Chemical Bonding and Molecular Structure: Class 11 Chemistry · Chapter 4

Chemistry Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

105 min
48 practice questions
33 MCQs

What you will learn in Chemical Bonding and Molecular Structure

Draw Lewis symbols and Lewis structures of molecules and ions, assign formal charges and recognise exceptions to the octet rule
Explain the formation of ionic compounds in terms of ionisation enthalpy, electron gain enthalpy and lattice enthalpy
Compare bonds by their length, angle, enthalpy and order, and describe them with resonance and dipole moments
Predict the shapes of molecules and ions with VSEPR theory, including the effect of lone pairs
Explain covalent bonds, σ and π bonds and molecular shapes with valence bond theory and sp, sp², sp³, sp³d and sp³d² hybridisation
Fill molecular orbital diagrams of homonuclear diatomic molecules and use them to find bond order and magnetic behaviour
Explain hydrogen bonding, its two types and its effect on physical properties

Original companion text aligned to NCERT Chemistry Class 11, Chapter 4, pp. 100–135. Not affiliated with or endorsed by NCERT.

Table salt shatters when you hit it and melts only above 800 °C. Sugar melts in a pan on the stove, and the gas in a cooking cylinder boils well below room temperature. Diamond, which is pure carbon, is the hardest natural material known. These differences come down to one thing: how the particles in each substance are held together.

Look around and you will not find loose atoms of oxygen, iron or carbon. Only the noble gases exist naturally as single atoms; every other element turns up with its atoms joined, in molecules or in huge arrays of ions, held by chemical bonds (रासायनिक आबंध). A chemical bond is the attraction that keeps atoms or ions together in a substance. This raises sharp questions. Why do atoms bond at all? Why does hydrogen exist as $H_{2}$ while helium stays single? Why is a water molecule bent but a carbon dioxide molecule straight?

You will meet four models, each filling a gap in the one before. Kössel–Lewis electron counting comes first, then VSEPR theory for shapes, valence bond theory for directed bonds and molecular orbital theory for magnetism. The unit ends with the hydrogen bond.

4.1 Kössel–Lewis Approach to Chemical Bonding

In 1916 two chemists working separately, Walther Kössel in Germany and Gilbert N. Lewis in the United States, gave the first explanation of valence in terms of electrons. Both began with one striking fact: the noble gases barely react, so their outer electron arrangement must be unusually stable.

Lewis's model split each atom into two parts. The kernel, the nucleus together with the inner electrons, carries a positive charge and does not change in reactions. Around it sits the outer shell, which he drew as a cube with room for one electron at each of its eight corners. A noble gas has every corner filled. Any other atom, Lewis argued, bonds so as to fill its own cube, either by handing electrons over or by sharing them.

Chemistry happens at the surface of an atom. Only the electrons of the outermost shell, the valence electrons, move or get shared; the inner electrons stay out of it. Lewis's shorthand writes the element's symbol with one dot for each valence electron. Such a drawing is a Lewis symbol (लूइस प्रतीक); Fig. 4.1 gives them for period 3.

Fig. 4.1Dots go singly on the four sides first, then pair up. Group valence is the number of dots up to four, then eight minus the dots.NCERT p. 101

Symbols Na, Mg, Al, Si, P, S, Cl and Ar in two rows, each with one dot per valence electron, from one dot for sodium to eight for argon. Dots go singly on four sides before pairing, and the electron count is written under each symbol.

Counting the dots predicts how many bonds an element usually forms, its group valence. With one to four dots, the atom uses all of them (Mg forms two bonds, Al three, Si four). With five or more, it is easier to collect the missing electrons, so the valence is eight minus the dots (P 3, S 2, Cl 1).

Kössel's starting point was the layout of the periodic table. Every noble gas is flanked by a halogen, which grabs electrons readily, and an alkali metal, which gives them up readily. Move one electron from the metal atom to the halogen atom and both end up with the electron count of their noble-gas neighbour: an outer $n s^{2} n p^{6}$ octet, or helium's $1 s^{2}$ pair in the case of $L i^{+}$ . The products are ions of opposite charge, and electrostatic attraction holds them together.

Potassium ( $[A r] 4 s^{1}$ ) loses its 4s electron and is left with argon's configuration. Bromine ( $[A r] 3 d^{10} 4 s^{2} 4 p^{5}$ ) gains one electron and reaches krypton's.

\begin{aligned} K & \overset{}{\to} K^{+} + e^{-} \\ B r + e^{-} & \overset{}{\to} B r^{-} \\ K^{+} + B r^{-} & \overset{}{\to} K B r \end{aligned}

Magnesium has two valence electrons, so it gives one to each of two chlorine atoms: $M g \overset{}{\to} M g^{2 +} + 2 e^{-}$ , and each $C l + e^{-} \overset{}{\to} C l^{-}$ . The compound is $M g C l_{2}$ (Fig. 4.2). Chemists called this ion-to-ion attraction an electrovalent bond; today we usually say ionic bond (आयनिक आबंध). The number of unit charges on an element's ion is its electrovalence: 2 for magnesium here (positive) and 1 for chlorine (negative). The scheme still underlies our picture of ionic crystals, but, as Kössel himself recognised, it says nothing useful about carbon compounds and many others.

3/3 $M g^{2 +}$ has neon's shell; each $C l^{-}$ has argon's octet.

Fig. 4.2Electron transfer gives every ion a noble-gas configuration.NCERT p. 101

A magnesium atom with two electrons drawn as crosses sits between two chlorine atoms with seven dots each. Step 1 shows the neutral atoms. Step 2 adds curved arrows carrying one magnesium electron to each chlorine. Step 3 shows Mg 2+ with no outer electrons and each chloride ion with eight electrons in brackets and a minus charge.

4.1.1 Octet Rule

Definition

Octet rule

Atoms combine by losing, gaining or sharing valence electrons so that each ends with eight electrons in its valence shell (two for hydrogen), like the nearest noble gas.

ExampleMagnesium loses two electrons, chlorine gains one, and carbon shares four pairs in $C H_{4}$ .

Kössel and Lewis proposed this electronic theory of chemical combination in 1916. It is a powerful rule of thumb, not a law of nature: Section 4.1.5 lists where it breaks down.

4.1.2 Covalent Bond

Transfer cannot explain $F_{2}$ : two identical atoms have no reason to swap an electron. Irving Langmuir took up Lewis's sharing idea in 1919, let go of the rigid cube, and named the result a covalent bond (सहसंयोजी आबंध): a bond in which two atoms hold electrons in common.

A fluorine atom ( $[H e] 2 s^{2} 2 p^{5}$ ) has seven valence electrons and wants eight. If two fluorine atoms each lend one electron to a common pair, each counts that pair towards its own total, and both reach eight. We write $F - F$ , the line standing for the shared pair. Each fluorine also keeps three pairs that it does not share: these lone pairs belong to one atom only.

Drawings like this are Lewis dot structures, and the same logic works whether the atoms are alike or different. One line always means one shared pair; normally each partner puts one electron into it; and in the finished drawing every atom has a noble-gas count, eight electrons, or two for hydrogen.

Atoms may share more than one pair. In methanal (formaldehyde, $H C H O$ ), carbon and oxygen share two pairs, a double bond. In hydrogen cyanide, $H C N$ , carbon and nitrogen share three pairs, a triple bond, as do the two atoms in $N_{2}$ . One shared pair is a single bond.

4.1.3 Lewis Representation of Simple Molecules (the Lewis Structures)

A Lewis structure does not show shape, but it shows which atoms are joined, by how many pairs, and where the lone pairs sit. That explains a great deal of chemistry, so it pays to draw them reliably. Use this recipe:

Count the valence electrons of all the atoms. For an anion add one electron per negative charge; for a cation remove one per positive charge. ( $S O_{4}^{2 -}$ : $6 + 4 \times 6 + 2 = 32$ ; $H_{3} O^{+}$ : $3 + 6 - 1 = 8$ .)
Sketch the skeleton. The least electronegative atom, never hydrogen, usually goes in the centre: S in $S O_{4}^{2 -}$ , C in $C O C l_{2}$ .
Join each outer atom to the centre with a single bond, two electrons each.
Complete the octets of the outer atoms with lone pairs, then put any electrons left over on the central atom.
Fix the centre. If the central atom still lacks an octet, turn a lone pair on a neighbour into a shared pair, making a double or triple bond.

4/4One O lone pair becomes a shared pair: C=O. All octets complete.

Fig. 4.3Every electron is placed and every atom ends with an octet. A tally like this works for any molecule or ion.NCERT Table 4.1 · p. 103

Phosgene skeleton: C in the centre, O above, two Cl below. Step 1 shows only symbols and the tally 4 + 6 + 14 = 24 electrons. Step 2 adds three single bonds, 18 electrons left. Step 3 adds three lone pairs on each outer atom and notes that carbon has only 6. Step 4 turns an oxygen lone pair into a second C–O bond, so every atom has eight.

Example 4.1NCERT Problem 4.1 · p. 104

Lewis structure of hydrogen cyanide

Draw the Lewis structure of hydrogen cyanide, $H C N$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 4.2NCERT Problem 4.2 · p. 104

Lewis structure of the nitronium ion

Draw the Lewis structure of the nitronium ion, $N O_{2}^{+}$ .

Try it yourself first. Then reveal the solution one step at a time.

4.1.4 Formal Charge

Where does the charge of $N O_{2}^{+}$ actually sit? The ion carries it as a whole, but a simple bookkeeping rule lets us tag each atom with a formal charge (औपचारिक आवेश). The rule treats every bond as perfectly fair sharing: an atom is credited with half the electrons of its bonds and all the electrons of its own lone pairs. Comparing that credit with the atom's valence electrons gives:

Formal charge

F.C. = V - L - \frac{1}{2} S

(4.1)

where

$V$: valence electrons of the free atom
$L$: electrons in lone pairs on the atom
$S$: electrons in the bonds to the atom (shared electrons)

In $N O_{2}^{+}$ (Example 4.2), nitrogen has $V = 5$ , $L = 0$ and $S = 8$ , so its formal charge is $5 - 0 - 4 = + 1$ . Each oxygen has $6 - 4 - 2 = 0$ . The formal charges add up to +1, the charge on the ion: a useful check on any structure.

Treat formal charges as accounting, not measurement. In $N O_{2}^{+}$ the oxygen atoms, being more electronegative, actually hold more electron density than nitrogen despite their zero formal charge. Formal charges earn their keep when you must choose between possible drawings: prefer the structure with the fewest and smallest formal charges, with any negative one on the more electronegative atom.

Example

Choosing a structure for dinitrogen oxide

Dinitrogen oxide, $N_{2} O$ (laughing gas), has 16 valence electrons. Three octet structures are possible: (a) $N = N = O$ , (b) $N \equiv N - O$ and (c) $N = O = N$ . Use formal charges to decide which describe the molecule best.

Try it yourself first. Then reveal the solution one step at a time.

4.1.5 Limitations of the Octet Rule

Most organic molecules and nearly all compounds of the second-period elements obey the octet rule. Outside that comfort zone it breaks down in three ways.

Compare

Three ways to break the octet rule

Incomplete octet

The central atom has fewer than eight electrons.
Common for atoms with fewer than four valence electrons: Be, B, Al.
Examples: $B e C l_{2}$ gas (4 electrons on Be); $B F_{3}$ and $B (C H_{3})_{3}$ (6 on B).

Odd-electron molecules

An odd total of electrons means at least one atom cannot have an octet.
Examples: $N O$ (11 valence electrons), $N O_{2}$ (17), $C l O_{2}$ (19).

Expanded octet

A central atom from period 3 onwards holds more than eight electrons.
Examples: $P C l_{5}$ (10 around P), $S F_{6}$ (12 around S), $I F_{7}$ (14 around I).

Expanded octets used to be explained by the empty d orbitals of period-3 atoms, a description you will meet in Section 4.6.3. The same element can also obey the rule: sulfur has an octet in $H_{2} S$ but twelve electrons around it in $S F_{6}$ .

The rule also has deeper flaws. Its very basis, the 'unreactive' noble gas, is not quite true: xenon forms $X e F_{4}$ , $X e F_{6}$ and $X e O_{3}$ , and krypton forms $K r F_{2}$ . It gives no clue to the shape of a molecule. And because it counts electrons with no reference to energy, it cannot compare the stabilities of molecules or the strengths of bonds.

