Review Topics Test 4
Chapter 5 Review Topics:
1. Lewis dot structures for elements.
2. Periodic Properties:
a. Atomic Radius
b. Ionization energy and the exceptions.
c. Electron affinity and the exceptions.
3. Know that the second IE is always greater than the first and that the third greater than second, etc..
4. Know the EN values for: F, C, H, Cl, N, and O.
5. Know the comparison of ionic radius for + ions and - ions with the atomic radius of the element.
6. Know the trend in ionic radius for isoelectronic species. (the higher the nuclear charge the smaller the isoelectronic ion).
7. Be able to determine whether a bond is polar covalent, nonpolar covalent or ionic based on the delta EN.
8. Be able to determine what is the order of increasing bond polarity based on EN trends. (e.g. compare the polarity of a series of bonds with different elements bonded to either O or F. In these cases. the less EN the element the more polar or more ionic the bond).In general, the greater the delta EN the more ionic or more polar the bond.
Chapter 7 Review Topics:
1. Lewis dot structures for atoms of elements.
2. Lewis dot structures for ions.
3. Lewis dot structures for ionic compounds (the cation will have no dots, only the charge, the anion will have 8 dots and the charge. Each will have parenthesis with the number that you have of each outside except if only one).
4. Lewis dot structures for covalent compounds. Remember that most elements will form octets except:
a. hydrogen-two electrons only
b. beryllium-four electrons only
c. boron and aluminum- six electrons only
d. elements from periods 3 or higher sometimes will end up with ten or twelve electrons and you will only know that after you are done with the Lewis dot structure.
5. Note that an effective way of doing Lewis dot structures is to:
a. Add up all of the valence electrons for all of the atoms in the formula (group numbers represent the valence electrons)
b. Distribute them in pairs on each bond first (between every two symbols.
c. Distribute then on the outside of each symbol surrounding the central atom next except for hydrogen.
d. Distribute extra electrons if any on the central atom. This would result in ten or twelve electrons on the central atom (period 3 or lower elements only).
e. At this point you may encounter one of the following:
are able to distribute all of the electrons as pairs and end up with single
ii. If your central element does not have 8 electrons you will take one pair of electrons from an adjacent atom and form a double bond. (Except B, Be and Al)
iii. If instead you have excess electrons then you will put them in pairs on the central atom as non-bonded electron pairs.
iv. If you are not able to end up with at least an octet on each symbol except for H, Be and B then you may have to resort to a triple bond. These are not super common and the typical elements that may end up with a triple bond include O, N, and C. (CN-, HCN, N2, C2H2)
v. Note that we are going to omit coordinate covalent bonds since this topic is normally not covered in text books.
f. Additional considerations for Lewis dot structures:
i. Halogens and H never double or triple bond.
ii.H is never in the middle
iii. Oxy acids have H bonded to O. (O will not be the central atom)
iv. If you end up with one double bond and one or two single bonds between the same elements then you will have two or three equivalent resonance structures. What you really have is each bond intermediate between a single and a double bond. Remember there can be only 0, 2, or
3. 0 if no double or triple bonds present, 2 if a single bond and a double bond present to the same atom, 3 if two single and two double bonds present to the same atom.
6. Note that single bonds are longer than double bonds and weaker and double bonds are longer than triple bonds and weaker.
7. Be able to determine formal charges for each atom. (VE-(NB+1/2BE)). The sum of the FC equals the charge of the species (0 for a neutral compound). Note that:
a. The formal charges should be as close to 0 as possible and a more EN atom should have a more negative formal charge.
b. Be able to use formal charges to determine which is the major contributor in a set of non equivalent resonance structures. The example that we did in class was OCN-.
Chapter 8 Review Topics
1. Electron geometries of molecules and polyatomic ions (from now on referred to as species) based on the number of electron groups around the central atom. (You must do Lewis dot structure first for this).
2. Molecular geometries or shapes of species based on the electron geometry and the number on non-bonded electron pairs on the central atom.
3. Bonding angles. Remember that if you have non-bonded electron pairs then the bonding angles get reduced from the normal one, but not so for the linear derived from the trigonal bipyramid or the square planar.
4. Polarity of species: If you have symmetrical structures then the species is nonpolar. In order to have a polar species you must have at least one polar bond and the species must not be symmetrical. This is usually the case when you have non-bonded electron pairs on the central atom. The two exceptions to this is when you have three non-bonded pairs on the trigonal bipyramid electron geometry, which results in a linear molecular geometry or shape. This will be nonpolar. Also two non-bonded electron pairs on the octahedral electron geometry, which results in a square planar molecular geometry or shape. This will also be nonpolar.
5. Valence Bond Theory: Hybridization. You must know which hybridization goes with which electron geometry:
a. linear-sp b. trigonal planar-sp2 c. tetrahedral-sp3 d. trigonal bipyramid-sp3d e. octahedral-sp3d2
6. Be able to recognize what the hybrid orbitals look like for the central atom given a formula. This is what results after hybridization and promotion. You should be able to do this for structures that are single bonded, and those that contain lone pairs.
