Chapter 9

John A. SchreifelsChemistry 211

Chapter 9-1

8–1

Chapter 9

Ionic and Covalent Bonding


Chapter 9-2

8–2

Overview

• Ionic Bonds– Describing Ionic Bonds– Electron Configuration of Ions– Ionic Radii

• Covalent Bonds– Describing Covalent Bonds– Polar Covalent Bonds: Electronegativity– Writing Lewis Electron-Dot Formulae– Bond Length and Order– Bond Energy


Chapter 9-3

8–3

IONIC BONDING

• Ionic bonds are electrically neutral groups that are held together by the attraction arising from the opposite charges of a cation and an anion.

• Substances that have ionic bonds in a solid form a salt having high melting point and high crystallinity.

• Bonding thought of as the result of the combination of neutral atoms with transfer of one or more electrons from one atom to the other.


Chapter 9-4

8–4

LEWIS SYMBOLS AND THE OCTET RULE

• It was observed that the electron configuration of many substances after ion formation was that of an inert gas octet rule.

• Octet rule: Main-group elements gain, lose, or share in chemical bonding so that they attain a valence octet (eight electrons in an atoms valence shell).

• E.g. The electron configuration of each reactant in the formation of KCl gives: – K+ is that of [Ar] – Cl is also that of [Ar].

• The other electrons in the atom are not as important in determining the reactivity of that substance.

• The octet rule is particularly important in compounds involving nonmetals.

Na Na+

Mg Mg2+

O O2-


Chapter 9-5

8–5

Energy in Ionic Bonding

• When potassium and chlorine atoms approach each other we have:

K(g) K+(g) + e Ei = +418 kJ

Cl(g)+ e Cl(g) Eea = 349 kJ

K(g)+Cl(g) K+(g) + Cl(g) E = + 69 kJ • Positive E = energy absorbed energetically not

allowed. • Driving force must be the formation of the crystalline

solid.

K+(g) + Cl(g) KCl(s)


Chapter 9-6

8–6

Formation of Crystalline Lattice• Energy of crystallization estimated from Coulomb’s

Law by – assuming ions are spheres.– Use ionic radii to determine charge separation.

rK+ = 133x1012 m; rCl = 181x1012 m

d = 133x1012 m + 181x1012 m = 314x1012 m

z1 = z2 = 1.602x1012 C(oulombs); actually one is the negative of the other.

k = 8.99x109 Jm/C2

dzz

kE 21

Coulomb’s Law

mol/kJ.mol/Jx.

molx.Emx

Cx.C/mJx.

34423103442

1231002612

21929

10314

10602110998

• This is related to the negative of the lattice energy, as discussed later.


Chapter 9-7

8–7

BORN-HABER CYCLE AND LATTICE ENERGIES

• Overall energetics for the formation of crystalline solids can be determined from a Born-Haber cycle that accounts for all of the steps towards the formation of solid salts from the elements. For the formation of KCl from its elements we have:

• Net energy change of 434 kJ/mol indicates energetically favored. • Energy for the fifth step is the negative of the • lattice energy: energy required to break ionic bonds and sublime (always

positive). E.g. Determine the lattice energy of BaCl2 if the heat of sublimation of Ba is 150.9 kJ/mol and the 1st and 2nd ionization energies are 502 and 966 kJ/mol, respectively. The heat for the synthesis of BaCl2(s) from its elements is 806.06 kJ/mol.

kJ. KCl(s) (g)Cl/ K(s) nergies ions and e Sum react kJ KCl(s) (g)Cl(g) K KCln of solid. Formatio

kJ. Cl eCl )eaEide ions (n of chlor. Formatio

kJ e K K(g) ssium on of pota. Ionizati

kJ Cl(g) (g) Cl/ lorine tion of ch. Dissocia kJ. K(g) (s) K he metal . Sub of t

4434 21 715 5

6348 4

418 3

122 212289 1

2

2


Chapter 9-8

8–8

Energy Level Diagram of Born Haber Cycle


Chapter 9-9

8–9

LATTICE ENERGIES AND PERIODICITY

• Lattice energy can also be determined from Coulomb’s law:– Directly proportional to charge on

each ion. – Inversely proportional to size of

compound (sum of ionic radii).

• Table (right) presents the lattice energies for alkali and alkaline earth ionic compounds. The lattice energies– decrease for compounds of a

particular cation with atomic number of the anion.

– decrease for compounds of a particular anion with atomic number of the cation.

dzz

kE 21

Coulomb’s Law

Lat. E, kJ/mol

Lat. E, kJ/mol

LiF 1030 MgCl2 2326

LiCl 834 SrCl2 2127

LiI 730 MgO 3795

NaF 910 CaO 3414

NaCl 788 SrO 3217

NaBr 732 ScN 7547

NaI 682

KF 808

KCl 701

KBr 671

CsCl 657

CsI 600


Chapter 9-10

8–10

Ionic Radii

• Ionic radius – a measure of the size of a spherical region around the nucleus of an atom where electrons are most likely to reside.– Cation loses electrons from its valence

shell. Electrons from other valence shells are closer to the nucleus.

