Chapter 4: Chemical Bonding and Molecular Structure

Step 1: Count the valence electrons.

Carbon ($$Z=6$$, group 14) has $$4$$ valence electrons and oxygen ($$Z=8$$, group 16) has $$6$$ valence electrons. The molecule is neutral, so the total number of valence electrons is

$$4 + 6 = 10 \text{ electrons (i.e. } 5 \text{ pairs)}.$$

Step 2: Draw a single bond and distribute the rest as lone pairs.

Place one C–O single bond; this uses $$2$$ electrons and leaves $$8$$ to be placed as lone pairs. Putting three lone pairs on oxygen and one lone pair on carbon accounts for all $$8$$.

Step 3: Check the octets and promote the bond if necessary.

Count the electrons around each atom in this single-bonded structure:

• Oxygen: $$2$$ (shared pair of the bond) $$+$$ $$6$$ (three lone pairs) $$= 8$$ — octet complete.
• Carbon: $$2$$ (shared pair of the bond) $$+$$ $$2$$ (one lone pair) $$= 4$$ — carbon is $$4$$ electrons short of an octet.

To complete carbon's octet, two lone pairs of oxygen are converted into shared pairs. This turns the single C–O bond into a triple bond, $$\mathrm{C}\!\equiv\!\mathrm{O}$$.

Recounting with the triple bond: each atom now has one lone pair plus the three shared pairs of the triple bond, i.e. $$2 + 6 = 8$$ electrons — both carbon and oxygen have a complete octet.

Lewis structure:

$$:\!\mathrm{C}\!\equiv\!\mathrm{O}\!:$$

(One lone pair on carbon, one lone pair on oxygen, and a triple bond joining them.)

Step 1: Count the total number of valence electrons.

Nitrogen ($$5e^-$$) $$+ 2 \times$$ oxygen ($$6e^-$$ each) $$+ 1$$ for the negative charge:

$$5 + (2\times 6) + 1 = 18 \text{ electrons (i.e. } 9 \text{ pairs)}.$$

Step 2: Choose the skeleton.

The less electronegative N is the central atom, with both O atoms terminal: $$\mathrm{O\!-\!N\!-\!O}$$.

Step 3: Draw single bonds and place lone pairs on the terminal oxygens.

Two N–O single bonds use $$4e^-$$. Place the remaining $$14e^-$$ as lone pairs — three lone pairs on each O (12$$e^-$$) and one lone pair on N ($$2e^-$$).

Step 4: Complete the octet on N.

N now has only $$6$$ electrons around it. To complete its octet, convert one lone pair on one of the oxygens into a shared pair, forming an $$\mathrm{N=O}$$ double bond on that side.

Lewis structure:

$$\left[\,\ddot{\mathrm{O}}\!=\!\ddot{\mathrm{N}}\!-\!\ddot{\mathrm{O}}\!:\,\right]^{-}$$

The double bond can equally well be drawn to the other oxygen; both arrangements are equivalent resonance forms, so the actual structure is a resonance hybrid with two equivalent N–O bonds of intermediate length.

Total valence electrons: C contributes $$4$$, three O contribute $$3\times6=18$$, and the $$2-$$ charge adds $$2$$ more, giving $$4+18+2 = 24$$ electrons.

If we attempt a single Lewis structure, the carbon must form one C=O double bond (so its octet is complete) and two C–O single bonds (these oxygens then carry the negative charges). The single Lewis structure is

$$\left[\begin{array}{c}\mathrm{O}\\ \|\\ \mathrm{O}\!-\!\mathrm{C}\!-\!\mathrm{O}\end{array}\right]^{2-}$$

However, all three C–O bonds in the carbonate ion are experimentally found to be of equal length ($$\approx 129\,\mathrm{pm}$$), which is intermediate between a C–O single bond ($$143\,\mathrm{pm}$$) and a C=O double bond ($$122\,\mathrm{pm}$$). A single Lewis structure cannot explain this.

This experimental observation is accounted for by resonance: the real structure is a resonance hybrid of the three equivalent canonical forms

$$\mathrm{(I)}\ \mathrm{O\!=\!C(-O^-)(-O^-)} \;\longleftrightarrow\; \mathrm{(II)}\ \mathrm{^-O\!-\!C(=O)(-O^-)} \;\longleftrightarrow\; \mathrm{(III)}\ \mathrm{^-O\!-\!C(-O^-)(=O)}$$

obtained by interchanging the position of the double bond. Each form contributes equally to the hybrid, so

• the three C–O bonds become equivalent with bond order $$\dfrac{1+1+2}{3} = \dfrac{4}{3} \approx 1.33;$$
• the negative charge is delocalised, with each oxygen carrying a $$-\dfrac{2}{3}$$ charge in the hybrid;
• the ion is planar with the central carbon $$sp^2$$-hybridised and bond angles of $$120^\circ$$.

Thus resonance correctly predicts the equal C–O bond lengths and the equal distribution of charge over the three oxygens in $$\mathrm{CO_3^{2-}}$$.

In $$\mathrm{CO_2}$$, carbon is the central atom and bonded to two oxygen atoms. The single Lewis structure that satisfies the octet of every atom is

$$\mathrm{O\!=\!C\!=\!O}$$

with each oxygen carrying two lone pairs and joined to carbon by a double bond. With two electron domains (two double bonds) and no lone pairs on carbon, VSEPR predicts a linear geometry with O=C=O bond angle $$= 180^\circ$$, and the C–O bond length is $$\approx 122\,\mathrm{pm}$$ (typical of C=O).

The experimental C–O bond length in $$\mathrm{CO_2}$$, however, is $$\approx 115\,\mathrm{pm}$$, which is appreciably shorter than a pure C=O double bond ($$\approx 121\,\mathrm{pm}$$) and approaches a C$$\equiv$$O triple bond ($$\approx 110\,\mathrm{pm}$$). This observation cannot be explained by the single Lewis structure above. It is accounted for by treating $$\mathrm{CO_2}$$ as a resonance hybrid of the following canonical structures:

$$\mathrm{(I)}\ \mathrm{O\!=\!C\!=\!O} \;\longleftrightarrow\; \mathrm{(II)}\ ^{-}\!\mathrm{O\!-\!C}\!\equiv\!\mathrm{O}^{+} \;\longleftrightarrow\; \mathrm{(III)}\ ^{+}\!\mathrm{O}\!\equiv\!\mathrm{C\!-\!O}^{-}$$

Structures (II) and (III) have one C=O double bond and one C$$\equiv$$O triple bond and impart partial triple-bond character to both C–O linkages of the hybrid. The bond order of each C–O linkage in the hybrid lies between 2 and 3, explaining the shorter observed bond length.

Overall, $$\mathrm{CO_2}$$ is a linear, non-polar molecule (the two equivalent C=O bond dipoles cancel) whose two C–O bonds are identical and possess partial triple-bond character due to resonance.

A chemical bond is the force of attraction that holds two atoms (or ions) together in a molecule or in a crystal. Atoms combine because the resulting aggregate has a lower energy than the separated atoms — bond formation is therefore an energetically favourable process.

Several theories have been proposed to explain how chemical bonds form:

1. Kossel–Lewis approach (electronic theory of valency). Inert (noble) gases possess very stable electronic configurations ($$ns^2np^6$$, an octet). Atoms of other elements try to attain such a stable noble-gas configuration by losing, gaining or sharing electrons. This forms the basis of the octet rule: "atoms tend to adjust their electronic arrangements so as to have eight electrons in their outermost shell".

On this basis bonds are formed in the following ways:

• Electrovalent (ionic) bond — by complete transfer of one or more electrons from one atom (typically a metal) to another (typically a non-metal). The atom losing electrons becomes a cation and the one gaining them becomes an anion, and the two oppositely-charged ions are then held together by electrostatic forces. Example: $$\mathrm{Na} + \mathrm{Cl} \rightarrow \mathrm{Na^+} + \mathrm{Cl^-} \rightarrow \mathrm{NaCl}$$.
• Covalent bond — by mutual sharing of one or more pairs of electrons between two atoms. Each shared pair is contributed by both atoms (one from each, in normal covalency). Example: $$\mathrm{H_2}$$, $$\mathrm{Cl_2}$$, $$\mathrm{CH_4}$$.
• Co-ordinate (dative) bond — a special case of a covalent bond in which the shared pair of electrons is contributed by only one of the two atoms. Example: the N$$\to$$B bond in $$\mathrm{H_3N\!\to\!BF_3}$$.

2. Modern view. Atoms approach each other and their valence orbitals overlap. The resulting redistribution of electron density between the nuclei lowers the potential energy of the system. A stable bond results when the decrease in potential energy due to attraction more than compensates for the increase due to inter-electronic and inter-nuclear repulsions, and the system attains a minimum-energy configuration at a definite internuclear distance — the equilibrium bond length.

Thus the driving force for chemical bond formation is the tendency of atoms to achieve a state of lower energy (and, in the Kossel–Lewis picture, a stable octet).

The Lewis dot symbol of an atom is its chemical symbol surrounded by dots equal to the number of valence electrons the atom possesses. The dots are placed on the four sides of the symbol — singly at first, and then paired up once each side already carries one dot. The number of valence electrons follows from the electronic configuration / group number:

Reading the dots around each symbol: $$\mathrm{Mg}$$ — 2 single dots; $$\mathrm{Na}$$ — 1 single dot; $$\mathrm{B}$$ — 3 single dots; $$\mathrm{O}$$ — 2 lone pairs $$+$$ 2 single dots (6 in all); $$\mathrm{N}$$ — 1 lone pair $$+$$ 3 single dots (5 in all); $$\mathrm{Br}$$ — 3 lone pairs $$+$$ 1 single dot (7 in all).

