Chapter 5: Chemical Bonding
Sometime around the beginning of the twentieth century, an American chemist named Gilbert Newton Lewis was struggling with a question his students kept asking him: what actually holds two atoms together? Chemists of that time could write formulas — \(\mathrm{H_2O}\), \(\mathrm{NaCl}\), \(\mathrm{CH_4}\) — and they knew from experiment that these combinations were fixed and reliable. But nobody could say why one atom of oxygen took exactly two atoms of hydrogen, and never three.
Lewis worked on the problem for years. He began sketching atoms as tiny cubes, with electrons sitting at the corners, and noticed something interesting: a cube has eight corners, and the unreactive noble gases have eight electrons in their outermost shell. In 1916 he published his answer. Atoms combine, he said, so that each one ends up with a stable outer arrangement of eight electrons — either by handing electrons over completely, or by sharing a pair between them. In the same year, working independently in Germany, Walther Kossel arrived at a closely related explanation for compounds like sodium chloride.
That simple idea — a shared pair of dots between two symbols — became one of the most useful pictures in all of chemistry. As shown in Fig. 5.0, it is where our study of the chemical bond begins.
Diagram
Fig. 5.0 – Opening illustration: on the left, a sodium atom transferring one electron to a chlorine atom to give Na⁺ and Cl⁻; on the right, two hydrogen atoms sharing a pair of dots to give H:H; caption “two ways to complete an outer shell”.
As you work through this chapter, you will gather everything you need to reason out each of these questions for yourself.
5.1 Introduction
Look around at the substances you meet every day — the water you drink, the common salt in your food, the oxygen you breathe. Almost none of them exist as free, separate atoms. Atoms are nearly always found joined to one another, and joined in fixed, definite ratios.
So a question arises at once: why are atoms held together in chemical compounds? There must be some force that holds them together. You have already learnt in lower classes that the forces holding atoms together in a compound are called chemical bonds.
Two Routes to a Chemical Bond
How are chemical bonds formed between two atoms? There are two ways in which a chemical bond can be formed — (i) by the loss and gain of electrons, and (ii) by the sharing of a pair of electrons between the two atoms.
These two routes look quite different, yet they arrive at the same destination. In either process of formation of a chemical bond, each atom attains a stable noble gas electronic configuration. The noble gases — helium, neon, argon and the rest — are famously unreactive because their outermost shells are already complete. An atom that manages to copy that arrangement reaches a similar low-energy, settled state, and that is exactly what bonding allows it to do. The two possible routes are compared in FC 5.1.
Flowchart
FC 5.1 – Two routes to a chemical bond: two atoms with incomplete outer shells → branch A: complete loss and gain of electrons → oppositely charged ions → ionic bond; branch B: sharing of an electron pair → shared pair held by both atoms → covalent bond; both branches converge → each atom attains a stable noble gas configuration
The Electrons That Take Part
Which electrons are involved in the formation of chemical bonds? Only the electrons present in the outermost shell of an atom take part in the formation of a chemical bond. These outermost electrons are called valence electrons, and the shell that holds them is the valence shell.
The inner electrons sit much closer to the nucleus and are held far more tightly, so they are not available for bonding. This is why, when we study bonding, we can ignore most of an atom’s electrons and pay attention only to the few in the outermost shell.
5.2 Kossel and Lewis Approach to Chemical Bonding
A number of attempts were made to explain the formation of the chemical bond in terms of electrons, but the first satisfactory explanation was given by W. Kossel and G. N. Lewis, working independently of each other. Both of them gave a logical explanation of valence which was based on the inertness of the noble gases. On the basis of this, they proposed a theory of valence known as the electronic theory of valence in 1916.
The starting point of their reasoning is worth pausing on. Valence — the combining capacity of an element — had been known from experiment for decades, but it was only a number in a table. Kossel and Lewis were the first to say that this number comes from the way electrons rearrange themselves.
The Kernel and the Outer Shell
According to Lewis, the atom can be pictured in terms of a positively charged kernel and an outer shell. The kernel is the nucleus together with all the inner electrons, and because the nucleus is positive while the inner electrons only partly cancel it, the kernel as a whole carries a positive charge.
The outer shell surrounding this kernel can accommodate a maximum of eight electrons. This set of eight electrons — this octet — represents a stable electronic arrangement, and it is the arrangement every atom tries to reach when it bonds.
Lewis stated that each atom achieves a stable octet during the formation of a chemical bond. In the case of sodium and chlorine, this can be achieved by the transfer of one electron from sodium to chlorine. Thus \(\mathrm{Na^+}\) (2, 8) and \(\mathrm{Cl^-}\) (2, 8, 8) ions are formed, which are then held together.
In the case of other molecules like \(\mathrm{H_2}\), \(\mathrm{F_2}\), \(\mathrm{Cl_2}\), \(\mathrm{HCl}\) and so on, the bond is formed by the sharing of a pair of electrons between the atoms. In this process too, each atom attains a stable outer octet of electrons.
The Octet Rule
In 1916, Kossel and Lewis proposed an important theory for explaining the formation of the chemical bond, known as the electronic theory of valence. This theory is mainly based on the octet rule developed by Lewis. The octet rule is based on the stability of the noble gases, which comes from the presence of eight electrons — an \(ns^2np^6\) arrangement — in the valence shell.
Here \(n\) stands for the number of the outermost shell. So for neon, \(n = 2\) and the valence shell arrangement is \(2s^2 2p^6\); for argon, \(n = 3\) and it is \(3s^2 3p^6\). In each case the count of outer electrons comes to \(2 + 6 = 8\).
The octet rule is found to be very useful in explaining the normal valence of elements and in the study of the chemical combination of atoms leading to the formation of a molecule. If chlorine needs one more electron to reach eight, its combining capacity should be one — and it is. If oxygen needs two, its valence should be two — and it is.
However, it should be noted that the octet rule is not valid for H and Li atoms. These atoms tend to have only two electrons in their valence shell, similar to that of helium (\(1s^2\)), which is called a duplet. Hydrogen and lithium sit so close to helium in the periodic table that copying helium costs them far less than reaching for eight electrons.
5.2.1 Ionic Bond
The first of the two routes to bonding is the complete transfer of electrons from one atom to another. The atom that loses electrons becomes a positively charged ion called a cation; the atom that gains them becomes a negatively charged ion called an anion. Opposite charges attract, and that attraction is the bond.
Formation of Sodium Chloride (NaCl)
The electronic configurations of sodium and chlorine are:
Na (\(Z = 11\)) \(1s^2 2s^2 2p^6 3s^1\) or 2, 8, 1
Cl (\(Z = 17\)) \(1s^2 2s^2 2p^6 3s^2 3p^5\) or 2, 8, 7
Sodium has one electron in its valence shell. It has a tendency to lose one electron to acquire the configuration of the nearest noble gas, neon (2, 8). Chlorine has seven electrons in its valence shell. It has a tendency to gain one electron and thereby acquire the configuration of the nearest noble gas, argon (2, 8, 8).
During the combination of sodium and chlorine atoms, the sodium atom transfers its valence electron to the chlorine atom. The sodium atom changes into the \(\mathrm{Na^+}\) ion, while the chlorine atom changes into the \(\mathrm{Cl^-}\) ion. The two ions are then held together by a strong electrostatic force of attraction. The formation of the ionic bond between Na and Cl is shown using Lewis dots in Fig. 5.1.
Diagram
Fig. 5.1 – Lewis dot representation of ionic bond formation. (a) Na· + ·Cl̈: → Na⊕ + :C̈l:⊖, with configurations 2,8,1 and 2,8,7 written below the reactants and 2,8 and 2,8,8 below the products; then Na⊕ + Cl⊖ → Na⊕Cl⊖ or NaCl, labelled “ionic bond”. (b) :C̈l· + ·C̈a· + ·C̈l: → :C̈l:⊖ + Ca²⊕ + :C̈l:⊖, with configurations 2,8,7 / 2,8,8,2 / 2,8,7 below the reactants and 2,8,8 / 2,8,8 / 2,8,8 below the products; then Cl⊖ + Ca²⊕ + Cl⊖ → CaCl₂ or Ca²⊕(Cl⊖)₂
Formation of Calcium Chloride (CaCl₂)
The same representation shows the formation of the compound calcium chloride from the elements calcium and chlorine. The electronic configurations are:
Calcium: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2\)
Cl: \(1s^2 2s^2 2p^6 3s^2 3p^5\)
Calcium has two electrons in its valence shell and must lose both to reach a noble gas arrangement, becoming \(\mathrm{Ca^{2+}}\). But each chlorine atom can accept only one electron. Two chlorine atoms are therefore required for every one calcium atom, which is exactly why the formula is \(\mathrm{CaCl_2}\) and not \(\mathrm{CaCl}\). This step-by-step transfer is drawn in the lower half of Fig. 5.1.
5.2.2 Ionic Solids and Lattice Enthalpy
Ionic solids are solids which contain cations and anions held together by ionic bonds. The Kossel treatment helps us to understand the formation of ionic bonds between ions of different elements.
The formation of ions depends on the ease with which an atom can lose or gain electrons. We can break the whole process into steps:
\[\mathrm{M}\text{(g)} \rightarrow \mathrm{M^+}\text{(g)} + \mathrm{e^-}\]
\[\mathrm{X}\text{(g)} + \mathrm{e^-} \rightarrow \mathrm{X^-}\text{(g)} \qquad \text{electron gain enthalpy}\]
\[\mathrm{M^+}\text{(g)} + \mathrm{X^-}\text{(g)} \rightarrow \mathrm{MX}\text{(g)} \rightarrow \mathrm{MX}\text{(s)}\]
The first step is measured by the ionisation enthalpy (\(\Delta_i H\)), the energy needed to pull an electron away from a gaseous atom. The second is measured by the electron gain enthalpy (\(\Delta_{eg} H\)), the energy change when a gaseous atom accepts an electron.
Ionization is always an endothermic process — energy must always be supplied to remove an electron from an atom, because the electron is being pulled away from a positive nucleus that is holding it. The electron gain process, on the other hand, can be exothermic or endothermic. Based on the ionisation enthalpy (\(\Delta_i H\)) and the electron gain enthalpy, we can predict which elements can form ionic compounds.
Elements having low ionization enthalpy can readily form an ionic bond with elements having a high negative value of electron gain enthalpy. Both these processes take place in the gaseous phase. All ionic compounds in the solid state have each cation surrounded by a specific number of anions, and vice versa.
The Crystal Lattice
The arrangement of cations and anions in a crystalline solid is ordered, and they are held together by coulombic forces of attraction. This regular, repeating three-dimensional arrangement is called a crystal lattice, and the lattice of sodium chloride is drawn in Fig. 5.2.
Diagram
Fig. 5.2 – NaCl crystal lattice: cubic array of alternating spheres, larger Cl⊖ ions and smaller Na⊕ ions, joined by lines showing that each ion is surrounded by six ions of opposite charge; label both ion types
During their formation, these compounds crystallize from the gaseous state (MX(g) → MX(s)) to the solid state. The structure in which they crystallize depends upon the size of the ions, their packing arrangement and other factors. The overall stability of the ionic solid depends upon the interactions between all these ions and the energy released during the formation of the crystal lattice.
Let us consider the formation of NaCl ionic solid.
\[\mathrm{Na}\text{(g)} \rightarrow \mathrm{Na^+}\text{(g)} + \mathrm{e^-} \qquad \Delta_i H = 495.8\ \text{kJ mol}^{-1}\]
\[\mathrm{Cl}\text{(g)} + \mathrm{e^-} \rightarrow \mathrm{Cl^-}\text{(g)} \qquad \Delta_{eg} H = -348.7\ \text{kJ mol}^{-1}\]
\[\mathrm{Na^+}\text{(g)} + \mathrm{Cl^-}\text{(g)} \rightarrow \mathrm{NaCl}\text{(g)} + 147.7\ \text{kJ mol}^{-1}\]
Read these three lines together and something important emerges. Making the pair of gaseous ions costs more energy than it returns, so simply completing the octet on each ion is not by itself enough to explain why sodium chloride exists.
The conversion of NaCl(g) → NaCl(s) is associated with the release of energy, which is \(-788\ \text{kJ mol}^{-1}\). This released energy is much more than the absorbed energy. Thus the stability of an ionic compound can be estimated by knowing the amount of energy released during lattice formation, and not just by the energy associated with completion of the octet around the ionic species in the gaseous state alone. The full sequence is set out in FC 5.2.
Flowchart
FC 5.2 – Energy steps in the formation of solid sodium chloride: Na(g) → Na⁺(g) + e⁻ (absorbs 495.8 kJ mol⁻¹) → Cl(g) + e⁻ → Cl⁻(g) (releases 348.7 kJ mol⁻¹) → gaseous ion pair Na⁺(g) + Cl⁻(g) → NaCl(g), net 147.7 kJ mol⁻¹ → NaCl(g) → NaCl(s), releases 788 kJ mol⁻¹ → stable ionic solid; side note at final step: large release here outweighs everything absorbed earlier
Lattice Enthalpy
The lattice enthalpy of NaCl is \(-788\ \text{kJ mol}^{-1}\), which means that 788 kJ of energy is required to separate 1 mole of NaCl into one mole of gaseous \(\mathrm{Na^+}\)(g) and \(\mathrm{Cl^-}\)(g) at an infinite distance from one another.
Table 5.1: Lattice enthalpy values of some ionic compounds
| Compound | Lattice enthalpy (kJ mol⁻¹) |
|---|---|
| LiCl | 853 |
| NaCl | 788 |
| BeF₂ | 3020 |
| CaCl₂ | 2258 |
| AlCl₃ | 5492 |
For the same anion and different cations:
- Cations having higher charge have large lattice energies than compounds having cations with lower charge. \(\mathrm{AlCl_3} > \mathrm{CaCl_2} > \mathrm{NaCl}\)
- As the size of the cation decreases, lattice energy increases. \(\mathrm{LiF} > \mathrm{NaF} > \mathrm{KF}\)
Both of these come from the same physical idea. The electrostatic attraction between two charges grows stronger when the charges are larger, and also when the centres of the two ions can approach more closely. A small, highly charged cation does both at once, and so grips its anions hardest.
Diagram
Fig. 5.3 – Bar chart of the lattice enthalpy values in Table 5.1: five bars for LiCl (853), NaCl (788), BeF₂ (3020), CaCl₂ (2258) and AlCl₃ (5492), y-axis “lattice enthalpy / kJ mol⁻¹”, each bar labelled with its value
5.2.3 Covalent Bond
In 1919 Lewis suggested that there are atoms which attain an inert gas configuration (that is, a \(1s^2\) or \(ns^2np^6\) configuration) by sharing one or more electron pairs with similar or dissimilar atoms. Each atom contributes one electron to the shared electron pair and has an equal claim on the shared electron pair.
Langmuir called the Lewis electron pair bond a covalent bond. Thus the concept of the covalent bond is known as the Lewis–Langmuir concept. A shared pair of electrons is represented as a dash (\(-\)) and is responsible for holding the two atoms together.
A covalent bond may be defined as follows.
The condition about electronegativity matters. Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself. If two atoms pull on the shared pair with roughly equal strength, neither can take it away completely, and sharing is the only possible outcome.
Formation of H₂ Molecule
The electronic configuration of the H atom is \(1s^1\). It needs one more electron to complete its valence shell. When two hydrogen atoms approach each other at a certain internuclear distance, they share their valence electrons. The shared pair of electrons belongs equally to both the hydrogen atoms. The two atoms are said to be linked by a single covalent bond, and a molecule \(\mathrm{H_2}\) is formed. This is shown in the upper part of Fig. 5.4.
\[\mathrm{H\cdot} + \mathrm{\cdot H} \rightarrow \mathrm{H\!:\!H} \quad \text{or} \quad \mathrm{H-H}\]
Formation of Cl₂ Molecule
The Lewis–Langmuir theory can explain the formation of the chlorine molecule, \(\mathrm{Cl_2}\). The Cl atom, with electronic configuration \([\mathrm{Ne}]3s^2 3p^5\), is one electron short of the argon configuration.
The formation of the \(\mathrm{Cl_2}\) molecule can be understood in terms of the sharing of a pair of electrons between two chlorine atoms. Each chlorine atom contributes one electron to the shared pair. In the process both the chlorine atoms attain the valence shell octet of the nearest noble gas, that is, argon. The dot picture is shown in the lower part of Fig. 5.4.
Diagram
Fig. 5.4 – Covalent bond formation by sharing. (a) H· + ·H → H:H or H–H, with the shared pair circled and labelled “shared pair of electrons”. (b) :C̈l· + ·C̈l: → :C̈l:C̈l: or Cl–Cl, with the shared pair circled and the three remaining lone pairs shown on each chlorine
The dots represent the electrons. Such structures are referred to as Lewis structures. The Lewis dot structure can be written for other molecules also, in which the combining atoms may be identical or different.
