The structure of IF7 is

Engineering

Chemistry

VSEPR Theory

Question

The structure of IF₇ is

square pyramid

trigonal bipyramid

pentagonal bipyramid

octahedral

The structure is pentagonal bipyramid having sp³d³ hybridisation as given below :

F_b – I – F_b = 72º (5 number) ; F_d – I – F_a = 90º (10 number)

F_b – I bond length = 1.858 ± 0.004 Å ; F_a – I bond length = 1.786 ± 0.007 Å.

Molecular Geometry of IF₇

To determine the structure of IF₇, we analyze its electron geometry and molecular shape using VSEPR theory.

Step 1: Count Valence Electrons

Iodine (I) has 7 valence electrons. Each fluorine (F) contributes 1 electron. Total valence electrons:

$7 + 7 \times 1 = 14$

Step 2: Determine Steric Number

Iodine forms 7 bonds with fluorine atoms and has 3 lone pairs (from 14 electrons total). Steric number = number of atoms bonded + number of lone pairs:

$7 + 3 = 10$

This corresponds to a pentagonal bipyramidal electron geometry.

Step 3: Identify Molecular Shape

With no lone pairs on the central atom in the bonding positions (all are bonding pairs), the molecular shape is also pentagonal bipyramidal.

Final Answer

The structure of IF₇ is pentagonal bipyramid.

Key Formulae and Theory

Steric Number (SN) = Number of bonded atoms + Number of lone pairs

For SN=7, electron geometry is pentagonal bipyramidal; if all are bonding pairs, molecular shape is the same.

Hybridization for pentagonal bipyramidal geometry is ${sp}^{3} d^{3}$ .

Detailed Solution

Molecular Geometry of IF₇

Step 1: Count Valence Electrons

Step 2: Determine Steric Number

Step 3: Identify Molecular Shape

Final Answer

Related Topics

Key Formulae and Theory

Detailed Solution

Molecular Geometry of IF7

Step 1: Count Valence Electrons

Step 2: Determine Steric Number

Step 3: Identify Molecular Shape

Final Answer

Related Topics

Key Formulae and Theory

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Molecular Geometry of IF₇