The Lewis symbol of an atom or ion shows the chemical symbol surrounded by dots equal to the number of valence electrons; the dots are placed on the four sides, singly first and then paired up. For an ion of charge $$+n$$ remove $$n$$ electrons; for an ion of charge $$-n$$ add $$n$$ electrons; and enclose the ion in square brackets with the charge written at the top right.

(a) $$\mathrm{S}$$ and $$\mathrm{S^{2-}}$$. Sulphur ($$Z=16$$, $$[\mathrm{Ne}]3s^2 3p^4$$) has $$6$$ valence electrons; $$\mathrm{S^{2-}}$$ has $$6+2 = 8$$.

$$\mathrm{S}:\quad\overset{\textstyle\cdot\,\cdot}{\underset{\textstyle\cdot\,\cdot}{\cdot\,\mathrm{S}\,\cdot}}\qquad(\text{6 dots} = \text{2 lone pairs} + \text{2 single dots})$$

$$\mathrm{S^{2-}}:\quad\left[\;\overset{\textstyle\cdot\,\cdot}{\underset{\textstyle\cdot\,\cdot}{:\,\mathrm{S}\,:}}\;\right]^{2-}\qquad(\text{8 dots} = \text{4 lone pairs})$$

(b) $$\mathrm{Al}$$ and $$\mathrm{Al^{3+}}$$. Aluminium ($$Z=13$$, $$[\mathrm{Ne}]3s^2 3p^1$$) has $$3$$ valence electrons; $$\mathrm{Al^{3+}}$$ has lost all three and so carries no dots.

$$\mathrm{Al}:\quad\overset{\textstyle\cdot}{\cdot\,\mathrm{Al}\,\cdot}\qquad(\text{3 single dots})$$

$$\mathrm{Al^{3+}}:\quad\left[\,\mathrm{Al}\,\right]^{3+}\qquad(\text{no dots})$$

(c) $$\mathrm{H}$$ and $$\mathrm{H^-}$$. Hydrogen ($$Z=1$$) has $$1$$ valence electron; $$\mathrm{H^-}$$ has $$2$$ (one lone pair — the He configuration).

$$\mathrm{H}:\quad\mathrm{H}\,\cdot\qquad(\text{1 dot})$$

$$\mathrm{H^-}:\quad\left[\,\mathrm{H}\!:\,\right]^{-}\qquad(\text{2 dots} = \text{1 lone pair})$$

For each species we (i) count the valence electrons, (ii) put a single bond between every pair of bonded atoms, (iii) distribute the remaining electrons as lone pairs so as to complete octets on the terminal atoms, and (iv) move lone pairs to form multiple bonds if the central atom is still short of an octet.

(i) $$\mathrm{H_2S}$$. Total valence electrons $$= 2(1) + 6 = 8$$. Two S–H single bonds use $$4e^-$$; the remaining $$4e^-$$ go on S as two lone pairs.

$$\mathrm{H}\!-\!\ddot{\mathrm{S}}\!-\!\mathrm{H}\;\;(\text{2 lone pairs on S})$$

(ii) $$\mathrm{SiCl_4}$$. Total valence electrons $$= 4 + 4(7) = 32$$. Four Si–Cl single bonds use $$8e^-$$; the remaining $$24e^-$$ are placed as three lone pairs on each Cl.

$$\mathrm{Cl_3 lone\;pair\;Si\!-\!Cl},$$ i.e. tetrahedral Si with four single bonds to Cl, each Cl carrying three lone pairs (no lone pair on Si).

(iii) $$\mathrm{BeF_2}$$. Total valence electrons $$= 2 + 2(7) = 16$$. Two Be–F single bonds use $$4e^-$$; the remaining $$12e^-$$ go as three lone pairs on each F. Be has only $$4$$ electrons in its valence shell — it is an exception to the octet rule (electron-deficient).

$$:\!\ddot{\mathrm{F}}\!-\!\mathrm{Be}\!-\!\ddot{\mathrm{F}}\!:\;\;\text{(linear, no lone pair on Be)}$$

(iv) $$\mathrm{CO_3^{2-}}$$. Total valence electrons $$= 4 + 3(6) + 2 = 24$$. Three C–O single bonds use $$6e^-$$; place the remaining $$18e^-$$ as lone pairs (three on each O). C still has only $$6$$ electrons, so move one lone pair from one O to form a C=O double bond. The two singly bonded oxygens each carry a negative charge.

$$\left[\;\!{}^-\!\ddot{\mathrm{O}}\!-\!\mathrm{C}(=\!\ddot{\mathrm{O}}\!:)\!-\!\ddot{\mathrm{O}}^-\;\right]^{2-}\quad\Longleftrightarrow\quad\text{(2 other equivalent resonance forms)}$$

(All three C–O bonds are equivalent in the resonance hybrid; bond order $$= 4/3$$.)

(v) $$\mathrm{HCOOH}$$ (formic acid). Total valence electrons $$= 2(1) + 4 + 2(6) = 18$$. The skeleton is

$$\mathrm{H\!-\!C(\!=\!O)\!-\!O\!-\!H}$$

i.e. C is bonded to H, doubly bonded to one O (carbonyl) and singly bonded to the second O (which is also bonded to H). The carbonyl O carries two lone pairs and the hydroxyl O carries two lone pairs; all atoms now have octets (H has duplet).

Octet rule. Kossel and Lewis pointed out that the noble (inert) gases possess very stable electronic configurations consisting of eight electrons in their outermost (valence) shell (except He, which has $$2$$). Atoms of other elements tend to attain the stable configuration of the nearest noble gas. In other words, "in the formation of a chemical bond, atoms lose, gain or share electrons so as to acquire eight electrons in their outermost shell". This is the octet rule.

Significance.

• It successfully accounts for the formation of a very large number of stable molecules and ions (e.g. $$\mathrm{NaCl}$$, $$\mathrm{CaO}$$, $$\mathrm{H_2O}$$, $$\mathrm{NH_3}$$, $$\mathrm{CH_4}$$, $$\mathrm{CO_2}$$).
• It rationalises the chemical formulae of common compounds — e.g. why Na is monovalent ($$3s^1$$), Mg divalent ($$3s^2$$), Al trivalent ($$3s^2 3p^1$$); why C is tetravalent and why O is divalent.
• It explains why certain elements such as Na (loses 1 e$${}^-$$) and Cl (gains 1 e$${}^-$$) react together in $$1:1$$ stoichiometry.
• It provides a simple book-keeping device — the Lewis dot structure — for predicting connectivity in molecules.

Limitations.

• Incomplete octet of the central atom (electron-deficient molecules). Molecules of elements having fewer than four valence electrons frequently leave the central atom short of an octet. e.g. $$\mathrm{LiCl}$$ (2 valence e$${}^-$$ around Li), $$\mathrm{BeH_2}$$ (4 e$${}^-$$ around Be), $$\mathrm{BCl_3}$$ (6 e$${}^-$$ around B).
• Expanded octet (hypervalent molecules). Elements of the third and higher periods can use their empty $$d$$-orbitals to accommodate more than 8 electrons in their valence shell, giving rise to molecules such as $$\mathrm{PF_5}$$ ($$10e^-$$), $$\mathrm{SF_6}$$ ($$12e^-$$), $$\mathrm{H_2SO_4}$$, $$\mathrm{IF_7}$$.
• Odd-electron species. The rule cannot apply to molecules with an odd number of valence electrons, e.g. $$\mathrm{NO}$$ ($$11e^-$$), $$\mathrm{NO_2}$$ ($$17e^-$$), $$\mathrm{ClO_2}$$.
• Failure for noble-gas compounds. Xe and Kr form compounds such as $$\mathrm{XeF_2}$$, $$\mathrm{XeF_4}$$, $$\mathrm{XeOF_2}$$, $$\mathrm{KrF_2}$$ even though their atoms already have an octet.
• It is silent about the shape of molecules; e.g. why $$\mathrm{H_2O}$$ is bent and not linear is not explained.
• It does not explain the relative stability of molecules — it does not say anything about the energetics of bond formation.
• Pure ionic and pure covalent character are extremes — most bonds have partial ionic/covalent character which the rule does not address.

An ionic (electrovalent) bond is formed by complete transfer of one or more electrons from a metallic atom to a non-metallic atom. The energetics can be summarised by the Born–Haber cycle. The factors that favour ionic bond formation are:

(i) Low ionisation enthalpy of the metal atom. The energy required to remove the electrons from the metal must be small. Therefore alkali and alkaline-earth metals (large atomic size, low effective nuclear charge for the outermost electron) readily form cations.

$$\mathrm{M}(g) \xrightarrow{\Delta_i H}\mathrm{M^+}(g) + e^-$$

(ii) High (negative) electron-gain enthalpy of the non-metal. A large amount of energy should be released when the gaseous non-metal atom accepts an electron; this favours formation of the anion. Halogens and chalcogens (high effective nuclear charge, small size) have high $$-\Delta_{eg}H$$.

$$\mathrm{X}(g) + e^- \xrightarrow{\Delta_{eg} H} \mathrm{X^-}(g)$$

(iii) High lattice enthalpy of the resulting compound. The release of a large amount of energy on the close packing of $$\mathrm{M^+}$$ and $$\mathrm{X^-}$$ ions into the crystal lattice (lattice enthalpy, $$\Delta_{lattice}H$$) is what ultimately drives the overall process to be exothermic. By the Kapustinskii / Born–Landé idea,

$$\Delta_{lattice}H \;\propto\; \dfrac{|Z^+||Z^-|}{r^+ + r^-}$$

So lattice enthalpy is large when the ions have large charges and small radii.

