Question

Iodine pentafluoride has which of the following hybridization? (a) \(\mathrm{d}^{2} \mathrm{sp}^{3}\) (b) \(\mathrm{dsp}^{3}\) (c) \(\mathrm{sp}^{3} \mathrm{~d}\) (d) \(\mathrm{sp}^{3} \mathrm{~d}^{2}\)

Short answer

The hybridization of iodine pentafluoride (IF5) is \\(\mathrm{sp}^{3}\mathrm{d}^{2}\\), which is option (d).

Step-by-step solution

Determine the Structure of Iodine Pentafluoride (IF5)

Start by analyzing the molecular formula of iodine pentafluoride, which is IF5. Iodine is the central atom surrounded by five fluorine atoms.

Count Valence Electrons

Iodine has 7 valence electrons, and each fluorine atom has 7 valence electrons. Considering all the atoms, IF5 has a total of 42 valence electrons: 7 from iodine and 35 from the five fluorine atoms (5 × 7 = 35).

Determine Electron Pair and Lone Pair

Five fluorine atoms form single bonds with the iodine atom using 10 valence electrons. Thus, we count these bonds as 5 electron pairs. Now, consider the lone pair on the iodine atom; iodine uses one more electron, completing its valence shell with six electron densities around it (5 bonding pairs + 1 lone pair).

Determine Hybridization from Electron Pair Geometry

The six electron densities around the iodine indicate an electron pair geometry that's octahedral. This suggests the hybridization involving six atomic orbitals, specifically \(\mathrm{sp}^{3}\mathrm{d}^{2}\).

Selection of the Correct Hybridization Option

The hybridization of iodine in IF5 is \(\mathrm{sp}^{3}\mathrm{d}^{2}\), which matches option (d).

Key concepts

Valence Electrons

Understanding the concept of valence electrons is crucial in chemistry as they play a key role in chemical bonding and molecular structure. Valence electrons are the outermost electrons of an atom and are the ones involved in forming bonds with other atoms. These electrons are significant because they determine how an atom will interact with other atoms.
For example, in iodine pentafluoride (IF₅), you start by identifying the number of valence electrons for each atom involved:

• Iodine: 7 valence electrons
• Fluorine: 7 valence electrons per atom

Since IF₅ has one iodine atom and five fluorine atoms, you calculate the total number of valence electrons as 42. This is done by adding 7 valence electrons from iodine to the collective 35 valence electrons from the five fluorines (5 x 7 = 35). Knowing the total number of valence electrons helps determine the molecular geometry and predict the reactivity of the molecule.

Electron Pair Geometry

Electron pair geometry is about the spatial arrangement of electron pairs around a central atom in a molecule. This geometry considers both bonding pairs (which are electrons shared with other atoms) and lone pairs (which are not involved in bonding).
In iodine pentafluoride (IF₅), the central iodine atom forms five bonds with fluorine atoms, which accounts for five electron pairs. However, iodine also has an extra lone pair of electrons that influences the overall shape.
When assessing electron pair geometry, both bond pairs and lone pairs need to be included. In the case of IF₅, the six electron densities around iodine (composed of five bonds and one lone pair) indicate an octahedral geometry. This suggests that if you are visualizing the geometry taking into account the electron pairs, iodine is considered to have a uniform distribution as if these electron pairs were placed at the corners of an octahedron.

Octahedral Geometry

Octahedral geometry is a type of three-dimensional molecular shape formed when there are six groups of electron pairs surrounding a central atom. This geometry is characterized by its symmetrical arrangement, where all the angles between the bonds are 90 degrees.
In molecules like iodine pentafluoride (IF₅), the iodine atom is at the center, with five fluorine atoms surrounding it in a way that resembles five corners of an octahedron. The sixth position is occupied by a lone pair, leading to the actual molecular shape being a "square pyramidal" (due to the influence of the lone pair).
Octahedral geometry plays an integral role in determining the chemical properties of a compound, providing insight into bonding angles, reactivity, and potential energy changes during reaction processes. Understanding such geometry helps in predicting how molecules might interact in chemical reactions.