The following are the important features of a covalent bond.
- Each bond is formed as a result of sharing of an electron pair between the two atoms.
- When a bond is formed, each combining atom contributes one electron to the shared pair.
- The combining atoms attain the outer shell noble gas configuration as a result of the sharing of electrons.
Thus in \(\mathrm{H_2O}\) and \(\mathrm{CCl_4}\) the formation of covalent bonds can be represented as shown in Fig. 5.5. In water, each H acquires a duplet while O acquires an octet; in carbon tetrachloride, all the atoms acquire an octet.
Diagram
Fig. 5.5 – Lewis structures of H₂O and CCl₄. Left: O at centre with two shared pairs to two H atoms (each shared pair circled) and two lone pairs on O; caption “H acquires duplet, O acquires octet”. Right: C at centre with four shared pairs to four Cl atoms, three lone pairs on each Cl; caption “all atoms acquire octet”
Formation of Multiple Bonds
When two atoms share one electron pair they are said to be joined by a single covalent bond. When two combining atoms share two pairs of electrons, the covalent bond between them is called a double bond. For example, there is a double bond between two carbon atoms in the ethylene molecule.
When two combining atoms share three electron pairs, a triple bond is formed, as in the case of two nitrogen atoms in the \(\mathrm{N_2}\) molecule, drawn in Fig. 5.6(a). Some other examples of multiple bonds are \(\mathrm{CO_2}\) and \(\mathrm{C_2H_2}\), drawn in Fig. 5.6(b) and Fig. 5.6(c).
Diagram
Fig. 5.6 – Multiple bonding, three panels. (a) N dots + N dots with three shared pairs circled, written below as N≡N. (b) O, C, O with two shared pairs on each side, written below as O=C=O. (c) H, C, C, H with three shared pairs between the carbons, written below as H–C≡C–H. Caption “Fig. 5.6 Multiple bonding”
Each additional shared pair pulls the two nuclei closer and holds them more firmly, which is why a triple bond is shorter and stronger than a double bond, and a double bond shorter and stronger than a single bond. You will meet this relationship again in a quantitative form later in the chapter.
5.2.4 Lewis Structures (Lewis Representations of Simple Molecules)
Lewis dot structures show a picture of bonding in molecules and ions in terms of the shared pairs of electrons and the octet rule. Although such a picture does not explain completely the bonding and behaviour of a molecule, it helps us to understand the formation and the properties of the molecule.
It is best to think of a Lewis structure as a first sketch — quick to draw, correct about how many bonds and how many lone pairs there are, but silent about the actual shape of the molecule or the energy of its bonds. Those questions are taken up by the theories that follow later in this chapter.
5.2.5 Steps to Write Lewis Dot Structures
- Add the total number of valence electrons of the combining atoms in the molecule.
- Write the skeletal structure of the molecule to show the atoms and the number of valence electrons forming the single bond between the atoms.
- Add the remaining electron pairs to complete the octet of each atom.
- If the octet is not complete, form multiple bonds between the atoms such that the octet of each atom is complete.
- In anions, add one electron for each negative charge.
- In cations, remove or subtract one electron from the valence electrons for each positive charge.
- In polyatomic atoms and ions, the least electronegative atom is the central atom. For example, ‘S’ is the central atom in \(\mathrm{SO_4^{2-}}\) and ‘N’ is the central atom in \(\mathrm{NO_3^{-}}\).
- After writing the number of electrons as shared pairs forming single bonds, the remaining electron pairs are used either for multiple bonds or remain as lone pairs.
The order of these steps is what makes the method reliable, and it is summarised in FC 5.3.
Flowchart
FC 5.3 – Writing a Lewis dot structure: count total valence electrons of all atoms → adjust for charge (anion: add one electron per negative charge; cation: subtract one electron per positive charge) → choose the least electronegative atom as the central atom → draw the skeletal structure with one single bond (one shared pair) to each surrounding atom → distribute the remaining electron pairs to complete the octet of the outer atoms → check the central atom: if octet complete → structure done; if octet incomplete → convert a lone pair on an outer atom into an extra shared pair (multiple bond) and check again
Table 5.2 includes the Lewis representation of some molecules.
Table 5.2: Lewis dot structures of some molecules/ions
| Molecule/Ion | Lewis representation |
|---|---|
| H₂ | H:H, H–H |
| O₂ | Two O atoms joined by two shared pairs (a double bond), with two lone pairs on each O |
| O₃ | Bent arrangement of three O atoms; one double bond and one single bond, with a positive formal charge on the central O and a negative formal charge on one terminal O |
| NF₃ | N at the centre with one lone pair, single-bonded to three F atoms, each F carrying three lone pairs |
| CO₃²⁻ | C at the centre, double-bonded to one O and single-bonded to two O atoms; whole ion enclosed in square brackets with a 2− charge |
| HNO₃ | N at the centre carrying a positive formal charge, double-bonded to one O, single-bonded to a negatively charged O, and single-bonded to an O–H group |
Diagram
Fig. 5.7 – Lewis dot structures drawn out for the entries of Table 5.2: O₂, O₃ (with ⊕ and ⊖ marked), NF₃, CO₃²⁻ (in brackets with 2− outside) and HNO₃ (with ⊕ on N and ⊖ on the singly bonded O); show every lone pair as a pair of dots
Diagram
Fig. 5.8 – Lewis dot structures of the nitrite ion: [Ö::N:Ö:]⊖ ; and the two equivalent forms [Ö=N–Ö:]⊖ and [:Ö–N=Ö]⊖, each in square brackets with the negative charge outside, lone pair on N shown as two dots
Diagram
Fig. 5.9 – Lewis structure of carbon monoxide: :C≡O: with three shared pairs circled between C and O, one lone pair on C and one lone pair on O
5.2.6 Formal Charge
The Lewis dot diagrams help us to get a picture of bonding in molecules which obey the octet rule. In the case of polyatomic molecules, double bonds or sometimes triple bonds are present, and such molecules can be represented by more than one Lewis structure. In the case of \(\mathrm{CO_3^{2-}}\) we can have three dot diagrams, shown in Fig. 5.10.
Diagram
Fig. 5.10 – Three Lewis dot structures of the carbonate ion: in each, C is at the centre with a double bond to a different one of the three oxygen atoms and single bonds to the other two, each singly bonded O carrying a negative charge; all three enclosed in square brackets with 2− outside and joined by the word “or”
Double bonds can be present between carbon and any one of the three oxygen atoms. Formal charges can help us in assigning bonds when several structures are possible.
While determining the best Lewis structure per molecule, the structure is chosen such that the formal charge is as close to zero as possible. Formal charge is assigned to an atom based on the electron dot structures of the molecule or ion — for example \(\mathrm{O_3}\), \(\mathrm{NH_4^{+}}\), \([\mathrm{N}=\mathrm{C}=\mathrm{O}]^-\), \([\mathrm{S}=\mathrm{C}=\mathrm{N}]^-\).
The formal charge on an atom in a Lewis structure of a polyatomic species can be determined using the following expression:
\[\text{Formal charge on an atom in a Lewis structure} = \left(\begin{array}{c}\text{total no. of valence}\\ \text{electrons in the free atom}\end{array}\right) - \left(\begin{array}{c}\text{total no. of non-bonding}\\ \text{or lone pair electrons}\end{array}\right) - \frac{1}{2}\left(\begin{array}{c}\text{total no. of bonding}\\ \text{or shared electrons}\end{array}\right)\]
or, in short form,
\[\text{FC} = \text{VE} - \text{NE} - \frac{\text{BE}}{2}\]
The logic behind the expression is simple. We first ask how many electrons the atom owned when it was free (VE). We then count how many it appears to own in the structure: all of its lone pair electrons belong to it alone (NE), while each shared pair is split down the middle, giving it half of the bonding electrons (BE/2). The difference between what it started with and what it appears to have now is the formal charge.
The structure having the lowest formal charge has the lowest energy.
1. Formal Charge in Ozone
Let us consider the ozone molecule \(\mathrm{O_3}\). Its Lewis structure is drawn in Fig. 5.11, where the three oxygen atoms are numbered 1, 2 and 3.
The formal charge on the central oxygen atom, no. 1
\[= 6 - 2 - \tfrac{1}{2}(6) = +1\]
The formal charge on the end oxygen atom marked as 2
\[= 6 - 4 - \tfrac{1}{2}(4) = 0\]
The formal charge on the end oxygen atom marked as 3
\[= 6 - 6 - \tfrac{1}{2}(2) = -1\]
Hence the \(\mathrm{O_3}\) molecule can be represented along with formal charges as shown in the lower part of Fig. 5.11.
Diagram
Fig. 5.11 – Ozone: bent O₃ with the central oxygen numbered 1, the doubly bonded terminal oxygen numbered 2 and the singly bonded terminal oxygen numbered 3, all lone pairs shown; below it the same structure redrawn with ⊕ on the central oxygen and ⊖ on the singly bonded terminal oxygen
The lowest energy structure can be selected using formal charges from the number of possible Lewis structures for a given species.
2. Formal Charge in Carbon Dioxide
Let us take the example of \(\mathrm{CO_2}\). It can be represented by the three structures A, B and C shown in Fig. 5.12.
\[\text{A: } \mathrm{O} = \mathrm{C} = \mathrm{O} \qquad \text{B: } \mathrm{O} \equiv \mathrm{C} - \mathrm{O} \qquad \text{C: } \mathrm{O} - \mathrm{C} \equiv \mathrm{O}\]
Diagram
Fig. 5.12 – Three Lewis structures of CO₂ with lone pairs shown and the two oxygen atoms labelled 1 and 2 in each: A = Ö=C=Ö, B = :O≡C–Ö:, C = :Ö–C≡O:
Assigning the formal charges on the carbon atom and the two oxygen atoms numbered 1 and 2:
Structure A
Number of electrons: 4 from carbon and 6 from each oxygen.
So the total number of electrons \(= 4 + 6 + 6 = 16\)
Formal charge on C \(= 4 - 0 - \tfrac{1}{2}(8) = 0\)
Formal charge on O-1 and O-2 \(= 6 - 4 - \tfrac{1}{2}(4) = 0\)
In this structure the formal charge on all atoms is zero.
Structure B
Formal charge on C \(= 4 - 0 - \tfrac{1}{2}(8) = 0\)
Formal charge on O-1 \(= 6 - 2 - \tfrac{6}{2} = 4 - 3 = +1\)
Formal charge on O-2 \(= 6 - 6 - \tfrac{2}{2} = -1\)
Structure C
Formal charge on C \(= 4 - 0 - \tfrac{1}{2}(8) = 0\)
Formal charge on 1st O \(= 6 - 6 - \tfrac{2}{2} = -1\)
Formal charge on 2nd O \(= 6 - 2 - \tfrac{6}{2} = 4 - 3 = +1\)
We find that in structure A the formal charge on all atoms is 0, while in structures B and C the formal charge on carbon is 0 but the oxygens have formal charges of \(-1\) or \(+1\). So the possible structure with the lowest energy will be structure A.
Diagram
Fig. 5.13 – Three Lewis dot structures A, B and C of the SCN⁻ ion with all lone pairs shown: A with two double bonds, B with an S–C single bond and a C≡N triple bond, C with an S≡C triple bond and a C–N single bond
Step II: Assign formal charges using
\[\text{FC} = \text{V.E} - \text{N.E} - \tfrac{1}{2}\,\text{B.E}\]
Structure A:
Formal charge on S \(= 6 - 4 - \tfrac{1}{2}(4) = 0\)
Formal charge on C \(= 4 - 0 - \tfrac{1}{2}(8) = 0\)
Formal charge on N \(= 5 - 4 - \tfrac{1}{2}(4) = -1\)
Structure B:
Formal charge on S \(= 6 - 6 - \tfrac{1}{2}(2) = -1\)
Formal charge on C \(= 4 - 0 - \tfrac{1}{2}(8) = 0\)
Formal charge on N \(= 5 - 2 - \tfrac{1}{2}(6) = 0\)
Structure C:
Formal charge on S \(= 6 - 2 - \tfrac{1}{2}(6) = +1\)
Formal charge on C \(= 4 - 0 - \tfrac{1}{2}(8) = 0\)
Formal charge on N \(= 5 - 6 - \tfrac{1}{2}(2) = -2\)
Generally the lowest energy structure has the smallest formal charges on the atoms.
5.2.7 Limitation of Octet Rule
The octet rule is based on the inert behaviour of the noble gases, which have their octet complete, that is, they have eight electrons in their valence shell. It is very useful to explain the structures and stability of organic molecules.
However, there are many molecules whose existence cannot be explained by the octet theory. The central atoms in these molecules do not have eight electrons in their valence shell, and yet they are stable — for example \(\mathrm{BeCl_2}\) and \(\mathrm{PF_5}\). Such molecules can be categorized as having:
- Incomplete octet
- Expanded octet
- Odd electrons
i. Molecules with Incomplete Octet
Examples: \(\mathrm{BF_3}\), \(\mathrm{BeCl_2}\), \(\mathrm{LiCl}\).
In these covalent molecules the central atoms B, Be and Li have less than eight electrons in their valence shell, but are stable.
- Li in \(\mathrm{LiCl}\) has only two electrons.
- Be in \(\mathrm{BeCl_2}\) has four electrons.
- B in \(\mathrm{BF_3}\) has six electrons in the valence shell.
These are drawn in Fig. 5.14. Notice that all three central atoms come from the first two groups of the periodic table and simply do not have enough valence electrons to reach eight, however they share them.
Diagram
Fig. 5.14 – Molecules with incomplete octet: Li–Cl with 2 electrons around Li, Cl–Be–Cl with 4 electrons around Be, and BF₃ with three B–F bonds and 6 electrons around B; mark the electron count around each central atom
ii. Molecules with Expanded Octet
Some molecules like \(\mathrm{SF_6}\), \(\mathrm{PCl_5}\) and \(\mathrm{H_2SO_4}\) have more than eight electrons around the central atom. In \(\mathrm{SF_6}\) there are 12 electrons around sulfur, in \(\mathrm{PCl_5}\) there are 10 electrons around phosphorus, and in \(\mathrm{H_2SO_4}\) there are 12 electrons around sulfur. All three are shown in Fig. 5.15.
It must be remembered that sulfur also forms many compounds in which the octet rule is obeyed. For example, in sulfur dichloride the sulfur atom has eight electrons around it, as the last part of Fig. 5.15 shows.
Diagram
Fig. 5.15 – Molecules with expanded octet and one that is not: SF₆ with six S–F bonds, labelled “12 electrons around sulfur”; PCl₅ with five P–Cl bonds, labelled “10 electrons around phosphorus”; H₂SO₄ with two S=O bonds and two S–O–H bonds, labelled “12 electrons around sulfur”; and :C̈l–S̈–C̈l: labelled “SCl₂, 8 electrons around sulfur”
iii. Odd Electron Molecules
Some molecules like NO (nitric oxide) and \(\mathrm{NO_2}\) (nitrogen dioxide) do not obey the octet rule. Both the N and O atoms have an odd number of electrons.
An octet needs an even count of electrons, so a molecule whose total valence electrons come to an odd number simply cannot pair them all up. One electron is always left unpaired, as drawn in Fig. 5.16.
Diagram
Fig. 5.16 – Odd electron molecules: Ṅ=Ö with a single unpaired dot shown on nitrogen; and Ö=N̈⊕–Ö:⊖ for nitrogen dioxide, with the odd electron marked
A Lewis structure tells us which atoms are joined and by how many pairs, but it says nothing about the angles between those bonds. The octet rule cannot tell us why water is bent while carbon dioxide is straight. That gap is filled by the theory taken up in the next section.
3. The octet rule fails to explain the difference in energies of molecules, though all the covalent bonds are formed in an identical manner, that is, by sharing a pair of electrons. The rule also fails to explain the difference in reactivities of different molecules.
If every shared pair were the same as every other, all covalent bonds would be equally strong and all covalent molecules equally reactive — which experiment plainly contradicts.
5.3 Valence Shell Electron Pair Repulsion Theory (VSEPR)
Properties of substances are dependent on the shape of the molecules. Two compounds built from the same kinds of atoms can behave quite differently simply because their atoms are arranged differently in space, and so the question of shape is not a decorative detail — it is central to chemistry.
The Lewis concept is unable to explain the shapes of molecules. A Lewis structure tells us faithfully which atoms are joined and how many pairs join them, but it is drawn flat on paper and says nothing about the angles at which the bonds actually radiate from the central atom. Shapes of all molecules cannot be described completely by any single theory.
One of the popular models used earlier to predict the shapes of covalent molecules is the valence shell electron pair repulsion theory, proposed by Sidgwick and Powell. It is based on the basic idea that the electron pairs on the atoms shown in the Lewis diagram repel each other. In the real molecule they arrange themselves in such a way that there is minimum repulsion between them.