4.2 Ionic or Electrovalent Bond

Kössel's picture says an ionic compound forms in two stages: the atoms become ions, then the ions pack into a crystal. Whether it forms depends on how easily the ions are made and on how much energy the crystal gives back. The steps for a metal M and a non-metal X are:

\begin{aligned} M (g) & \overset{}{\to} M^{+} (g) + e^{-} \\ X (g) + e^{-} & \overset{}{\to} X^{-} (g) \\ M^{+} (g) + X^{-} (g) & \overset{}{\to} M X (s) \end{aligned}

The first step costs the ionisation enthalpy, which is always positive. The second involves the electron gain enthalpy, $Δ_{e g} H$ : the enthalpy change for adding one electron to an isolated gaseous atom in its ground state, negative for some elements (energy released) and positive for others. Electron affinity describes the same change with the opposite sign. Ionic bonds therefore form most readily between a low ionisation enthalpy element and one with a large negative electron gain enthalpy: a metal and a non-metal.

So in a typical ionic compound the cation comes from a metal and the anion from a non-metal. A well-known exception is the ammonium ion, $N H_{4}^{+}$ : all its atoms are non-metals, yet it plays the part of a metal ion in salts such as ammonium chloride and ammonium nitrate.

Now look at the numbers for potassium chloride (Fig. 4.4). Ionising potassium costs $419 k J m o l^{- 1}$ ; adding an electron to chlorine releases $349 k J m o l^{- 1}$ . Making the pair of gaseous ions therefore still costs $+ 70 k J m o l^{- 1}$ : reaching octets is uphill. The crystal tips the balance. Bringing the ions together into the solid releases about $718 k J m o l^{- 1}$ , so the overall change is strongly downhill.

4/4The lattice releases about $718 k J m o l^{- 1}$ : overall about $648 k J m o l^{- 1}$ downhill.

Fig. 4.4Making the ions costs energy; the lattice repays it many times over.NCERT p. 106

An energy ladder starting at K(g) + Cl(g). Step 1 shows the starting level only. Step 2 adds an arrow up of 419 kJ/mol for ionising potassium. Step 3 adds an arrow down of 349 kJ/mol to K+ and Cl− ions, still above the start. Step 4 adds an arrow down of 718 kJ/mol to KCl(s), 648 kJ/mol below the start overall.

In the solid, each ion is surrounded by ions of opposite charge in a regular three-dimensional lattice. In KCl (the rock-salt arrangement) each $K^{+}$ touches six $C l^{-}$ ions and each $C l^{-}$ six $K^{+}$ ; the arrangement a salt adopts depends on the sizes of its ions. So the stability of an ionic compound is judged by its lattice enthalpy, not by octets alone.

4.2.1 Lattice Enthalpy

Definition

Lattice enthalpy

The energy required to pull one mole of an ionic solid apart into gaseous ions that no longer interact.

Example $K C l (s) \overset{}{\to} K^{+} (g) + C l^{-} (g)$ needs about $718 k J m o l^{- 1}$ .

A single ion in a crystal is pulled by every oppositely charged ion and pushed by every like-charged one, near and far, in a pattern set by the crystal structure. Calculating a lattice enthalpy therefore means summing these interactions over the whole lattice; one ion pair is not enough. Two trends are worth remembering. Lattice enthalpy rises steeply with the charges on the ions and falls as the ions get larger. Magnesium oxide, with $M g^{2 +}$ and $O^{2 -}$ ions, has a lattice enthalpy of roughly $3800 k J m o l^{- 1}$ , more than five times that of KCl.

4.3 Bond Parameters

A bond can be described by a handful of measurable quantities: its length, the angles it makes, the energy needed to break it and its order. Two further ideas rest on these: resonance and polarity.

4.3.1 Bond Length

The bond length is how far apart the nuclei of two bonded atoms sit when the molecule is in its lowest-energy state; spectroscopy and X-ray or electron diffraction measure it. A bond length can be split into two contributions, one from each atom: the atoms' covalent radii. An element's covalent radius is found by halving the length of a single bond between two of its own atoms (half the Cl–Cl distance gives the radius of Cl). Then, for atoms A and B:

Bond length from covalent radii

R = r_{A} + r_{B}

(4.2)

where

$R$: the bond length
$r_{A}, r_{B}$: the covalent radii of A and B

Try it on bromine monochloride: the covalent radii of Br ( $114 p m$ ) and Cl ( $99 p m$ ) add up to $213 p m$ , close to the measured $214 p m$ . Atoms that are not bonded, such as atoms of neighbouring molecules in a solid, stay much farther apart, because their full electron clouds bump into each other. Half of that contact distance between like atoms is the van der Waals radius, a measure of the size of a non-bonded atom. Bromine's is about $185 p m$ , far larger than its covalent radius (Fig. 4.5).

Fig. 4.5The covalent radius comes from a bonded pair; the van der Waals radius from atoms in neighbouring molecules that just touch.NCERT Fig. 4.2 · p. 107

Two Br2 molecules end to end, each atom drawn as a small core circle inside a larger van der Waals circle. A bracket over the bonded pair marks 228 pm, twice the covalent radius; a bracket between the touching atoms of neighbouring molecules marks 370 pm, twice the van der Waals radius.

4.3.2 Bond Angle

Where two bonds meet at an atom they make a bond angle; strictly, it is the angle between the orbitals holding the two bonding pairs at the central atom of a molecule or complex ion. Bond angles come from spectroscopic measurements and are quoted in degrees. Together with bond lengths they fix the positions of the atoms, which is what the shape of a molecule means: H–N–H in ammonia is 107°, and H–S–H in hydrogen sulfide only about 92°.

4.3.3 Bond Enthalpy

Definition

Bond enthalpy

The energy needed, per mole, to break a particular kind of bond between two atoms, with every species in the gas phase. Unit: $k J m o l^{- 1}$ .

ExampleBreaking a mole of Cl–Cl bonds needs $243 k J m o l^{- 1}$ .

\begin{aligned} C l_{2} (g) \overset{}{\to} 2 C l (g) \\ Δ_{a} H = 243 k J m o l^{- 1} \\ H B r (g) \overset{}{\to} H (g) + B r (g) \\ Δ_{a} H = 366 k J m o l^{- 1} \end{aligned}

The larger the bond enthalpy, the stronger the bond. Multiple bonds are especially strong: the triple bond in carbon monoxide needs about $1076 k J m o l^{- 1}$ , the most of any diatomic molecule.

In a molecule with several identical bonds, each successive break needs a different energy, because the surroundings of the remaining bonds change. The first C–H bond of methane needs about $439 k J m o l^{- 1}$ , the last far less. So we use the average (mean) bond enthalpy: the total energy to break all bonds of that kind, divided by their number. Atomising methane, $C H_{4} (g) \overset{}{\to} C (g) + 4 H (g)$ , needs about $1664 k J m o l^{- 1}$ , so the average C–H bond enthalpy is $416 k J m o l^{- 1}$ . Tables list such averages.

4.3.4 Bond Order

Definition

Bond order

In the Lewis picture, the number of shared electron pairs between two atoms: 1 for a single bond, 2 for a double bond, 3 for a triple bond.

So $C l_{2}$ has bond order 1, $O_{2}$ 2 and $N_{2}$ 3; carbon monoxide, with three shared pairs, also has 3. Isoelectronic species (same number of electrons, similar structure) have the same bond order: $C N^{-}$ , $C_{2}^{2 -}$ (the acetylide ion) and $N O^{+}$ are isoelectronic with $N_{2}$ and all have bond order 3.

One correlation runs through the whole of chemistry: as bond order increases, bond enthalpy increases and bond length decreases. Table 4.1 shows it for carbon–carbon and nitrogen–nitrogen bonds.

Table 4.1Typical lengths and average enthalpies of single, double and triple bonds

Bond	Bond order	Length / pm	Enthalpy / kJ mol⁻¹
$C - C$	1	154	348
$C = C$	2	134	614
$C \equiv C$	3	120	839
$N - N$	1	145	163
$N = N$	2	125	418
$N \equiv N$	3	110	946

Quick check

1 of 3

Covalent radii: C

77 p m

, Cl

99 p m

. Estimate the C–Cl bond length in

C C l_{4}

, in pm.

Show answer

Answer 176 pm

R = r_{C} + r_{C l} = 77 + 99 = 176 p m

, close to the measured value. Doubling one radius instead of adding the two gives a wrong answer.

4.3.5 Resonance Structures

Some species refuse to be drawn with one Lewis structure. Take the formate ion, $H C O O^{-}$ , the anion of formic acid (the acid in ant stings). Any octet structure has one C=O and one C–O bond, which should differ in length: roughly $122 p m$ and $143 p m$ . Experiment finds the two carbon–oxygen bonds identical, about $125 p m$ , in between.

The way out is resonance. When a single Lewis structure cannot describe a species, we write two or more structures that keep every nucleus in the same place and differ only in where the electrons are drawn. These are the canonical or resonance structures, linked by a double-headed arrow $⟷$ . The species itself has one structure, the resonance hybrid (अनुनाद संकर), which blends their features. In formate each C–O bond is about one and a half bonds, and the negative charge is shared equally by the two oxygens (Fig. 4.6).

3/3The real ion: two equal C–O bonds of about $125 p m$ , half a negative charge on each O.

Fig. 4.6Neither canonical structure exists; the hybrid has two identical C–O bonds.NCERT Fig. 4.3 · p. 109

Formate ion: C bonded to H and two O atoms. Step 1, structure I: double bond to the upper O, single bond to the lower O, which carries the minus charge and three lone pairs. Step 2, structure II: the same nuclei with the double bond and charge swapped. Step 3, hybrid: each C–O drawn as a solid plus a dashed line, 125 pm, with half a negative charge on each O.

Definition

Resonance hybrid

The single real structure of a species that needs two or more canonical Lewis structures to describe it. It is lower in energy than any of the canonical structures.

Resonance has two measurable effects. The real species is lower in energy, and so more stable, than any single canonical structure suggests. And properties that would differ between bonds in one structure, such as length, come out identical. Think of a mixed colour: green paint is not blue half the time and yellow the rest, and it was never a tin of each. The same goes for a resonance hybrid:

It has one structure all the time; nothing flips back and forth.
The canonical structures are drawings we need because a Lewis structure cannot show a partial bond; none of them is a real molecule.
Resonance is not tautomerism. Keto and enol tautomers really interconvert in an equilibrium because a hydrogen atom moves; canonical structures differ only in where electrons are drawn.

Example 4.3NCERT Problem 4.3 · p. 110

Resonance in the nitrate ion

All three N–O bonds in the nitrate ion, $N O_{3}^{-}$ , are equally long. Explain this using resonance.

Try it yourself first. Then reveal the solution one step at a time.

Example 4.4NCERT Problem 4.4 · p. 110

Bond lengths in benzene

In benzene, $C_{6} H_{6}$ , the six carbon atoms form a flat ring and every carbon–carbon bond is $139 p m$ long. A typical C–C bond is $154 p m$ and a C=C bond $134 p m$ . What do these data tell you about the structure of benzene?

Try it yourself first. Then reveal the solution one step at a time.

4.3.6 Polarity of Bonds

Pure ionic and pure covalent bonds are limiting cases; real bonds lie somewhere between. If both atoms are the same, as in $H_{2}$ , $N_{2}$ or $C l_{2}$ , neither nucleus wins the tug of war: the shared pair sits midway and the bond is non-polar. If one atom is more electronegative, it drags the pair towards itself. In hydrogen chloride the density shifts towards chlorine, so chlorine is slightly negative ( $δ -$ ) and hydrogen slightly positive ( $δ +$ ): a polar covalent bond.

A polar bond separates charge, so it has a dipole moment (द्विध्रुव आघूर्ण), written $μ$ . It is the separated charge multiplied by the distance between the centres of positive and negative charge.

Dipole moment

μ = Q \times r

(4.3)

where

$Q$: the magnitude of the charge at each centre
$r$: the distance between the centres

Dipole moments are usually given in debye (D): $1 D = 3.336 \times 10^{- 30} C m$ . A dipole moment is a vector. Physicists draw it from the negative centre to the positive one. Chemists draw a crossed arrow, cross at the positive end and head at the negative end, showing where electron density has shifted. The two conventions point opposite ways.

In a molecule with several polar bonds, each bond has its own bond dipole, and the dipole moment of the molecule is their vector sum. Shape therefore matters as much as polarity (Fig. 4.7). In water the two O–H bond dipoles meet at 104.5°; they do not cancel, and their resultant gives water a dipole moment of 1.85 D. In carbon dioxide the two C=O dipoles point in exactly opposite directions and cancel, so $C O_{2}$ has zero dipole moment although each bond is polar. In boron trifluoride, three B–F dipoles in a plane at 120° also cancel.

