Chemical bonding

Class 11 Chemistry, unit 3, book pages 39–79

Practise 71 MCQs Take the unit quiz Review 22 cards

Atoms combine because bonding lowers their energy, and the type and strength of a bond decide the properties of the substance. This unit starts with electronegativity, which tells us whether a bond is ionic, polar covalent or nonpolar covalent, and with dipole moment and bond energy. It then explains the shapes of molecules with VSEPR theory, the formation of covalent bonds with valence bond theory and hybridization, the co-ordinate (dative) bond, and the forces between molecules including hydrogen bonding. It ends with molecular orbital theory, which explains bond order and the paramagnetism of oxygen.

Learning outcomes (21)

Define electronegativity as the power of an atom to attract electrons to itself.
Explain the factors influencing the electronegativities of elements in terms of nuclear charge, shielding by inner shells and subshells.
Explain the trends in electronegativity across a period and down a group of the periodic table.
Use the differences in Pauling electronegativity values to predict the formation of ionic and covalent bonds.
Describe covalent bonding in molecules using the concept of hybridization to describe $s p$ , $s p^{2}$ and $s p^{3}$ orbitals.
Use bond energy values and the concept of bond length to compare the reactivity of covalent molecules.
Describe the shapes and bond angles in molecules using VSEPR theory (including describing sketching).
Predict the shapes and bond angles in molecules and ions.
Explain hybridization and types of hybridization.
Explain valence bond theory.
Explain the importance of VSEPR theory in the field of drug design by discussing how the shape and bond angles of the molecules help chemists predict their interactions in the body.
Explain the salient features of molecular orbital theory.
Explain the paramagnetic nature of the oxygen molecule in the light of MOT.
Calculate the bond order of $N X_{2}$ , $O X_{2}$ , $F X_{2}$ and $He X_{2}$ .
Describe the types of van der Waals' forces.
Describe hydrogen bonding limited to molecules including ammonia and water.
Use the concept of hydrogen bonding to explain the anomalous properties of water.
Use the concept of electronegativity to explain bond polarity and dipole moments of molecules.
State that, in general, ionic, covalent and metallic bonding are stronger than intermolecular forces.
Recognize that molecular ions and polyatomic ions can have expanded octets, for example sulphate and nitrate.
Analyze the formation of the dative bond in $CO$ , ozone ( $O X_{3}$ ) and the $H X_{3} O X^{+}$ ion (resonance structures are not required).

3.1ElectronegativityOften examined

Electronegativity is the power of a covalently bonded atom to attract the shared pair of electrons towards itself in a molecule.

Electronegativity: The power of a covalently bonded atom to attract the shared pair of electrons towards itself in a molecule.

It is a dimensionless property and has no unit, because it is only a tendency. It shows the net result of the tendencies of the atoms of different elements to attract the bond-forming electron pairs. Electronegativity is measured on several scales; the most commonly used scale was designed by Linus Pauling, and its values are the ones used in this book.

Factors affecting electronegativity

The electronegativity of an atom depends on three factors.

Size of the atom
Nuclear charge
Screening (shielding) by inner electrons

How the three factors work

Nuclear charge. As the number of protons in the nucleus increases, the nuclear charge increases and the size of the atom decreases. The bonded electrons are then closer to the nucleus and are attracted more strongly, so electronegativity increases.
Size of the atom. A smaller atom holds the shared pair closer to its nucleus, so a smaller atom is more electronegative.
Screening by inner electrons. When the number of inner-shell electrons increases, the distance between the nucleus and the valence electrons increases, and the inner electrons shield the nucleus. The attraction of the bonded electrons towards the nucleus falls, so electronegativity decreases.

Table 3.1 gives the Pauling electronegativity values of the elements. The book prints it as a periodic table; here the same values are grouped by period and block.

Table 3.1: Pauling electronegativity values of the elements (values as printed in the book, grouped by block)
Period	s-block	d-block	p-block
1	H 2.20
2	Li 0.98, Be 1.57		B 2.04, C 2.55, N 3.04, O 3.44, F 3.98
3	Na 0.93, Mg 1.31		Al 1.61, Si 1.90, P 2.19, S 2.58, Cl 3.16
4	K 0.82, Ca 1.00	Sc 1.36, Ti 1.54, V 1.63, Cr 1.66, Mn 1.55, Fe 1.83, Co 1.88, Ni 1.91, Cu 1.90, Zn 1.65	Ga 1.81, Ge 2.01, As 2.18, Se 2.55, Br 2.96
5	Rb 0.82, Sr 0.95	Y 1.22, Zr 1.33, Nb 1.6, Mo 2.16, Tc 1.9, Ru 2.2, Rh 2.28, Pd 2.20, Ag 1.93, Cd 1.69	In 1.78, Sn 1.96, Sb 2.05, Te 2.1, I 2.66
6	Cs 0.79, Ba 0.89	La 1.1, Hf 1.3, Ta 1.5, W 2.36, Re 1.9, Os 2.2, Ir 2.20, Pt 2.28, Au 2.54, Hg 2.00	Tl 1.62, Pb 2.33, Bi 2.02, Po 2.0, At 2.2
7	Fr 0.7, Ra 0.9

3.1.2Trend of electronegativity in the periodic tableOften examined

Down a group

Electronegativity decreases from top to bottom in a group. The atomic number increases as we move down the group, so the nuclear charge also increases, but the effect of this increase is overcome by the addition of an extra shell, which increases the shielding. The attraction between the shared electrons and the nucleus decreases, so the value of electronegativity falls.

For example, in the halogen group the electronegativity decreases from fluorine to astatine, as shown in Figure 3.1.

Figure 3.1: Trend of electronegativity of the halogens down the group. The bars get shorter from fluorine to astatine.

Across a period

Electronegativity increases from left to right across a period. Moving from left to right, the effective nuclear charge increases, while the shielding stays constant because the electrons are added to the same shell. The size of the atom therefore decreases and the power of the atom to attract the shared pair of electrons increases.

For example, in the third period the electronegativity rises from sodium to chlorine, as shown in Figure 3.2.

Figure 3.2: Trend of electronegativity of the third period elements. The bars get taller from sodium to chlorine.

Trends of electronegativity
Direction in the periodic table	What happens	Reason
Down a group	Electronegativity decreases	An extra shell is added, so the shielding increases and the atom becomes larger; this outweighs the larger nuclear charge
Left to right across a period	Electronegativity increases	The effective nuclear charge increases, the shielding stays constant and the atom becomes smaller

3.1.3Nature of bond in terms of Pauling electronegativity valuesOften examined

The difference in electronegativity between two bonded atoms tells us the type of bond between them. According to the book:

Nature of bond from Pauling electronegativity differences
Electronegativity difference	Type of bond
Higher than 1.8	Ionic bond
Between 1.8 and 0.4	Polar covalent bond
0.4 or lower	Nonpolar covalent bond
0 (both atoms the same)	The pair of electrons is shared equally

Example 3.1: bond types of oxygen, potassium chloride and hydrogen chloride

Consider the following substances and determine their bond types. (i) Oxygen molecule,

O = O

. (ii) Potassium chloride,

KCl

. (iii) Hydrogen chloride,

HCl

(i) Oxygen molecule. It has two oxygen atoms, and oxygen has an electronegativity of 3.44, so each atom has 3.44. Find the difference.
$Δ E N = 3.44 - 3.44 = 0$
The difference is $0$ , so neither atom pulls the electrons to itself more strongly than the other. The molecule has a nonpolar covalent bond.
(ii) Potassium chloride. Potassium has an electronegativity of 0.82 and chlorine has 3.16. Find the difference.
$Δ E N = 3.16 - 0.82 = 2.34$
The difference $2.34$ is greater than $1.8$ , so the bond is ionic.
(iii) Hydrogen chloride. Hydrogen has an electronegativity of 2.20 and chlorine has 3.16. Find the difference.
$Δ E N = 3.16 - 2.20 = 0.96$
Chlorine is more electronegative than hydrogen, so chlorine pulls the shared electrons closer to itself. The difference $0.96$ lies between $0.4$ and $1.8$ , so the bond is a polar covalent bond.

O X_{2} : nonpolar covalent; KCl : ionic; HCl : polar covalent

Worked example: placing bonds on the electronegativity scale

Find the electronegativity difference and the bond type of C–H, N–H, H–F and Na–Cl, using the values of Table 3.1 (H 2.20, C 2.55, N 3.04, F 3.98, Na 0.93, Cl 3.16).

Subtract the smaller value from the larger one for each bond.
$C-H N-H H-F Na-Cl : 2.55 - 2.20 = 0.35 : 3.04 - 2.20 = 0.84 : 3.98 - 2.20 = 1.78 : 3.16 - 0.93 = 2.23$
Compare with the limits: $0.4$ or lower is nonpolar covalent, between $0.4$ and $1.8$ is polar covalent, higher than $1.8$ is ionic.
C–H ( $0.35 \leq 0.4$ ) is nonpolar covalent. N–H ( $0.84$ ) is polar covalent. Na–Cl ( $2.23 > 1.8$ ) is ionic.
H–F is the tricky one. $1.78$ is just below $1.8$ , so by the book's rule it is a polar covalent bond (the book rounds it to $1.8$ in Table 3.2, but the value used for the rule is $1.78$ ).

The four bonds on the electronegativity-difference scale. H–F (accent) lies just inside the polar covalent region.

C-H: nonpolar covalent; N-H, H-F: polar covalent; Na-Cl: ionic

3.1.4Covalent character in a compound

The covalent character of a compound depends on the polarising power of the cation, which in turn depends on the oxidation state of the cation. The higher the polarising power of the cation, the more covalent character the compound has. That is why covalent character increases from left to right in periods.

Aluminium chloride and magnesium chloride

Why is aluminium chloride more covalent than magnesium chloride?

Find the oxidation state of each metal.
$AlCl X_{3} : Al = + 3 MgCl X_{2} : Mg = + 2$
A cation with a higher charge has a higher polarising power. $Al X^{3 +}$ therefore polarises the chloride ion more than $Mg X^{2 +}$ does.
Because of its high polarising power, aluminium can polarise the anion, so that electrons are shared between the two ions. Covalent character develops in the compound.
So $AlCl X_{3}$ is more covalent in nature than $MgCl X_{2}$ .

covalent character: AlCl X_{3} > MgCl X_{2}

3.2Dipole moment

The dipole moment is the product of the magnitude of the positive or negative charge and the distance between them. Mathematically,

μ = q \cdot r

Dipole moment

$μ$ is the dipole moment, $q$ is the charge and $r$ is the charge separation distance.

The unit of dipole moment given in the book is the debye. One debye is equal to $3.335 \times 1 0^{- 30} C m$ (C m is the coulomb metre).

The difference between the electronegativities of two atoms in a compound determines the overall dipole moment and the overall polarity of the compound. A dipole moment is a vector: it has a magnitude and a direction, drawn as an arrow from the positive end to the negative end.

The dipole moment of H–Cl has a magnitude and a direction. The arrow points from the positive end (hydrogen) to the negative end (chlorine).

The greater the electronegativity difference between two atoms, the more polar the bond is (the more ionic its nature). A small difference in electronegativity makes the bond less polar, that is, more covalent in nature.

3.3Polar and nonpolar covalent bondOften examined

In a polar covalent bond, because of the electronegativity difference, the more electronegative atom attracts the shared electrons towards itself and bears a partial negative charge ( $δ -$ ), while the other atom bears a partial positive charge ( $δ +$ ). Such molecules are called polar molecules. Polar molecules have dipoles: electric charges of equal magnitude and opposite sign.

Symmetrical polyatomic molecules with a linear, trigonal planar or tetrahedral shape are nonpolar. In these molecules the bond dipoles exert equal and opposite effects and cancel, so there is no overall dipole. Carbon dioxide is the example given in the book.

The two C=O bond dipoles of carbon dioxide are equal and opposite, so the net dipole moment is zero and CO₂ is nonpolar.

Because of the electronegativity difference between carbon and oxygen, each C=O bond in $CO X_{2}$ is polar. However, $CO X_{2}$ is a nonpolar molecule, because the two dipoles are equal and opposite and cancel each other.

Polar and nonpolar molecules
Molecule type	Bond dipoles	Molecule
Symmetrical (linear, trigonal planar, tetrahedral), same atoms on the central atom	Cancel	Nonpolar, for example $CO X_{2}$ , $CCl X_{4}$
Unsymmetrical shape, or different atoms on the central atom	Do not cancel	Polar, for example $HCl$ , $H X_{2} O$

Concept assessment exercise 3.1: polar or nonpolar?

Are the following molecules polar or nonpolar? In each case give a reason for your answer. Electronegativity values: F = 4.0, Cl = 3.0, Br = 2.8, S = 2.5, C = 2.5, H = 2.1. (a) Chlorine,

Cl X_{2}

. (b) Hydrogen fluoride,

HF

. (c) The V-shaped molecule sulfur dichloride,

SCl X_{2}

. (d) The tetrahedral molecule chloromethane,

CH X_{3} Cl

. (e) The tetrahedral molecule tetrabromomethane,

CBr X_{4}

(a) $Cl X_{2}$ . Both atoms are chlorine, so the electronegativity difference is zero and the bonding pair is shared equally. There is no dipole: nonpolar.
$Δ E N = 3.0 - 3.0 = 0$
(b) $HF$ . Fluorine is much more electronegative than hydrogen, so the shared pair is pulled towards fluorine ( $δ -$ ) and away from hydrogen ( $δ +$ ). A diatomic molecule has only this one bond, so nothing cancels it: polar.
$Δ E N = 4.0 - 2.1 = 1.9$
(c) $SCl X_{2}$ . Each S–Cl bond is polar because chlorine is more electronegative than sulfur. The molecule is V-shaped, so the two bond dipoles do not point in opposite directions and do not cancel: they add to a net dipole. Polar.
$Δ E N = 3.0 - 2.5 = 0.5$
(d) $CH X_{3} Cl$ . The C–Cl bond (difference $0.5$ ) is more polar than the C–H bonds (difference $0.4$ ), and the three different kinds of atoms make the molecule unsymmetrical. The dipoles do not cancel, and the net dipole points towards chlorine. Polar.
(e) $CBr X_{4}$ . Each C–Br bond is slightly polar (difference $0.3$ ), but the molecule is a perfectly symmetrical tetrahedron with four identical bonds, so the four bond dipoles cancel. The net dipole moment is zero: nonpolar.
$Δ E N = 2.8 - 2.5 = 0.3$

Bond dipoles (grey) and the net dipole (accent) for the molecules in Concept assessment exercise 3.1. SCl₂ and CH₃Cl have a net dipole; the bond dipoles of CBr₄ cancel.

(a) nonpolar, (b) polar, (c) polar, (d) polar, (e) nonpolar

3.4Bond energy (bond enthalpy) and bond lengthOften examined

The energy required to break the bond of the same type in one mole of a substance is known as the bond energy (bond enthalpy). Its unit is $kJ mo l^{- 1}$ .

Bond energy (bond enthalpy): The energy required to break the bond of the same type in one mole of a substance. Unit: $kJ mo l^{- 1}$ .

Bond energy depends upon four factors:

Electronegativity difference between the two bonded atoms
Size of the atom
Bond order
Bond length

The greater the electronegativity difference between two bonded atoms, the more polar the bond is and the greater the bond energy. Table 3.2 gives the bond energies of the hydrogen halides (hydrides of group VII).

Table 3.2: Bond energy of the hydrogen halides
Bond	Bond energy ( $kJ mo l^{- 1}$ )	Electronegativity difference between bonded atoms
H–F	562	1.8
H–Cl	431	0.9
H–Br	366	0.7
H–I	299	0.4

From these data we conclude that the greater the electronegativity difference, the higher the bond energy. For example, HF has a greater bond energy than HCl. Figure 3.3 shows the bond enthalpy of HX from HF to HI.

Figure 3.3: Bond enthalpy of the hydrogen halides from HF to HI. The bond energy falls as the electronegativity difference falls.

Bond length

Bond length is the distance between the nuclei of two covalently bonded atoms. It is measured in angstroms ( $\overset{˚}{A}$ ) or picometres ( $pm$ ), where $1 pm = 1 0^{- 12} m$ . Bond length depends upon the size of the atoms: the smaller the atoms, the shorter the bond length and the greater the bond energy.

Bond length: The distance between the nuclei of two covalently bonded atoms. Measured in $\overset{˚}{A}$ or $pm$ ; $1 pm = 1 0^{- 12} m$ .

For example, the bond energy of H–H is $436 kJ mo l^{- 1}$ and that of H–Br is $366 kJ mo l^{- 1}$ . This is due to the shorter bond length of the hydrogen molecule compared with HBr, since the bromine atom is larger than the hydrogen atom.

Bond energy and bond length of the halogens

Table 3.3: Bond energy and bond length of the halogens
Bond	Bond energy ( $kJ mo l^{- 1}$ )	Bond length ( $pm$ )
F–F	158	149
Cl–Cl	242	199
Br–Br	193	228
I–I	151	266

The atomic radius increases from fluorine to iodine. The bond energy decreases from chlorine to iodine, but the fluorine molecule has a lower bond energy than chlorine. The reason is the small size of the fluorine atoms: their lone pairs repel each other, so a smaller amount of energy is required to break the bond between the two fluorine atoms.

Figure 3.4 (a): Bond length of the halogens from Cl₂ to I₂. The bond gets longer as the atoms get larger.

Figure 3.4 (b): Bond enthalpy of the halogens from F₂ to I₂. The bond energy of Cl₂ is the highest; F₂ is lower than Cl₂ (accent).

Bond energy and reactivity

Bond energy is also used to predict the reactivity of covalent molecules. The greater the bond energy, the stronger the bond, and the more stable and less reactive the molecule will be. For example, among the alkyl halides, alkyl fluorides are less reactive than alkyl chlorides, because the bond energy of C–F is greater than that of C–Cl.

Summary
Factor	Effect on bond energy
Greater electronegativity difference	Bond energy is greater
Shorter bond length (smaller atoms)	Bond energy is greater
Greater bond energy	Stronger bond, more stable and less reactive molecule

3.5Shapes of molecules: valence shell electron pair repulsion theoryOften examined

The valence shell electron pair repulsion (VSEPR) theory is used to determine the shape and the bond angle in a molecule.

VSEPR theory: Electrons are negatively charged, so the electron pairs in the valence shell of the central atom repel each other. The pairs move as far apart as possible, and the arrangement that minimises the repulsion gives the shape of the molecule.

As electrons are negatively charged particles, they repel each other when they are close together. So a pair of electrons in the bonds surrounding the central atom in a molecule will repel the other electron pairs. These repulsions force the pairs apart until the repulsive forces are minimised.

The shape and the bond angles of a covalently bonded molecule depend on the number of pairs of electrons around the central atom, and on whether these pairs are lone pairs or bonding pairs. Lone pairs have a more concentrated electron charge cloud than bonding pairs; their clouds are wider and slightly closer to the nucleus of the central atom. This results in a different amount of repulsion between different types of electron pairs.

The order of repulsion is:

lone pair-lone pair (most repulsion) > lone pair-bond pair > bond pair-bond pair (least repulsion)

3.5.1Shapes of molecules containing two electron pairsOften examined

Molecules in which the central atom is surrounded by two electron pairs and no lone pairs are linear in shape, with a bond angle of $18 0^{\circ}$ .

AX $_{2}$ : $BeF X_{2}$

The central atom, beryllium, contributes two valence electrons, and each fluorine atom contributes one electron to the bond with beryllium. The Lewis electron structure is shown below.

Lewis structure and shape of BeF₂. The two bonding pairs are 180° apart, so the molecule is linear.

In $BeF X_{2}$ there are two electron pairs around the central beryllium atom. The arrangement that minimises the repulsion places the two groups $18 0^{\circ}$ apart.

Two electron pairs
Formula type	Bonding pairs	Lone pairs	Shape	Bond angle	Example
AX $_{2}$	2	0	Linear	$18 0^{\circ}$	$BeF X_{2}$

3.5.2Shapes of molecules containing three electron pairsOften examined

Molecules in which the central atom is surrounded by three electron pairs have a trigonal arrangement of the pairs. With no lone pair the shape is trigonal planar and the angle is $12 0^{\circ}$ .

Three bonding pairs, AX $_{3}$ : $BCl X_{3}$

The central atom, boron, contributes three valence electrons, and each chlorine atom contributes one electron to the bond (chlorine has seven valence electrons in all). The Lewis electron structure and the shape of $BCl X_{3}$ are shown below.