(iv) Achievement of a noble-gas configuration. The formation of cation/anion should lead to a stable octet (or pseudo-noble-gas) configuration.

(v) Large electronegativity difference between the two atoms (commonly $$\geq 1.7$$ on the Pauling scale) favours electron transfer rather than sharing.

Net energy balance: an ionic bond is formed only when

$$\Delta_i H + \Delta_{eg} H + \Delta_{lattice}H \;<\; 0,$$

i.e. when the lattice enthalpy released is large enough to outweigh the (positive) ionisation enthalpy and any positive contribution from the electron-gain step.

VSEPR (Valence-Shell Electron-Pair Repulsion) theory says that the geometry around a central atom is decided by the total number of electron pairs (bond pairs + lone pairs) in its valence shell; these arrange themselves so as to be as far apart as possible. Lone pairs occupy more space than bond pairs.

(a) $$\mathrm{BeCl_2}$$. Central Be has $$2$$ valence electrons; with two Cl atoms forming two Be–Cl single bonds, the steric number is $$2$$ (2 bond pairs, no lone pairs). The two pairs arrange themselves linearly.

$$\therefore \text{Linear, Cl–Be–Cl bond angle} = 180^\circ.$$

(b) $$\mathrm{BCl_3}$$. B has $$3$$ valence electrons and forms three B–Cl single bonds, giving steric number $$3$$ (3 bp, 0 lp).

$$\therefore \text{Trigonal planar, Cl–B–Cl bond angle} = 120^\circ.$$

(c) $$\mathrm{SiCl_4}$$. Si has $$4$$ valence electrons and forms four Si–Cl single bonds, steric number $$4$$ (4 bp, 0 lp).

$$\therefore \text{Regular tetrahedral, Cl–Si–Cl bond angle} = 109.5^\circ.$$

(d) $$\mathrm{AsF_5}$$. As has $$5$$ valence electrons and forms five As–F single bonds, steric number $$5$$ (5 bp, 0 lp).

$$\therefore \text{Trigonal bipyramidal — three equatorial F's at } 120^\circ\text{ and two axial F's at } 90^\circ\text{ to the equatorial plane.}$$

(e) $$\mathrm{H_2S}$$. S has $$6$$ valence electrons; two of them form two S–H bonds and the remaining four sit as two lone pairs. Steric number $$= 2+2 = 4$$.

The four electron pairs are arranged tetrahedrally, but the molecular shape (ignoring lone pairs) is bent (angular). Lone-pair–lone-pair and lone-pair–bond-pair repulsions compress the H–S–H angle below the regular tetrahedral $$109.5^\circ$$; experimentally the angle $$\approx 92^\circ$$.

(f) $$\mathrm{PH_3}$$. P has $$5$$ valence electrons; three of them form three P–H bonds and the remaining two sit as one lone pair. Steric number $$= 3+1 = 4$$.

The four pairs adopt a tetrahedral arrangement, but the molecular shape is trigonal pyramidal. The lone pair compresses the H–P–H angle below $$109.5^\circ$$; experimentally $$\approx 93.5^\circ$$.

In both molecules the central atom is $$sp^3$$-hybridised, so the underlying electron-pair geometry is tetrahedral. The molecular shape, however, depends on the relative numbers of bond pairs (bp) and lone pairs (lp):

According to VSEPR theory the magnitude of repulsion between different pairs of electrons follows the order

$$\text{lp–lp} \;>\; \text{lp–bp} \;>\; \text{bp–bp}.$$

Starting from the ideal tetrahedral angle of $$109.5^\circ$$, each lone pair compresses the bond-pair–bond-pair angle from the ideal value because the lone-pair–bond-pair repulsion is stronger than bond-pair–bond-pair repulsion. The greater the number of lone pairs on the central atom, the greater the compression of the bond angle.

$$\mathrm{NH_3}$$ has one lone pair, while $$\mathrm{H_2O}$$ has two. The two lone pairs in water exert greater repulsive force on the bond pairs (and also repel each other) than the single lone pair in ammonia. Consequently, the H–O–H bond angle is reduced more than the H–N–H bond angle:

$$109.5^\circ \;\xrightarrow{\text{1 lp}}\; 107^\circ\ (\mathrm{NH_3}) \;\xrightarrow{\text{2 lp}}\; 104.5^\circ\ (\mathrm{H_2O}).$$

Hence the bond angle in water ($$104.5^\circ$$) is smaller than that in ammonia ($$107^\circ$$).

Bond order is defined as the number of chemical bonds (or shared electron pairs) between a pair of atoms. In Lewis structures it equals the number of bonds drawn between the two atoms. In molecular-orbital theory it is given by

$$\text{Bond order} \;=\; \dfrac{1}{2}\,(N_b - N_a),$$

where $$N_b$$ and $$N_a$$ are the numbers of electrons in bonding and antibonding molecular orbitals respectively.

Examples — $$\mathrm{H_2}$$: B.O. $$= 1$$; $$\mathrm{O_2}$$: B.O. $$= 2$$; $$\mathrm{N_2}$$: B.O. $$= 3$$.

Relation between bond order and bond strength. Other factors being similar (same pair of atoms, similar bond environment),

• Bond strength (bond-dissociation enthalpy) increases with bond order, because a larger number of shared electron pairs increases the electron density between the nuclei and hence the attraction. e.g. for the C–C linkage: B.O. $$1 \rightarrow 348\,\mathrm{kJ/mol}$$ (C–C); B.O. $$2 \rightarrow 614\,\mathrm{kJ/mol}$$ (C=C); B.O. $$3 \rightarrow 839\,\mathrm{kJ/mol}$$ (C$$\equiv$$C).
• Bond length decreases with increasing bond order. e.g. $$r_{\mathrm{C-C}} = 154\,\mathrm{pm}$$, $$r_{\mathrm{C=C}} = 134\,\mathrm{pm}$$, $$r_{\mathrm{C\equiv C}} = 120\,\mathrm{pm}$$.

Hence we may say: the higher the bond order, the shorter and stronger the bond. In MO language, a higher bond order corresponds to a greater excess of bonding over antibonding electrons and thus greater stability of the molecule.

Bond length is defined as the equilibrium distance between the nuclei of two atoms which are chemically bonded together in a molecule. It is the inter-nuclear distance at which the potential-energy curve of the diatomic species has its minimum and is expressed in $$\text{\AA}$$ or pm ($$1\,\text{\AA} = 100\,\mathrm{pm}$$).

Experimentally bond lengths are determined by spectroscopic methods, X-ray diffraction and electron-diffraction studies. For a covalent bond between two like atoms A–A, half the inter-nuclear distance is called the covalent radius of A:

$$r_A = \dfrac{d_{\mathrm{A-A}}}{2}.$$

For a heteronuclear bond A–B, the bond length is approximately the sum of the covalent radii of A and B:

$$d_{\mathrm{A-B}} \;\approx\; r_A + r_B.$$

For ionic compounds, the inter-nuclear distance equals the sum of the ionic radii of the cation and anion: $$d = r^+ + r^-$$.

Bond length decreases with increasing bond order (e.g. C–C: $$154\,\mathrm{pm}$$, C=C: $$134\,\mathrm{pm}$$, C$$\equiv$$C: $$120\,\mathrm{pm}$$) and increases with the size of the bonded atoms.

When a single Lewis structure cannot adequately describe a molecule or ion, the actual structure is regarded as the average of several alternative Lewis structures — called canonical (resonating) structures — which differ only in the position of electrons (lone pairs and π-electrons), not in the positions of the nuclei. The real species is called the resonance hybrid and the phenomenon is termed resonance. The double-headed arrow ($$\leftrightarrow$$) is used between canonical forms.

Important aspects of resonance.

• Canonical structures are hypothetical; the real molecule never exists as any one of them. The actual species (resonance hybrid) is a delocalised structure intermediate between the canonical forms.
• The canonical structures must have the same number and arrangement of atoms and the same number of unpaired electrons; only the electron distribution differs.
• The resonance hybrid is more stable than any individual canonical structure. The extra stability is called the resonance energy (or delocalisation energy).
• The greater the number of equivalent (low-energy) canonical structures, the greater the resonance stabilisation.

Application to $$\mathrm{CO_3^{2-}}$$. Three equivalent Lewis structures can be drawn for the carbonate ion by interchanging the position of the C=O double bond among the three oxygens:

$$\mathrm{(I)}\;\mathrm{O\!=\!C(-O^-)(-O^-)}\;\longleftrightarrow\;\mathrm{(II)}\;\mathrm{^-O\!-\!C(=O)(-O^-)}\;\longleftrightarrow\;\mathrm{(III)}\;\mathrm{^-O\!-\!C(-O^-)(=O)}$$

None of these canonical structures alone is consistent with experiment, which shows that all three C–O bonds are equivalent (same length, $$\approx 129\,\mathrm{pm}$$, intermediate between C–O single ($$143\,\mathrm{pm}$$) and C=O double ($$122\,\mathrm{pm}$$) bonds), and that the negative charge is distributed symmetrically. The resonance hybrid explains both observations:

• Each C–O bond has bond order $$\dfrac{1+1+2}{3}=\dfrac{4}{3}\approx 1.33$$;
• Each oxygen carries an equal fractional charge of $$-\dfrac{2}{3}$$;
• The ion is planar, with C $$sp^2$$-hybridised and bond angles of $$120^\circ$$;
• $$\mathrm{CO_3^{2-}}$$ is more stable (lower energy) than any single canonical form would suggest, owing to resonance energy.