Here is a puzzle. Fluorine is more electronegative than nitrogen and hydrogen is less, so N–F bonds are polar in the opposite sense to N–H bonds. Yet $N H_{3}$ has a dipole moment of 1.47 D and $N F_{3}$ only 0.23 D. Both molecules are pyramids with a lone pair on nitrogen, and that lone pair is a region of negative charge on the far side of the nitrogen. Now add the vectors. In ammonia the bonds pull density towards nitrogen and the lone pair lies beyond it, so the two contributions point the same way and add. In nitrogen trifluoride the bonds pull density down towards the fluorines while the lone pair points up, so they largely cancel.

4/4The lone pair reinforces the bond dipoles in NH₃ but opposes them in NF₃.

Fig. 4.7Shape decides whether bond dipoles add up or cancel. Arrows use the chemists' crossed-arrow convention.NCERT p. 111

Crossed arrows show bond dipoles. Step 1: H–Cl with δ+ on H, δ− on Cl and electron density shifted towards Cl. Step 2: bent water, two O–H dipoles pointing to O and a resultant of 1.85 D. Step 3: linear O=C=O with equal, opposite dipoles and zero net dipole. Step 4: NH3 and NF3 pyramids; in NH3 the bond and lone-pair arrows point the same way (1.47 D), in NF3 opposite ways (0.23 D).

Table 4.2Dipole moments of some molecules in the gas phase

Molecule	Shape	μ / D
$H_{2} O$	bent	1.85
$H_{2} S$	bent	0.97
$S O_{2}$	bent	1.63
$C O_{2}$	linear	0
$N H_{3}$	trigonal pyramidal	1.47
$N F_{3}$	trigonal pyramidal	0.23
$B F_{3}$	trigonal planar	0
$C H_{3} C l$	tetrahedral	1.89
$C H_{2} C l_{2}$	tetrahedral	1.60
$C H C l_{3}$	tetrahedral	1.04
$C C l_{4}$	tetrahedral	0

Notice the chloromethanes. All four are roughly tetrahedral, yet the dipole moment changes as chlorine atoms replace hydrogens, and it vanishes in perfectly symmetric $C C l_{4}$ .

Example

How ionic is the H–Cl bond?

HCl has a bond length of $127 p m$ and a dipole moment of $1.11 D$ . What would its dipole moment be if the bond were completely ionic, with charges $\pm e$ on the two nuclei? What percentage ionic character does the real value suggest? ( $e = 1.602 \times 10^{- 19} C$ .)

Try it yourself first. Then reveal the solution one step at a time.

Just as covalent bonds have some ionic character, ionic bonds have some covalent character. A small or highly charged cation pulls the electron cloud of a nearby anion towards itself, distorting (polarising) it. Electron density then builds up between the two nuclei, as in a covalent bond. Kasimir Fajans summarised when this matters most:

Smaller cation, larger anion: more covalent. $L i I$ is more covalent than $L i F$ , because the large iodide ion is easily distorted.
Higher charge on the cation: more covalent. $A l C l_{3}$ , with $A l^{3 +}$ , is far more covalent than $N a C l$ : it sublimes near 180 °C, whereas NaCl melts only at 801 °C.
Same size and charge: a cation with a d-electron shell, $(n - 1) d^{n} n s^{0}$ , polarises more than one with a noble-gas shell, $n s^{2} n p^{6}$ . $H g^{2 +}$ and $C a^{2 +}$ are almost the same size, yet $H g C l_{2}$ melts at 276 °C and $C a C l_{2}$ at 772 °C.

The polarising power of the cation, the polarisability of the anion and the resulting distortion together decide how much covalent character an ionic bond has.

4.4 The Valence Shell Electron Pair Repulsion (VSEPR) Theory

Lewis structures say nothing about shape, yet shape decides a molecule's polarity, how it reacts and even how it smells. The valence shell electron pair repulsion (VSEPR) theory predicts shapes from little more than a count of electron pairs. Nevil Sidgwick and Herbert Powell proposed it in 1940; Ronald Nyholm and Ronald Gillespie refined it in 1957. It rests on a short list of assumptions:

Only the electron pairs in the valence shell of the central atom matter, bonding pairs and lone pairs alike.
Being clouds of negative charge, the pairs push one another apart and settle where the push is least: as far from each other as they can get, as if spread over the surface of a sphere.
All the electrons of a double or triple bond lie between the same two atoms, so a multiple bond counts once, as a single 'super pair'.
For a species with resonance structures, any one structure gives the same prediction.
Not all pairs push equally hard: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.

That last ranking was Nyholm and Gillespie's contribution. A bonding pair is held between two nuclei and stretched along the bond; a lone pair is held by one nucleus only, so it spreads out close to the central atom and crowds its neighbours. Shapes that contain lone pairs are therefore distorted, with bond angles a few degrees smaller than the ideal values.

We write a species as ${A B}_{x} E_{y}$ : A is the central atom, B a bonded atom and E a lone pair on A. The total $x + y$ fixes the arrangement of electron pairs (Fig. 4.8); the shape describes only where the atoms are. We look first at central atoms without lone pairs, then at those with them.

3/5Tetrahedral: 109.5°.

Fig. 4.8Electron pairs spread as far apart as they can: these arrangements give the largest possible separations.NCERT Fig. 4.6 · p. 113

A central atom A with outer atoms, wedges for bonds towards you and dashes for bonds away. Step 1: two pairs, linear, 180°. Step 2: three, trigonal planar, 120°. Step 3: four, tetrahedral, 109.5°. Step 4: five, trigonal bipyramidal, 90° to the axis and 120° in the equator. Step 5: six, octahedral, 90°.

No lone pairs ( ${A B}_{x}$ ). The shape is the arrangement itself: $H g C l_{2}$ is linear, $B C l_{3}$ trigonal planar, $C C l_{4}$ tetrahedral, $P F_{5}$ trigonal bipyramidal and $S F_{6}$ octahedral. Multiple bonds change nothing: $C O_{2}$ , with two C=O 'super pairs', is linear, and $S O_{3}$ is trigonal planar.

One or more lone pairs ( ${A B}_{x} E_{y}$ ). The pairs keep the same arrangement, but the shape names only the atoms. Methane, ammonia and water all have four pairs arranged tetrahedrally (Fig. 4.9). Ammonia has one lone pair, so its shape is a trigonal pyramid, and the stronger lone pair–bond pair repulsion closes the H–N–H angle from 109.5° to 107°. Water has two lone pairs, so it is bent, and the even stronger lone pair–lone pair repulsion narrows H–O–H to 104.5°.

2/3One lone pair: trigonal pyramid, 107°.

Fig. 4.9Each lone pair takes up more room than a bond pair and squeezes the bond angles.NCERT Table 4.8 · p. 116

A tetrahedral arrangement drawn three times. Step 1: methane, four C–H bonds, 109.5°. Step 2: ammonia, the top position holds a lone-pair lobe, shape pyramidal, angle 107°. Step 3: water, two lone-pair lobes and two O–H bonds, shape bent, angle 104.5°.

In a trigonal bipyramid the five positions are not alike. The three equatorial positions are 120° apart; each axial position makes 90° with all three equatorial ones. A lone pair placed axially would meet three bond pairs at 90°; placed equatorially, it meets only two (Fig. 4.10). Since 90° repulsions are the costly ones, lone pairs go equatorial. Thus $S e F_{4}$ ( $A B_{4} E$ ) is a see-saw, $B r F_{3}$ ( $A B_{3} E_{2}$ ) is T-shaped, and with three equatorial lone pairs $X e F_{2}$ ( $A B_{2} E_{3}$ ) is linear. In an octahedron, two lone pairs sit opposite each other, which makes $A B_{4} E_{2}$ species square planar.

Fig. 4.10An equatorial lone pair has fewer 90° neighbours, so AB₄E molecules are see-saw shaped.NCERT Table 4.8 · p. 116

Two trigonal bipyramids with four atoms and one lone-pair lobe. Left: the lone pair is axial and dashed lines join it to three bond pairs at 90°, labelled less stable. Right: the lone pair is equatorial with only two dashed 90° contacts, labelled more stable.

Table 4.3VSEPR arrangements and shapes

Type	Bond + lone pairs	Arrangement	Shape	Examples
$A B_{2}$	2 + 0	linear	linear	$C O_{2}$ , $H g C l_{2}$
$A B_{3}$	3 + 0	trigonal planar	trigonal planar	$S O_{3}$ , $B C l_{3}$
$A B_{2} E$	2 + 1	trigonal planar	bent	$S n C l_{2}$ (gas), $N O_{2}^{-}$
$A B_{4}$	4 + 0	tetrahedral	tetrahedral	$C C l_{4}$ , $N H_{4}^{+}$
$A B_{3} E$	3 + 1	tetrahedral	trigonal pyramidal	$P C l_{3}$ , $H_{3} O^{+}$
$A B_{2} E_{2}$	2 + 2	tetrahedral	bent	$O F_{2}$ , $H_{2} S$
$A B_{5}$	5 + 0	trigonal bipyramidal	trigonal bipyramidal	$P F_{5}$ , $A s F_{5}$
$A B_{4} E$	4 + 1	trigonal bipyramidal	see-saw	$S e F_{4}$ , $T e C l_{4}$
$A B_{3} E_{2}$	3 + 2	trigonal bipyramidal	T-shaped	$B r F_{3}$
$A B_{2} E_{3}$	2 + 3	trigonal bipyramidal	linear	$X e F_{2}$ , $I_{3}^{-}$
$A B_{6}$	6 + 0	octahedral	octahedral	$S F_{6}$ , $P F_{6}^{-}$
$A B_{5} E$	5 + 1	octahedral	square pyramidal	$I F_{5}$
$A B_{4} E_{2}$	4 + 2	octahedral	square planar	$X e F_{4}$ , $I C l_{4}^{-}$

In practice VSEPR gets the shapes of most p-block molecules right, even in close cases. Bear in mind that it is a working rule rather than a derivation: why electron-pair repulsion should control shape so well is still argued over. Explore every combination in Fig. 4.11.

Fig. 4.11Choose the numbers of bond pairs and lone pairs, then drag the model to turn it.

Example

Predicting shapes step by step

Predict the shapes of $C l O_{3}^{-}$ , $I C l_{2}^{-}$ , $I F_{4}^{+}$ and $B r F_{4}^{-}$ .

Try it yourself first. Then reveal the solution one step at a time.

Try this

Balloon models of electron pairs

You’ll need

6 round balloons of one size
1 larger balloon
a marker pen

Predict first

You twist four equal balloons together at their knots. How will they settle?

in a flat square, 90° apartpointing to the corners of a tetrahedronall bunched on one side

Blow up six round balloons to the same size and tie them off.
Twist two together at their knots and let go. Note the angle between them.
Twist in a third balloon at the centre, then a fourth, fifth and sixth. Sketch the shape each time.
Make a four-balloon bunch in which one balloon is noticeably larger, standing for a lone pair, and watch the other three.

Quick check

1 of 4

Iodine trifluoride,

I F_{3}

, has three bond pairs and two lone pairs on iodine. What is its shape?

trigonal planartrigonal pyramidalT-shapedsee-saw

Show answer

Answer (C) T-shaped

Five pairs make a trigonal bipyramid; both lone pairs take equatorial positions, leaving the three F atoms in a T. Trigonal planar would ignore the lone pairs.

4.5 Valence Bond Theory

Lewis structures and VSEPR are bookkeeping and geometry; neither explains why a shared pair holds two atoms together. They also miss differences that matter. Fluorine and chlorine molecules each have one shared pair, yet the F–F bond ( $142 p m$ , $158 k J m o l^{- 1}$ ) is weaker than the Cl–Cl bond ( $199 p m$ , $243 k J m o l^{- 1}$ ), even though it is shorter. Two theories built on quantum mechanics tackle such questions: valence bond (VB) theory and molecular orbital (MO) theory.

VB theory dates from 1927, when Walter Heitler and Fritz London applied quantum mechanics to the hydrogen molecule; Linus Pauling and others then extended it. Its full form is mathematical, so here we keep to the ideas: atomic orbitals and configurations from Unit 2, plus overlap and hybridisation, introduced below.

Picture two hydrogen atoms, A and B, drifting together (Fig. 4.12). At large separation nothing happens. As they close in, each nucleus begins to tug on the other atom's electron as well as its own, while the two electrons repel each other and so do the two nuclei. On balance the extra attraction is larger, so the energy of the pair drops. The drop stops at $74 p m$ , where pulls and pushes balance and the energy is as low as it gets: this separation is the H–H bond length. Squeeze the nuclei closer still and their repulsion takes over, driving the energy sharply up.

Sitting at the bottom of this energy well, the molecule has less energy than two free atoms, which is what 'stable' means here. The depth of the well is the bond enthalpy: to pull a mole of $H_{2}$ molecules apart you must supply $436 k J$ , so $H_{2} (g) \overset{}{\to} 2 H (g)$ has $Δ H = + 436 k J m o l^{- 1}$ .