– Cation also has more protons than electrons which adds to the pull on the remaining electrons and decreases the radius.

– Anion has more electrons than protons; the pull of the nucleus is less per valence electron. Also, the electron – electron repulsion is greater. These lead to larger radius for an anion.


Chapter 9-11

8–11

Ionic Radii - Trends

– Ionic radii increase down any column because of the addition of electron shells.

– In general, across any period the cations decrease in radius. When you reach the anions, there is an abrupt increase in radius, and then the radius again decreases.


Chapter 9-12

8–12

Ionic Radii - Isoelectronic Ions

• Isoelectronic substances have the same number of electrons and electron configuration.

-216

-1718

19

220 S Cl Ar K Ca

All have 18 electrons

• Largest radius = ion with smallest number of protons.• Smallest radius = ion (atom) with largest number of

protons.


Chapter 9-13

8–13

The Covalent Bond

• Repulsive forces of the electrons offset by the attractive forces between the electrons and the two nuclei.

• Most stable bond energy and bond distance characterizes bonds between two atoms.

• Strengths of Covalent Bonds: • Bonds form because their formation produces lower energy

state than when atoms are separated. • Breaking bonds increases the overall energy of the system.

Energy for breaking bonds has a positive sign (negative means that energy is given off).

H - H (g) 2H(g) H = 436 kJ.• Ionic vs. Covalent Bonds

– Ionic compounds have high melting and boiling points and tend to be crystalline;

– Covalently bound compounds tend to have lower melting points since the attractive forces between the molecules are relatively weak.


Chapter 9-14

8–14

Lewis Structures

• Lewis structure valence electrons represented by dots and are placed where they would be in any bonding that might exist.

• Lewis structures of second row elements:– H2 BH3 – CH4 NH3

– H2O HF• Each has 8 electrons around the central atom; thus

we can predict the number of bonds that will form from the position in the periodic table. E.g. The structure of chlorine is :

..

..Cl.... :Cl:

– Bonding electrons = shared electrons. – Non-bonding or lone pair = unshared electrons


Chapter 9-15

8–15

Lewis Structures(cont’d)• Octet can be filled by donation of electrons from each atom or

one atom can supply both electrons. E.g. H+ + NH3 . "co-ordinate covalent bond". E.g.2

• Multiple bonds may form as a result when the two atoms forming the bond do not have enough electrons. – O=O– NN

• Multiple bonds are shorter and stronger than single bonds because of the extra electrons holding the two atoms together.

4NH 43 BFFBF


Chapter 9-16

8–16

Polar Bonds: Electronegativity

• Electronegativity is a measure of the atom’s ability to gain or lose electrons. It is directly related to its ionization tendency and its ability to form the inert gas configuration. Obtained by:

where Ei = ionization energy and Eea = the electron affiinity. E.g. Li has a very low ionization energy and electron affinity, while Cl has a both a high ionization energy and high electron affinity. Electronegativity will be high for Cl and low for Li.

– Fluorine has the highest electronegativity of 4.0. • Electronegativities (see Fig. 9.15)

– increase from bottom to top of periodic table and – increase to a maximum towards the top right.

• Combination of elements with intermediate electronegativities forms bonds that are intermediate between covalent and ionic.

• can provide an insight as to the type of bond that would be expected. • Ionic bonds formed when 2 • covalent bonds forms when 1.• Polar covalent forms when 1 2, the bonding is "intermediate"

between the two.

2

EE eai


Chapter 9-17

8–17

Polar Bonds: Electronegativity2

E.g.1 Determine the polarity of the N – H in NH3.

E.g. 2 Predict the type of bond formed in CCl4. • The magnitude of indicates if electrons are polarized around

one element in preference to the other. • Polar bond polar. With intermediate , a small charge on the

atom due to that bond develops. + and designates which is the positive and negative side respectively.E.g.3 Determine the relative polarities of HF, HCl, HBr and HI.


Chapter 9-18

8–18

Lewis Structures of Polyatomic Molecules

• Procedure for more complicated molecules: – Determine the total number of valence electrons from each

atom. – Distributed atoms around the central atom (least

electronegative. Hydrogen atoms are usually attached to any oxygen.

– Satisfy the octet of the atoms bonded to the central atom.– Satisfy the octet of the central atom by distributing the

remaining electrons as electron pairs around it. (multiple bonds may be necessary)

E.g. Determine the Lewis structure of H2SO4.

E.g. Draw the Lewis dot structures of NCl3, CSe2, and CO.


Chapter 9-19

8–19

FORMAL CHARGES

• Formal Charge (of an atom in a Lewis formula) the hypothetical charge obtained by assuming that bonding electrons are equally shared between the two atoms involved in the bond. Lone pair electrons belong only to the atom to which they are bound.