No, structures (1) and (2) cannot be regarded as canonical forms (contributing structures) of a resonance hybrid representing $$\mathrm{H_3PO_3}$$.

One of the cardinal rules of resonance is that the positions of the atomic nuclei must be identical in all canonical structures; only the electron distribution (lone pairs and π-electrons) may differ. The two given structures violate this rule:

• In structure (1), all three hydrogen atoms are attached to oxygen atoms (i.e. $$\mathrm{H-O-P(OH)_2(O)}$$ with three P–OH groups).
• In structure (2), only two hydrogens are bonded to oxygen, while the third hydrogen is bonded directly to the phosphorus atom (i.e. $$\mathrm{H-P(=O)(OH)_2}$$).

Hence the two structures differ in the position of a hydrogen atom (P–H vs. O–H), which means they are not resonance forms but two different tautomers/isomers of $$\mathrm{H_3PO_3}$$. (Experimentally, $$\mathrm{H_3PO_3}$$ exists predominantly as structure (2) and behaves as a dibasic acid because only the two O–H hydrogens are ionisable; the direct P–H hydrogen is not acidic.)

Therefore the two given representations are not canonical forms of $$\mathrm{H_3PO_3}$$.

In each case the central atom is bonded to terminal O atoms; the resonance (canonical) forms differ only in the placement of the multiple bond among the equivalent terminal oxygens — the positions of the nuclei stay the same.

(i) $$\mathrm{SO_3}$$. Total valence electrons $$= 6 + 3(6) = 24$$. S is the central atom and is bonded to three oxygens. Three equivalent canonical structures are obtained by shifting the position of the S=O double bond among the three oxygens; in each form there is one S=O double bond and two S–O$${}^-$$ single bonds, with the central S carrying a $$+2$$ formal charge:

$$\mathrm{(I)}\;\mathrm{O\!=\!S(-O^-)(-O^-)}\;\longleftrightarrow\;\mathrm{(II)}\;\mathrm{^-O\!-\!S(=O)(-O^-)}\;\longleftrightarrow\;\mathrm{(III)}\;\mathrm{^-O\!-\!S(-O^-)(=O)}$$

The resonance hybrid is the trigonal-planar $$\mathrm{SO_3}$$ molecule in which all three S–O bonds are equivalent, with an average bond order of $$\dfrac{2+1+1}{3} = \dfrac{4}{3}$$.

(ii) $$\mathrm{NO_2}$$. Total valence electrons $$= 5 + 2(6) = 17$$ (odd-electron species). Two equivalent canonical structures are obtained by interchanging the position of the N=O double bond:

$$\mathrm{(I)}\ \mathrm{O\!=\!\dot{N}\!-\!\ddot{O}}\;\longleftrightarrow\;\mathrm{(II)}\ \mathrm{\ddot{O}\!-\!\dot{N}\!=\!O}$$

Each canonical form has one N=O double bond, one N–O single bond, and an odd electron on N. The hybrid is a bent molecule with equivalent N–O bonds (intermediate between single and double, B.O. $$= 1.5$$).

(iii) $$\mathrm{NO_3^-}$$. Total valence electrons $$= 5 + 3(6) + 1 = 24$$. N is bonded to three O atoms. Three equivalent canonical structures (one N=O double bond and two N–O$${}^-$$ single bonds) are obtained by rotating the position of the double bond:

$$\mathrm{(I)}\;\mathrm{O\!=\!N(-O^-)(-O^-)}\;\longleftrightarrow\;\mathrm{(II)}\;\mathrm{^-O\!-\!N(=O)(-O^-)}\;\longleftrightarrow\;\mathrm{(III)}\;\mathrm{^-O\!-\!N(-O^-)(=O)}$$

In every canonical form N has a formal charge of $$+1$$ and the two singly bonded oxygens each carry $$-1$$. The resonance hybrid is trigonal planar with all three N–O bonds equivalent (B.O. $$= 4/3$$) and the charge $$-1$$ delocalised equally over all three oxygens (each O carries $$-\tfrac{2}{3}$$).

The dipole moment of a molecule is the vector sum of the bond dipoles of all the bonds in the molecule:

$$\vec{\mu}_{\text{molecule}} \;=\; \sum_i \vec{\mu}_i.$$

Both C–O bond in $$\mathrm{CO_2}$$ and O–H bond in $$\mathrm{H_2O}$$ are polar individually (O is more electronegative than C, and O is more electronegative than H). Whether the molecule as a whole has a net dipole moment therefore depends on its geometry, because the bond dipoles must be added as vectors.

$$\mathrm{CO_2}$$. Experimentally the dipole moment of $$\mathrm{CO_2}$$ is $$0\,\mathrm{D}$$. The only way two equal C=O bond dipoles can give a zero resultant is if they point in exactly opposite directions along the same line. Hence $$\mathrm{CO_2}$$ must be linear ($$\angle\mathrm{O\!=\!C\!=\!O} = 180^\circ$$), with $$\vec{\mu}_{C\!=\!O} + \vec{\mu}_{C\!=\!O} = 0$$. Schematically:

$$\overset{\delta-}{\mathrm{O}}\!\Longleftarrow\!\overset{\delta+}{\mathrm{C}}\!\Longrightarrow\!\overset{\delta-}{\mathrm{O}}\;\;\Longrightarrow\;\;\vec{\mu}_{\text{net}} = 0.$$

$$\mathrm{H_2O}$$. Experimentally, water has a substantial dipole moment of $$1.85\,\mathrm{D}$$ (non-zero). If $$\mathrm{H_2O}$$ were linear, the two equal O–H bond dipoles (pointing from H towards O) would also cancel and give $$\vec{\mu}=0$$. Since water is observed to have a non-zero dipole moment, the molecule cannot be linear; it must be bent (angular), so that the two O–H bond vectors add to give a net resultant directed along the bisector of the H–O–H angle. (The two lone pairs on oxygen are responsible for the bending; $$\angle\mathrm{H\!-\!O\!-\!H} \approx 104.5^\circ$$.)

Thus the contrasting dipole moments — zero for $$\mathrm{CO_2}$$ and $$1.85\,\mathrm{D}$$ for $$\mathrm{H_2O}$$ — are direct evidence that $$\mathrm{CO_2}$$ is linear while $$\mathrm{H_2O}$$ is bent.

Dipole moment ($$\mu$$) is a vector quantity defined for a polar bond/molecule as the product of the magnitude of the partial charge $$q$$ and the distance $$d$$ separating the centres of positive and negative charge:

$$\mu = q \times d.$$

It is measured in Debye units ($$1\,\mathrm{D} = 3.33564 \times 10^{-30}\,\mathrm{C\,m}$$). Its main uses are:

• Predicting polarity of molecules. Molecules with non-zero dipole moments are polar (e.g. $$\mathrm{HCl}$$, $$\mu = 1.03\,\mathrm{D}$$); those with $$\mu = 0$$ are non-polar (e.g. $$\mathrm{H_2}$$, $$\mathrm{CO_2}$$, $$\mathrm{BF_3}$$, $$\mathrm{CH_4}$$).
• Distinguishing geometries / determining molecular shape. If a molecule has a non-zero dipole moment it cannot have a symmetrical shape that would cancel the bond dipoles. Example: $$\mathrm{CO_2}$$ has $$\mu=0$$ — so it is linear; $$\mathrm{H_2O}$$ has $$\mu = 1.85\,\mathrm{D}$$ — so it is bent. $$\mathrm{NH_3}$$ ($$\mu = 1.46\,\mathrm{D}$$) is pyramidal, while $$\mathrm{BF_3}$$ ($$\mu = 0$$) is planar.
• Distinguishing cis/trans (geometrical) isomers. e.g. cis-1,2-dichloroethene has non-zero $$\mu$$, while the trans isomer has $$\mu = 0$$ because the two C–Cl bond dipoles cancel.
• Estimating per-cent ionic character of a bond. By comparing the observed dipole moment with the hypothetical fully-ionic value:

$$\%\ \text{ionic character} \;=\; \dfrac{\mu_{\text{observed}}}{\mu_{\text{ionic}}}\times 100,\quad \text{where } \mu_{\text{ionic}} = e \times d.$$

e.g. for HCl with $$\mu_{obs} = 1.03\,\mathrm{D}$$ and $$\mu_{\text{ionic}} = 6.12\,\mathrm{D}$$ (calc.), the per-cent ionic character is $$\approx 17\%$$.

• Comparing relative polarities and hence physical properties such as boiling point, solubility, dielectric constant.
• Determining symmetry of a molecule: highly symmetric molecules ($$\mathrm{CCl_4}$$, $$\mathrm{SF_6}$$) have $$\mu = 0$$.

Electronegativity ($$\chi$$). The tendency of an atom in a chemical bond to attract towards itself the shared pair of electrons. It is a relative number (dimensionless) and has no units (Pauling, Mulliken, Allred–Rochow scales).

Electron gain enthalpy ($$\Delta_{eg}H$$). The enthalpy change accompanying the addition of an electron to an isolated, gaseous, neutral atom in its ground state to form a gaseous anion:

$$\mathrm{X}(g) + e^-(g) \;\longrightarrow\; \mathrm{X^-}(g);\quad \Delta H = \Delta_{eg}H.$$

It has the units of energy (kJ/mol).

Differences.