Lewis structure and shape of BCl₃: three bonding pairs, no lone pair, trigonal planar with 120° angles.

There are three electron pairs around the central atom. To minimise repulsion the groups are placed $12 0^{\circ}$ apart, and the geometry is trigonal planar.

Three bonding pairs, AX $_{3}$ : $CO X_{3} X^{2 -}$

In the carbonate ion the central carbon atom has four valence electrons and each oxygen atom has six valence electrons. There are three bonding pairs around the carbon atom, so the molecular geometry of $CO X_{3} X^{2 -}$ is trigonal planar.

Lewis structure and shape of the carbonate ion: three bonding pairs around carbon, trigonal planar.

Two bonding pairs and one lone pair, AX $_{2}$ E: $SO X_{2}$

The central atom, sulphur, has six valence electrons, and so does each oxygen atom. The Lewis electron structure and the shape of $SO X_{2}$ are shown below.

Lewis structure and shape of SO₂: two bonding pairs and one lone pair (accent). The shape is bent (V-shaped) with an angle less than 120°.

There are three electron pairs around the central atom: two double bonds and one lone pair. This is two bonding pairs and one lone pair, so the structure is designated AX $_{2}$ E.

A lone pair occupies more space near the central atom than a bonding pair, so bonding pairs and lone pairs repel each other. In $SO X_{2}$ there is one bond pair–bond pair interaction and two lone pair–bond pair interactions. Thus the shape is bent (V-shaped) and the angle is less than $12 0^{\circ}$ .

Three electron pairs
Formula type	Bonding pairs	Lone pairs	Shape	Bond angle	Example
AX $_{3}$	3	0	Trigonal planar	$12 0^{\circ}$	$BCl X_{3}$ , $CO X_{3} X^{2 -}$
AX $_{2}$ E	2	1	Bent (V-shaped)	Less than $12 0^{\circ}$	$SO X_{2}$

3.5.3Shapes of molecules containing four electron pairsOften examined

Four bonding pairs, AX $_{4}$ : $CH X_{4}$

Molecules in which the central atom is surrounded by four electron pairs and no lone pair are tetrahedral in shape, with a bond angle of $109. 5^{\circ}$ . In $CH X_{4}$ the carbon atom is surrounded by four electron pairs, all of them bonding pairs.

Lewis structure and shape of CH₄: four bonding pairs, tetrahedral, 109.5°. A solid wedge comes towards the viewer and a hatched bond goes away.

Three bonding pairs and one lone pair, AX $_{3}$ E: $NH X_{3}$

Molecules in which the central atom is surrounded by four electron pairs, three bonding pairs and one lone pair, are trigonal pyramidal in shape. In ammonia the central atom, nitrogen, has five valence electrons and each hydrogen atom donates one valence electron. The nitrogen atom is surrounded by four electron pairs, of which three are bonding pairs and one is a lone pair. With three bonding pairs and one lone pair the structure is designated AX $_{3}$ E.

Lewis structure and shape of NH₃: three bonding pairs and one lone pair (accent), trigonal pyramidal, 107°.

There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal. The H–N–H bond angles are less than the ideal angle of $109. 5^{\circ}$ because of lone pair–bond pair repulsion.

Two bonding pairs and two lone pairs, AX $_{2}$ E $_{2}$ : $H X_{2} O$

Molecules in which the central atom is surrounded by four electron pairs, two bonding pairs and two lone pairs, are bent in shape. The structure is designated AX $_{2}$ E $_{2}$ , with a total of four electron pairs. Because of the lone pair–lone pair, lone pair–bond pair and bond pair–bond pair interactions, we expect a significant deviation from the idealised tetrahedral angle.

Lewis structure and shape of H₂O: two bonding pairs and two lone pairs (accent), bent, 104.5°.

There are two nuclei about the central atom, so the molecular shape is bent (V-shaped). The H–O–H angle is even less than the H–N–H angle in $NH X_{3}$ , as we expect because of the presence of two lone pairs on the central atom rather than one.

Four electron pairs: the same arrangement, three different shapes
Molecule	Formula type	Bonding pairs	Lone pairs	Shape	Bond angle
$CH X_{4}$	AX $_{4}$	4	0	Tetrahedral	$109. 5^{\circ}$
$NH X_{3}$	AX $_{3}$ E	3	1	Trigonal pyramidal	$10 7^{\circ}$
$H X_{2} O$	AX $_{2}$ E $_{2}$	2	2	Bent (V-shaped)	$104. 5^{\circ}$

3.5.4Shapes of molecules containing five electron pairsOften examined

In some molecules the central atom does not have a noble gas configuration. Molecules in which the central atom is surrounded by five electron pairs and no lone pair are trigonal bipyramidal in shape. For example, \ce{PCl5} has five bonding pairs of electrons and no lone pairs.

The repulsion between the electron pairs makes the trigonal bipyramid the most stable structure. Three of the chlorine atoms lie in the same plane as the phosphorus atom, and the \ce{Cl-P-Cl} bond angles within this plane are 120^\circ. Two of the chlorine atoms lie above and below this plane, at 90^\circ to it.

Lewis structure and shape of PCl₅: five bonding pairs, trigonal bipyramidal. The three equatorial chlorine atoms are 120° apart and the two axial ones are 90° to the plane.

Five electron pairs
Formula type	Bonding pairs	Lone pairs	Shape	Bond angles	Example
AX_5	5	0	Trigonal bipyramidal	90^\circ and 120^\circ	\ce{PCl5}

3.5.5Shapes of molecules containing six electron pairsOften examined

Molecules in which the central atom is surrounded by six electron pairs and no lone pair are octahedral in shape. All the angles are 90^\circ. Sulphur hexafluoride, \ce{SF6}, has twelve electrons around the central sulphur atom. We say that the sulphur atom has an expanded octet.

Lewis structure and shape of SF₆: six bonding pairs, octahedral, all angles 90°. Sulphur has twelve electrons around it (an expanded octet).

Six electron pairs
Formula type	Bonding pairs	Lone pairs	Shape	Bond angle	Example
AX_6	6	0	Octahedral	90^\circ	\ce{SF6}

3.5.6Expanded octet (hypervalency) and summary of shapesOften examined

The octet rule applies well to atoms in the second row of the periodic table, where a full valence shell has eight electrons with the configuration s^2p^6. Even elements in the third and fourth rows are known to follow this rule sometimes, but not always. In larger atoms, from the 3rd period onwards, the valence shell also contains additional subshells (d, f and g subshells).

Therefore the atoms of 3rd, 4th ... period elements can have higher valence shell counts by "expanding" into these additional subshells. When an atom has more than eight electrons in its valence shell, it is said to be hypervalent. This is also called an expanded octet. Figure 3.5 shows molecular ions with an expanded octet.

Expanded octet (hypervalency): The condition in which the central atom has more than eight electrons in its valence shell. It is possible for elements of the 3rd period and beyond, because they have additional d subshells available.

Figure 3.5: Ions with an expanded octet. Sulphur in sulphate has 12 electrons around it and phosphorus in phosphate has 10. Nitrogen in nitrate keeps an octet (8 electrons).

Table 3.4 summarises all the shapes described so far.

Table 3.4: Summary of shapes of molecules (the first column is headed "number of valence electrons" in the book)
Electrons around the central atom	Bonding pairs	Lone pairs	Total electron pairs	Arrangement of pairs	Shape and angle	Examples
4	2	0	2	Linear	Linear, 180^\circ	\ce{BeCl2}, \ce{HCN}, \ce{CO2}
6	3	0	3	Trigonal planar	Trigonal planar, 120^\circ	\ce{BF3}, \ce{AlCl3}, \ce{NO3^-}
6	2	1	3	Trigonal planar	Bent, 118^\circ	\ce{SnCl2}
8	4	0	4	Tetrahedral	Tetrahedral, 109.5^\circ	\ce{CH4}, \ce{SiCl4}, \ce{SO4^2-}, \ce{NH4^+}
8	3	1	4	Tetrahedral	Trigonal pyramid, 107^\circ	\ce{NH3}, \ce{PCl3}, \ce{ClO3^-}
8	2	2	4	Tetrahedral	Bent, 104.5^\circ	\ce{H2O}, \ce{H2S}
10	5	0	5	Trigonal bipyramid	Trigonal bipyramid, 90^\circ and 120^\circ	\ce{PCl5}, \ce{PI5}
12	6	0	6	Octahedral	Octahedral, 90^\circ	\ce{SF6}

3.5.7Working out the shapes of moleculesOften examined

The shape of a molecule or ion is worked out in a few steps.

Method

Add together the number of valence electrons of the central atom and the number of electrons contributed to bonding by the surrounding atoms. Include the ion charge if there is one (subtract for a positive charge, add for a negative charge).
Divide the total number of electrons by 2. This gives the number of electron pairs distributed around the central atom.
Arrange the electron pairs round the central atom to keep them as far apart as possible (2 pairs linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral).
Count the atoms bonded to the central atom: this is the number of bonding pairs. The remaining pairs are lone pairs.
Name the shape from the atoms only (Table 3.4).

Example 1: the geometry of \ce{NH3}

Predict the geometry of \ce{NH3}.

Valence electrons of the central nitrogen atom: 5.
5
Electrons contributed by the three hydrogen atoms: 3\times1=3.
3
Total number of electrons.
5+3=8
Number of electron pairs.
8\div2=4
Four electron pairs are arranged tetrahedrally.
Number of atoms around the central atom: 3, so there are three bonding pairs. Number of lone pairs: 4-3=1.
\text{lone pairs}=4-3=1
Three bonding pairs and one lone pair: the molecule is trigonal pyramidal.

\ce{NH3}:\ \text{trigonal pyramidal (AX}_3\text{E)}

Example 2: the geometry of \ce{NH4^+}

Predict the geometry of \ce{NH4^+}.

Valence electrons of the central nitrogen atom: 5.
5
Electrons contributed by the four hydrogen atoms: 4\times1=4.
4
The ion has a charge of +1, so one electron is removed. Total number of electrons.
5+4-1=8
Number of electron pairs.
8\div2=4
Four electron pairs are arranged tetrahedrally.
Number of atoms around the central atom: 4, so there are four bonding pairs and no lone pair (4-4=0).
Four bonding pairs and no lone pair: the ion is tetrahedral.

\ce{NH4^+}:\ \text{tetrahedral (AX}_4\text{)}

Worked example: the geometry of \ce{H2S}

Predict the geometry of \ce{H2S} and say how its bond angle compares with 109.5^\circ.

Valence electrons of the central sulphur atom (group 16): 6.
6
Electrons contributed by the two hydrogen atoms: 2\times1=2.
2
Total number of electrons.
6+2=8
Number of electron pairs.
8\div2=4
Four electron pairs are arranged tetrahedrally. Two atoms are bonded to sulphur, so there are two bonding pairs and 4-2=2 lone pairs.
\text{AX}_2\text{E}_2
Name the shape from the atoms only: bent (V-shaped). The two lone pairs repel the bonding pairs more than they repel each other, so the H–S–H angle is less than 109.5^\circ, as in \ce{H2O} (104.5^\circ).

Lewis structure and shape of \ce{H2S}: two bonding pairs and two lone pairs (accent), bent.

\ce{H2S}:\ \text{bent (AX}_2\text{E}_2\text{), angle}<109.5^\circ

3.5.8Importance of VSEPR theory in the field of drugs

By predicting molecular shapes and bond angles, VSEPR theory helps medicinal chemists to understand how drug molecules interact in the body.

The shape and dimensions of the molecules used as drugs also depend upon the length and the angles of the bonds between their atoms. Any change in the bond angles changes the overall three-dimensional shape of a molecule, and so changes its binding ability or its permeability through the membranes of the target cell.

Molecular shape and enzymatic reactions

Enzymes recognise and bind to drug substrates on the basis of their molecular shape. VSEPR theory helps to predict this three-dimensional structure, which allows designers to tailor molecules for the best interaction with the active sites of enzymes. Understanding molecular geometry through VSEPR theory helps in creating drugs with better substrate recognition and more targeted interactions in enzymatic reactions.

Enzyme–substrate interaction in drug design

Consider the interaction between a drug and the enzyme acetylcholinesterase in neuropharmacology. VSEPR theory guides the design of drugs whose shapes fit precisely into the active site of the enzyme, which improves their efficacy and specificity.

Now imagine a drug molecule designed to interact with a specific enzyme. Its VSEPR-derived geometry must be considered to ensure the best binding. If the active site of the enzyme has a pocket with a tetrahedral arrangement of binding sites, a drug with a complementary tetrahedral geometry might fit more effectively.

Molecular geometry in drug delivery systems

VSEPR theory predicts molecular shapes and so plays a vital role in designing effective drug carriers. Liposomes, spherical lipid-based vesicles, are widely used as drug carriers because of their biocompatibility and versatility. VSEPR theory informs the design of liposomal carriers by predicting the best arrangement of electron pairs, which influences their three-dimensional structure. For instance, tailoring liposomes into specific shapes, such as elongated or targeted geometries, improves their ability to move through biological barriers and deliver drugs efficiently to the desired tissues.

Understanding molecular geometry also helps scientists to understand the shapes of more complex molecules, such as proteins and DNA. The shapes of these molecules play an important role in determining the jobs they perform in our bodies.

Cisplatin

Cisplatin is a neutral complex, \ce{Pt(NH3)2Cl2}, with a square planar structure. It is one of the most effective anticancer agents and is widely used in the treatment of solid tumours. It is generally considered a cytotoxic drug that kills cancer cells by damaging DNA and inhibiting DNA synthesis. The DNA damage caused by cisplatin activates various signalling pathways that prevent or promote cell death.

Cisplatin, Pt(NH₃)₂Cl₂: a neutral complex with a square planar shape. The two NH₃ groups are next to each other.

Concept assessment exercise 3.2: shapes of molecules

(1a) Predict the shapes of: (i) tetrachloromethane, \ce{CCl4}; (ii) beryllium chloride, \ce{BeCl2}; (iii) phosphorus(III) chloride, \ce{PCl3}. (1b) Draw dot-and-cross diagrams for the following molecules and then predict their shapes: (i) hydrogen sulphide, \ce{H2S}; (ii) phosphine, \ce{PH3}. (2) (a) Draw a dot-and-cross diagram for a molecule of selenium hexafluoride, \ce{SeF6}. (b) Predict the shape of selenium hexafluoride. (c) Draw the shape of the phosphorus(V) chloride molecule.

Question 1(a): predict the shapes

(i) \ce{CCl4}. Carbon has 4 valence electrons and the four chlorine atoms contribute 4 electrons for bonding.
4+4=8\ \text{electrons}=4\ \text{pairs}
Four pairs are all bonding pairs (four atoms on carbon) and there is no lone pair, so the shape is tetrahedral with a bond angle of 109.5^\circ (AX_4).
(ii) \ce{BeCl2}. Beryllium has 2 valence electrons and each chlorine contributes 1 electron for bonding.
2+2=4\ \text{electrons}=2\ \text{pairs}
Two bonding pairs and no lone pair: the shape is linear with a bond angle of 180^\circ (AX_2).
(iii) \ce{PCl3}. Phosphorus has 5 valence electrons and the three chlorine atoms contribute 3 electrons for bonding.
5+3=8\ \text{electrons}=4\ \text{pairs}
Four pairs: three bonding pairs and 4-3=1 lone pair (AX_3E). The shape is trigonal pyramidal with an angle of about 107^\circ.

Question 1(b): dot-and-cross diagrams

(i) \ce{H2S}. Sulphur has 6 valence electrons. Two of them are shared with the two hydrogen atoms (one pair each) and the other four form two lone pairs.
Total electron pairs on sulphur: 2 bonding pairs + 2 lone pairs =4 pairs, arranged tetrahedrally (AX_2E_2). Two lone pairs push the bonding pairs together, so the shape is bent (V-shaped). The book's Table 3.4 gives 104.5^\circ for \ce{H2S}, like \ce{H2O}.
(ii) \ce{PH3}. Phosphorus has 5 valence electrons. Three are shared with three hydrogen atoms and the other two form one lone pair.
Total electron pairs on phosphorus: 3 bonding pairs + 1 lone pair =4 pairs (AX_3E). The shape is trigonal pyramidal, like \ce{NH3}.

Dot-and-cross diagrams of H₂S and PH₃. Dots are the electrons of the central atom and crosses are the electrons of hydrogen.

Question 2: selenium hexafluoride and phosphorus(V) chloride

(a) Selenium has 6 valence electrons and each of the six fluorine atoms contributes 1 electron to a bond. Every fluorine atom also has three lone pairs.
6+6=12\ \text{electrons}=6\ \text{pairs}
All six pairs are bonding pairs, so selenium has twelve electrons around it: an expanded octet (see the diagram below).
(b) Six bonding pairs and no lone pair (AX_6): the shape is octahedral with all angles 90^\circ.

SeF₆: six shared pairs and no lone pair on selenium (12 electrons around Se), so the shape is octahedral with 90° angles.

(c) Phosphorus(V) chloride is \ce{PCl5}. Phosphorus has 5 valence electrons and the five chlorine atoms contribute 5.
5+5=10\ \text{electrons}=5\ \text{pairs}
Five bonding pairs and no lone pair (AX_5): the shape is trigonal bipyramidal. Three chlorine atoms lie in a plane at 120^\circ to each other and two lie above and below the plane at 90^\circ (the figure above).

Lewis structure and shape of PCl₅: five bonding pairs, trigonal bipyramidal. The three equatorial chlorine atoms are 120° apart and the two axial ones are 90° to the plane.

\ce{CCl4}:\text{tetrahedral};\ \ce{BeCl2}:\text{linear};\ \ce{PCl3}:\text{trigonal pyramidal};\ \ce{H2S}:\text{bent};\ \ce{PH3}:\text{trigonal pyramidal};\ \ce{SeF6}:\text{octahedral}

3.6Valence bond theoryOften examined

Valence bond theory (VBT) was proposed by Heitler and London (1927) and later developed by Pauling. It has successfully explained the bond energies, bond lengths and shapes of covalent molecules.

According to valence bond theory, a covalent bond is formed by the overlapping of a partially filled atomic orbital of one atom with a partially filled atomic orbital of the other atom. The two overlapping atomic orbitals retain their identity. The term overlap means that the two orbitals share the same common region in space.

Main points of VBT

A bond between two atoms is formed by the overlap of half-filled atomic orbitals of the two atoms. These atomic orbitals retain their identity.
The electrons of the overlapping orbitals have opposite spin.
The number of bonds formed is equal to the number of unpaired electrons present in the outermost shell of the atom.
The overlapping of two orbitals forms a single bond, which is a sigma bond. By the overlapping of additional orbitals, multiple bonds (double and triple) are formed.
In order to form a bond, the overlapping orbitals must have the same symmetry with respect to the bond axis. The bond axis is the line joining the nuclei of the two bonded atoms.
Energy is released by the overlapping of orbitals. The greater the overlap between the orbitals, the greater is the energy released and the stronger is the bond formed.

3.6Types of overlapping and nature of covalent bondsOften examined

The two main types of covalent bonds obtained by overlapping are the sigma bond (\sigma) and the pi bond (\pi).

Sigma bond

When there is a single bond between two atoms, the bond is a sigma bond. A sigma bond is formed by the head-on overlapping of atomic orbitals. It is formed by s–s, s–p_x, s–p_y, s–p_z and p_x–p_x overlap (Figure 3.6).

Figure 3.6: Formation of a sigma bond by (a) s–s overlap, (b) s–p overlap and (c) p–p overlap. In every case the orbitals overlap head on, along the line joining the nuclei.

Pi bond

A pi bond is formed by the parallel (sideways) overlapping of p orbitals (between p_y–p_y and p_z–p_z).

Figure 3.7: Parallel (sideways) overlapping of p orbitals forms a pi (π) bond. The overlap is above and below the bond axis.

When there is a double bond between two atoms, one should be a sigma bond and the other a pi bond. When there is a triple bond between two atoms, there is one sigma bond and two pi bonds. Ethene is an example (Figure 3.8).

Figure 3.8: Structure of ethene. The C=C double bond is one σ bond along the axis (accent) and one π bond above and below it; the four C–H bonds are σ bonds.

