Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The I-I bond distance in \(\mathrm{I}_{2}\) (gas phase) is \(267 \mathrm{pm}\), in the \(\left[\mathrm{I}_{3}\right]^{+}\) ion is \(268 \mathrm{pm},\) and in \(\left[\mathrm{I}_{3}\right]^{-}\) is \(290 \mathrm{pm}\) (for the \(\left[\mathrm{AsPh}_{4}\right]^{+}\) salt). (a) Draw Lewis structures for these species. Do these representations account for the variation in bond distance? (b) Use MO theory to describe the bonding and deduce the I-I bond order in each species. Are your results consistent with the structural data?

Short Answer

Expert verified
Lewis structures and MO theory justify bond length variations. Increased electron repulsion and delocalization lead to longer bonds in \( \left[\mathrm{I}_3\right]^- \). The theoretical and experimental data are consistent.

Step by step solution

01

Drawing Lewis Structures

For \( \mathrm{I}_2 \), draw a single bond (I-I) connecting the two iodine atoms, with each iodine having three lone pairs. \( \left[\mathrm{I}_3\right]^+ \) is triangular, with a bond between the two terminal iodine atoms and the central one, typically showing a single bond with two lone pairs on each terminal iodine and one lone pair on the central iodine. \( \left[\mathrm{I}_3\right]^- \) is depicted with resonance between a chemical bond order of 0.5 between each I-I, due to delocalization, with two lone pairs on each outer iodine atom and three on the central one. These structures help understand why \( \left[\mathrm{I}_3\right]^- \) has a longer bond compared to \( \left[\mathrm{I}_3\right]^+ \).
02

Understanding Bond Distance Variation through Lewis Structures

The bond distances can be explained by the electron count: more electron density in I-I bonds (as in \( \mathrm{I}_2 \)) results in shorter bond lengths. In \( \left[\mathrm{I}_3\right]^- \), excess negative charge leads to longer bonds due to electron-electron repulsions. In \( \left[\mathrm{I}_3\right]^+ \), the lack of extra electrons leads to shorter bonds closer to \( \mathrm{I}_2 \) values. Hence, Lewis structures do provide insight into these bond length variations.
03

Bonding Analysis using Molecular Orbital Theory

Start with \( \mathrm{I}_2 \): it has a bond order of 1 with a σ-bond formed by overlap of p orbitals. For \( \left[\mathrm{I}_3\right]^+ \), the bond order is calculated as 0.5 per I-I bond due to electron removal resulting in a delocalized bonding scenario. In \( \left[\mathrm{I}_3\right]^- \), the electron addition results in a bond order of \(1/3\) per I-I bond, with more delocalization. The bond order directly influences bond length, explaining the longer bond in \( \left[\mathrm{I}_3\right]^- \).
04

