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Chapter 9: Problem 68

What hybridization do you expect for the atom that is underlined in each of the following species? (a) \(\underline{\mathrm{O}}_{2}^{-} ;(\mathbf{b}) \underline{\mathrm{N}} \mathrm{H}_{4}^{+} ;\) (c) \(\mathrm{SCN}^{-}\) (d) \(\underline{\mathrm{Br}} \mathrm{Cl}_{3}\)

Short Answer

Expert verified
The hybridization for the underlined atoms in each species is as follows: (a) \(\underline{\mathrm{O}}_{2}^{-}\): sp (b) \(\underline{\mathrm{N}} \mathrm{H}_{4}^{+}\): sp3 (c) \(\mathrm{SCN}^{-}\) (Carbon is the central atom): sp (d) \(\underline{\mathrm{Br}} \mathrm{Cl}_{3}\): sp2

Step by step solution

01

Count the number of ligands surrounding the central atom

In this case, the central atom is Oxygen. We have one other oxygen atom and one extra electron due to the negative charge. The steric number equals 2.
02

Determine the hybridization

For two ligands, we'll have an electronic geometry of linear (shape). To obtain this geometry, we use one s and one p orbital to hybridize. Thus, hybridization is sp. (b) \(\underline{\mathrm{N}} \mathrm{H}_{4}^{+}\)
03

Count the number of ligands surrounding the central atom

In this case, the central atom is Nitrogen. We have four Hydrogen atoms around it and one electron less due to the positive charge. The steric number equals 4.
04

Determine the hybridization

For four ligands, we'll have an electronic geometry of tetrahedral (shape). To obtain this geometry, we use one s and three p orbitals to hybridize. Thus, hybridization is sp3. (c) \(\mathrm{SCN}^{-}\)
05

Count the number of ligands surrounding the central atom

In this case, the central atom is Carbon. We have one Sulfur atom and one Nitrogen atom attached to it. The steric number equals 2.
06

Determine the hybridization

For two ligands, we'll have an electronic geometry of linear (shape). To obtain this geometry, we use one s and one p orbital to hybridize. Thus, hybridization is sp. (d) \(\underline{\mathrm{Br}} \mathrm{Cl}_{3}\)
07

Count the number of ligands surrounding the central atom

In this case, the central atom is Bromine. We have three Chlorine atoms attached to it. The steric number equals 3.
08

Determine the hybridization

For three ligands, we'll have an electronic geometry of trigonal planar (shape). To obtain this geometry, we use one s and two p orbitals to hybridize. Thus, hybridization is sp2.

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Key Concepts

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

Steric Number
The steric number is a crucial concept in determining an atom's hybridization in a molecule. It's the sum of the number of atoms bonded to the central atom and the number of lone electron pairs on that atom. This number helps us understand the arrangement of electrons and atoms around the central nucleus.

Here's how to find it:
  • Count the number of atoms directly bonded to the central atom. Each one contributes a count of one.
  • Count the lone electron pairs on the central atom. These, too, contribute a value of one each to the steric number.
For example, in the \( ext{NH}_4^+\) ion, nitrogen is surrounded by four hydrogen atoms with no lone pairs, resulting in a steric number of 4.
In \( ext{BrCl}_3\), bromine is bonded to three chlorine atoms and has two lone pairs, giving it a steric number of 5.
Electronic Geometry
Electronic geometry describes the spatial arrangement of all regions of electron density (bonds and lone pairs) around the central atom in a molecule. It is determined by the steric number.

The most common electronic geometries include:
  • Linear, with a steric number of 2, forming a straight line.
  • Trigonal planar, with a steric number of 3, arranging atoms at the corners of an equilateral triangle.
  • Tetrahedral, with a steric number of 4, arranging atoms at the corners of a tetrahedron.
For instance, a steric number of 4 implies a tetrahedral electronic geometry. This guides us in visualizing the positions of atoms in three-dimensional space, affecting bond angles and molecular shape.
sp Hybridization
The concept of sp hybridization comes into play when the steric number is 2 and involves the mixing of one s orbital and one p orbital. This process creates two sp hybrid orbitals that are linearly arranged, 180 degrees apart from each other, providing a linear electronic geometry.
This kind of hybridization is seen in molecules like CO鈧� and SCN鈦�. In these cases, the involved atom shares its electrons with two others to form a linear structure. The resulting shape minimizes repulsion between the electron pairs, following the VSEPR theory.
Understanding sp hybridization helps us predict how atoms bond and arrange themselves, offering insights into molecular shape and stability.
sp3 Hybridization
sp3 hybridization happens when the steric number is 4. It involves the mixing of one s orbital with three p orbitals, producing four sp3 hybrid orbitals.

This kind of hybridization leads to a tetrahedral electronic geometry, where atoms are positioned to minimize repulsion at angles of approximately 109.5 degrees.
  • This configuration is common in organic molecules, such as methane (CH鈧�), where carbon forms four bonds and achieves a stable arrangement.
  • In NH鈧勨伜, nitrogen undergoes sp3 hybridization to bond with four hydrogen atoms in a symmetric fashion.
Understanding sp3 hybridization is essential for predicting molecular geometry and how it might interact with other molecules, fundamental knowledge for chemistry students grappling with molecular shapes and reactions.

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Most popular questions from this chapter

Benzaldehyde, \(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}\), is a fragrant substance responsible for the aroma of almonds. Its Lewis structure is O=Cc1cccc(C=O)c1 (a) What is the hybridization at each of the carbonatoms of the molecule? (b) What is the total number of valence electrons in benzaldehyde? (c) How many of the valence electrons are used to make \(\sigma\) bonds in the molecule? (d) How many valence electrons are used to make \(\pi\) bonds? (e) How many valence electrons remain in nonbonding pairs in the molecule?

Consider the Lewis structure for acetic acid, which is known as vinegar: CCC(=O)O (a) What are the approximate bond angles about each of the two carbon atoms, and what are the hybridizations of the orbitals on each of them? (b) What are the hybridizations of the orbitals on the two oxygen atoms, and what are the approximate bond angles at the oxygen that is connected to carbon and hydrogen? (c) What is the total number of \(\sigma\) bonds in the entire molecule, and what is the total number of \(\pi\) bonds?

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, \(\mathrm{O}_{3}\).

(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a \(\sigma\) bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two s orbitals combine to form a \(\pi\) bond? Explain.

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