E.g. determine the formal charge on all elements: PCl3, PCl5, and HNO3.

• formal charge (FC) allows the prediction of the more likely resonance structure.

• To determine the more likely resonance structure:– FC should be as close to zero as possible.– Negative charge should reside on the most electronegative and

positive charge on the least electronegative element.

E.g. draw the resonance structures of H2SO4; determine the formal charge on each element and decide which is the most likely structure.


Chapter 9-20

8–20

Lewis Structures and Resonance

• Quantum theory indicates that any position possible for an electron.

• Equivalent electron positions often possible:

• E.g. SO2: O=S-O and :O-S=O. – Each structure equally likely.

– the true form of the molecule is a hybrid of these and is called resonance and the hybrid form is called a resonance

hybrid.


Chapter 9-21

8–21

Exceptions to the Octet Rule

• Although many molecules obey the octet rule, there are exceptions where the central atom has more than eight electrons.– Generally, if a nonmetal is in the third period or

greater it can accommodate as many as twelve electrons, if it is the central atom.

– These elements have unfilled “d” subshells that can be used for bonding.

E.g determine the Lewis dot structure of XeF4, ICl3, and SF4


Chapter 9-22

8–22

Bond Dissociation Enthalpies

• Bond dissociation energy, D – the energy required to break one mole of a type of bond in an isolated molecule in the gas phase.

• Useful for estimation of heat of unknown reactions.• Average bond energies listed in tables (e.g. C – H

bond); rest pf structure not very important• HO-H bond in H2O and CH3O-H bond are 492 and

435 kJ/mol. • Hess’s law can be used with bond dissociation

energies to estimate the enthalpy change of a reaction. The breaking in a C – H bond would be C – H(g) C(g) + H(g) H = D = 410 kJ. – Sign always positive since energy must be supplied to break

bond.


Chapter 9-23

8–23

Using Bond Dissociation Enthalpies

• E.g. Estimate the heat of formation of H2O(g) from bond dissociation energies. Thus determine:

• H2(g) + ½ O2(g) H2O(g) = ?• From the book (Table 9.5):

H – H (g) 2H(g) H = D1 = 436 kJ ½ O=O O(g) H = D2 = 494/2 = 247 kJ2H(g) + O(g) H – O – H (g) H = 2D3 = 2*459 kJ H2(g) + ½ O2(g) H2O(g) = 235 kJ

Actual = 241.8 kJ• Can be determined by suming all the energies for the bonds

broken and subtract from if the sum of the energies for the bonds formed.

E.g. 2 Estimate the energy change for the chlorination of ethylene:– CH2=CH2(g) + Cl2(g) CH2ClCH2Cl

ofH

ofH


Chapter 9-24

8–24

Using Bond Dissociation Enthalpies

• It may be necessary to include a phase change since many reactions or reactants are not in the gas phase.

E.g. Determine the heat of formation of CCl4(l).

• Solution: The reaction is:

C(gr) + 2Cl2(g) CCl4(l) = ?

• Write the reactions and sum energies:

ofH

C(gr) C(g) H1 = 715 kJ

2Cl – Cl(g) 4Cl(g) H2 = 486

C(g) + 4Cl(g) CCl4(g) H3 = 1320

CCl4(g) CCl4(l) H4 = 43

C(gr) + 2Cl2(g) CCl4(l) H = 162 kJ

Actual is 139 kJ.


Chapter 9-25

8–25

Electronegativities9_12

Li1.0

Na0.9

K0.8

Rb0.8

Cs0.7

Fr0.7

Be1.5

Mg1.2

B2.0

Al1.5

C2.5

Si1.8

N3.0

P2.1

O3.5

S2.5

F4.0

Cl3.0

Ca1.0

Sr1.0

Ba0.9

Ra0.9

Sc1.3

Y1.2

La–Lu1.1–1.2

Ti1.5

Zr1.4

Hf1.3

V1.6

Nb1.6

Ta1.5

Cr1.6

Mo1.8

W1.7

Mn1.5

Tc1.9

Re1.9

Fe1.8

Ru2.2

Os2.2

Co1.8

Rh2.2

Ir2.2

Ni1.8

Pd2.2

Pt2.2

Cu1.9

Ag1.9

Au2.4

Zn1.6

Cd1.7

Hg1.9

Ga1.6

In1.7

Tl1.8

Ge1.8

Sn1.8

Pb1.8

As2.0

Sb1.9

Bi1.9

Se2.4

Te2.1

Po2.0

Br2.8

I2.5

At2.2

IA IIA

IIIB IVB VB VIB VIIB IB IIB

IIIA IVA VA VIA VIIA

VIIIB

H2.1

Ac–No1.1–1.7

Return to slide 16


Chapter 9-26

8–26

Return to Slide 23

Chapter 9

Documents