3/4Minimum at $74 p m$ : $436 k J m o l^{- 1}$ below the separate atoms.

Fig. 4.12The bond length is where the energy is lowest; the depth of the well is the bond enthalpy.NCERT Fig. 4.8 · p. 118

Potential energy against distance between two H nuclei: steep at short distance, a minimum at 74 pm and −436 kJ/mol, then rising towards zero. A marker and a pair of atoms show each stage. Step 1: far apart, energy zero. Step 2: closer, attraction arrows, energy falling. Step 3: at the minimum, overlapping atoms. Step 4: too close, repulsion arrows, energy rising steeply.

4.5.1 Orbital Overlap Concept

At the bottom of the well the two 1s orbitals occupy some of the same space: they overlap. In the shared region the two electrons, which must have opposite spins, pair up. This is VB theory's picture of a covalent bond: half-filled valence orbitals on two atoms overlap, and their electrons pair with opposite spins. More overlap means a stronger bond.

4.5.2 Directional Properties of Bonds

Spherical 1s orbitals overlap equally well from any direction, so for $H_{2}$ the question of direction never arises. It does for bigger molecules: methane has four bonds at exactly 109.5°, ammonia is a low pyramid and water is bent. A complete theory has to produce these angles, and VB theory does so by adding hybridisation to overlap (Section 4.6).

4.5.3 Overlapping of Atomic Orbitals

Each lobe of an orbital carries a sign, + or −: the sign of the wave function there, rather like the crest and trough of a water wave. It has nothing to do with electric charge. When two lobes of the same sign merge, the waves reinforce: the overlap is positive and can make a bond. Lobes of opposite sign cancel, giving negative overlap. And if an orbital meets the + and − halves of another equally, as an s orbital does when it approaches a p orbital from the side, the effects cancel exactly and the overlap is zero (Fig. 4.13). Bonds need positive overlap.

Now test this on methane. Suppose carbon first moves one 2s electron into the empty 2p orbital, giving $2 s^{1} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}$ and four unpaired electrons; the cost is repaid by forming four bonds instead of two. If these four pure orbitals overlapped with hydrogen 1s orbitals, three C–H bonds would lie along the x, y and z axes at 90° to one another, and the fourth, from the spherical 2s orbital, would point nowhere in particular. That is not methane, whose four bonds are identical and 109.5° apart. Ammonia and water fare no better: pure p orbitals predict 90°, not 107° and 104.5°.

4/5Overlap along the axis: a σ bond.

Fig. 4.13Signs are phases of the wave function, not charges. Matching phases give bonding overlap.NCERT Fig. 4.9 · p. 119

Orbital lobes marked + or − on a horizontal axis. Step 1: two s orbitals of the same sign overlap, positive overlap. Step 2: lobes of opposite sign meet end to end, negative overlap. Step 3: an s orbital meets a p orbital lying across the axis, zero overlap. Step 4: two p orbitals overlap head-on with the same sign, a σ bond. Step 5: two parallel p orbitals overlap above and below the axis, a π bond.

4.5.4 Types of Overlapping and Nature of Covalent Bonds

Overlap comes in two geometries. In a sigma (σ) bond the orbitals meet head-on along the internuclear axis, called axial overlap. This can happen between two s orbitals (as in $H_{2}$ ), between an s and a p orbital (as in $H F$ ), or between two p orbitals pointing at each other (as in $F_{2}$ ). The electron density of a σ bond is symmetrical about the axis.

A pi (π) bond needs two p orbitals standing side by side, parallel to each other and at right angles to the axis. They overlap edge to edge above the axis and below it, so the bond's electron density forms two regions on either side of the line joining the nuclei, with none on the line itself.

4.5.5 Strength of Sigma and pi Bonds

Compare the numbers: a C–C single bond needs $348 k J m o l^{- 1}$ to break, a C=C double bond $614 k J m o l^{- 1}$ , well short of twice as much. The second bond is a π bond, and sideways overlap is smaller than head-on overlap, so a π bond is weaker than a σ bond. Between two atoms the first bond is always σ, and any extra bonds are π: a double bond is σ + π, a triple bond σ + 2π.

4.6 Hybridisation

To rescue VB theory from the methane problem (Section 4.5.3), Linus Pauling let an atom blend its valence orbitals before it bonds. In hybridisation (संकरण), orbitals of one atom that are close in energy are combined into the same number of new orbitals, all alike in shape and energy, and bonds are then formed by these hybrid orbitals. Carbon's one 2s and three 2p orbitals, for example, give four $s p^{3}$ hybrids.

Definition

Hybridisation

The mixing of atomic orbitals of similar energy on one atom to form an equal number of equivalent hybrid orbitals, which are then used for bonding.

Four facts about hybrid orbitals are worth remembering:

You get out as many hybrids as you put atomic orbitals in.
Within a set the hybrids are identical, apart from the direction each one points.
Each has one fat lobe that reaches further than a pure orbital, so it overlaps better and makes a stronger bond.
A set points in directions that keep the electron pairs as far apart as possible, so naming the hybridisation names the geometry.

And four conditions:

Only orbitals of the valence shell take part.
Their energies must be close: a 1s orbital, far below, never mixes with 2s and 2p.
No electron has to be promoted first; promotion is just bookkeeping.
Orbitals that hold lone pairs can join in, not only half-filled ones.

4.6.1 Types of Hybridisation

sp hybridisation. One s and one p orbital give two sp hybrids, each half s and half p in character. The two point in exactly opposite directions, 180° apart, each with a large lobe in front and a small one behind; the arrangement is also called diagonal hybridisation. An sp atom with two partners therefore gives a linear molecule. Gaseous beryllium fluoride shows this. Beryllium ( $1 s^{2} 2 s^{2}$ ) has no unpaired electron until one 2s electron moves to 2p; the 2s and that 2p orbital then blend into two sp hybrids, each overlapping end-on with a singly occupied 2p orbital of fluorine, so F–Be–F is a straight line.

sp² hybridisation. One s and two p orbitals make three hybrids, each one-third s, spread in a plane at 120° to one another. In boron trifluoride, boron ( $2 s^{2} 2 p^{1}$ ) uses its three hybrids for three B–F σ bonds, so the molecule is a flat triangle with 120° angles. One 2p orbital of boron is left out of the mixing; it stays empty and sticks out above and below the plane.

sp³ hybridisation. All three p orbitals join the s orbital, giving four hybrids that are one-quarter s and three-quarters p. They aim at the corners of a tetrahedron, 109.5° apart. In methane each overlaps the 1s orbital of a hydrogen atom, producing four identical C–H σ bonds (Fig. 4.14).

Fig. 4.14Promotion gives four half-filled orbitals; mixing makes them identical.NCERT Fig. 4.12 · p. 122

Three rows of orbital boxes for carbon. Ground state: 2s paired, two of three 2p boxes with one electron. Excited state: one electron in 2s and in each 2p box. sp3: four equal boxes with one electron each, ready to form four C–H σ bonds.

3/3Four hybrids, 109.5°: tetrahedral (CH₄).

Fig. 4.15More p character means more hybrids and smaller angles between them.NCERT Fig. 4.10 · p. 121

Hybrid orbitals drawn as large lobes with tiny back lobes around a nucleus. Step 1: two sp lobes in a line, 180°, 50 % s character. Step 2: three sp2 lobes in a plane, 120°, 33 % s. Step 3: four sp3 lobes towards the corners of a tetrahedron, 109.5°, one drawn dashed behind, 25 % s.

The same set of four explains ammonia and water once some hybrids hold lone pairs. Nitrogen has five valence electrons: two fill one sp³ hybrid as a lone pair and three sit singly in the others, ready to bond to three hydrogens. Oxygen has six: two lone-pair hybrids and two singly occupied ones, so it bonds to two hydrogens. The lone pairs then squeeze the bonds just as VSEPR says, from 109.5° to 107° in pyramidal $N H_{3}$ and to 104.5° in bent (V-shaped) $H_{2} O$ .

4.6.2 Other Examples of sp³, sp² and sp Hybridisation

Ethane, $C_{2} H_{6}$ : both carbons sp³. The carbons join through an sp³–sp³ σ bond, $154 p m$ long, and each uses its other three hybrids for sp³–1s σ bonds to hydrogen, $109 p m$ long.

Ethene, $C_{2} H_{4}$ : both carbons sp². Each carbon's three sp² hybrids build the flat σ framework: one C–C bond and two C–H bonds ( $108 p m$ ). Left over on each carbon is one p orbital standing straight up from the molecular plane; the two overlap sideways in a π bond, with density above and below the plane. So the C=C double bond ( $134 p m$ ) is one σ bond plus one π bond. All six atoms are coplanar, with H–C–H about 117° and H–C–C about 121°.

Ethyne, $C_{2} H_{2}$ : both carbons sp. The two sp hybrids on each carbon point in opposite directions, one at the other carbon and one at a hydrogen, so all four atoms lie on a line. Each carbon keeps two p orbitals at right angles to each other and to the axis. They pair up sideways with their partners to make two π bonds that together wrap around the axis. The C≡C bond, σ + 2π, is $120 p m$ long (Fig. 4.16).

2/4The p orbitals overlap sideways: one π bond above and below the plane.

Fig. 4.16Hybrids build the σ framework; leftover p orbitals overlap sideways to make π bonds.NCERT Fig. 4.15 · p. 123

Step 1: ethene in perspective, σ bonds from sp2 hybrids and an upright p orbital on each carbon. Step 2: those p orbitals merge into π clouds above and below the C–C axis. Step 3: linear ethyne with σ bonds from sp hybrids and two p orbitals at right angles on each carbon. Step 4: they overlap in two pairs, giving two π bonds around the axis.

Example

Hybridisation and bond counting in acrylonitrile

Acrylonitrile, $C H_{2} = C H - C \equiv N$ , is used to make acrylic fibres. Give the hybridisation of each carbon (call them C1, C2, C3 from the left) and count the σ and π bonds.

Try it yourself first. Then reveal the solution one step at a time.

4.6.3 Hybridisation of Elements involving d Orbitals

Atoms beyond period 2 have d orbitals within reach. In period 3 the empty 3d level lies close in energy to 3s and 3p on one side and to 4s and 4p on the other. That allows two kinds of d-containing hybrid: those built from $n s$ , $n p$ and $n d$ orbitals, and those built from $(n - 1) d$ , $n s$ and $n p$ . A mixture of 3p, 3d and 4s is ruled out, because the 3p–4s energy gap is too wide. Table 4.4 lists the common types.

Table 4.4Common hybridisation schemes

Hybrid	Orbitals mixed	Geometry	Angles	Examples
$s p$	s + p	linear	180°	$B e F_{2}$ (gas), $C_{2} H_{2}$
$s p^{2}$	s + 2p	trigonal planar	120°	$B F_{3}$ , $C_{2} H_{4}$
$s p^{3}$	s + 3p	tetrahedral	109.5°	$C H_{4}$ , $N H_{4}^{+}$
$d s p^{2}$	(n−1)d + s + 2p	square planar	90°	$[P t (N H_{3})_{4}]^{2 +}$
$s p^{3} d$	s + 3p + d	trigonal bipyramidal	90°, 120°	$P C l_{5}$ , $A s F_{5}$
$s p^{3} d^{2}$	s + 3p + 2d	octahedral	90°	$S F_{6}$ , $P F_{6}^{-}$
$d^{2} s p^{3}$	2(n−1)d + s + 3p	octahedral	90°	$[F e (C N)_{6}]^{3 -}$

$P C l_{5}$ (sp³d). Phosphorus ( $[N e] 3 s^{2} 3 p^{3}$ ) has only three unpaired electrons, but moving one 3s electron into an empty 3d orbital gives five. The 3s, three 3p and one 3d orbitals blend into five sp³d hybrids aimed at the corners of a trigonal bipyramid, and each overlaps a half-filled 3p orbital of a chlorine atom. Unlike the bonds of methane, these five are of two kinds (Fig. 4.17). Three equatorial bonds form a flat triangle around phosphorus, 120° apart; two axial bonds point straight up and down, at 90° to that triangle. An axial bond pair has three neighbours at 90°, an equatorial one only two, so the axial bonds are pushed out a little, to about $212 p m$ against $202 p m$ . Being longer, they are weaker, and on warming $P C l_{5}$ readily gives up chlorine.Modern calculations show that d orbitals play a much smaller part in such bonds than this model suggests, but the sp³d and sp³d² labels remain a convenient link between hybrids and VSEPR shapes.