The arrangement of electrons is called the electron pair geometry. These pairs may be shared in a covalent bond, or they may be lone pairs.
Rules of VSEPR
- Electron pairs arrange themselves in such a way that repulsion between them is minimum.
- The molecule acquires minimum energy and maximum stability.
- Lone pairs of electrons also contribute in determining the shape of the molecule.
- Repulsion of other electron pairs by the lone pair (L.P) is stronger than that of the bonding pair (B.P).
The trend for repulsion between electron pairs is
\[\text{L.P} - \text{L.P} \;>\; \text{L.P} - \text{B.P} \;>\; \text{B.P} - \text{B.P}\]
Lone pair – lone pair repulsion is maximum because this electron pair is under the influence of only one nucleus, while the bonded pair is shared between two nuclei. A bonding pair is being pulled from both ends and is therefore drawn into a narrower region between the two atoms, so it takes up less angular room. A lone pair, held by one nucleus only, spreads out more and pushes harder on everything near it. This difference in “bulk” is drawn in Fig. 5.17.
Diagram
Fig. 5.17 – Comparison of a bonding pair and a lone pair: on the left, an electron cloud stretched narrowly between two nuclei A and B; on the right, a fatter electron cloud attached to a single nucleus A; below, the repulsion order lp–lp > lp–bp > bp–bp shown as three pairs of arrows of decreasing size
Thus the number of lone pairs and bonded pairs of electrons decides the shape of molecules. Molecules having no lone pair of electrons have a regular geometry.
Shapes When There Are No Lone Pairs
When every electron pair around the central atom is a bonding pair, all the repulsions are of the same kind and the pairs spread out into a perfectly symmetrical arrangement. The five basic arrangements are collected in Table 5.3 and drawn in Fig. 5.18.
Table 5.3: Geometry of molecules (having no lone pair of electrons)
| Number of electron pairs | Arrangement of electron pairs | Molecular geometry | Examples |
|---|---|---|---|
| 2 | Linear (180°) | Linear | BeBr₂, CO₂ |
| 3 | Trigonal planar (120°) | Trigonal planar | BF₃, BCl₃, BH₃ |
| 4 | Tetrahedral | Tetrahedral | CH₄, NH₄⁺, SiCl₄ |
| 5 | Trigonal bipyramidal | Trigonal bipyramidal | PCl₅, SbF₅, AsF₅ |
| 6 | Octahedral | Octahedral | SF₆, TeF₆, SeF₆ |
Diagram
Fig. 5.18 – The five regular geometries, ball-and-stick, one per row: linear CO₂ with the 180° angle marked; trigonal planar BCl₃ with the 120° angle marked; tetrahedral CH₄; trigonal bipyramidal PCl₅ showing the two axial and three equatorial positions; octahedral SF₆
Shapes When Lone Pairs Are Present
Depending on the number of lone pairs and bonded pairs of electrons, molecules can be represented as \(\mathrm{AB_2E}\), \(\mathrm{AB_3E}\), \(\mathrm{AB_2E_2}\), \(\mathrm{AB_4E}\), \(\mathrm{AB_3E_2}\), \(\mathrm{AB_5E}\) and \(\mathrm{AB_4E_2}\), where A is the central atom, B is a bonded atom and E is a lone pair of electrons. Examples of the above types of molecules are given in Table 5.4.
Table 5.4: Geometry of some molecules (having one or more lone pairs of electrons)
| Molecule type | No. of lone pairs | No. of bonding pairs | Bond angle shown | Shape | Examples |
|---|---|---|---|---|---|
| AB₂E | 1 | 2 | 119.3° | Bent | SO₂, O₃ |
| AB₃E | 1 | 3 | 106.7° | Trigonal pyramidal | NH₃, PCl₃ |
| AB₂E₂ | 2 | 2 | 104.5° | Bent | H₂O, OF₂, H₂S, SCl₂, etc. |
| AB₄E | 1 | 4 | — | See saw | SF₄ |
| AB₃E₂ | 2 | 3 | 86.2° | T-shape | ClF₃, BrF₃, ICl₃, etc. |
| AB₅E | 1 | 5 | — | Square pyramid | BrF₅, IF₅ |
| AB₄E₂ | 2 | 4 | — | Square planar | XeF₄ |
Diagram
Fig. 5.19 – Ball-and-stick shapes for the seven types in Table 5.4, each with its lone pair(s) shown as a shaded cloud on the central atom: bent SO₂ (119.3°), trigonal pyramidal NH₃ (106.7°), bent H₂O (104.5°), see-saw SF₄, T-shaped ClF₃ (86.2°), square pyramidal BrF₅, square planar XeF₄
Reading these two tables together gives a reliable working method: first count all the electron pairs to fix the electron pair geometry, then remove the lone pairs from the picture to see the shape the atoms actually make. The procedure is set out in FC 5.4.
Flowchart
FC 5.4 – Predicting a molecular shape by VSEPR: write the Lewis structure of the molecule or ion → count the electron pairs around the central atom (bonding pairs + lone pairs) → read off the electron pair geometry from Table 5.3 (2 → linear, 3 → trigonal planar, 4 → tetrahedral, 5 → trigonal bipyramidal, 6 → octahedral) → branch: no lone pairs → shape is the same as the electron pair geometry, angles regular; one or more lone pairs → delete the lone pair positions and name the shape from the remaining atoms, then reduce the bond angle because lone pair repulsion is stronger
The VSEPR theory is therefore able to predict and also explain the geometry of a large number of compounds, particularly of p-block elements.
Explaining the Bond Angles in NH₃ and H₂O
Let us explain the bond angles in \(\mathrm{NH_3}\) and \(\mathrm{H_2O}\) in detail, since these two molecules show very clearly how a lone pair squeezes a bond angle.
1. Ammonia, NH₃
The expected geometry is tetrahedral and the bond angle 109°28′. The central nitrogen atom has in all 8 electrons in its valence shell, out of which 6 are involved in forming three N–H covalent bonds; the remaining pair forms the lone pair.
There are two types of repulsions between the electron pairs:
- Lone pair – bond pair
- Bond pair – bond pair
The lone pair – bond pair repulsions are stronger, and so the bonded pairs are pushed inside, thus reducing the bond angle to 107°18′ and the shape of the molecule becomes pyramidal.
2. Water Molecule, H₂O
The central atom oxygen has six electrons in its valence shell. On bond formation with two hydrogen atoms there are 8 electrons in the valence shell of oxygen. Out of these, two pairs are bonded pairs and two are lone pairs.
Due to lone pair – lone pair repulsion, the lone pairs are pushed towards the bond pairs, and the bond pair – lone pair repulsion becomes stronger, thereby reducing the H–O–H bond angle from the tetrahedral one to 104°35′, and the geometry of the molecule is angular. Fig. 5.20 places the two molecules side by side so the progressive squeezing can be seen.
Diagram
Fig. 5.20 – Ammonia and water compared: NH₃ drawn as a trigonal pyramid with one shaded lone pair on N and the 107°18′ angle marked; H₂O drawn as a bent molecule with two shaded lone pairs on O and the 104°35′ angle marked; a note that the angle falls further as the number of lone pairs rises
Beyond the Lewis Picture
The Kossel and Lewis approach to chemical bonding is the first step in understanding the nature of the chemical bond, and VSEPR builds usefully on top of it. But neither tells us what a bond actually is in terms of the orbitals you studied in the chapter on atomic structure.
More advanced theories of bonding were put forth to account for the newly discovered properties of compounds in the light of the quantum mechanical theory of atomic structure. Two important approaches regarding the nature of the chemical bond are valence bond theory and molecular orbital theory, and these are taken up in the sections that follow.
5.4 Valence Bond Theory
In order to explain covalent bonding, Heitler and London developed the valence bond theory on the basis of wave mechanics. This theory was further extended by Pauling and Slater.
The shift in thinking here is worth noticing. Lewis asked how many electrons an atom must share; valence bond theory asks where those electrons are, and answers in terms of the atomic orbitals you met in the chapter on atomic structure. A bond is no longer a pair of dots on paper — it is two orbitals reaching into the same region of space.
5.4.1 Postulates of Valence Bond Theory
A covalent bond is formed when the half-filled valence orbital of one atom overlaps with a half-filled valence orbital of another atom.
The electrons in the half-filled valence orbitals must have opposite spins.
During bond formation the half-filled orbitals overlap and the opposite spins of the electrons get neutralized. The increased electron density decreases the nuclear repulsion, and energy is released during the overlapping of the orbitals.
The greater the extent of overlap, the stronger is the bond formed. However, complete overlap of orbitals does not take place, due to internuclear repulsions.
If an atom possesses more than one unpaired electron, then it can form more than one bond. So the number of bonds formed will be equal to the number of half-filled orbitals in the valence shell, that is, the number of unpaired electrons.
The distance at which the attractive and repulsive forces balance each other is the equilibrium distance between the nuclei of the bonded atoms. At this distance the total energy of the bonded atoms is minimum and stability is maximum.
Electrons which are paired in the valence shell cannot participate in bond formation. However, in an atom, if there is one or more vacant orbital present, then these electrons can unpair and participate in bond formation, provided the energies of the filled and vacant orbitals differ only slightly from each other.
During bond formation the ‘s’ orbital, which is spherical, can overlap in any direction. The ‘p’ orbitals can overlap only in the x, y or z directions. (Similarly, ‘d’ and ‘f’ orbitals are oriented in certain directions in space and overlap only in those directions.) Thus the covalent bond is directional in nature.
Postulate viii is worth dwelling on, because it explains something VSEPR could only assume. A bond points somewhere because the orbital that makes it points somewhere. That is why molecules have definite shapes at all.
5.4.2 Interacting Forces During Covalent Bond Formation
By now we have understood that a covalent bond is formed by the overlap of two half-filled atomic orbitals, that the bonded atoms are more stable than the free atoms, and that the energy of the bonded atoms is less than that of the free atoms. So a lowering of energy takes place during bond formation. How does this happen?
This happens due to interactive forces which develop between the nuclei of the two atoms and also their electrons. These forces may be attractive and repulsive: repulsion between the nucleus of A and the nucleus of B and between the electrons of the two atoms, and attraction between the nucleus of atom A and the electrons of atom B, and likewise between the nucleus of B and the electrons of A.
The balance between the attractive and repulsive forces decides whether the bond will be formed or not. When the attractive forces are stronger than the repulsive forces, overlap takes place between the two half-filled orbitals, a bond is formed, and the energy of the system is lowered.
The Potential Energy Curve
This lowering of energy during bond formation is depicted in the potential energy diagram. To understand this, let us consider the formation of an \(\mathrm{H_2}\) molecule from atoms of hydrogen, each containing one unpaired electron.
When the two atoms are far away from each other there are no interactions between them. The energy of the system is the sum of the potential energies of the two atoms, which is arbitrarily taken as zero. This choice of zero is only a convenient starting mark — what matters is not the absolute value but how the energy changes as the atoms approach, and this is plotted in Fig. 5.21.
Diagram
Fig. 5.21 – Potential energy diagram for the formation of the hydrogen molecule: y-axis “total energy”, x-axis “distance between nuclei”; curve starting at zero on the right at large separation, falling to a minimum, then rising steeply on the left; mark the minimum point and label the corresponding distance as “equilibrium distance”
Repulsive forces destabilize the system, with an increase in the energy of the system, while attractive forces decrease the energy. Experimentally it has been found that during the formation of the hydrogen molecule the magnitude of the newly developed attractive forces contributes more than that of the newly developed repulsive forces. As a result, the potential energy of the system begins to decrease.
As the atoms come closer to one another, the energy of the system decreases. The overlap increases only up to a certain distance between the two nuclei, where the attractive and repulsive forces balance each other and the system attains minimum energy, as seen in Fig. 5.21. At this stage a bond is formed between the two atoms of hydrogen.
If the two atoms are further pushed closer to each other, the repulsive forces become more predominant and the energy of the system starts increasing, and stability decreases.
If atoms containing electrons with parallel spins are brought close to each other, the potential energy of the system increases and bond formation does not take place.
5.4.3 Overlap of Atomic Orbitals
The formation of a bond has been explained on the basis of the overlap of atomic orbitals having the same energy and symmetry. In the preceding section we have seen that the strength of the bond depends on the extent of overlap of the orbitals. The greater the overlap, the stronger is the bond.
The orbitals holding the electrons vary in shape, energy and symmetry. So the extent of overlap depends on the shape and size of the orbital.
On the basis of the above considerations we have 2 types of bonds.
- sigma bond (\(\sigma\))
- pi bond (\(\pi\))
i. Sigma Bond
When the overlap of the bonding orbitals is along the internuclear axis, it is called sigma overlap or a sigma bond.
The \(\sigma\) bond is formed by the overlap of the following orbitals.
- Two ‘s’ orbitals
- One ‘s’ and one \(p_z\) orbital
- Two ‘p’ orbitals
a. s–s overlap: e.g. H₂
The \(1s^1\) orbitals of two hydrogen atoms overlap along the internuclear axis to form a \(\sigma\) bond between the atoms in the \(\mathrm{H_2}\) molecule, as drawn in Fig. 5.22.
Electronic configuration of H : \(1s^1\)
Diagram
Fig. 5.22 – s–s overlap: two spherical 1s orbitals, each with a dot for its electron, approaching and merging along a horizontal internuclear axis to give the H₂ molecule; label “1s orbitals of H” and “H₂ molecule”
b. p–p overlap
This type of overlap takes place when two p orbitals from different atoms overlap along the internuclear axis, for example in the \(\mathrm{F_2}\) molecule, shown in Fig. 5.23.
Electronic configuration of fluorine: \(1s^2, 2s^2, 2p_x^2, p_y^2, p_z^1\). The \(2p_z\) orbitals of the fluorine atoms overlap along the internuclear axis to form a p–p \(\sigma\) overlap.
\(\mathrm{F_2}\) molecule: \(1s^2,\ 2s^2,\ 2p_x^2,\ p_y^2,\ p_z^1\)
Notice why it must be the \(p_z\) orbital that does the bonding: it is the only p orbital of fluorine that is half filled, and by convention the z axis is taken as the internuclear axis.
Diagram
Fig. 5.23 – p–p sigma overlap: two dumbbell-shaped p orbitals lying along the same horizontal axis, their near lobes overlapping end to end to give the F₂ molecule; label each “p” and mark the internuclear axis
c. s–p σ bond
In this type of overlap, one half-filled ‘s’ orbital of one atom and one half-filled ‘p’ orbital of another atom overlap along the internuclear axis — for example in the HF molecule, drawn in Fig. 5.24.
Electronic configuration:
H \(\rightarrow 1s^1\) ; F : \(1s^2,\ 2s^2,\ 2p_x^2,\ p_y^2,\ p_z^1\)
The \(1s^1\) orbital of hydrogen and the \(2p_z^1\) orbital of fluorine overlap to form an s–p \(\sigma\) overlap.
Diagram
Fig. 5.24 – s–p sigma overlap: a spherical s orbital on the left overlapping end to end with one lobe of a dumbbell-shaped p orbital on the right, along the internuclear axis; label “s” and “p”
ii. p–p Overlap / π Overlap / π Bond
When two half-filled orbitals of two atoms overlap sideways (laterally), it is called \(\pi\) overlap, and it is perpendicular to the internuclear axis. The picture is given in Fig. 5.25.
Diagram
Fig. 5.25 – π overlap: two dumbbell p orbitals standing parallel and side by side, their upper lobes overlapping above the axis and their lower lobes below, giving a two-part electron cloud; label “p orbital + p orbital = π overlap”
Because a sideways approach never lets the two lobes meet as fully as an end-on approach does, a \(\pi\) bond involves less overlap than a \(\sigma\) bond between the same pair of atoms, and is therefore the weaker of the two.
5.4.4 Hybridization
The valence bond theory explained well the formation of the covalent bond by the overlap of orbitals in the case of simple molecules like \(\mathrm{H_2}\), \(\mathrm{F_2}\), \(\mathrm{H-F}\) and so on. Accordingly, the maximum number of covalent bonds which an atom can form equals the number of unpaired electrons present in its valence shell.
But the theory does not explain how beryllium forms two covalent bonds, or how boron shows trivalency and carbon shows tetravalency, in spite of their electronic configurations — for example in \(\mathrm{BeH_2}\), \(\mathrm{BF_3}\), \(\mathrm{CH_4}\), \(\mathrm{CCl_4}\) and so on.
Be : \(1s^2,\ 2s^2\)
B : \(1s^2,\ 2s^2,\ 2p^1\)
C : \(1s^2,\ 2s^2,\ 2p_x^1,\ 2p_y^1\)
Count the unpaired electrons in each of these ground state configurations and the difficulty is plain: beryllium has none at all, boron has one, and carbon has two. On the strength of postulate v, they should form zero, one and two bonds. Experiment says two, three and four.
