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# 11.3: Electronic and Molecular Geometry

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

The student will:

• identify the most likely type of bonding (ionic, covalent, or polar covalent) for each compound given binary formulas and an electronegativity chart.
• draw Lewis structures for simple molecules that violate the octet rule.
• from a list of binary compounds, identify those that require electron promotion in the explanation of their bonding.
• identify the type of hybridization in various molecules.
• explain the necessity for the concept of hybridized orbitals.

## Vocabulary

• electron promotion
• orbital hybridization
• polar covalent bonds

## Introduction

In your study of chemistry, you may have noticed occasionally that after you have learned the “rules” for how matter behaves, you are introduced to the exceptions. For example, shortly after you learned the filling order rule for electron configuration, you were introduced to the exceptions caused by the slightly lower potential energy when the d orbitals were exactly half-filled or completely filled. Shortly after you learned that metals lose all their valence electrons when forming ions, you were introduced to some variable oxidation number metals that give up a different number of electrons.

In this section, you will be introduced to exceptions to the octet rule. Most of the exceptions to rules in chemistry are due to the fact that the behavior of matter is controlled by the tendency toward lowest potential energy. Therefore, when a small variation to the rule produces slightly lower potential energy, matter will follow that path.

## Electronegativity

Recall from the chapter on “Chemical Periodicity” that the ability for an atom in a molecule to attract shared electrons is called electronegativity. The electronegativity of atoms has been quantified in several ways. One method that is widely accepted is that of Linus Pauling. The Pauling electronegativity values are shown below.

When two atoms combine, the difference between their electronegativities is an indication of the type of bond that will form. If the difference between the electronegativities of the two atoms is small, neither atom can take the shared electrons completely away from the other atom, and the resulting bond will be covalent. If the difference between the electronegativities is large, the more electronegative atom will take the bonding electrons completely away from the other atom (electron transfer will occur), and the bond will be ionic. There is no exact number for the difference in electronegativity that will produce an ionic bond, but many chemists suggest $1.7$ as the approximate difference that will produce an ionic bond.

If we compare a few electronegativities, we can see that $1.7$ is a reasonable suggestion. The difference between the electronegativities of hydrogen and oxygen is $1.2$, and we know the bonds in the water molecule are covalent. In magnesium chloride, the electronegativity difference is $1.8$, and we know this molecule contains ionic bonds.

This video describes ionic, covalent, and polar covalent bonds and relates the types of bonds formed to the differences in electronegativity of the bonding atoms (2g): http://www.youtube.com/watch?v=tSG4R4YZUW8 (5:17).

## The Partial Ionic Character of Covalent Bonds

So far, we have discussed two extreme types of bonds. One case is when two identical atoms bond. They have exactly the same electronegativities, thus the two bonded atoms pull exactly equally on the shared electrons. The other case is when the bonded atoms have a very large difference in their electronegativities. In this case, the more electronegative atom will take the electrons completely away from the other atom, forming an ionic bond.

What about the molecules whose electronegativities are not the same but the difference is not as large as $1.7$? For these molecules, the electrons are still shared by the two atoms, but they are not shared equally. The shared electrons are pulled closer to the more electronegative atom. This results in an uneven distribution of electrons over the molecule and causes slight charges on opposite ends of the molecule. Such molecules are said to have a dipole or are called polar molecules (see figure below). These charges are not full $+1$ and $-1$ charges, but rather they are fractions of charges. For small fractions of charges, we use the symbols $\delta +$ and $\delta -$.

Atoms whose electronegativities are not the same but are not different enough to cause complete electron transfer will form polar covalent bonds. These polar covalent bonds will be discussed in more detail later on in this chapter. With the introduction of the polar covalent bond, we now know of three different types of bonds.

In the figure below, molecule A represents a perfectly covalent bond that would form between identical atoms. Molecule B is a polar covalent bond formed between atoms whose electronegativities are not the same but whose electronegativity difference is less than $1.7$, while molecule C is an ionic bond formed between atoms whose electronegativity difference is greater than $1.7$.

For an introduction to polar bonds see these two videos. The first video defines polar bonds. The second one consists of lecture and animation about polar bonds and polar molecules (2b, 2f): http://www.youtube.com/watch?v=rRoLSBeOjME (1:49), http://www.youtube.com/watch?v=LKAjTE7B2x0 (4:43).