Consistency Check with Structural Data

The calculated bond orders using MO theory align well with the observed bond lengths. A higher bond order in \( \mathrm{I}_2 \) results in the shortest bond length. The fractional bond order in \( \left[\mathrm{I}_3\right]^+ \) corresponds to an intermediate bond length, and the lower bond order in \( \left[\mathrm{I}_3\right]^- \) explains the longest bond length observed. Thus, the theoretical bond orders are consistent with the experimental data.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Lewis Structures
Lewis structures are a simple yet powerful tool to represent the bonds between atoms in a molecule. They are visual representations that show how atoms in a molecule are bonded together, where shared electrons form bonds and unshared electron pairs are shown as dots.
For iodine compounds, such as \[\mathrm{I}_2, \left[\mathrm{I}_3\right]^+,\text{ and } \left[\mathrm{I}_3\right]^-,\]these diagrams help visualize their bonding structure:
  • In \( \mathrm{I}_2 \), the Lewis structure features a single bond (I-I) connecting the two iodine atoms, with each iodine also having three lone pairs.
  • For \(\left[\mathrm{I}_3\right]^+\), the structure is a linear molecule with one bond between each terminal iodine and a central iodine, each terminal iodine also having two lone pairs and the central having one.
  • The \(\left[\mathrm{I}_3\right]^-\) structure is notable for its resonance between bonds, influencing bond lengths, where more lone pairs are on the central iodine compared to the others.
These representations allow us to estimate why varying electronegativity and resonance can lead to different bond lengths.
Bond Order
Bond order in chemistry indicates the number of chemical bonds between a pair of atoms. Generally, a higher bond order implies a stronger, shorter bond.
In terms of iodine compounds,\[\mathrm{I}_2 \text{ has a bond order of } 1.\]This suggests a single sigma bond formed due to p orbital overlap. For \(\left[\mathrm{I}_3\right]^+\), with one electron removed, the bond order reduces to 0.5 per each I-I connection, indicating reduced bonding strength due to electron delocalization.
In contrast, \(\left[\mathrm{I}_3\right]^-\) has an even lower bond order of \(1/3\) per bond due to extra electron addition and increased delocalization, explaining its longest bond length among the group. Understanding bond order aids in predicting and explaining the molecular properties such as stability and bond length.
Bond Length
Bond length refers to the distance between the nuclei of two bonded atoms. It varies depending on bond order, atom size, and electronegativity.
For iodine compounds, the bond lengths correspond with the bond orders such that:
  • In \(\mathrm{I}_2\), the bond length is 267 pm, correlating with a bond order of 1, indicating the shortest and strongest bond.
  • The \(\left[\mathrm{I}_3\right]^+\) has a bond length of 268 pm, slightly longer due to a bond order of 0.5, showing intermediate bond strength.
  • Finally, \(\left[\mathrm{I}_3\right]^-\) has the longest bond length of 290 pm corresponding to a much-reduced bond order of \(1/3\). This demonstrates weaker bonding resulting from electron overload and extra delocalization.
Bond length offers invaluable insights into molecular geometry and its reactivity or bonding properties.
Iodine Chemistry
Iodine chemistry is a fascinating area that explores the behavior and interaction of iodine in different chemical scenarios. Iodine, typically characterized by its heavy atomic mass and violet color, forms various interesting compounds such as molecular iodine \(\mathrm{I}_2\) and polyatomic species like \(\left[\mathrm{I}_3\right]^+\) and \(\left[\mathrm{I}_3\right]^-\).
These compounds showcase unique features like bond lengths and orders that are influenced by their electron configurations:
  • Molecular iodine \(\mathrm{I}_2\) shows simple diatomic characteristics with intense color and moderate reactivity.
  • The triiodide ion \(\left[\mathrm{I}_3\right]^-\) is often used as an indicator in redox reactions due to its electron-rich nature.
  • Conversely, \(\left[\mathrm{I}_3\right]^+\) is more scarce but significant in studying electron-deficient bonding scenarios.
Understanding these variations in iodine chemistry opens doors to more comprehensive applications in synthesis, analysis, and industrial chemistry, underscoring iodine's versatile nature and utility in chemical science.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) \(\mathrm{BCl}_{3}\) has \(D_{3 \mathrm{h}}\) symmetry. Draw the structure of \(\mathrm{BCl}_{3}\) and give values for the bond angles. \(\mathrm{NCl}_{3}\) has \(C_{3 \mathrm{v}}\) symmetry. Is it possible to state the bond angles from this information? (b) Derive the symmetry labels for the atomic orbitals on \(\mathrm{B}\) in \(\mathrm{BCl}_{3}\) and on \(\mathrm{N}\) in \(\mathrm{NCl}_{3}\).

Suggest an appropriate hybridization scheme for the central atom in each of the following species: (a) \(\mathrm{SiF}_{4}\); (b) \(\left[\mathrm{PdCl}_{4}\right]^{2-} ;(\mathrm{c}) \mathrm{NF}_{3} ;(\mathrm{d}) \mathrm{F}_{2} \mathrm{O} ;(\mathrm{e})\left[\mathrm{CoH}_{5}\right]^{4-}\); \((\mathrm{f})\left[\mathrm{FeH}_{6}\right]^{4-} ;(\mathrm{g}) \mathrm{CS}_{2} ;(\mathrm{h}) \mathrm{BF}_{3}\).

The structures below show (on the left) an octahedral and (on the right) a trigonal prismatic \(\mathrm{XY}_{6}\) molecule. (a) To what point groups do these molecules belong? (b) The bonding MOs in octahedral \(\mathrm{XY}_{6}\) have \(a_{1 g}, e_{g}\) and \(t_{1 u}\) symmetries. Confirm that these symmetries are consistent with the point group that you have assigned. (c) Can there be a triply degenerate set of bonding orbitals for the trigonal prismatic \(\mathrm{XY}_{6}\) molecule? Rationalize your answer.

What is meant by a ligand group orbital?

(a) Is \(\mathrm{CO}_{2}\) linear or bent? (b) What hybridization is appropriate for the \(C\) atom? (c) Outline a bonding scheme for \(\mathrm{CO}_{2}\) using the hybridization scheme you have suggested. (d) What \(C-O\) bond order does your scheme imply? (e) Draw a Lewis structure for \(\mathrm{CO}_{2}\). Is this structure consistent with the results you obtained in parts (c) and (d)?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free