In order to explain the observed valency in these and such other compounds, the concept of hybridization was put forward.
Promotion to the Excited State
It was suggested that one electron in the ‘2s’ orbital is promoted to the empty ‘2p’ orbital. Thus in the excited state, Be, B and C have two, three and four half-filled orbitals, respectively. Their excited state configurations are shown as orbital boxes in Fig. 5.26.
Diagram
Fig. 5.26 – Electronic configurations in the excited state, orbital box notation with columns headed 1s, 2s and 2p (three boxes): Be with a paired arrow in 1s, one up arrow in 2s and one up arrow in the first 2p box; B with a paired arrow in 1s, one up arrow in 2s and up arrows in two 2p boxes; C with a paired arrow in 1s, one up arrow in 2s and up arrows in all three 2p boxes
In the excited state, Be, B and C have 2, 3 and 4 half-filled orbitals. So Be, B and C can form 2, 3 and 4 bonds respectively.
Why Promotion Alone Is Not Enough
This concept helps us to understand how Be forms 2 bonds whereas B and C form 3 and 4 bonds respectively — but it cannot explain how all the bonds have the same bond length and bond strength.
For example, in \(\mathrm{BeF_2}\), beryllium will use one s and one half-filled p orbital to overlap with two half-filled ‘p’ orbitals on fluorine. So in the molecule there will be one s–p bond and one p–p bond, which will not be of equal strength; but actually both Be–F bonds are of the same strength.
A similar situation is seen in \(\mathrm{BF_3}\), \(\mathrm{BH_3}\), \(\mathrm{CH_4}\) and \(\mathrm{CCl_4}\). In \(\mathrm{CH_4}\) all bonds are of equal strength, although the overlaps are between the s, \(p_x\), \(p_y\) and \(p_z\) orbitals of carbon and the ‘s’ orbital of hydrogen; experimentally all C–H bond lengths, bond strengths and bond angles are found to be identical.
This can be explained using another concept — hybridization in the valence bond theory. This concept helps to explain the observed structural properties of many molecules.
Steps Considered in Hybridization
- Formation of the excited state
- Mixing and recasting of orbitals
i. Formation of the excited state: The paired electrons in the ground state are uncoupled and one electron is promoted to a vacant orbital having slightly higher energy. Now the total number of half-filled orbitals is equal to the valency of the element in the stable compound.
For example, in \(\mathrm{BeF_2}\) the valency of Be is two. In the excited state one electron from the 2s orbital is uncoupled and promoted to a 2p orbital:
Ground state: \(1s^2\ 2s^2\)
Excited state: \(1s^2\ 2s^1\ 2p^1\)
ii. Mixing and recasting: In this step the ‘s’ and ‘p’ orbitals, having slightly different energies, mix with each other. Redistribution of electron density and energy takes place, and two new orbitals having exactly the same shape and energy are formed.
These new orbitals arrange themselves in space in such a way that there is minimum repulsion and maximum separation between them. So during the formation of sp hybrid orbitals, as in Be, the two sp hybrid orbitals are at 180°.
The whole sequence is summarised in FC 5.5.
Flowchart
FC 5.5 – Steps in hybridization: ground state atom with paired valence electrons → promotion: one paired electron moved to a vacant orbital of slightly higher energy → excited state with the required number of half-filled orbitals → mixing and recasting of the s and p orbitals of comparable energy → equal number of new hybrid orbitals, identical in shape and energy → hybrid orbitals arrange themselves for minimum repulsion → axial overlap with orbitals of other atoms → molecule of definite shape and equal bonds
Conditions for Hybridization
- Orbitals belonging to the same atom can participate in hybridization.
- Orbitals having nearly the same energy can undergo hybridization. So 2s and 2p orbitals undergo hybridization, but 3s and 2p orbitals do not.
The second condition follows from the energy gap. Mixing works only when the orbitals being mixed start out close in energy; a 3s orbital lies far above a 2p orbital, so no useful mixing is possible between them.
Characteristic Features of Hybrid Orbitals
- The number of hybrid orbitals formed is exactly the same as the number of participating atomic orbitals.
- They have the same energy and shape.
- Hybrid orbitals are oriented in space in such a way that there is minimum repulsion, and they are thus directional in nature.
- The hybrid orbitals are different in shape from the participating atomic orbitals, but they bear the characteristics of the atomic orbitals from which they are derived.
- Each hybrid orbital can hold two electrons with opposite spins.
- A hybrid orbital has two lobes on the two sides of the nucleus. One lobe is large and the other small.
- Covalent bonds formed by hybrid orbitals are stronger than those formed by pure orbitals, because the hybrid orbital has electron density concentrated on the side with the larger lobe while the other lobe is small, allowing greater overlap of the orbitals.
Feature vii ties the whole idea back to postulate iv: greater overlap means a stronger bond, and the lopsided shape of a hybrid orbital is exactly what makes greater overlap possible.
5.4.5 Types of Hybridization and Geometry of Molecules
Different types of hybrid orbitals are obtained from the atomic orbitals that participate in hybridization.
s and p orbitals can hybridize to form the following hybrid orbitals:
- \(sp^3\)
- \(sp^2\)
- \(sp\)
i. sp³ Hybridization
In this type, one ‘s’ and three ‘p’ orbitals having comparable energy mix and recast to form four \(sp^3\) hybrid orbitals. It should be remembered that the ‘s’ orbital is spherically symmetrical, while the \(p_x\), \(p_y\) and \(p_z\) orbitals have two lobes and are directed along the x, y and z axes respectively.
The four \(sp^3\) hybrid orbitals formed are equivalent in energy and shape. They have one large lobe and one small lobe. They are at an angle of 109°28′ with each other in space and point towards the corners of a tetrahedron, as drawn in Fig. 5.27. \(\mathrm{CH_4}\), \(\mathrm{NH_3}\) and \(\mathrm{H_2O}\) are examples where the orbitals on the central atom undergo \(sp^3\) hybridization.
Diagram
Fig. 5.27 – Formation of sp³ hybrid orbitals: on the left, a spherical 2s orbital and the three dumbbell orbitals 2pₓ, 2p_y, 2p_z shown separately on x, y, z axes; an arrow labelled “sp³ hybrids” leading to four identical lobed orbitals pointing to the corners of a tetrahedron, shown both separately and together
Formation of methane (CH₄) molecule
The ground state electronic configuration of carbon is \(1s^2,\ 2s^2,\ 2p_x^1,\ p_y^1,\ p_z^0\). In order to form four equivalent bonds with hydrogen, the 2s and 2p orbitals undergo hybridization. The three stages are shown as orbital boxes in Fig. 5.28.
One electron from the 2s orbital of the carbon atom is excited to the \(2p_z\) orbital. Then the four orbitals 2s, \(p_x\), \(p_y\) and \(p_z\) mix and recast to form four new \(sp^3\) hybrid orbitals having the same shape and equal energy. They are maximum apart and have tetrahedral geometry. Each hybrid orbital contains one unpaired electron.
Each of these \(sp^3\) hybrid orbitals with one electron overlaps axially with the 1s orbital of a hydrogen atom to form one C–H sigma bond. Thus in the \(\mathrm{CH_4}\) molecule we have four C–H bonds formed by \(sp^3\)–s overlap.
Diagram
Fig. 5.28 – Electronic configuration of carbon in orbital box notation with columns 1s, 2s, 2p: ground state (1s paired, 2s paired, two singly occupied 2p boxes and one empty); excited state (1s paired, 2s single, three singly occupied 2p boxes); sp³ hybrid orbitals row showing four singly occupied equivalent boxes labelled “four sp³ hybrid orbitals”. Below, four 1s orbitals of H overlapping axially with the four sp³ hybrid orbitals of C to give tetrahedral CH₄ with the 109.5° angle marked
The same \(sp^3\) hybridization operates in ammonia and water, except that there some of the hybrid orbitals hold lone pairs instead of forming bonds, as Fig. 5.29 shows. This is the orbital picture behind the shapes VSEPR predicted for these two molecules in the previous section.
Diagram
Fig. 5.29 – Formation and orbital pictures of NH₃ and H₂O. Top row: 2s and 2p orbitals of nitrogen shown separately, then the four sp³ hybrids together, then NH₃ with three sp³–1s N–H sigma bonds and the fourth sp³ hybrid carrying the lone pair on nitrogen. Bottom row: the same sequence for oxygen, giving H₂O with two sp³–1s O–H sigma bonds and two sp³ hybrids each carrying a lone pair
ii. sp² Hybridization
This hybridization involves the mixing of one s and two p orbitals to give three \(sp^2\) hybrid orbitals of the same energy and shape. The three orbitals are maximum apart, oriented at an angle of 120°, and are in one plane. The third p orbital does not participate in hybridization and remains at right angles to the plane of the \(sp^2\) hybrid orbitals. \(\mathrm{BF_3}\) and \(\mathrm{C_2H_4}\) molecules are examples of \(sp^2\) hybridization. The formation of these orbitals is drawn in Fig. 5.30.
Diagram
Fig. 5.30 – Formation of sp² hybrid orbitals: 2s, 2pₓ and 2p_y orbitals shown separately on their axes; an arrow labelled “sp² hybrids” leading to three identical lobed orbitals in one plane at 120° to one another, shown separately and together, with the unhybridized p_z orbital drawn perpendicular to that plane
Formation of C₂H₄ molecule
This molecule contains two carbon atoms, each bound to two hydrogen atoms. Each carbon atom undergoes \(sp^2\) hybridization. One ‘s’ orbital and two ‘p’ orbitals on carbon hybridize to form three \(sp^2\) hybrid orbitals of equal energy and symmetry.
Two \(sp^2\) hybrid orbitals overlap axially with the ‘s’ orbitals of two hydrogen atoms to form \(sp^2\)–s \(\sigma\) bonds. The third \(sp^2\) hybrid orbital of each carbon overlaps axially with that of the other carbon. The unhybridized ‘p’ orbitals on the two carbon atoms overlap laterally to form a lateral \(\pi\) overlap.
Thus the \(\mathrm{C_2H_4}\) molecule has four \(sp^2\)–s \(\sigma\) bonds, one \(sp^2\)–\(sp^2\) \(\sigma\) bond and one p–p \(\pi\) bond. The arrangement is shown in Fig. 5.31, along with the orbital box diagram of the carbon atom in this state.
Diagram
Fig. 5.31 – Bonding in ethene. Left: orbital box notation for carbon with columns 1s, 2s, 2p — ground state, excited state, and “sp² hybridized” row showing three singly occupied sp² boxes plus one singly occupied pure p orbital. Right: two carbon atoms each with three sp² hybrids in a plane forming σ bonds to two H atoms and to each other, with the two unhybridized p_z orbitals above and below the plane overlapping sideways to give the π bond; label the σ bonds and the π bond
Formation of boron trifluoride (BF₃) molecule
Need of hybridisation: The observed valency of boron in \(\mathrm{BF_3}\) is three and its geometry is triangular planar, which can be explained on the basis of \(sp^2\) hybridisation.
sp² hybridisation of the boron atom: In the \(\mathrm{BF_3}\) molecule the central boron atom undergoes \(sp^2\) hybridisation. The ground state electronic configuration of boron (\(Z = 5\)) is
\[1s^2\ 2s^2\ 2p_x^1\ 2p_y^0\ 2p_z^0\]
To explain the valency of boron in \(\mathrm{BF_3}\), one electron from the 2s orbital of the boron atom is uncoupled and promoted to the vacant \(2p_y\) orbital. Thus the excited state electronic configuration of boron becomes
\[1s^2\ 2s^1\ 2p_x^1\ 2p_y^1\ 2p_z^0\]
The three orbitals, that is 2s, \(2p_x\) and \(2p_y\) of boron, undergo \(sp^2\) hybridisation to form three \(sp^2\) hybrid orbitals of equivalent energy, which are oriented along the three corners of an equilateral triangle making an angle of 120°.
Thus boron in the hybridised state has the following electronic configuration: \(1s^2\) and three singly occupied hybrid orbitals.
Orbital overlap: Each \(sp^2\) hybrid orbital of the boron atom, having an unpaired electron, overlaps axially with the half-filled \(2p_z\) orbital of a fluorine atom containing an electron with opposite spin, to form three B–F sigma bonds by \(sp^2\)–p type of overlap.
Bond angle: Each F–B–F bond angle in the \(\mathrm{BF_3}\) molecule is 120°.
Geometry: The geometry of the \(\mathrm{BF_3}\) molecule is trigonal planar. This is drawn in Fig. 5.32.
Diagram
Fig. 5.32 – Formation of BF₃: on the left, the s + pₓ + p_y combination shown as overlapping orbital outlines; centre, the three sp² hybrid orbitals in one plane at 120°; right, the complete BF₃ molecule with three sp²–p sigma bonds to three fluorine atoms and the 120° bond angle marked
iii. sp Hybridization
In this type, one ‘s’ and one ‘p’ orbital undergo mixing and recasting to form two sp hybrid orbitals of the same energy and shape. The two hybrid orbitals are placed at an angle of 180°. The other two p orbitals do not participate in hybridization and are at right angles to the hybrid orbitals. Examples are \(\mathrm{BeCl_2}\) and the acetylene molecule, \(\mathrm{HC}\equiv\mathrm{CH}\). The formation is shown in Fig. 5.33.
Diagram
Fig. 5.33 – Formation of sp hybrid orbitals: 2s and 2pₓ orbitals shown separately on their axes; an arrow leading to two identical lobed orbitals pointing in exactly opposite directions along one line at 180°; an inset showing the two hybrids in cross-section and a schematic of the hybrids drawn together
Formation of acetylene molecule
This molecule contains two carbon and two hydrogen atoms. Each carbon atom undergoes sp hybridization. One s and one p orbital mix and recast to give two sp hybrid orbitals arranged at 180° to each other.
Out of the two sp hybrid orbitals of a carbon atom, one overlaps axially with the ‘s’ orbital of hydrogen, while the other sp hybrid orbital overlaps with the sp hybrid orbital of the other carbon atom to form the sp–sp \(\sigma\) bond. The C–H \(\sigma\) bond is formed by sp–s overlap.
The remaining two unhybridized p orbitals on each carbon overlap laterally to form two p–p \(\pi\) bonds. So there are three bonds between the two carbon atoms: one C–C \(\sigma\) bond (sp–sp overlap) and two C–C \(\pi\) bonds (p–p overlap). The fourth bond, an sp–s \(\sigma\) bond between C and H, satisfies the fourth valency of carbon. All of this is drawn in Fig. 5.34.
Diagram
Fig. 5.34 – Bonding in acetylene. Left: orbital box notation for carbon with columns 1s, 2s, 2p — ground state, excited state, and “sp hybridized” row showing two singly occupied sp boxes plus two singly occupied pure p orbitals. Right: the linear H–C≡C–H skeleton with the three σ bonds marked, one π bond cloud above and below the axis, and the second π bond cloud in front of and behind the axis, drawn perpendicular to the first
Flowchart
FC 5.6 – Deciding the type of hybridization of a central atom: write the ground state configuration → promote a paired electron if the observed valency demands more unpaired electrons → count the orbitals that must be mixed (one s + however many p orbitals are needed) → branch: s + 1p → two sp hybrids → linear, 180°; s + 2p → three sp² hybrids → trigonal planar, 120°, one pure p left for a π bond; s + 3p → four sp³ hybrids → tetrahedral, 109°28′, no p left, so only σ bonds
Diagram
Fig. 5.35 – Formation of BeCl₂. Top: orbital box notation for beryllium with columns 1s, 2s, 2p — ground state (1s paired, 2s paired, 2p empty), excited state (1s paired, 2s single, one 2p single), and “sp hybridized” row showing two singly occupied sp boxes plus two empty pure p orbitals. Bottom: the linear molecule Cl–Be–Cl showing the two sp–p σ bonds and the 180° arrangement, with the Cl p_z orbitals labelled
5.4.6 Importance and Limitation of Valence Bond Theory
Importance of Valence Bond Theory (V.B)
V.B. theory introduced five new concepts in chemical bonding.
- Delocalization of electrons over the two nuclei.
- Shielding effect of electrons.
- Covalent character of a bond.
- Partial ionic character of a covalent bond.
- The concept of resonance, and the connection between resonance energy and molecular stability.
The last of these will be taken up in detail at the end of this chapter.
5.4.7 Limitations of Valence Bond Theory
V.B. theory explains only the formation of a covalent bond in which the shared pair of electrons comes from two bonding atoms. However, it offers no explanation for the formation of a co-ordinate covalent bond, in which both the electrons are contributed by one of the bonded atoms.
