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What is the hybridization of the central atom in (a) \(\mathrm{PBr}_{5}\), (b) \(\mathrm{CH}_{2} \mathrm{O},\) (c) \(\mathrm{O}_{3},(\mathbf{d}) \mathrm{NO}_{2} ?\)

Short Answer

Expert verified
The hybridization of the central atoms in the given molecules are as follows: (a) \( PBr_5 \): sp3d (b) \( CH_2O \): sp2 (c) \( O_3 \): sp3 (d) \( NO_2 \): sp2

Step by step solution

01

Determining the central atom

For each of the molecules, we need to identify the central atom which is the atom that is the least electronegative and has the highest connectivity to other atoms: (a) PBr5: Phosphorus (P) is the central atom (b) CH2O: Carbon (C) is the central atom (c) O3: Oxygen (O) is the central atom (d) NO2: Nitrogen (N) is the central atom
02

Calculate the electron domains

For each central atom, calculate the number of electron domains which is the sum of its bonds and lone pairs of electrons. (a) PBr5: Phosphorus (P) has 5 bonds (one with each Br atom) and no lone pairs. So, electron domains = 5. (b) CH2O: Carbon (C) has 2 bonds (one with each H atom) and 1 double bond (with the O atom). So, electron domains = 3. (c) O3: Oxygen (O) has 4 electron domains. Two of them are bonds with Z type atoms (the other two O atoms) and two lone pairs. (d) NO2: Nitrogen (N) has three electron domains: 1 double bond (with one O atom), 1 single bond (with the other O atom), and 1 unpaired electron (odd electron species). So, electron domains = 3.
03

Determine the Hybridization

Based on the number of electron domains, we can determine the hybridization of each central atom. (a) PBr5: 5 electron domains correspond to sp3d hybridization. (b) CH2O: 3 electron domains correspond to sp2 hybridization. (c) O3: 4 electron domains correspond to sp3 hybridization. (d) NO2: 3 electron domains correspond to sp2 hybridization. To conclude, the hybridization of the central atoms in the given molecules are as follows: (a) PBr5: sp3d (b) CH2O: sp2 (c) O3: sp3 (d) NO2: sp2

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Key Concepts

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

Electron Domains
When discussing chemical hybridization, the concept of electron domains is incredibly important. Electron domains encapsulate the regions around a central atom where electrons are predominantly located. These include both bonds (single, double, or triple) and lone electron pairs.
  • For example, if a central atom forms two single bonds and has one lone pair, it has three electron domains in total.
  • Each bond counts as one electron domain, regardless of its type (single or multiple).
  • Lone pairs of electrons also contribute a domain each.
Understanding electron domains is crucial in predicting molecular geometry and hybridization of a molecule. By tallying these domains, we can deduce the hybridization type by matching the count of domains to known hybrid configurations. For instance:
  • 3 electron domains align with an sp2 hybridization.
  • 5 electron domains align with an sp3d hybridization.
This makes electron domains a foundational element in predicting the overall shape and properties of molecules.
Central Atom
The central atom in a molecule is a key player in determining molecular geometry and hybridization. It is typically the atom with the lowest electronegativity, making it capable of forming more bonds with other atoms in the molecule.
  • In a polyatomic molecule, the central atom is most likely to be bonded to multiple atoms around it.
  • For instance, in the \(\mathrm{PBr}_{5}\) molecule, phosphorus (P) is the central atom because it can connect with all five bromine atoms.
Identifying the central atom helps clarify the number of surrounding bonds and lone pairs, directly affecting total electron domains. Knowing the central atom can link directly to understanding the spatial and electronic structure of a molecule. As we examine further:
  • The central atom serves as the backbone for understanding the full molecular structure.
  • It impacts the molecular stability and reactivity based on electronegativity and atomic size.
Grasping which atom is central aids in exploring the greater context of chemical composition and behavior in compounds.
Molecular Geometry
Molecular geometry is the three-dimensional arrangement of atoms within a molecule. This spatial configuration results from the number and type of electron domains around the central atom. Understanding molecular geometry is critical because it reveals the overall shape of the molecule, which affects its chemical and physical properties.
  • For example, a molecule like \(\mathrm{CH}_{2} \mathrm{O}\) with \(\mathrm{sp}^{2}\) hybridization results in a trigonal planar geometry.
  • In contrast, \(\mathrm{PBr}_{5}\) with \(\mathrm{sp}^{3}\ d\) hybridization forms a trigonal bipyramidal geometry.
Different arrangements of electron domains lead to various geometric structures. Aspects like lone pairs can push bonding atoms apart, altering the idealized geometrical shape. Recognizing this, molecular geometries can be predicted:
  • Linear, trigonal, and tetrahedral are just a few examples of potential geometrical outcomes.
  • Each shape has defined bond angles, helping predict molecule interaction with other entities.
In sum, mastering molecular geometry provides insights into molecular function and intermolecular interactions, making it indispensable in chemistry.

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Most popular questions from this chapter

(a) If the valence atomic orbitals of an atom are sp hybridized, how many unhybridized \(p\) orbitals remain in the valence shell? How many \(\pi\) bonds can the atom form? (b) Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single \(\sigma\) bond or around a double \((\sigma\) plus \(\pi)\) bond, or would they be the same?

Name the proper three-dimensional molecular shapes for each of the following molecules or ions, showing lone pairs as needed: \((\mathbf{a}) \mathrm{ClO}_{2}^{-}(\mathbf{b}) \mathrm{SO}_{4}^{2-}(\mathbf{c}) \mathrm{NF}_{3}(\mathbf{d}) \mathrm{CCl}_{2} \mathrm{Br}_{2}(\mathbf{e}) \mathrm{SF}_{4}^{2+}\)

Dihydroxybenzene, \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}\), exists in three forms (isomers) called ortho, meta, and para: Which of these has a nonzero dipole moment?

(a) Write a single Lewis structure for \(\mathrm{N}_{2} \mathrm{O},\) and determine the hybridization of the central \(\mathrm{N}\) atom. (b) Are there other possible Lewis structures for the molecule? (c) Would you expect \(\mathrm{N}_{2} \mathrm{O}\) to exhibit delocalized \(\pi\) bonding?

The following is part of a molecular orbital energy-level diagram for MOs constructed from 1 s atomic orbitals. (a) What labels do we use for the two MOs shown? (b) For which of the following molecules or ions could this be the energy-level diagram: $$ \mathrm{H}_{2} \mathrm{He}_{2}, \mathrm{H}_{2}^{+}, \mathrm{He}_{2}^{+}, \mathrm{or} \mathrm{H}_{2}^{-} ? $$ (c) What is the bond order of the molecule or ion? (d) If an electron is added to the system, into which of the MOs will it be added? [Section 9.7\(]\)

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