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Chapter 2: Problem 2

Use the Lewis structure model to deduce the type of nitrogen-nitrogen bond present in (a) \(\mathrm{N}_{2} \mathrm{H}_{4},\) (b) \(\mathrm{N}_{2} \mathrm{F}_{4}\) (c) \(\mathrm{N}_{2} \mathrm{F}_{2}\) and (d) \(\left[\mathrm{N}_{2} \mathrm{H}_{5}\right]^{+}\)

Short Answer

Expert verified
N鈧侶鈧� and N鈧侳鈧� have single N鈥揘 bonds; N鈧侳鈧� has a double N=N bond; [N鈧侶鈧匽鈦� has a single N鈥揘 bond.

Step by step solution

01

Determine Valence Electrons

To determine the type of nitrogen-nitrogen bond, first, calculate the total number of valence electrons available for each molecule. - **N鈧侶鈧�**: Nitrogen has 5 valence electrons each, and hydrogen has 1. Thus, for N鈧侶鈧�, we have \( 2 \times 5 + 4 \times 1 = 14 \) valence electrons.- **N鈧侳鈧�**: Nitrogen has 5 valence electrons each, and fluorine has 7. Thus, for N鈧侳鈧�, we have \( 2 \times 5 + 4 \times 7 = 34 \) valence electrons.- **N鈧侳鈧�**: Similarly, for N鈧侳鈧�, we have \( 2 \times 5 + 2 \times 7 = 24 \) valence electrons.- **[N鈧侶鈧匽鈦�**: Nitrogen has 5 valence electrons each, and hydrogen has 1. Additionally, due to the positive charge, subtract 1 electron. Thus, we have \( 2 \times 5 + 5 \times 1 - 1 = 14 \) valence electrons.
02

Sketch Basic Lewis Structures

Based on total valence electrons, draw the basic Lewis structures for each molecule: - **N鈧侶鈧�**: Connect two nitrogen atoms with a single bond (N鈥揘) and satisfy each nitrogen with three hydrogen atoms each. - **N鈧侳鈧�**: Connect two nitrogen atoms with a single bond (N鈥揘) and attach two fluorine atoms to each nitrogen atom. - **N鈧侳鈧�**: Connect the two nitrogen atoms with a double bond (N=N) and attach one fluorine atom to each nitrogen atom. - **[N鈧侶鈧匽鈦�**: Connect the two nitrogen atoms with a single bond (N鈥揘) and attach a set of hydrogens, distributed evenly to give each nitrogen atom a single bonded hydrogen, one nitrogen will have three hydrogens forming the cationic structure.
03

Analyze Bond Types in Structures

Observe the Lewis structures to identify nitrogen-nitrogen bond orders based on shared electron pairs: - **N鈧侶鈧�**: Displays a single N鈥揘 bond with shared lone electron pairs creating a stable single bond. - **N鈧侳鈧�**: Has a single N鈥揘 bond where each nitrogen is saturated with its other bonded atoms. - **N鈧侳鈧�**: Has a double N=N bond to meet the octet rule with shared electrons. - **[N鈧侶鈧匽鈦�**: Contains a single N鈥揘 bond as all valence electrons are paired with hydrogen while maintaining a positive charge.