The oxygen molecule is expected to be diamagnetic according to this theory. The two atoms in the oxygen molecule should have completely filled electronic shells, which give no unpaired electrons to the molecule, making it diamagnetic. However, experimentally the molecule is found to be paramagnetic, having two unpaired electrons. Thus this theory fails to explain the paramagnetism of the oxygen molecule.
Here the words diamagnetic and paramagnetic need a moment. A substance in which all the electrons are paired is diamagnetic and is weakly repelled by a magnetic field; a substance that has one or more unpaired electrons is paramagnetic and is attracted into a magnetic field. Oxygen’s behaviour in a magnetic field is therefore direct experimental evidence that unpaired electrons are present in the molecule — and the Lewis structure \(\mathrm{O}=\mathrm{O}\), with every electron neatly paired, cannot account for them.
- Valence bond theory does not explain the bonding in electron-deficient molecules like \(\mathrm{B_2H_6}\), in which the central atom possesses fewer electrons than are required for an octet of electrons.
These three failures are not small ones, and they are the reason a second and more general treatment of the covalent bond was developed. That treatment is the subject of the next section.
5.5 Molecular Orbital Theory
You are familiar with the valence bond theory, which describes the formation of covalent bonds by the overlap of half-filled atomic orbitals. This theory is successful in giving a satisfactory electronic description for a large number of molecules. In some cases, however, it gives an incorrect electronic description. Therefore another bonding description, called molecular orbital theory (MO theory), has been introduced.
It has been found that MO theory gives a more accurate description of the electronic structure of molecules.
Orbitals, Old and New
The concept of an orbital is introduced by quantum mechanics. The quantum mechanical wave equation is a differential equation, and its solution is called a wave function, usually written \(\psi\). The square of the wave function gives a measure of the probability of finding an electron within a certain region of space of an atom. That is nothing but an atomic orbital.
MO theory does not concentrate on individual atoms. It considers the molecule as a whole, rather than an atom, for the bonding. Accordingly, a molecular orbital is the property of a molecule, similar to what an atomic orbital is to an atom.
Hence a molecular orbital can be depicted through the square of a wave function that gives the probability of finding an electron within a certain region of space in a molecule. Like atomic orbitals, molecular orbitals have energy levels and definite shapes. They also contain a maximum of two electrons with opposite spins.
The difference from valence bond theory is fundamental. In VB theory an electron pair stays localised between two particular atoms; in MO theory the electrons belong to the molecule as a whole and may spread over several nuclei at once.
5.5.1 Formation of Molecular Orbitals
According to MO theory, the formation of molecular orbitals from atomic orbitals is expressed in terms of the Linear Combination of Atomic Orbitals (LCAO).
The formation of molecular orbitals can be understood by considering the interference of the electron waves of the combining atoms. Interference of electron waves can be constructive or destructive — that is, the waves can reinforce each other or cancel each other. This is the same behaviour you have seen with waves of any kind: crest meeting crest builds a bigger wave, while crest meeting trough cancels out.
So two atomic orbitals combine in two ways to form molecular orbitals:
- By addition of their wave functions.
- By subtraction of their wave functions.
Addition of the atomic orbital wave functions results in the formation of a molecular orbital which is lower in energy than the atomic orbitals, and is termed a Bonding Molecular Orbital (BMO).
Subtraction of the atomic orbitals results in the formation of a molecular orbital which is higher in energy than the atomic orbitals, and is termed an Antibonding Molecular Orbital (AMO).
\[\sigma = \psi_A + \psi_B \qquad \sigma^* = \psi_A - \psi_B\]
In a bonding molecular orbital, a large electron density is observed between the nuclei of the bonding atoms, more than in the individual atomic orbitals. On the other hand, in the antibonding orbital the electron density is nearly zero between the nuclei.
So placing an electron in a bonding orbital leads to the formation of a covalent bond, while placing an electron in an antibonding orbital makes the bond unstable. The relative energies are shown in Fig. 5.36.
Diagram
Fig. 5.36 – Formation of molecular orbitals: energy increasing upwards; atomic orbital ψA on the left and ψB on the right at the same level, each with an electron arrow; sloping lines meeting at a lower central box labelled “bonding orbital, σ = ψA + ψB, lower energy than the atomic orbitals” holding a paired arrow, and at a higher central box labelled “antibonding orbital, σ* = ψA − ψB, higher energy than the atomic orbitals”
5.5.2 Conditions for the Combination of Atomic Orbitals
Atomic orbitals can be combined linearly to give molecular orbitals only if the following conditions are fulfilled.
i. The combining atomic orbitals must have comparable energies. So the 1s orbital of one atom can combine with the 1s orbital of another atom, but not with a 2s orbital, because the energy of the 2s orbital is much higher than that of the 1s orbital.
ii. The combining atomic orbitals must have the same symmetry along the molecular axis. Conventionally the z axis is taken as the internuclear axis. So even if atomic orbitals have the same energy, if their symmetry is not the same they cannot combine.
For example, the 2s orbital of an atom can combine only with the \(2p_z\) orbital of another atom, and not with the \(2p_x\) or \(2p_y\) orbital of that atom, because the symmetries are not the same. \(p_z\) is symmetrical along the z axis while \(p_x\) is symmetrical along the x axis.
The reason is easy to picture. If a spherical s orbital approaches a \(p_x\) orbital lying across the path of approach, it overlaps the positive lobe and the negative lobe by exactly equal amounts, so the constructive and destructive parts cancel and no net bonding results.
iii. The combining atomic orbitals must overlap to the maximum extent. The greater the overlap, the greater is the electron density between the nuclei, and so the stronger is the bond formed.
5.5.3 Types of Molecular Orbitals
In diatomic molecules, the molecular orbitals formed by the combination of atomic orbitals are of two types: (i) \(\sigma\) and (ii) \(\pi\).
According to this nomenclature, \(\sigma\) designates a molecular orbital which is symmetrical around the bond axis, and \(\pi\) designates molecular orbitals which are unsymmetrical.
This is clear if we consider a linear combination of (i) two ‘s’ orbitals and (ii) two p orbitals.
i. The s orbitals are spherically symmetric along the x, y and z axes. The combination of two ‘1s’ orbitals centred on the two nuclei of two atoms leads to two \(\sigma\) molecular orbitals which are symmetrical along the bond axis. One of these is \(\sigma\) bonding and the other is \(\sigma^*\) antibonding, as shown in Fig. 5.37.
Diagram
Fig. 5.37 – Linear combination of two s orbitals: two isolated spherical 1s orbitals of H atoms in the centre with a dashed line marked “energy of isolated H atoms”; an upward branch labelled “Subtract (1s − 1s)” leading to two separated lobes with a node between them, labelled “antibonding MO, σ*1s”; a downward branch labelled “Add (1s + 1s)” leading to a single merged sausage-shaped cloud, labelled “bonding MO, σ1s”
ii. If we consider ‘z’ to be the internuclear axis, then the linear combination of \(p_z\) orbitals from two atoms can form the \(\sigma 2p_z\) bonding and \(\sigma^*(2p_z)\) antibonding molecular orbitals.
The \(p_x\) and \(p_y\) orbitals are not symmetrical along the bond axis. They have a positive lobe above the axis and a negative lobe below the axis. Hence the linear combination of such orbitals leads to the formation of molecular orbitals with positive and negative lobes above and below the bond axis.
These are designated as \(\pi\) bonding and \(\pi^*\) antibonding orbitals. The electron density in such \(\pi\) orbitals is concentrated above and below the bond axis. The \(\pi^*\) molecular orbital has a node between the nuclei, as drawn in Fig. 5.38.
A node is simply a region where the probability of finding the electron falls to zero. Its presence between the two nuclei is exactly why an antibonding orbital pushes the atoms apart instead of holding them together — there is no electron density in the region where glue is needed.
Diagram
Fig. 5.38 – Formation of π and π* molecular orbitals: upper row, two parallel p orbitals side by side with an arrow leading to four separated lobes with a nodal plane cutting between the nuclei, labelled π*p; lower row, the same two parallel p orbitals with an arrow leading to two continuous clouds, one above and one below the bond axis, labelled πp
5.5.4 Energy Levels and Electronic Configuration
We have seen earlier that on the combination of two 1s orbitals, two molecular orbitals \(\sigma 1s\) and \(\sigma^* 1s\) are formed. Similarly, two 2s orbitals yield \(\sigma 2s\) and \(\sigma^* 2s\) molecular orbitals.
The three 2p orbitals on one atom combine with the three 2p orbitals on another atom, forming six molecular orbitals, designated as \(\sigma 2p_z\), \(\pi p_x\), \(\pi p_y\) and \(\sigma^* 2p_z\), \(\pi^* p_x\), \(\pi^* p_y\).
The molecular orbitals in homonuclear diatomic molecules have been determined experimentally. (A homonuclear diatomic molecule is one made of two atoms of the same element — \(\mathrm{H_2}\), \(\mathrm{N_2}\), \(\mathrm{O_2}\) and so on.)
For diatomic molecules of second row elements except \(\mathrm{O_2}\) and \(\mathrm{F_2}\), the rank order of energies is
\[\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z\]
For \(\mathrm{O_2}\) and \(\mathrm{F_2}\), the increasing order of energies was found to be
\[\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z\]
The only difference between the two lists is the position of \(\sigma 2p_z\) relative to the \(\pi 2p\) pair, but that single swap changes the whole filling pattern, so it must be checked before writing any configuration. The two orders are set side by side in Fig. 5.39.
Diagram
Fig. 5.39 – The two molecular orbital energy orders compared: two vertical energy ladders side by side, energy increasing upwards, both starting σ1s, σ1s, σ2s, σ2s; the left ladder (for B₂ to N₂) then shows the degenerate π2pₓ = π2p_y pair below σ2p_z; the right ladder (for O₂ and F₂) shows σ2p_z below the degenerate π2pₓ = π2p_y pair; both then continue with π2pₓ = π2p_y and σ*2p_z at the top; highlight the swapped levels
The sequence of filling the molecular orbitals gives the electronic configuration of molecules. The electronic configuration of molecules provides the following information.
a. Stability of molecules: If the number of electrons in bonding MOs is greater than the number in antibonding MOs, the molecule is stable.
b. Magnetic nature of molecules: If all the MOs in a molecule are completely filled with two electrons each, the molecule is diamagnetic, that is, repelled by a magnetic field. However, if at least one MO is half filled (having one electron), the molecule is paramagnetic, that is, attracted by a magnetic field.
c. Bond order of molecule: The bond order of the molecule can be calculated from the number of electrons in bonding MOs (\(N_b\)) and antibonding MOs (\(N_a\)).
\[\text{Bond order} = \frac{N_b - N_a}{2}\]
Flowchart
FC 5.7 – Working out the MO description of a diatomic molecule: count the total number of electrons in the molecule or ion (add electrons for a negative charge, remove for a positive) → choose the correct energy order (O₂ and F₂ use σ2p_z below π2p; the others use π2p below σ2p_z) → fill the molecular orbitals from the lowest energy upwards, two electrons per orbital with paired spins, applying Hund’s rule to the degenerate π pair → write the electronic configuration → count N_b and N_a → bond order = (N_b − N_a)/2 → branch: all electrons paired → diamagnetic; one or more unpaired → paramagnetic
5.5.5 Key Ideas of MO Theory
MOs in molecules are similar to AOs of atoms. A molecular orbital describes a region of space in the molecule representing the probability of finding an electron.
MOs are formed by combining AOs of different atoms. The number of MOs formed is equal to the number of AOs combined.
Atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals.
MOs that are lower in energy than the starting AOs are bonding (\(\sigma\)) MOs, and those higher in energy are antibonding (\(\sigma^*\)) MOs.
The electrons are filled in MOs beginning with the lowest energy.
Only two electrons occupy each molecular orbital, and they have opposite spins — that is, their spins are paired.
The bond order of the molecule can be calculated from the number of bonding and antibonding electrons.
Points v and vi should look familiar: they are the Aufbau principle and Pauli’s exclusion principle, which you applied to atoms, now applied to molecules. Hund’s rule carries over too, which is why the two degenerate \(\pi^*\) orbitals of oxygen take one electron each before either takes a second.
5.5.6 MO Description of Simple Diatomic Molecules
1. Hydrogen Molecule, H₂
The hydrogen molecule is formed by the linear combination of two hydrogen atoms, each having one electron in its 1s orbital. The linear combination of two 1s orbitals gives two molecular orbitals, \(\sigma 1s\) and \(\sigma^* 1s\). The two electrons from the hydrogen atoms occupy the \(\sigma 1s\) molecular orbital, and \(\sigma^* 1s\) remains vacant, as shown in Fig. 5.40.
The electronic configuration of the hydrogen molecule is written as \(\sigma 1s^2\).
Bond order \(=\) (bonding electrons \(-\) antibonding electrons) \(\div\ 2\)
For hydrogen, bond order \(= (2 - 0) \div 2 = 1\)
So in the \(\mathrm{H_2}\) molecule there exists one covalent bond between the two hydrogen atoms. The bond length is 74 pm and the bond dissociation energy is \(438\ \text{kJ mol}^{-1}\). As there are no unpaired electrons present, the \(\mathrm{H_2}\) molecule is diamagnetic.
Diagram
Fig. 5.40 – MO diagram for H₂: energy axis on the left; a 1s level for each H atom at the two sides; dashed lines meeting at a lower central level labelled σ holding two arrows of opposite spin, and an upper central level labelled σ* left empty; label the sides H and H and the centre H₂
2. Lithium Molecule, Li₂
Each lithium atom has 3 electrons with the electronic configuration \(1s^2, 2s^1\), so the \(\mathrm{Li_2}\) molecule will have 6 electrons. The linear combination of atomic orbitals gives four molecular orbitals: \(\sigma 1s\), \(\sigma^* 1s\), \(\sigma 2s\), \(\sigma^* 2s\).
The electronic configuration of the \(\mathrm{Li_2}\) molecule will be
\[(\sigma 1s)^2\ (\sigma^* 1s)^2\ (\sigma 2s)^2\]
Bond order \(= (4 - 2) \div 2 = 1\)
This means that in the \(\mathrm{Li_2}\) molecule there is one bond between the two lithium atoms. Such \(\mathrm{Li_2}\) molecules are found in the vapour state. As there are no unpaired electrons, the molecule is diamagnetic. The energy level diagram is given in Fig. 5.41.
Notice how the filled \(\sigma 1s\) and \(\sigma^* 1s\) pair cancel each other out: two electrons stabilising and two destabilising leave no net contribution. Only the 2s electrons actually build the bond.
Diagram
Fig. 5.41 – MO diagram for Li₂: energy increasing upwards; 1s and 2s levels shown for each Li atom on the two sides; centre showing σ1s (filled), σ1s (filled), σ2s (filled) and σ2s (empty); label the sides Li and Li and the centre Li₂
3. N₂ Molecule
N : \(1s^2\ 2s^2\ 2p^3\)
The electronic configuration of the \(\mathrm{N_2}\) molecule (14 electrons) is
\[(\sigma 1s)^2\ (\sigma^* 1s)^2\ (\sigma 2s)^2\ (\sigma^* 2s)^2\ (\pi_x)^2\ (\pi_y)^2\ (\sigma 2p_z)^2\]
Bond order of the \(\mathrm{N_2}\) molecule \(= \dfrac{10 - 4}{2} = 3\)
The \(\mathrm{N_2}\) molecule is diamagnetic. Its MO diagram is given in Fig. 5.42. Note that nitrogen follows the first of the two energy orders, so the \(\pi\) orbitals are filled before \(\sigma 2p_z\).
Diagram
Fig. 5.42 – MO diagram for N₂: energy increasing upwards; 1s, 2s and 2p levels for each N atom on the two sides; centre showing in order σ1s, σ1s, σ2s, σ2s, the degenerate pair π2pₓ and π2p_y (both filled), σ2p_z (filled), then the empty π2pₓ, π2p_y and σ*2p_z; all occupied orbitals drawn with paired arrows
4. O₂ Molecule
\(_8\mathrm{O}\) : \(1s^2\ 2s^2\ 2p^4\)
The electronic configuration of the \(\mathrm{O_2}\) molecule (16 electrons) is
\[(\sigma 1s)^2\ (\sigma^* 1s)^2\ (\sigma 2s)^2\ (\sigma^* 2s)^2\ (\sigma 2p_z)^2\ (\pi 2p_x)^2\ (\pi 2p_y)^2\ (\pi^* 2p_x)^1\ (\pi^* 2p_y)^1\]
Bond order of the \(\mathrm{O_2}\) molecule \(= \dfrac{10 - 6}{2} = 2\)
The \(\mathrm{O_2}\) molecule is paramagnetic due to the presence of 2 unpaired electrons in the \(\pi^*\) orbitals. The MO diagram is given in Fig. 5.43.
The last two electrons are the whole point. There are two \(\pi^*\) orbitals of exactly equal energy and only two electrons left to place, so by Hund’s rule they go one into each, with parallel spins, rather than pairing up in one orbital. MO theory therefore predicts two unpaired electrons — and that is exactly what experiment finds.
