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Chapter 13: Problem 107

The correct order of bond angle of \(\mathrm{NO}_{2}^{+}, \mathrm{NO}_{2}\) and \(\mathrm{NO}_{2}\) is (a) \(\mathrm{NO}_{2}^{+}<\mathrm{NO}_{2}<\mathrm{NO}_{2}^{-}\) (b) \(\mathrm{NO}_{2}^{+}=\mathrm{NO}_{2}^{-}<\mathrm{NO}_{2}\) (c) \(\mathrm{NO}_{2}^{+}>\mathrm{NO}_{2}^{-}>\mathrm{NO}_{2}\) (d) \(\mathrm{NO}_{2}^{\top}>\mathrm{NO}_{2}<\mathrm{NO}_{2}^{-}\)

Short Answer

Expert verified
Option (c) \(\mathrm{NO}_{2}^{+} > \mathrm{NO}_{2} > \mathrm{NO}_{2}^{-}\).

Step by step solution

01

Understanding the Problem

The problem asks us to order the bond angles of three molecules: \(\mathrm{NO}_{2}^{+}, \mathrm{NO}_{2}\), and \(\mathrm{NO}_{2}^{-}\). Bond angles are determined by the arrangement of electron pairs around the central atom. We must identify which has the largest and smallest bond angle.
02

Analyzing Molecular Geometries

Each molecule differs based on its number of electron pairs and lone electron pairs. \(\mathrm{NO}_{2}^{+}\) has a linear geometry with no lone pairs, \(\mathrm{NO}_{2}\) has a bent geometry with one lone pair, and \(\mathrm{NO}_{2}^{-}\) also has a bent geometry but with two lone pairs.
03

Applying VSEPR Theory

VSEPR (Valence Shell Electron Pair Repulsion) theory suggests that the presence of lone pairs decreases the bond angle by repelling the bonding pairs inward. Fewer lone pairs mean wider angles. Hence, \(\mathrm{NO}_{2}^{+}\) with no lone pairs will have the largest bond angle.
04

Determining Relative Angles

Given that \(\mathrm{NO}_{2}\) has one lone pair and \(\mathrm{NO}_{2}^{-}\) has two, \(\mathrm{NO}_{2}\) will have a larger bond angle compared to \(\mathrm{NO}_{2}^{-}\). Therefore, the bond angles decrease in this order: \(\mathrm{NO}_{2}^{+} > \mathrm{NO}_{2} > \mathrm{NO}_{2}^{-}\).
05

Concluding Order

The correct order from largest to smallest bond angle is: \(\mathrm{NO}_{2}^{+} > \mathrm{NO}_{2} > \mathrm{NO}_{2}^{-}\), which corresponds to option (c).

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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 all about the spatial arrangement of atoms in a molecule. This geometry helps determine a molecule's shape and affects properties like polarity, reactivity, and phase of matter.
Understanding molecular geometry involves:
  • Atom Positioning: The central atom is surrounded by other atoms, with lone or bonding pairs of electrons influencing their positions.
  • VSEPR Theory: It explains how electron pairs, whether bonding or lone, layout to minimize repulsion, determining structure.
In this exercise, we have different molecular geometries:
  • oindent ext{NO}_2^+: with a linear shape due to no lone pairs, allowing atoms to stretch out evenly.
  • NO2: bent because of one lone pair, pushing atoms closer together compared to a linear shape.
  • NO2-: also bent but with more compression due to two lone pairs, resulting in a more pronounced angle.
Bond Angles
Bond angles refer to the angles between two bonding pairs around a central atom. These angles influence the 3D shape and essentially define the molecule's geometry. In our studied molecules, bond angles are crucial.
  • No Lone Pairs: As seen in \( \mathrm{NO}_2^+ \), bond angles are at their widest since electron repulsion is minimized directly by the shortage of lone pairs.
  • One Lone Pair: Present in \( \mathrm{NO}_2 \). The bond angle narrows as the lone pair pushes the bonding pairs closer, distorting the ideal bond angle.
  • Two Lone Pairs: Seen in \( \mathrm{NO}_2^- \). This scenario further decreases the bond angle due to increased electron pair repulsion between the lone pairs and bonding pairs.
Such differences in bond angles are vividly illustrated by each molecule鈥檚 electronic setup.
Electron Pairs
Electron pairs play a vital role in defining molecular structure and its subsequent geometry. These pairs exist either as shared (bonding pairs) contributing to chemical bonds or as unshared (lone pairs) localized on the central atom.
  • Bonding Pairs: Found between atoms, stabilizing the molecule by forming covalent bonds.
  • Lone Pairs: Found on a central atom and do not participate directly in bonding, but affect geometry by altering angles.
  • VSEPR Consideration: The VSEPR Theory posits that electron pairs, being negatively charged, repel each other and strive to be as far apart as possible. Placement affects the ultimate geometry.
In \( \mathrm{NO}_2^+ \), the absence of lone pairs lets bonding pairs spread maximum apart. Meanwhile, \( \mathrm{NO}_2 \) and \( \mathrm{NO}_2^- \) possess lone pairs, adjusting geometry through their repulsions.
Lone Pairs
Lone pairs are those electron pairs residing near the central atom but not partaking into bonding. These lone pairs have a significant impact on molecular geometry and bond angles.
  • Presence: Affects electronic and molecular geometries by repelling bonding pairs.
  • Increasing Repulsion: More lone pairs increase repulsive forces within a molecule, shrinking bond angles compared to their lone pair-free counterparts.
  • Example in \( \mathrm{NO}_2 \): Here, a singular lone pair distorts the geometric plan, causing deviation from linearity to a bent form.
  • Example in \( \mathrm{NO}_2^- \): Two lone pairs amplify the deviation further, pressing bonding pairs more.
Evaluating lone pairs helps in comprehending the 3D shape and how electron distribution affects complete molecular behavior.

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Most popular questions from this chapter

Which of the following species exhibits the diamagnetic behaviour? (a) \(\mathrm{O}_{2}^{+}\) (b) \(\mathrm{O}_{2}\) (c) NO (d) \(\mathrm{O}_{2}^{2-}\)

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, \(\mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}, \mathrm{Be}^{2++}\) ? [2007] (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) \(\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2+}\)

Which of the following molecules has trigonal planar geometry? (a) \(\mathrm{BF}_{3}\) (b) \(\mathrm{NH}_{3}\) (c) \(\mathrm{PCl}_{3}\) (d) \(\mathrm{IF}_{3}\)

The molecular shapes of \(\mathrm{SF}_{4}, \mathrm{CF}_{4}\) and \(\mathrm{XeF}_{4}\) are \([2005]\) (a) the same with 2,0 and 1 lone pairs of electrons on the central atom, respectively (b) the same with 1,1 and 1 lone pair of electrons on the central atom, respectively (c) different with 0,1 and 2 lone pairs of electrons on the central atom, respectively (d) different with 1,0 and 2 lone pairs of electrons on the central atom, respectively

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