Sigma and pi bonds
Bond	Overlap	Where is the electron density?	Number in a double bond / triple bond
Sigma (\sigma)	Head on, along the bond axis	Between the two nuclei, on the line joining them	1 / 1
Pi (\pi)	Parallel (sideways) overlap of p orbitals	Above and below the plane of the bond axis	1 / 2

Strength of sigma and pi bonds

A sigma bond is stronger than a pi bond. In a sigma bond the electron density is in between the line joining the two nuclei, while in a pi bond the electron density is above and below the plane. A pi bond is therefore more diffuse, and less energy is required to break it.

3.6.1Applications of valence bond theoryOften examined

Single bond formation

(a) Formation of the \ce{H2} molecule

The electron configuration of a hydrogen atom is 1s^1. The half-filled s orbitals of two hydrogen atoms overlap to form the H–H bond, where the electron density is in between the line joining the two nuclei, so the bond formed is a sigma (\sigma) bond.

Figure 3.9: Orbital diagram of the H₂ molecule. The two half-filled 1s orbitals overlap to form an H–H σ bond.

(b) Formation of the \ce{F2} molecule

The electronic configuration of fluorine is 1s^2\,2s^2\,2p^5. The half-filled 2p_z orbital of each fluorine atom overlaps head on to form the F–F bond, which is a sigma bond.

Figure 3.10: Orbital diagram of the F₂ molecule. The half-filled 2pz orbitals of the two fluorine atoms overlap head on to form an F–F σ bond.

(c) Formation of the HF molecule

In the formation of the HF molecule, the half-filled 1s orbital of the hydrogen atom overlaps with the half-filled 2p_z orbital of the fluorine atom to form an H–F \sigma bond.

Figure 3.11: Orbital diagram of the H–F molecule. The half-filled 1s orbital of hydrogen overlaps with the half-filled 2pz orbital of fluorine to form an H–F σ bond.

Double bond formation: (d) the \ce{O2} molecule

In the oxygen molecule each oxygen atom has two half-filled 2p orbitals (2p_y and 2p_z), so according to valence bond theory two bonds are formed. One bond is due to the head-on overlap of two half-filled p orbitals, which is a sigma bond. The other is formed by the parallel overlap of half-filled p orbitals, which is a pi bond. The electronic configuration of an oxygen atom and the overlapping of the orbitals in the \ce{O2} molecule are shown in Figure 3.12.

Figure 3.12: Formation of σ and π bonds in the O₂ molecule. One pair of half-filled 2p orbitals overlaps head on (σ) and the other pair overlaps sideways (π).

Triple bond formation: (e) the \ce{N2} molecule

The nitrogen molecule contains two nitrogen atoms. Each nitrogen atom has three half-filled p orbitals, so according to VBT three bonds are formed. The half-filled 2p_x orbitals of the two atoms overlap head on and form a sigma bond, while the half-filled 2p_y and 2p_z orbitals of the two atoms overlap in parallel and form two pi bonds. Hence a triple bond is formed between the two nitrogen atoms. The configuration of the nitrogen atom and the overlapping in the nitrogen molecule are shown in Figure 3.13.

Figure 3.13: Orbital diagram of the N₂ molecule. One 2p pair overlaps head on (σ) and the other two pairs overlap sideways (two π bonds, one drawn dashed because its lobes are in front of and behind the page).

Applications of valence bond theory
Molecule	Orbitals that overlap	Bonds formed
\ce{H2}	1s–1s	One \sigma bond
\ce{F2}	2p_z–2p_z (head on)	One \sigma bond
HF	1s of H with 2p_z of F (head on)	One \sigma bond
\ce{O2}	2p–2p head on, and 2p–2p sideways	One \sigma and one \pi bond (double bond)
\ce{N2}	2p_x–2p_x head on; 2p_y–2p_y and 2p_z–2p_z sideways	One \sigma and two \pi bonds (triple bond)

3.7HybridizationOften examined

Hybridization is the concept of mixing atomic orbitals of different energies and different shapes to produce a new set of orbitals of the same energy and the same shape. The new orbitals are called hybrid orbitals, and the process is known as hybridization.

Hybridization: The mixing of atomic orbitals of different energies and different shapes to produce a new set of orbitals of the same energy and the same shape, called hybrid orbitals.

During hybridization, the electrons present in the ground state are promoted to the excited state, so the number of unpaired electrons increases. The energy required for the excitation of the electron is compensated by the energy released during bond formation.

There are several types of hybridization, depending on the nature of the atomic orbitals involved (s, p, d and f). Hybrid orbitals are very useful in explaining molecular geometry and atomic bonding properties.

Types of hybridization

There are many types of hybridization, for example sp, sp^2, sp^3, dsp^2, dsp^3, d^2sp^3 and d^3sp^3. The book discusses only three of them: sp, sp^2 and sp^3.

The three types of hybridization in the book
Type	Orbitals mixed	Hybrid orbitals formed	Shape and angle	Example
sp	one s + one p	2	Linear, 180^\circ	\ce{BeCl2}, \ce{C2H2}
sp^2	one s + two p	3	Trigonal planar, 120^\circ	\ce{BCl3}, \ce{C2H4}
sp^3	one s + three p	4	Tetrahedral, 109.5^\circ	\ce{CH4}, \ce{C2H6}, \ce{NH3}, \ce{H2O}

3.7.1sp hybridizationOften examined

The intermixing of **one s and one p orbital of different energies and different shapes to produce two new hybrid orbitals of the same energy and the same shape is known as sp hybridization**. An sp hybrid orbital has 50\% s-character and 50\% p-character. The sp hybrid orbitals can form only sigma bonds by overlapping with other atomic orbitals.

Examples of sp hybridization

1. Beryllium chloride, \ce{BeCl2}

The electronic configuration of Be in the ground state is 1s^2\,2s^2. Since there are no unpaired electrons, it undergoes excitation by promoting one of its 2s electrons into an empty 2p orbital. In the excited state the configuration of Be is 1s^2\,2s^1\,2p^1. In the excited state the beryllium atom undergoes sp hybridization by mixing a 2s and one 2p orbital, and two new hybrid orbitals are formed, which are arranged linearly.

When the two beryllium sp hybrid orbitals overlap with p orbitals of chlorine atoms, the bonding is as shown in Figure 3.14. Thus \ce{BeCl2} is linear in shape with a bond angle of 180^\circ.

Figure 3.14: Hybridization in BeCl₂, a linear molecule. One 2s electron of beryllium is promoted to 2p, the 2s and one 2p orbital mix to give two sp hybrid orbitals (accent), and each overlaps with a p orbital of chlorine.

2. Acetylene, \ce{C2H2}

In acetylene the central atoms are carbon. The ground state electronic configuration of carbon is 1s^2\,2s^2\,2p_x^1\,2p_y^1. There are only two unpaired electrons in the ground state. However, the valency of carbon is four, that is, it forms four bonds. To form four bonds, there must be four unpaired electrons. Hence carbon promotes one of its 2s electrons into the empty 2p_z orbital in the excited state. Thus, in the excited state, the electronic configuration of carbon is 1s^2\,2s^1\,2p_x^1\,2p_y^1\,2p_z^1.

Each carbon atom undergoes sp hybridization by using a 2s and one 2p orbital in the excited state to give two half-filled sp orbitals, which are arranged linearly. The two carbon atoms form a \sigma sp–sp bond with each other by using these orbitals. However, there are also two unhybridised p orbitals, that is, 2p_y and 2p_z, on each carbon atom, which are perpendicular to the sp hybrid orbitals. These orbitals form two \pi bonds between the two carbon atoms.

Thus a triple bond (one \sigma sp–sp bond and two \pi bonds) is formed between the carbon atoms. Each carbon atom forms a sigma bond with a hydrogen atom by the overlapping of sp–s. Thus the acetylene molecule is linear with a bond angle of 180^\circ.

Figure 3.15: Hybridized orbital diagram and linear structure of the acetylene molecule. Each carbon uses two sp orbitals (σ bonds) and two unhybridised p orbitals, which form the two π bonds.

Bonding in acetylene
Bond in \ce{C2H2}	Type	Orbitals overlapping
C–C (first bond)	\sigma	sp–sp
C–C (second and third bonds)	two \pi	2p_y–2p_y and 2p_z–2p_z
C–H (two)	\sigma	sp–s

3.7.2sp² hybridizationOften examined

The intermixing of **one s and two p orbitals having different energy and different shape to produce three hybrid orbitals of the same energy and the same shape is known as sp^2 hybridization**.

Examples of sp^2 hybridization

1. Boron trichloride, \ce{BCl3}

The electronic configuration of B in the ground state is 1s^2\,2s^2\,2p^1, with only one unpaired electron. The formation of three bonds with chlorine atoms requires three unpaired electrons, so one of the 2s electrons is promoted into the 2p_y orbital by absorbing energy. The excited state configuration of boron is 1s^2\,2s^1\,2p_x^1\,2p_y^1. In the excited state, boron undergoes sp^2 hybridization by using a 2s and two 2p orbitals to give three half-filled sp^2 hybrid orbitals, which are oriented in trigonal planar symmetry.

In \ce{BCl3} the three sp^2 hybridised orbitals of the boron atom overlap with the p orbitals of three chlorine atoms and form three \sigma sp^2–p bonds. Thus the shape of \ce{BCl3} is trigonal planar with bond angles equal to 120^\circ.

Figure 3.16: BCl₃, a trigonal planar molecule. One 2s electron of boron is promoted, the 2s and two 2p orbitals mix to give three sp² orbitals (accent) at 120°, and each overlaps with a p orbital of chlorine.

2. Ethylene (ethene), \ce{C2H4}

During the formation of the ethylene molecule, each carbon atom undergoes sp^2 hybridization in its excited state by mixing 2s and two 2p orbitals to give three half-filled sp^2 hybrid orbitals oriented in trigonal planar symmetry. There is also one half-filled unhybridised 2p orbital on each carbon, perpendicular to the plane of the sp^2 hybrid orbitals.

The carbon atoms form a \sigma bond with each other by using sp^2 hybrid orbitals. A \pi bond is also formed between them by the parallel overlapping of the unhybridised 2p_z orbitals. Thus there is a double bond (\sigma and \pi) between the two carbon atoms. Each carbon atom also forms two \sigma bonds between its sp^2 orbitals and the s orbitals of two hydrogen atoms.

Thus the ethylene molecule is planar, with H–C–H and H–C–C bond angles equal to 120^\circ. All the atoms are present in one plane.

Figure 3.17: C₂H₄, a planar molecule. Each carbon uses three sp² orbitals (accent) for σ bonds; the unhybridised 2p orbitals overlap sideways to give the π bond. All bond angles are 120°.

Bonding in ethene
Bond in \ce{C2H4}	Type	Orbitals overlapping
C=C (first bond)	\sigma	sp^2–sp^2
C=C (second bond)	\pi	2p_z–2p_z (sideways)
C–H (four)	\sigma	sp^2–s

3.7.3sp³ hybridizationOften examined

The intermixing of **one s and three p orbitals having different energies and different shapes to produce four hybrid orbitals of the same energy and the same shape is known as sp^3 hybridization**.

Mixing of one 2s and three 2p orbitals gives four equivalent sp³ hybrid orbitals (accent) of the same energy and shape.

Examples of sp^3 hybridization

1. Methane, \ce{CH4}

During the formation of the methane molecule the carbon atom undergoes sp^3 hybridization. In the excited state, by mixing one 2s and three 2p orbitals, four half-filled sp^3 hybrid orbitals are formed, which are oriented in tetrahedral symmetry in space around the carbon atom.

Each of these sp^3 hybrid orbitals forms a sigma bond with the s orbital of a hydrogen atom. Thus carbon forms four \sigma sp^3–s bonds with four hydrogen atoms. The methane molecule is tetrahedral in shape with a bond angle of 109.5^\circ.

Figure 3.18: CH₄, a tetrahedral molecule. Four sp³ orbitals of carbon (accent boxes) overlap with the 1s orbitals of four hydrogen atoms to form four σ sp³–s bonds.

2. Ethane, \ce{C2H6}

Just as in the methane molecule, each carbon atom in ethane undergoes sp^3 hybridization in the excited state to give four hybrid orbitals in tetrahedral geometry. The two carbon atoms form a \sigma sp^3–sp^3 bond with each other by the overlapping of sp^3 hybrid orbitals along the internuclear axis. The six carbon–hydrogen bonds are formed from overlaps between the remaining sp^3 orbitals on the two carbons and the 1s orbitals of the hydrogen atoms. All of these are sigma bonds.

Figure 3.19: C₂H₆, a molecule built of two tetrahedral carbon atoms joined by a σ sp³–sp³ bond. All bonds are σ bonds.

3. Ammonia, \ce{NH3}

The ground state electronic configuration of the nitrogen atom is 1s^2\,2s^2\,2p_x^1\,2p_y^1\,2p_z^1. Since there are three unpaired electrons in the 2p sublevel, the nitrogen atom can form three bonds with three hydrogen atoms.

In ammonia the nitrogen atom undergoes sp^3 hybridization of a 2s and three 2p orbitals to give four sp^3 orbitals, which are arranged in tetrahedral symmetry. This arrangement gives more stability to the molecule because the repulsions are minimised. Among them, three are half filled and one is full.

The three half-filled sp^3 orbitals of the nitrogen atom overlap with the s orbitals of three hydrogen atoms. There is also a lone pair on the nitrogen atom, in the full sp^3 hybrid orbital. It occupies more space than the bond pairs. The H–N–H bond angle is therefore not equal to the normal tetrahedral angle of 109.5^\circ: the reported bond angle is 107^\circ. The observed decrease in the bond angle is due to the repulsion caused by the lone pair over the bond pairs. That is why the ammonia molecule is trigonal pyramidal in shape.

Figure 3.20: NH₃, a trigonal pyramidal molecule. Three half-filled sp³ orbitals of nitrogen overlap with the 1s orbitals of hydrogen; the fourth sp³ orbital holds the lone pair (accent).

4. Water, \ce{H2O}

The electronic configuration of oxygen is 1s^2\,2s^2\,2p_x^2\,2p_y^1\,2p_z^1. There are two unpaired electrons in the oxygen atom, which may form bonds with hydrogen atoms. The experimental bond angle reported was 104.5^\circ. To account for this, sp^3 hybridization before the bond formation was proposed.

During the formation of the water molecule the oxygen atom undergoes sp^3 hybridization by mixing a 2s and three 2p orbitals to furnish four sp^3 hybrid orbitals oriented in tetrahedral geometry. Among them two are half filled and the remaining two are completely filled.

The oxygen atom now forms two \sigma sp^3–s bonds with hydrogen atoms by using the half-filled hybrid orbitals. The bond angle between H–O–H in the water molecule is 104.5^\circ. This is again due to the repulsions caused by the two lone pairs on the bond pairs. Thus the water molecule gets an angular shape (V shape).

Figure 3.21: Hybridization in the H₂O molecule. Oxygen uses four sp³ orbitals: two overlap with hydrogen 1s orbitals and two hold lone pairs (accent), so the molecule is bent.

Molecules with sp^3 hybridization
Molecule	Hybridization of the central atom	Bonding pairs	Lone pairs	Shape	Bond angle
\ce{CH4}	sp^3	4	0	Tetrahedral	109.5^\circ
\ce{C2H6} (each C)	sp^3	4	0	Tetrahedral	109.5^\circ
\ce{NH3}	sp^3	3	1	Trigonal pyramidal	107^\circ
\ce{H2O}	sp^3	2	2	Bent (V-shaped)	104.5^\circ

3.8Co-ordinate covalent bondOften examined

A co-ordinate covalent bond (or dative covalent bond) is formed when one atom provides both the electrons needed for the covalent bond; that is, the shared pair of electrons is donated by one atom only. For dative covalent bonding we need:

one atom that has a lone pair of electrons (the donor), and
a second atom that has an unfilled empty orbital to accept the lone pair (an electron-deficient atom or molecule).

Co-ordinate (dative) covalent bond: A covalent bond in which the shared pair of electrons is donated by one atom only. The donor has a lone pair; the acceptor has an empty orbital. It is shown by an arrow from the donor to the acceptor.

Hydronium ion, \ce{H3O^+}

During the formation of the hydronium ion, water, which has a lone pair of electrons, donates a pair of electrons and forms a dative bond with the electron-deficient hydrogen ion (proton).

\ce{H2O + H+ -> H3O+}

Figure 3.22: Formation of a co-ordinate covalent bond in the hydronium ion. Water donates a lone pair (accent) to the electron-deficient hydrogen ion H⁺.

Ozone, \ce{O3}, and carbon monoxide, \ce{CO}

In the ozone (\ce{O3}) molecule, the central oxygen atom donates its electron pair to the other oxygen atom and forms a dative bond.

A co-ordinate covalent bond is also present in the carbon monoxide molecule. In this molecule the oxygen atom forms a co-ordinate covalent bond with the carbon atom, so the triple bond \ce{C#O} is made of two ordinary covalent bonds and one dative bond.

The dative bond (accent arrow) in ozone and carbon monoxide. The arrow points from the atom that supplies the pair to the atom that receives it.

Donors and acceptors
Species	Donor	Acceptor
\ce{H3O+}	Oxygen of \ce{H2O} (lone pair)	\ce{H+} (empty 1s orbital)
\ce{O3}	Central oxygen atom	Terminal oxygen atom
\ce{CO}	Oxygen	Carbon

Concept assessment exercise 3.3: co-ordinate bonds

(1) Draw dot-and-cross diagrams to show the formation of a co-ordinate bond between: (i) boron trifluoride, \ce{BCl3}, and ammonia, \ce{NH3}, to form the compound \ce{Cl3BNH3}; (ii) phosphine, \ce{PH3}, and a hydrogen ion, \ce{H+}, to form the ion \ce{PH4+}. (2) Draw the displayed formulae of the products formed in part (1). Show the co-ordinate bond by an arrow.

(i) Find the donor and the acceptor. In \ce{BCl3} boron has only three shared pairs, that is 6 electrons, so it has an empty 2p orbital and is electron deficient: it is the acceptor. In \ce{NH3} nitrogen has a lone pair: it is the donor.
Show the reaction. The lone pair of nitrogen is shared with boron. Both electrons of this shared pair come from nitrogen (drawn as the accent pair), so boron now has 8 electrons around it.
\ce{BCl3 + :NH3 -> Cl3B <- NH3}
(ii) In \ce{PH3} phosphorus has a lone pair (the donor). The hydrogen ion \ce{H+} has no electron and an empty 1s orbital (the acceptor). Phosphorus shares its lone pair with \ce{H+}.
\ce{PH3 + H+ -> PH4+}
In the \ce{PH4+} ion all four P–H bonds look the same, but one of them was formed from the lone pair of phosphorus. The + charge is outside the brackets.

Dot-and-cross diagram for the formation of Cl₃B←NH₃. Boron in BCl₃ has only six electrons and an empty orbital; the lone pair of nitrogen (accent) forms the co-ordinate bond.

Dot-and-cross diagram for the formation of the PH₄⁺ ion. The lone pair of phosphorus (accent) is shared with H⁺.

Question 2: displayed formulae

Replace each shared pair by a line. Show the co-ordinate bond as an arrow pointing from the donor (N or P) to the acceptor (B or H).

Displayed formulae of the products. The co-ordinate bond is shown by an arrow (accent) pointing from the donor atom to the acceptor atom.

\ce{Cl3B<-NH3}\qquad\text{and}\qquad \ce{PH4+}

3.9Intermolecular forces (forces between molecules)Often examined

Intermolecular forces are the forces that are present between covalent molecules. Physical properties such as melting point, boiling point and density can be explained on the basis of intermolecular forces.

Intermolecular forces: Forces of attraction between covalent molecules (forces between molecules). They are much weaker than the covalent bonds inside a molecule (intramolecular forces).

Types of intermolecular forces

There are three types of intermolecular forces:

Permanent dipole–dipole forces
Instantaneous dipole induced dipole forces
Hydrogen bonding

1. Permanent dipole–dipole forces

These forces are present between polar molecules. Polar molecules are those in which, because of the electronegativity difference, one atom bears a partial positive charge and the other, which is more electronegative, bears a partial negative charge.

Some molecules with polar bonds have an overall dipole (for example \ce{H^{\delta+}-Cl^{\delta-}}), although not all do. In \ce{CCl4} the individual dipoles cancel each other out, so \ce{CCl4} is a nonpolar molecule. There are attractions between the permanent dipoles of neighbouring molecules, for example between \ce{H-Cl} molecules: \ce{H^{\delta+}-Cl^{\delta-}\cdots H^{\delta+}-Cl^{\delta-}}.

The stronger the dipole–dipole forces, the higher the melting point, boiling point, heat of vaporisation and heat of sublimation of the compound.

