Logout Open In App
Open In App
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

Chapter 4: Problem 8

In which of the following compounds will the bond angle be maximum? (a) \(\mathrm{NH}_{3}\) (b) \(\mathrm{NH}_{4}^{+}\) (c) \(\mathrm{PCl}_{3}\) (d) \(\mathrm{SCl}_{2}\)

Short Answer

Expert verified
The maximum bond angle is in \(\mathrm{NH}_{4}^{+}\) with 109.5 degrees.

Step by step solution

01

Understand the Electron Pair Geometry

The bond angle depends on the repulsion between electron pairs according to VSEPR (Valence Shell Electron Pair Repulsion) theory. We need to consider the molecular shape and any lone pairs that might affect the bond angles.
02

Analyze each compound

- (a) In \(NH_3\), the compound has a trigonal pyramidal shape with one lone pair, leading to a bond angle of about 107 degrees.- (b) In \(NH_4^+\), the compound has a tetrahedral shape with no lone pairs, resulting in bond angles of 109.5 degrees.- (c) In \(PCl_3\), the compound is trigonal pyramidal with a lone pair, similar to NH3, leading to slightly smaller angles.- (d) In \(SCl_2\), the compound has a bent shape with two lone pairs, resulting in bond angles significantly less than 109.5 degrees.
03

Choose the compound with the maximum bond angle

Tetrahedral shapes generally have the largest bond angle of 109.5 degrees due to symmetric repulsions of bonded atoms. Among the given options, \(\mathrm{NH}_{4}^{+}\) is the only tetrahedral compound.

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.

Start your free trial

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.

Bond Angle
Bond angles play a crucial role in understanding the geometry of molecules. They are defined as the angle between two bonds that have a common atom. For example, in a water molecule, the bond angle refers to the angle between the two hydrogen-oxygen bonds.

In the context of VSEPR theory, bond angles are influenced by the repulsions between electron pairs around the central atom. These can include:
  • Bonds between atoms.
  • Lone pairs of electrons, which are not involved in bonding.
  • The general presence of both bonding pairs and lone pairs.
The presence of lone pairs usually results in smaller bond angles because lone pairs repel more strongly than bonding pairs. As seen in the exercise,
  • In \(\mathrm{NH}_{4}^{+}\), the bond angle is 109.5 degrees due to the absence of lone pairs—this maximizes symmetrically due to the tetrahedral shape.
  • In \(\mathrm{NH}_{3}\), the bond angle is slightly lower at about 107 degrees because of one lone pair, which reduces the bond angle slightly from 109.5 degrees.
  • In \(\mathrm{SCl}_{2}\), the bond angle is significantly smaller due to two lone pairs affecting the geometry greatly.
Electron Pair Geometry
Electron pair geometry is a foundational concept that dictates the three-dimensional arrangement of electron pairs around a central atom.
According to VSEPR theory, electron pairs arrange themselves as far apart as possible to minimize repulsion. This leads to different geometries:
  • Linear: Occurs with two electron pairs, forming a straight line with a bond angle of 180 degrees.
  • Trigonal planar: Involves three pairs, creating a plane with 120-degree angles.
  • Tetrahedral: Has four electron pairs, creating a three-dimensional arrangement with bond angles of 109.5 degrees.
  • Trigonal bipyramidal: Consists of five electron pairs, creating varied angles of 120 and 90 degrees.
  • Octahedral: Comprises six pairs, producing 90-degree angles.
In the exercise, \(\mathrm{NH}_{4}^{+}\) exhibits a tetrahedral geometry with four bonding pairs and no lone pairs, which provides the typical bond angle of 109.5 degrees. This is consistent with the VSEPR prediction for a molecule with a tetrahedral electron pair geometry.
Molecular Shape
Molecular shape is how a molecule appears based on the positions of nuclei, considering only bonded atoms. It is crucial for understanding how a molecule behaves in interactions.
When lone pairs are present, they influence the molecular shape significantly. This is because lone pairs exert more repulsion than bonding pairs, thus distorting the tracer path traced by the atoms. Some common molecular shapes include:
  • Linear: Even without lone pairs, a linear shape maintains straight bonds.
  • Trigonal Planar: Seen in molecules with no lone pairs, adopting a planar structure.
  • Trigonal Pyramidal: Occurs with one lone pair, changing from planar to a three-sided pyramid.
  • Bent: Occurs in molecules with two bonding pairs and one or more lone pairs.
Consider the exercise's listed molecules:
  • \(\mathrm{NH}_{3}\) has a trigonal pyramidal shape, as it has one lone pair which affects its typical trigonal planar configuration.
  • \(\mathrm{PCl}_{3}\), similar to \(\mathrm{NH}_{3}\), possesses a lone pair, resulting in a pyramidal shape.
  • \(\mathrm{SCl}_{2}\) shows a bent shape due to two lone pairs, which reduces the bond angles significantly from the ideal tetrahedral angle.
Understanding these shapes helps explain the properties and behaviors of different chemical compounds in various contexts.

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 molecules/ions does not contain unpaired electrons? (a) \(\mathrm{O}_{2}^{2-}\) (b) \(\mathrm{B}_{2}\) (c) \(\mathrm{N}_{2}^{+}\) (d) \(\mathrm{O}_{2}\)

Both \(\mathrm{BF}_{3}\) and \(\mathrm{NF}_{3}\) are covalent but \(\mathrm{BF}_{3}\) molecule is non-polar while \(\mathrm{NF}_{3}\) is polar because: (a) Atomic size of boron is smaller than nitrogen (b) \(\mathrm{BF}_{3}\) is planar but \(\mathrm{NF}_{3}\) is pyramidal (c) Boron is a metal while nitrogen is gas (d) BF bond has no dipole moment while NF bond has dipole

What is the number of sigma and pi bonds present in a molecule of sulphuric acid? (a) \(6 \sigma, 2 \pi\) (b) \(6 \sigma, 0 \pi\) (c) \(2 \sigma, 4 \pi\) (d) \(2 \sigma, 2 \pi\)

Among the following compound which one has maximum number of lone pairs of electrons on central atom: (a) \(\left[\mathrm{ClO}_{3}\right]^{-}\) (b) \(\mathrm{XeF}_{A}\) (c) \(\mathrm{SF}_{4}\) (d) \(\left[\mathrm{I}_{3}\right]\)

The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing power of the cationic species, \(\mathrm{K}^{+}, \mathrm{Ca}^{2+}\), \(\mathrm{Mg}^{2+}, \mathrm{Be}^{2+} ?\) (a) \(\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2}+<\mathrm{Mg}^{2+}\) (b) \(\mathrm{K}^{+}<\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}\) (c) \(\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}\) (d) \(\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2+}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation
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