Problem 45 - A Complete Resource Book in Chemistry for JEE Main

Question

The hybridization of \(\mathrm{I}\) in \(\mathrm{IF}_{3}\) is (a) \(\mathrm{sp}^{3} \mathrm{~d}\) (b) \(\mathrm{sp}^{3}\) (c) \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) (d) \(\mathrm{sp}^{3} \mathrm{~d}^{3}\)

Step-by-Step Solution

Answer

The hybridization of I in \(\mathrm{IF}_3\) is \(\mathrm{sp}^3 \mathrm{d}\).

Step 1: Determine the Total Number of Valence Electrons

Iodine (I) is in group 17 and has 7 valence electrons. Each fluorine (F) atom also has 7 valence electrons, but since fluorine atoms form bonds using one electron each, we count one electron per bond. Thus, there are 3 fluorine atoms, contributing 3 electrons through bonding. Therefore, the total number of valence electrons is 7 (from I) + 3 (one from each bond with F) = 10 valence electrons.

Step 2: Calculate the Number of Electron Pairs

Since we have a total of 10 valence electrons, we can divide this number by 2 to find the number of electron pairs. We get \(\frac{10}{2} = 5\) electron pairs around the central iodine atom.

Step 3: Count Bonding and Lone Pairs on Iodine

Among the 5 pairs determined, 3 pairs are used for bonding with the 3 fluorine atoms, and the remaining 2 pairs are non-bonding or lone pairs on the iodine atom.

Step 4: Deduce the Hybridization

Hybridization is determined by the steric number, which is the sum of the number of atoms bonded to the central atom and the number of lone pairs on the central atom. Here, the steric number is 3 (bonds) + 2 (lone pairs) = 5. Therefore, the hybridization involves \(\mathrm{sp}^3\mathrm{d}\) orbitals, as it is typically consistent with a steric number of 5.

Key Concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a key role in chemical bonding. These electrons are involved in forming bonds with other atoms by either being shared or transferred. For iodine, which is in group 17 of the periodic table, it has 7 valence electrons. This helps determine how iodine can interact and bond with other atoms, such as fluorine in the compound \((\mathrm{IF}_{3}\)).

• Understanding Valence Electrons in Iodine: As a halogen, iodine needs just one more electron to achieve a stable octet, which is a complete set of eight electrons in its outer shell.
• Contributions from Fluorine: Fluorine atoms also have 7 valence electrons and contribute one electron each when bonding in \((\mathrm{IF}_{3}\)). This simplifies the calculation for total valence electrons involved in bonding.

Together, this makes it possible to calculate the total number of valence electrons available for bonding within the molecule, which is a foundational step for determining the molecule's hybridization.

Electron Pairs

Once we know the total number of valence electrons in a molecule, we need to understand how these electrons are organized. This is done by counting electron pairs. An electron pair consists of two electrons that occupy an orbital.To find out the number of electron pairs:

• Divide the total number of valence electrons by 2: For \((\mathrm{IF}_{3}\))\, there are 10 total valence electrons, so there are 5 electron pairs.
• Distinguish between bonding and lone pairs: Out of the 5 pairs, 3 are used in bonding with fluorine atoms; the other 2 are non-bonding lone pairs on iodine.

Understanding the distribution of electron pairs is crucial in determining the molecular shape and predicting how the molecule will interact with other substances. These electron pairs influence the molecular geometry by repelling each other, leading to specific shapes and hybridization states that minimize the repulsion.

Steric Number

The steric number of an atom in a molecule communicates the basic idea of how many regions of electron density are around a central atom. It is essentially the sum of bonded atoms and lone pairs on the central atom, and this value is vital for determining the hybridization type of the molecule.For iodine in \((\mathrm{IF}_{3}\))\, the steric number calculation is straightforward:

• Counting bond pairs and lone pairs: Iodine forms bonds with three fluorine atoms and also has two lone electron pairs.
• Add them together for the steric number: 3 bonds + 2 lone pairs = 5A steric number of 5 suggests an \(\mathrm{sp}^3\mathrm{d}\) hybridization. This hybridization reflects that five orbitals from iodine have hybridized to accommodate electron sharing and lone pairs, which ultimately determines the shape and geometry of the molecule. Recognizing this helps in understanding the molecule's spatial arrangement, stability, and how it might react with other compounds.