Chapter 6

Ionic Bonds and Some Main-Group Chemistry

Ions and Their Electron Configurations

• Chapter 2– Metals lose electrons to form cations

• Main group elements tend to lose the same # of electrons as their group number

– Nonmetals when combined with metals gain electrons to form anions

• The number gained = 8 – group #

• WHY???????

Ions and Their Electron Configurations

• Main Group Elements - the number of electrons gained or lost = the number of electrons necessary to obtain the nearest noble gas or pseudo-noble gas configuration (filled d-orbital).

– Main group metal – the # electrons lost often = # of valence electrons

– Nonmetal – the # electrons gained = # of electrons necessary to achieve 8 in the valence shell

• Na, Ne, Na+ O, Ne, O2-

• K, Ar, K+ Cl, Ar, Cl-

• Ga, Ga3+ Sn, Sn4+

Ions and Their Electron Configurations

• Transition element ions formed by– 1st losing their valence electrons– 2nd losing d-orbital electrons to obtain the

charge

• Fe2+ Fe3+ Co2+

Co4+

• Ag+ Zn2+

Problem

• Select the correct set of quantum numbers (n, l, ml, ms) for the first electron removed in the formation of a cation for strontium, Sr. 

– A.  5, 1 , 0, -½– B.  5, 1, 0, ½– C.  5, 0, 1, ½– D.  5, 1, 1, ½– E.  5, 0, 0, -½

Ionic Radii

• Effect of charge on ionic radii evaluated using atomic size and Zeff

– Atomic size increases with increasing energy levels

– Zeff • Cation – more protons than electrons – stronger attraction• Anion – less protons than electrons – weaker attraction

• Cation < Neutral atom < anion

Ionic Radii

Ionization Energy

• Ionization Energy (Ei) – the energy required by 1 mole of gaseous atoms for 1 mole of electrons to be ejected.

– Na EI Na+ + e-

– Energy necessary to make a cation• More energy required for smaller atoms

Ionization Energy

Ionization Energy Trend

• Increases across a period – held tighter

• Decreases down a group – larger (n)

Higher Ionization Energies

• Ionization energy not limited to a single electron– 2nd electron removed requires even more

energy• Removal of first leads to more protons than

electrons

– To break into a noble gas configuration requires a lot of energy because of their stability

Problem

• Which one of the following equations correctly represents the process relating to the ionization energy of X? 

– A.  X(s) → X+(g) + e-

– B.  X2(g) → X+(g) + X-(g)– C.  X(g) + e- → X-(g)– D.  X-(g) → X(g) + e-– E.  X(g) → X+(g) + e-

Problem

• Which of the following elements has the smallest first ionization energy? 

– A.  Rb– B.  Mg– C.  I– D.  As– E.  F

Problem

• Which of the following elements has the largest second ionization energy (IE2)? 

– A.  Li– B.  B– C.  O– D.  F– E.  Na

Electron Affinity

• Electron Affinity (Eea) - Describes the energy associated with the absorption of 1 mole of electrons by 1 mole of gaseous atoms

– Cl + e- Cl-

– Can be positive or negative• Positive – energy had to be added to force the atom to

absorb the electron (non-favored)• Negative – energy was given off when the electron was

added to the atom (favored)

Electron Affinity

Electron Affinity

• Trend– Eea becomes more negative across a period

– Eea becomes more positive down a group• Larger atoms don’t care as much

Problem

• Which one of the following equations correctly represents the process involved in the electron affinity of X? 

– A.  X(g) → X+(g) + e-– B.  X+(g) → X+(aq)– C.  X+(g) + e- → X(g)– D.  X(g) + e- → X-(g)– E.  X+(g) + Y-(g) → XY(s)

Problem

• Select the element with the most negative electron affinity (i.e., accepts an electron most readily). 

– A.  H– B.  Li– C.  C– D.  F– E.  Ne

Ionic Bonds and the Formation of Ionic Solids

• Ionic bonds:

• Form when an element with a small Ei

value comes in contact with an

element with a negative Eea value.

Ionic Bonds and the Formation of Ionic Solids

• The energetics of ionic reactions can be viewed

on a Born–Haber Cycle which shows how each

step contributes to the overall reaction energy.

• That energy is called the lattice energy (U) of the

solid. By convention, the lattice energy refers to

the breakup of the crystal into ions.

Ionic Bonds and the Formation of Ionic Solids

Born–Haber Cycle for NaCl:

Ionic Solids and the Formation of Ionic Compounds

Born–Haber Cycle for MgCl2:

Ionic Bonds and the Formation of Ionic Solids

• Calculate the lattice energy (in kJ/mol) for the formation of

CaH2 from its elements.

• Calculate the overall energy change (in kJ/mol) for the

formation of CaCl from its elements.

• Calculate the overall energy change (in kJ/mol) for the

formation of CaCl2 from its elements.

• Which is more likely to form, CaCl or CaCl2?

Ionic Bonds and the Formation of Ionic Solids

• Trends in Lattice Energy– Increases with increasing charge– Increases with decreasing ion size

The Octet Rule

• General Conclusions– Group 1A form +1 by losing ns1 electron to obtain

noble gas configuration– Group 2A form +2 by losing ns2 electrons to obtain

noble gas configuration– Group 3A tend to form +3 by losing ns2np1 electrons

to obtain noble gas configuration– Group 7A gain 1 electron to form ns2np6 hence giving

them a noble gas configuration– Group 8A neither gain nor lose electrons because of

stability of configuration

The Octet Rule

• Octet Rule – main group elements tend to undergo reactions that leave them with 8 outer shell electrons

• Why does the rule work?

The Octet Rule

• Electrons of metals are most likely lost due to the fact that:– core electrons are shielding them from the

nucleus

– Zeff is lower

– Small ionization energies– Upon loss they obtain noble gas configuration

The Octet Rule

• Nonmetals are most likely to gain electrons due to the fact that:– no additional shielding occurs from the

nucleus occurs due to core electrons

– Zeff is higher

– More negative electron affinities– Upon gain they obtain noble gas configuration

Optional Homework

• Text – 6.30, 6.36, 6.38, 6.40, 6.42, 6.44, 6.46, 6.48, 6.58, 6.64, 6.66

• Chapter 6 Homework

Required Homework

• Assignment 6