ch 19

Question 1

deviations from raoult’s law
list the /possible interactions

Answer 1

no interactions
unfavorable interactins(repulsion)
favorable interactions

Question 2

deviations from raoult’s law
no interactions =

Answer 2

ideal solution; two nonpolars
VP is as expected
delta H = 0

Question 3

deviations from raoult’s law
unfavorable interactions

Answer 3

repulsions
one pushes other out of the solution –> gas
polar+ nonpolar
VP is higher than expected, positive deviation
delta H = positive –> endothermic

Question 4

deviations from raoult’s law
favorable interactions

Answer 4

solution interactions are stronger than components
more interactions as a solution
“keep” components in solution
VP lower than expected
delta H negative –> exothermic

Question 5

complex ion

Answer 5

transition metals surrounded by ligands
covalent bond between transition metals and ligands
transition metal is e- acceptor
ligands are e- donors

Question 6

[Co(NH3)6] (NO3)3
label the components

Answer 6

[Co(NH3)6]NO3)3 = coordination compound
[Co(NH3)6] = complex ion
Co = transition metal
NH3 = ligand
NO3 = counter ion
counter ion and complex ion connected by ionic bond –> break in solution

Question 7

coordination number

Answer 7

bonds to central transition metal

Question 8

types of ligands

Answer 8

ligands must have a lone pair

neutral = NH3 and H2O
charged= Cl-
monodentate
polydentate

Question 9

coordination number and relative shape

Answer 9

2= linear
4= tetrahedral or square planar
6 = octahedral

Question 10

monodentate vs polydentate

Answer 10

monodentate = one binding site for ligand ( one transition metal to bind to)
polydentate = more than 1 bonding site
- polydentate must have multiple lone pairs and at least 3 to 4 atoms between lone pair bonding sites of ligands

Question 11

ethylenediamine (en)

Answer 11

H2NCH2CH2NH2
C2H6N2

Question 12

naming coordination compounds

Answer 12

1. cation first
2. ligands before metal
3. if complex ion is anion, add “ate” to metal
4. oxidation state is written in roman numerals
5. alphabetical order of ligands

Question 13

naming ligands in coordination compounds;
H2O
NH3
CO

Answer 13

H2O = aqua
NH3 = ammine
CO - carbonyl

Question 14

naming ligands in coordination compounds;
anion
Cl =
OH=
CN=

Answer 14

Cl = chloro
OH - hydroxo
CN = cyano
end in “o”

Question 15

naming ligands in coordination compounds;
ide =
ate =
ite =

Answer 15

ide = o
ate = ato
ite = ito

Question 16

prefixes for # ligands
2,3,4…

Answer 16

2 = di
3 = tri
4 = tetra

Question 17

prefixes for polydentate ligands 2,3,4

Answer 17

2 - bis
3 - tris
4 - tetra

Question 18

name the folllowing
K4[Ni(CN)4]
K2[Cr(H2O)2(C2O4)2]
[Fe(H2O)5OH]Cl2

Answer 18

potassium tetracyanonickelate(0)
potassium diaquabisoxylatochromate(II)
pentaaquahydroxoiron(III) chloride

Question 19

what is the formula
tris(en)cobalt(III) sulfate

Answer 19

[Co(en)3]2(SO4)3

Question 20

isomers

Answer 20

compounds with the same number and type of atom but are structurally different
structural and stereoisomers

Question 21

structural isomers

Answer 21

different bonds
different formula
different name
[MASB]C
[MASC]B

difference in linkage

Question 22

stereoisomers

Answer 22

same bonds, different orientation
same formula
different name

Question 23

geometric stereoisomers

Answer 23

cis and trans isomers
cis = AA
BB
trans = AB
BA

Question 24

optical stereoisomers

Answer 24

one molecule is mirror image
cannot superimpose

Question 25

which one of the following complexes can have geometric isomers?
A. [Pt(NH3)2Cl2] = square planar
B. [Zn(NH3)2Cl2] = tetrahedral
C. [Cu(NH3)4]2+ = square planar
D. [Cu(NH3)5Cl]2+ = octahedral
E. All

Answer 25

A. [Pt(NH3)2Cl2] = square planar

tetrahedrals can’t have geometric isomers since their geomtry isn’t flat (not on same plane)

square planar C has 4 of the same ligands –> can’t make cis/trans

octahedral has 5 same ligands and 1 Cl –> ca’t make cis/trans

Question 26

crystal field theory

Answer 26

consider d-orbital orientations and covalent bonds to transition metals

eg = e- density on axis –> dx^2-y^2 and dz^2
t2g = e- density between axes –> dxy dxz dyz

caused by e-/e- repulsions between e- in d-orbitals and e- in covalent bonds

Question 27

d- orbital diagram for octahedrals

Answer 27

6 covalent bonds
90 degrees
ligaments bond ON axis
^
| _____ _____ eg
| _______ ______ _____t2g

Question 28

d-orbital diagram for tetrahedral

Answer 28

4 covalent bonds
109.5 degrees
ligands bond between axis

• __ ____ ____ t2g
• ___ ____ eg

______ _____ ______ t2g

Question 29

d-orbital diagram for square planars

Answer 29

bonds lie on axis
* _____ dx^2-y^2
* ___ dxy
* ___dz^2
* ___ dxz ___ dyz

___ dx^2 - y^2

Question 30

look over the shapes of each of the d-orbital levels

Question 31

splitting energy

Answer 31

2 competing effects
1. energy costs from pairing
2. energy costs from promotion

when the splitting energy is small… electrons would rather be promoted(unpaired electrons) (high spin)

when the splitting energy is large – electrons would rather be paired (low spin)

Question 32

list the general splitting energies for
tetrahedral
square planar
octahedral

Answer 32

tetrahedral = small splitting energy
square planar = large splitting energy
octahedral = look at spectrochemical series

Question 33

large splitting energy =

Answer 33

paired electrons

Question 34

small splitting energy

Answer 34

unpaired electrons
rather be promoted

Question 35

color emission of compounds

Answer 35

result of e- transition between eg and t2g
electrons can absorb photon for promotion
we see reflected light (opposite of color wheel)

Question 36

calculate energy and wavelength using what equation

Answer 36

E=hc/wavelength
h = 6.626x10^-34 J*s
E = Joules per photon