2. Instantaneous dipole induced dipole forces

These are the forces present between nonpolar molecules.

Instantaneous dipole induced dipole forces: The forces of attraction between nonpolar molecules, caused by a temporary (instantaneous) dipole in one molecule that induces a temporary dipole in a neighbouring molecule.

Even molecules with no polar bonds have temporary dipoles. This is because of the uneven distribution of electrons due to the constant movement of electrons. This induces a temporary dipole in a neighbouring molecule, and the temporary dipole induced dipole forces of attraction are produced.

Figure 3.23: Instantaneous dipole induced dipole forces. In (a) the charge clouds are symmetrical; in (b) more electron density is on the right-hand side of atom A (instantaneous dipole), which induces a dipole in atom B.

The strength of permanent dipole–dipole forces and instantaneous dipole induced dipole forces (together called van der Waals' forces) depends on the following factors:

Molecular size
Atomic size
Molecular mass
Number of electrons in the molecule

The bigger the molecule, that is, the more electrons it has, the larger its size or the higher its relative molecular mass, the greater the van der Waals' forces. Table 3.5 gives the boiling points of the first four alkanes.

Table 3.5: Boiling points of the first four alkanes
Name	Molecular formula	Boiling point (^\circ\mathrm{C})
Methane	\ce{CH4}	−164
Ethane	\ce{C2H6}	−89
Propane	\ce{C3H8}	−42
Butane	\ce{C4H10}	−1

For example, Table 3.5 shows that the boiling point of propane (\ce{C3H8}) is greater than that of ethane (\ce{C2H6}). The reason is that the molecular mass of \ce{C3H8} is 44, which is greater than the M_r of \ce{C2H6}, which is 30.

Worked example: boiling points of the alkanes against M_r

Using Table 3.5, work out M_r and the number of electrons of the first four alkanes, plot boiling point against M_r and explain the trend.

Add the relative atomic masses (\ce{C}=12, \ce{H}=1).
\ce{CH4}=16,\ \ce{C2H6}=30,\ \ce{C3H8}=44,\ \ce{C4H10}=58
Count the electrons: each carbon has 6 and each hydrogen 1.
10,\ 18,\ 26,\ 34
The boiling point rises by 75, 47 and 41\,^\circ\mathrm{C} from one alkane to the next, so it increases steadily with M_r.
-164\to-89\to-42\to-1
Alkanes are nonpolar, so only instantaneous dipole induced dipole forces act. These are stronger when the molecule is bigger, with more electrons and a higher M_r, so more energy is needed to separate the molecules.
Boiling does not break the covalent bonds inside the molecules, only the forces between them.

Boiling points of the first four alkanes

The boiling point rises as the relative molecular mass increases (plotted in kelvin, T=\theta+273, so that all values are positive).

\text{larger }M_r\Rightarrow\text{stronger forces between molecules}\Rightarrow\text{higher boiling point}

3. Hydrogen bonding

Hydrogen bonding is a special case of permanent dipole–dipole forces. It is present between molecules in which hydrogen is covalently bonded to a highly electronegative element such as F, N or O. There is a strong attraction between the lone pair on the N, O or F atom of one molecule and the exposed hydrogen nucleus of another molecule. This is simply a strong intermolecular force: it is not a bond.

Hydrogen bonding: A strong intermolecular force of attraction between a hydrogen atom covalently bonded to F, N or O in one molecule and the lone pair on an F, N or O atom of another molecule.

Covalent compounds that form hydrogen bonds with water molecules are soluble in water. For example, ammonia is soluble in water. Figure 3.24 shows the hydrogen bonding between ammonia and water molecules, and between like molecules.

Figure 3.24: Hydrogen bonding between water molecules, ammonia molecules, and ammonia and water. The dotted line joins the hydrogen (δ+) to the lone pair of N or O (δ−) of the next molecule.

The three types of intermolecular forces
Force	Present between	Example
Permanent dipole–dipole	Polar molecules	\ce{HCl}
Instantaneous dipole induced dipole	Nonpolar molecules (and all molecules)	\ce{CH4}, alkanes, halogens
Hydrogen bonding	H bonded to F, N or O, and a lone pair on F, N or O of another molecule	\ce{H2O}, \ce{NH3}, \ce{HF}

3.9.2Peculiar behaviour of hydrogen bondingOften examined

Ice is less dense than water

Most solids are denser than their liquids, because in the solid state the molecules are more closely packed. But this is not true of water. In ice there is a three-dimensional hydrogen-bonded network of water molecules. This produces a rigid lattice in which each oxygen atom is surrounded by a tetrahedron of hydrogen atoms. This more open arrangement, due to the relatively long hydrogen bonds, allows the water molecules to be slightly farther apart than in the liquid (Figure 3.25). Hence water in the solid state occupies more space, so the density of ice is less than that of liquid water. That is why ice floats on water.

Figure 3.25: A model of ice. Each oxygen atom is surrounded tetrahedrally by four hydrogen atoms: two joined to it by short covalent bonds and two joined by long hydrogen bonds (dotted, accent) from neighbouring molecules. This open arrangement makes ice less dense than water.

The strength of intermolecular forces

\text{hydrogen bonding} > \text{permanent dipole–dipole} > \text{instantaneous dipole induced dipole forces}

Table 3.6: Comparison of boiling points of different compounds
Molecule	\ce{CH4}	\ce{HCl}	\ce{H2O}
Boiling point (^\circ\mathrm{C})	−162	−85	100
Intermolecular forces	Instantaneous dipole induced dipole forces	Permanent dipole–dipole forces	Hydrogen bonding

Table 3.6: Comparison of boiling points of different compounds
Molecule	Boiling point (^\circ\mathrm{C})	Intermolecular forces
\ce{CH4}	−162	Instantaneous dipole induced dipole forces
\ce{HCl}	−85	Permanent dipole–dipole forces
\ce{H2O}	100	Hydrogen bonding

Methane has only instantaneous dipole induced dipole forces and boils at -162\,^\circ\mathrm{C}. Hydrogen chloride has permanent dipole–dipole forces and boils at -85\,^\circ\mathrm{C}. Water has hydrogen bonding and boils at 100\,^\circ\mathrm{C}.

The melting and boiling points of covalent molecules depend upon the intermolecular forces, not on the covalent bonds in the molecule. Covalent bonds are very strong (values in hundreds of \mathrm{kJ\,mol^{-1}}). The forces between molecules are much weaker than the covalent bond. Figure 3.26 shows the boiling points of the hydrides of groups 14, 15, 16 and 17 (groups 4, 5, 6 and 7 in the book's numbering).

Figure 3.26: Boiling points of the hydrides of group 14, 15, 16 and 17 elements (groups 4, 5, 6 and 7 in the book's numbering) against the period of the central element, plotted in kelvin (T=\theta+273 with \theta in ^\circ\mathrm{C}) so that all values are positive. Group 16 is the accent line, group 17 ink, group 15 blue and group 14 green. Water, hydrogen fluoride and ammonia lie far above the trend of the other hydrides of their group.

Concept assessment exercise 3.4: boiling points and hydrogen bonds

(1a) The boiling points of the halogens are: fluorine -188\,^\circ\mathrm{C}, chlorine -35\,^\circ\mathrm{C}, bromine +59\,^\circ\mathrm{C}, iodine +184\,^\circ\mathrm{C}. Explain the trend in these boiling points. (1b) The table lists the formulae and boiling points of some alkanes: methane \ce{CH4} -164\,^\circ\mathrm{C}; ethane \ce{CH3CH3} -88\,^\circ\mathrm{C}; propane \ce{CH3CH2CH3} -42\,^\circ\mathrm{C}; butane \ce{CH3CH2CH2CH3} 0\,^\circ\mathrm{C}. Explain this trend. (2) Bromine, \ce{Br2}, and iodine monochloride, \ce{ICl}, have the same number of electrons. But the boiling point of iodine monochloride is nearly 40\,^\circ\mathrm{C} higher than the boiling point of bromine. Explain this difference. (3) Draw diagrams to show hydrogen bonding between the following molecules: (a) ethanol and water, (b) ammonia and water, (c) two HF molecules.

Question 1(a): the halogens

Halogen molecules (\ce{F2}, \ce{Cl2}, \ce{Br2}, \ce{I2}) are nonpolar, so the only forces between them are instantaneous dipole induced dipole forces.
Down the group the molecules get larger, with more electrons and a higher relative molecular mass, so the electron cloud is easier to distort and the instantaneous dipole induced dipole forces become stronger.
More energy is needed to separate the molecules, so the boiling point rises from -188\,^\circ\mathrm{C} (\ce{F2}, a gas) to +184\,^\circ\mathrm{C} (\ce{I2}, a solid that sublimes easily).

Question 1(b): the alkanes

Alkanes are nonpolar, so they have only instantaneous dipole induced dipole forces.
Each member is larger than the one before: relative molecular masses 16, 30, 44 and 58, with more electrons in each molecule.
The forces between the molecules therefore become stronger from methane to butane, so the boiling point rises (-164 \to -88 \to -42 \to 0\,^\circ\mathrm{C}).

Question 2: Br₂ and ICl

Both molecules have the same number of electrons (2\times35=70 for \ce{Br2} and 53+17=70 for \ce{ICl}) and similar masses, so their instantaneous dipole induced dipole forces are similar.
\ce{Br2} is a nonpolar molecule (two identical atoms). \ce{ICl} is polar, because chlorine is more electronegative than iodine, so \ce{ICl} also has permanent dipole–dipole forces between its molecules.
The extra permanent dipole–dipole attraction needs more energy to overcome, so \ce{ICl} has the higher boiling point (about 40\,^\circ\mathrm{C} more).

Question 3: hydrogen bonds

In each diagram a hydrogen atom bonded to O, N or F (it carries \delta+) is joined by a dotted line to the lone pair on the O, N or F atom (\delta-) of the other molecule.
(a) Ethanol and water: the O–H hydrogen of ethanol bonds to the oxygen of water. (b) Ammonia and water: the hydrogen of water bonds to the lone pair on nitrogen. (c) Two HF molecules: the hydrogen of one HF bonds to the fluorine of the other.

Hydrogen bonding in (a) ethanol and water, (b) ammonia and water and (c) two HF molecules. In each, the dotted line joins H(δ+) to the lone pair of the O, N or F of the other molecule.

\ce{F2}<\ce{Cl2}<\ce{Br2}<\ce{I2};\quad \ce{CH4}<\ce{C2H6}<\ce{C3H8}<\ce{C4H10};\quad \ce{ICl}>\ce{Br2}\ (\text{dipole–dipole})

3.9.3Comparison of the strength of ionic, covalent, metallic bonding and intermolecular forces

The strength of a bond depends on the specific elements involved.

Ionic bonds contain strong electrostatic forces of attraction between positive and negative ions.
Metallic bonds occur between the atoms within a metal, where electrons are shared and free to move throughout the material. This creates a "sea" of delocalised electrons that hold the metal atoms together.
Covalent bonds are formed by the sharing of electrons. This sharing results in the formation of a strong bond between the atoms. The electrons are held in the region between the nuclei of the bonded atoms, and the positive nuclei are attracted to these shared electrons. This leads to a stable and strong bonding interaction.

Metallic bonding: positive metal ions held together by a sea of delocalised electrons that are free to move through the metal.

Intermolecular forces are weaker than ionic, covalent and metallic bonding and occur between molecules, not within a single molecule. These forces hold molecules together in a substance, but they are much weaker than the forces within a molecule (the covalent bonds).

Table 3.7: Melting and boiling points of ionic, simple covalent, giant covalent and metallic substances
Substance	Bonding / structure	Melting point (^\circ\mathrm{C})	Boiling point (^\circ\mathrm{C})
Sodium chloride	Ionic (electrostatic forces)	801	1465
Diamond	Giant covalent (covalent bond)	4000	4830
Iron	Metallic (electrostatic forces)	1538	2862
Methane	Simple covalent (intermolecular forces)	−182	−161.6

It is obvious from Table 3.7 that the melting point and boiling point of methane are lower than those of the other substances, because it contains weak intermolecular forces, which require less energy to break.

Strength of bonds compared with intermolecular forces
Type of attraction	Between	Strength
Ionic bonding	Positive and negative ions	Strong
Covalent bonding	Atoms sharing electron pairs	Strong
Metallic bonding	Metal ions and delocalised electrons	Strong
Intermolecular forces	Separate molecules	Much weaker

3.10Molecular orbital theoryOften examined

The molecular orbital theory (MOT) was proposed by Hund and Mulliken in 1932. According to this theory, the linear combination of atomic orbitals forms new orbitals called molecular orbitals, which are characteristic of the whole molecule. In this process the identity of both atomic orbitals is lost. After the combination of atomic orbitals, different types of molecular orbitals are formed, which differ in energy. The one of lower energy is called the bonding molecular orbital (BMO), while the other, of higher energy, is called the anti-bonding molecular orbital (ABMO).

Figure 3.27: Energy diagram of bonding and anti-bonding molecular orbitals. Two 1s atomic orbitals combine to give a bonding orbital σ1s of lower energy (overlapping lobes) and an anti-bonding orbital σ*1s of higher energy (separated lobes).

Molecular orbital theory: A bonding theory in which the atomic orbitals of the bonded atoms combine linearly to form molecular orbitals that belong to the whole molecule. Each combination gives a lower-energy bonding orbital and a higher-energy anti-bonding orbital.

The paramagnetic or diamagnetic nature of a molecule can be explained on the basis of molecular orbital theory:

Molecules having unpaired electrons in their molecular orbitals are paramagnetic in nature and can be attracted by a magnetic field.
Molecules having no unpaired electrons in their molecular orbitals are diamagnetic in nature and cannot be attracted by a magnetic field. They repel a magnetic field.

Magnetic behaviour
Property	Paramagnetic	Diamagnetic
Unpaired electrons in the molecular orbitals	Present	Absent
In a magnetic field	Attracted	Repelled (not attracted)
Example	\ce{O2}	\ce{H2}, \ce{N2}, \ce{F2}

3.10.1Bonding and antibonding molecular orbitalsOften examined

Bonding molecular orbital (BMO)

A bonding molecular orbital is formed by the overlap (addition) of atomic orbitals. A BMO has a high electron density in the region between the two nuclei, which is responsible for the stability of the bond. Each electron in the BMO contributes to the attraction between the two atoms. Bonding molecular orbitals are designated \sigma and \pi.

Anti-bonding molecular orbital (ABMO)

An anti-bonding molecular orbital is formed by the subtraction overlap of atomic orbitals. It has zero electron density in between the two nuclei, and the electron density is concentrated on the opposite sides. Each electron in an ABMO contributes to the repulsion between the two atoms. Anti-bonding molecular orbitals are designated \sigma^* and \pi^*.

BMO and ABMO compared
	Bonding MO	Anti-bonding MO
Formed by	Overlap (addition) of atomic orbitals	Subtraction overlap of atomic orbitals
Energy	Lower than the atomic orbitals	Higher than the atomic orbitals
Electron density between the nuclei	High	Zero
Effect of electrons	Attraction between the atoms (stability)	Repulsion between the atoms
Designation	\sigma, \pi	\sigma^, \pi^

The number of molecular orbitals formed is equal to the number of atomic orbitals that are combined. The electrons in molecular orbitals are influenced by all the nuclei, but in an atomic orbital an electron is influenced by one nucleus.

The filling of electrons into the molecular orbitals occurs according to:

Aufbau principle
Pauli's exclusion principle
Hund's rule

Two types of overlapping in molecular orbital theory

Head-on approach (linear overlapping or linear combination)
Sideways approach (parallel overlapping)

Head-on approach or linear overlapping takes place by the combination of s–s, s–p and p_x–p_x orbitals. When two s atomic orbitals overlap with each other, two molecular orbitals are formed: \sigma s of low energy and \sigma^* s of high energy.

Sideways approach or parallel overlapping of atomic orbitals takes place between p_y–p_y and p_z–p_z. When two p orbitals p_y or p_z overlap with each other, two molecular orbitals are formed: \pi p_y=\pi p_z of low energy and \pi^*p_y=\pi^*p_z of high energy.

When the three 2p atomic orbitals (p_x, p_y and p_z) of one atom overlap with the three 2p atomic orbitals of the other atom, six molecular orbitals (three bonding and three anti-bonding) are formed. According to their energy, these molecular orbitals are (in the order of increasing energy):

\sigma 2p_x < \pi 2p_y=\pi 2p_z < \pi^* 2p_y=\pi^* 2p_z < \sigma^* 2p_x

3.10.2Bond order (number of bonds)Often examined

Bond order is the total number of bonds formed when the atomic orbitals of two atoms overlap with each other. It is half of the difference between the number of bonding electrons and the number of anti-bonding electrons.

\text{Bond order}=\dfrac{N_b-N_a}{2}

Bond order

N_b is the number of electrons in bonding molecular orbitals (BMOs) and N_a is the number of electrons in anti-bonding molecular orbitals (ABMOs).

Bond order: Half of the difference between the number of electrons in bonding molecular orbitals and the number of electrons in anti-bonding molecular orbitals; it is the number of bonds between the two atoms.

Bond order of the hydrogen molecule

Calculate the number of bonds formed in the hydrogen molecule.

Count the electrons in the bonding orbital. Both electrons of \ce{H2} are in \sigma1s.
N_b=2
Count the electrons in the anti-bonding orbital. \sigma^*1s is empty.
N_a=0
Apply the formula.
\text{Bond order}=\dfrac{2-0}{2}=1

\text{Bond order of }\ce{H2}=1

3.10.3Relative energies of the molecular orbitalsOften examined

The relative energies of the molecular orbitals formed from 2s and 2p atomic orbitals are determined by spectroscopic measurements.

(a) Heavier diatomic molecules: \ce{O2}, \ce{F2} and their ions

The following increasing order of energy shows the molecular orbitals of diatomic molecules such as \ce{O2} and \ce{F2} and their positive and negative ions:

\sigma1s<\sigma^*1s<\sigma2s<\sigma^*2s<\sigma2p_x<\pi2p_y=\pi2p_z<\pi^*2p_y=\pi^*2p_z<\sigma^*2p_x

(b) Lighter diatomic molecules: \ce{H2}, \ce{He2}, \ce{B2}, \ce{C2} and \ce{N2}

The lighter molecules such as \ce{H2}, \ce{He2}, \ce{B2}, \ce{C2} and \ce{N2} show a slightly different energy order:

\sigma1s<\sigma^*1s<\sigma2s<\sigma^*2s<\pi2p_y=\pi2p_z<\sigma2p_x<\pi^*2p_y=\pi^*2p_z<\sigma^*2p_x

The elements of the second period with atomic number 3 to 7 (before oxygen) have a relatively small difference in energy between the 2s and 2p orbitals. As the energy difference between the 2s and 2p atomic orbitals is small, the \sigma2s and \sigma^*2s molecular orbitals do not retain pure s-character (hybridization of the atomic orbitals). The mixing of 2s and 2p orbitals is the primary cause of the difference between the molecular orbitals of nitrogen and oxygen, which is influenced by the initial atomic orbital energies. Hence for these lighter molecules the energy of \sigma2p_x is greater than that of \pi2p_y and \pi2p_z.

The two energy orders
Molecules	Order of \sigma2p_x and \pi2p	Full order of energy
\ce{O2}, \ce{F2} and their ions	\sigma2p_x is below \pi2p_y=\pi2p_z	\sigma1s<\sigma^1s<\sigma2s<\sigma^2s<\sigma2p_x<\pi2p<\pi^2p<\sigma^2p_x
\ce{H2}, \ce{He2}, \ce{B2}, \ce{C2}, \ce{N2}	\sigma2p_x is above \pi2p_y=\pi2p_z	\sigma1s<\sigma^1s<\sigma2s<\sigma^2s<\pi2p<\sigma2p_x<\pi^2p<\sigma^2p_x

3.10.4Molecular orbital diagram of homonuclear diatomic moleculesOften examined

After discussing the basic principles of molecular orbital theory, we can take up the electronic structures and bonding properties of some homonuclear diatomic molecules.

1. Hydrogen molecule, \ce{H2}

The hydrogen molecule is formed from the overlap of the 1s atomic orbitals of two hydrogen atoms. They give rise to two molecular orbitals, \sigma1s and \sigma^*1s. The molecule has two electrons, which occupy the lower-energy \sigma1s orbital, as shown in Figure 3.28.

Figure 3.28: Energy diagram of the BMO and ABMO of H₂. The two electrons occupy the lower σ1s orbital with opposite spins.