7. Be able to recognize the hybridization on C when it is single bonded only, when it has a double bond and when it has a triple bond.
8. Know that a single bond is a sigma bond ( bond) and a double bond is a single bond plus an additional bond ( bond). Be able to determine the number of each type of bond on a C that is single bonded (4 ), double bonded (3 and 1) or triple bonded (2 and 2 ).
Chapter 9 Review Topics
1. Be able to draw molecular orbital diagrams for diatomic species.
2. Be able to calculate bond order (bonding e’s minus antibonding e’s)/2
3. The higher the bond order the more stable the structure.
4. When doing molecular orbital diagrams for heteronuclear diatomic molecules, the more electronegative element has lower energy for the atomic orbitals than the less electronegative. As a result the delta E is greater for the separation between bonding and antibonding molecular orbitals and this contributes to having more polar or ionic character.
Chapter 10 Review Topics
1. Three acid-base theories:
a. Arrhenius-Formula contains H for acids and OH with metal or NH4+ for bases. The definition is an acid ionizes to produce H+ in aqueous solution and a base ionizes to form OH- in aqueous solution.
b. Bronsted-Lowry- An H+ is always transferred from an acid to a base. An acid is therefore a proton donor and a base is a proton acceptor. NH3 and the amines are Bronsted Lowry bases but not Arrhenius bases.
c. Lewis- An acid is an electron acceptor and a base is an electron donor. Just remember BF3 (H, Cl, Br or I can be instead of F) is a Lewis acid because it lacks a pair of electrons in the octet. NH3 is a Lewis base because it has an extra electron pair it can share. The product is BF3-NH3.
2. Know how to complete the ionization reactions of acids and bases in water including NH3 and the amines. (review from module 6).
3. Know how to complete the neutralization reactions of acids and bases including when the base is NH3 or an amine. (review from module 6).
4. Know how to recognize the conjugate acid of any base and the conjugate base of any acid. The acid will always have one more H+ than the base. The base will have one less H+ than the acid. So we have conjugate acid base pairs. In an equilibrium process the acid can be on the left or right and the base can be on the left or right. Then the other side has the conjugate base of the acid and the conjugate acid of the base. Water acts like an acid in the presence of a base and as a base in the presence of an acid. The reaction is always the transfer of an H+ from the acid to the base.
5. Know that the stronger the acid the weaker the conjugate base and vice versa. Also the stronger the base the weaker the conjugate acid and vice versa. As a result, conjugate bases of the strong acids (the anions of the strong acids) are always neutral.
6. Know the order of increasing acid strengths:
a. For binary acids with anions in the same group the larger the anion (conjugate base) the stronger the acid because the anion will be more stable. HI>HBr>HCl>HF
b. For binary acids with anions in the same period, the more electronegative the element of the anion the stronger the acid. HCl>H2S
c. For ternary oxyacids with the anion in the same group the more electronegative the element of the anion the stronger the acid because it pulls electrons more away from the acidic H(s). HClO4>HBrO4>HIO4
d. For ternary oxyacids with the anion in the same period the more electronegative the element of the anion the stronger the acid for the same reason as 6c. HClO4>H2SO4>H3PO4>H3ASO4
e. For ternary oxyacids with the same anion the more oxygens the stronger the acid. HClO4>HClO3>HClO2>HClO
7. The first H in polyprotic acids is more acidic than the second and the second is more acidic than the third. H3PO4>H2PO4->HPO42-
8. Know that an acid is leveled in water when it is stronger than H3O+. This means that water does not distinguish between these acids as far as strengths, treats them all the same.
9. Acidic salts contain a metal and hydrogen(s) plus an anion.
Review Topics Test 4-Chapter 15 Sections 15-5 to 15-9
1. Be able to convert grams of a substance which is part of a thermochemical equation to energy (J, kJ, cal, of kcal) and vice versa
2. Recognize the most common physical state of each element.
3. Know that the delta H of formation of the most common physical state of an element is 0.
4. Know that the delta H of a reaction is the sum of the delta H(’s ) of formation of the product(s), each multiplied by the number of moles minus the sum of the delta H(‘s) of formation of the reactant(s), each multiplied by the number of moles.
5. Be able to identify a formation reaction, formation of a formula from the elements in the most common state, each with the correct amount of moles needed to form one mole of the substance.
6. Be able to combine reactions for which the delta H of reaction is known to obtain the delta H of a reaction for which it is not known by flipping the reaction (sign of delta H changes), or multiplying by a coefficient (delta H is also multiplied).
7. Be able to use bond energies to predict delta H for a reaction. It will be the sum of the bond energies of the bonds present in the reactants minus the bond energies of the bonds present in the products.