Diagram
Fig. 5.43 – MO diagram for O₂: energy increasing upwards; 2s and 2p levels for each O atom on the two sides; centre showing in order σ1s, σ1s, σ2s, σ2s, σ2p_z, the degenerate filled pair π2pₓ and π2p_y, then the degenerate pair π2pₓ and π2p_y each carrying a single arrow with parallel spins, and the empty σ*2p_z at the top; highlight the two unpaired electrons
5. F₂ Molecule
\(_9\mathrm{F}\) : \(1s^2\ 2s^2\ 2p^5\)
The electronic configuration of the \(\mathrm{F_2}\) molecule (18 electrons) is
\[(\sigma 1s)^2\ (\sigma^* 1s)^2\ (\sigma 2s)^2\ (\sigma^* 2s)^2\ (\sigma 2p_z)^2\ (\pi 2p_x)^2\ (\pi 2p_y)^2\ (\pi^* 2p_x)^2\ (\pi^* 2p_y)^2\]
Bond order of the \(\mathrm{F_2}\) molecule \(= \dfrac{10 - 8}{2} = 1\)
The \(\mathrm{F_2}\) molecule is diamagnetic. Its MO diagram is given in Fig. 5.44.
Diagram
Fig. 5.44 – MO diagram for F₂: energy increasing upwards; 2s and 2p levels for each F atom on the two sides; centre showing σ1s, σ1s, σ2s, σ2s, σ2p_z, the filled degenerate pair π2pₓ and π2p_y, the filled degenerate pair π2pₓ and π2p_y, and the empty σ*2p_z; caption “Bond Order = 1”; label the sides F and F and the centre F–F
Read down the last two columns and the pattern is clear. From \(\mathrm{N_2}\) onwards, every extra pair of electrons goes into antibonding orbitals, so the bond order falls steadily from 3 to 2 to 1 — and the bonds in these molecules get correspondingly longer and weaker.
5.6 Parameters of Covalent Bond
A covalent bond is characterised by different parameters. These parameters help us to understand how strong the bond is between the two atoms, what the distance between them is, and what the shape of the molecule is.
Four such parameters are taken up here — bond angle, bond enthalpy, bond length and bond order — followed by a fifth property, the polarity of the bond. Together they turn the qualitative pictures of the previous sections into numbers that can be measured in a laboratory and compared across molecules.
5.6.1 Bond Angle
The electrons which participate in bond formation are present in orbitals. The angle between the orbitals holding the bonding electrons is called the bond angle. It is marked for the water molecule in Fig. 5.45.
Diagram
Fig. 5.45 – Bond angle in the water molecule: bent H–O–H with the two lone pairs shown as dots on oxygen and a curved arrow between the two O–H bonds labelled “bond angle”
Bond angle can be determined experimentally using spectroscopic techniques. The value of the bond angle gives an idea about the arrangement of orbitals around the central atom and the shape of the molecule. Some measured values are collected in Table 5.5.
Table 5.5: Bond angles of some molecules
| Molecule | Bond angle | |
|---|---|---|
| 1 | H₂O | 104°28′ |
| 2 | NH₃ | 107 |
| 3 | BF₃ | 120 |
Values quoted for the same molecule may differ very slightly from one source to another, because they come from different experimental techniques and different conditions of measurement. What matters for reasoning is the pattern: water is squeezed further below the tetrahedral angle than ammonia is, exactly as VSEPR predicted from the number of lone pairs.
5.6.2 Bond Enthalpy
For diatomic molecules the dissociation energy is the same as the bond enthalpy. The bond enthalpy for the \(\mathrm{H_2}\) molecule is \(435.8\ \text{kJ mol}^{-1}\).
The bond enthalpy is a measure of the strength of the bond between two atoms, and can be measured experimentally. The N–N bond in \(\mathrm{N_2}\) is stronger than the O–O bond in \(\mathrm{O_2}\). The larger the bond dissociation energy, the stronger is the bond in the molecule. For the heteronuclear diatomic molecule HCl, the bond enthalpy was found to be \(431.0\ \text{kJ mol}^{-1}\).
Bond Enthalpy in Polyatomic Molecules
In the case of polyatomic molecules, the bond enthalpy and the bond dissociation energy are not identical. Bond enthalpy is the average of the sum of successive bond dissociation energies. For example, consider the dissociation of water, which happens in two separate steps:
\[\mathrm{H_2O}\text{(g)} \rightarrow \mathrm{H}\text{(g)} + \mathrm{\dot{O}H}\text{(g)} \qquad \Delta H_1 = 502\ \text{kJ mol}^{-1}\]
\[\mathrm{\dot{O}H}\text{(g)} \rightarrow \mathrm{H}\text{(g)} + \mathrm{O}\text{(g)} \qquad \Delta H_2 = 427\ \text{kJ mol}^{-1}\]
The average bond enthalpy of the O–H bond in \(\mathrm{H_2O}\) is therefore
\[\Delta_a H = \frac{502 + 427}{2} = 464.5\ \text{kJ mol}^{-1}\]
The two-step route and the averaging are set out together in FC 5.8.
Flowchart
FC 5.8 – Successive dissociation of water and the average O–H bond enthalpy: H₂O(g) → first O–H bond breaks, absorbing ΔH₁ = 502 kJ mol⁻¹ → H(g) + ȮH(g) → second O–H bond breaks, absorbing ΔH₂ = 427 kJ mol⁻¹ → H(g) + O(g); branch from the two values → take the arithmetic mean (502 + 427)/2 → average O–H bond enthalpy ΔaH = 464.5 kJ mol⁻¹
In both the above equations the bond between O and H is broken, but the amount of energy required to break the bond is different — that is, the enthalpies of the two O–H bonds in water are different.
This difference arises due to the fact that the cleavage of the two O–H bonds in water takes place in two steps. In the first step one O–H bond breaks, leaving behind the OH radical. Now the electronic environment around the oxygen to which hydrogen is attached is different from that around the oxygen in the \(\mathrm{H_2O}\) molecule, and this causes a change in the successive bond dissociation energy.
The same difference is observed in the enthalpy values of the O–H bond in \(\mathrm{C_2H_5OH}\). Oxygen here is attached to the \(\mathrm{C_2H_5}\) group, therefore the hydrogen of O–H is in a different environment from the hydrogen of H–O–H.
In the same way, the bond enthalpy value of any covalent bond is slightly different for each bond of that kind in a given molecule, and also in different molecules. The average values of bond enthalpy, \(\Delta_a H\), are determined from the experimentally measured values of a large number of compounds containing a particular bond. Average bond enthalpy data are given in Table 5.6. In general, a stronger bond implies a larger bond enthalpy.
Table 5.6: Bond enthalpies
| Bond | \(\Delta_a H\) / kJ mol⁻¹ |
|---|---|
| C – H | 400 – 415 |
| N – H | 390 |
| O – H | 460 – 464 |
| C – C | 345 |
| C – N | 290 – 315 |
| C – O | 355 – 380 |
| C – Cl | 330 |
| C – Br | 275 |
| O – O | 175 – 184 |
| C = C | 610 – 630 |
| C ≡ C | 835 |
| C = O | 724 – 757 |
| C ≡ N | 854 |
5.6.3 Bond Length
Bond lengths are measured by X-ray and electron diffraction techniques.
Each atom of the bonded pair contributes to the bond length. Bond length depends upon the size of the atoms and the multiplicity of the bonds.
It increases with an increase in the size of the atom, and decreases with an increase in the multiplicity of the bond. It is generally measured in picometre (pm) or in the Angstrom unit (Å). Since the order of bond lengths is
\[\text{single} > \text{double} > \text{triple}\]
the C–C single bond is longer and the C≡C triple bond is shorter.
A picometre is \(10^{-12}\) metre, and an Angstrom is \(10^{-10}\) metre, so \(1\ \text{Å} = 100\ \text{pm}\). Being able to move between the two units is useful, since older data tables are often written in Å.
Some typical average bond lengths of C–C single, double and triple bonds and others are shown in Table 5.7.
Table 5.7: Average bond lengths for some single, double and triple bonds
| Type of bond | Covalent bond length (pm) | Type of bond | Covalent bond length (pm) |
|---|---|---|---|
| O – H | 96 | H₂ (H–H) | 74 |
| C – H | 107 | F₂ (F–F) | 144 |
| N – O | 136 | Cl₂ (Cl–Cl) | 199 |
| C – O | 143 | Br₂ (Br–Br) | 228 |
| C – N | 143 | I₂ (I–I) | 267 |
| C – C | 154 | N₂ (N≡N) | 109 |
| C = O | 121 | O₂ (O=O) | 121 |
| N = O | 122 | HF (H–F) | 92 |
| C = C | 133 | HCl (H–Cl) | 127 |
| C = N | 138 | HBr (H–Br) | 141 |
| C ≡ N | 116 | HI (H–I) | 160 |
| C ≡ C | 120 |
Diagram
Fig. 5.46 – Bar chart of the hydrogen halide bond lengths from Table 5.7: four bars for HF (92 pm), HCl (127 pm), HBr (141 pm) and HI (160 pm), y-axis “bond length / pm”, each bar labelled with its value and with the halogen atom drawn above it as a circle of increasing size
5.6.4 Bond Order
According to the Lewis theory, the bond order is equal to the number of bonds between the two atoms in a molecule. The bond order in \(\mathrm{H_2}\), \(\mathrm{O_2}\) and \(\mathrm{N_2}\) is 1, 2 and 3 respectively.
This is the same quantity you calculated from molecular orbitals in the previous section, arrived at by a much simpler route. For these three molecules the two methods agree exactly, which is a useful check on both.
Isoelectronic molecules and ions have identical bond order. For example, \(\mathrm{N_2}\), CO and \(\mathrm{NO^+}\) each have bond order 3. \(\mathrm{F_2}\), whereas \(\mathrm{O_2^{2-}}\), has bond order 1.
Isoelectronic species are those which have the same total number of electrons. Since bond order in MO theory is decided entirely by how those electrons are distributed among bonding and antibonding orbitals, species with the same electron count naturally end up with the same bond order.
Stabilities of molecules can be determined by knowing the bond order. As bond order increases, bond enthalpy increases and bond length decreases.
5.6.5 Polarity of a Covalent Bond
Covalent bonds are formed between two atoms of the same or different elements. Thus a covalent bond is formed between atoms of the same element in H–H, F–F, Cl–Cl and so on. The shared pair of electrons is attracted equally by both atoms and is situated midway between the two atoms. Such a covalent bond is termed a non-polar covalent bond.
H : H \(\rightarrow\) H–H, electron pair at the centre, non-polar covalent bond
When a covalent bond is formed between two atoms of different elements which have different electronegativities, the shared electron pair does not remain at the centre. The electron pair is pulled towards the more electronegative atom, resulting in the separation of charges. This gives rise to a dipole. The more electronegative atom acquires a partial negative charge and the other atom gets a partial positive charge. Such a bond is called a polar covalent bond. Examples of polar molecules include HF, HCl and so on.
H : F \(\rightarrow\) \(^{\delta+}\mathrm{H}-\mathrm{F}^{\delta-}\) H–F, polar covalent bond
The symbol \(\delta\) (delta) is used because the charge separation is only partial. Unlike the full transfer of an electron that creates \(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\), here the electron pair is merely pulled off-centre, so each atom carries a fraction of a unit charge. The contrast is drawn in Fig. 5.47.
Diagram
Fig. 5.47 – Non-polar and polar covalent bonds compared: on the left, H–H with the shared electron pair shown exactly at the midpoint, labelled “non-polar”; on the right, H–F with the shared pair drawn much closer to F, δ+ marked above H and δ− above F, and a crossed arrow pointing from H towards F, labelled “polar”
Fluorine is more electronegative than hydrogen, therefore the shared electron pair is more towards fluorine, and the atoms acquire partial positive and partial negative charges respectively.
Polarity of the covalent bond increases as the difference in the electronegativity between the bonded atoms increases. When the difference in electronegativities of the combining atoms is about 1.7, the ionic percentage in the covalent bond is 50%.
That last figure is worth pausing on, because it shows that ionic and covalent bonding are not two separate boxes but two ends of one continuous scale. At an electronegativity difference of about 1.7 the bond is half ionic and half covalent; below it the covalent character dominates, above it the ionic character does.
5.7 Dipole Moment
In the previous section we saw that when two atoms of different electronegativity are joined, the shared pair shifts and partial charges appear. Dipole moment is the quantity that measures how much it has shifted, so that different bonds and different molecules can be compared with a number rather than a description.
\[\mu = Q \times r\]
where \(Q\) is the charge and \(r\) is the distance of separation.
The unit of dipole moment is the debye (D).
\[1\ \text{D} = 3.33564\times10^{-30}\ \text{C m}\]
where C stands for coulomb and m for metre.
Dipole moment, being a vector quantity, is represented by a small arrow with its tail on the positive centre and its head pointing towards the negative centre.
\[^{\delta+}\mathrm{H} \longrightarrow \mathrm{F}^{\delta-} \qquad (\mu = 1.91\ \text{D})\]
The crossed arrow drawn above the Lewis structure indicates the direction of the shift of electron density towards the more electronegative atom, as shown in Fig. 5.48.
Diagram
Fig. 5.48 – Representation of a bond dipole: the H–F molecule with δ+ over H and δ− over F, and above it a crossed arrow (a line with a small vertical cross-bar at the tail end and an arrowhead at the other end) pointing from H towards F; label the tail “positive centre” and the head “negative centre”
Dipole Moments of Polyatomic Molecules
Each polar bond in a polyatomic molecule has its own dipole. The resultant dipole of the molecule is decided by
- the shape of the molecule, that is, the spatial arrangement of the bonds
- the contribution of the individual dipoles and those of the lone pairs of electrons, if any
The dipole moment of a polyatomic molecule is the vector sum of the dipole moments of the various bonds and lone pairs.
Because it is a vector sum and not an ordinary sum, direction can cancel size. Two equal dipoles pointing in exactly opposite directions add up to nothing at all, no matter how polar each individual bond may be. This single idea explains most of the surprises in Table 5.8.
Consider \(\mathrm{BeF_2}\) and \(\mathrm{BF_3}\), drawn in Fig. 5.49.
BeF₂ is a linear molecule, and the dipoles are in opposite directions and are of equal strength, so the net dipole moment of \(\mathrm{BeF_2}\) is 0.
\[\mathrm{F} \longleftarrow \mathrm{Be} \longrightarrow \mathrm{F} \qquad \text{net dipole in } \mathrm{BeF_2} = 0\]
BF₃ has its three bond dipoles set at angles to one another. In \(\mathrm{BF_3}\) the bond angle is 120° and the molecule is symmetrical. The three B–F bonds are oriented at 120° with each other, and the sum of any two is equal and opposite to the third; therefore the sum of the three B–F dipoles \(= 0\).
Diagram
Fig. 5.49 – Cancellation of bond dipoles. Left: linear F–Be–F with an arrow on each bond pointing outwards from Be towards F, the two arrows equal and opposite, labelled “net dipole in BeF₂ = 0”. Right: trigonal planar BF₃ with three equal arrows pointing from B towards each F at 120° to one another, labelled “net dipole in BF₃ = 0”
In the case of angular molecules, both the lone pairs and the electronegativity difference contribute to the dipole moment.
Lone Pairs and Dipole Moment
In some molecules the central atom has unshared or lone pairs of electrons. These lone pairs also contribute to the overall dipole of the molecule.
Nitrogen in \(\mathrm{NH_3}\) and oxygen in \(\mathrm{H_2O}\) possess lone pairs. These reinforce the dipoles due to the N–H and O–H bonds, and both these molecules are highly polar, as Fig. 5.50 shows.
\[\mathrm{NH_3}: \mu = 1.48\ \text{D} \qquad \mathrm{H_2O}: \mu = 1.85\ \text{D}\]
Nitrogen has only one lone pair, while oxygen has two lone pairs, which is reflected in the higher dipole moment of water.
Diagram
Fig. 5.50 – Lone pair contribution to dipole moment. Left: pyramidal NH₃ with δ− on N, δ+ on each H, one lone pair drawn as a lobe on N with an arrow pointing along it, and a large resultant arrow pointing upward; labelled µ = 1.48 D. Right: bent H₂O with δ− on O, δ+ on each H, two lone pair lobes on O with arrows along them, and a large resultant arrow pointing upward; labelled µ = 1.85 D
Comparing NH₃ and NF₃
Consider \(\mathrm{NH_3}\) and \(\mathrm{NF_3}\).
Both have a pyramidal shape with a lone pair of electrons on the nitrogen atom. Here hydrogen is less electronegative than nitrogen, while fluorine is more electronegative than nitrogen.
