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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) Does \(C S_{2}\) have a dipole moment? If so, in which direction does the net dipole point? (b) Does \(\mathrm{SO}_{2}\) have a dipole moment? If so, in which direction does the net dipole point?

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6},\) is a planar molecule that has the following carbon-carbon bond lengths: $$ \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{134 \mathrm{pm}} \mathrm{CH}=\mathrm{CH}_{2} $$ (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle \(\mathrm{C}-\mathrm{C}\) bond length in butadiene \((148 \mathrm{pm})\) is a little shorter than the average \(\mathrm{C}-\mathrm{C}\) single bond length (154 pm). Does this imply that the middle \(\mathrm{C}-\mathrm{C}\) bond in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\) single bond? (d) Based on your answer for part (c), discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\mathrm{C}\) bond.

From their Lewis structures, determine the number of \(\sigma\) and \(\pi\) bonds in each of the following molecules or ions: (a) hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\) (b) hydrogen cyanide, HCN; (c) sulphur trioxide, \(\mathrm{SO}_{3} ;\) (d) ozone, \(\mathrm{O}_{3}\).

The energy-level diagram in Figure 9.36 shows that the sideways overlap of a pair of \(p\) orbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding \(\pi\) orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the \(\pi_{2 p}\) to the \(\pi_{2 p}^{*}\) molecular orbital. (a) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the HOMO in ethylene? (b) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the LUMO in ethylene? (c) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene stronger or weaker in the excited state than in the ground state? Why? (d) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene easier to twist in the ground state or in the excited state?

Consider the \(\mathrm{H}_{2}^{+}\) ion. (a) Sketch the molecular orbitals of the ion and draw its energy-level diagram. (b) How many electrons are there in the \(\mathrm{H}_{2}^{+}\) ion? (c) Write the electron configuration of the ion in terms of its MOs. (d) What is the bond order in \(\mathrm{H}_{2}^{+}\) ? (e) Suppose that the ion is excited by light so that an electron moves from a lower-energy to a higher- energy MO. Would you expect the excited-state \(\mathrm{H}_{2}^{+}\) ion to be stable or to fall apart? (f) Which of the following statements about part (e) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

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