Total number of electrons =2 (the electronic configuration of the hydrogen atom is 1s^1).
Configuration of the molecule.
\ce{H2}=(\sigma1s)^2,\ (\sigma^*1s)^0
Bond order.
\dfrac{2-0}{2}=1
Magnetic character: there is no unpaired electron, so \ce{H2} is diamagnetic.

2. Helium molecule (hypothetical), \ce{He2}

The energy level diagram for \ce{He2} is similar to that of \ce{H2}, except that it has two more electrons, which occupy the anti-bonding \sigma^*1s orbital, as shown in Figure 3.29.

Figure 3.29: Energy diagram of the BMO and ABMO of He₂. Both σ1s and σ*1s are filled, so the anti-bonding electrons cancel the bonding electrons.

Total number of electrons =4 (the electronic configuration of helium is 1s^2).
Configuration of the molecule.
\ce{He2}=(\sigma1s)^2,\ (\sigma^*1s)^2
Bond order.
\dfrac{2-2}{2}=0
Magnetic character: there is no unpaired electron in the molecular orbitals of \ce{He2}, so it is diamagnetic.
As the bond order of \ce{He2} is zero, the molecule is unstable and does not exist. Helium exists only as monoatomic molecules.

3. Nitrogen molecule, \ce{N2}

Nitrogen has three unpaired electrons in the 2p orbitals. The electronic configuration of nitrogen is 1s^2\,2s^2\,2p_x^1\,2p_y^1\,2p_z^1. The total number of electrons in the \ce{N2} molecule is 14. Figure 3.30 represents the molecular orbitals of the nitrogen molecule.

Figure 3.30: Energy diagram of the BMO and ABMO of N₂ (valence orbitals only; the filled σ1s and σ1s core orbitals are not drawn). Here π2py = π2pz lie below σ2px. All bonding levels up to σ2px are full and the anti-bonding π levels are empty.

The distribution of electrons in the molecular orbitals of the nitrogen molecule is:

\ce{N2}=(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\pi2p_y)^2=(\pi2p_z)^2(\sigma2p_x)^2

Count the bonding electrons: \sigma1s (2), \sigma2s (2), \pi2p_y and \pi2p_z (4) and \sigma2p_x (2).
N_b=2+2+4+2=10
Count the anti-bonding electrons: \sigma^*1s (2) and \sigma^*2s (2).
N_a=2+2=4
Bond order.
\dfrac{10-4}{2}=3
Magnetic character: there is no unpaired electron in the molecular orbitals of \ce{N2}, so it is diamagnetic.

The bond order of \ce{N2} is three: the two nitrogen atoms are bonded through a triple bond. The \ce{N2} molecule is very stable and has a very high bond energy of 946\,\mathrm{kJ\,mol^{-1}}. It is diamagnetic in nature and has a very short bond length.

4. Oxygen molecule, \ce{O2}

The electronic configuration of oxygen is 1s^2\,2s^2\,2p_x^2\,2p_y^1\,2p_z^1. Oxygen contains two unpaired electrons in the 2p orbitals, that is 2p_y and 2p_z. The total number of electrons in the \ce{O2} molecule is 16. Figure 3.31 represents the molecular orbitals of the oxygen molecule.

Figure 3.31: Energy diagram of the BMO and ABMO of O₂ (valence orbitals only). The two electrons in π*2py and π*2pz are unpaired (accent levels), so O₂ is paramagnetic.

\ce{O2}=(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_x)^2(\pi2p_y)^2=(\pi2p_z)^2(\pi^*2p_y)^1=(\pi^*2p_z)^1

Bonding electrons: \sigma1s (2), \sigma2s (2), \sigma2p_x (2), \pi2p_y and \pi2p_z (4).
N_b=10
Anti-bonding electrons: \sigma^*1s (2), \sigma^*2s (2), \pi^*2p_y (1) and \pi^*2p_z (1).
N_a=6
Bond order.
\dfrac{10-6}{2}=2

Paramagnetic nature of the oxygen molecule

It is found from the molecular orbital diagram of the oxygen molecule that the electrons present in the two \pi^* orbitals are unpaired. The two unpaired electrons reside in the degenerate anti-bonding orbitals \pi^*2p_y and \pi^*2p_z. These unpaired electrons not only go around the atom in their orbitals but also spin, which creates a magnetic field. The unpaired electrons spin in the same direction, which increases the magnetic field effect. Due to the presence of these unpaired electrons, \ce{O2} is found to be paramagnetic in nature.

The paramagnetic behaviour of the \ce{O2} molecule is not explained by valence bond theory. It is obvious that the two oxygen atoms are bonded through a double bond, so we conclude that the molecule should be very stable, as it has a high bond energy, 498\,\mathrm{kJ\,mol^{-1}}, with a bond length of 1.21\,\mathrm{\AA}.

5. Fluorine molecule, \ce{F2}

The electronic configuration of fluorine is 1s^2\,2s^2\,2p_x^2\,2p_y^2\,2p_z^1. The total number of electrons in the \ce{F2} molecule is 18. Figure 3.32 represents the molecular orbitals of the fluorine molecule.

Figure 3.32: Energy diagram of the BMO and ABMO of F₂ (valence orbitals only). Every level is full except σ*2px; all electrons are paired, so F₂ is diamagnetic.

\ce{F2}=(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_x)^2(\pi2p_y)^2=(\pi2p_z)^2(\pi^*2p_y)^2=(\pi^*2p_z)^2

Bonding electrons.
N_b=2+2+2+4=10
Anti-bonding electrons: \sigma^*1s (2), \sigma^*2s (2), \pi^*2p_y (2) and \pi^*2p_z (2).
N_a=8
Bond order.
\dfrac{10-8}{2}=1
Magnetic character: all the electrons in the molecular orbitals of \ce{F2} are paired, so it is diamagnetic.

The bond order of \ce{F2} is one: the two F atoms are bonded through a single bond. The F–F bond distance (1.43\,\mathrm{\AA}) is longer than the bond distances in \ce{O2} (1.21\,\mathrm{\AA}) and \ce{N2} (1.09\,\mathrm{\AA}). The bond energy of the \ce{F2} molecule is quite low (159\,\mathrm{kJ\,mol^{-1}}). It is diamagnetic in nature because it has no unpaired electrons.

Summary: molecular orbital configurations of the molecules in the book
Molecule	Electrons	Electron configuration	Bond order	Magnetic character
\ce{H2}	2	(\sigma1s)^2	\dfrac{2-0}{2}=1	Diamagnetic
\ce{He2}	4	(\sigma1s)^2(\sigma^*1s)^2	\dfrac{2-2}{2}=0	Diamagnetic (does not exist)
\ce{N2}	14	(\sigma1s)^2(\sigma^1s)^2(\sigma2s)^2(\sigma^2s)^2(\pi2p_y)^2=(\pi2p_z)^2(\sigma2p_x)^2	\dfrac{10-4}{2}=3	Diamagnetic
\ce{O2}	16	(\sigma1s)^2(\sigma^1s)^2(\sigma2s)^2(\sigma^2s)^2(\sigma2p_x)^2(\pi2p_y)^2=(\pi2p_z)^2(\pi^2p_y)^1=(\pi^2p_z)^1	\dfrac{10-6}{2}=2	Paramagnetic
\ce{F2}	18	(\sigma1s)^2(\sigma^1s)^2(\sigma2s)^2(\sigma^2s)^2(\sigma2p_x)^2(\pi2p_y)^2=(\pi2p_z)^2(\pi^2p_y)^2=(\pi^2p_z)^2	\dfrac{10-8}{2}=1	Diamagnetic

The higher the bond order, the shorter and stronger the bond
Molecule	Bond order	Bond length	Bond energy (\mathrm{kJ\,mol^{-1}})
\ce{N2}	3	1.09\,\mathrm{\AA} (shortest)	946 (very high)
\ce{O2}	2	1.21\,\mathrm{\AA}	498
\ce{F2}	1	1.43\,\mathrm{\AA} (longest)	159 (quite low)

Worked example: molecular orbital diagram of \ce{B2}

Write the molecular orbital configuration of \ce{B2}, find its bond order and say whether it is paramagnetic or diamagnetic.

Boron has 5 electrons, so \ce{B2} has 10.
2\times5=10
\ce{B2} is one of the lighter molecules, so \pi2p lies below \sigma2p_x. Fill the orbitals in order of energy.
\sigma1s<\sigma^*1s<\sigma2s<\sigma^*2s<\pi2p_y=\pi2p_z<\sigma2p_x
The first eight electrons fill \sigma1s, \sigma^*1s, \sigma2s and \sigma^*2s. The last two go one each into the degenerate \pi2p_y and \pi2p_z orbitals with parallel spins (Hund's rule).
\ce{B2}=(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\pi2p_y)^1=(\pi2p_z)^1
Count bonding electrons (\sigma1s 2, \sigma2s 2, \pi2p 2) and anti-bonding electrons (\sigma^*1s 2, \sigma^*2s 2).
N_b=6,\quad N_a=4
Bond order.
\dfrac{6-4}{2}=1
Two unpaired electrons are present, so \ce{B2} is paramagnetic.

Valence molecular orbital diagram of \ce{B2} (the filled 1s levels are not drawn). The last two electrons (accent) occupy the two \pi orbitals singly.

\text{Bond order}=1;\ \ce{B2}\ \text{is paramagnetic}

Formula sheet

Dipole moment

\mu = q \times r

q is the magnitude of the charge on either end of the bond and r is the distance between the charges (the bond length).

Unit: debye (\mathrm{D})

Bond order

\text{bond order}=\frac{N_b-N_a}{2}

N_b is the number of electrons in bonding molecular orbitals and N_a the number in anti-bonding molecular orbitals.

no unit

Key terms

Electronegativity: The power of an atom to attract the shared electrons in a bond towards itself.
Dipole moment: The product of the charge separated and the distance between the charges, \mu = q \times r; a measure of the polarity of a bond or molecule.
Bond energy: The energy needed to break one mole of a particular bond in gaseous molecules; a larger value means a stronger bond.
Bond length: The average distance between the nuclei of two bonded atoms.
VSEPR theory: Electron pairs around a central atom repel one another and so take positions that keep them as far apart as possible, which fixes the shape of the molecule.
Expanded octet: A central atom with more than eight electrons in its valence shell, as in \ce{PCl5} and \ce{SF6}.
Valence bond theory: A covalent bond forms when half-filled atomic orbitals overlap and the two electrons of opposite spin are shared.
Sigma bond: A bond formed by head-on overlap of orbitals along the line joining the nuclei.
Pi bond: A bond formed by sideways overlap of parallel p orbitals above and below the bond axis.
Hybridization: The mixing of atomic orbitals of different energies on one atom to give the same number of equivalent hybrid orbitals.
Co-ordinate (dative) bond: A covalent bond in which both shared electrons come from one atom, the donor, and the other atom, the acceptor, supplies an empty orbital.
Intermolecular forces: The attractions between molecules, which are much weaker than ionic, covalent and metallic bonds.
van der Waals' forces: The weak intermolecular forces: dipole-dipole, dipole-induced dipole and London dispersion forces.
Hydrogen bond: A strong dipole-dipole attraction between a hydrogen atom bonded to \ce{F}, \ce{O} or \ce{N} and a lone pair on another such atom.
Molecular orbital theory: Atomic orbitals combine to form molecular orbitals that belong to the whole molecule and are filled by the electrons of all the atoms.
Bonding molecular orbital (BMO): A molecular orbital of lower energy than the atomic orbitals, formed by in-phase overlap, which holds electron density between the nuclei.
Anti-bonding molecular orbital (ABMO): A molecular orbital of higher energy than the atomic orbitals, formed by out-of-phase overlap, which has a node between the nuclei.
Bond order: Half the difference between the number of electrons in bonding and in anti-bonding molecular orbitals.
Paramagnetic: Attracted by a magnetic field because the species has unpaired electrons.
Diamagnetic: Slightly repelled by a magnetic field because all electrons are paired.

Multiple choice questionsp. 74–77

The book prints no answer key for this exercise, so every key below was worked out from the unit's teaching pages. Each reason gives the rule that decides the answer and says why the tempting option is wrong.

Which of the following molecules is nonpolar?

A\ce{CCl2F2}
B\ce{CHCl3}
C\ce{C2Cl4}Answer
D\ce{C2H5Cl}

Why

\ce{C2Cl4} is nonpolar. Each carbon carries two identical chlorine atoms, and the molecule is planar and symmetrical, so the four \ce{C-Cl} bond dipoles point in directions that cancel. The net dipole moment is zero.

tetrachloroethene (\ce{C2Cl4})

\ce{CCl2F2}, \ce{CHCl3} and \ce{C2H5Cl} are polar. Around the central carbon their atoms are not all alike, so the bond dipoles do not cancel. The book's rule is that symmetric molecules are nonpolar and unsymmetrical ones are polar.

Which is the most likely shape of hydrazine, \ce{N2H4}?

AEach N drawn with two H atoms above and below, and a 90° angle at the N
BA chain H-N-N-H with one H above each N, and a 90° angle at the N
CEach N drawn with two H atoms above and below, and a 120° angle at the N
DEach N drawn with two H atoms, and a 107° angle at the NAnswer

Why

Each nitrogen in hydrazine has three bonding pairs (two N-H and one N-N) and one lone pair. This is the AX_3E arrangement of ammonia, so each N is pyramidal with an angle close to 107°. Option D shows this.

A and B give a 90° angle, which is far too small for an N atom with four electron pairs. C gives 120°, which is the angle of a trigonal planar atom with no lone pair, and that is not the arrangement around N here.

Note. The four options are drawn structures in the book. The options above are described in words; the answer is the one with the 107° angle at N.

iii

Which molecule contains only six bonding electrons?

A\ce{C2H4}
B\ce{C2F6}
C\ce{H2O}
D\ce{NF3}Answer

Why

Count the electrons in the bonds: each single bond holds 2 electrons, and a double bond holds 4. Lone pairs are not bonding electrons.

\ce{NF3}: three N-F single bonds.
3\times2=6\ \text{bonding electrons}
\ce{H2O}: two O-H bonds.
2\times2=4
\ce{C2H4}: one C=C double bond and four C-H bonds.
4+4\times2=12
\ce{C2F6}: one C-C bond and six C-F bonds.
7\times2=14

Only \ce{NF3} has six bonding electrons. The lone pair on N is not counted.

Which molecule is trigonal planar in shape?

A\ce{NF3}
B\ce{C2Cl4}Answer
C\ce{C3H6}
D\ce{C3H8}

Why

In \ce{C2Cl4} each carbon is bonded to three atoms (two Cl and the other carbon through a double bond) and has no lone pair. Three electron pairs repel each other to 120° and lie in one plane, so the arrangement around each carbon is trigonal planar.

tetrachloroethene (\ce{C2Cl4})

\ce{NF3} has a lone pair on N (AX_3E), so it is trigonal pyramidal. The carbons in \ce{C3H8} each have four bonding pairs and are tetrahedral. \ce{C3H6} is not trigonal planar around every atom, so it is not the answer.

Note. The formula \ce{C3H6} also fits propene, where only two of the three carbons are trigonal planar. The answer is B because it is the only molecule that is trigonal planar around every central carbon.

Which of the following orbitals overlap with each other in the \ce{PH3} molecule?

Asp^2-s
Bsp-s
Csp^3-p
Dsp^3-sAnswer

Why

\ce{PH3} has the same AX_3E arrangement as ammonia. The phosphorus atom uses sp^3 hybrid orbitals. Three of them are half-filled and overlap head-on with the 1s orbital of each hydrogen atom, giving three \sigma sp^3-s bonds. The fourth sp^3 orbital holds the lone pair.

sp^2 and sp hybrids do not describe the phosphorus atom in this molecule. Hydrogen has no p orbitals, so an sp^3-p overlap cannot occur.

The diagram shows the variation of the boiling points of the hydrogen halides. What explains the higher boiling point of hydrogen fluoride?

AThe bond energy of HF molecules is greater than in other hydrogen halides.
BThe effect of nuclear shielding is much reduced in fluorine, which polarizes the HF molecule.
CThe electronegativity of fluorine is much higher than for other elements in the group.
DThere is hydrogen bonding between HF molecules.Answer

Why

Boiling separates molecules from each other. It does not break the covalent bonds inside them, so the bond energy (option A) is not the reason. In HF the hydrogen is bonded to the very electronegative fluorine, so each H atom is attracted to a lone pair on a neighbouring F atom. This hydrogen bonding is much stronger than the dipole-dipole and van der Waals forces between HCl, HBr and HI molecules, so much more energy is needed to separate HF molecules.

Option C is true of fluorine but is not the reason for the high boiling point; it only explains why the hydrogen bonds are strong. Option B describes an effect that does not change the forces between molecules.

vii

Which graph correctly describes a trend found in the halogen group?

ABond length in \ce{X2} decreases from \ce{Cl2} to \ce{I2}.
BStrength of van der Waals' forces increases from \ce{Cl2} to \ce{I2}.Answer
CBoiling point of \ce{X2} decreases from \ce{Cl2} to \ce{I2}.
DBond energy of \ce{HX} increases from \ce{HCl} to \ce{HI}.

Why

The strength of van der Waals' forces grows with the size and number of electrons of a molecule, so it rises from \ce{Cl2} to \ce{I2} (graph B).

Graph A is wrong: the atoms get bigger down the group, so the bond length in \ce{X2} increases. Table 3.3 gives \ce{Cl-Cl} 199 pm, \ce{Br-Br} 228 pm and \ce{I-I} 266 pm.
Graph C is wrong: the boiling point of the halogens rises down the group, so it does not fall.
Graph D is wrong: the bond energy of the hydrogen halides falls from HCl (431 kJ mol^{-1}) to HI (299 kJ mol^{-1}), as in Table 3.2.

viii

sp^3 hybridization is not important in describing the bonding in:

A\ce{NH4+}
B\ce{CCl4}
C\ce{H2O}
D\ce{AgCl}Answer

Why

\ce{NH4+}, \ce{CCl4} and \ce{H2O} each have four electron pairs around the central atom. Their bonding is described by four sp^3 hybrid orbitals, as the book does for methane, ammonia and water.

\ce{AgCl} is an ionic compound. The bonding is the electrostatic attraction between \ce{Ag+} and \ce{Cl-} ions, so no overlap of hybrid orbitals is involved.

In the formation of \ce{N2+} from \ce{N2}, the electron is removed from:

A\sigma 2p_x orbitalAnswer
B\sigma^* 2p_x orbital
C\pi 2p_y orbital
D\pi^* 2p_y orbital

Why

Nitrogen has 14 electrons, so \ce{N2} is (\sigma 1s)^2(\sigma^*1s)^2(\sigma 2s)^2(\sigma^*2s)^2(\pi 2p_y)^2(\pi 2p_z)^2(\sigma 2p_x)^2. For N_2 the book's energy order places \sigma 2p_x above the \pi 2p orbitals, so \sigma 2p_x is the highest occupied orbital.

The \pi^* and \sigma^* orbitals of \ce{N2} are empty, so they cannot lose an electron. Removing one electron from \sigma 2p_x leaves 7 bonding and 2 antibonding electrons, a bond order of 2.5.

In which process are hydrogen bonds broken?

A\ce{H2(l) -> H2(g)}
B\ce{NH3(l) -> NH3(g)}Answer
C\ce{2HI(g) -> H2(g) + I2(g)}
D\ce{CH4(g) -> C(g) + 4H(g)}

Why

Ammonia molecules are linked to each other by hydrogen bonds (N-H...N). Boiling the liquid separates these molecules, so the hydrogen bonds break as \ce{NH3(l)} turns into \ce{NH3(g)}.

In A only weak van der Waals forces act between \ce{H2} molecules. C and D break covalent bonds (H-I and C-H), which is a chemical change, and hydrogen bonds are not covalent bonds.

