Ch 301 Exam 2 Review

Ch 301 Exam 2 Review - Exam 2 Review CH301 Click to edit...

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Click to edit Master subtitle style Exam 2 Review CH301 Dr. Sparks
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Ionic Vs. Covalent bonding Ionic compounds Metal + non-metal Electrons are transferred Soluble in polar solvents NOT soluble in non-polar solvents Are electrolytes: can conduct electricity Covalent compounds Usually two non-metals Electrons are shared polar covalent are soluble in polar solvents non-polar covalent are soluble in non-polar solvents Not generally good electrolytes
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Coulomb’s law for ionic compounds - coulomb’s law describes the strength of interaction between two ions Coulomb’s law: V = 2.31 x 10-19 Jnm (Q1 x Q2) r V = describes the strength of interaction Q1 and Q2 = the charges on the ions that are participating in the ionic compound r = the radius or distance between the nuclei of the two ions The charges on the ions are directly proportional to the strength of the interaction = so higher charges means stronger interactions The radius of the ions is inversely proportional to the strength of the interaction = so the smaller the radii, the stronger the interactions
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Which of the following ionic compounds has the strongest coulombic attractions? MgCl2 MgBr2 CaBr2 CaCl2 + 2 - 1 + 2 - 1 + 2 - 1 + 2 - 1 Radii: Mg2+ Ca2+ Cl1- Br1- Smaller radii = stronger attractions
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KCl has an ionic energy of 468 Kj. The ionic radius of CaO is 2.3 times that of KCl. What is the ionic energy of CaO? V = 2.31 x 10-19 ( Q1 x Q2 ) r We know: V for KCl = 468 Kj Q1 x Q2 for KCl = 1 x 1 = 1 2 = 4 R for CaO is 2.3 x R for KCl Q is directly proportional R is inversely proportional If the (Q1 x Q2) value is 4 times bigger than for KCl, then we should multiply the V of KCl by 4 If the radius is 2.3 times bigger than for KCl, then we should divide the V of KCl by 2.3 V = (468 Kj x 4) = 814 Kj 2.3
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Lattice Energy The amount of energy that is given off when a lattice of ions is formed Lattice E = K [ Q1xQ2 ] Notice that the relationships between charge and lattice energy as well as radius and lattice energy are exactly the same as they were for coulombic attractions. r
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Click to edit Master subtitle style Electron Dot Structures
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Dot Structures for an atom 1. Draw out the electron configuration for the valence electrons of the atom: O 2 s 2 p O = 2s 22p4 2. Draw in pairs and unpaired electrons around the symbol for the atom: O 2 sets of paired electrons 2 unpaired electrons
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Dot Structures for ions: O2 - 1. Draw out the electron configuration for the valence electrons of the atom in its neutral state (O): 2 s 2 p O = 2s 22p4 Now add two electrons to account for the 2- charge on the oxygen. For a cation, you would take away electrons.
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2. Draw in pairs and unpaired electrons around the symbol for the atom: O 2 s 2 p O = 2s 22p4 O2 - 4 sets of paired electrons 3. Add brackets and the overall charge of the ion.
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Ch 301 Exam 2 Review - Exam 2 Review CH301 Click to edit...

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