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Chapter 2: Problem 40

Consider the Lewis structures for the following molecules: $$\begin{equation} \mathrm{CO}_{2}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{2}^{-}, \text {and } \mathrm{NO}_{3}^{-} \end{equation}$$ Which molecule or molecules exhibit \(s p^{2}\) hybridization around the central atom? (A) \(\mathrm{CO}_{2}\) and \(\mathrm{CO}_{3}^{2-}\) (B) \(\mathrm{NO}_{2}^{-}\) and \(\mathrm{NO}_{3^{-}}\) (C) \(\mathrm{CO}_{3}^{2-}\) and \(\mathrm{NO}_{3}^{-}\) (D) \(\mathrm{CO}_{3}^{2-}, \mathrm{NO}_{2}^{-},\) and \(\mathrm{NO}_{3}^{-}\)

Short Answer

Expert verified
The molecules that exhibit \(sp^{2}\) hybridization are \(\mathrm{CO}_{3}^{2-}\), \(\mathrm{NO}_{2}^{-}\), and \(\mathrm{NO}_{3}^{-}\), i.e., option (D).

Step by step solution

01

Understanding the Lewis Structure and hybridization

We have four molecules given, \(\mathrm{CO}_{2}\), \(\mathrm{CO}_{3}^{2-}\), \(\mathrm{NO}_{2}^{-}\), and \(\mathrm{NO}_{3}^{-}\). We need to draw the Lewis Structure for each molecule and understand the orbitals involved. From the structures, we should count the total number of the 蟽-bonds and the lone pairs on the central atom to determine the type of hybridization. For \(sp^{2}\) hybridization, there should be 3 sigma bonds or lone pairs.
02

Analyzing \(\mathrm{CO}_{2}\)

For \(\mathrm{CO}_{2}\), the carbon atom is the central atom, and it forms two double bonds with the two oxygen atoms. This means there are two 蟽-bonds with no lone pair on Carbon. Hence, the hybridization on carbon in \(\mathrm{CO}_{2}\) is \(sp\). Therefore, \(\mathrm{CO}_{2}\) doesn't show \(sp^{2}\) hybridization.
03

Analyzing \(\mathrm{CO}_{3}^{2-}\)

For \(\mathrm{CO}_{3}^{2-}\), the carbon atom is the central atom. It forms two single bonds with two oxygen atoms and one double bond with one oxygen atom. Hence, there are three 蟽-bonds generating \(sp^{2}\) hybridization. Hence, \(\mathrm{CO}_{3}^{2-}\) does have \(sp^{2}\) hybridization.
04

Analyzing \(\mathrm{NO}_{2}^{-}\)

For \(\mathrm{NO}_{2}^{-}\), the Nitrogen atom is the central atom, and it forms one single bond and one double bond with the two oxygen atoms. With one lone pair on Nitrogen, there ensues three pairs in total, resulting in \(sp^{2}\) hybridization for \(\mathrm{NO}_{2}^{-}\). Therefore, \(\mathrm{NO}_{2}^{-}\) does have \(sp^{2}\) hybridization.
05

Analyzing \(\mathrm{NO}_{3}^{-}\)

For \(\mathrm{NO}_{3}^{-}\), the Nitrogen atom is the central atom. Here, Nitrogen forms three single bonds with each of the three oxygen atoms. Hence there are three 蟽-bonds resulting in \(sp^{2}\) hybridization. Therefore, \(\mathrm{NO}_{3}^{-}\) also has \(sp^{2}\) hybridization.
06

Final Answer

Now, looking at all the molecules, the ones that do show \(sp^{2}\) hybridization are: \(\mathrm{CO}_{3}^{2-}\), \(\mathrm{NO}_{2}^{-}\), and \(\mathrm{NO}_{3}^{-}\)