## Electron Promotion

Consider the molecule formed by the combination of beryllium and hydrogen, beryllium hydride ($\text{BeH}_2$). The electronegativity difference between these two atoms is only $0.6$, with beryllium being slightly more electropositive. Therefore, the bonds in $\text{BeH}_2$ are polar covalent bonds. The Lewis structure for the molecule is:

The first thing we might notice about the $\text{BeH}_2$ molecule is that it does not obey the octet rule. Beryllium has two valence electrons and each hydrogen has one, so there are not enough valence electrons to produce an octet around the beryllium atom.

There is also another unusual thing about $\text{BeH}_2$. The orbital representation for the electron configuration of beryllium shows the $2s$ orbital is full. Earlier in this chapter, we said that the atoms involved in covalent bonding must each have a half-filled orbital. How can beryllium form two bonds when it does not have any half-filled orbitals? The explanation to this question involves the concept of electron promotion. As illustrated below, by promoting an electron from the $2s$ orbital to one of the $2p$ orbitals, beryllium theoretically acquires two half-filled orbitals. These half-filled orbitals are therefore able to form two covalent bonds. The amount of necessary to promote an electron from a $2s$ orbital to a $2p$ orbital is small compared to the amount of energy released when beryllium forms two covalent bonds. As a result, in exchange for a small input of energy, a large quantity of energy released.

## Orbital Hybridization

We have not, however, completed the explanation as to why $\text{BeH}_2$ does not follow the octet rule. Bonds formed by overlapping s orbitals are, in general, shorter, stronger, and more flexible than bonds formed by overlapping p orbitals from the same shell. Therefore, when considering the two bonds in $\text{BeH}_2$, it would be reasonable to assume that we could determine which bond was formed by hydrogen overlapping with beryllium’s half-filled $2s$ orbital and which was formed by hydrogen overlapping beryllium’s half-filled $2p$ orbital. Examination of the two bonds, however, shows them to be identical in length, strength, and flexibility. The values for the length, strength, and flexibility also fall in between the values expected for overlapping s and p orbitals. The concept used to explain these observations is called orbital hybridization.

In orbital hybridization, all the orbitals involved in bonding are hybridized and form a set of identical orbitals that have properties intermediate to the properties of the orbitals from which they were created.

As seen in the image above, the half-filled $2s$ and $2p$ orbitals hybridize to form two completely new orbitals named sp hybridized orbitals. These two new orbitals are identical to each other, and both have characteristics somewhere between the characteristics of $2s$ and $2p$ orbitals. When these half-filled sp hybridized orbitals overlap with the hydrogen atoms, two identical bonds are produced.

## Covalent Bonds of Group 3A

The only member of Group 3A that forms a significant number of covalent compounds is boron. Numerous boron compounds exist, with boron forming three bonds in many of the compounds. Since the ground state electron configuration of boron, $1s^22s^22p^1$, has a filled $2s$ orbital and one half-filled $2p$ orbital, the concept of electron promotion is needed to explain boron's formation of three bonds.

One of the electrons in the $2s$ orbital is promoted to an empty $2p$ orbital, producing three half-filled orbitals that allows boron to form three covalent bonds (illustrated in the figure above). The three bonds in boron compounds such as $\text{BH}_3$ are found to be identical in length, strength, and flexibility. The $2s$ orbital and two $2p$ orbitals hybridize to form three identical $sp^2$ hybridized orbitals.

The three hybridized orbitals can now overlap with other atoms and form three identical bonds, making compounds such as $\text{BH}_3$ and $\text{BCl}_3$. In these boron compounds, the central atom is surrounded by only six electrons, and therefore this is another exception to the octet rule.

It should be recognized that the $sp^2$ hybridized orbitals are not the same as $sp$ hybridized orbitals. If you mix one bucket of blue paint with one bucket of yellow paint, you get green paint. If you mix two buckets of blue with one bucket of yellow, you will also get green paint, but it will not be the same shade of green as before. Hybridizing one $s$ and one $p$ orbital does not produce the same orbitals as hybridizing one $s$ and two $p$ orbitals.

## Covalent Bonds of Group 4A

The members of Group 4A that form covalent bonds are carbon and silicon. Their electron configurations are $1s^22s^22p^2$ and $1s^22s^22p^63s^23p^2$, respectively. In the outer energy level of both atoms, the $s$ orbital is filled, two of the $p$ orbitals are half-filled, and the third $p$ orbital is empty. In the majority of the covalent compounds containing carbon and silicon, these atoms are found to have four bonds. As a result, electron promotion must also occur in these atoms. If one of the electrons in the $s$ orbital is promoted to the empty $p$ orbital, four half-filled orbitals are produced, as shown in the image below.

