Question

Based on Lewis structures, predict the ordering of \(\mathrm{N}-\mathrm{O}\) bond lengths in \(\mathrm{NO}^{+}, \mathrm{NO}_{2}^{-}\), and \(\mathrm{NO}_{3}^{-} .\)

Short answer

Based on their Lewis structures and bond orders, the N-O bond lengths have the following order: \(NO^{+} < NO_{2}^{-} < NO_{3}^{-}\). The shortest bond length is in \(NO^{+}\), followed by \(NO_{2}^{-}\), and the longest bond length is in \(NO_{3}^{-}\).

Step-by-step solution

Draw the Lewis structures of the molecules/ions

For each given species, we need to draw their respective Lewis structures. Let's start with NO^+. 1. NO^+ (nitrosonium ion): N has 5 valence electrons and O has 6 valence electrons. As there is one positive charge on the ion, one electron is removed from the system, leaving us with a total of 10 electrons. Therefore, the Lewis structure will be: O = N^+ 2. NO2^- (nitrite ion): N has 5 valence electrons and O has 6 valence electrons. The negative charge adds an additional electron, giving us a total of 18 electrons. The Lewis structure can be drawn as follows (with a resonance structure): O = N - O^- O^-. | N = O 3. NO3^- (nitrate ion): N has 5 valence electrons and O has 6 valence electrons. Again, considering an additional electron due to the negative charge, we have a total of 24 electrons. The Lewis structure can be drawn as follows: O^-. | N - O^- \ O^- Now that we have the Lewis structures, we can determine the bond order.

Determine the bond order

Bond order is the number of chemical bonds between a pair of atoms, which can be calculated as ½*(number of bonding electrons - number of anti-bonding electrons). For our species: 1. NO^+: There's a double bond between N and O, so the bond order is 2. 2. NO2^-: There is a resonance structure for the nitrite ion, with a π bond shared between the two oxygen atoms, so the bond order is 1.5 (3/2). 3. NO3^-: There's a resonance structure in the nitrate ion as well, with three single bonds and three π bonds shared between three oxygen atoms, which gives a bond order of 1(⅓) + 2(⅔) = 4/3 = 1.333.

Predict the ordering of the N-O bond lengths

Now that we have the bond orders, we can use them to predict the bond lengths. A higher bond order means stronger bonds and shorter bond lengths. So, the ordering of the bond lengths based on the bond orders will be: NO^+ (2) < NO2^- (1.5) < NO3^- (1.333) In conclusion, the N-O bond length in NO^+ is the shortest, followed by NO2^-, and the longest N-O bond is in NO3^-.

Key concepts

Bond Order

When discussing molecules, understanding bond order is crucial. Bond order is a concept that quantifies the number of bonds between a pair of atoms. It is calculated using the formula: \[ \text{Bond Order} = \frac{1}{2} \times (\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons}) \]A higher bond order indicates a stronger, shorter bond, while a lower bond order corresponds to a longer, weaker bond.

• For the nitrosonium ion, \(NO^+\), the bond order is 2 because there is one double bond and no resonance structures contributing additional bonds.
• In the nitrite ion, \(NO_2^-\), we encounter resonance, which will be explored further, resulting in a bond order of 1.5. This means one and a half bonds effectively form between nitrogen and oxygen.
• For the nitrate ion, \(NO_3^-\), the bond order is approximately 1.333 due to resonance leading to partial bonds spread across three oxygen atoms.

Recognizing bond order helps us predict bond strengths and lengths throughout different molecules.

Resonance Structures

Resonance structures are an essential part of understanding molecular stability and electron distribution. These structures are various forms a molecule can take by redistributing electrons in multiple ways, without moving the atoms.

• Resonance is not an oscillation between forms; rather, it represents the average of the electron distribution.
• In \(NO_2^-\), the nitrite ion, resonance allows a π bond to delocalize across the two oxygen atoms, which results in an average bond order between nitrogen and each oxygen.
• In \(NO_3^-\), the nitrate ion, resonance affects the placement of π bonds between one nitrogen and three oxygen atoms. This creates an even spread of partial double bonds across all N-O bonds, thus reducing the individual bond strength to an order of around 1.333.

Understanding resonance is key to explaining the behavior and properties of molecules, especially when calculating bond orders and predicting molecular geometry.

Molecular Geometry

Molecular geometry discusses the shape of molecules, which is determined by the spatial arrangement of atoms. This concept relies on the VSEPR (Valence Shell Electron Pair Repulsion) theory, which states that electron pairs repel and molecules adjust their shapes to minimize this repulsion.Let's explore the geometries of \(NO^+\), \(NO_2^-\), and \(NO_3^-\):

• For \(NO^+\), the molecule is linear, with the nitrogen atom bonded directly to the oxygen atom with a double bond.
• Approximation of the geometry of \(NO_2^-\) gives us a bent shape due to one lone pair on the nitrogen, creating an angle less than 120° between the oxygen atoms.
• In \(NO_3^-\), the nitrate ion, we have a trigonal planar geometry as the three oxygens are symmetrically arranged around the nitrogen, creating 120° bond angles due to the absence of lone pairs on nitrogen.

Grasping molecular geometry is essential for predicting physical properties and reactivity of molecules, assisting us in drawing conclusions about their bond lengths and strengths.