The resultant dipole moment of \(\mathrm{NH_3}\) is \(4.90\times10^{-30}\ \text{C m}\), while that of \(\mathrm{NF_3}\) is \(0.80\times10^{-30}\ \text{C m}\).
This difference is because in the case of \(\mathrm{NH_3}\) the orbital dipole due to the lone pair is in the same direction as the resultant dipole moment of the N–H bonds, and hence gets added. In \(\mathrm{NF_3}\), on the other hand, the orbital moment is in the direction opposite to the resultant dipole moment of the three N–F bonds. The orbital dipole due to the lone pair decreases the effect of the resultant N–F bond moments, which results in its low dipole moment. The two cases are drawn side by side in Fig. 5.51.
The direction of each bond dipole is what makes the difference. In \(\mathrm{NH_3}\) the nitrogen is the more electronegative partner, so every N–H dipole points towards nitrogen — the same way the lone pair points. In \(\mathrm{NF_3}\) fluorine is the more electronegative partner, so every N–F dipole points away from nitrogen — against the lone pair.
Diagram
Fig. 5.51 – NH₃ and NF₃ compared. Left: pyramidal NH₃ with δ− on N and δ+ on each H, three bond dipole arrows pointing up towards N, the lone pair arrow also pointing up, and a large “resultant dipole moment” arrow at the top; labelled 4.90 × 10⁻³⁰ C m. Right: pyramidal NF₃ with δ+ on N and δ− on each F, three bond dipole arrows pointing down towards the F atoms, the lone pair arrow pointing up in opposition, and a short “resultant dipole moment” arrow; labelled 0.80 × 10⁻³⁰ C m
Tetrahedral Molecules
\(\mathrm{CH_4}\): the central atom carbon has no lone pair, and the molecule is non-polar. Its four C–H bonds are identical and point to the corners of a regular tetrahedron, so their dipoles cancel completely.
In \(\mathrm{CHCl_3}\) the dipoles are not equal and do not cancel, hence \(\mathrm{CHCl_3}\) is polar with a non-zero dipole moment, \(\mu = 1.04\). Replacing three of the hydrogen atoms of methane by the much more electronegative chlorine atoms destroys the symmetry, and the vector sum no longer comes to zero. Both molecules are drawn in Fig. 5.52.
Diagram
Fig. 5.52 – Tetrahedral molecules compared. Left: CH₄ drawn as a regular tetrahedron with four identical C–H bond dipole arrows that cancel, labelled “non-polar, µ = 0”. Right: CHCl₃ with one C–H bond at the top and three C–Cl bonds below, arrows on the C–Cl bonds pointing towards the chlorine atoms and a small arrow on the C–H bond, with a net downward resultant arrow, labelled µ = 1.04
The reasoning used for both these molecules works for any molecule at all, and it is set out step by step in FC 5.9.
Flowchart
FC 5.9 – Deciding whether a molecule is polar: examine each bond in the molecule → branch: all bonds non-polar (atoms of equal electronegativity) → molecule is non-polar, µ = 0; at least one polar bond present → find the shape of the molecule by VSEPR → draw a dipole arrow on each polar bond, plus one for each lone pair on the central atom → add the arrows as vectors → branch: the arrows cancel exactly (symmetrical molecule, no lone pairs on the central atom, all outer atoms identical) → non-polar, µ = 0; the arrows do not cancel → polar, µ ≠ 0
Dipole moments of some molecules are shown in Table 5.8.
Table 5.8: Dipole moments and geometry of some molecules
| Type of molecule | Example | Dipole moment µ (D) | Geometry |
|---|---|---|---|
| Molecule AB | HF | 1.91 | linear |
| HCl | 1.03 | linear | |
| HBr | 0.79 | linear | |
| H₂ | 0 | linear | |
| Molecule AB₂ | H₂O | 1.85 | bent |
| H₂S | 0.95 | bent | |
| CO₂ | 0 | linear | |
| Molecule AB₃ | NH₃ | 1.47 | trigonal pyramidal |
| NF₃ | 0.23 | trigonal pyramidal | |
| BF₃ | 0 | trigonal planar | |
| Molecule AB₄ | CH₄ | 0 | tetrahedral |
| CHCl₃ | 1.04 | tetrahedral | |
| CCl₄ | 0 | tetrahedral |
5.7.1 Covalent Character of Ionic Bond
Several ionic compounds possess partial covalent character and show properties similar to covalent compounds. For example, LiCl is ionic, but it is more soluble in organic solvents than in water.
The reason is that no cation holds its electrons and no anion keeps its own electron cloud entirely to itself. A small, highly charged cation pulls at the electron cloud of the neighbouring anion and distorts it — a process called polarisation. The more the anion is polarised, the more its electron density is drawn into the space between the two ions, and that shared electron density is precisely what a covalent bond is. The effect is drawn in Fig. 5.53.
Diagram
Fig. 5.53 – Polarisation of an anion by a cation: (a) an ideal ion pair — a small positive sphere beside a large, perfectly spherical negative ion, labelled “purely ionic”; (b) the same pair with the anion’s electron cloud pulled out of shape towards the cation and electron density building up in the region between the two nuclei, labelled “polarised — partial covalent character”
To explain the partial covalent character in ionic bonds, Fajans put forth the following rules.
1. The smaller the size of the cation and the larger the size of the anion, the greater is the covalent character rendered to the ionic bond. For example, \(\mathrm{Li^+Cl^-}\) is more covalent than \(\mathrm{Na^+Cl^-}\). Similarly, \(\mathrm{Li^+I^-}\) is more covalent than \(\mathrm{Li^+Cl^-}\).
A small cation concentrates its positive charge into a tiny volume and can therefore pull hard on a neighbouring electron cloud. A large anion holds its outermost electrons loosely and is easily distorted. Both conditions favour polarisation.
2. The greater the charge on the cation, the more is the covalent character of the ionic bond. For example, the covalent character of \(\mathrm{AlCl_3}\), \(\mathrm{MgCl_2}\) and \(\mathrm{NaCl}\) decreases in the following order:
\[\mathrm{Al^{3+}(Cl^-)_3} > \mathrm{Mg^{2+}(Cl^-)_2} > \mathrm{Na^+Cl^-}\]
3. A cation with the outer electronic configuration of the \(s^2p^6d^{10}\) type possesses a greater polarising power compared to a cation having the same size and same charge but having an outer electronic configuration of the \(s^2p^6\) type.
This is because the d electrons of the \(s^2p^6d^{10}\) shell screen the nuclear charge less effectively compared to the s and p electrons of the \(s^2p^6\) shell. Hence the effective nuclear charge in a cation having the \(s^2p^6d^{10}\) configuration is greater than that of the one having the \(s^2p^6\) configuration.
For example, \(\mathrm{Cu^+Cl^-}\) is more covalent than \(\mathrm{Na^+Cl^-}\). Here
\(\mathrm{Cu^+}\) : \(1s^2\ 2s^2 2p^6\ 3s^2 3p^6 3d^{10}\)
\(\mathrm{Na^+}\) : \(1s^2\ 2s^2 2p^6\)
Screening, or shielding, is the way inner electrons stand between the nucleus and the outer ones and reduce the pull the outer electrons feel. Because d orbitals are diffuse and poorly shaped for this job, the ten d electrons of \(\mathrm{Cu^+}\) leave much more of the nuclear charge exposed than a full \(s^2p^6\) shell would, so \(\mathrm{Cu^+}\) polarises a chloride ion far more strongly than \(\mathrm{Na^+}\) of similar size does.
5.8 Resonance
Many polyatomic molecules can be represented by more than one Lewis structure. Consider, for example, the three structures written for \(\mathrm{CO_3^{2-}}\), which you have already met in Fig. 5.10. Each structure differs from the other only in the position of the electrons, without changing the positions of the atoms.
That last point is the heart of the matter. The atoms stay exactly where they are; only the double bond moves from one oxygen to another. Nothing about the shape of the ion changes from one drawing to the next.
None of these individual structures is adequate to explain the properties. The actual structure of \(\mathrm{CO_3^{2-}}\) is a combination of three Lewis structures and is called the resonance hybrid.
Resonance signifies that there is more than one possible way in which the electrons can be assigned in a Lewis structure. The various structures are called canonical forms.
Resonance in the Ozone Molecule
Now consider the example of \(\mathrm{O_3}\). It is the resonance hybrid of the structures shown in Fig. 5.54.
Structures I and II are the canonical forms, while structure III is the resonance hybrid. In canonical form I, one oxygen–oxygen bond is a double bond of length 121 pm and the other is a single bond of length 148 pm; in form II the two are interchanged. In the resonance hybrid III, however, both bonds are found to be of the same length, 128 pm — a value lying between that of a single bond and that of a double bond.
Diagram
Fig. 5.54 – Resonance in the O₃ molecule: structure I with a double bond of 121 pm on the left and a single bond of 148 pm on the right, all lone pairs and the ⊕ and ⊖ formal charges marked; a double-headed arrow (↔︎) leading to structure II, the mirror image with 148 pm on the left and 121 pm on the right; below them structure III labelled “Resonance Hybrid”, drawn with two identical partial bonds of 128 pm each, shown as a solid line plus a dotted line
Note the arrow used between the canonical forms. It is a double-headed arrow, \(\leftrightarrow\), and it must never be confused with the pair of half-arrows \(\rightleftharpoons\) used for a reversible reaction. The double-headed arrow does not mean the two structures interconvert; it means neither one alone is the truth, and the real molecule is a blend of both.
Resonance Energy
We have seen that many polyatomic ions and molecules can be represented by different canonical forms, and each form has a different energy. The energy of the resonating forms is different from that of the most stable structure, the resonance hybrid. The difference in energy between the stable contributing structure and the resonating forms is usually defined as the resonance energy.
To summarize, it can be stated that:
a. The energy of the resonance hybrid structure is less than the energy of any single canonical form. Hence resonance stabilizes certain polyatomic molecules or ions.
b. The average of all the resonating structures contributes to the overall bonding characteristic features of the molecule.
This will be clear from the example of ozone. Ozone can be represented by two canonical forms, I and II, shown earlier. III is the resonance hybrid. The energy of III is less than that of I and II, and this energy difference is drawn in Fig. 5.55.
Diagram
Fig. 5.55 – Resonance energy: a vertical energy axis with a higher horizontal level labelled “canonical forms I and II” and a lower horizontal level labelled “resonance hybrid III”; a double-headed vertical arrow between the two levels labelled “resonance energy”
The greater the resonance energy, the more stable the species. This is why ions such as carbonate and nitrate, and molecules such as ozone, are far less reactive than a single Lewis structure with its exposed double bond would lead you to expect — the electrons are spread out rather than concentrated, and spreading electrons out always lowers energy.
5.9 Summary
A chemical bond is the force of attraction that holds atoms together in a compound. Bonds are formed in two ways — by the complete loss and gain of electrons, and by the sharing of an electron pair. Only the valence electrons, those in the outermost shell, take part.
Kossel and Lewis proposed the electronic theory of valence in 1916. Lewis pictured an atom as a positively charged kernel surrounded by an outer shell holding up to eight electrons.
Octet rule: while forming a bond, an atom loses, gains or shares electrons so that its valence shell contains eight electrons, giving it the configuration of the nearest inert gas. H and Li are exceptions; they aim for a duplet, like helium (\(1s^2\)).
An ionic bond is the electrostatic attraction between the oppositely charged ions produced by complete electron transfer. Sodium (2, 8, 1) loses one electron and chlorine (2, 8, 7) gains one to give \(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\); calcium loses two, so two chloride ions are needed and the formula is \(\mathrm{CaCl_2}\).
Ion formation depends on ionisation enthalpy (\(\Delta_i H\)), which is always positive, and electron gain enthalpy (\(\Delta_{eg} H\)), which may be negative or positive. Elements with low ionisation enthalpy combine readily with elements having a high negative electron gain enthalpy.
Lattice enthalpy is the energy required to separate one mole of a solid ionic compound completely into its gaseous ions. For NaCl the value is 788 kJ mol⁻¹. It is the large energy released on lattice formation, not merely the completion of octets, that makes an ionic solid stable.
Lattice enthalpy rises as the charge on the cation rises (\(\mathrm{AlCl_3} > \mathrm{CaCl_2} > \mathrm{NaCl}\)) and as the size of the cation falls (\(\mathrm{LiF} > \mathrm{NaF} > \mathrm{KF}\)).
A covalent bond is the attractive force arising from the mutual sharing of electrons between atoms of similar or slightly different electronegativity. Sharing one pair gives a single bond, two pairs a double bond and three pairs a triple bond.
Lewis dot structures are drawn by counting all the valence electrons, adjusting for charge, choosing the least electronegative atom as the central atom, drawing single bonds, completing the octets of the outer atoms, and forming multiple bonds if the central atom’s octet is still incomplete.
Formal charge helps choose between competing Lewis structures:
\[\text{FC} = \text{VE} - \text{NE} - \frac{\text{BE}}{2}\]
The structure in which the formal charges are closest to zero has the lowest energy — which is why \(\mathrm{O}=\mathrm{C}=\mathrm{O}\) is the preferred structure for carbon dioxide.
The octet rule has limits. Some molecules have an incomplete octet (\(\mathrm{LiCl}\), \(\mathrm{BeCl_2}\), \(\mathrm{BF_3}\)), some an expanded octet (\(\mathrm{PCl_5}\), \(\mathrm{SF_6}\), \(\mathrm{H_2SO_4}\)), and some an odd number of electrons (\(\mathrm{NO}\), \(\mathrm{NO_2}\)). The rule also fails to explain molecular shapes, bond energies and differences in reactivity.
VSEPR theory says the electron pairs around a central atom arrange themselves for minimum repulsion, giving minimum energy and maximum stability. Lone pairs repel more strongly than bonding pairs, in the order L.P–L.P > L.P–B.P > B.P–B.P.
With no lone pairs the shape matches the electron pair geometry: 2 pairs linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral. Lone pairs squeeze the bond angle — hence \(\mathrm{CH_4}\) 109°28′, \(\mathrm{NH_3}\) 107°18′ and pyramidal, \(\mathrm{H_2O}\) 104°35′ and angular.
Valence bond theory, developed by Heitler and London and extended by Pauling and Slater, describes a bond as the overlap of two half-filled orbitals holding electrons of opposite spin. The greater the overlap, the stronger the bond. The potential energy curve for \(\mathrm{H_2}\) shows the bond forming at the distance of minimum energy, where attraction and repulsion balance.
End-to-end overlap along the internuclear axis gives a sigma (\(\sigma\)) bond, from s–s, s–p or p–p orbitals. Sideways overlap of parallel p orbitals gives the weaker pi (\(\pi\)) bond, with electron density above and below the axis.
Hybridization is the mixing of valence orbitals of the same atom and their recasting into an equal number of new equivalent hybrid orbitals. It proceeds by promotion to the excited state followed by mixing and recasting, and requires orbitals of the same atom with nearly equal energies.
\(sp^3\) hybridization gives four orbitals at 109°28′, tetrahedral, as in \(\mathrm{CH_4}\), \(\mathrm{NH_3}\) and \(\mathrm{H_2O}\). \(sp^2\) gives three orbitals at 120° in a plane with one p orbital left over, as in \(\mathrm{BF_3}\) and \(\mathrm{C_2H_4}\). \(sp\) gives two orbitals at 180° with two p orbitals left over, as in \(\mathrm{BeCl_2}\) and \(\mathrm{C_2H_2}\).
Valence bond theory introduced delocalization, shielding, covalent character, partial ionic character and resonance. It fails, however, to explain co-ordinate bonds, the paramagnetism of \(\mathrm{O_2}\), and bonding in electron-deficient molecules such as \(\mathrm{B_2H_6}\).
Molecular orbital theory treats the molecule as a whole. Atomic orbitals combine linearly (LCAO): addition of wave functions gives a bonding molecular orbital of lower energy, subtraction gives an antibonding molecular orbital of higher energy with a node between the nuclei. The combining orbitals must have comparable energies, the same symmetry about the molecular axis, and maximum overlap.
Bond order follows from the electron count:
\[\text{Bond order} = \frac{N_b - N_a}{2}\]
All electrons paired means the molecule is diamagnetic; any unpaired electron makes it paramagnetic.
For second row diatomics the energy order places \(\pi 2p\) below \(\sigma 2p_z\), except in \(\mathrm{O_2}\) and \(\mathrm{F_2}\), where the order is reversed. This treatment gives bond orders of 1 for \(\mathrm{H_2}\) and \(\mathrm{Li_2}\), 3 for \(\mathrm{N_2}\), 2 for \(\mathrm{O_2}\) and 1 for \(\mathrm{F_2}\), and correctly predicts that \(\mathrm{O_2}\) is paramagnetic with two unpaired electrons in its \(\pi^*\) orbitals.
