Ch 301 Exam 2 Review

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

This preview shows pages 1–11. Sign up to view the full content.

Click to edit Master subtitle style Exam 2 Review CH301 Dr. Sparks

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
Click to edit Master subtitle style Electron Dot Structures

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 49

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

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online