The \ce{CN-} ion is widely used in the synthesis of organic compounds. What is the pattern of electron pairs in this ion?

	bonding pairs of electrons	lone pairs on carbon atom	lone pairs on nitrogen atom
A	2	1	1
B	2	2	1
C	3	1	1
D	3	1	2

A2 bonding pairs, 1 lone pair on carbon, 1 lone pair on nitrogen
B2 bonding pairs, 2 lone pairs on carbon, 1 lone pair on nitrogen
C3 bonding pairs, 1 lone pair on carbon, 1 lone pair on nitrogenAnswer
D3 bonding pairs, 1 lone pair on carbon, 2 lone pairs on nitrogen

Why

Count the valence electrons: C 4, N 5, and 1 more for the negative charge.
4+5+1=10\ \text{electrons}=5\ \text{pairs}
Draw the structure. The C and N atoms share a triple bond (3 bonding pairs). The remaining 2 pairs are lone pairs, one on C and one on N.
3+1+1=5\ \text{pairs}

The pattern is 3 bonding pairs, 1 lone pair on carbon and 1 lone pair on nitrogen (option C). Option B has the right number of pairs but only 2 bonding pairs, which is a double bond, not the triple bond of the ion. Option D has 6 pairs (12 electrons), more than the 10 electrons in the ion.

Note. The book prints the ion as CN1-; it is read as \ce{CN-}.

xii

Which chlorine compound has bonding that can be described as ionic with some covalent character?

A\ce{NaCl}
B\ce{MgCl2}Answer
C\ce{AlCl3}
D\ce{SiCl4}

Why

Covalent character grows with the polarising power of the cation, which is high for small, highly charged cations. \ce{NaCl} is almost purely ionic, because Na^+ has a low polarising power.

\ce{MgCl2} contains the small, doubly charged Mg^{2+} ion. It polarises the chloride ions enough to share some electron density, so it is ionic with some covalent character. \ce{AlCl3} is more covalent still, because Al^{3+} polarises more strongly (the book's comparison). \ce{SiCl4} is a covalent molecule with no ionic character.

xiii

Gaseous nitrogen is less reactive than gaseous fluorine. What is the reason for this difference in reactivity?

AThe boiling point of nitrogen is lower than that of fluorine.
BThe relative molecular mass of nitrogen is lower than that of fluorine.
CThe atomic radius of nitrogen is greater than that of fluorine.
DThe bond strength in the molecule is greater in nitrogen than in fluorine.Answer

Why

The reactivity of a covalent molecule depends on how strongly its atoms are bonded. The N\equivN triple bond has a bond energy of 946 kJ mol^{-1}, while the F-F bond has 158 kJ mol^{-1} (Table 3.3). The strong N\equivN bond must be broken before nitrogen can react, so \ce{N2} is much less reactive.

A, B and C do not explain the reactivity. The boiling point and the molecular mass do not decide how strongly the atoms are bonded, and the larger atomic radius of nitrogen does not make its molecule strongly bonded.

Short questionsp. 77–79

The book does not label questions 2 to 11 as short or long. The questions that are short answers (question 6, question 10 and question 11) are grouped here; the multi-part questions that need diagrams and longer working are in the long group. The book prints no marks or answers for them, so the marks are typical FBISE marks.

(a) Describe the shape of the carbon dioxide molecule. (b) Bromine is a liquid at room temperature. Weak van der Waals forces hold the bromine molecules together. Describe how van der Waals forces arise.

Solution

(a) Describe the shape of the carbon dioxide molecule.

Carbon has 4 valence electrons and each oxygen has 6, so \ce{CO2} has 16 valence electrons, or 8 pairs. Carbon forms two double bonds, one to each O, so it has no lone pair. The two double bonds count as two regions of electron density around C.

Central atom with two electron-pair regions and no lone pairs (AX_2).
\text{AX}_2\ \Rightarrow\ \text{linear}
The regions repel to be as far apart as possible, so the O=C=O bond angle is 180°.

carbon dioxide \ce{CO2} is linear, with the carbon sp hybridised.

The book lists \ce{CO2} as an example of a linear molecule with four valence electrons (Table 3.4).

\text{linear},\ \angle\text{O=C=O}=180^\circ

(b) Describe how van der Waals forces arise between bromine molecules.

These are instantaneous dipole-induced dipole forces, which act between nonpolar molecules such as \ce{Br2}. Electrons are always moving, so at any instant the electron cloud of a molecule is uneven. One end becomes slightly negative (δ-) and the other slightly positive (δ+), giving a temporary dipole.

Figure 3.23: Instantaneous dipole induced dipole forces between two atoms.

The temporary dipole distorts the electron cloud of a neighbouring molecule and induces a dipole in it. The two dipoles attract each other, the δ+ end of one molecule facing the δ- end of the other. The dipoles keep changing, but the attraction is always present.

A \ce{Br2} molecule has 70 electrons and a large, easily distorted electron cloud, so these forces are strong enough to keep bromine a liquid at room temperature.

Explain the magnetic properties of \ce{O2}, \ce{O2+} and \ce{N2} by applying molecular orbital theory.

Solution

A molecule is paramagnetic if its molecular orbitals contain unpaired electrons, and diamagnetic if all of its electrons are paired. The bond order is (number of electrons in bonding MOs minus number in antibonding MOs) divided by 2.

\ce{O2} has 16 electrons. Its configuration is (\sigma 1s)^2(\sigma^*1s)^2(\sigma 2s)^2(\sigma^*2s)^2(\sigma 2p_x)^2(\pi 2p_y)^2(\pi 2p_z)^2(\pi^*2p_y)^1(\pi^*2p_z)^1. The two electrons in the degenerate \pi^* orbitals are unpaired, so \ce{O2} is paramagnetic.
\text{bond order}=\dfrac{10-6}{2}=2
\ce{O2+} has 15 electrons, so one electron is removed from a \pi^*2p orbital. The other \pi^* orbital keeps one unpaired electron, so \ce{O2+} is paramagnetic.
\text{bond order}=\dfrac{10-5}{2}=2.5
\ce{N2} has 14 electrons, all paired in its MOs, so it is diamagnetic.
\text{bond order}=\dfrac{10-4}{2}=3

\ce{O2}:\ \text{paramagnetic (2 unpaired)};\quad \ce{O2+}:\ \text{paramagnetic (1 unpaired)};\quad \ce{N2}:\ \text{diamagnetic}

Differentiate between a sigma bond and a pi bond.

Solution

A sigma bond is formed by head-on (end-to-end) overlap of orbitals along the line joining the two nuclei. A pi bond is formed by sideways, parallel overlap of p orbitals lying above and below that line. A single bond is always one sigma bond, a double bond is one sigma and one pi bond, and a triple bond is one sigma and two pi bonds.

Sigma and pi bonds compared
Feature	Sigma (σ) bond	Pi (π) bond
How it forms	Head-on (linear) overlap of orbitals	Sideways (parallel) overlap of p orbitals
Orbitals involved	s-s, s-p, p-p (p_x-p_x)	p-p only (p_y-p_y and p_z-p_z)
Where the electron density lies	Between the two nuclei, on the bond axis	Above and below the plane of the bond axis
Strength	Stronger, because the overlap is greater and the density lies between the nuclei	Weaker, because the sideways overlap is smaller
Present in	Every covalent bond (single, double or triple)	Only in double and triple bonds

The book explains the difference in strength by the position of the electron density: in a sigma bond it lies between the nuclei, while in a pi bond it lies above and below the plane, so less energy is needed to break a pi bond.

Long questionsp. 77–79

The book does not label questions 2 to 11 as short or long. The multi-part questions that need diagrams, derivations or longer explanations are grouped here: questions 2, 3, 4, 5, 7, 8 and 9. The book prints no marks or answers for them, so the marks are typical FBISE marks.

The table shows the atomic number and boiling points of some noble gases. (a) Explain this trend in boiling points. (b) Xenon forms a number of covalently bonded compounds with fluorine. (i) Draw a dot-and-cross diagram for xenon tetrafluoride, \ce{XeF4}. (ii) Suggest a shape for \ce{XeF4} and explain why you chose this shape.

Atomic numbers and boiling points of noble gases (atomic numbers corrected from the book)
Gas	helium	neon	argon	krypton	xenon
Atomic number	2	10	18	36	54
Boiling point / °C	−253	−246	−186	−152	−107

Solution

(a) Explain the trend in boiling points.

The noble gases are monatomic, so the only forces between their atoms are instantaneous dipole-induced dipole (van der Waals) forces. These forces are stronger when the atom has more electrons and a larger electron cloud, because a larger cloud is more easily distorted to form a temporary dipole.

Going down the group the atomic number rises from 2 (helium) to 54 (xenon). The atoms become larger and have more electrons, so the van der Waals forces become stronger. More energy is needed to separate the atoms, so the boiling point rises from −253 °C for helium to −107 °C for xenon.

(b)(i) Dot-and-cross diagram for \ce{XeF4}.

Count the valence electrons: Xe has 8 and each F has 7.
8+4\times7=36\ \text{electrons}
Four Xe-F bonds use 8 electrons (4 bonding pairs). Each F has 3 lone pairs (6 electrons), and Xe keeps 2 lone pairs (4 electrons).
8+4\times6+4=36
Draw the diagram. Xe's electrons are dots and F's electrons are crosses. Each bond shows one electron from each atom.

Dot-and-cross diagram of XeF4: four Xe-F bonds and two lone pairs on Xe.

(b)(ii) Suggest a shape for \ce{XeF4} and explain it.

Count the electron pairs around Xe: 4 bonding pairs and 2 lone pairs, so 6 pairs in total.
4\ \text{bonding pairs}+2\ \text{lone pairs}=6\ \text{pairs}
Six electron pairs are arranged octahedrally (Table 3.4). Lone pairs repel more strongly than bonding pairs, so the two lone pairs take up opposite positions (trans to each other). The four F atoms are left in one plane.

The shape is square planar, with F-Xe-F bond angles of 90°. The two lone pairs sit above and below the plane of the four fluorine atoms.

\text{square planar},\ \angle\text{F-Xe-F}=90^\circ

Note. The book prints the atomic numbers as -12, -110, -118, -136 and -154. The minus signs are misprints; the values used are the atomic numbers 2, 10, 18, 36 and 54. In the book the parts of (b) are labelled i and iii, with no ii; the shape question is answered here as (ii).

Aluminium chloride, \ce{AlCl3}, and ammonia, \ce{NH3}, are both covalent molecules. (a) Draw a diagram of an ammonia molecule, showing its shape. Show any lone pairs. State the bond angle HNH. (b) What type of forces are present in an ammonia molecule? Draw a diagram to show forces between ammonia molecules. (c) (i) Explain how a co-ordinate bond is formed. (ii) Draw a dot-and-cross diagram to show the bonding in the compound formed between ammonia and aluminium chloride.

Solution

(a) Draw ammonia showing its shape, lone pair and HNH angle.

Count the valence electrons: N has 5 and each H has 1.
5+3\times1=8\ \text{electrons}=4\ \text{pairs}
Four pairs: 3 bonding pairs and 1 lone pair (AX_3E). The shape is trigonal pyramidal. The lone pair on N is drawn above the plane of the three N-H bonds.

Dot-and-cross diagram of NH3 with the lone pair on N.

The HNH angle is 107°. This is less than the tetrahedral angle of 109.5° because the lone pair occupies more space and repels the bonding pairs more strongly than they repel each other.

\text{trigonal pyramidal},\ \angle\text{HNH}=107^\circ

(b) Forces in ammonia and between ammonia molecules.

Within the molecule the N-H bonds are polar covalent bonds. The electronegativity difference is 3.04-2.20=0.84, which lies between 0.4 and 1.8, so N is δ- and H is δ+. Because the molecule is trigonal pyramidal, the bond dipoles do not cancel, and NH3 is a polar molecule.

Between molecules, hydrogen bonding is the strongest force, because H is bonded to N, the electronegative atom. Permanent dipole-dipole forces and instantaneous dipole-induced dipole (van der Waals) forces also act.

Hydrogen bonds between ammonia molecules.

In the diagram the δ+ hydrogen of one molecule is attracted to the lone pair on the δ- nitrogen of the next molecule. The dashed line is the hydrogen bond. It is an intermolecular force, not a covalent bond.

(c)(i) How a co-ordinate bond is formed.

A co-ordinate (dative) covalent bond forms when one atom supplies both electrons of the shared pair. The N atom of \ce{NH3} has a lone pair of electrons. The Al atom of \ce{AlCl3} has only six electrons in its outer shell and so has an empty orbital that can accept a pair.

The lone pair of N moves into the empty orbital of Al, forming an N-Al bond. Both electrons come from N, so the bond is drawn as an arrow from N to Al. N becomes positive and Al becomes negative in the adduct.

(c)(ii) Dot-and-cross diagram of the ammonia and aluminium chloride compound.

Al has 3 electrons and forms three Al-Cl bonds. Each Cl shares one electron with Al (crosses), and Al's electrons are dots.
N supplies both electrons of the N-Al bond (two dots), and the N-H bonds have one electron from each atom.

Dot-and-cross diagram of H3N-AlCl3 showing the dative bond from N to Al.

Note. The book writes the compound as H3NAlCl3. It is the adduct H3N-AlCl3, in which the lone pair of NH3 is donated to aluminium (Cl3Al-NH3).

Electronegativity values can be used to predict the polarity of bonds. (a) Explain the term electronegativity. (b) The electronegativity values for some atoms are H = 2.1, C = 2.5, F = 4.0, Cl = 3.0, I = 2.5. Use these values to predict the polarity of each bond by copying the bonded atoms and adding δ+ or δ- above each atom: (i) H-I (ii) F-I (iii) C-Cl. (c) Describe the shape of the ICl3 molecule and mention the bond angle. (d) (i) Explain the trend in boiling points from HCl to HI. (ii) Explain why the boiling point of HF is so much higher than that of HCl. (e) Tetrachloromethane, CCl4, is a nonpolar molecule. Draw a diagram to show its shape and explain why it is nonpolar.

Solution

(a) Explain the term electronegativity.

Electronegativity is the power of a covalently bonded atom to attract the shared pair of electrons towards itself in a molecule. It has no unit, because it describes a tendency and is not a measured quantity. The most common scale is the Pauling scale used in the book.

(b)(i) Polarity of H-I.

Electronegativity difference (I minus H).
\Delta EN=2.5-2.1=0.4
On the book's scale a difference of 0.4 or less is a nonpolar covalent bond, so the pair is shared equally and no δ charges are added.

\ce{H-I}\ \text{is nonpolar covalent (no }\delta\text{ charges)}

(b)(ii) Polarity of F-I.

Electronegativity difference.
\Delta EN=4.0-2.5=1.5
A difference between 0.4 and 1.8 is a polar covalent bond. F is more electronegative, so it takes the δ- charge and I takes the δ+.

\overset{\delta-}{\ce{F}}\!-\!\overset{\delta+}{\ce{I}}

(b)(iii) Polarity of C-Cl.

Electronegativity difference.
\Delta EN=3.0-2.5=0.5
A difference of 0.5 lies between 0.4 and 1.8, so the bond is polar covalent. Cl is more electronegative, so Cl is δ- and C is δ+.

\overset{\delta+}{\ce{C}}\!-\!\overset{\delta-}{\ce{Cl}}

Count the electron pairs: I has 7 valence electrons and each Cl adds 1.
7+3=10\ \text{electrons}=5\ \text{pairs}
Five pairs: 3 bonding pairs and 2 lone pairs. Five pairs are arranged as a trigonal bipyramid. The two lone pairs take the equatorial positions, where they are 120° apart and repel less than they would in axial positions.
The three Cl atoms therefore sit at the two axial positions and one equatorial position.

The shape is T-shaped. The Cl-I-Cl bond angles are about 90° between the equatorial and axial Cl atoms, and 180° between the two axial Cl atoms. The book does not draw ICl3; the shape follows from its five-pair rules.

\text{T-shaped},\ \angle\text{Cl-I-Cl}\approx90^\circ

(d)(i) Trend in boiling points from HCl to HI.

All hydrogen halides are polar, so permanent dipole-dipole forces act between them. From HCl to HI the electronegativity difference falls from 0.9 to 0.4 (Table 3.2), so the dipole effect weakens. At the same time the molecules become larger and have more electrons (Cl has 17 electrons, I has 53), so the van der Waals forces become much stronger.

The stronger van der Waals forces outweigh the weaker dipole effect, so more energy is needed to boil HI than HCl. The boiling point rises from −85 °C for HCl to −35 °C for HI.

(d)(ii) Why the boiling point of HF is much higher than that of HCl.

In HF the hydrogen is bonded to fluorine, the most electronegative element, so the H atoms are strongly δ+ and are attracted to the lone pairs on neighbouring F atoms. This is hydrogen bonding, the strongest intermolecular force of these molecules.

Hydrogen bonds are much stronger than the dipole-dipole forces between HCl molecules, so a lot more energy is needed to separate HF molecules. This is why HF boils at +20 °C while HCl boils at −85 °C.

(e) Shape and nonpolarity of CCl4.

Four bonding pairs and no lone pairs on C (AX_4). The shape is tetrahedral with Cl-C-Cl angles of 109.5°.
Each C-Cl bond is polar (the difference is 0.5, with Cl δ-). The four bond dipoles are equal in size and point to the four corners of a tetrahedron, so their sum is zero.

Tetrachloromethane: four equal C-Cl bond dipoles cancel.

Because the four dipoles cancel, the molecule has no overall dipole and is nonpolar, even though each bond is polar. The book's rule is that symmetrical molecules are nonpolar.

\mu_{\text{net}}=0

(a) Hydrogen sulphide, H2S, is a covalent compound. Explain the type of hybridization and write the bond angle in HSH. Show on your diagram the partial charges on each atom as δ+ or δ- and an arrow showing the exact direction of the dipole in the molecule as a whole. (b) Oxygen, O, sulphur, S, and selenium, Se, are in the same group. (i) Explain why hydrogen selenide, H2Se, has a higher boiling point than hydrogen sulphide, H2S. (ii) Explain why the boiling point of water is so much higher than the boiling point of hydrogen sulphide.

Solution

(a) Hybridization, bond angle and dipole of H2S.

Count the electron pairs: S has 6 valence electrons and each H adds 1.
6+2=8\ \text{electrons}=4\ \text{pairs}
Four pairs: 2 bonding pairs and 2 lone pairs (AX_2E_2). The shape is bent.

S uses four sp^3 hybrid orbitals, formed from one 3s and three 3p orbitals. Two of the hybrids are half-filled and form the S-H σ bonds with the H 1s orbitals. The other two hold the lone pairs.

The H-S-H bond angle is about 92°. The S-H bonds are polar (S is more electronegative than H, with S δ- and H δ+), and the bent shape means the two bond dipoles do not cancel. The net dipole points from the midpoint of the H-H line towards S.

Partial charges and the net dipole of H2S.

\text{sp}^3,\ \text{bent},\ \angle\text{HSH}\approx92^\circ

(b)(i) Why H2Se has a higher boiling point than H2S.

Both molecules are bent covalent compounds of similar polarity, since S and Se have similar electronegativities (2.58 and 2.55). The difference in boiling point comes from van der Waals forces.

H2Se has 36 electrons and a molar mass of about 81 g mol^{-1}, against 18 electrons and about 34 g mol^{-1} for H2S. Its larger electron cloud is more easily distorted, so the instantaneous dipole-induced dipole forces are stronger. More energy is needed to boil H2Se.

(b)(ii) Why water boils so much higher than H2S.

In water the O atom is very electronegative (3.44), so the O-H bonds are strongly polar and each H atom is attracted to a lone pair on a neighbouring O atom. These are hydrogen bonds, the strongest intermolecular force here.

S is much less electronegative, so the H atoms of H2S cannot form hydrogen bonds. Only dipole-dipole and van der Waals forces act between H2S molecules. Water therefore boils at 100 °C, far above H2S, which boils at about −60 °C (Figure 3.26).

Note. Table 3.4 gives 104.5° for bent AX2E2 molecules, using water as the example. The measured H-S-H angle is about 92°, so the value of 104.5° does not apply to H2S.

Water is extensively hydrogen bonded. This gives it anomalous (peculiar) properties. (a) Explain why ice is less dense than liquid water. Also state two other anomalous properties of water. (b) Propanone has the structure shown. When propanone dissolves in water, it forms a hydrogen bond with water. Draw a diagram to show a propanone molecule and a water molecule forming a hydrogen bond. (c) (i) Propanone has a double bond. One bond is a σ bond and the other a π bond. Explain the difference between a σ bond and a π bond in terms of how they are formed. (ii) Copy the diagram, then complete it to show the shapes of the electron clouds in the σ bond and the π bond between the carbon atoms in ethene. Label your diagram.

Solution

(a) Why ice is less dense than water, and two other anomalous properties.

In ice each O atom is hydrogen bonded to four neighbouring water molecules, arranged tetrahedrally, in a rigid three-dimensional lattice. The hydrogen bonds are relatively long, so the molecules are held farther apart than in liquid water. Ice therefore occupies more space for the same mass, its density is lower, and it floats on water.

