Question

Sketch the resonance structures for the nitrite ion, \(\mathrm{NO}_{2}^{-} .\) Describe the electron-pair and molecular geometries of the ion. From these geometries, decide on the O-N-O bond angle, the average NO bond order, and the \(\mathrm{N}\) atom hybridization.

Short answer

The nitrite ion, \( \mathrm{NO}_{2}^{-} \), has a bent molecular geometry with a bond angle of about \(115^\circ\). The average \( \mathrm{N-O} \) bond order is 1.5, and the nitrogen is \( \mathrm{sp}^2 \) hybridized.

Step-by-step solution

Count the Valence Electrons

Nitrite ion, \( \mathrm{NO}_{2}^{-} \), consists of 1 nitrogen atom, 2 oxygen atoms, and 1 extra electron due to the negative charge. Nitrogen has 5 valence electrons, each oxygen has 6, and the extra electron adds 1 more, combining to a total of 18 valence electrons.

Draw the Resonance Structures

For \( \mathrm{NO}_{2}^{-} \), two major resonance structures can be drawn. In each, one oxygen forms a double bond (\( \mathrm{N} = \mathrm{O} \)) and the other a single bond (\( \mathrm{N}-\mathrm{O}^{-} \)) with nitrogen. The double bond location alternates between the oxygens in these structures. Each structure has a different oxygen holding the negative charge.

Describe Electron-Pair Geometry

Considering the Lewis structures and with 3 electron domains around nitrogen (2 bonds and 1 lone pair), the electron-pair geometry is trigonal planar.

Describe Molecular Geometry

With one lone pair affecting the geometry, the molecular shape of \( \mathrm{NO}_{2}^{-} \) is bent. This is due to the lone pair repulsion altering the angles between the bonds.

Estimate the O-N-O Bond Angle

In a perfect trigonal planar arrangement, the angles would be \(120^\circ\). However, due to lone pair repulsion, the \( \mathrm{O-N-O} \) bond angle is slightly less than \(120^\circ\), approximately \(115^\circ\).

Determine the Average N-O Bond Order

The presence of resonance structures means delocalization of electrons. Each nitrogen-oxygen bond averages one and a half bond orders, due to resonance between single and double bonds. Thus, the average \( \mathrm{N-O} \) bond order is 1.5.

Identify the Hybridization of Nitrogen

Given its trigonal planar electron-pair geometry involving 3 domains, nitrogen is \( \mathrm{sp}^2 \) hybridized to accommodate the bonds and lone pair.

Key concepts

Valence Electrons

Understanding valence electrons is crucial when studying molecular structures like the nitrite ion \( \mathrm{NO}_{2}^{-} \). Valence electrons are the outermost electrons of an atom and are involved in forming chemical bonds. For the nitrite ion, you need the valence electrons from nitrogen and oxygen and also include the extra electron due to its negative charge.

• Nitrogen contributes 5 valence electrons.
• Each oxygen atom provides 6 valence electrons, totaling 12 for both oxygens.
• The additional negative charge in the nitrite ion adds 1 more electron.

When combined, you have a total of 18 valence electrons.
Counting these electrons correctly is the first step in understanding the potential bonding and resonance structures of \( \mathrm{NO}_{2}^{-} \). Accurately mapping out the valence electrons helps in deriving the molecular geometry and predicting chemical behavior.

Trigonal Planar Geometry

The term trigonal planar geometry refers to the arrangement where three electron domains are situated around a central atom in a flat plane. For the nitrite ion, \( \mathrm{NO}_{2}^{-} \), it is crucial to recognize that this includes a combination of bonding pairs and lone pairs of electrons. Nitrogen, in this case, is the central atom surrounded by three domains: two nitrogen-oxygen bonds and one lone pair.

• It forms a triangle-like shape when visualized in three dimensions, hence the term "trigonal."
• All atoms are positioned around the central atom evenly.

The trigonal planar arrangement would typically form angles of \( 120^\circ \) between the bonds.
However, in \( \mathrm{NO}_{2}^{-} \), the lone pair on nitrogen slightly distorts this perfect geometry, resulting in a 'bent' shape for the actual molecular geometry. Recognizing this geometry is essential in understanding how the ion behaves in chemical reactions and interactions.

Hybridization

Hybridization is the concept of mixing atomic orbitals to form new, hybridized orbitals that can form chemical bonds. In the nitrite ion \( \mathrm{NO}_{2}^{-} \), nitrogen is involved in trigonal planar geometry. To accommodate its bonding and lone pair of electrons, nitrogen undergoes \( \mathrm{sp}^2\) hybridization. This means:

• One s orbital combines with two p orbitals to create three \( \mathrm{sp}^2 \) hybrid orbitals.
• Each \( \mathrm{sp}^2 \) orbital offers lower energy and greater directional character than pure atomic orbitals do.

These \( \mathrm{sp}^2\) orbitals arrange themselves trigonal planar around nitrogen, optimized for forming the bonds and accommodating the lone pair effectively.
Understanding hybridization helps explain not just the structure of the molecule but its reactivity and interaction with other molecules. For \( \mathrm{NO}_{2}^{-} \), this hybridization is key to understanding how the molecule stabilizes its resonance structures.

Bond Angle

The bond angle in a molecule describes the angle formed between two adjacent bonds that share a common atom. For \( \mathrm{NO}_{2}^{-} \), we need to take into account both the average bond order and structural influences. In an ideal trigonal planar configuration, bond angles are \( 120^\circ \).
However, with \( \mathrm{NO}_{2}^{-} \), the presence of a lone pair on nitrogen causes repulsion that slightly compresses the \( \mathrm{O-N-O} \) bond angle to less than \( 120^\circ \), approximately \( 115^\circ \).

• The lone pair exerts more repulsion than bonded pairs, reducing the bond angle.
• This adjustment ensures the molecule minimizes repulsive forces for stability.

Knowing the effect of lone pairs on bond angles is critical for predicting the true shape of molecules and understanding phenomena like bond strain and molecular fit during interactions.