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Specify the electron-pair and molecular geometry for each underlined atom in the following list. Describe the hybrid orbital set used by this atom in each molecule or ion. (a) \(\underline{\mathrm{BBr}}_{3}\) (b) \(\underline{\mathrm{CO}_{2}}\) (c) \(\underline{\mathrm{CH}_{2} \mathrm{Cl}_{2}}\) (d) \(\underline{\mathrm{CO}_{3}^{2-}}\)

Short Answer

Expert verified
(a) Trigonal planar, sp2; (b) Linear, sp; (c) Tetrahedral, sp3; (d) Trigonal planar, sp2.

Step by step solution

01

Determine the geometry for BBr3

In BBr3, the central atom is boron (B). To find the electron-pair geometry, consider the electron pairs around boron. Boron typically forms three bonds without lone pairs, which leads to three regions of electron density. According to the VSEPR theory, this configuration corresponds to a trigonal planar geometry. Hence, both the electron-pair and molecular geometry for BBr3 are trigonal planar. The hybridization of boron in BBr3 is sp2, which corresponds to three regions of electron density.
02

Analyze the geometry for CO2

In CO2, carbon (C) is the central atom. Carbon forms two double bonds with oxygen atoms, creating two regions of electron density. According to VSEPR theory, two regions of electron density lead to a linear geometry. Therefore, both the electron-pair and molecular geometry are linear for carbon in CO2. The hybridization of carbon in CO2 is sp, accommodating the linear geometry with two areas of electron density.
03

Determine the geometry for CH2Cl2

In CH2Cl2, carbon (C) is the central atom. Carbon forms four single bonds: two with hydrogen atoms and two with chlorine atoms, resulting in four regions of electron density. VSEPR theory says that four areas of electron density result in a tetrahedral geometry. Both the electron-pair and molecular geometry for carbon in CH2Cl2 are tetrahedral, with carbon's hybridization being sp3, suitable for four bonds or electron pairs.
04

Examine the geometry for CO3^2−

In CO3^2−, carbon (C) is the central atom. It forms three equivalent bonds with oxygen, with resonance delocalizing the double bond. This results in three regions of electron density around the carbon atom (one double bond in resonance). According to VSEPR, this leads to a trigonal planar geometry. The carbon in CO3^2− uses sp2 hybridization, which aligns with three regions of electron density and a planar shape.

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Key Concepts

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

VSEPR Theory
Valence Shell Electron Pair Repulsion (VSEPR) Theory is a concept used to predict the geometry of individual molecules. This theory is primarily based on minimizing the repulsions between electron pairs around a central atom. It assumes that electron pairs will arrange themselves in a way that minimizes repulsion, thus determining the molecule's geometry.

In practice:
  • Every bond – single, double, or triple – constitutes one "region" of electron density.
  • Lone pairs, though non-bonding, also count as regions of electron density.
  • The more regions of electron density a molecule has, the angles between these regions will change to decrease repulsion, resulting in different shapes.
For instance, in \(\mathrm{CO}_2\), which has two areas of electron density, the molecule adopts a linear shape, as two points align in a straight path to be as far apart as possible.
Hybridization
Hybridization is a concept in chemistry that describes the mixing of orbitals to form new, equivalent hybrid orbitals that are suitable for pairing electrons in chemical bonds.

This mixing of atomic orbitals leads to the formation of hybrid orbitals of equal energy and shape. The type of hybridization reflects the electron-pair geometry:
  • sp hybridization indicates linear configuration with two orbitals involved.
  • sp\(^2\) hybridization occurs in trigonal planar shapes involving three orbitals.
  • sp\(^3\) hybridization is found in tetrahedral shapes with four orbitals.
For example, in \(\mathrm{CH_2Cl_2}\), carbon undergoes sp\(^3\) hybridization, allowing it to form four sigma bonds with the hydrogen and chlorine atoms arranged tetrahedrally.
Electron-Pair Geometry
Electron-pair geometry considers all areas of electron density around a central atom, including bonds and lone pairs, to predict the 3D arrangement of a molecule. This geometry is crucial in diagnostics of overall molecule shape and behavior.

For example:
  • The molecule \(\mathrm{BBr_3}\), boron atom is surrounded by three bromine atoms, leading to three regions of electron density. The electron-pair geometry is therefore trigonal planar.
  • In molecules where there are lone pairs, like in ammonia (\(\mathrm{NH_3}\)), the geometry shifts slightly due to the lone pair repulsion.
Electron-pair geometry is a primary determinant of the molecular structure and is the foundation for understanding spatial arrangements in molecules.
Chemical Bonding
Chemical bonding involves the attraction between atoms due to electron sharing or exchange to achieve stability. Bonds are the fundamental connections within molecules and dictate the molecular geometry.

There are different types of chemical bonds:
  • Single bonds involve one shared pair of electrons (sigma bonds).
  • Double and triple bonds include pi bonds alongside sigma bonds, increasing the electron density between the nuclei.
In molecules like \(\mathrm{CO_3^{2-}}\), resonance occurs where double bonds distribute across atoms, giving each bond character beyond single or double definitions. This bonding affects the hybridization state and consequently the geometry, shaping the molecule in a way to balance repulsion while maximizing bonding interactions.

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Most popular questions from this chapter

Melamine is an important industrial chemical, used to make fertilizers and plastics. (a) The carbon-nitrogen bond lengths in the ring are all the same length (about \(140 \mathrm{pm}\) ). Explain. (b) Melamine is made by the decomposition of urea, \(\left(\mathrm{H}_{2} \mathrm{N}\right)_{2} \mathrm{CO}\) \(6\left(\mathrm{H}_{2} \mathrm{N}\right)_{2} \mathrm{CO}(\mathrm{s}) \rightarrow \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{N}_{6}(\mathrm{s})+6 \mathrm{NH}_{3}(\mathrm{g})+3 \mathrm{CO}_{2}(\mathrm{g})\) Calculate the enthalpy change for this reaction. Is it endo- or exothermic? \(\left[\Delta_{f} H^{\circ} \text { for melamine(s) }=\right.\) \(-66.1 \mathrm{kJ} / \mathrm{mol}\) and for urea(s) \(=-333.1 \mathrm{kJ} / \mathrm{moll}\)

Sketch the resonance structures for the nitrate ion, \(\mathrm{NO}_{3}^{-} .\) Is the hybridization of the \(\mathrm{N}\) atom the same or different in each structure? Describe the orbitals involved in bond formation by the central \(\mathrm{N}\) atom.

Draw the Lewis structure for 1,1 -dimethylhydrazine \([\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NNH}_{2}, \mathrm{a}\) compound used as a rocket fuel]. What is the hybridization for the two nitrogen atoms in this molecule? What orbitals overlap to form the bond between the nitrogen atoms?

Nitrogen, \(\mathrm{N}_{2}\), can ionize to form \(\mathrm{N}_{2}^{+}\) or add an electron to give \(\mathrm{N}_{2}^{-}\). Using molecular orbital theory, compare these species with regard to (a) their magnetic character, (b) net number of \(\pi\) bonds, (c) bond order, (d) bond length, and (e) bond strength.

What is the hybridization of the carbon atom in phosgene, \(\mathrm{Cl}_{2} \mathrm{CO}\) ? Give a complete description of the \(\sigma\) and \(\pi\) bonding in this molecule.

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