Fig. 4.17Axial bonds meet three neighbours at 90°, equatorial bonds only two, so the axial bonds are longer.NCERT Fig. 4.17 · p. 125

Phosphorus at the centre of a trigonal bipyramid. Three equatorial Cl atoms lie in a shaded plane at 120° to one another, one bond labelled 202 pm. Two axial Cl atoms sit above and below at 90° to the plane, the upper bond labelled 212 pm.

$S F_{6}$ (sp³d²). Sulfur ( $[N e] 3 s^{2} 3 p^{4}$ ) has two unpaired electrons; promoting one 3s and one 3p electron into 3d orbitals gives six. The six sp³d² hybrids point to the corners of a regular octahedron, and each forms a σ bond with a fluorine 2p orbital. All six S–F bonds are equivalent, with 90° between neighbours.

Quick check

1 of 3

What is the hybridisation of carbon in

C O_{2}

s p^{2}

s p^{3}

s p

s p^{3} d

Show answer

Answer (C) $s p$

Carbon forms two σ bonds and two π bonds (two C=O), so it uses two sp hybrids and two unhybridised p orbitals; the molecule is linear. Counting four bonds and choosing

s p^{3}

forgets that π bonds use unhybridised p orbitals.

4.7 Molecular Orbital Theory

VB theory ties electrons to particular bonds between two atoms. Molecular orbital (MO) theory, developed by Friedrich Hund and Robert S. Mulliken in 1932, treats them as belonging to the molecule as a whole. Its main ideas are:

Atomic orbitals that are similar in energy and match in symmetry combine to give molecular orbitals, which spread over two or more nuclei (polycentric, unlike the monocentric atomic orbitals).
Orbitals are conserved in number: two atomic orbitals always give two molecular orbitals, a bonding one, lower in energy and more stable, and an antibonding one, higher in energy.
Each molecular orbital describes where an electron is likely to be found around the nuclei of the molecule.
The molecule's electrons fill these orbitals by the rules used for atoms: lowest energy first (aufbau), at most two per orbital with opposite spins (Pauli), and singly into equal-energy orbitals before pairing (Hund).

4.7.1 Formation of Molecular Orbitals: Linear Combination of Atomic Orbitals (LCAO)

An exact molecular orbital would come from solving the Schrödinger equation for the molecule, but that equation has no exact solution once more than one electron is involved. Chemists therefore build approximate molecular orbitals out of atomic ones, the wave functions $ψ$ whose values give the amplitude of the electron wave. In the linear combination of atomic orbitals (LCAO) method a molecular orbital (आण्विक कक्षक) is the sum or the difference of atomic wave functions. For two hydrogen atoms A and B, with 1s wave functions $ψ_{A}$ and $ψ_{B}$ :

LCAO for two 1s orbitals

\begin{aligned} σ & = ψ_{A} + ψ_{B} \\ σ^{*} & = ψ_{A} - ψ_{B} \end{aligned}

(4.4)

where

$σ$: the bonding molecular orbital (addition)
$σ^{*}$: the antibonding molecular orbital (subtraction)

Why is one combination bonding and the other antibonding? Adding the waves reinforces them in the region between the nuclei (Fig. 4.18). Electron density piles up there and attracts both nuclei towards the middle, so an electron in σ has less energy than in either 1s orbital. Subtracting makes the waves cancel midway: the antibonding orbital has a nodal plane between the nuclei, where the density is zero, and its electrons sit mostly behind the nuclei, which are left to repel each other. An electron in $σ^{*}$ therefore has more energy than in a 1s orbital and weakens the bond. Roughly, the bonding orbital drops as far below the atomic level as the antibonding orbital rises above it.

4/4 $σ 1 s$ lies below the atomic orbitals, $σ^{*} 1 s$ above them.

Fig. 4.18Adding waves piles density between the nuclei; subtracting leaves a node there.NCERT Fig. 4.19 · p. 126

Two nuclei A and B on an axis. Step 1: separate 1s wave curves, ψA and ψB. Step 2: their sum, one curve over both nuclei with density between them, σ. Step 3: their difference, positive over A, negative over B, crossing zero at a dashed nodal plane, σ*. Step 4: energy diagram with σ1s below the two 1s levels and σ*1s above.

Definition

Molecular orbital

An orbital that extends over two or more nuclei of a molecule, formed by combining atomic orbitals. Bonding orbitals are lower in energy than the parent atomic orbitals; antibonding orbitals are higher.

4.7.2 Conditions for the Combination of Atomic Orbitals

Atomic orbitals combine effectively only if three conditions hold:

Similar energies. For two identical atoms, a 2s orbital combines with the other atom's 2s, never with its far deeper 1s. (Between very different atoms the matching is looser.)
The same symmetry about the bond axis (taken as z). A $2 p_{x}$ orbital can combine with the other atom's $2 p_{x}$ , but not with its $2 p_{y}$ or $2 p_{z}$ , although all three have the same energy.
Good overlap. The more the orbitals overlap, the more electron density collects between the nuclei.

4.7.3 Types of Molecular Orbitals

Molecular orbitals of diatomic molecules are classed by symmetry and labelled σ, π, δ and so on. Spin a σ orbital about the bond axis and it looks the same from every side. Two 1s orbitals give the pair $σ 1 s$ and $σ^{*} 1 s$ ; two $2 p_{z}$ orbitals, which point along the axis, give $σ 2 p_{z}$ and $σ^{*} 2 p_{z}$ . A π orbital fails this test, because it has a lobe of one sign above the axis and of the other sign below. Side-by-side $2 p_{x}$ orbitals give $π 2 p_{x}$ and $π^{*} 2 p_{x}$ , and $2 p_{y}$ orbitals the matching pair. In the bonding π orbital the density lies above and below the axis; the antibonding $π^{*}$ orbital also has a node between the nuclei (Fig. 4.19).

Fig. 4.19σ orbitals look the same all round the axis; π orbitals have lobes above and below it.NCERT Fig. 4.20 · p. 128

Three rows of atomic orbitals combining, lobes marked + and −. Row a: two 1s orbitals give a σ1s cloud around both nuclei and a σ*1s pair of lobes with a node between. Row b: two 2pz orbitals give σ2pz with a big central lobe and σ*2pz with a node in the middle. Row c: two side-by-side 2px orbitals give π2px lobes above and below the axis and π*2px with four lobes and a node.

4.7.4 Energy Level Diagram for Molecular Orbitals

The 2s and 2p orbitals of two atoms, eight atomic orbitals in all, give eight molecular orbitals. The bonding ones are $σ 2 s$ , $σ 2 p_{z}$ , $π 2 p_{x}$ and $π 2 p_{y}$ ; the antibonding ones are $σ^{*} 2 s$ , $σ^{*} 2 p_{z}$ , $π^{*} 2 p_{x}$ and $π^{*} 2 p_{y}$ . Spectroscopy fixes their order. For $O_{2}$ and $F_{2}$ the order of increasing energy is:

O₂, F₂

\begin{aligned} σ 1 s < σ^{*} 1 s < σ 2 s \\ < σ^{*} 2 s < σ 2 p_{z} \\ < (π 2 p_{x} = π 2 p_{y}) \\ < (π^{*} 2 p_{x} = π^{*} 2 p_{y}) \\ < σ^{*} 2 p_{z} \end{aligned}

For the lighter molecules $L i_{2}$ , $B e_{2}$ , $B_{2}$ , $C_{2}$ and $N_{2}$ , experiment shows a different order:

Li₂ to N₂

\begin{aligned} σ 1 s < σ^{*} 1 s < σ 2 s \\ < σ^{*} 2 s \\ < (π 2 p_{x} = π 2 p_{y}) \\ < σ 2 p_{z} \\ < (π^{*} 2 p_{x} = π^{*} 2 p_{y}) \\ < σ^{*} 2 p_{z} \end{aligned}

The key difference is that in these lighter molecules $σ 2 p_{z}$ lies above the pair of $π 2 p$ orbitals (Fig. 4.20). The reason, a mixing of 2s and 2p orbitals that are close in energy in the lighter atoms, is beyond this book.

1/2 $π 2 p$ below $σ 2 p_{z}$ (with 2s–2p mixing).

Fig. 4.20Only the positions of

σ 2 p_{z}

and the

π 2 p

pair swap between the two diagrams.NCERT Fig. 4.21 · p. 129

A molecular orbital energy diagram with 2s and 2p atomic levels on both sides and molecular levels in the middle, from σ2s up to σ*2pz. Step 1, for B2, C2 and N2: the σ2pz level, drawn in colour, lies above the pair of π2p levels. Step 2, for O2 and F2: the σ2pz level has moved below the π2p pair.

4.7.5 Electronic Configuration and Molecular Behaviour

The distribution of electrons among the molecular orbitals is the electronic configuration of the molecule. It predicts several properties.

Stability. Count the electrons in bonding orbitals, $N_{b}$ , and in antibonding orbitals, $N_{a}$ . When $N_{b}$ is larger the bonding effect dominates and the molecule holds together; when $N_{a}$ is larger it falls apart.

Bond order in MO theory

Bond order = \frac{1}{2} (N_{b} - N_{a})

(4.5)

Bond order. Positive: a stable molecule. Zero or negative: no stable molecule.
Type of bond. Bond orders 1, 2 and 3 match the single, double and triple bonds of Lewis structures; MO theory also allows halves, such as 2.5 for $O_{2}^{+}$ .
Bond length. The higher the bond order, the shorter the bond.
Magnetism. A molecule with every orbital doubly filled is diamagnetic, weakly pushed out of a magnetic field. One or more singly filled orbitals make it paramagnetic, drawn into the field.

Quick check

1 of 3

Taking z as the molecular axis, which pair of atomic orbitals can combine to form molecular orbitals?

2 p_{z}

and

2 p_{x}

1 s

and

2 p_{y}

2 p_{y}

and

2 p_{z}

2 p_{x}

and

2 p_{x}

Show answer

Answer (D) $2 p_{x}$ and $2 p_{x}$

2 p_{x}

with

2 p_{x}

have the same symmetry about the axis and give

π 2 p_{x}

and

π^{*} 2 p_{x}

. The other pairs have different symmetries, so their overlap is zero.

4.8 Bonding in Some Homonuclear Diatomic Molecules

Now apply the rules: count the electrons, fill the orbitals in the right order, then read off bond order and magnetism. KK below stands for the filled inner shells, $(σ 1 s)^{2} (σ^{*} 1 s)^{2}$ , which add nothing to bonding. Check each case in Fig. 4.21 as you read.

InteractiveMolecular orbital filler

Fig. 4.21Pick a molecule or ion and add electrons one at a time to see bond order and magnetism change.

Hydrogen, $H_{2}$ . Two electrons fill $σ 1 s$ , giving $(σ 1 s)^{2}$ and bond order $\frac{1}{2} (2 - 0) = 1$ , a single bond. All electrons are paired, so $H_{2}$ is diamagnetic. Its bond length is $74 p m$ and its bond enthalpy $436 k J m o l^{- 1}$ .

Helium, $H e_{2}$ . Four electrons give $(σ 1 s)^{2} (σ^{*} 1 s)^{2}$ . The antibonding pair cancels the bonding pair: bond order 0, so $H e_{2}$ does not exist as a stable molecule. Beryllium, $B e_{2}$ , fails the same way: $K K (σ 2 s)^{2} (σ^{*} 2 s)^{2}$ , bond order 0.

Lithium, $L i_{2}$ . Six electrons: $K K (σ 2 s)^{2}$ . Four electrons are in bonding orbitals and two in antibonding ones, so the bond order is $\frac{1}{2} (4 - 2) = 1$ . MO theory therefore predicts a stable, diamagnetic molecule, and that is what spectroscopists find: $L i_{2}$ molecules occur in hot lithium vapour.

Carbon, $C_{2}$ . Twelve electrons, filled in the lighter-molecule order:

\begin{aligned} C_{2} : & K K (σ 2 s)^{2} (σ^{*} 2 s)^{2} \\ (π 2 p_{x})^{2} (π 2 p_{y})^{2} \end{aligned}

Bond order $\frac{1}{2} (8 - 4) = 2$ , with every electron paired, so $C_{2}$ should be diamagnetic; such molecules exist in carbon vapour at high temperature. Look at where the bonding comes from: the $σ 2 s$ and $σ^{*} 2 s$ pairs cancel, and all the net bonding is supplied by the four electrons in the two π orbitals. The double bond of $C_{2}$ is thus two π bonds with no σ bond, unlike the σ + π double bond of ethene. Nitrogen, $N_{2}$ , adds a pair in $σ 2 p_{z}$ : bond order 3, diamagnetic.

