Download transition metals KEY

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Optional Homework 8 – do not turn in!!
1.) Write the electron configurations for the following species:
a. Ag
1s22s22p63s23p64s23d104p65s24d9  5s14d10
b. Ag+1 1s22s22p63s23p64s23d104p6
c. Cd+2 1s22s22p63s23p64s23d104p6
5s04d10
5s04d10
d. Ir+3 1s22s22p63s23p64s23d104p65s24d105p66s24f145d7 
6s04f145d6
2.) Find the charge of all species in each of the following coordination compounds, if it is a polyatomic
ion, I just want the overall charge of that ion, not individual oxidation numbers of every element!:
HINT: you may need to go back and review your polyatomic ion charges!!
a. K[CuCl2]
K:
+1
Cu:
+1
Cl: -
1
b. (NH4)2[Ni(CN)4]
NH4:
+1
Ni:
+2
CN:
-1
c. [Ti(H2O)6]2(SO4)3
Ti:
+3
H2O:
0
SO4:
-2
d. Al4[V(CN)6]3
+3
Al:
V:
+2
CN:
-1
3.) Identify the complex ion and the ligands in the compound K3[Fe(CN)5CO]. Find the oxidation
number of the metal ion in complex ion: be sure to label the inner sphere (ligands bonded
covalently to the metal) and outer sphere ligands (counter ions) in the complex!
K3[Fe(CN)5CO]:
K is an outer sphere ligand
CN is an inner sphere ligand (-1)
CO is an inner sphere ligand (0)
Iron is the transition metal oxidation state = +2
4.) Find the charge on the nitrosyl ligand (NO) in the Co(III) compound: Na2[Co(CN)5NO]
Na = +1 x 2 = +2
CN = -1 x 5 = -5
Co = +3
Na2[Co(CN)5NO]
So the NO ligand must be neutral!
5.) If the coordination complex [Cr(NH3)5Cl]Cl2 dissociated into its ions, what ions would be formed?
Write a balanced equation showing this dissociation!
[Cr(NH3)5Cl]Cl2
→ ???
[Cr(NH3)5Cl]Cl2
→
[Cr(NH3)5Cl]+2 (aq) + 2Cl-1 (aq)
6.) Using the ideas of Lewis and resonance structures, determine which of the following ligands can
participate in linkage isomerism (meaning can they cause a compound to be a linkage isomer).
DRAW the resonance structures in order to explain! Remember that a linkage isomer can bond to
the metal ion from two distinctly different places on the molecule. And in order to have a bond –
you need electrons!!!
a. NO2-1
O N O
O N O
There are lone pairs present on
both the N and the O so the species can bond through either the N or O which
means this species can be a linkage isomer
b. SO2
O
S
O
O
S
O
O
S
O
There are lone pair electrons
on the S and the O in the molecule, therefore this species can participate in
linkage isomerism
c. NO3-1
-1
-1
O
O
N
-1
O
O
O
N
O
O
O
N
O
There are only
lone pair electrons on the oxygens in this species, therefore, the only way for this
ion to bond is through the oxygen. Since all of the oxygens are equivalent to one
another, there is no linkage isomerism possible
7.) Define/explain the crystal field splitting energy ()?
Crystal field splitting occurs when incoming ligands “invade” the space of certain d orbitals.
This causes those d orbitals to become higher in energy than their non-interfered with
counterparts. Depending on the shape (octahedral, tetrahedral, or square planar) the d
orbitals will split apart differently. The crystal field splitting energy is the new different in
energy between the higher energy d orbitals and the lower energy d orbitals as a result of
the split.
8.) Sketch a structure for the following complex ions:
a. [PdCl4] -2 (square planar)
Cl
Cl
Pd
Cl
-2
Cl
b. [Be(OH)4]-2 (tetrahedral)
-2
OH
HO Be OH
OH
c. [AgCl2]-1
Cl
Ag
Cl
d. [Co(NH3)5CO3]+1
NH3
H3N Co NH3
H3N
NH3
CO3


9.) Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high
temperatures, as described by the following balanced equation. How many grams of N2 are formed
when 18.1 g of NH3 are reacted with 90.4 g of CuO? What starting material, if any is left over? How
many grams of that material are left over?