Two other anomalous properties are:

Water has an unusually high boiling point (100 °C). Its hydrides in the same group, H2S, H2Se and H2Te, boil at much lower temperatures (Figure 3.26), because hydrogen bonds must be broken to boil water.
Water dissolves covalent compounds that can form hydrogen bonds with it, for example ammonia and ethanol.

(b) Hydrogen bond between propanone and water.

propanone Propanone has a C=O group with two methyl groups on the carbon.

The O atom of the C=O group has lone pairs and is δ-. The H atoms of water are δ+. The lone pair on the carbonyl oxygen attracts a hydrogen atom of water, giving a hydrogen bond O-H...O=C.

Hydrogen bond between a propanone molecule and a water molecule.

(c)(i) How σ and π bonds are formed.

A σ bond is formed by head-on overlap of orbitals along the line joining the two nuclei (s-s, s-p or p-p). Its electron density lies between the nuclei on that line, so the overlap is large and the bond is strong.

A π bond is formed by sideways, parallel overlap of the unhybridised p orbitals (p_y with p_y, or p_z with p_z). Its electron density lies above and below the plane of the nuclei. This sideways overlap is less effective, so the π bond is weaker than the σ bond. A double bond contains one σ and one π bond.

(c)(ii) Electron clouds in ethene.

Each carbon in ethene forms three sp^2 hybrid orbitals, which give the σ bonds to the other carbon and to two H atoms. The third, unhybridised, p orbital on each carbon lies perpendicular to the plane of the molecule.

Ethene: the σ electron cloud lies along the C-C axis, and the π clouds lie above and below the plane.

The σ cloud is the region between the two carbon nuclei. The two π clouds are the lobes above and below the plane, formed by the parallel overlap of the two p orbitals.

82025 Annual I, Set D, Q.2 (ii)

Energies of orbitals can be explained by molecular orbital theory. It has been observed that in the case of nitrogen molecule σ2p_x is higher in energy than π2p_y and π2p_z. (a) Draw molecular orbital energy diagram for nitrogen molecule. (b) Give reason why the σ2p_x energy is greater than π2p_y and π2p_z.

Solution

(a) Molecular orbital energy diagram of N2.

Nitrogen has 7 electrons, so N2 has 14 electrons. The 1s pair is omitted from the drawing; the valence orbitals hold 10 electrons.
2\times7=14
Fill the MOs in order of increasing energy (Aufbau principle), two electrons per orbital with opposite spins (Pauli), and one electron per orbital in degenerate levels before pairing (Hund's rule).
The bond order is half the difference between bonding and antibonding electrons.
\dfrac{10-4}{2}=3

Figure 3.30: Energy diagram of the bonding and antibonding molecular orbitals of N2 (2s and 2p levels; the 1s pair is left out).

The electron configuration is (\sigma 1s)^2(\sigma^*1s)^2(\sigma 2s)^2(\sigma^*2s)^2(\pi 2p_y)^2(\pi 2p_z)^2(\sigma 2p_x)^2. There are no unpaired electrons, so N2 is diamagnetic.

\text{bond order}=3

(b) Why σ2p_x lies above π2p_y and π2p_z.

For the second-period atoms up to nitrogen, the 2s and 2p atomic orbitals differ by only a small amount of energy. The 2s and 2p orbitals therefore mix. This mixing changes the energies of the molecular orbitals: the σ orbitals no longer keep a pure s character, and σ2p_x is pushed up in energy.

As a result the energy of \sigma 2p_x is greater than that of \pi 2p_y and \pi 2p_z, which is the order shown in part (a). This is the main cause of the difference in the MO energy order between nitrogen and oxygen.

92026 Annual I, Set B, Q.6 (b) OR

Carbon can make a bond with hydrogen to form ethyne. Bond energy of C-H is the same, although 2s and 2p orbitals are involved, which have a difference in energies. Explain the formation of ethyne molecule on the basis of hybridization with the help of diagram.

Solution

Ground state of carbon: 1s² 2s² 2p², with two unpaired electrons. Carbon needs four unpaired electrons to form four bonds, so one 2s electron is promoted into the empty 2p orbital.
2s^2\,2p^2\ \to\ 2s^1\,2p_x^1\,2p_y^1\,2p_z^1
sp hybridisation: one 2s and one 2p orbital mix to give two sp hybrid orbitals of the same energy and shape. Each is 50% s and 50% p, and they point in opposite directions (180°). The 2p_y and 2p_z orbitals stay unhybridised.
Energy is needed for promotion and hybridisation, but it is released when the bonds form, so the overall process is favourable.

Promotion of a 2s electron and sp hybridisation of carbon in ethyne.

Each carbon forms one σ bond with the other carbon, using the sp orbital of each (sp-sp overlap). Each carbon also forms a σ bond with one H atom, using its other sp orbital and the H 1s orbital (sp-s overlap). The 2p_y orbitals of the two carbons overlap sideways to give one π bond, and the 2p_z orbitals give a second π bond.

Bonds in ethyne: the σ framework and the two π bonds.

Because the two sp orbitals of each carbon have the same energy and shape, the two C-H bonds are formed in the same way, so they have the same bond energy. The 2s and 2p energy difference is removed by hybridisation. The molecule is linear, with H-C-C bond angles of 180°.

\text{H-C}\equiv\text{C-H}:\ \text{linear, 180}^\circ

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 21 mark

Which pair of elements will most likely form an ionic bond?

A\ce{C} and \ce{H}
B\ce{B} and \ce{Cl}
C\ce{Na} and \ce{O}Answer
D\ce{H} and \ce{Cl}

Why

Use the electronegativity difference (Table 3.1 values). The bond is ionic if the difference is higher than 1.8, polar covalent between 0.4 and 1.8, and nonpolar covalent if 0.4 or lower.

Pair	Electronegativities	Difference	Bond
C and H	2.55, 2.20	0.35	Nonpolar covalent
B and Cl	2.04, 3.16	1.12	Polar covalent
Na and O	0.93, 3.44	2.51	Ionic
H and Cl	2.20, 3.16	0.96	Polar covalent

Na is a metal of very low electronegativity and O is highly electronegative, so Na loses its electron to O (\ce{Na+} and \ce{O^2-}). The tempting wrong answer is B (B and Cl), because \ce{BCl3} looks like a metal chloride, but boron is a non-metal and the difference of 1.12 is below 1.8, so \ce{BCl3} is covalent.

2025 Annual I, Set B, MCQ 31 mark

Which statement is FALSE about comparative bond energy values of hydrogen halide molecules?

A\ce{H-Cl > H-F}Answer
B\ce{H-F > H-Cl}
C\ce{H-Cl > H-Br}
D\ce{H-Br > H-I}

Why

The bond energy of the hydrogen halides falls from HF to HI as the electronegativity difference falls (Table 3.2): H–F 562, H–Cl 431, H–Br 366, H–I 299\ \mathrm{kJ\,mol^{-1}}. The correct order is \ce{H-F > H-Cl > H-Br > H-I}.

Statements B, C and D agree with this order. Statement A, \ce{H-Cl > H-F}, reverses it (431 is not greater than 562), so A is false. The tempting wrong choice is B: \ce{H-F > H-Cl} looks surprising because fluorine is so small, but the large electronegativity difference (1.8 against 0.9) makes H–F the strongest bond. The weak F–F bond of the halogens (lone-pair repulsion) is a different case and does not apply to H–F.

2026 Annual I, Set B, MCQ 101 mark

According to VSEPR, shape of \ce{CH4} is:

ALinear
BPlanar
CTetrahedralAnswer
DBent

Why

In \ce{CH4} the carbon atom has 4 valence electrons and each hydrogen atom adds 1, so there are 4+4=8 electrons, that is 4 electron pairs. All four are bonding pairs (AX_4, no lone pair). To keep the repulsion between the pairs as small as possible they point to the corners of a tetrahedron, so the shape is tetrahedral with H–C–H angles of 109.5^\circ.

Option B (planar) is the tempting wrong answer because \ce{CH4} is drawn flat on paper with four bonds at 90^\circ. A flat arrangement would give bond angles of only 90^\circ and more repulsion; the real angle is 109.5^\circ in three dimensions. Linear (A) and bent (D) need two bonding pairs, as in \ce{BeCl2} and \ce{H2O}.

2025 Annual I, Set B, Q.2 (ii)1.5+1.5 = 3 marks

Briefly explain the structure of \ce{PCl3} and \ce{NH4+} ion on the basis of VSEPR theory.

Solution

VSEPR theory says that the electron pairs around the central atom repel one another and take up positions as far apart as possible. The shape is named from the atoms only.

\ce{PCl3} (1.5 marks)

Valence electrons of P: 5. The three chlorine atoms contribute 1 each.
5+3\times1=8\ \text{electrons}
Number of electron pairs.
8\div2=4
The four pairs are arranged tetrahedrally. Three atoms are bonded to P, so there are 3 bonding pairs and 4-3=1 lone pair (AX_3E).
With three bonding pairs and one lone pair the shape is trigonal pyramidal. The lone pair repels the bonding pairs more strongly than they repel each other, so the Cl–P–Cl angles are less than 109.5^\circ.

\ce{NH4+} (1.5 marks)

Valence electrons of N: 5. The four hydrogen atoms contribute 4 and the +1 charge removes one electron.
5+4-1=8\ \text{electrons}
Number of electron pairs.
8\div2=4
Four atoms are bonded to N, so all 4 pairs are bonding pairs and there is no lone pair (AX_4). All the pairs repel equally, so the ion is tetrahedral with H–N–H angles of 109.5^\circ.

Shapes of \ce{PCl3} (trigonal pyramidal, lone pair on P) and \ce{NH4+} (tetrahedral).

	\ce{PCl3}	\ce{NH4+}
Electron pairs	4	4
Bonding pairs, lone pairs	3, 1	4, 0
Formula type	AX_3E	AX_4
Shape	Trigonal pyramidal	Tetrahedral
Bond angle	Less than 109.5^\circ	109.5^\circ

\ce{PCl3}:\ \text{trigonal pyramidal};\quad \ce{NH4^+}:\ \text{tetrahedral}

2025 Annual I, Set B, Q.2 (iii)1+1+1 = 3 marks

Keeping in view the factors affecting bond energy, justify the order of bond energies in the following: (a) \ce{Cl2 > F2} (b) \ce{H2 > HBr} (c) \ce{N2 > O2}

Solution

Bond energy depends on the electronegativity difference between the bonded atoms, the size of the atoms (bond length) and the bond order. Table 3.3 and Table 3.2 give the values used below.

(a) \ce{Cl2 > F2} (1 mark)

Bond energy: \ce{Cl-Cl}=242 and \ce{F-F}=158\ \mathrm{kJ\,mol^{-1}}. Fluorine atoms are very small, so the bond is short (149\ \mathrm{pm}) and the lone pairs on the two fluorine atoms are close together and repel each other. This weakens the F–F bond and less energy is needed to break it. In \ce{Cl2} the atoms are larger, the bond is longer (199\ \mathrm{pm}) and the lone pair repulsion is smaller, so the bond energy is higher.

(b) \ce{H2 > HBr} (1 mark)

Bond energy: \ce{H-H}=436 and \ce{H-Br}=366\ \mathrm{kJ\,mol^{-1}}. A bromine atom is much larger than a hydrogen atom, so the H–Br bond is longer than the H–H bond. The shorter the bond length, the greater the bond energy, so \ce{H2} has the higher bond energy.

(c) \ce{N2 > O2} (1 mark)

Bond energy: \ce{N2}=946 and \ce{O2}=498\ \mathrm{kJ\,mol^{-1}}. The bond order of \ce{N2} is 3 (triple bond) and that of \ce{O2} is 2 (double bond). The higher the bond order, the shorter (1.09\ \mathrm{\AA} against 1.21\ \mathrm{\AA}) and stronger the bond, so \ce{N2} has the higher bond energy.

Order	Factor that explains it
\ce{Cl2 > F2}	Lone pair repulsion between the very small F atoms lowers the F–F bond energy
\ce{H2 > HBr}	Shorter bond length (smaller atoms)
\ce{N2 > O2}	Higher bond order (3 against 2)

2025 Annual I, Set B, Q.3 OR3+3+1 = 7 marks

Explain formation of covalent bonds in \ce{O2} molecule on the basis of molecular orbital theory and valence bond theory. Justify that molecular orbital theory is superior to valence bond theory.

Solution

1. Molecular orbital theory (3 marks)

According to molecular orbital theory (MOT) the atomic orbitals of the two oxygen atoms combine linearly to form molecular orbitals that belong to the whole molecule. Each combination gives a lower-energy bonding orbital (BMO) and a higher-energy anti-bonding orbital (ABMO). The number of molecular orbitals equals the number of atomic orbitals combined.

Each oxygen atom (1s^2\,2s^2\,2p_x^2\,2p_y^1\,2p_z^1) has 8 electrons, so \ce{O2} has 16 electrons.
Fill the molecular orbitals in order of increasing energy: \sigma1s<\sigma^*1s<\sigma2s<\sigma^*2s<\sigma2p_x<\pi2p_y=\pi2p_z<\pi^*2p_y=\pi^*2p_z<\sigma^*2p_x. Two electrons go in each orbital (Pauli), and the last two electrons go singly into the degenerate \pi^*2p_y and \pi^*2p_z orbitals (Hund's rule).
\ce{O2}=(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_x)^2(\pi2p_y)^2=(\pi2p_z)^2(\pi^*2p_y)^1=(\pi^*2p_z)^1
Bonding electrons: 2+2+2+4=10. Anti-bonding electrons: 2+2+1+1=6.
N_b=10,\quad N_a=6
Bond order.
\text{bond order}=\dfrac{N_b-N_a}{2}=\dfrac{10-6}{2}=2

Molecular orbital energy diagram of \ce{O2} (valence orbitals only). The two electrons in π*2py and π*2pz are unpaired (accent levels).

The bond order is 2, so the two oxygen atoms are joined by a double bond (bond energy 498\ \mathrm{kJ\,mol^{-1}}, bond length 1.21\ \mathrm{\AA}). The two unpaired electrons in the \pi^* orbitals make \ce{O2} paramagnetic.

2. Valence bond theory (3 marks)

According to valence bond theory (VBT) a covalent bond forms by the overlap of half-filled atomic orbitals, and the electrons of the overlapping orbitals have opposite spins. Each oxygen atom has two half-filled 2p orbitals (2p_y and 2p_z), so two bonds are formed.

One pair of half-filled 2p orbitals overlaps head on (along the axis joining the nuclei) to form a **\sigma bond**.
The other pair of half-filled 2p orbitals overlaps sideways (parallel overlap) to form a **\pi bond**.
One \sigma bond and one \pi bond make the double bond \ce{O=O}.

Formation of σ and π bonds in the \ce{O2} molecule on the basis of valence bond theory.

3. Why MOT is superior to VBT (1 mark)

VBT shows all the electrons of \ce{O2} as paired (a \sigma bond and a \pi bond, each a pair of opposite spins), so it predicts that \ce{O2} is diamagnetic. In fact liquid oxygen is attracted by a magnet: \ce{O2} is paramagnetic. MOT explains this, because two electrons occupy the two degenerate \pi^*2p_y and \pi^*2p_z orbitals singly with parallel spins. MOT also gives the bond order (2) from the electron count, and it explains why \ce{He2} does not exist (bond order 0). VBT cannot explain the paramagnetic behaviour of \ce{O2}, so MOT is the better theory.

	Valence bond theory	Molecular orbital theory
Bond formed by	Overlap of half-filled atomic orbitals	Linear combination of atomic orbitals into molecular orbitals of the whole molecule
Bonds in \ce{O2}	One \sigma and one \pi bond	Bond order 2
Magnetic character of \ce{O2}	All electrons paired (wrong)	Two unpaired electrons: paramagnetic (correct)

\ce{O2}:\ \text{bond order}=2,\ \text{paramagnetic (MOT)}

2026 Annual I, Set B, Q.6 (a) OR3 marks

Explain in detail the type of hybridization, bond angles and geometrical shapes of following molecule: \ce{BF3}

Solution

Hybridization is the mixing of atomic orbitals of different energies and shapes to give the same number of new orbitals of equal energy and shape. In \ce{BF3} the central boron atom is sp^2 hybridised.

Step 1: ground state and promotion

Ground state of boron: 1s^2\,2s^2\,2p^1. There is only one unpaired electron, but three bonds with fluorine need three unpaired electrons.
\ce{B}:\ 1s^2\,2s^2\,2p_x^1
One 2s electron is promoted to the empty 2p_y orbital. The energy needed is paid back by the energy released when the bonds form.
\ce{B^*}:\ 1s^2\,2s^1\,2p_x^1\,2p_y^1

Step 2: sp2 hybridization

One 2s and two 2p orbitals of boron mix to give **three equivalent half-filled sp^2 hybrid orbitals**. They point to the corners of an equilateral triangle in one plane, 120^\circ apart (trigonal planar symmetry).

Step 3: overlap with fluorine

Fluorine (1s^2\,2s^2\,2p^5) has one half-filled 2p orbital. Each sp^2 orbital of boron overlaps head on with a half-filled 2p orbital of one fluorine atom to give a \sigma sp^2–p bond. Three \sigma bonds are formed and boron has no lone pair.

Hybridization in \ce{BF3}: promotion of a 2s electron, three sp² orbitals (accent) at 120°, and overlap with a 2p orbital of each fluorine atom.

Summary for \ce{BF3}
Feature	\ce{BF3}
Hybridization of B	sp^2 (one s + two p)
Bonds	Three \sigma sp^2–p bonds
Bond angle	120^\circ (F–B–F)
Geometrical shape	Trigonal planar (AX_3, no lone pair)

\ce{BF3}:\ sp^2,\ 120^\circ,\ \text{trigonal planar}

2026 Annual I, Set B, Q.6 (b) OR3 marks

Explain in detail the type of hybridization, bond angles and geometrical shapes of following molecule: \ce{C2H2}

Solution

In acetylene (ethyne) each carbon atom is sp hybridised. The molecule is linear, with a \ce{C#C} triple bond.

Step 1: promotion

Ground state of carbon: only two unpaired electrons, but carbon forms four bonds.
\ce{C}:\ 1s^2\,2s^2\,2p_x^1\,2p_y^1
One 2s electron is promoted to the empty 2p_z orbital, giving four unpaired electrons.
\ce{C^*}:\ 1s^2\,2s^1\,2p_x^1\,2p_y^1\,2p_z^1

Step 2: sp hybridization

Each carbon mixes its 2s orbital with one 2p orbital to give **two half-filled sp hybrid orbitals**, arranged in a straight line (180^\circ). The other two 2p orbitals (2p_y and 2p_z) stay unhybridised and are perpendicular to the sp orbitals and to each other.

Step 3: bonds formed

The two carbon atoms overlap one sp orbital each head on to form a C–C \sigma sp–sp bond.
The other sp orbital of each carbon overlaps head on with the 1s orbital of a hydrogen atom to form a \sigma sp–s bond. There are two C–H \sigma bonds.
The unhybridised 2p_y orbitals overlap sideways, and so do the 2p_z orbitals, to form **two \pi bonds**. Together with the \sigma bond they make the carbon–carbon triple bond.

Hybridization and bonding in \ce{C2H2}: two sp orbitals on each carbon (σ bonds) and two unhybridised p orbitals (two π bonds).

Bond in \ce{C2H2}	Type	Orbitals overlapping
C–C (first bond)	\sigma	sp–sp
C–C (second and third bonds)	two \pi	2p_y–2p_y and 2p_z–2p_z
C–H (two)	\sigma	sp–s

Because the two carbon atoms and the two hydrogen atoms lie in a straight line, the H–C–C bond angle is **180^\circ and the molecule is linear**.

Summary for \ce{C2H2}
Feature	\ce{C2H2}
Hybridization of each C	sp (one s + one p)
Bond angle	180^\circ
Geometrical shape	Linear

\ce{C2H2}:\ sp,\ 180^\circ,\ \text{linear}

2025 Annual I, Set D, MCQ 21 mark

A molecule with formula \ce{H2CO} has three sigma bond pairs and one pi bond pair in valence shell of central atom (carbon). Its structure is:

ATriangular pyramid
BTrigonal planarAnswer
CBent
DTetrahedral

Why

In \ce{H2C=O} the carbon atom is bonded to two hydrogen atoms (two \sigma bonds) and to oxygen by a double bond (one \sigma and one \pi bond). That is three \sigma bond pairs, as the question says. In VSEPR the \pi bond lies in the same region as the \sigma bond of the double bond, so a double bond counts as one group of electrons, as in \ce{SO2} and \ce{CO3^2-} in the book.