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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 diagrams that represent the bonding between atoms in a molecule as well as the lone pairs of electrons that may exist. These structures help us understand the layout of the molecule, including how atoms are connected. When drawing Lewis structures, follow these steps:
  • Count the total number of valence electrons for the molecule.
  • Arrange the atoms, determining which is the central atom (typically the least electronegative).
  • Draw single bonds (蟽-bonds) between the central atom and surrounding atoms.
  • Distribute the remaining electrons as lone pairs to complete the octet for each atom.
  • If necessary, form double or triple bonds to fulfill the octet rule for each atom.
Understanding Lewis structures is critical because they provide insight into the molecular geometry and hybridization of molecules.
sigma bonds
Sigma bonds (\(\sigma\)-bonds) are the strongest type of covalent chemical bond formed by the "head-on" overlap of atomic orbitals. They are typically considered the first bond formed between two atoms and allow for free rotation around the bond axis.
  • These bonds are commonly found in all the basic types of covalent bonds: single, double, and triple bonds.
  • In a double bond, there is one 蟽-bond and one 蟺-bond, while in a triple bond, there is one 蟽-bond and two 蟺-bonds.
  • In organic molecules, these are formed between hybridized orbitals such as 畏-orbitals and the remaining non-hybridized p orbitals.
Sigma bonds are crucial for determining the geometry and hybridization of the central atom in a molecule.
central atom hybridization
Central atom hybridization involves the mixing of atomic orbitals to form new hybrid orbitals that can overlap to form bonds in a compound. Hybridization helps explain molecular shapes and bond angles and depends on the number of sigma bonds and lone pairs around the central atom.
  • For an atom exhibiting \(sp^2\) hybridization, it means the atom has three regions of electron density (i.e., bonds or lone pairs).
  • An example is carbon in \(\text{CO}_3^{2-}\), where it bonds with three oxygen atoms using three \(\sigma\)-bonds.
  • The concept explains how atoms like nitrogen or carbon, which involve \(sp^2\) hybridization, aim to minimize repulsion to attain a stable configuration.
Understanding these concepts allows scientists to predict molecular geometries and reactions.
electron pairs
Electron pairs can be either bonding pairs involved in a covalent bond or nonbonding (lone) pairs that are not involved in bonding. Understanding electron pairs is vital as they affect the molecular geometry, the type of hybridization, and overall stability.
  • Bonding pairs are shared between atoms as part of a bond, creating the \(\sigma\)=bond.
  • Lone pairs reside on a single atom and occupy space, affecting the geometry and bond angles.
  • Each lone pair counts as one region of electron density.
Accurate measurement and consideration of these pairs are necessary to predict the exact shape and properties of molecules, influencing chemical reactivity and interaction.
molecular geometry
Molecular geometry is the three-dimensional arrangement of atoms within a molecule. It is shaped by electron pairs arranged around the central atom. The geometry dictates the molecule's physical and chemical properties.
  • The VSEPR (Valence Shell Electron Pair Repulsion) theory helps predict the geometry by minimizing electron pair repulsions.
  • For \(sp^2\) hybridized central atoms, a typical molecular geometry is trigonal planar, with bond angles of approximately 120 degrees.
  • Molecules like \(\text{CO}_3^{2-}\) and \(\text{NO}_3^{-}\) exhibit such geometries.
Knowledge of molecular geometry is crucial in fields like pharmacology and materials science, where molecular shape impacts function and reactivity.

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Most popular questions from this chapter

A sample of \(\mathrm{H}_{2} \mathrm{S}\) gas is placed in an evacuated, sealed container and heated until the following decomposition reaction occurs at \(1000 \mathrm{K} :\) $$2 \mathrm{H}_{2} \mathrm{S}(g) \rightarrow 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \qquad K_{\mathrm{c}}=1.0 \times 10^{-6}$$ If, at a given point in the reaction, the value for the reaction quotient \(Q\) is determined to be \(2.5 \times 10^{-8},\) which of the following is occurring? (A) The concentration of the reactant is decreasing while the concentration of the products is increasing. (B) The concentration of the reactant is increasing while the concentration of the products is decreasing. (C) The system has passed the equilibrium point, and the concentration of all species involved in the reaction will remain constant. (D) The concentrations of all species involved are changing at the same rate.

The bond length between any two nonmetal atoms is achieved under which of the following conditions? (A) Where the energy of interaction between the atoms is at its minimum value (B) Where the nuclei of each atom exhibits the strongest attraction to the electrons of the other atom (C) The point at which the attractive and repulsive forces between the two atoms are equal (D) The closest point at which a valence electron from one atom can transfer to the other atom

Consider the Lewis structures for the following molecules: $$\begin{equation} \mathrm{CO}_{2}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{2}^{-}, \text {and } \mathrm{NO}_{3}^{-} \end{equation}$$ Which molecules are best represented by multiple resonance structures? (A) \(\mathrm{CO}_{2}\) and \(\mathrm{CO}_{3}^{2-}\) (B) \(\mathrm{NO}_{2}^{-}\) and \(\mathrm{NO}_{3}^{-}\) (C) \(\mathrm{CO}_{3}^{2-}\) and \(\mathrm{NO}_{3^{-}}^{-}\) (D) \(\mathrm{CO}_{3}^{2-}, \mathrm{NO}_{2}^{-},\) and \(\mathrm{NO}_{3}^{-}\)

\(2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) The data below was gathered for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at 310 \(\mathrm{K}\) via the equation above. \(\begin{array}{|c|c|}\hline \text { Time(s) } & {\left[\mathrm{N}_{2} \mathbf{O}_{5}\right](M)} \\ \hline 0 & {0.250} \\ \hline 500 . & {0.190} \\\ \hline 1000 . & {0.145} \\ \hline 2000 . & {0.085} \\ \hline\end{array}\) (a) How does the rate of appearance of NO_{2} \text { compare to the rate of } disappearance of \(\mathrm{N}_{2} \mathrm{O}_{5}\) ? Justify your answer. (b) The reaction is determined to be first order overall. On the axes below, create a graph of some function of concentration vs. time that will produce a straight line. Label and scale your axes appropriately. (c) (i) What is the rate constant for this reaction? Include units. (ii) What would the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) be at \(t=1500 \mathrm{s} ?\) (iii) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) ? (d) Would the addition of a catalyst increase, decrease, or have no effect on the following variables? Justify your answers. (i) Rate of disappearance of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (ii) Magnitude of the rate constant (iii) Half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\)

20.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) is placed in a beaker and titrated with a solution of \(1.0 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2},\) resulting in the creation of a precipitate. If the experiment were repeated and the \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) was diluted to 40.0 \(\mathrm{mL}\) with distilled water prior to the titration, how would that affect the volume of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) needed to reach the equivalence point? (A) It would be cut in half. (B) It would decrease by a factor of 1.5. (C) It would double. (D) It would not change.
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