Question

The shape of $\text{ IF}_{6}^{-}\text{ }$ is:
A) Tetragonally distorted octahedron
B) Pyramidal
C) Octahedral
D) Square pyramidal

Answer 1

Hint: The VSEPR theory is used to determine the shape of a molecule by calculating the number of electron pairs (bond pair or lone pair) in the molecule. The $\text{ IF}_{6}^{-}\text{ }$ is a negatively charged molecule such that the central atom iodine holds on one extra electron. The shape of the molecule is calculated by determining the total number of electrons associated with the molecule and dividing it by 2.

Complete step by step answer:
The VSEPR theory is used to determine the shape of the molecule. The shape can be determined by considering the number of electrons pair and their repulsion between the electron pairs.
The VSEPR rules to determine the shape are as follows:
1) Determine the central atom in the molecule
2) Write the electronic configuration and count all the valence electrons
3) Add one electron for each bonding atom on the central atom
4) Add or subtract the one electron from the total of electrons for the charge.
5) Divide the total number of electrons by 2 to find out the total number of electron pairs in the molecule. The electrons pair can be a bonding pair or non-bonding pair of electrons.
6) Use the numbers to determine the shape.
7) Consider the electron pair repulsion to achieve a stable shape for a molecule.
Here, we are interested to find out the shape of $\text{ IF}_{6}^{-}\text{ }$. The iodine is a central atom which is surrounded by the 6 fluorine atoms.
The electronic configuration of iodine is as follows:
$\text{ I=}\left[ \text{Kr} \right]\text{ 4}{{\text{d}}^{\text{10}}}\text{ 5}{{\text{s}}^{\text{2}}}\text{ 5}{{\text{p}}^{\text{5}}}\text{ }$
The valence shell of the iodine atom has 7 valence electrons.

One fluorine atom surrounding the iodine atom provides an electron to the iodine atom. Therefore total number of electrons shared by the six fluorine atoms is equal to,
$\text{ Electron from 6F atoms = 6}\times \text{1 }{{\text{e}}^{-}}\text{ = 6 }{{\text{e}}^{-}}\text{ }$
Let's determine the charge on the iodine atom. The molecule is given as $\text{ IF}_{6}^{-}\text{ }$. Such that the molecule acquires the negative charge, then the charge on the iodine atom is negative.
Thus, the iodine contains one extra valence electron from the negative charge hence subtract one electron from the total number of electrons. We have,
$\text{ }\begin{matrix}
   {} & \text{Valence electron of I} & {} & 7 \\
   + & \text{electron from 6F } & + & 6 \\
   + & \text{1 electron (for -ve charge)} & + & 1 \\
   {} & \overline{\text{Total electrons on IF}_{\text{6}}^{\text{-}}} & {} & \overline{14\text{ }{{\text{e}}^{-}}} \\
\end{matrix}\text{ }$

Thus, the $\text{ IF}_{6}^{-}\text{ }$ molecule has a total $14\text{ }{{\text{e}}^{-}}$.
On dividing the electrons we get the number of electrons pair. As,
$\text{ Number of }{{\text{e}}^{-}}\text{ pair = }\dfrac{\text{Total }{{\text{e}}^{-}}}{2}\text{ =}\dfrac{14\text{ }{{\text{e}}^{-}}}{2}=7\text{ }$

The $\text{ IF}_{6}^{-}\text{ }$ molecule has the 6 electron pair. Since the iodine is surrounded by the 6 fluorine atom, by looking at the number we can say that the six fluorine atoms are at the corner of regular octahedron, that is out of 8 valence electrons (one electron from negative charge), 6 are involved in the bond formation with fluorine and one electron pair is remain as the lone pair on the iodine atoms.

This lone pair of electrons is responsible for the electron pair repulsion between the bond pair and lone pair. This repulsion results in the distortion of octahedral geometry. As this distortion is observed in one of the trigonal planes of the octahedron, it is also called as the tetragonally distorted octahedral geometry.
So, the correct answer is “Option A”.

Note: Note that, while calculating the total number of electrons associated with the molecule you end up getting the ‘odd’ number of electrons, then that means you have made a wrong calculation. The number of electrons involved in the molecule is always even-number.