on end Cl = 0;

The molecule ${\mathrm{l}}_2{\mathrm{Cl}}_6$ has a planar shape and the molecule ${\mathrm{Al}}_2{\mathrm{Cl}}_6$ does not have a planar shape.