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- 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 $ .

- Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals of equal energies and suitable for the qualitative description of atomic bonding properties.

- Following is the list of hybridization and their geometry-

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 |