Problem 52 - Chemistry: The Central Science in SI Units, Global Edition

Question

Write the Lewis structure for each of the following species, describe its geometry, and indicate the oxidation state of the nitrogen: \((\mathbf{a}) \mathrm{N}_{2} \mathrm{O}_{3},\) (b) \(\mathrm{NOCl}\) (c) \(\mathrm{NO}_{2} \mathrm{Cl}\) (d) \(\mathrm{N}_{2} \mathrm{O}_{4}\).

Step-by-Step Solution

Answer

In summary, for each species: (a) N2O3: Lewis structure is ONNOO, geometry is Bent for both N atoms and Linear for the central O atom, and the oxidation state of N is +3. (b) NOCl: Lewis structure is ONCl, geometry is Bent for N, and the oxidation state of N is +1. (c) NO2Cl: Lewis structure is ONOCl, geometry is Bent for N, and the oxidation state of N is +1. (d) N2O4: Lewis structure is ONNOON, geometry is Bent for both N atoms and Linear for both central O atoms, and the oxidation state of N is +2.

Step 1: Draw the Lewis Structure for Each Species

To draw the Lewis structure for each species, we need to follow these rules: 1. Count the total number of valence electrons in the compound. 2. Connect the atoms with single bonds, and distribute the remaining electrons. 3. Check if the atoms fulfill the octet rule (or duet rule for hydrogen). (a) N2O3 - 6(valence electrons of N) × 2(N atoms) + 6(valence electrons of O) × 3(O atoms) = 24 valence electrons - The structure is ONNOO - Each N atom gets 1, and O atom gets 2 lone electron pairs (b) NOCl - 5(valence electrons of N) × 1(N atom) + 6(valence electrons of O) × 1(O atom) + 7(valence electrons of Cl) × 1(Cl atom) = 18 valence electrons - The structure is ONCl - N gets 1, and O gets 2 lone electron pairs (c) NO2Cl - 5(valence electrons of N) × 1(N atom) + 6(valence electrons of O) × 2(O atoms) + 7(valence electrons of Cl) × 1(Cl atom) = 24 valence electrons - The structure is ONOCl - N gets 0, and O gets 2 lone electron pairs (d) N2O4 - 5(valence electrons of N) × 2(N atoms) + 6(valence electrons of O) × 4(O atoms) = 34 valence electrons - The structure is ONNOON - Each N atom gets 1, and each O atom gets 2 lone electron pairs

Step 2: Determine the Molecular Geometry for Each Species

We'll use the VSEPR(V) model to determine the molecular geometry for each species. The geometries are given below: (a) N2O3: Bent for both N atoms, Linear for the central O atom (b) NOCl: Bent for N (c) NO2Cl: Bent for N (d) N2O4: Bent for both N atoms, Linear for both central O atoms

Step 3: Calculate the Oxidation State of Nitrogen in Each Species

To determine the oxidation state of nitrogen in each species, we'll use this formula: \(\displaystyle \mathrm{Oxidation\ State} \ =\ \mathrm{Group\ Number} -\mathrm{Lone\ pair\ electrons} -\dfrac{1}{2}\mathrm{Bonding\ electrons}\) (a) N2O3: - 5 - 4 - (½ × 4) = +3 for both N atoms (b) NOCl: - 5 - 4 - (½ × 4) = +1 for N (c) NO2Cl: - 5 - 0 - (½ × 8) = +1 for N (d) N2O4: - 5 - 4 - (½ × 6) = +2 for both N atoms In conclusion, the Lewis structures, geometries, and oxidation states of nitrogen in each species are: (a) N2O3: ONNOO, Bent for both N atoms, Linear for the central O atom, +3 oxidation state for both N atoms (b) NOCl: ONCl, Bent for N, +1 oxidation state for N (c) NO2Cl: ONOCl, Bent for N, +1 oxidation state for N (d) N2O4: ONNOON, Bent for both N atoms, Linear for both central O atoms, +2 oxidation state for both N atoms

Key Concepts

Molecular Geometry

Understanding the molecular geometry of a compound is crucial when predicting its chemical behavior. Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. This arrangement affects various properties such as polarity, reactivity, and interaction with other molecules. In our exercise, using the VSEPR (Valence Shell Electron Pair Repulsion) model helps in predicting molecular geometry based on electron pair repulsion.