Oxygen, $O_{2}$ . Sixteen electrons, filled in the heavier-molecule order:

\begin{aligned} O_{2} : & K K (σ 2 s)^{2} (σ^{*} 2 s)^{2} \\ (σ 2 p_{z})^{2} (π 2 p_{x})^{2} \\ (π 2 p_{y})^{2} (π^{*} 2 p_{x})^{1} \\ (π^{*} 2 p_{y})^{1} \end{aligned}

Counting the inner shells, ten electrons are in bonding orbitals and six in antibonding ones, so the bond order is $\frac{1}{2} (10 - 6) = 2$ , a double bond. By Hund's rule the last two electrons enter the two $π^{*}$ orbitals singly with parallel spins: $O_{2}$ has two unpaired electrons and is paramagnetic. Liquid oxygen really does cling between the poles of a strong magnet. The Lewis structure O=O, with every electron paired, cannot explain this; MO theory can. Table 4.5 extends the method to ions of nitrogen and oxygen.

Table 4.5MO predictions for nitrogen and oxygen species

Species	Electrons	Bond order	Unpaired e⁻	Magnetic?
$N_{2}$	14	3	0	dia
$N_{2}^{+}$	13	2.5	1	para
$O_{2}^{+}$	15	2.5	1	para
$O_{2}$	16	2	2	para
$O_{2}^{-}$ (superoxide)	17	1.5	1	para
$O_{2}^{2 -}$ (peroxide)	18	1	0	dia

In the oxygen series, bond order falls from 2.5 to 1 as electrons are added to the antibonding $π^{*}$ orbitals, and the O–O bond lengthens from about $112 p m$ in $O_{2}^{+}$ to about $149 p m$ in the peroxide ion.

Example

Why is B₂ paramagnetic?

Boron vapour contains $B_{2}$ molecules, which are paramagnetic. Write the configuration of $B_{2}$ using each of the two orbital orders, decide which order agrees with experiment, and find the bond order.

Try it yourself first. Then reveal the solution one step at a time.

4.9 Hydrogen Bonding

Water boils at 100 °C, but hydrogen sulfide, a heavier molecule of the same bent shape, boils at −60 °C. Table 4.6 shows the same pattern for the hydrides of nitrogen and fluorine. Something extra holds molecules of $N H_{3}$ , $H_{2} O$ and $H F$ together.

Table 4.6Boiling points of some hydrides

Hydride	b.p. / °C	Next in group	b.p. / °C
$N H_{3}$	−33	$P H_{3}$	−88
$H_{2} O$	100	$H_{2} S$	−60
$H F$	20	$H C l$	−85

The extra attraction comes from hydrogen atoms attached to nitrogen, oxygen or fluorine. These three atoms are small and strongly electronegative, so an N–H, O–H or F–H bond is very polar and its hydrogen end carries an unusually large partial positive charge. Such a hydrogen is attracted to a lone pair on an N, O or F atom of another molecule. The link is a hydrogen bond (हाइड्रोजन आबंध), drawn as a dotted line. It is far weaker than a covalent bond, usually a tenth of the strength or less, but there are a great many of them.

Solid and liquid hydrogen fluoride consist of zigzag chains, $\dots H - F \dots H - F \dots$ , in which every hydrogen sits between two fluorine atoms, joined to one by a short covalent bond and to the other by a longer hydrogen bond (Fig. 4.22).

Definition

Hydrogen bond

The attractive force between a hydrogen atom covalently bonded to F, O or N and a lone pair on another F, O or N atom, in a different molecule or the same one.

1/3Between molecules: each H points at a lone pair on a neighbouring O.

Fig. 4.22Dotted lines are hydrogen bonds; solid lines are covalent bonds.NCERT Fig. 4.22 · p. 131

Dotted lines are hydrogen bonds, solid lines covalent bonds. Step 1: three water molecules, each H of one pointing at the O of the next, labelled δ+ and δ−. Step 2: a zigzag chain of HF molecules, each H bridging two F atoms. Step 3: salicylaldehyde, a benzene ring whose OH hydrogen is dotted to the aldehyde oxygen of the same molecule.

4.9.1 Cause of Formation of Hydrogen Bond

Why only F, O and N? In a bond H–X, the more electronegative X is, the further the shared pair is pulled from hydrogen; and hydrogen, with no inner electrons, is then left almost exposed. The result is a strongly polar unit, $δ +$ on H and $δ -$ on X. The $δ +$ hydrogen of one unit is attracted electrostatically to the $δ -$ atom of another, and that attraction is the hydrogen bond.

How much hydrogen bonding there is depends on the state: most in the solid, where every molecule is held in place, less in the liquid, and least in the gas, where molecules are far apart. Its effects are everywhere. Ice is less dense than liquid water, and floats, because each molecule in ice is hydrogen-bonded to four others in an open, cage-like network. The two strands of DNA are zipped together by hydrogen bonds.

4.9.2 Types of H-Bonds

Intermolecular hydrogen bonds link one molecule to another, either molecules of one substance (water, ammonia, ethanoic acid) or of two (ammonia dissolved in water, ethanol mixed with water). They raise boiling points and explain why such liquids mix so well.

Intramolecular hydrogen bonds stay inside one molecule. They form when an O–H or N–H hydrogen sits next to an O or N atom of the same molecule, closing a ring. In 2-hydroxybenzaldehyde (salicylaldehyde) the –OH hydrogen reaches across to the oxygen of the neighbouring –CHO group; in 2-nitrophenol it reaches an oxygen of the nitro group. A molecule that bonds to itself holds its neighbours less tightly, so 2-nitrophenol boils at about 216 °C, well below 4-nitrophenol (about 279 °C), whose –OH groups can only bond to other molecules.

Try this

Why does ice float?

You’ll need

a small clear plastic cup
a marker pen
water
a freezer

Predict first

You half-fill a cup, mark the water level and freeze it. Where will the top of the ice be?

below the markexactly at the markabove the mark

Half-fill the cup with water and mark the water level on the outside.
Stand the cup upright in a freezer overnight. Use plastic, never glass: water expands as it freezes and can crack a glass.
Compare the top of the ice with your mark.
Drop an ice cube into a glass of water and see where it floats.

Recap

Chemical Bonding and Molecular Structure: key points and summary

Kössel explained ions by electron transfer to noble-gas configurations; Lewis explained covalent bonds by sharing pairs. Lewis symbols show valence electrons (Section 4.1).
Lewis structures follow a recipe: count electrons, build the skeleton, add bonds and lone pairs, then make multiple bonds. Formal charge $V - L - \frac{1}{2} S$ picks the best structure (Section 4.1.4).
The octet rule fails for incomplete octets, odd-electron molecules and expanded octets, and says nothing about shape or energy (Section 4.1.5).
Ionic compounds form when the lattice enthalpy repays the cost of making gaseous ions; lattice enthalpy grows with ionic charge and falls with ionic size (Section 4.2).
Bond length $R = r_{A} + r_{B}$ ; bond angle and bond enthalpy describe bonds; higher bond order means shorter, stronger bonds (Section 4.3).
When one Lewis structure fails, canonical structures with fixed nuclei combine into a single, more stable resonance hybrid (Section 4.3.5).
Polar bonds have dipole moments $μ = Q \times r$ (in debye); a molecule's dipole is the vector sum of bond and lone-pair dipoles. Fajans' rules give ionic bonds covalent character (Section 4.3.6).
VSEPR: electron pairs spread as far apart as possible, with lp–lp > lp–bp > bp–bp; lone pairs squeeze angles and take equatorial sites (Section 4.4).
VB theory: half-filled orbitals overlap and pair electrons; head-on overlap gives σ bonds, sideways overlap weaker π bonds (Section 4.5).
Hybridisation (sp, sp², sp³, sp³d, sp³d²) explains linear, trigonal, tetrahedral, trigonal bipyramidal and octahedral geometries and multiple bonds in $C_{2} H_{4}$ and $C_{2} H_{2}$ (Section 4.6).
MO theory: atomic orbitals combine (LCAO) into bonding and antibonding molecular orbitals; similar energy, same symmetry and good overlap are needed (Section 4.7).
Bond order $= \frac{1}{2} (N_{b} - N_{a})$ ; singly occupied orbitals make a species paramagnetic, as in $O_{2}$ and $B_{2}$ (Section 4.7.5).
$H e_{2}$ and $B e_{2}$ have bond order 0 and do not exist; $L i_{2}$ , $C_{2}$ and $N_{2}$ are diamagnetic (Section 4.8).
Hydrogen bonds link H on F, O or N to a lone pair on F, O or N, between molecules or within one, with large effects on boiling points and on ice (Section 4.9).

Practice · 40 questions

Practice questions with answers (NCERT “Exercises”)