Carbon therefore has three groups, no lone pair (AX_3), and the shape is trigonal planar with angles of about 120^\circ, like \ce{BCl3} and \ce{C2H4} (sp^2). The tempting wrong answer is D (tetrahedral), from counting the \pi pair as a fourth pair; it does not add a new direction. Triangular pyramid (A) needs a lone pair, and bent (C) needs two bonded atoms.

2025 Annual I, Set D, MCQ 41 mark

Which molecule possesses stronger London dispersion forces?

An-Butane
Bn-PentaneAnswer
CIso butane
DIso pentane

Why

London (instantaneous dipole induced dipole) forces get stronger as the molecule gets bigger, with more electrons and a higher molecular mass. They also need close contact between molecules.

\ce{C5H12} has more electrons and a higher molecular mass (M_r=72) than \ce{C4H10} (M_r=58), so the pentanes beat the butanes. Of the two pentanes, n-pentane is a long, straight chain that lies close against its neighbours over its whole length, while iso-pentane is branched and more compact, so it has less surface in contact. The straight-chain n-pentane has the stronger forces (boiling point 36\ ^\circ\mathrm{C} against 28\ ^\circ\mathrm{C} for iso-pentane).

The tempting wrong answer is D (iso-pentane): it has the same formula and mass as n-pentane, but its branching reduces the contact between molecules.

2026 Annual I, Set D, MCQ 101 mark

According to VSEPR theory bond angle in methane is:

A90^{\circ}
B109.5^{\circ}Answer
C120^{\circ}
D180^{\circ}

Why

Methane has four bonding pairs and no lone pair on carbon (AX_4). The four pairs repel each other equally and move as far apart as possible, to the corners of a regular tetrahedron. The angle between any two C–H bonds is therefore **109.5^\circ**.

120^\circ (C) is the tempting wrong answer: it is the angle for three electron pairs, as in \ce{BCl3}. 180^\circ is for two pairs (\ce{BeCl2}) and 90^\circ is for the octahedral arrangement of six pairs. In \ce{NH3} and \ce{H2O} the lone pairs squeeze the angle below 109.5^\circ (107^\circ and 104.5^\circ), but \ce{CH4} has none.

2025 Annual I, Set D, Q.2 (ii)3 marks

Draw molecular orbital diagram for \ce{N2} and calculate its bond order.

Solution

Each nitrogen atom has 7 electrons (1s^2\,2s^2\,2p_x^1\,2p_y^1\,2p_z^1), so \ce{N2} has 14 electrons. The filled 1s levels are not drawn; the valence orbitals hold 10 electrons.
2\times7=14
Fill the molecular orbitals in order of energy (Aufbau principle), two electrons in each with opposite spins (Pauli). In \ce{N2} the \pi2p_y=\pi2p_z orbitals lie below \sigma2p_x.
\ce{N2}=(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\pi2p_y)^2=(\pi2p_z)^2(\sigma2p_x)^2

Molecular orbital energy diagram of \ce{N2} (valence orbitals only). All bonding levels up to σ2px are full and the anti-bonding π* levels are empty.

Bonding electrons: \sigma1s (2), \sigma2s (2), \pi2p_y and \pi2p_z (4), \sigma2p_x (2).
N_b=2+2+4+2=10
Anti-bonding electrons: \sigma^*1s (2) and \sigma^*2s (2).
N_a=2+2=4
Bond order.
\text{bond order}=\dfrac{N_b-N_a}{2}=\dfrac{10-4}{2}=3

The bond order of 3 means a triple bond. All electrons are paired, so \ce{N2} is diamagnetic. It has a very short bond (1.09\ \mathrm{\AA}) and a very high bond energy (946\ \mathrm{kJ\,mol^{-1}}).

\text{bond order of }\ce{N2}=\dfrac{10-4}{2}=3

2025 Annual I, Set D, Q.2 (vi)3 marks

Justify the difference in the boiling points of \ce{CH4} (-162\ ^{\circ}\mathrm{C}), \ce{HCl} (-85\ ^{\circ}\mathrm{C}) and \ce{H2O} (100\ ^{\circ}\mathrm{C}).

Solution

A liquid boils when its molecules have enough energy to overcome the forces between them. The stronger the intermolecular forces, the higher the boiling point. The three molecules have different types of intermolecular force.

Intermolecular forces and boiling points (Table 3.6)
Molecule	Boiling point	Polarity	Intermolecular force
\ce{CH4}	-162\ ^\circ\mathrm{C}	Nonpolar (C–H dipoles cancel in the tetrahedron)	Instantaneous dipole induced dipole (London) forces only: weakest
\ce{HCl}	-85\ ^\circ\mathrm{C}	Polar (Cl is \delta-, H is \delta+)	Permanent dipole–dipole forces, and London forces
\ce{H2O}	100\ ^\circ\mathrm{C}	Polar, O–H bonds	Hydrogen bonding (H bonded to O), the strongest

Explanation (1 mark each)

\ce{CH4} is nonpolar. Only weak instantaneous dipole induced dipole forces act between its molecules, so very little energy is needed to separate them and it boils at the lowest temperature, -162\ ^\circ\mathrm{C}.
\ce{HCl} is polar because of the electronegativity difference (3.16-2.20=0.96). Permanent dipole–dipole attractions between \ce{H^{\delta+}-Cl^{\delta-}} molecules are stronger than London forces alone, so it boils higher, at -85\ ^\circ\mathrm{C}.
In \ce{H2O} hydrogen is bonded to the highly electronegative oxygen, which has lone pairs. Each hydrogen is strongly attracted to a lone pair of a neighbouring molecule (hydrogen bonding), the strongest of the three intermolecular forces. Much more energy is needed to separate the molecules, so water boils at 100\ ^\circ\mathrm{C}.

\ce{H2O}\ (\text{H-bonding}) > \ce{HCl}\ (\text{dipole-dipole}) > \ce{CH4}\ (\text{London})

2025 Annual I, Set D, Q.31+3+3 = 7 marks

What is hybridization? Write down three differences between sp^{3} and sp^{2} hybridizations. Explain bonding in \ce{BF3} on the basis of hybridization.

Solution

1. Hybridization (1 mark)

Hybridization is the concept of mixing atomic orbitals of different energies and different shapes to produce a new set of orbitals of the same energy and the same shape, called hybrid orbitals. The number of hybrid orbitals formed equals the number of atomic orbitals mixed. An electron is usually promoted first to increase the number of unpaired electrons; the energy needed is repaid by the energy released when the bonds form.

2. Differences between sp3 and sp2 hybridization (3 marks)

Any three rows earn the marks
Point	sp^3 hybridization	sp^2 hybridization
Orbitals mixed	One s and three p orbitals	One s and two p orbitals
Number of hybrid orbitals	Four	Three
Shape and bond angle	Tetrahedral, 109.5^\circ	Trigonal planar, 120^\circ
Unhybridised p orbital	None left	One left on the atom, which can form a \pi bond
Examples	\ce{CH4}, \ce{C2H6}, \ce{NH3}, \ce{H2O}	\ce{BCl3}, \ce{BF3}, \ce{C2H4}

3. Bonding in BF3 (3 marks)

Step 1: ground state and promotion

Ground state of boron: 1s^2\,2s^2\,2p^1. There is only one unpaired electron, but three bonds with fluorine need three unpaired electrons.
\ce{B}:\ 1s^2\,2s^2\,2p_x^1
One 2s electron is promoted to the empty 2p_y orbital. The energy needed is paid back by the energy released when the bonds form.
\ce{B^*}:\ 1s^2\,2s^1\,2p_x^1\,2p_y^1

Step 2: sp2 hybridization

Step 3: overlap with fluorine

Hybridization in \ce{BF3}: promotion of a 2s electron, three sp² orbitals (accent) at 120°, and overlap with a 2p orbital of each fluorine atom.

Summary for \ce{BF3}
Feature	\ce{BF3}
Hybridization of B	sp^2 (one s + two p)
Bonds	Three \sigma sp^2–p bonds
Bond angle	120^\circ (F–B–F)
Geometrical shape	Trigonal planar (AX_3, no lone pair)

\ce{BF3}:\ sp^2,\ 120^\circ,\ \text{trigonal planar}

2026 Annual I, Set D, Q.6 (a) OR3 marks

Explain in detail the type of hybridization, bond angles and geometrical shapes of following molecule: \ce{C2H4}

Solution

In ethylene (ethene) each carbon atom is sp^2 hybridised. The molecule is planar and has a \ce{C=C} double bond.

Step 1: promotion

Ground state of carbon has two unpaired electrons, but carbon needs four to form four bonds.
\ce{C}:\ 1s^2\,2s^2\,2p_x^1\,2p_y^1
One 2s electron is promoted to the empty 2p_z orbital.
\ce{C^*}:\ 1s^2\,2s^1\,2p_x^1\,2p_y^1\,2p_z^1

Step 2: sp2 hybridization

Each carbon mixes its 2s orbital with two 2p orbitals to give **three half-filled sp^2 hybrid orbitals** in one plane, 120^\circ apart (trigonal planar). One half-filled 2p_z orbital stays unhybridised, perpendicular to that plane.

Step 3: bonds formed

The two carbon atoms overlap one sp^2 orbital each head on to form a C–C \sigma sp^2–sp^2 bond.
The other two sp^2 orbitals of each carbon overlap with the 1s orbitals of two hydrogen atoms to give four C–H \sigma sp^2–s bonds.
The unhybridised 2p_z orbitals of the two carbons overlap sideways (parallel overlap) above and below the plane to form one \pi bond. The \sigma and the \pi bond together are the C=C double bond.

Hybridization and bonding in \ce{C2H4}: each carbon uses three sp² orbitals (accent) for σ bonds, and the unhybridised 2p orbitals overlap sideways to give the π bond.

Bond in \ce{C2H4}	Type	Orbitals overlapping
C=C (first bond)	\sigma	sp^2–sp^2
C=C (second bond)	\pi	2p_z–2p_z (sideways)
C–H (four)	\sigma	sp^2–s

All the atoms lie in one plane. The H–C–H and H–C–C bond angles are **120^\circ and the shape around each carbon is trigonal planar** (the molecule as a whole is planar).

Summary for \ce{C2H4}
Feature	\ce{C2H4}
Hybridization of each C	sp^2 (one s + two p)
Bond angle	120^\circ
Geometrical shape	Planar; trigonal planar at each carbon

\ce{C2H4}:\ sp^2,\ 120^\circ,\ \text{planar (trigonal planar at each C)}

2026 Annual I, Set D, Q.6 (b) OR3 marks

Explain in detail the type of hybridization, bond angles and geometrical shapes of following molecule: \ce{NH3}

Solution

In ammonia the central nitrogen atom is sp^3 hybridised. Of the four sp^3 orbitals, three are used for bonds and one holds a lone pair, so the molecule is trigonal pyramidal.

Step 1: electron pairs (VSEPR check)

Nitrogen has 5 valence electrons and the three hydrogen atoms give 3.
5+3=8\ \text{electrons}=4\ \text{pairs}
Three pairs are bonding pairs and one is a lone pair (AX_3E). The four pairs are arranged tetrahedrally.

Step 2: sp3 hybridization

The ground state of nitrogen is 1s^2\,2s^2\,2p_x^1\,2p_y^1\,2p_z^1. It already has three unpaired electrons, so no promotion is needed. The 2s orbital (full) mixes with the three 2p orbitals to give **four sp^3 hybrid orbitals, directed to the corners of a tetrahedron. Three of them are half-filled and one is full**.

Step 3: bonds and lone pair

The three half-filled sp^3 orbitals of nitrogen overlap head on with the 1s orbitals of three hydrogen atoms to form three \sigma sp^3–s bonds. The fourth sp^3 orbital holds the lone pair.

Hybridization in \ce{NH3}: three half-filled sp³ orbitals of nitrogen overlap with the 1s orbitals of hydrogen; the fourth sp³ orbital holds the lone pair (accent).

Bond angle and shape

A lone pair occupies more space than a bonding pair, so it repels the three bonding pairs more strongly than they repel each other. The H–N–H angle is therefore reduced from the tetrahedral 109.5^\circ to **107^\circ. Describing the positions of the atoms only, the molecule is trigonal pyramidal**.

Summary for \ce{NH3}
Feature	\ce{NH3}
Hybridization of N	sp^3 (one s + three p)
Bonding pairs and lone pairs	3 and 1
Bond angle	107^\circ (less than 109.5^\circ)
Geometrical shape	Trigonal pyramidal

\ce{NH3}:\ sp^3,\ 107^\circ,\ \text{trigonal pyramidal}

Exam focus

Explain the hybridization, bond angles and shape of a molecule (\ce{BF3}, \ce{C2H2}, \ce{C2H4}, \ce{NH3}), showing promotion, the hybrid orbitals and the overlap with a diagram. (long)
Asked in both 2026 sets as the Q.6 OR (Set B: \ce{BF3} and \ce{C2H2}; Set D: \ce{C2H4} and \ce{NH3}) and in 2025 Set D Q.3 (\ce{BF3} with the sp3 against sp2 differences). Book long question 9 is the ethyne case. Learn the five molecules in the book's hybridization table.
Go to answer
State what hybridization is and give three differences between sp^3 and sp^2 (orbitals mixed, number of hybrid orbitals, shape and angle). (short)
2025 Set D Q.3 (1+3+3 marks). The table of sp, sp^2 and sp^3 in the hybridization section gives the points.
Go to answer
Draw the molecular orbital diagram of \ce{N2} or \ce{O2}, write its configuration and find the bond order and magnetic character. (long)
2025 Set D Q.2 (ii) (\ce{N2} diagram and bond order) and 2025 Set B Q.3 OR (\ce{O2} by MOT and VBT, 7 marks). Book short question 10 and long question 8 ask the same. Remember that \ce{O2} is the only paramagnetic molecule of the set.
Go to answer
Justify why molecular orbital theory is better than valence bond theory (paramagnetism of \ce{O2}). (short)
The last mark of 2025 Set B Q.3 OR. VBT shows \ce{O2} with all electrons paired, but \ce{O2} is paramagnetic, which only MOT explains.
Go to answer
Predict the shape of a molecule or ion by VSEPR (count the electron pairs, bonding and lone pairs): \ce{CH4}, \ce{PCl3}, \ce{NH4+}, \ce{H2CO}. (short)
Asked as MCQ in both 2026 sets (Set B MCQ 10 shape of \ce{CH4}, Set D MCQ 10 its angle), in 2025 Set B Q.2 (ii) (\ce{PCl3} and \ce{NH4+}) and 2025 Set D MCQ 2 (\ce{H2CO}). Use the method of the worked examples and remember that a double bond counts as one group.
Go to answer
Justify the order of bond energies using electronegativity difference, bond length and bond order (\ce{Cl2 > F2}, \ce{H2 > HBr}, \ce{N2 > O2}, hydrogen halides). (short)
2025 Set B Q.2 (iii) (3 marks) and 2025 Set B MCQ 3 (order \ce{H-F > H-Cl > H-Br > H-I}). Tables 3.2 and 3.3 give the values; the exam tip on the weak F–F bond is the usual 1 mark.
Go to answer
Explain the difference in boiling points of \ce{CH4}, \ce{HCl} and \ce{H2O} from their intermolecular forces, and compare London forces in alkanes. (short)
2025 Set D Q.2 (vi) (Table 3.6) and 2025 Set D MCQ 4 (London forces in n-pentane against iso-pentane). The same forces are tested in unit 5 papers (2026 MCQ 16 and Q.2 (v) OR). Rule: hydrogen bonding > dipole–dipole > London; bigger and straighter molecules have stronger London forces.
Go to answer
Decide the type of bond (ionic, polar covalent, nonpolar covalent) from the electronegativity difference. (mcq)
2025 Set B MCQ 2 (Na and O against B and Cl). Use the cut-offs 0.4 and 1.8. Electronegativity itself and its trend are also asked in 2026 Q.2 (xii), which is solved in unit 10.
Go to answer
Differentiate between a sigma bond and a pi bond, and count the \sigma and \pi bonds in a molecule. (short)
Not asked as a separate question in the 2025 or 2026 papers, but book short question 11 asks it, and the \sigma/\pi count is needed for the hybridization answers above and for \ce{H2CO} (2025 Set D MCQ 2).
Go to answer

3.1ElectronegativityOften examined

Factors affecting electronegativity

3.1.2Trend of electronegativity in the periodic tableOften examined

Down a group

Across a period

3.1.3Nature of bond in terms of Pauling electronegativity valuesOften examined

3.1.4Covalent character in a compound

3.2Dipole moment

3.3Polar and nonpolar covalent bondOften examined

3.4Bond energy (bond enthalpy) and bond lengthOften examined

Bond length

Bond energy and bond length of the halogens

Bond energy and reactivity

3.5Shapes of molecules: valence shell electron pair repulsion theoryOften examined

3.5.1Shapes of molecules containing two electron pairsOften examined

AX2​: BeFX2​

3.5.2Shapes of molecules containing three electron pairsOften examined

Three bonding pairs, AX3​: BClX3​

Three bonding pairs, AX3​: COX3​X2−

Two bonding pairs and one lone pair, AX2​E: SOX2​

3.5.3Shapes of molecules containing four electron pairsOften examined

Four bonding pairs, AX4​: CHX4​

Three bonding pairs and one lone pair, AX3​E: NHX3​

Two bonding pairs and two lone pairs, AX2​E2​: HX2​O

3.5.4Shapes of molecules containing five electron pairsOften examined

3.5.5Shapes of molecules containing six electron pairsOften examined

3.5.6Expanded octet (hypervalency) and summary of shapesOften examined

3.5.7Working out the shapes of moleculesOften examined

3.5.8Importance of VSEPR theory in the field of drugs

Molecular shape and enzymatic reactions

Enzyme–substrate interaction in drug design

Molecular geometry in drug delivery systems

Cisplatin

Question 1(a): predict the shapes

Question 1(b): dot-and-cross diagrams

Question 2: selenium hexafluoride and phosphorus(V) chloride

3.6Valence bond theoryOften examined

Main points of VBT

3.6Types of overlapping and nature of covalent bondsOften examined

Sigma bond

Pi bond

Strength of sigma and pi bonds

3.6.1Applications of valence bond theoryOften examined

Single bond formation

(a) Formation of the \ce{H2} molecule

(b) Formation of the \ce{F2} molecule

(c) Formation of the HF molecule

Double bond formation: (d) the \ce{O2} molecule

Triple bond formation: (e) the \ce{N2} molecule

3.7HybridizationOften examined

Types of hybridization

3.7.1sp hybridizationOften examined

Examples of sp hybridization

1. Beryllium chloride, \ce{BeCl2}

2. Acetylene, \ce{C2H2}

3.7.2sp² hybridizationOften examined

Examples of sp^2 hybridization

1. Boron trichloride, \ce{BCl3}

2. Ethylene (ethene), \ce{C2H4}

3.7.3sp³ hybridizationOften examined

Examples of sp^3 hybridization

1. Methane, \ce{CH4}

2. Ethane, \ce{C2H6}

3. Ammonia, \ce{NH3}

4. Water, \ce{H2O}

3.8Co-ordinate covalent bondOften examined

Hydronium ion, \ce{H3O^+}

Ozone, \ce{O3}, and carbon monoxide, \ce{CO}

Question 2: displayed formulae

3.9Intermolecular forces (forces between molecules)Often examined

Types of intermolecular forces

1. Permanent dipole–dipole forces

2. Instantaneous dipole induced dipole forces

3. Hydrogen bonding

3.9.2Peculiar behaviour of hydrogen bondingOften examined

Ice is less dense than water

The strength of intermolecular forces

Question 1(a): the halogens

Question 1(b): the alkanes

Question 2: Br₂ and ICl

AX $_{2}$ : $BeF X_{2}$

Three bonding pairs, AX $_{3}$ : $BCl X_{3}$

Three bonding pairs, AX $_{3}$ : $CO X_{3} X^{2 -}$

Two bonding pairs and one lone pair, AX $_{2}$ E: $SO X_{2}$

Four bonding pairs, AX $_{4}$ : $CH X_{4}$

Three bonding pairs and one lone pair, AX $_{3}$ E: $NH X_{3}$

Two bonding pairs and two lone pairs, AX $_{2}$ E $_{2}$ : $H X_{2} O$