2 NH3 (g) + 3 CuO (s)  N2 (g) + 3 Cu (s) + 3 H2O (g)
Ahhhh, a good ol’ limiting reagent problem!
18.1 g NH 3 x
mol
1 mol N 2
x
= 0.531 mol N 2
17.031 g 2 mol NH 3
1 mole CuO
1 mol N 2
x
= 0.379 mol N 2  Limiting Reagent!
79.545 g CuO
3 mol CuO
28.013 g N 2
0.379 mol N 2 x
= 10.6 g N 2 produced
1 mol N 2
90.4 g CuO x
Since CuO is the limiting reagent, NH3 is in excess. To determine this amount of excess, determine
how much extra NH3, will react with the starting amount of CuO and then subtract the amount
used from the amount of NH3 you started with.
90.4 g CuO x
1 mole CuO
2 mol NH 3 17.04 grams NH 3
x
x
 12.9 grams NH 3 used
79.545 g CuO
3 mol CuO
1 mole NH 3
We started with 18.1 grams of NH3 and we used up 12.9 grams of it, that leaves 5.2
grams of NH3
left over!
10.) Given the following information, draw the Lewis structure, determine the molecular shape, VSEPR
shape, hybridization around the central atom, and predicted bond angle for each species.
a. PH3
P: 1 x 5 = 5
H: 1 x 3 = 3
8 valence e-1
H P H
H
tetrahedral shape for both
angle 109.5o
sp3
b. CF4
C: 1 x 4 = 4
F: 4 x 7 = 28
32 valence e-1
F
F C F
F
tetrahedral for both
angle: 109.5o
sp3
c. SO4 -2
S: 1 x 6 = 6
O: 4 x 6 = 24
30 + 2 = 32 valence e-1
-2
O
O
S
O
O
tetrahedral for both
angle: 109.5o
sp3
d. PF5
P: 1 x 5 = 5
F: 5 x 7 = 35
40 valence e-1
F
F
P
F
F F
trigonal bipyramidal for both angles: 180o, 90o, and 120o sp3d
e. KrF2
Kr: 1 x 8 = 8
F: 2 x 7 = 14
22 valence e-1
F
Kr F
V: trigonal bipyramidal
angle: 180o
sp3d
M: linear
(between the atoms)
f. TeF5 -1
Te:1 x 6 = 6
F: 5 x 7 = 35
41 + 1 = 42 valence e-1
-1
F
F
Te
F
F
F
V: octahedral angle: 90o and 180o sp3d2
M: square based pyramid
11.) Does the Qc for the formation of 1 mole of NO from its elements differ from the Qc of the
decomposition of 1 mole of NO into its elements? Explain and give the relationship between the
two Qc values.
½ N2 + ½ O2 → 1NO
[NO]
Q=
[N 2 ]1/2 [O 2 ]1/2
1NO → ½ N2 + ½ O2
[N 2 ]1/2 [O 2 ]1/2
Q=
[NO]
The two Q values are inverses of one another! When the equation is reversed, the products
become the reactants and the reactants become the products, so their location in the Q
expression changes from denominator to numerator so the Q value is “upside” down or
inverted!
12.) Balance the reaction and write the Qc
a. ____2____NaHCO3 (s)  ________Na2CO3 (s) + ________CO2 (g) + _______H2O (g)
Q = [CO2][H2O]
13.) Given the following Qc expressions, write the balanced chemical equation:
a. Q = [CO2]2 [H2O]2
[C2H4][O2]3
The Q expression is products over reactants and the exponents are the coefficients used in the
balanced equation:
C2H4 + 3O2 → 2CO2 + 2H2O
__C__ 9.
a.
b.
c.
d.
e.
What is the correct answer to the following expression: 3.33 x 10-5 + 8.13 x 10-7?
3 x 10-5
3.4 x 10-5
3.41 x 10-5
3.411 x 10-5
3.4113 x 10-5
__B__ 10.
a.
b.
c.
d.
Rank the subatomic particles from lowest to highest mass?
electrons < neutrons < protons
electrons < protons  neutrons
neutrons < electrons < protons
electrons = protons < neutrons
_C__
A substance that contains unpaired electrons is attracted to a magnetic field. This
substance is said to be ________.
Ionic
Ferromagnetic
Paramagnetic
Polar
Diamagnetic
11.
a.
b.
c.
d.
e.