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Transcript
```Fullerton College Office of Special Programs
SI Review 111B
Handout #14 Solutions
1. To determine the oxidation state of the metal ion, we consider the charges of all known ions and
work backwards to the charge on the metal ion:
a.
[Co(NH3)5Br]2+
The total charge is equal to the total charges of all ions…
Total Charge = (Co?) + 5(NH3) + (Br-)
+2 = (Co?) + 5(0) + (-1)
Co = +3
Since there are six monodentate ligands, the metal ion has a coordination number of six.
b.
[Fe(CN)6]4Total Charge = (Fe?) + 6(CN-)
-4 = (Fe?) +6(-1)
Fe = +2
Since there are six monodentate ligands, the metal ion has a coordination number of six.
c.
[Co(C2O4)3]4Total Charge = (Co?) + 3(C2O42-)
-4 = (Co?) + 3(-2)
Co = +2
Since there are three bidentate ligands, the metal ion has a coordination number of six.
d.
[PdCl4]2Total Charge = (Pd?) + 4(Cl-)
-2 = (Pd?) + 4(-1)
Pd = +2
Since there are four monodentate ligands, the metal ion has a coordination number of
four.
2. For this problem, the geometries have been given to you. It is important to remember that these
geometries are experimentally determined. Although we have the ability to predict the geometry,
our predictions are less reliable than predictions concerning the shape of main group molecules.
a.
[Pt(en)Cl2], dichloro(ethylenediamine)platinum(II)
Cl _
Cl
_ __
ENGAGE in STEM is funded through the U.S Dept.Ptof_ Education, Hispanic Serving Institutions (H.S.I),
_
__
STEM & Articulation Programs, Cooperative
Arrangement Agreement Grant
N
N
(There is only one structure for a square planar geometry and this formula.)
b.
Cl
__
Pd_
_
__
CO
OH2
CO __
CO __
H2O
c.
_
__
_ Pd__
Cl
OC
[Fe(CO)4Cl2)]+, tetracarbonyldichloroiron(III)
OC
___
OC
Cl ____ Fe ____ OC
___
Cl
OC
OC
___
OC
Cl ____ Fe ____ Cl
d.
trans-tetracarbonyldichloroiron(III)
___
CO
cis-tetracarbonyldichloroiron(III)
OC
[Cr(CO)3(NH3)3]3+, tricarbonyltriamminechromium(III)
OC
___
OC
H3N ____ Fe ____ NH3
___
H3N
OC
OC
___
OC
H3N ____ Fe ____ OC
___
H3N
mer-tricarbonyltriamminechromium(III)
NH3
fac-tricarbonyltriamminechromium(III)
3. Linkage isomers are a form of structural isomer, where the formulas are the same, but the
bonding is different. There are two linkage isomers for [Mn(NH3)5(NO2)]2+,
pentaamminenitromanganese(II), above, and pentaamminenitritomanganese(II), below. Note,
the formulas for these isomers are generally written so as to identify the difference in bonding
pentaamminnitromanganese(II), [Mn(NH3)5(NO2)]2+:
___
NH3
NH3
H3N ____ Mn ____ NO2
___
H3N
NH3
pentamminenitritomanganese(II), [Mn(NH3)5(ONO)]2+:
___
NH3
NH3
H3N ____ Mn ____ ONO
___
H3N
NH3
4. In writing the formula of a coordination compound (complex), be sure to follow the guidelines
provided in class.
a.
hexaaquanickel(II) chloride
[Ni(H2O)6]Cl2
b.
pentacarbonylchloromanganese(I)
Mn(CO)5Cl
c.
NH4[V(H2O)2Br4]
d.
tris(ethylenediamine)cobalt(III) trioxalatoferrate(III)
[Co(en)3][Fe(C2O4)3]
5. In naming coordination compounds (complexes), be sure to follow the guidelines provided in class.
a.
[Cu(CN)4]2-
tetracyanocuprate(II)
b.
[Mn(CO)3(NO2)3]2+
tricarbonyltrinitritomanganese(V)
c.
Na[Cr(H2O)2(C2O4)2]