Ionic Bonds and Some Main-Group Chemistry 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 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