A covalent bond is described by measurable parameters. Bond angle is the angle between the bonding orbitals. Bond enthalpy is the energy needed to break one mole of a bond in the gaseous state; in polyatomic molecules the average value is used, so that for the O–H bond in water \(\Delta_a H = (502 + 427)/2 = 464.5\ \text{kJ mol}^{-1}\). Bond length is the equilibrium internuclear distance, measured by X-ray and electron diffraction.
Bond length increases with atomic size and decreases with bond multiplicity. As bond order increases, bond enthalpy increases and bond length decreases. Isoelectronic species have identical bond order — \(\mathrm{N_2}\), \(\mathrm{CO}\) and \(\mathrm{NO^+}\) all have bond order 3.
A bond between identical atoms is non-polar; between atoms of different electronegativity the shared pair shifts, partial charges \(\delta+\) and \(\delta-\) appear, and the bond is polar. At an electronegativity difference of about 1.7 the bond is 50% ionic.
Dipole moment measures this shift:
\[\mu = Q \times r, \qquad 1\ \text{D} = 3.33564\times10^{-30}\ \text{C m}\]
It is a vector, drawn from the positive centre towards the negative centre.
In a polyatomic molecule the dipole moment is the vector sum of all bond dipoles and lone pair dipoles. Symmetrical molecules such as \(\mathrm{BeF_2}\), \(\mathrm{BF_3}\), \(\mathrm{CO_2}\), \(\mathrm{CH_4}\) and \(\mathrm{CCl_4}\) have \(\mu = 0\) despite having polar bonds, while \(\mathrm{H_2O}\) (1.85 D), \(\mathrm{NH_3}\) (1.47 D) and \(\mathrm{CHCl_3}\) (1.04 D) do not. The lone pair dipole adds to the bond dipoles in \(\mathrm{NH_3}\) but opposes them in \(\mathrm{NF_3}\), which is why \(\mathrm{NF_3}\) has so small a value.
Ionic bonds can acquire partial covalent character through polarisation of the anion by the cation. By Fajans’ rules this increases with a smaller cation, a larger anion, a higher cation charge, and an \(s^2p^6d^{10}\) outer configuration on the cation rather than \(s^2p^6\).
Resonance is the representation of a species by two or more Lewis structures, called canonical forms, that differ only in the positions of electrons while the atoms stay fixed. The real species is the resonance hybrid, and it is more stable than any canonical form by an amount called the resonance energy. In ozone both O–O bonds are equal at 128 pm, between the 121 pm of a double bond and the 148 pm of a single bond.
Flowchart
FC 5.9 – The chapter’s line of argument: atoms with incomplete valence shells → octet rule (Kossel and Lewis) → branch A: electron transfer → ionic bond → ionic solid, lattice enthalpy; branch B: electron sharing → covalent bond → Lewis dot structures → formal charge → limitations of the octet rule → VSEPR gives molecular shapes → valence bond theory gives orbital overlap, σ and π bonds, hybridization → limitations of VB theory (paramagnetism of O₂) → molecular orbital theory gives bond order and magnetic behaviour → both branches converge → measurable bond parameters: bond angle, bond enthalpy, bond length, bond order, polarity and dipole moment → the two branches meet again: covalent character in ionic bonds (Fajans’ rules) and resonance in covalent species
5.10 Check Your Concepts
Sodium is a soft metal that reacts violently with water, and chlorine is a poisonous gas, yet sodium chloride is safe to eat. Explain, in terms of electronic configuration, why the compound behaves so differently from the elements that form it.
Calcium combines with chlorine to give \(\mathrm{CaCl_2}\) and not \(\mathrm{CaCl}\). Using the electronic configurations of the two elements, explain why two chlorine atoms are required for every calcium atom.
Assertion (A): Magnesium oxide has a much higher lattice enthalpy than sodium chloride. Reason (R): Both the charges on the ions and the sizes of the ions favour a stronger electrostatic attraction in magnesium oxide. State whether A and R are correct, and whether R is the correct explanation of A.
Arrange \(\mathrm{NaCl}\), \(\mathrm{MgCl_2}\) and \(\mathrm{AlCl_3}\) in increasing order of covalent character. Name the rule you have applied and state which property of the cation decides the order.
\(\mathrm{Cu^+}\) and \(\mathrm{Na^+}\) are of comparable size and carry the same charge, yet \(\mathrm{CuCl}\) is more covalent than \(\mathrm{NaCl}\). Explain, referring to the outer electronic configurations of the two cations and to the screening of nuclear charge.
The central atoms in \(\mathrm{LiCl}\), \(\mathrm{BeCl_2}\) and \(\mathrm{BF_3}\) all have incomplete octets, yet all three molecules are stable. State the number of electrons around the central atom in each, and explain why the octet rule cannot be applied to them.
Nitric oxide (NO) and nitrogen dioxide (\(\mathrm{NO_2}\)) cannot obey the octet rule however they are drawn. What feature do these two molecules share that makes this unavoidable?
The xenon atom in \(\mathrm{XeF_4}\) is surrounded by four bonding pairs and two lone pairs. Predict the electron pair geometry and the shape of the molecule, and explain why the two lone pairs take up positions opposite each other rather than adjacent.
Match the molecule type in column I with the correct shape in column II.
| Column I | Column II |
|---|---|
| a. AB₃E | i. square planar |
| b. AB₂E₂ | ii. square pyramid |
| c. AB₃E₂ | iii. trigonal pyramidal |
| d. AB₅E | iv. bent |
| e. AB₄E₂ | v. T-shape |
According to the postulates of valence bond theory, the electrons in the two overlapping half-filled orbitals must have opposite spins. State what happens to the potential energy of the system if two atoms carrying electrons of parallel spin are brought close together.
Describe what happens to the total energy of a system of two hydrogen atoms as they approach one another from a large separation. State clearly at which stage the bond is formed, and what happens if the two nuclei are pushed still closer.
Assertion (A): A covalent bond formed by a hybrid orbital is stronger than one formed by a pure atomic orbital. Reason (R): A hybrid orbital has one large lobe and one small lobe, which allows a greater extent of overlap. State whether A and R are correct, and whether R is the correct explanation of A.
The 2s and 2p orbitals of an atom undergo hybridization, but its 3s and 2p orbitals do not. State the condition for hybridization that is violated in the second case, and explain why it matters.
For the molecule \(\mathrm{CH_3-C}\equiv\mathrm{CH}\), state the type of hybridization of each carbon atom, and give the total number of sigma and pi bonds present.
The 2s orbital of one atom can combine with the \(2p_z\) orbital of another atom, but not with its \(2p_x\) orbital, even though both p orbitals have the same energy. Explain which condition for the combination of atomic orbitals is responsible.
A species is found to have a bond order of zero. State what this tells you about the species, and give one example from the second period for which this happens.
State the three limitations of valence bond theory listed in this chapter, and identify which of them is removed by molecular orbital theory.
Water contains two O–H bonds, yet the energy required to break the first is 502 kJ mol⁻¹ and the second only 427 kJ mol⁻¹. Explain the difference, and state what is meant by the average bond enthalpy of the O–H bond.
The C≡N bond is 116 pm long while the C≡C bond is 120 pm long, although both are triple bonds. Explain the difference in terms of the two factors that decide bond length.
Name one molecule that contains four polar bonds and yet has a dipole moment of zero. Explain how this is possible, and state what change to the molecule would give it a non-zero dipole moment.
\(\mathrm{NH_3}\) and \(\mathrm{NF_3}\) are both trigonal pyramidal and both have one lone pair on nitrogen, yet their dipole moments are very different. Explain, referring to the direction of the bond dipoles and of the lone pair dipole in each molecule.
Distinguish between canonical forms and a resonance hybrid, giving three points of difference. Illustrate your answer using ozone, and state which of the two is more stable and by what name that energy difference is known.
Draw Lewis dot diagrams for the following:
- hydrogen (\(\mathrm{H_2}\))
- water (\(\mathrm{H_2O}\))
- carbon dioxide (\(\mathrm{CO_2}\))
- methane (\(\mathrm{CH_4}\))
- lithium fluoride (\(\mathrm{LiF}\))
Draw a diagram for bonding in ethene with \(sp^2\) hybridisation.
Draw Lewis electron dot structures of:
- \(\mathrm{HF}\)
- \(\mathrm{C_2H_6}\)
- \(\mathrm{C_2H_4}\)
- \(\mathrm{CF_3Cl}\)
- \(\mathrm{SO_2}\)
Draw orbital diagrams of:
- the fluorine molecule
- the hydrogen fluoride molecule
Distinguish between a sigma bond and a pi bond.
Display the electron distribution around the oxygen atom in the water molecule and state the shape of the molecule; also write the H–O–H bond angle.
State the octet rule. Explain its inadequacies with respect to
- incomplete octet
- expanded octet
Explain in brief, with one example each:
- ionic bond
- covalent bond
- co-ordinate bond
Give reasons for the need for hybridisation.
Explain the geometry of the methane molecule on the basis of hybridisation.
In the ammonia molecule the bond angle is 107°, and in the water molecule it is 104°35′, although in both the central atoms are \(sp^3\) hybridized. Explain.
Give reasons for:
- a sigma (\(\sigma\)) bond is stronger than a pi (\(\pi\)) bond
- HF is a polar molecule
- carbon is tetravalent in nature
Which type of hybridization is present in the ammonia molecule? Write the geometry and the bond angle present in ammonia.
Identify the type of orbital overlap present in
- \(\mathrm{H_2}\)
- \(\mathrm{F_2}\)
- the H–F molecule. Explain diagrammatically.
F–Be–F is a linear molecule but H–O–H is angular. Explain.
The \(\mathrm{BF_3}\) molecule is planar but \(\mathrm{NH_3}\) is pyramidal. Explain.
In the case of bond formation in the acetylene molecule:
- How many covalent bonds are formed?
- State the number of sigma and pi bonds formed.
- Name the type of hybridisation.
Define:
- bond enthalpy
- bond length
Predict the shape and bond angles in the following molecules:
- \(\mathrm{CF_4}\)
- \(\mathrm{NF_3}\)
- \(\mathrm{HCN}\)
- \(\mathrm{H_2S}\)
Indicate the factor on which the stability of an ionic compound is measured.
Arrange the following compounds on the basis of lattice energies in decreasing (descending) order: \(\mathrm{BeF_2}\), \(\mathrm{AlCl_3}\), \(\mathrm{LiCl}\), \(\mathrm{CaCl_2}\), \(\mathrm{NaCl}\)
Give the total number of electrons around sulphur (S) in the \(\mathrm{SF_6}\) compound.
A covalent bond is directional in nature. Justify.
What are the interacting forces present during the formation of a molecule of a compound?
Give the type of overlap by which a pi (\(\pi\)) bond is formed.
Mention the steps involved in hybridization.
Write the formula to calculate the bond order of a molecule.
Why is the \(\mathrm{O_2}\) molecule paramagnetic?
What do you mean by formal charge? Explain its significance with the help of a suitable example.
5.11 Practice With Numericals
Calculate the total number of valence electrons that must be accommodated while writing the Lewis dot structure of (i) the carbonate ion \(\mathrm{CO_3^{2-}}\) and (ii) the ammonium ion \(\mathrm{NH_4^{+}}\).
Calculate the total number of valence electrons to be accommodated in the Lewis dot structure of the sulphate ion \(\mathrm{SO_4^{2-}}\), and state which atom is placed at the centre and why.
For the ion \([\mathrm{N}=\mathrm{C}=\mathrm{O}]^-\), calculate the formal charge on the nitrogen, carbon and oxygen atoms using \(\text{FC} = \text{VE} - \text{NE} - \tfrac{1}{2}\text{BE}\). Verify that the formal charges add up to the overall charge on the ion.
Write the molecular orbital electronic configuration of the \(\mathrm{C_2}\) molecule (12 electrons). Calculate its bond order and state whether it is diamagnetic or paramagnetic.
Write the molecular orbital electronic configuration of \(\mathrm{O_2^{+}}\). Calculate its bond order, state its magnetic nature, and say whether its bond is shorter or longer than that in \(\mathrm{O_2}\).
Write the molecular orbital electronic configuration of the peroxide ion \(\mathrm{O_2^{2-}}\) (18 electrons). Calculate its bond order and state its magnetic nature.
Write the molecular orbital electronic configuration of \(\mathrm{Ne_2}\) (20 electrons), calculate its bond order, and hence state whether the \(\mathrm{Ne_2}\) molecule can exist.
The bond enthalpy of \(\mathrm{H_2}\) is \(435.8\ \text{kJ mol}^{-1}\) and that of HCl is \(431.0\ \text{kJ mol}^{-1}\). State which bond is stronger, and calculate the difference in energy required to break one mole of each bond.
Using Table 5.6, calculate the difference in bond enthalpy between a C≡C bond and a C–C bond, and comment on what the difference tells you about the strength of a pi bond compared with a sigma bond.
The resultant dipole moment of ammonia is \(4.90\times10^{-30}\ \text{C m}\). Express this value in debye, given that \(1\ \text{D} = 3.33564\times10^{-30}\ \text{C m}\), and check your answer against Table 5.8.
The H–Cl bond length is 127 pm and the observed dipole moment of HCl is 1.03 D. Calculate the dipole moment, in debye, that HCl would have if the bond were 100% ionic. (Charge on an electron \(= 1.6\times10^{-19}\ \text{C}\); \(1\ \text{D} = 3.33564\times10^{-30}\ \text{C m}\).) Comment on what the comparison tells you about the H–Cl bond.
Using the values in Table 5.1, calculate how many times greater the lattice enthalpy of \(\mathrm{AlCl_3}\) is than that of \(\mathrm{NaCl}\), and state the two reasons for the difference.
Using Table 5.7, calculate the difference in bond length between a C–C single bond and a C≡C triple bond. Express the C≡C bond length in Angstrom units, given that \(1\ \text{Å} = 100\ \text{pm}\).
5.12 Case-Based Questions
- Using data from the table, answer the following
| Examples | C₂H₆ Ethane | C₂H₄ Ethene | C₂H₂ Ethyne |
|---|---|---|---|
| Structure | –C–C– | C=C | –C≡C– |
| Type of bond between carbons | single | double | triple |
| Bond length (nm) | 0.154 | 0.134 | 0.120 |
| Bond enthalpy (kJ mol⁻¹) | 348 | 612 | 837 |
a. What happens to the bond length when unsaturation increases?
b. Which is the most stable compound?
c. Indicate the relation between bond strength and bond enthalpy.
d. Comment on the overall relation between bond length, bond enthalpy, bond strength and stability.5.13 Objective Questions
- Which molecule is linear?
- \(\mathrm{SO_3}\)
- \(\mathrm{CO_2}\)
- \(\mathrm{H_2S}\)
- \(\mathrm{Cl_2O}\)
- When the following bond types are listed in decreasing order of strength (strongest first), which is the correct order?
- covalent > hydrogen > van der Waals’
- covalent > van der Waals’ > hydrogen
- hydrogen > covalent > van der Waals’
- van der Waals’ > hydrogen > covalent
- Valence Shell Electron Pair Repulsion (VSEPR) theory is used to predict which of the following:
- energy levels in an atom
- the shapes of molecules and ions
- the electronegativities of elements
- the type of bonding in compounds
- Which of the following is true for \(\mathrm{CO_2}\)?
| C=O bond | CO₂ molecule | |
|---|---|---|
| A | polar | non-polar |
| B | non-polar | polar |
| C | polar | polar |
| D | non-polar | non-polar |
- The \(\mathrm{O_2}\) molecule is paramagnetic. It is explained on the basis of:
- hybridisation
- VBT
- MOT
- VSEPR
- The angle between two covalent bonds is minimum in:
- \(\mathrm{CH_4}\)
- \(\mathrm{C_2H_2}\)
- \(\mathrm{NH_3}\)
- \(\mathrm{H_2O}\)
- Complete the flow chart
| Molecular formula | Structural formula | Shape / Geometry | Bond angle |
|---|---|---|---|
| BeCl₂ | ______ | ______ | 180° |
| ______ | O=C=O | Linear | ______ |
| C₂H₂ | ______ | ______ | ______ |
- Complete the following table
| Molecule | Type of hybridisation | Type of bonds | Geometry | Bond angle |
|---|---|---|---|---|
| CH₄ | ______ | 4 C–H, 4σ bonds | Tetrahedral | ______ |
| NH₃ | sp³ | 3 N–H, 3σ bonds, 1 lone pair | ______ | ______ |
| H₂O | ______ | ______ | angular | 104.5° |
| BF₃ | sp² | ______ | ______ | 120° |
| C₂H₄ | ______ | ______ | ______ | 120° |
| BeF₂ | ______ | 2 Be–F | Linear | ______ |
| C₂H₂ | sp | (3σ + 2π): 1 C–C σ, 2 C–H σ, 2 C–C π | ______ | ______ |