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Chapter 4: Problem 85

How does the nitrogen-oxygen bond length in the nitrate ion compare to the nitrogen-oxygen bond length in the nitrite ion?

Short Answer

Expert verified
Explain the reasoning behind your answer. Answer: The nitrogen-oxygen bond length in the nitrate ion (NO3-) is shorter compared to the nitrogen-oxygen bond length in the nitrite ion (NO2-). This is due to the higher average bond order of 1.67 in the nitrate ion, compared to the average bond order of 1.5 in the nitrite ion. Higher bond order corresponds to shorter bond lengths.

Step by step solution

01

Draw the Lewis structures and resonance forms for nitrate and nitrite ions

Firstly, it's important to draw the Lewis structures and resonance forms for both ions: For the nitrate ion (NO3-): 1. Nitrogen (N) is in the center with a single bond to each of the three oxygens (O). 2. Each oxygen has two lone pairs. 3. There is one extra electron (as it has -1 charge) that is distributed among the N-O bonds as double bonds in resonance structures. Resonance forms for nitrate ion: [N(+),O(-)][O(+),N,O(-)][O(-), N, O(+)] For the nitrite ion (NO2-): 1. Nitrogen (N) is in the center with a single bond and a double bond to the two oxygens (O). 2. Each oxygen has two lone pairs. 3. There is one extra electron (as it has -1 charge) that is distributed among the N-O bonds as double bonds in resonance structures. Resonance forms for nitrite ion: [N(+),O(-)][O(-),N,O(+)]
02

Calculate the average bond orders

To determine the bond length, we need to find the average bond order in each ion. Bond order is the number of bonds between two atoms, averaged across all resonance structures. For the nitrate ion (NO3-): The average bond order is equal to the total number of bonds shared with nitrogen divided by the number of oxygen atoms. Average bond order for nitrate ion: (3 single bonds + 1 double bond) / 3 Oxygen atoms = (3 + 2) / 3 = 5/3 鈮� 1.67 For the nitrite ion (NO2-): The average bond order is equal to the total number of bonds shared with nitrogen divided by the number of oxygen atoms. Average bond order for nitrite ion: (1 single bond + 1 double bond) / 2 Oxygen atoms = (1 + 2) / 2 = 3/2 = 1.5
03

Compare the bond lengths

Since the nitrate ion has a higher average bond order (1.67) than the nitrite ion (1.5), the nitrogen-oxygen bond in the nitrate ion is shorter as higher bond order corresponds to a shorter bond length. In conclusion, the nitrogen-oxygen bond length in the nitrate ion (NO3-) is shorter compared to the nitrogen-oxygen bond length in the nitrite ion (NO2-) due to a higher average bond order.

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Key Concepts

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

Lewis Structures
Lewis structures are a way to visually represent molecules and ions to showcase their atoms and bonding. These diagrams are essential for understanding how atoms combine and arrange themselves.
When drawing Lewis structures, follow these steps:
  • Determine the total number of valence electrons available. For ions, adjust for any charge present by adding or subtracting electrons accordingly.
  • Select the central atom, typically the least electronegative element, except hydrogen, which is never the central atom in these structures.
  • Place the remaining atoms around the central atom and draw single bonds to connect them. Each bond represents a pair of electrons.
  • Distribute the rest of the valence electrons as lone pairs around the outer atoms to satisfy the octet rule, placing any extra electrons on the central atom if needed.
For instance, when sketching the Lewis structure for the nitrate ion, NO鈧冣伝, nitrogen is the central atom with single bonds to the three oxygen atoms. An additional electron pair is drawn to represent the extra charge, distributed to balance the resonance forms.
Resonance Structures
Resonance structures refer to different possible arrangements of electrons in a molecule that have the same atomic connectivity. These variations highlight the delocalization of electrons across certain atoms.
Despite their usefulness for visualizing electron distribution, a single resonance structure cannot fully depict a molecule's reality. Instead, the true electron configuration is a hybrid of all valid resonance forms.
To identify resonance structures:
  • Ensure that each resonance form obeys the rules of valence, particularly adhering to the octet rule wherever possible.
  • Shift electrons, specifically lone pairs and pi bonds, while keeping the positions of atoms constant across different structures.
  • Represent the different forms using a double-headed arrow between structures to alert readers to the resonance.
For example, the nitrate ion (NO鈧冣伝) showcases resonance through varied electron placements in double bonds between nitrogen and oxygen. This distribution allows each oxygen a chance to share its pi electrons with nitrogen, forming equivalent structures.
Bond Order
Bond order is a critical concept in molecular chemistry, providing insight into bond stability, length, and strength. It's defined as the number of bonds between two atoms in a molecule.
To find the bond order:
  • Count the total number of bonds present between the atoms in all resonance structures.
  • Divide this total by the number of resonance structures to get an average order.
Higher bond orders imply shorter and stronger bonds due to increased electronic overlap between bonded atoms.
In our exercise, the nitrate ion features a bond order of approximately 1.67. This is in contrast to the nitrite ion's bond order of 1.5, leading to shorter nitrogen-oxygen bonds in the nitrate ion. This relationship between bond order and bond length is key in understanding why some bonds are more tightly held together than others.

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Most popular questions from this chapter

Draw Lewis structures for fulminic acid (HCNO) that show all resonance forms.

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