chapter 6 ionic bonds and some main- group chemistry
TRANSCRIPT
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Chapter 6
Ionic Bonds and Some Main-Group Chemistry
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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???????
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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+
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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+
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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
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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
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Ionic Radii
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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
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Ionization Energy
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Ionization Energy Trend
• Increases across a period – held tighter
• Decreases down a group – larger (n)
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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
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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-
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Problem
• Which of the following elements has the smallest first ionization energy?
– A. Rb– B. Mg– C. I– D. As– E. F
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Problem
• Which of the following elements has the largest second ionization energy (IE2)?
– A. Li– B. B– C. O– D. F– E. Na
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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)
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Electron Affinity
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Electron Affinity
• Trend– Eea becomes more negative across a period
– Eea becomes more positive down a group• Larger atoms don’t care as much
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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)
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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
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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.
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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.
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Ionic Bonds and the Formation of Ionic Solids
Born–Haber Cycle for NaCl:
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Ionic Solids and the Formation of Ionic Compounds
Born–Haber Cycle for MgCl2:
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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?
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Ionic Bonds and the Formation of Ionic Solids
• Trends in Lattice Energy– Increases with increasing charge– Increases with decreasing ion size
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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
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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?
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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
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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
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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
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Required Homework
• Assignment 6