With four half-filled orbitals, carbon and silicon can form the four covalent bonds necessary to create $\text{CH}_4$, $\text{CCl}_4$, and so on. As you may have already suspected, the four bonds in carbon and silicon compounds are all identical. Therefore, orbital hybridization also occurs in this family. The four hybridized orbitals, as shown below, are called $sp^3$ hybridized orbitals.

Since the $sp^3$ hybridized orbitals are a mix of one $s$ and three $p$ orbitals, their characteristics are not the same as $sp^2$ hybridized orbitals or $sp$ hybridized orbitals. These four identical orbitals can be overlapped by four hydrogen atoms and form four identical bonds. Consequently, the covalent compounds in Group 4A do obey the octet rule.

## Covalent Bonds of Group 5A

In Group 5A, the electron configuration of the outermost energy level is $s^2p^3$. These atoms have three half-filled $p$ orbitals available for bonding and would appear to form three covalent bonds with no need for either electron promotion or hybridization. The first member of the family, nitrogen, commonly forms three bonds in compounds such as $\text{NH}_3, \ \text{NCl}_3$, and $\text{N}_2\text{O}_3$. The second member of the family, phosphorus, also forms similar compounds, like $\text{PH}_3$ and $\text{PCl}_3$. Phosphorus, however, can also form compounds with five covalent bonds, such as $\text{PCl}_5$. In order to explain five identical bonds for the phosphorus atom, we will need to use the concepts of electron promotion and orbital hybridization.

As shown in the figure above, phosphorus promotes one of its $3s$ electrons into an orbital in the $3d$ sub-level. This gives phosphorus five half-filled orbitals and allows it to form the five bonds in $\text{PCl}_5$ and other phosphorus compounds with five covalent bonds. It also makes it clear why nitrogen does not form compounds with five bonds, because nitrogen does not have a $d$ sub-energy level available. The five bonds in these compounds are identical, which means that one $s$ orbital, three $p$ orbitals, and one $d$ orbital hybridize. These hybridized orbitals will be called $sp^3d$ or $dsp^3$ hybridized orbitals. Molecules like $\mathrm{PCl}_5$ and $\mathrm{PF}_5$ will have the central atom surrounded by $10$ electrons, so this electron promotion produces another group of molecules that do not obey the octet rule.

## Covalent Bonds of Group 6A

The outermost energy level of the members of Group 6A has the electron configuration $s^2p^4$. This outer energy level has a filled $s$ orbital, one filled $p$ orbital, and two half-filled $p$ orbitals. With two half-filled orbitals available for overlap, all of the members of this family can and do form two covalent bonds. There are, however, some compounds that form more than two covalent bonds. Oxygen has no $d$ orbitals, so promotion into the d sub-energy level is not possible. Sulfur and larger members of the family, however, do have $d$ orbitals that allow for electron promotion.

As you can see in the image above, sulfur can promote two electrons into the $d$ orbitals and produce six half-filled orbitals available for overlap. The six orbitals are hybridized and form $d^2sp^3$ or $sp^3d^2$ hybridized orbitals. Such orbitals are necessary for the formation of compounds such as $\text{SF}_6$.

## Covalent Bonds of Group 7A

All of the members of Group 7A have outermost energy level electron configurations $s^2p^5$. This configuration has one half-filled $p$ orbital, which allows the members of this family to form one covalent bond. The majority of compounds formed by this family contain one bond. Fluorine has no $d$ orbitals and can only form one bond. Unlike fluorine, chlorine and the larger members of the family have empty $d$ orbitals that allow them to undergo electron promotion. Chlorine can promote one $p$ electron to a $d$ orbital to produce three half-filled orbitals, which allow compounds like $\text{ClF}_3$ to form. Chlorine can also promote two electrons to form five half-filled orbitals, which results in compounds like $\text{ClF}_5$.

## Covalent Bonds of Group 8A

Even members of the noble gases, under extreme conditions, can form some covalent compounds. Since the ground state electron configuration for these atoms has completely filled outer energy levels, the only way that they can form bonds is by electron promotion and orbital hybridization.

In the diagram above, figure A shows xenon promoting one electron from a $p$ to a $d$ orbital. This allows xenon to form compounds like $\text{XeF}_2$ that contain two covalent bonds. In figure B, xenon is promoting two electrons from $p$ to $d$ orbitals, producing four orbitals available for bonding and forming compounds like $\text{XeF}_4$. In both Groups 7A and 8A, when electrons are promoted and orbitals are hybridized, even though some of the orbitals are filled and some are half-filled, hybridization is considered to involve all the outer energy level orbitals that contain electrons. Therefore, in figure A, the orbital hybridization would be $sp^3d$, and in figure B, the hybridization would be $sp^3d^2$.