Although the two properties run roughly in parallel (an atom that strongly attracts the bonded pair also usually releases much energy on capturing a free electron), they are not the same — one is a relative tendency in a bonded state and the other is the measurable enthalpy of a specific process for the free atom.

A polar covalent bond is a covalent bond between two atoms of different electronegativities. Because the more electronegative atom attracts the shared electron pair more strongly, the shared pair is displaced towards it, producing a partial negative charge ($$\delta-$$) on the more electronegative atom and an equal partial positive charge ($$\delta+$$) on the less electronegative atom. The bond therefore has both covalent character (sharing) and ionic character (charge separation).

Example: HCl. Hydrogen ($$\chi = 2.1$$) and chlorine ($$\chi = 3.0$$) differ in electronegativity by $$\Delta\chi = 0.9$$. The shared electron pair in H–Cl is displaced towards the more electronegative chlorine atom, giving

$$\overset{\delta+}{\mathrm{H}}\!-\!\overset{\delta-}{\mathrm{Cl}}.$$

The bond thus possesses a permanent dipole moment of $$1.03\,\mathrm{D}$$ directed from H ($$\delta+$$) to Cl ($$\delta-$$). Other examples are $$\mathrm{H_2O}$$, $$\mathrm{NH_3}$$, $$\mathrm{HF}$$ and the C=O bond.

The greater the difference in electronegativity ($$\Delta\chi$$) between the two atoms, the greater the polarity of the bond. When $$\Delta\chi$$ is very large (commonly $$\Delta\chi \geq 1.7$$ on the Pauling scale), the bond is regarded as essentially ionic; when $$\Delta\chi = 0$$ the bond is purely (non-polar) covalent.

Hannay–Smyth gave an empirical relation for the percentage ionic character of a bond:

$$\%\,\text{ionic character} = 16\,(\chi_A - \chi_B) + 3.5\,(\chi_A - \chi_B)^2.$$

For HCl, $$\Delta\chi = 0.9$$, giving $$\%\,\text{ionic character} \approx 17\%$$, i.e. HCl is largely covalent but with significant ionic character — a typical polar covalent bond.

The ionic character of a bond rises with the difference in electronegativity ($$\Delta\chi$$) between the bonded atoms. Using Pauling electronegativities:

The Cl–F bond, although having the same $$\Delta\chi$$ as S–O numerically, is more polar than S–O if one is being precise about the small difference; however many texts put $$\mathrm{SO_2}$$ before $$\mathrm{ClF_3}$$ because the dipole moments of $$\mathrm{SO_2}$$ (1.62 D, bent) and $$\mathrm{ClF_3}$$ (0.6 D, T-shaped) reflect their geometry. Following the NCERT convention (purely on $$\Delta\chi$$):

$$\mathrm{N_2} \;<\; \mathrm{SO_2} \;<\; \mathrm{ClF_3} \;<\; \mathrm{K_2O} \;<\; \mathrm{LiF}.$$

(Increasing ionic character means increasing electronegativity difference; $$\mathrm{N_2}$$ is purely covalent ($$\Delta\chi=0$$), whereas $$\mathrm{LiF}$$ is the most ionic of the lot.)

Total valence electrons in $$\mathrm{CH_3COOH}$$ are

$$4\,(\text{H}) + 2\times 4\,(\text{C}) + 2\times 6\,(\text{O}) = 4 + 8 + 12 = 24.$$

The skeletal connectivity (from the textbook) is correct: $$\mathrm{H_3C\!-\!C(=O)\!-\!O\!-\!H}$$. The errors in the printed figure are (i) one of the C–H bonds is drawn as a double bond, which would make H pentavalent (impossible since H has only one valence electron — it can form only a single bond), and (ii) the lone pairs on the two oxygens may be missing or misplaced.

The correct Lewis structure:

$$\mathrm{H}\!-\!\overset{\displaystyle H}{\underset{\displaystyle H}{\mathrm{C}}}\!-\!\mathrm{C}\!(=\!\ddot{\mathrm{O}}\!:)\!-\!\ddot{\mathrm{O}}\!-\!\mathrm{H}$$

That is: the methyl carbon ($$\mathrm{C_1}$$) is $$sp^3$$-hybridised and forms three single C–H bonds and one C–C single bond. The carboxyl carbon ($$\mathrm{C_2}$$) is $$sp^2$$-hybridised, forms a C=O double bond with one oxygen (the carbonyl oxygen carrying two lone pairs) and a C–O–H single bond with the other oxygen (the hydroxyl oxygen carrying two lone pairs). All atoms have completed their octets (H — duplet, C — octet, O — octet).

Electron book-keeping check: bonds drawn = $$3\,(C_1\!-\!H) + 1\,(C_1\!-\!C_2) + 1\,(C_2\!=\!O) + 1\,(C_2\!-\!O) + 1\,(O\!-\!H) = 8$$ bonds $$\Rightarrow$$ $$8\times 2 = 16$$ shared electrons. The two oxygens have $$2 + 2 = 4$$ lone pairs $$= 8$$ electrons. Total $$= 16 + 8 = 24$$ — matches.

According to VSEPR theory the four pairs of bonding electrons around the central carbon atom in $$\mathrm{CH_4}$$ will arrange themselves in space so as to be as far apart as possible (i.e. to minimise the inter-pair electrostatic repulsion). For four pairs the geometry that gives the largest angle between any two pairs is the regular tetrahedron, in which the H–C–H angle is

$$\theta_{\text{tetra}} = 109.5^\circ\;\;(\cos^{-1}\!\left(-\tfrac{1}{3}\right)).$$

If the molecule were square planar with the four H atoms at the corners of a square and C at its centre, the H–C–H angle for adjacent hydrogens would be

$$\theta_{\text{sq-pl}} = 90^\circ,$$

which is considerably smaller than the tetrahedral angle. This would force the four bonding pairs (and the H atoms) much closer together, leading to greater bond-pair–bond-pair repulsion and a higher energy.

The tetrahedral arrangement, by contrast, gives the maximum possible separation of the four bond pairs ($$109.5^\circ$$) and hence the minimum mutual repulsion and the lowest energy.

From the orbital-overlap (hybridisation) view, the carbon atom in $$\mathrm{CH_4}$$ uses four equivalent $$sp^3$$ hybrid orbitals, which by their very mathematical form are directed towards the four corners of a regular tetrahedron — not to the corners of a square — and overlap with the four hydrogen $$1s$$ orbitals to form four equivalent, maximally-overlapped C–H sigma bonds.

Therefore the tetrahedral geometry is energetically more stable than the hypothetical square-planar geometry, and $$\mathrm{CH_4}$$ is observed to be tetrahedral, not square planar.

Each individual Be–H bond is polar because beryllium ($$\chi = 1.5$$) and hydrogen ($$\chi = 2.1$$) differ in electronegativity (here $$\Delta\chi \approx 0.6$$); the bond dipole is directed from Be towards H ($$\overset{\delta+}{\mathrm{Be}}\!-\!\overset{\delta-}{\mathrm{H}}$$).

$$\mathrm{BeH_2}$$ is a linear molecule (Be is $$sp$$-hybridised, with two bond pairs and no lone pair on Be; the H–Be–H bond angle is $$180^\circ$$). Hence the two equal Be–H bond dipoles point in exactly opposite directions along the same line:

$$\overset{\delta-}{\mathrm{H}}\!\Longleftarrow\!\overset{\delta+}{\mathrm{Be}}\!\Longrightarrow\!\overset{\delta-}{\mathrm{H}}$$

The net dipole moment of the molecule is the vector sum of the individual bond dipoles:

$$\vec{\mu}_{\text{net}} \;=\; \vec{\mu}_{\mathrm{Be\!-\!H_1}} + \vec{\mu}_{\mathrm{Be\!-\!H_2}} \;=\; \vec{\mu}\,(-\hat{x}) + \vec{\mu}\,(+\hat{x}) \;=\; \vec{0}.$$

Since the two equal and oppositely directed Be–H bond dipoles exactly cancel each other, the molecule as a whole has zero dipole moment, in spite of the individual bonds being polar.

$$\mathrm{NH_3}$$ has the higher dipole moment, although the individual N–F bond is more polar than the N–H bond.

Both $$\mathrm{NH_3}$$ and $$\mathrm{NF_3}$$ are trigonal pyramidal ($$sp^3$$-hybridised N with one lone pair and three bonded atoms). In each case the resultant dipole moment of the molecule has two contributions:

• the vector sum of the three N–X bond dipoles ($$X = $$ H or F), which points along the C$$_3$$ axis;
• the dipole of the lone pair on nitrogen, which points along the same axis (from N outwards along the symmetry axis on the side opposite the three X atoms).

The direction of the bond dipole is decided by the electronegativity difference:

• In $$\mathrm{NH_3}$$, $$\chi_{\mathrm{N}} = 3.0 > \chi_{\mathrm{H}} = 2.1$$, so each bond dipole points from H towards N. Thus the resultant of the three N–H bond dipoles is directed along the C$$_3$$ axis towards the N atom; this resultant is in the same direction as the lone-pair moment. The two contributions add:

$$\mu(\mathrm{NH_3}) = \mu_{\text{bp(sum)}} + \mu_{\text{lp}} \;=\; 1.46\,\mathrm{D}.$$

• In $$\mathrm{NF_3}$$, $$\chi_{\mathrm{F}} = 4.0 > \chi_{\mathrm{N}} = 3.0$$, so each bond dipole points from N towards F. The resultant of the three N–F bond dipoles is directed along the C$$_3$$ axis away from the N atom; this is opposite to the lone-pair moment. The two contributions therefore subtract:

$$\mu(\mathrm{NF_3}) = |\,\mu_{\text{bp(sum)}} - \mu_{\text{lp}}\,| \;=\; 0.24\,\mathrm{D}.$$

Hence $$\mu(\mathrm{NH_3}) > \mu(\mathrm{NF_3})$$.

