Select the correct statement(s) about IF7.a)I atom is sp3d3-hybridisedb)I atom is in highest oxidation statec)There are five I—F longest and two I—F shortest bondsd)It has pentagonal bipyramidal structureCorrect answer is option 'A,B,C,D'. Can you explain this answer?

Question

Select the correct statement(s) about IF7.a)I atom is sp3d3-hybridisedb)I atom is in highest oxidation statec)There are five I&#8212;F longest and two I&#8212;F shortest bondsd)It has pentagonal bipyramidal structureCorrect answer is option 'A,B,C,D'. Can you explain this answer?

Answer

Hybridisation of l-True
(b) Oxidation state of I = + 7 (highest) -True.
(c) It has two types of bonds two axial (shortest) and five equatorial (longest) - True.

(d) Pentagonal bipyramidai structure - True.

Answer

Hybridisation of l-True
(b) Oxidation state of I = + 7 (highest) -True.
(c) It has two types of bonds two axial (shortest) and five equatorial (longest) - True.

(d) Pentagonal bipyramidai structure - True.

Answer

IF7: Structure and Bonding

Introduction
IF7 is the chemical formula for iodine heptafluoride. It is a compound consisting of iodine and fluorine atoms. Let's analyze the given statements about IF7 and determine which ones are correct.

Statement A: I atom is sp3d3-hybridized
This statement is correct. In IF7, the iodine atom forms seven sigma bonds with fluorine atoms. To accommodate these bonds, the iodine atom undergoes sp3d3 hybridization. This hybridization involves the mixing of one s orbital, three p orbitals, and three d orbitals to form seven hybrid orbitals. These hybrid orbitals then overlap with the fluorine atom orbitals to form sigma bonds.

Statement B: I atom is in the highest oxidation state
This statement is correct. In IF7, iodine is in its highest oxidation state, which is +7. The fluorine atoms are highly electronegative, and each fluorine atom contributes one electron towards the bond with iodine. Since there are seven fluorine atoms, iodine gains a total of seven electrons, resulting in a +7 oxidation state.

Statement C: There are five IF longest and two IF shortest bonds
This statement is correct. In IF7, the molecule adopts a pentagonal bipyramidal structure. There are five fluorine atoms bonded to the central iodine atom in the equatorial plane, forming the longest bonds. Additionally, there are two fluorine atoms bonded to the iodine atom in the axial positions, forming the shortest bonds. This arrangement gives rise to the pentagonal bipyramidal structure.

Statement D: It has a pentagonal bipyramidal structure
This statement is correct. As mentioned in statement C, IF7 adopts a pentagonal bipyramidal structure. The central iodine atom is surrounded by five fluorine atoms in the equatorial plane and two fluorine atoms in the axial positions. This arrangement results in a molecule with pentagonal symmetry.

Summary
In summary, all the given statements (A, B, C, and D) about IF7 are correct. The iodine atom in IF7 is sp3d3-hybridized, it is in the highest oxidation state (+7), there are five longest and two shortest IF bonds, and the molecule has a pentagonal bipyramidal structure.

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