For example, the geometry of the compound \((\mathrm{N}_{2} \mathrm{O}_{3})\) is determined by the presence of lone pairs and bonded atoms around the nitrogen atoms. The nitrogen molecules adopt a bent geometry due to the electron pairs repelling each other, resulting in distinct angles. Similarly, \((\mathrm{NOCl})\) and \((\mathrm{NO}_{2} \mathrm{Cl})\) also have a bent geometry around the nitrogen atom.

For \((\mathrm{N}_{2} \mathrm{O}_{4})\), the molecule exhibits a bent geometry for the nitrogen atoms, while the central oxygen atoms align linearly, optimizing the spatial arrangement to minimize repulsion.

Oxidation State

The oxidation state of an element within a compound indicates the degree of oxidation, meaning the hypothetical charge the atom would have if all bonds were purely ionic. Determining the oxidation state can help in understanding the electron distribution in the molecule.

In nitrogen compounds such as \(\mathrm{N}_{2} \mathrm{O}_{3}\), each nitrogen atom has an oxidation state of +3. This is calculated by subtracting the number of lone pair electrons and half of the shared (bonding) electrons from the group number of nitrogen. Similarly, for \(\mathrm{NOCl}\), the nitrogen atom has an oxidation state of +1 because there are fewer electrons shared compared to \(\mathrm{N}_{2} \mathrm{O}_{3}\).

In the compound \(\mathrm{NO}_{2} \mathrm{Cl}\), nitrogen also has an oxidation state of +1 due to its bonding configuration. Lastly, \(\mathrm{N}_{2} \mathrm{O}_{4}\) shows each nitrogen with an oxidation state of +2, indicating more electrons are shared with other atoms.

Valence Electrons

Valence electrons play a critical role in forming chemical bonds. These are the electrons located in the outermost shell of an atom and determine how atoms will interact with each other. The number of valence electrons influences the shape and stability of the resultant molecule.

For instance, each nitrogen atom has 5 valence electrons, while oxygen has 6, and chlorine has 7. These valence electrons are involved in creating Lewis structures, which give a visual representation of the molecule by indicating bonds and lone pairs of electrons.

In molecules like \(\mathrm{N}_{2} \mathrm{O}_{3}\), the total number of valence electrons is calculated as 24. These electrons are then distributed to fulfill bonding needs and the octet rule, ensuring that all atoms achieve a stable electron configuration. The same method applies to \(\mathrm{NOCl}\), \(\mathrm{NO}_{2} \mathrm{Cl}\), and \(\mathrm{N}_{2} \mathrm{O}_{4}\). This demonstrates how valence electrons define the bonding and structure of a molecule.

VSEPR Model

The VSEPR model, or Valence Shell Electron Pair Repulsion model, is a theory used to predict the shape of individual molecules based on electron-pair interactions. This model suggests that electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion.

By applying the VSEPR model, chemists can predict how the various lone and bonded electron pairs affect the spatial arrangement of the atoms. For \(\mathrm{N}_{2} \mathrm{O}_{3}\), the electron pairs repulse each other leading to a bent geometry for nitrogen atoms and linear for the central oxygen atom.

The model also explains why \(\mathrm{NOCl}\) and \(\mathrm{NO}_{2} \mathrm{Cl}\) molecules have a bent geometry due to the additional electron clouds around the nitrogen atoms. For \(\mathrm{N}_{2} \mathrm{O}_{4}\), the packing of the electrons causes a bent formation in nitrogen, with the central oxygens remaining linear.