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

Short Answer

Expert verified
The hybridization of \(\text{I}\) in \(\text{IF}_3\) is \(\text{sp}^3\text{d}\) (option a).

Step by step solution

01

Understand the Concept of Hybridization

Hybridization is the mixing of atomic orbitals to form a new set of equivalent orbitals. The hybridization type is determined by the number of electron domains (bond pairs and lone pairs) around the central atom.
02

Determine the Number of Bonds and Lone Pairs

In the molecule \(\text{IF}_3\), Iodine (I) is the central atom. It forms three sigma bonds with three Fluorine (F) atoms. Additionally, iodine carries two lone pairs of electrons.
03

Calculate the Total Electron Domains

Calculate the total number of electron domains by adding the number of sigma bonds and lone pairs. For \(\text{IF}_3\), this is 3 (bonds) + 2 (lone pairs) = 5 electron domains.
04

Assign the Hybridization Based on Electron Domains

The hybridization corresponds to the number of electron domains:- sp: 2 electron domains- sp\(^2\): 3 electron domains- sp\(^3\): 4 electron domains- sp\(^3\)d: 5 electron domainsThus, \(\text{IF}_3\) has a hybridization of \(\text{sp}^3\text{d}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Geometry
Molecular geometry is the three-dimensional shape a molecule adopts, which determines its reactivity and properties. In the case of iodine trifluoride (1F_32), the geometry is affected by both its sigma bonds and lone pairs of electrons. 3IF_34 has three fluorine atoms bonded to a central iodine atom. However, iodine also has two lone pairs that must be considered.
When determining molecular geometry, it's essential to consider both bonded atoms and lone pairs, as they all occupy space around the central atom. The arrangement of these gives rise to the base geometrical shape. For 2IF_33, considering the lone pairs, the shape is described as a "T-shaped" geometry. This is because the lone pairs repel each other and the bonded pairs, causing the bonded atoms to be pushed closer together in a T-shaped arrangement.
This "T-shaped" geometry falls under the umbrella of "trigonal bipyramidal" electron arrangement, which includes five total electron domains, where two of these are lone pairs.
Valence Bond Theory
Valence Bond Theory explains how atoms come together to form molecules, considering atomic orbitals and their interactions during bond formation. This theory is pivotal in understanding how molecules like 1F_32 form. In 1F_32, the iodine atom forms covalent bonds with fluorine atoms by sharing electrons.
According to this theory, when atoms approach each other, their half-filled atomic orbitals overlap, leading to the formation of a bond. The overlap of orbitals allows electrons to be shared between atoms, creating a more stable electronic arrangement. In 1F_32, iodine utilizes its available p orbitals and one additional d orbital from the same energy level to form new hybrid orbitals. This happens during hybridization, crucial for arranging electrons in a manner that minimizes repulsion and stabilizes the structure.
  • This hybridization enhances the overlap between orbitals, resulting in stronger sigma bonds and the formation of a stable molecule.
Electron Domain
The concept of electron domains is essential when predicting the shape and hybridization of a molecule. An electron domain is any area in a molecule where electrons are likely found, which includes both bonds and lone pairs of electrons on the central atom.
In terms of hybridization and geometry, knowing the total number of electron domains helps to determine the hybridization type of a given atom. For 1F_32, iodine has a total of five electron domains comprising three sigma bonds and two lone pairs.
  • The presence of these five electron domains suggests a trigonal bipyramidal electron arrangement, which is characteristic of sp^3d hybridization.
  • Electron domains influence molecular polarity and spatial arrangement. More domains can lead to more complex shapes and angles, often making molecules with many domains non-linear and asymmetrical.
If you were to consider a molecule with fewer domains, you might expect simpler geometries like linear or trigonal planar arrangements.
Sigma Bonds
Sigma bonds are a type of covalent bond characterized by the direct overlap of orbitals along the axis connecting two atomic nuclei. These are the strongest type of covalent bonds due to the effective orbital overlap along the bonding axis.
In 1F_32, each iodine-fluorine bond is a sigma bond formed by the overlap of iodine's hybridized orbital with the fluorine atom's orbital. Each bond localization strengthens the entire molecular structure.
  • Sigma bonds allow for free rotation around the bond axis, which is a defining characteristic of these bonds.
  • Within 1F_32, these sigma bonds are essential for maintaining the stability of the molecule's T-shaped geometry.
  • While they are strong and stable, the molecule's overall shape and properties are influenced by the placement and effect of lone pairs, not just the sigma bonds alone.
This makes sigma bonds vital in understanding molecular stability and enables predictions of how molecules might react or interact in chemical processes.

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Most popular questions from this chapter

The decreasing order of the boiling points of the following hydrides is: 1\. \(\mathrm{NH}_{3}\) 2\. \(\mathrm{PH}_{3}\) 3\. \(\mathrm{AsH}_{3}\) 4\. \(\mathrm{SbH}_{3}\) 5\. \(\mathrm{H}_{2} \mathrm{O}\) (a) \(5>4>1>3>2\) (b) \(5>1>2>3>4\) (c) \(2>4>3>1>5\) (d) \(4>3>1>2>5\)

In \(\mathrm{X}-\mathrm{Hi}-\mathrm{Y}\), where both \(\mathrm{X}\) and \(\mathrm{Y}\) are electronegative elements: (a) Electron density of \(\mathrm{X}\) will increase and the eletron density on \(\mathrm{H}\) will decrease (b) On both species electron density will increase (c) On both species electron density will decrease (d) On \(\mathrm{X}\) the electron density will decrease and on \(\mathrm{H}\) it will increases

Which of the following is diamagnetic? (a) \(\mathrm{H}_{2}^{+}\) (b) \(\mathrm{O}_{2}\) (c) \(\mathrm{Li}_{2}\) (d) \(\mathrm{He}_{2}\)

\(\mathrm{H}_{2} \mathrm{O}\) is dipolar, whereas \(\mathrm{BeF}_{2}\) is not. It is because: (a) The electronegativity of \(\mathrm{F}\) is greater than that of \(\mathrm{O}\) (b) \(\mathrm{H}_{2} \mathrm{O}\) involves hydrogen bonding whereas \(\mathrm{BeF}_{2}\) is a discrete molecule (c) \(\mathrm{H}_{2} \mathrm{O}\) is linear and \(\mathrm{BeF}_{2}\) is angular (d) \(\mathrm{H}_{2} \mathrm{O}\) is angular and \(\mathrm{BeF}_{2}\) is linear

Which one of the following pairs of species have the same bond order? (a) \(\mathrm{CN}^{-}\) and \(\mathrm{NO}^{+}\) (b) \(\mathrm{CN}^{-}\) and \(\mathrm{CN}^{+}\) (c) \(\mathrm{O} 2^{-}\) and \(\mathrm{CN}^{-}\) (d) \(\mathrm{NO}^{+}\) and \(\mathrm{CN}^{+}\)

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