Table below summarizes the different types of hybridized orbitals discussed in this section.

Summary of Hybridized Orbitals
Orbitals Hybridized Hybridized Orbital Names Number of Orbitals
$s + p$ $sp$ $2$
$s + p + p$ $sp^2$ $3$
$s + p + p + p$ $sp^3$ $4$
$s + p + p + p + d$ $sp^3d$ $5$
$s + p + p + p + d + d$ $sp^3d^2$ $6$

This animation reviews the differences in ionic, covalent, and polar covalent bonds (2a): http://www.youtube.com/watch?v=Kj3o0XvhVqQ (1:26).

## Lesson Summary

• A difference of approximately $1.7$ or more in the electronegativities of bonded atoms will produce an ionic bond.
• Bonds between atoms whose electronegativity difference is between $0$ and $1.7$ will be polar covalent.
• A relatively small number of stable molecules do not obey the octet rule.
• Electron promotion occurs within an energy level when an electron from a lower sub-energy level is promoted to a higher sub-energy level.
• The concept involved in orbital hybridization is that all the orbitals of an atom that are involved in bonding are hybridized to orbitals that are identical and have properties intermediate to the properties of the orbitals that are hybridized.
• Covalent bonding in Group 2A involves electron promotion and $sp$ hybridization.
• Covalent bonding in Group 3A involves electron promotion and $sp^2$ hybridization.
• Covalent bonding in Group 4A involves electron promotion and $sp^3$ hybridization.
• Covalent bonding in Group 5A involves electron promotion and $sp^3d$ hybridization.
• Covalent bonding in Group 6A involves electron promotion and $sp^3d^2$ hybridization.
• Covalent bonding in Group 8A involves electron promotion and can involve $sp^3d$ or $sp^3d^2$ hybridization.

The following websites provide more information on electron promotion and orbital hybridization.

This video is a ChemStudy film called “Shapes and Polarities of Molecules.”

## Review Questions

1. Explain the differences among a covalent bond, a polar covalent bond, and an ionic bond.
2. Explain why a pair of atoms form a covalent bond rather than an ionic bond.
3. Predict which of the following bonds will be more polar and explain why: $\mathrm{P-Cl}$ or $\mathrm{S-Cl}$.
4. What laboratory evidence necessitates the theory of hybridized orbitals?
5. Which of the following molecules has a central atom with $sp^2$ hydridized orbitals?
1. $\text{C}_2\text{H}_2$
2. $\text{CH}_2\text{Cl}_2$
3. $\text{BF}_3$
4. $\text{CH}_3\text{CH}_2\text{OH}$
5. $\text{HF}$
6. Fill in the type of hybridization necessary for the following molecules in Table below.
Table for Review Question 6
Molecule Type of Hybridization
$\text{H}_2\text{O}$
$\text{NH}_3$
$\text{BeCl}_2$
$\text{NaH}$
$\text{BF}_3$
$\text{PCl}_5$
$\text{BrF}_5$
$\text{SF}_6$
$\text{XeF}_2$

Answer questions 7-10 by following the example below.

Example: The formula for boron trichloride is $\text{BCl}_3$.

a. Draw the electron configuration for the central atom before any electron promotion occurs.

b. Draw the electron configuration for the central atom after any electron promotion occurs.

c. Draw the electron configuration after orbital hybridization occurs.

1. The formula for beryllium bromide is $\text{BeBr}_2$.
1. Draw the electron configuration for the central atom before any electron promotion occurs.
2. Draw the electron configuration for the central atom after any electron promotion occurs.
3. Draw the electron configuration after orbital hybridization occurs.
2. The formula for sulfur hexafluoride is $\text{SF}_6$.
1. Draw the electron configuration for the central atom before any electron promotion occurs.
2. Draw the electron configuration for the central atom after any electron promotion occurs.
3. Draw the electron configuration after orbital hybridization occurs.
3. The formula for silicon tetrahydride is $\text{SiH}_4$.
1. Draw the electron configuration for the central atom before any electron promotion occurs.
2. Draw the electron configuration for the central atom after any electron promotion occurs.
3. Draw the electron configuration after orbital hybridization occurs.
4. The formula for phosphorus pentaiodide is $\text{PI}_5$.
1. Draw the electron configuration for the central atom before any electron promotion occurs.
2. Draw the electron configuration for the central atom after any electron promotion occurs.
3. Draw the electron configuration after orbital hybridization occurs.

Feb 23, 2012

Jan 07, 2015