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 II-O-II bond angle in \(\mathrm{II}_{2} \mathrm{O}\) is \(104.5^{\circ}\) and not \(109^{\circ} 28^{\prime}\) because of (1) high electronegativity of oxygen (2) bond pair-bond pair repulsion (3) Lonc pair-lone pair repulsion (4) Lonc pair-bond pair repulsion

Short Answer

Expert verified
The bond angle in \(\text{H}_{2}\text{O}\) is \[104.5^{\text{circ}}\] because of lone pair-lone pair repulsion (option 3).

Step by step solution

01

Identify the Geometry

Recognize that the molecule \(\text{H}_{2}\text{O}\) has a bent or V-shaped molecular geometry.
02

Determine Ideal Bond Angle

Understand that the ideal bond angle for a tetrahedral arrangement without lone pairs is \[109.5^{\text{circ}}\].
03

Consider Lone Pair Effects

Hydrogen peroxide (\(\text{H}_{2}\text{O}\)) has two lone pairs on the oxygen atom. These lone pairs repel more strongly than bonding pairs, reducing the bond angle.
04

Examine Repulsions

Lone pair-lone pair repulsion is stronger than bond pair-bond pair and lone pair-bond pair repulsion, causing the bond angle to decrease from the ideal tetrahedral angle.
05

Confirm the Correct Answer

Given the information, the bond angle in \(\text{H}_{2}\text{O}\) is \[104.5^{\text{circ}}\], primarily due to lone pair-lone pair repulsion. Therefore, the correct answer is option (3).

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.

Molecular Geometry
The molecular geometry of a molecule describes the three-dimensional arrangement of its atoms. For \(\text{H}_{2}\text{O}\), the molecular geometry is bent or V-shaped. This is because the oxygen atom in water forms two bonds with hydrogen atoms and has two lone pairs of electrons. According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, a molecule will arrange its electron pairs to minimize repulsion. In water's case, the two lone pairs and two bond pairs spread out to least interact with each other. This results in a bent shape with a bond angle less than the ideal tetrahedral angle.
Lone Pair Repulsion
Lone pair repulsion refers to the repulsive force between the non-bonding pairs of electrons in a molecule. These lone pairs are located on the oxygen atom in \(\text{H}_{2}\text{O}\). Lone pairs occupy more space around the central atom than bonding pairs, causing greater repulsion. This increased repulsion between lone pairs forces the bonding pairs closer together. As a result, the bond angle in \(\text{H}_{2}\text{O}\) is reduced from the ideal tetrahedral angle of 109.5° to 104.5°.
Bond Pair Repulsion
Bond pair repulsion involves the repulsive force between bonding electron pairs in a molecule. Although bond pair repulsion is present, it is not as strong as lone pair repulsion. In \(\text{H}_{2}\text{O}\), the bond pairs between the oxygen and hydrogen atoms have a repulsive interaction. This repulsion is less intense compared to lone pair-lone pair and lone pair-bond pair interactions. Thus, while bond pair repulsion does affect the structure of the molecule, it is not the primary reason for the reduced bond angle.
Electronegativity Effects
Electronegativity is the ability of an atom to attract electrons towards itself. Oxygen, with its high electronegativity, attracts electrons more strongly than hydrogen. This results in a partial negative charge on the oxygen and a partial positive charge on the hydrogens. While the electronegativity of oxygen influences bond polarity and the molecule’s overall dipole moment, it does not directly cause the reduction of the bond angle in \(\text{H}_{2}\text{O}\). The primary factor responsible for the bond angle of 104.5° is the repulsion between lone pairs on the oxygen atom, not the difference in electronegativity.

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

Which of the following statement is false? (1) The boiling point of a compound is raised by intermolecular hydrogen bonding. (2) o-nitrophenol is more volatile than p-nitrophenol because of intramolecular hydrogen bonding. (3) Lower alcohols such as methanol and ethanol arc miscible in water due to hydrogen bonding. (4) Intramolecular hydrogen bond enhances the boiling point.

Which orbital is used by oxygen atom to form a sigma bond with other oxygen atom in \(\mathrm{O}_{2}\) molecule? (1) sp hybrid orbital (2) \(\mathrm{sp}^{2}\) hybrid orbital (3) \(\mathrm{sp}^{3}\) hybrid orbital (4) pure p-orbital

The shape of \(\mathrm{NO}_{3}^{-}\) is planar. It is formed by the overlapping of oxygen orbitals with which orbitals of nitrogen are (1) sp \(^{3}\) hybridised (2) sp \(^{2}\) hybridised (3) sp hybridiscd (4) 3 purc p-orbitals

Which of the following statement is falsc? (1) The measurement of dipole moment of a gaseous diatomic molecule is a direct indication of bond polarity. (2) For a polyatomic molecule, its dipole moment is determined only from bond moments. (3) \(\mathrm{SO}_{2}\) is nonlinear and as a consequence it should have dipole moment, (4) Lone pair of electrons present on central atom can give rise to dipole moment.

The order of relative strengths of bonds formed by sp, \(\mathrm{sp}^{2}\) and \(\mathrm{sp}^{3}\) hybrid orbitals is (1) \(\mathrm{sp}>\mathrm{sp}^{2}>\mathrm{sp}^{3}\) (2) \(\mathrm{sp}<\mathrm{sp}^{2}<\mathrm{sp}^{3}\) (3) \(\mathrm{sp}=\mathrm{sp}^{2}=\mathrm{sp}^{3}\) (4) \(\mathrm{sp}<\mathrm{sp}^{2}>\mathrm{sp}^{3}\)

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