Problem 15 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The shape of \(\mathrm{IF}_{6}\) is : (a) Trigonally distorted octahedron (b) Pyramidal (c) Octahedral (d) Square antiprism

Step-by-Step Solution

Answer

The shape of IF₆ is (a) Trigonally distorted octahedron.

Step 1: Determine Hybridization

Identify the central atom in the molecule IF e from the formula IF e. The central atom is iodine (I). For a molecule like IF e, calculate the hybridization of the central atom by considering the valence electrons and the attached atoms. Iodine (I) has 7 valence electrons. With 6 fluorine (F) atoms attached, it uses 6 of its electrons, leaving 1 electron pair, thus forming 6 bonds and having 1 lone pair. This indicates the hybridization is sp³d².

Step 2: Analyze Geometrical Shape Based on Hybridization

The hybridization sp³d² suggests that in an ideal state, if there were no lone pairs, the geometry would be octahedral. However, the presence of a lone pair will distort this geometry.

Step 3: Consider Effect of Lone Pairs

In molecular geometry, lone pairs occupy more space than bonded pairs, leading to a distortion from the ideal geometry. For an octahedral arrangement (sp³d²), the presence of a lone pair results in a shape commonly known as a 'square pyramidal' or 'trigonal distortions' depending on distribution.

Step 4: Identify Correct Shape

Given the options, a trigonal distortion due to the lone pair produces a slightly distorted octahedral shape around the Iodine atom. This results in the shape being most closely identified with (a) Trigonally distorted octahedron.

Key Concepts

Hybridization

Hybridization is a concept used to describe the mixing of atomic orbitals in an atom to form new, hybrid orbitals. These hybrid orbitals have different energies and shapes compared to the original atomic orbitals.

In the molecule \(\mathrm{IF}_6\), iodine is the central atom with six fluorine atoms attached. To determine the hybridization, we count the valence electrons of iodine, which is 7. As it forms 6 bonds with fluorine atoms and retains one electron as a lone pair, the iodine atom is involved in forming 6 sigma bonds. This situation corresponds to \(sp^3d^2\) hybridization, which generally suggests an octahedral geometry.

A helpful way to think about hybridization is visualizing the transformation of atomic orbitals (\(s\), \(p\), and sometimes \(d\) orbitals) into new, equivalent hybrid orbitals. This conceptual tool helps us explain the geometry and bond angles observed in molecules.

Lone Pairs Effect

Lone pairs are electrons that are not shared with another atom and do not participate in bonding. In \(\mathrm{IF}_6\), the lone pair on the iodine atom affects the molecule's shape significantly.

Lone pairs are larger than bonding pairs in terms of their electron cloud, creating more repulsion and occupying more space. This extra repulsion forces the surrounding atoms to rearrange themselves, leading to molecular geometry distortions.

In an octahedral molecule like \(\mathrm{IF}_6\), the presence of a lone pair can lead to trigonal distortion. The lone pair pushes the fluorine atoms away, causing a deviation from the ideal geometry. Hence, it becomes a trigonal distortion rather than perfectly octahedral.

Trigonal Distortion

Trigonal distortion is a specific type of geometric distortion caused by lone pairs in a molecule originally meant to have an octahedral shape.

When a molecule like \(\mathrm{IF}_6\) has \(sp^3d^2\) hybridization, the ideal shape is octahedral. However, the presence of a lone pair alters this geometry. Instead of a perfect octahedron, the arrangement changes to accommodate additional electronic repulsions.

This distortion is called trigonal distortion or sometimes referred to as square pyramidal distortion. It slightly dethrones the initial symmetry of the octahedral shape by adjusting the angles between the surrounding atoms. Understanding this effect helps in predicting and visualizing the real molecular structure more accurately.

Octahedral Shape

The octahedral shape is a common molecular geometry that arises from \(sp^3d^2\) hybridization. This configuration involves six electron pairs (either bonding or non-bonding) surrounding a central atom.

In theory, an octahedral shape is highly symmetrical, with bond angles of 90 degrees between adjacent bonds. The central atom, like iodine in \(\mathrm{IF}_6\), pairs with six peripheral atoms, each positioned at the corners of an octahedron.

However, due to factors like lone pairs or asymmetric bonding arrangements, deviations from this ideal shape can occur. Molecules with lone pairs often exhibit slight distortions from the ideal octahedral angle to minimize repulsion.

Valence Electrons

Valence electrons are the outermost electrons of an atom and are crucial for determining the atom's chemical properties, including its ability to bond with other atoms.

Iodine in \(\mathrm{IF}_6\) has 7 valence electrons. Understanding the number of valence electrons helps predict the atom's behavior in molecular interactions like bonding or forming lone pairs.

By knowing an atom's valence electrons, we can gauge its hybridization and assess how it fits into the molecule’s overall geometry. In the case of \(\mathrm{IF}_6\), these electrons determine the central iodine’s bonding capacity and play a part in any subsequent geometric distortions such as trigonal distortion.