Hybridisation. The process of intermixing of atomic orbitals of comparable energies belonging to the same atom to give rise to a set of new orbitals (called hybrid orbitals) of equivalent energy and identical shape is called hybridisation. The number of hybrid orbitals obtained is equal to the number of atomic orbitals mixed. Hybrid orbitals overlap better than pure orbitals because they are more directional (they protrude more on one side of the nucleus); this leads to stronger bonds.

Salient features.

• Only orbitals of nearly the same energy on the same atom can mix.
• The number of hybrid orbitals formed = number of atomic orbitals mixed.
• All hybrid orbitals are equivalent in shape and energy.
• The hybrid orbitals are oriented in space so that the inter-orbital repulsion is minimised — this dictates the geometry of the molecule.

(i) $$sp$$ hybridisation. One $$s$$ and one $$p$$ orbital intermix to give two equivalent $$sp$$ hybrid orbitals. Each has $$50\%\,s$$ and $$50\%\,p$$ character. The two $$sp$$ orbitals are linear, directed at $$180^\circ$$ to each other. Shape: each hybrid orbital is large on one side of the nucleus and small on the other (a slightly distorted dumb-bell). Examples — $$\mathrm{BeCl_2}$$, $$\mathrm{CO_2}$$, $$\mathrm{C_2H_2}$$.

$$s + p_x \;\longrightarrow\; \text{two } sp \text{ hybrids at } 180^\circ.$$

(ii) $$sp^2$$ hybridisation. One $$s$$ orbital and two $$p$$ orbitals mix to give three equivalent $$sp^2$$ hybrid orbitals, each with $$33\%\,s$$ and $$67\%\,p$$ character. They are trigonal planar — directed to the corners of an equilateral triangle in a plane, at angles of $$120^\circ$$ to each other. The third $$p$$ orbital (unhybridised) is perpendicular to this plane. Examples — $$\mathrm{BCl_3}$$, $$\mathrm{C_2H_4}$$, $$\mathrm{SO_3}$$.

$$s + p_x + p_y \;\longrightarrow\; \text{three } sp^2 \text{ hybrids at } 120^\circ.$$

(iii) $$sp^3$$ hybridisation. One $$s$$ orbital and all three $$p$$ orbitals mix to give four equivalent $$sp^3$$ hybrid orbitals, each with $$25\%\,s$$ and $$75\%\,p$$ character. They are directed to the four corners of a regular tetrahedron, with inter-orbital angles of $$109.5^\circ$$. Examples — $$\mathrm{CH_4}$$, $$\mathrm{NH_3}$$, $$\mathrm{H_2O}$$, $$\mathrm{SiCl_4}$$.

$$s + p_x + p_y + p_z \;\longrightarrow\; \text{four } sp^3 \text{ hybrids at } 109.5^\circ.$$

In each case the larger lobe of every hybrid orbital points outward from the nucleus along the bonding direction, allowing efficient overlap with the orbital of the bonding partner to form a strong sigma bond.

In $$\mathrm{AlCl_3}$$. Aluminium (group 13) has the ground-state configuration $$[\mathrm{Ne}]3s^2 3p^1$$. In the excited state it becomes $$[\mathrm{Ne}]3s^1 3p^2$$, and the one $$3s$$ and two $$3p$$ orbitals mix to form three equivalent $$sp^2$$ hybrid orbitals, which overlap with the $$p$$-orbitals of the three Cl atoms to form three Al–Cl sigma bonds. Hence in $$\mathrm{AlCl_3}$$, Al is $$sp^2$$-hybridised and the molecule is trigonal planar with Cl–Al–Cl bond angles of $$120^\circ$$. The Al atom has an empty unhybridised $$3p_z$$ orbital, so it is electron-deficient (only six electrons in its valence shell).

Formation of $$\mathrm{AlCl_4^-}$$. The chloride ion $$\mathrm{Cl^-}$$ donates a lone pair to the empty $$3p_z$$ orbital of Al, forming a co-ordinate (dative) Al$$\leftarrow$$Cl bond. To accommodate the four bond pairs symmetrically and minimise their mutual repulsion, the original three $$sp^2$$ hybrid orbitals and the now-occupied $$3p_z$$ orbital mix to give four equivalent $$sp^3$$ hybrid orbitals.

$$\underbrace{sp^2 + 3p_z}_{\text{re-mix}} \;\longrightarrow\; 4\,sp^3 \text{ orbitals}.$$

Hence Al in $$\mathrm{AlCl_4^-}$$ is $$sp^3$$-hybridised and the ion is tetrahedral with Cl–Al–Cl bond angles of $$109.5^\circ$$. All four Al–Cl bonds are equivalent (the dative bond is indistinguishable from the original covalent bonds).

Net change: hybridisation of Al changes from $$sp^2$$ (trigonal planar) in $$\mathrm{AlCl_3}$$ to $$sp^3$$ (tetrahedral) in $$\mathrm{AlCl_4^-}$$.

$$\mathrm{BF_3}$$. Boron has the ground-state configuration $$1s^2 2s^2 2p^1$$. In the excited state ($$1s^2 2s^1 2p^2$$) one $$2s$$ and two $$2p$$ orbitals mix to form three equivalent $$sp^2$$ hybrid orbitals. These overlap with $$p$$-orbitals of the three F atoms, giving three B–F sigma bonds. Hence $$\mathrm{BF_3}$$ is $$sp^2$$-hybridised, trigonal planar (F–B–F $$= 120^\circ$$). B has an empty unhybridised $$2p_z$$ orbital and only six electrons in its valence shell — it is electron-deficient and a Lewis acid.

$$\mathrm{NH_3}$$. Nitrogen has configuration $$1s^2 2s^2 2p^3$$. The $$2s$$ and three $$2p$$ orbitals mix to give four $$sp^3$$ hybrid orbitals. Three of them overlap with the $$1s$$ orbitals of the three H atoms (forming three N–H sigma bonds), and the fourth contains the lone pair on N. Hence $$\mathrm{NH_3}$$ is $$sp^3$$-hybridised, trigonal pyramidal.

Adduct $$\mathrm{F_3B\!\cdot\!NH_3}$$ (or $$\mathrm{F_3B\!\leftarrow\!NH_3}$$). The lone pair on the N atom of ammonia is donated to the empty $$2p_z$$ orbital of boron, forming a dative B$$\leftarrow$$N bond. As a result, boron now has four bond pairs in its valence shell. To accommodate them symmetrically, the three $$sp^2$$ orbitals of B and its empty $$2p_z$$ remix to give four $$sp^3$$ hybrid orbitals.

$$\therefore \text{Hybridisation of B changes from } sp^2\;(\text{trigonal planar}) \;\longrightarrow\; sp^3\;(\text{tetrahedral}).$$

Nitrogen, however, was already $$sp^3$$ in $$\mathrm{NH_3}$$ (with one lone pair occupying the fourth $$sp^3$$ orbital). After donation of this lone pair to B, the same orbital now contains a shared pair forming the B$$\leftarrow$$N bond, but the hybridisation of N remains unchanged ($$sp^3$$); only the geometry around N becomes slightly more like ideal tetrahedral.

Summary. Boron: $$sp^2 \to sp^3$$ (yes, change). Nitrogen: $$sp^3 \to sp^3$$ (no change).

(i) $$\mathrm{C_2H_4}$$ (ethene) — formation of the C=C double bond.

Each carbon atom undergoes $$sp^2$$ hybridisation: one $$2s$$ and two $$2p$$ orbitals ($$2p_x$$ and $$2p_y$$, say) mix to give three equivalent $$sp^2$$ hybrids in a plane at $$120^\circ$$ to one another. The third $$p$$ orbital ($$2p_z$$) remains unhybridised and lies perpendicular to the plane of the $$sp^2$$ hybrids.

Two of the three $$sp^2$$ hybrids of each carbon form $$\sigma$$-bonds with the $$1s$$ orbitals of two hydrogen atoms (one on each side). The remaining $$sp^2$$ hybrid of each C overlaps head-on (along the C–C internuclear axis) with the corresponding $$sp^2$$ hybrid of the other carbon, forming a strong C–C sigma ($$\sigma_{sp^2-sp^2}$$) bond.

The two unhybridised $$2p_z$$ orbitals (one on each carbon), being parallel to one another and perpendicular to the molecular plane, now overlap sideways above and below the plane to form a pi ($$\pi_{p-p}$$) bond. The combination of one $$\sigma$$ and one $$\pi$$ bond gives the carbon–carbon double bond:

$$\mathrm{C=C}\;=\;\sigma_{sp^2\text{-}sp^2} + \pi_{p\text{-}p}.$$

Diagram (described): two trigonal-planar carbons with H's at $$120^\circ$$; lobes of the $$2p_z$$ orbitals point up and down on both atoms and overlap laterally to form a $$\pi$$ cloud above and below the C–C axis. The H–C–H and H–C=C angles are each approximately $$120^\circ$$ and the whole molecule is planar.

(ii) $$\mathrm{C_2H_2}$$ (ethyne) — formation of the C$$\equiv$$C triple bond.

Each carbon undergoes $$sp$$ hybridisation: one $$2s$$ and one $$2p$$ orbital ($$2p_x$$, along the internuclear axis) mix to give two equivalent $$sp$$ hybrids lying along the axis at $$180^\circ$$. The other two $$p$$ orbitals ($$2p_y$$ and $$2p_z$$) remain unhybridised and are perpendicular to the molecular axis and to each other.