Short answerEasyNCERT Exercises 4.1
Explain why two hydrogen atoms form a molecule but two helium atoms do not, using (a) the duplet idea and (b) the energy of the system as the atoms approach.
Show solution
(a) By sharing their single electrons, both H atoms reach helium's duplet; a helium atom already has one, so sharing gains nothing. (b) Approaching H atoms feel new attractions (each nucleus for the other's electron) that outweigh the new repulsions, so the energy falls to a minimum at $74 p m$ : a bond. Helium has no half-filled orbital, and $H e_{2}$ has MO bond order 0, so there is no minimum.
Short answerEasyNCERT Exercises 4.2
Write the Lewis symbols of Ca, K, Al, S, P and I, and give the usual group valence of each.
Show solution
Table 4.7
Atom Dots Group valence
Ca 2 2
K 1 1
Al 3 3
S 6 8 − 6 = 2
P 5 8 − 5 = 3
I 7 8 − 7 = 1
Short answerEasyNCERT Exercises 4.3
Write Lewis symbols for O and $O^{2 -}$ , Mg and $M g^{2 +}$ , N and $N^{3 -}$ . In which pairs has the atom gained electrons, and in which lost them?
Show solution
O has 6 dots; $O^{2 -}$ has 8 dots in brackets with a 2− charge. N has 5 dots; $N^{3 -}$ has 8 dots in brackets with a 3− charge. Both gained electrons to reach neon's octet. Mg has 2 dots; $M g^{2 +}$ has no valence dots (its new outer shell, the n = 2 shell, is a complete octet). Magnesium lost electrons.
4
Short answerMediumNCERT Exercises 4.4
Draw Lewis structures for the following.
1. $P H_{3}$
  Show solution
  8 electrons. P forms three P–H bonds and keeps one lone pair.
2. $C S_{2}$
  Show solution
  16 electrons. S=C=S, with two lone pairs on each sulfur and none on carbon.
3. $S O_{3}^{2 -}$
  Show solution
  $6 + 18 + 2 = 26$ electrons. S forms single bonds to three O atoms and has one lone pair; each O has three lone pairs. Formal charges: S +1, each O −1, total −2.
4. $H C H O$
  Show solution
  12 electrons. Carbon in the centre with two C–H bonds and a C=O double bond; oxygen has two lone pairs.
Short answerEasyNCERT Exercises 4.5
State the octet rule. Give one example where it is useful and one example of each of its three kinds of failure.
Show solution
Atoms tend to lose, gain or share electrons until they have eight in the valence shell (two for H). It explains $M g O$ (transfer) and $C H_{4}$ (sharing) and most organic structures. Failures: incomplete octet, $B F_{3}$ (6 electrons on B); odd-electron molecule, $N O$ (11 electrons); expanded octet, $P C l_{5}$ (10 around P). It also ignores noble-gas compounds such as $X e F_{4}$ , shape and energy.
Short answerMediumNCERT Exercises 4.6
Lithium fluoride is ionic, but beryllium iodide is largely covalent. Use the factors that favour ionic bonding to explain why.
Show solution
Ionic bonding is favoured by a low ionisation enthalpy of the metal, a large negative electron gain enthalpy of the non-metal and a large lattice enthalpy. Li ionises easily, F gains an electron readily, and the small ions give a large lattice enthalpy. Forming $B e^{2 +}$ costs far more energy, and the tiny, doubly charged ion strongly polarises the large iodide ion (Fajans' rules), so the bond is mostly covalent.
7
Short answerMediumNCERT Exercises 4.7
Use VSEPR theory to predict the shape of each species.
1. $S n C l_{4}$
  Show solution
  4 bond pairs, no lone pair ( $A B_{4}$ ): tetrahedral.
2. $S b C l_{5}$
  Show solution
  5 bond pairs ( $A B_{5}$ ): trigonal bipyramidal.
3. $S e C l_{2}$
  Show solution
  Se has 6 valence electrons; 2 bonds leave 4, two lone pairs ( $A B_{2} E_{2}$ ): bent.
4. $A s C l_{3}$
  Show solution
  As has 5 valence electrons; 3 bonds leave one lone pair ( $A B_{3} E$ ): trigonal pyramidal.
Short answerMediumNCERT Exercises 4.8
$N H_{4}^{+}$ , $N H_{3}$ and $N H_{2}^{-}$ all have four electron pairs around nitrogen. Predict the order of their H–N–H bond angles and explain it.
Show solution
Lone pairs: 0 in $N H_{4}^{+}$ , 1 in $N H_{3}$ , 2 in $N H_{2}^{-}$ . All have a tetrahedral arrangement of pairs. Lone pairs repel more strongly than bond pairs (lp–lp > lp–bp > bp–bp), so each extra lone pair squeezes the bond angle: $N H_{4}^{+}$ (109.5°) > $N H_{3}$ (107°) > $N H_{2}^{-}$ (smallest, near water's 104.5°).
Short answerEasyNCERT Exercises 4.9
How does bond strength depend on bond order? Illustrate with the carbon–oxygen bonds in $C H_{3} O H$ , $H C H O$ and $C O$ .
Show solution
The C–O bond orders are 1, 2 and 3. As bond order rises, more electron pairs hold the nuclei, so bond enthalpy increases and bond length decreases. The bonds shorten from about $143$ to $121$ to $113 p m$ , and the bond in CO (about $1076 k J m o l^{- 1}$ ) is the strongest.
Short answerEasyNCERT Exercises 4.10
Define bond length. Using covalent radii N $70 p m$ and H $37 p m$ , estimate the N–H bond length and compare it with the measured value in ammonia, $101 p m$ .
Show solution
Bond length is the separation of the two nuclei in a bond at its energy minimum. Estimate: $R = 70 + 37 = 107 p m$ , about 6 % above the measured $101 p m$ . Adding covalent radii is only approximate; polar bonds such as N–H are usually a little shorter than the sum.
Short answerMediumNCERT Exercises 4.11
Both N–O bonds in the nitrite ion, $N O_{2}^{-}$ , are about $124 p m$ long. Explain this using resonance, and state the main points to remember about canonical structures.
Show solution
18 electrons. An octet structure has one N=O and one N–O⁻ bond, with a lone pair on N; drawing the double bond to the other O gives a second, equivalent canonical structure. The ion is the hybrid: each N–O bond has order 1.5 and each O carries half the negative charge, so both bonds are equal and intermediate between N–O and N=O. Remember: canonical structures keep the nuclei fixed; none exists on its own; the ion does not flip between them; the hybrid is more stable than any of them and averages the bond properties.
Short answerMediumNCERT Exercises 4.12
Two structures are written for the formula HNCO: $H - N = C = O$ and $N \equiv C - O - H$ . Can they be canonical structures of one resonance hybrid? Explain.
Show solution
No. Canonical structures must have every nucleus in the same place and differ only in electron positions. Here the hydrogen is bonded to nitrogen in one and to oxygen in the other, so the nuclei are arranged differently. They are two different compounds, isocyanic acid and cyanic acid, related as isomers (tautomers), not as resonance forms.
13
Short answerMediumNCERT Exercises 4.13
Write the resonance structures of each species.
1. $S O_{2}$ (octet structures)
  Show solution
  $O = S - O$ and $O - S = O$ , each with one lone pair on S; the singly bonded O carries −1 and S +1.
2. $C H_{3} C O O^{-}$
  Show solution
  Two structures: C=O to one oxygen and C–O⁻ to the other, then swapped. Both C–O bonds are equal in the hybrid.
3. $N_{3}^{-}$ (azide)
  Show solution
  Three: $N = N = N$ (formal charges −1, +1, −1), $N \equiv N - N$ (0, +1, −2) and $N - N \equiv N$ (−2, +1, 0). The first, with the smallest charges, contributes most.
14
Short answerMediumNCERT Exercises 4.14
Show with Lewis symbols how electrons move when each pair of elements reacts to form an ionic compound, and give the formula.
1. Li and O
  Show solution
  Two Li atoms each lose one electron to one O atom: $2 L i^{+}$ and $O^{2 -}$ , giving $L i_{2} O$ .
2. Mg and N
  Show solution
  Three Mg atoms lose six electrons; two N atoms gain three each: $3 M g^{2 +}$ and $2 N^{3 -}$ , giving $M g_{3} N_{2}$ .
3. Al and S
  Show solution
  Two Al atoms lose six electrons; three S atoms gain two each: $2 A l^{3 +}$ and $3 S^{2 -}$ , giving $A l_{2} S_{3}$ .
Short answerMediumNCERT Exercises 4.15
$C S_{2}$ has zero dipole moment, but sulfur dichloride, $S C l_{2}$ , has a measurable dipole moment. What does this tell you about the shapes of the two molecules?
Show solution
Both contain polar bonds. In $C S_{2}$ the two C=S bond dipoles must cancel, so the molecule is linear (S=C=S, 180°). In $S C l_{2}$ they do not cancel, so the molecule is bent: sulfur has two lone pairs ( $A B_{2} E_{2}$ ).
Short answerEasyNCERT Exercises 4.16
Give four uses of dipole moment measurements, with an example of each.
Show solution
1. Deciding whether a molecule is polar: $C C l_{4}$ (0) is non-polar, $C H C l_{3}$ (1.04 D) polar. 2. Finding shape: zero for $B F_{3}$ shows it is planar; 1.47 D for $N H_{3}$ shows it is pyramidal. 3. Estimating ionic character: the ratio of the measured to the fully ionic dipole moment (about 18 % for HCl). 4. Telling isomers apart: cis-1,2-dichloroethene is polar, the trans isomer has zero dipole moment.
Short answerEasyNCERT Exercises 4.17
Electronegativity and electron gain enthalpy both describe an atom's appetite for electrons. Explain how they differ: what each refers to, whether it can be measured, and its unit.
Show solution
Electronegativity is the ability of an atom in a bonded state to attract the shared pair of electrons towards itself. It is a relative number on a scale (such as Pauling's), has no unit and cannot be measured directly. Electron gain enthalpy is the enthalpy change when an isolated gaseous atom gains an electron. It is a measurable energy, in $k J m o l^{- 1}$ .
Short answerEasyNCERT Exercises 4.18
Explain what makes the H–Br bond polar. Show the partial charges and the direction of the dipole moment in both conventions.
Show solution
Bromine is more electronegative than hydrogen, so the shared pair lies closer to Br: $\overset{δ +}{H} ⁣ - ⁣ \overset{δ -}{B r}$ . The chemists' crossed arrow points from H (cross) to Br (head), showing the shift of electron density. The physicists' dipole vector points the other way, from Br to H.
19
Quick answerMediumNCERT Exercises 4.19
Arrange the bonds in order of increasing ionic character.
1. Cl–Cl in $C l_{2}$
2. Cs–F in $C s F$
3. H–Cl in $H C l$
4. I–Cl in $I C l$
5. Na–Cl in $N a C l$
Show answer
Answer
Cl–Cl in $C l_{2}$
I–Cl in $I C l$
H–Cl in $H C l$
Na–Cl in $N a C l$
Cs–F in $C s F$
Electronegativity differences (Pauling): Cl–Cl 0, I–Cl about 0.5, H–Cl about 1.0, Na–Cl about 2.2, Cs–F about 3.2. The larger the difference, the more ionic the bond.
Short answerMediumNCERT Exercises 4.20
A student drew ethanal, $C H_{3} C H O$ , with a single C–O bond, three lone pairs on oxygen and one lone pair on the aldehyde carbon. Find what is wrong and give the correct structure.
Show solution
Count: $2 \times 4 + 4 \times 1 + 6 = 18$ electrons. The student's drawing uses 6 bonds (12 electrons) plus 4 lone pairs (8), which is 20: two too many. Formal charges agree: C would be $4 - 2 - 3 = - 1$ and O $6 - 6 - 1 = - 1$ , a total of −2 for a neutral molecule. Correct structure: remove the lone pair on carbon and make the C–O bond a double bond, $C H_{3} - C H = O$ , with two lone pairs on oxygen. Every atom now has an octet (duplet for H) and zero formal charge.
Short answerMediumNCERT Exercises 4.21
Why is $C C l_{4}$ tetrahedral and not square planar? Give a VSEPR argument, a hybridisation argument and one piece of experimental evidence.
Show solution
VSEPR: four bond pairs are farthest apart at 109.5° (tetrahedron), not 90° (square), so repulsion is least. Hybridisation: carbon is sp³; its four hybrids point to the corners of a tetrahedron. A square planar centre would need $d s p^{2}$ hybrids, which carbon cannot form. Evidence: $C C l_{4}$ has zero dipole moment, and $C H_{2} C l_{2}$ exists as only one compound. A square planar carbon would give two isomers of $C H_{2} C l_{2}$ (Cl atoms adjacent or opposite).
Short answerEasyNCERT Exercises 4.22
Si–F bonds are very polar, yet $S i F_{4}$ has zero dipole moment. Explain.
Show solution
$S i F_{4}$ is tetrahedral ( $A B_{4}$ ). Its four equal Si–F bond dipoles point to the corners of a regular tetrahedron, and their vector sum is zero, so the molecule is non-polar.
Short answerMediumNCERT Exercises 4.23
Water's dipole moment is 1.85 D, but that of oxygen difluoride, $O F_{2}$ , is only about 0.30 D, although fluorine is more electronegative than oxygen. Explain using the lone pairs.
Show solution
Both are bent with two lone pairs on O. The lone pairs contribute dipoles pointing away from O, on the side opposite the bonds. In $H_{2} O$ oxygen is the negative end of each O–H bond, so the resultant of the bond dipoles points the same way as the lone-pair contribution: they add. In $O F_{2}$ the O–F bond dipoles point towards F, opposite to the lone-pair contribution, which cancels most of their effect.
Short answerEasyNCERT Exercises 4.24
Explain hybridisation with carbon as the example, then compare the sp, sp² and sp³ sets: number of orbitals, angle between them, s character and one example each.
Show solution
Hybridisation is the mixing of atomic orbitals of similar energy on one atom to give an equal number of equivalent hybrid orbitals.
Table 4.8
Set Orbitals Angle s character Example
sp 2 180° 50 % $B e F_{2}$
sp² 3 120° 33 % $B F_{3}$
sp³ 4 109.5° 25 % $C H_{4}$
Short answerMediumNCERT Exercises 4.25
Does the hybridisation of boron change in the reaction $B F_{3} + F^{-} \overset{}{\to} B F_{4}^{-}$ ? What happens to the shape?
Show solution
In $B F_{3}$ boron is sp² hybridised (trigonal planar, 120°), with an empty 2p orbital. $F^{-}$ donates a lone pair into boron's empty orbital. Boron now has four bond pairs and becomes sp³ hybridised: $B F_{4}^{-}$ is tetrahedral (109.5°).
Short answerMediumNCERT Exercises 4.26
When ammonia picks up a proton, $N H_{3} + H^{+} \overset{}{\to} N H_{4}^{+}$ , does the hybridisation of nitrogen change? What happens to the shape and the bond angle?
Show solution
No: nitrogen is sp³ in both. In $N H_{3}$ one hybrid holds a lone pair; in $N H_{4}^{+}$ that pair bonds to the proton. The shape changes from trigonal pyramidal to tetrahedral, and the angle opens from 107° to 109.5°, because there is no longer a lone pair to squeeze the bonds.
27
Short answerMediumNCERT Exercises 4.27
Describe the σ and π bonds in each molecule, saying which orbitals overlap.
1. methanal, $H C H O$
  Show solution
  C is sp²: two sp²–1s σ bonds to H and one sp²–2p σ bond to O. The unhybridised 2p of C overlaps sideways with a 2p of O, forming the π bond of C=O.
2. hydrogen cyanide, $H C N$
  Show solution
  C is sp: an sp–1s σ bond to H and an sp–2p σ bond to N, in a straight line. C's two unhybridised 2p orbitals overlap sideways with two 2p orbitals of N, forming two π bonds at right angles.
28
NumericalMediumNCERT Exercises 4.28
Count the σ and π bonds in each molecule.
1. buta-1,3-diene, $C H_{2} = C H - C H = C H_{2}$ : σ, π
  Show answer
  Answer (1) 9, (2) 2
  Six C–H and three C–C σ bonds; two π bonds.
2. benzene, $C_{6} H_{6}$ : σ, π
  Show answer
  Answer (1) 12, (2) 3
  Six C–H and six C–C σ bonds; three π bonds.
29
Quick answerMediumNCERT Exercises 4.29
Taking x as the internuclear axis, which of these orbital pairs can form a σ bond?
$2 s$ and $2 s$ $2 p_{x}$ and $2 p_{x}$ $2 p_{z}$ and $2 p_{z}$ $1 s$ and $2 p_{y}$
Select all that apply
Show answer
Answer
(A) $2 s$ and $2 s$
(B) $2 p_{x}$ and $2 p_{x}$
2s–2s and $2 p_{x}$ – $2 p_{x}$ overlap head-on along x. Two $2 p_{z}$ orbitals lie across the axis and overlap sideways (π). 1s with $2 p_{y}$ gives zero overlap.
30
Short answerMediumNCERT Exercises 4.30
Give the hybridisation of each carbon atom, from left to right.
1. $C H_{3} - C \equiv N$
  Show solution
  sp³, sp.
2. $C H_{2} = C = C H_{2}$
  Show solution
  sp², sp, sp² (the middle carbon has two double bonds).
3. $C H_{3} C O C H_{3}$
  Show solution
  sp³, sp² (the C=O carbon), sp³.
4. $H C O O H$
  Show solution
  sp² (one C=O double bond).
Short answerEasyNCERT Exercises 4.31
What are bond pairs and lone pairs? Count both around the central atom in $H_{2} S$ , $P C l_{3}$ and $X e F_{4}$ .
Show solution
A bond pair is shared between two atoms; a lone pair belongs to one atom and is not used in bonding. $H_{2} S$ : 2 bond pairs, 2 lone pairs. $P C l_{3}$ : 3 bond pairs, 1 lone pair. $X e F_{4}$ : 4 bond pairs, 2 lone pairs.

