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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 σ-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 σ-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 σ-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 σ-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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