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Draw a picture of \(2 \mathrm{NO}_{2}\) molecules colliding with the proper orientation to produce \(1 \mathrm{~N}_{2} \mathrm{O}_{4}\) molecule. $$ 2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) $$

Short Answer

Expert verified
A drawing of 2 NO2 molecules can be depicted as two individual nitrogen atoms each bonded to two oxygen atoms (one single and one double bond). When they collide, the final sketch is a single N2O4 molecule. This molecule has a structure with two nitrogen atoms bound together and bonded to two oxygen atoms each, one with a double bond and one with a single bond.

Step by step solution

01

Draw the initial molecules

The first step towards solving this problem will be to draw two NO2 (Nitrogen Dioxide) molecules. NO2 has one nitrogen atom and two oxygen atoms. Nitrogen is in the center, bonded with two oxygen atoms, one on each side. These bonds aren't exactly alike because nitrogen is not in the exact middle. One bond is a single bond and the other is a double bond due to the extra unpaired electron on nitrogen. This structure gives NO2 a bent shape.
02

Show the collision

The next step is to illustrate the collision of the two NO2 molecules. When the two NO2 molecules collide, the unpaired extra electrons of Nitrogen atoms get paired together. It is important to depict the two molecules positioned in such a way that the Nitrogen atoms are close and in a situation where they can react, suggesting a collision.
03

Draw the final molecule

After the collision, the two NO2 molecules combine to form one N2O4 molecule. Draw a new molecule, N2O4, which consists of two nitrogen atoms in the middle and four oxygen atoms, two on each side of the nitrogen molecules. There are a single bond and a double bond between the nitrogen and oxygen atoms on each side. The nitrogen atoms are also connected with a single bond to each other.

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Key Concepts

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

Chemical Bonding
Understanding chemical bonding is crucial for visualizing how atoms join together to form molecules. In the context of the exercise involving the reaction of two NO2 molecules to form N2O4, the bonding principles tell us a lot about the stability and reactivity of these molecules.

Nitrogen dioxide, NO2, exhibits two types of bonds: a single bond and a double bond with oxygen atoms. This difference arises because nitrogen has five valence electrons and tries to reach a stable configuration by forming bonds with oxygen atoms. One oxygen forms a double bond, sharing two pairs of electrons with nitrogen, while the other forms a single bond, sharing one pair of electrons. An interesting point to note is that the lone pair of electrons on nitrogen creates a bent molecular shape and is also responsible for NO2's tendency to dimerize — to seek out and bond with another NO2 molecule to complete its octet, leading to the formation of N2O4.

Chemical bonding is dictated by the need for stability, and the visualization of these bonds aids students in understanding how molecules interact and combine during chemical reactions.
Molecular Geometry
The molecular geometry is a crucial concept in understanding the shapes and spatial arrangements of atoms in a molecule. NO2's molecular geometry can be described as bent or V-shaped due to the presence of a lone electron pair on the nitrogen atom, which repels the bonding pairs of electrons, pushing them away and creating the bent shape.

When NO2 molecules collide to form N2O4, the orientation of these molecules is a significant factor. If NO2 molecules collide with the wrong orientation, the reaction to form N2O4 may not occur efficiently. Proper orientation facilitates the coupling of the lone electron on each nitrogen, forming a stable N-N bond.

By visualizing the V-shaped NO2 and understanding that upon dimerization the N-N single bond is formed, students can grasp how the initial bent shapes convert into the linear central part of N2O4, flanked by symmetrical oxygen atoms bonded to each nitrogen. This linear arrangement alters the original V-shape of NO2 into the planar molecule of N2O4, illustrating how molecular geometry can change through chemical reactions.
Reaction Mechanisms
Reaction mechanisms detail the step-by-step sequence of events at the molecular level through which reactants transform into products. In our nitrogen dioxide example, the mechanism is a simple combination of two NO2 molecules, but even in this simplicity, it is full of instructional value.

The process begins when two NO2 molecules collide with the correct orientation, which is essential for reaction mechanisms as the spatial arrangement dictates the likelihood of a successful reaction. The unpaired electron on each nitrogen atom needs to be in proximity to pair up and form the N-N bond. This pairing represents a critical step in the mechanism and highlights the importance of the collision theory, which postulates that molecules must collide with sufficient energy and a favorable orientation to react.

Following the formation of the N-N bond, the molecule reorganizes into the N2O4 structure. Visualizing this mechanism helps students understand that chemical reactions are not just about reactants and products but about the precise dance of electrons and atoms that leads to the formation of new substances. Moreover, this focus on the detailed steps promotes a deeper comprehension of how molecules interact and the factors influencing reaction rates and outcomes.

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Most popular questions from this chapter

About \(80 \%\) of the ammonia produced by industry goes into the production of fertilizers. Ammonia is produced in industry by the reaction of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) in the following reversible reaction: \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) exothermic (a) Which conditions of pressure and temperature favor the formation of the most amounts of ammonia? (b) Which conditions of temperature will increase the rate of formation of ammonia? (c) Use your answers to parts (a) and (b) to determine whether the reaction should be run at extremely high temperatures, extremely low temperatures, or an intermediate temperature.

Consider the following reaction and its equilibrium constant at \(100^{\circ} \mathrm{C}\) : $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad K_{\mathrm{eq}}=6.5 $$ If \(0.250 \mathrm{~mol}\) of each reactant and product is mixed into a \(1.0\) - \(\mathrm{L}\) container, will the reaction proceed in the forward or reverse direction, or is it already at equilibrium?

Does the value of the equilibrium constant increase or decrease when the temperature of each of the following equilibrium systems is decreased? (a) \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) endothermic (b) \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\) exothermic

Consider the following system at equilibrium: $$ \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) $$ For each of the following changes, predict whether the equilibrium concentration of \(\mathrm{NO}_{2}\) will increase or decrease. (a) a decrease in the concentration of NO (b) an increase in the concentration of \(\mathrm{SO}_{3}\) (c) an increase in the concentration of \(\mathrm{SO}_{2}\) (d) a decrease in the concentration of \(\mathrm{SO}_{2}\)

Consider the following factors: increase in temperature, increase in concentration, and addition of a catalyst. Which increases the fraction of collisions that are effective?

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