One $$sp$$ hybrid of each carbon forms a $$\sigma$$-bond with the $$1s$$ orbital of a hydrogen atom (giving the two C–H bonds). The remaining $$sp$$ hybrid of each carbon overlaps head-on (along the C–C axis) with that of the other carbon to give a C–C sigma ($$\sigma_{sp\text{-}sp}$$) bond.

The two unhybridised $$2p_y$$ orbitals (parallel to each other) overlap sideways to form one $$\pi$$ bond above and below the C–C axis; simultaneously the two unhybridised $$2p_z$$ orbitals overlap sideways to form a second $$\pi$$ bond at right angles to the first. Thus the C$$\equiv$$C triple bond comprises one $$\sigma$$ and two $$\pi$$ bonds:

$$\mathrm{C}\!\equiv\!\mathrm{C}\;=\;\sigma_{sp\text{-}sp} + \pi_{p_y\text{-}p_y} + \pi_{p_z\text{-}p_z}.$$

The two $$\pi$$ electron clouds are mutually perpendicular and together form a cylindrically symmetric envelope around the C–C axis. The molecule is linear (H–C$$\equiv$$C–H, bond angle $$180^\circ$$).

In a covalent compound, the valence electrons of an atom in its bonded state can be divided into two categories.

(i) Bond pairs (shared pairs). A pair of electrons that is shared between two atoms and is responsible for the bond between them is called a bond pair (or shared pair). Bond pairs are localised between two nuclei.

Example. In the methane molecule $$\mathrm{CH_4}$$ there are four C–H bonds. Each C–H bond consists of one shared pair of electrons (one from C and one from H). Hence $$\mathrm{CH_4}$$ has $$4$$ bond pairs and no lone pairs on the central carbon.

$$\mathrm{CH_4}:\;\mathrm{H}\!-\!\mathrm{C}\!-\!\mathrm{H}\text{ (with two more H above and below)},\quad\text{4 bond pairs on C, 0 lone pairs}.$$

(ii) Lone pairs (non-bonding pairs). A pair of valence electrons that is not involved in bonding (i.e. it remains on one atom and is not shared with another) is called a lone pair (or non-bonding pair). Lone pairs occupy more space than bond pairs because they are attracted by only one nucleus.

Example. In the water molecule $$\mathrm{H_2O}$$, the oxygen atom has $$6$$ valence electrons. Two of them are used in two O–H bond pairs, while the remaining four form two lone pairs on oxygen:

$$\mathrm{H}\!-\!\ddot{\mathrm{O}}\!-\!\mathrm{H} \quad\text{(2 bond pairs and 2 lone pairs on O)}.$$

The presence of the two lone pairs on oxygen reduces the H–O–H bond angle from the ideal tetrahedral $$109.5^\circ$$ to $$\approx 104.5^\circ$$.

Similarly, in $$\mathrm{NH_3}$$ there are three N–H bond pairs and one lone pair on nitrogen.

Consider two hydrogen atoms $$\mathrm{H_A}$$ and $$\mathrm{H_B}$$, each having one electron ($$e_A$$ and $$e_B$$) in a $$1s$$ orbital. According to valence-bond theory (proposed by Heitler and London, 1927; later modified by Pauling and Slater), a covalent bond is formed when two singly-occupied atomic orbitals with electrons of opposite spin overlap.

When the two H atoms are far apart, their interaction is negligible and the potential energy of the system is taken as zero. As they approach each other, the following interactions arise:

• Attractive forces: (i) between the nucleus of one atom and the electron of the other ($$\mathrm{H_A}$$ nucleus – $$e_B$$ and $$\mathrm{H_B}$$ nucleus – $$e_A$$), and (ii) between the two electrons of opposite spin (in a paired state, the exchange interaction is attractive).
• Repulsive forces: (i) between the two nuclei, and (ii) between the two electrons.

At large separations, attractive forces dominate over repulsive forces. The potential energy of the system decreases as the atoms approach until an internuclear distance is reached at which the attractive and repulsive forces just balance and the potential energy attains its minimum. This distance is called the equilibrium bond length ($$r_e = 74\,\mathrm{pm}$$ for $$\mathrm{H_2}$$).

At this point the two $$1s$$ orbitals overlap head-on, producing a region of high electron density between the two nuclei. The two electrons (with opposite spins, in accordance with the Pauli exclusion principle) pair up in the overlap region, forming a $$\sigma_{1s\text{-}1s}$$ bond.

If the atoms are brought still closer than $$r_e$$, the nuclear–nuclear and electron–electron repulsions increase rapidly and the potential energy rises steeply.

The energy released when two H atoms combine to form a $$\mathrm{H_2}$$ molecule (the depth of the well below the zero of energy) is the bond dissociation enthalpy: $$\Delta H_{\mathrm{H\!-\!H}} = 435.8\,\mathrm{kJ/mol}.$$

$$\mathrm{H}(g)\;+\;\mathrm{H}(g)\;\longrightarrow\;\mathrm{H_2}(g);\quad \Delta H = -435.8\,\mathrm{kJ/mol}.$$

The $$\mathrm{H_2}$$ molecule is more stable than the two isolated H atoms by exactly this amount of energy. Thus the formation of $$\mathrm{H_2}$$ is explained by the overlap of two singly-occupied $$1s$$ orbitals containing electrons of opposite spin, resulting in a $$\sigma$$ bond between the two H atoms.

According to Molecular Orbital Theory, atomic orbitals (AOs) on different atoms combine linearly to give molecular orbitals (MOs). Mathematically, for two AOs $$\psi_A$$ and $$\psi_B$$,

$$\psi_{\mathrm{MO}}^{\pm} = c_1 \psi_A \pm c_2 \psi_B,$$

which generates a bonding ($$+$$ combination) and an antibonding ($$-$$ combination) MO. For this linear combination of atomic orbitals (LCAO) to be effective, the following conditions must be met.

(i) Comparable energies. The atomic orbitals to be combined must have nearly the same energy. e.g. the $$1s$$ orbital of one atom combines effectively with the $$1s$$ orbital of another atom, but not with a $$2s$$ or $$2p$$ orbital that is far apart in energy.

(ii) Appreciable overlap. The atomic orbitals must overlap to a sufficient extent. The greater the overlap, the lower the energy of the bonding MO (and the more stable the bond).

(iii) Same symmetry about the molecular (internuclear) axis. The orbitals being combined must have the same symmetry about the internuclear axis. Taking the x-axis as the bond axis: a $$2p_x$$ orbital can combine with another $$2p_x$$ (or with $$1s$$, $$2s$$, since they are spherically symmetric), but a $$2p_x$$ cannot combine with a $$2p_y$$ or $$2p_z$$ (different symmetry — net overlap is zero, since positive and negative lobes cancel out).

If any of these conditions is not satisfied, the constructive overlap is negligible and no effective molecular orbital is formed.

Each Be atom has the ground-state electronic configuration $$1s^2 2s^2$$, i.e. $$4$$ electrons. A hypothetical $$\mathrm{Be_2}$$ molecule would therefore contain $$8$$ electrons. Filling the molecular orbitals in order of increasing energy (for $$\mathrm{Be_2}$$ only the $$\sigma_{1s}$$, $$\sigma^*_{1s}$$, $$\sigma_{2s}$$, $$\sigma^*_{2s}$$ MOs are populated):

$$\mathrm{Be_2}:\;(\sigma_{1s})^2\,(\sigma^*_{1s})^2\,(\sigma_{2s})^2\,(\sigma^*_{2s})^2.$$

Number of bonding electrons $$N_b = 2 + 2 = 4.$$ Number of antibonding electrons $$N_a = 2 + 2 = 4.$$ Bond order:

$$\text{B.O.} = \dfrac{1}{2}(N_b - N_a) = \dfrac{1}{2}(4-4) = 0.$$

A zero bond order means that the bonding effect exerted by the electrons in the bonding MOs is exactly cancelled by the destabilising effect of the electrons in the antibonding MOs. Consequently no net stabilisation of the diatomic species over the two free Be atoms is achieved, and the $$\mathrm{Be_2}$$ molecule is unstable and does not exist as a discrete, bound molecule under ordinary conditions.

$$\mathrm{O_2}$$ and its ions are described by molecular-orbital (MO) theory. Apart from the inner shell, the $$\sigma_{2s}$$ and $$\sigma^*_{2s}$$ orbitals are completely filled in every one of these species; they hold one bonding pair and one antibonding pair, which cancel exactly and contribute nothing to the bond order. They are therefore left out of the count below, and we consider only the electrons in the MOs formed from the $$2p$$ atomic orbitals, listed here in increasing order of energy:

$$\sigma_{2p_z},\; \pi_{2p_x}\!\equiv\!\pi_{2p_y},\; \pi^*_{2p_x}\!\equiv\!\pi^*_{2p_y},\; \sigma^*_{2p_z}.$$

In the table, $$N_b$$ is the number of electrons in the bonding $$2p$$ orbitals ($$\sigma_{2p_z}$$ and the two $$\pi_{2p}$$) and $$N_a$$ the number in the antibonding $$2p$$ orbitals ($$\pi^*_{2p}$$ — the $$\sigma^*_{2p_z}$$ stays empty for all four species). The bond order is $$\tfrac{1}{2}(N_b - N_a)$$.