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Key Concepts

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

Understanding Nitrogen-Nitrogen Bonds
Nitrogen-nitrogen bonds are the connections between two nitrogen atoms in a molecule. These bonds can be classified based on how many pairs of electrons are shared between the atoms. There are three main types:
  • Single bond (N鈥揘): One pair of electrons is shared. This is the weakest type of nitrogen-nitrogen bond and typically found in compounds like hydrazine ( N鈧侶鈧�).
  • Double bond (N=N): Two pairs of electrons are shared, providing a stronger bond. This is found in molecules like difluorodiazine ( N鈧侳鈧�).
  • Triple bond (N鈮): This bond involves three shared pairs of electrons and is the strongest, as seen in nitrogen gas ( N鈧�). However, it does not appear in the given exercise compounds.
Each bond type affects the molecule's stability and reactivity. Recognizing these bonds is crucial for understanding molecular behavior and interactions.
The Role of Valence Electrons
Valence electrons are the outermost electrons of an atom and play a fundamental role in bond formation. When drawing Lewis structures, the first step is to determine the total number of valence electrons available for bonding. This is done by adding the valence electrons of each atom and considering any charge present.
  • For nitrogen, there are 5 valence electrons per atom.
  • For hydrogen, there is 1 valence electron per atom.
  • Fluorine has 7 valence electrons per atom.
In the given compounds, calculating the available valence electrons helps predict possible molecular structures:
  • N鈧侶鈧� has 14 valence electrons.
  • N鈧侳鈧� has 34 valence electrons.
  • N鈧侳鈧� has 24 valence electrons.
  • [ N鈧侶鈧匽鈦� has 14 valence electrons after accounting for its positive charge.
Understanding valence electrons aids in anticipating bond formation and molecular polarity.
Exploring Molecular Geometry
Molecular geometry describes the three-dimensional arrangement of atoms in a molecule. It is influenced by the types of bonds and the number of lone pairs around the central atom. Here are some typical geometries resulting from the compounds in the exercise:
  • N鈧侶鈧� (hydrazine) is described as having a "bent" or "angular" geometry due to its single N鈥揘 bond and surrounding hydrogen atoms.
  • N鈧侳鈧� tends to have a "staggered" or "twisted" conformation to minimize repulsion between large fluorine atoms.
  • N鈧侳鈧�, with its double N=N bond, typically exhibits a planar geometry, maintaining equal spacing amongst atoms.
  • [ N鈧侶鈧匽鈦�, considering the positive charge and single N鈥揘 bond, forms a somewhat asymmetrical geometry with the attached hydrogens.
These shapes are crucial to predicting how a molecule might interact with others, and can influence properties like solubility and boiling points.
Deciphering Bond Orders
The bond order refers to the number of chemical bonds between a pair of atoms. In nitrogen-containing compounds, it tells us how strongly two nitrogen atoms are linked.
  • A single bond has a bond order of 1, as there's only one pair of shared electrons, seen in N鈧侶鈧� and N鈧侳鈧�.
  • A double bond, with a bond order of 2, indicates two pairs of shared electrons, as observed in N鈧侳鈧�.
Higher bond orders generally correspond to shorter and stronger bonds. By evaluating these bond orders, chemists can deduce a wealth of information about molecular stability, reactivity, and overall chemical behavior. This insight is vital for applications in synthesis and materials science.

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Most popular questions from this chapter

State whether you expect the following species to possess stereoisomers and, if so, draw their structures and give them distinguishing labels: (a) \(\mathrm{BF}_{2} \mathrm{Cl}\) (b) \(\mathrm{POCl}_{3}\) (c) MePF \(_{4} ;(\mathrm{d})\left[\mathrm{PF}_{2} \mathrm{Cl}_{4}\right]^{-}\)

Confirm that the octet rule is obeyed by each of the atoms in the following molecules: (a) \(\mathrm{CF}_{4},\) (b) \(\mathrm{O}_{2}\) , (c) AsBr \(_{3}\) (d) \(\mathrm{SF}_{2}\)

(a) Use MO theory to determine the bond order in each of \(\left[\mathrm{He}_{2}\right]^{+}\) and \(\left[\mathrm{He}_{2}\right]^{2+}\) (b) Does the MO picture of the bonding in these ions suggest that they are viable species?

One member of cach of the following sets of compounds is not isoelectronic with the others. Which one in each set is the odd one out? (a) \(\left[\mathrm{NO}_{2}\right]^{+}, \mathrm{CO}_{2},\left[\mathrm{NO}_{2}\right]^{-}\) and \(\left[\mathrm{N}_{3}\right]^{-}\) (b) \([\mathrm{CN}]^{-}, \mathrm{N}_{2}, \mathrm{CO},[\mathrm{NO}]^{+}\) and \(\left[\mathrm{O}_{2}\right]^{2-}\) (c) \(\left[\mathrm{SiF}_{6}\right]^{2-},\left[\mathrm{PF}_{6}\right]^{-},\left[\mathrm{AlF}_{6}\right]^{3-}\) and \(\left[\mathrm{BrF}_{6}\right]^{-}\)

(a) Write down the ions that are present in the compound \(\left[\mathrm{PCl}_{4}\right]\left[\mathrm{PCl}_{3} \mathrm{F}_{3}\right] .\) What shape do you expect each ion to adopt? In theory, does cither ion possess stereoisomers? (b) Use the VSEPR model to rationalize why \(\mathrm{BCl}_{3}\) and \(\mathrm{NCl}_{3}\) do not adopt similar structures. Is either molecule expected to be polar? Rationalize your answer.
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