Atom	Dots	Group valence
Ca	2	2
K	1	1
Al	3	3
S	6	8 − 6 = 2
P	5	8 − 5 = 3
I	7	8 − 7 = 1

Set	Orbitals	Angle	s character	Example
sp	2	180°	50 %	$B e F_{2}$
sp²	3	120°	33 %	$B F_{3}$
sp³	4	109.5°	25 %	$C H_{4}$

Short answerEasyNCERT Exercises 4.32

Compare σ and π bonds under five headings: type of overlap, where the electron density lies, strength, whether the bond can exist alone, and rotation about the bond.

Show solution

Table 4.9

Feature	σ bond	π bond
Overlap	head-on, along the axis	sideways, parallel p orbitals
Electron density	on the axis, symmetric about it	above and below the axis
Strength	stronger (more overlap)	weaker
Exists alone?	yes, the first bond	only with a σ bond
Rotation about bond	free	restricted

Short answerMediumNCERT Exercises 4.33
Use valence bond theory to describe how a molecule of hydrogen fluoride forms.
Show solution
H has a half-filled 1s orbital; F ( $2 s^{2} 2 p^{5}$ ) has a half-filled $2 p_{z}$ orbital. As the atoms approach along z, these overlap head-on (s–p overlap) and the two electrons pair with opposite spins. The result is a σ bond. Energy is released as the atoms approach because attractions outweigh repulsions, down to a minimum at the bond length (about $92 p m$ ); the bond is polar, with δ− on F.
34
Short answerMediumNCERT Exercises 4.34
For two identical atoms with z as the axis, which of these atomic orbital pairs combine effectively? For each pair that does not, name the condition it breaks.
1. 1s and 1s
  Show solution
  Yes: same energy and symmetry ( $σ 1 s$ , $σ^{*} 1 s$ ).
2. 1s and 2s
  Show solution
  No: very different energies.
3. $2 p_{z}$ and $2 p_{x}$
  Show solution
  No: different symmetry about the z axis.
4. $2 p_{x}$ and $2 p_{x}$
  Show solution
  Yes: same energy and symmetry ( $π 2 p_{x}$ , $π^{*} 2 p_{x}$ ).
Short answerMediumNCERT Exercises 4.35
Show with molecular orbital filling why $N e_{2}$ is not expected to exist, while $N e_{2}^{+}$ could.
Show solution
$N e_{2}$ has 20 electrons, filling every orbital up to $σ^{*} 2 p_{z}$ : $N_{b} = N_{a} = 10$ , so the bond order is 0 and no stable molecule forms. $N e_{2}^{+}$ has 19 electrons, with only one in $σ^{*} 2 p_{z}$ : bond order $\frac{1}{2} (10 - 9) = \frac{1}{2}$ , so a weakly bound ion can exist (it has been observed in the gas phase).
Short answerHardNCERT Exercises 4.36
Arrange $C_{2}^{+}$ , $C_{2}$ , $C_{2}^{-}$ and $C_{2}^{2 -}$ in order of increasing stability, and state whether each is paramagnetic or diamagnetic.
Show solution
Use the lighter-molecule order ( $π 2 p$ below $σ 2 p_{z}$ ). $C_{2}$ (12 electrons): $K K (σ 2 s)^{2} (σ^{*} 2 s)^{2} (π 2 p)^{4}$ , bond order 2, diamagnetic. $C_{2}^{+}$ (11): $(π 2 p)^{3}$ , bond order 1.5, paramagnetic. $C_{2}^{-}$ (13): one electron in $σ 2 p_{z}$ , bond order 2.5, paramagnetic. $C_{2}^{2 -}$ (14): $(σ 2 p_{z})^{2}$ , bond order 3, diamagnetic (isoelectronic with $N_{2}$ ). Increasing stability: $C_{2}^{+} < C_{2} < C_{2}^{-} < C_{2}^{2 -}$ .
Short answerEasyNCERT Exercises 4.37
Orbital lobes are often marked + and −. What do these signs mean, and why do they matter for bonding?
Show solution
They are the signs (phases) of the wave function in each lobe, not electric charges. Overlap of lobes with the same sign is positive and builds electron density between the nuclei (bonding, $ψ_{A} + ψ_{B}$ ). Overlap of opposite signs is negative and gives a node (antibonding, $ψ_{A} - ψ_{B}$ ). Equal + and − contributions cancel to zero overlap, so no bond forms.
Short answerMediumNCERT Exercises 4.38
Describe the hybridisation of arsenic in $A s F_{5}$ , and explain why its axial bonds are longer than its equatorial bonds.
Show solution
As ( $[A r] 3 d^{10} 4 s^{2} 4 p^{3}$ ) promotes one 4s electron into a 4d orbital, giving five half-filled orbitals that mix into five sp³d hybrids pointing to the corners of a trigonal bipyramid. Each overlaps a half-filled 2p orbital of F. Each axial bond pair is repelled by three equatorial bond pairs at 90°, while each equatorial pair has only two neighbours at 90°. The greater repulsion stretches the axial bonds.
Short answerEasyNCERT Exercises 4.39
What is a hydrogen bond? Rank it against covalent bonds and van der Waals forces by strength, with rough magnitudes.
Show solution
A hydrogen bond is the attraction between H bonded to F, O or N and a lone pair on another F, O or N atom. Strength: covalent bonds (hundreds of $k J m o l^{- 1}$ ) ≫ hydrogen bonds (roughly 5–40 $k J m o l^{- 1}$ ) > van der Waals forces (weaker still). A hydrogen bond is thus stronger than van der Waals forces but much weaker than a covalent bond.
40
NumericalMediumNCERT Exercises 4.40
Calculate the bond order of each species.
1. $F_{2}$
  Show answer
  Answer 1
  18 electrons: $N_{b} = 10$ , $N_{a} = 8$ , so $\frac{1}{2} (10 - 8) = 1$ .
2. $F_{2}^{+}$
  Show answer
  Answer 1.5
  17 electrons: one fewer antibonding electron, so $\frac{1}{2} (10 - 7) = 1.5$ .
3. $B_{2}$
  Show answer
  Answer 1
  10 electrons: $\frac{1}{2} (6 - 4) = 1$ .
4. $N_{2}^{+}$
  Show answer
  Answer 2.5
  13 electrons: one bonding electron removed from $N_{2}$ , so $\frac{1}{2} (9 - 4) = 2.5$ .

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Quick check: MCQs on Chemical Bonding and Molecular Structure

4.1.5 Limitations of the Octet Rule4 questions
4.2.1 Lattice Enthalpy3 questions
4.3.4 Bond Order3 questions
4.3.6 Polarity of Bonds3 questions
4.4 The Valence Shell Electron Pair Repulsion (VSEPR) Theory4 questions
4.5.5 Strength of Sigma and pi Bonds3 questions
4.6.3 Hybridisation of Elements involving d Orbitals3 questions
4.7.5 Electronic Configuration and Molecular Behaviour3 questions
4.8 Bonding in Some Homonuclear Diatomic Molecules4 questions
4.9.2 Types of H-Bonds3 questions

Glossary · 15

Key terms (English–Hindi)

bond enthalpyआबंध एन्थैल्पीaabandh enthalpy: The energy needed to break one mole of a given type of bond, with all species in the gas phase.
bond orderआबंध कोटिaabandh koti: The number of bonds between two atoms: shared pairs in a Lewis structure, or $\frac{1}{2} (N_{b} - N_{a})$ in MO theory.

covalent bondसहसंयोजी आबंधsahsanyoji aabandh: A bond formed when two atoms share one or more pairs of electrons.First seen in the chapter →

dipole momentद्विध्रुव आघूर्णdvidhruv aaghurn: Charge × separation of the centres of positive and negative charge, $μ = Q \times r$ , usually in debye (D).First seen in the chapter →

formal chargeऔपचारिक आवेशaupcharik aavesh: Valence electrons of the free atom minus its lone-pair electrons minus half its bonding electrons in a Lewis structure.First seen in the chapter →

hybridisationसंकरणsankaran: Mixing of atomic orbitals of similar energy on one atom to give an equal number of equivalent hybrid orbitals.First seen in the chapter →
hydrogen bondहाइड्रोजन आबंधhydrogen aabandh: The attraction between H bonded to F, O or N and a lone pair on another F, O or N atom.First seen in the chapter →

ionic (electrovalent) bondआयनिक आबंधaayanik aabandh: The electrostatic attraction between oppositely charged ions formed by transfer of electrons.First seen in the chapter →

lattice enthalpyजालक एन्थैल्पीjaalak enthalpy: The energy needed to separate one mole of an ionic solid completely into gaseous ions.
Lewis symbolलूइस प्रतीकlewis prateek: An element's symbol surrounded by one dot for each valence electron.First seen in the chapter →

molecular orbitalआण्विक कक्षकaanvik kakshak: An orbital that spreads over the whole molecule, formed by combining atomic orbitals; it can be bonding or antibonding.First seen in the chapter →

octet ruleअष्टक नियमashtak niyam: Atoms tend to lose, gain or share electrons until they have eight in their valence shell (two for hydrogen).

pi (π) bondपाई आबंधpai aabandh: A bond formed by sideways overlap of parallel p orbitals, with electron density above and below the bond axis.First seen in the chapter →

resonance hybridअनुनाद संकरanunaad sankar: The one real structure of a species that needs two or more canonical Lewis structures to describe it.First seen in the chapter →

sigma (σ) bondसिग्मा आबंधsigma aabandh: A bond formed by head-on overlap of orbitals along the line joining the two nuclei.First seen in the chapter →

4.1 Kössel–Lewis Approach to Chemical Bonding

4.1.1 Octet Rule

4.1.2 Covalent Bond

4.1.3 Lewis Representation of Simple Molecules (the Lewis Structures)

4.1.4 Formal Charge

4.1.5 Limitations of the Octet Rule

4.2 Ionic or Electrovalent Bond

4.2.1 Lattice Enthalpy

4.3 Bond Parameters

4.3.1 Bond Length

4.3.2 Bond Angle

4.3.3 Bond Enthalpy

4.3.4 Bond Order

4.3.5 Resonance Structures

4.3.6 Polarity of Bonds

4.4 The Valence Shell Electron Pair Repulsion (VSEPR) Theory

4.5 Valence Bond Theory

4.5.1 Orbital Overlap Concept

4.5.2 Directional Properties of Bonds

4.5.3 Overlapping of Atomic Orbitals

4.5.4 Types of Overlapping and Nature of Covalent Bonds

4.5.5 Strength of Sigma and pi Bonds

4.6 Hybridisation

4.6.1 Types of Hybridisation

4.6.2 Other Examples of sp³, sp² and sp Hybridisation

4.6.3 Hybridisation of Elements involving d Orbitals

4.7 Molecular Orbital Theory

4.7.1 Formation of Molecular Orbitals: Linear Combination of Atomic Orbitals (LCAO)

4.7.2 Conditions for the Combination of Atomic Orbitals

4.7.3 Types of Molecular Orbitals

4.7.4 Energy Level Diagram for Molecular Orbitals

4.7.5 Electronic Configuration and Molecular Behaviour

4.8 Bonding in Some Homonuclear Diatomic Molecules

4.9 Hydrogen Bonding

4.9.1 Cause of Formation of Hydrogen Bond

4.9.2 Types of H-Bonds

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