(The two $$\pi^*_{2p}$$ electrons of $$\mathrm{O_2}$$ occupy the two degenerate $$\pi^*$$ orbitals singly with parallel spins, by Hund's rule — giving $$2$$ unpaired electrons. The third $$\pi^*$$ electron in $$\mathrm{O_2^-}$$ pairs up in one of these orbitals, leaving $$1$$ unpaired; in $$\mathrm{O_2^{2-}}$$ both $$\pi^*$$ orbitals are full, so none is unpaired.)

Stability order (the greater the bond order, the stronger the bond and the more stable the species):

$$\mathrm{O_2^+}\,(2.5) \;>\; \mathrm{O_2}\,(2.0) \;>\; \mathrm{O_2^-}\,(1.5) \;>\; \mathrm{O_2^{2-}}\,(1.0).$$

Magnetic properties. $$\mathrm{O_2^+}$$, $$\mathrm{O_2}$$ and $$\mathrm{O_2^-}$$ each have unpaired electrons ($$1$$, $$2$$ and $$1$$ respectively) and are therefore paramagnetic; $$\mathrm{O_2^{2-}}$$ has all its electrons paired and is diamagnetic.

An atomic orbital is described by a wave function $$\psi$$ that is a solution of the Schrödinger equation. The wave function itself is not directly observable, but its square $$\psi^2$$ gives the probability density of finding the electron at a point.

In pictorial representations of orbitals (e.g. of $$p$$ orbitals), the lobes are marked with a $$+$$ or $$-$$ sign. These signs do not represent electric charge but the algebraic sign (phase) of the wave function $$\psi$$ in that region of space — i.e. whether $$\psi$$ is positive or negative there.

The sign of the wave function is important when atomic orbitals combine to form molecular orbitals:

• When two lobes of the same sign overlap, the wave functions interfere constructively (their values add), producing an increased probability of finding electrons between the nuclei. This forms a bonding molecular orbital.
• When two lobes of opposite sign overlap, the wave functions interfere destructively (they cancel), producing a node between the nuclei. This forms an antibonding molecular orbital.
• If lobes of opposite signs are equal in extent (symmetric arrangement), the net overlap is zero and no bond is formed — the two orbitals are said to have different symmetry about the internuclear axis.

Thus the $$+$$ and $$-$$ signs on the lobes of an orbital are the algebraic signs of $$\psi$$; they determine whether the overlap of two orbitals will be constructive (bonding) or destructive (antibonding) and hence whether an effective molecular orbital can be formed.

Hybridisation in $$\mathrm{PCl_5}$$. Phosphorus has the ground-state configuration $$1s^2 2s^2 2p^6 3s^2 3p^3$$ (i.e. $$[\mathrm{Ne}]3s^2 3p^3$$). In its excited state one $$3s$$ electron is promoted to an empty $$3d$$ orbital, giving the configuration

$$3s^1\,3p_x^1\,3p_y^1\,3p_z^1\,3d_{z^2}^1.$$

These five singly-occupied orbitals — one $$3s$$, three $$3p$$ and one $$3d_{z^2}$$ — mix to form five equivalent $$sp^3d$$ hybrid orbitals. According to VSEPR theory, these five hybrid orbitals adopt a trigonal-bipyramidal arrangement: three lie in a plane at $$120^\circ$$ to one another (the equatorial orbitals) and the other two lie above and below this plane along the principal axis, at $$90^\circ$$ to the equatorial plane (the axial orbitals). The five $$sp^3d$$ hybrid orbitals overlap with the half-filled $$3p_z$$ orbitals of five Cl atoms to form five P–Cl sigma bonds.

The geometry of $$\mathrm{PCl_5}$$ is therefore trigonal bipyramidal with three equatorial Cl atoms ($$120^\circ$$ apart) and two axial Cl atoms ($$180^\circ$$ apart, each at $$90^\circ$$ to the equatorial plane).

Why axial bonds are longer than equatorial bonds. The five P–Cl bonds in $$\mathrm{PCl_5}$$ are not all equivalent:

• An equatorial Cl–P–Cl angle is $$120^\circ$$; each equatorial bond pair is repelled by only two other equatorial bond pairs at $$120^\circ$$ and two axial bond pairs at $$90^\circ$$.
• An axial bond pair is repelled by three equatorial bond pairs at $$90^\circ$$ and one axial bond pair at $$180^\circ$$.

Repulsion between bond pairs at $$90^\circ$$ is much stronger than at $$120^\circ$$ (or at $$180^\circ$$, where it is least). Hence each axial bond pair experiences three strong $$90^\circ$$ repulsions from the three equatorial bond pairs, while each equatorial bond pair experiences only two such $$90^\circ$$ repulsions (from the two axial bond pairs). The greater the repulsion, the more the bond pair is pushed away from the central atom — i.e. the longer is the bond.

Consequently, the two axial P–Cl bonds ($$\approx 219\,\mathrm{pm}$$) are slightly longer than the three equatorial P–Cl bonds ($$\approx 202\,\mathrm{pm}$$). This is also why axial bonds are weaker than equatorial bonds, and why $$\mathrm{PCl_5}$$ readily dissociates into trigonal-planar $$\mathrm{PCl_3}$$ (equatorial bonds) and $$\mathrm{Cl_2}$$ on heating.

Hydrogen bond. When a hydrogen atom is covalently bonded to a small, highly electronegative atom (F, O, N), the H atom acquires a substantial partial positive charge ($$\delta+$$) and the electronegative atom acquires a partial negative charge ($$\delta-$$). This $$\delta+$$ hydrogen atom of one molecule is then electrostatically attracted to a lone pair on the electronegative atom (F, O, N) of another molecule (or of another part of the same molecule). This attractive interaction, written as

$$\mathrm{X\!-\!H\,\cdots\,Y},\quad \text{with X, Y = F, O or N},$$

is called a hydrogen bond. The dashed line ($$\cdots$$) denotes the hydrogen bond and the solid line a normal covalent bond. Examples: $$\mathrm{O\!-\!H\,\cdots\,O}$$ in water, $$\mathrm{N\!-\!H\,\cdots\,N}$$ in ammonia, $$\mathrm{F\!-\!H\,\cdots\,F}$$ in HF.

Two types: intermolecular (between different molecules — e.g. in water, alcohols, $$o$$-/$$p$$-isomers of nitrophenol etc.) and intramolecular (within the same molecule — e.g. in $$o$$-nitrophenol, salicylaldehyde).

Comparison with van der Waals forces. Hydrogen bonds are stronger than van der Waals forces but weaker than ordinary covalent bonds. Typical bond energies (kJ/mol):

Hence a hydrogen bond is roughly $$10$$ times stronger than ordinary van der Waals interactions but $$10$$–$$20$$ times weaker than a normal covalent bond. The strong intermolecular H-bonding in water is responsible for many of its anomalous properties: high boiling point ($$100^\circ\mathrm{C}$$), large $$\Delta_{vap}H$$, expansion on freezing, high surface tension, and so on.

Bond order (B.O.) is defined as half the difference between the number of electrons in bonding ($$N_b$$) and antibonding ($$N_a$$) molecular orbitals:

$$\text{B.O.} \;=\; \dfrac{1}{2}\,(N_b - N_a).$$

It represents the number of bonds (single, double, triple) between two atoms. The higher the bond order, the shorter and stronger the bond and the more stable the molecule. A bond order of zero means no net bonding and the molecule does not exist.

(i) $$\mathrm{N_2}$$ ($$14$$ electrons). MO configuration (for N$$_2$$ the order $$\sigma_{2s}$$, $$\sigma^*_{2s}$$, $$\pi_{2p}$$, $$\sigma_{2p}$$, $$\pi^*_{2p}$$, $$\sigma^*_{2p}$$ is used because the $$\pi_{2p}$$ lies below the $$\sigma_{2p}$$ for $$Z\leq 7$$):

$$\mathrm{N_2}:\;(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\sigma_{2p_z})^2.$$

$$N_b = 2+2+2+2+2 = 10$$;\;\; $$N_a = 2+2 = 4$$.

$$\therefore\;\text{B.O.}(\mathrm{N_2}) = \dfrac{10-4}{2} = 3.$$

(ii) $$\mathrm{O_2}$$ ($$16$$ electrons).

$$\mathrm{O_2}:\;(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi^*_{2p_x})^1(\pi^*_{2p_y})^1.$$

$$N_b = 2+2+2+2+2 = 10$$;\;\; $$N_a = 2+2+1+1 = 6$$.

$$\therefore\;\text{B.O.}(\mathrm{O_2}) = \dfrac{10-6}{2} = 2.$$

(iii) $$\mathrm{O_2^+}$$ ($$15$$ electrons). One electron is removed from a $$\pi^*$$ antibonding MO compared to $$\mathrm{O_2}$$.

$$N_b = 10,\;\; N_a = 5.\quad\therefore\;\text{B.O.}(\mathrm{O_2^+}) = \dfrac{10-5}{2} = 2.5.$$

(iv) $$\mathrm{O_2^-}$$ ($$17$$ electrons). One extra electron compared to $$\mathrm{O_2}$$, placed in a $$\pi^*$$ antibonding MO.

$$N_b = 10,\;\; N_a = 7.\quad\therefore\;\text{B.O.}(\mathrm{O_2^-}) = \dfrac{10-7}{2} = 1.5.$$

Summary. $$\text{B.O.}(\mathrm{N_2}) = 3,\;\text{B.O.}(\mathrm{O_2}) = 2,\;\text{B.O.}(\mathrm{O_2^+}) = 2.5,\;\text{B.O.}(\mathrm{O_2^-}) = 1.5.$$