Question

# If true enter 1, else enter 0. In ${{\left[ {{I}_{8}} \right]}^{2-}}$ ion, two I-I bonds are larger whereas five I-I bonds are smaller.

Hint: The class of polyhalogen anions entirely composed of iodine atoms is called polyiodides. ${{\left[ {{I}_{8}} \right]}^{2-}}$ belongs to the octa atomic species of polyiodides class. The structure and bonding of any molecule can be predicted based on hybridization.
Formula used:

- For determining the type of hybridization of the given structure, the following formula is used-
$H=\frac{1}{2}(V+M+A-C)$
where H is the number of orbitals involved in hybridization.
V is the number of monovalent atoms.
C is the charge on the cation.
A is the charge on the anion.

- Iodine is a p-block element and it has 7 valence electrons. From this, we can calculate the hybridization $s{{p}^{3}}{{d}^{3}}$ .

- The $s{{p}_{3}}{{d}_{3}}$hybridization has five bonds in a plane, one bond above the plane and one bond below the plane giving a pentagonal bipyramidal geometry.

- Seven hybrid orbitals are not equivalent hybrid orbitals because five of them are directed towards the corners of a regular pentagon while the remaining two are directed above and below the plane.

- In pentagonal bipyramidal geometry, the bond angle is $72{}^\circ$ and $90{}^\circ$ .

So, the given statement is true.

 Type of hybridization No.of hybrid orbitals involved Bond angle Structure sp 2 $180{}^\circ$ Linear $s{{p}^{2}}$ 3 $120{}^\circ$ Plane triangle $s{{p}^{3}}$ 4 $109.5{}^\circ$ Tetrahedral $s{{p}^{2}}d$ 4 $90{}^\circ$ Square planar $s{{p}^{3}}{{d}^{2}}$ 5 $90{}^\circ \And 120{}^\circ$ Trigonal bipyramidal $s{{p}^{3}}{{d}^{2}}$ 6 $90{}^\circ$ Octahedral $s{{p}^{3}}{{d}^{3}}$ 7 $90{}^\circ \And 72{}^\circ$ Pentagonal bipyramidal