What is Inorganic Chemistry?
Inorganic Chemistry Chem 151A and L
Week 1-1
Coordination Chemistry: (Chapter 11, pages 387-391) What is Inorganic Chemistry? Inorganic is everything other than carbon. You are prejudiced against inorganic chemistry due to a lack of knowledge but the majority of the chemistry in the world is inorganic. Just not as important to biology. Nevertheless, biology has critical uses of inorganic chemistry that are essential. Inorganic chemistry is made of coordinate bonds or “dative” bond These bonds can be increased in strength by using the chelate effect (connected ligands). Chela is greek for crab. Alfred Werner: First Nobel prize for inorganic chemistry, 1913
Werner used conductivity to figure out what was bound to the cobalt and what was a free ion.
These results also allowed Werner to postulate that the bonds were static and the complexes had fixed geometries.
Magnetism: (Chapter 11, pages 459-468) Measuring magnetism is critical to the inorganic chemist. (magnetic susceptibility,Chi, C) A diamagneitc substance placed in an external magnetic field has an induced circulation of electrons producing a net magnetic moment aligned in opposition to the field (repulsion). This diamagnetism is due to the paired electrons in the bond. The Chi value of a diamag. material is small, negative (≈ 1x10-6 cgs) and temperature independent.
Week 1-2
Week 1-3
Fig11.1 drago
Unpaired electrons produce paramagnetism. The Chi is positive and greater than diamagnetism (≈1x10-4 cgs). The material is attracted to the field and it is only induced when an external field is applied. Paramagnet in a Field: When the field is removed all electrons are randomized by thermal motion. This is why paramagnetism decreases as temperature increases. Curie Law: Chi(C) = Constant/T
or
1/Chi = T/Constant
Fig 11.57
Important point: the magnetism can be directly related to the number of unpaired electrons. We measure Chi (C m) yet m gives us information regarding the number of free electrons.
Xm = N2m2/3RT
or
m = (3RTXm/ N2)1/2 = 2.84(XmT)1/2
Where N is avogadros number, R ideal gas constant, T is temp, m is expressed in Bohr magnetons (BM) Paramagnetism originates with spins and orbital motion, spin-spin, orbital-orbital and spin-orbital interactions. However, the spin-spin interactions are the most prominent so we can derive the spin only equation. m = (4S(S+1))1/2 = (n(n+2))1/2 where n is the number of un-paired electrons. This is how experimentalists determine the number of free electrons. Magnetic Properties
Week 1-4
Structure of Inorganic Complexes: pg. 391-393 and 138-153 Week 1-5
Valence bond Theory Cobalt Hexamine Chloride- [Co(NH3)6]Cl3 Co has 27 electrons and Co3+ has 24 electrons
Pauling suggested the following method to understand the ligand bonding. Fill in 3 d-electrons per Hunds rule.
o 1s22s22p6 3s
ooo ooooo o ooo 3p
3d
4s
4p
Six NH3 approach the Co3+ core which will rearrange the Co+3 electrons.
o 1s22s22p6 3s
ooo ooooo o ooo 3p
3d
4s
4p
The six NH3 ligand fill up the outer 6 empty orbitals However, Pauling’s description did not account for the fact that all Co-NH3 bonds are equivalent. Hybridization was already applied to carbon where you get sp3 so Pauling applied this to cobalt hexamine. Hybridization of atomic orbitals for cobalt hexamine means we need six equal, empty orbitals, d2sp3.
The d orbitals chosen will be the orbitals along x,y,z axis (dz2 and dx2-y2)
Week 1-6
These 6 hybridized orbitals will house all 12 electrons from NH3 and lead to a perfect octahedron, 6 d2sp3 orbitals If this theory is to work it has to be applied to other compounds. Cobalt Hexaflouro Chloride The first observation is that this compound is paramagnetic so the electrons in the d-orbitals can not be paired.
ooooo o ooo ooooo 3d
4s
4p
4d
Hybridization is now sp3d2 which is the same as d2sp3. Measure ≈ 5 BM for magnetizm, this yields 4 unpaired electrons by the equation, m = (n(n+2))1/2. This theory then established two classifications of complexes. Innersphere complexes: [Co(NH3)6]2+ will hybridize the 3d orbitals Outersphere complexes: [Co(F6)]4- will hybridize the 4 d orbitals. Pauling explained this as weak ligands (F-) which stay in the outersphere and strong ligands (NH3) which interact with the innersphere.
Week 1-7
Paulings Theory now needed to explain other systems in which we know the magnetism. [Ni(CN)4]2- was known to exist and CN- was known to be a strong ligand. Ni2+ has 8 d orbital electrons
ooooo o ooo ooooo 3d
4s
4p
4d
Now rearrange d-electrons due to the strong ligands.
ooooo o ooo ooooo 3d
4s
4p
4d
Fill in 8 electrons for CN- so that have hybridization of dsp2 Pauling predicted square planar and diamagnetic with hybridization of the dx2-y2 orbital, which is correct. Note the empty p orbital could react with another approaching ligand.
Now Consider [NiCl4]2- where Cl- is a weak ligand.
Week 1-8
ooooo o ooo ooooo 3d
4s
4p
4d
Pauling predicts an sp3 hybridization which indicates tetrahedral geometry and two un-paired electrons. This is correct so theory is working. Consider [Ni(P(Et3))Cl2] This complex is diamagnetic and has a coordination of 4 so it can be either Tetrahedral
ooooo o ooo ooooo 3d
4s
4p
4d
or square planar
ooooo o ooo ooooo 3d
4s
4p
4d
This tells us the complex has to be square planar because its diamagnetic. Consider [Ni(NH3)6]2+ which is a paramagnetic compound, but why?
ooooo o ooo ooooo 3d
4s
4p
4d
Week 1-9
Ammonia acted as a stong ligand with Co3+ (sp3d2) but now with Ni2+, ammonia is acting as a weak ligand. To do this Pauling had to hybridize the d2sp3 orbitals and moved the two electrons in the 3d orbitals to the 4d orbitals.
ooooo o ooo ooooo 3d
4s
4p
4d
This was explained as a outersphere promoter. The strong ammonia ligands promote the two electrons to the outer sphere. Valence bond theory explains the magnetic moment and geometry but Pauling really found support for his theory when he was able to explain reactivites. For example: Co2+ = Co3+ + 1 electron.
-1.84V
Positive deltaG, Non-spontaneous reaction (deltaG = -nFEo) [Co(CN)6]4- = [Co(CN)6]3- + 1 electron
+0.83 V
Negative delta, this complex spontaneously oxidizes in solution. This is because the strong ligand, CN- promotes the electron to the 4d shell and thus is easily oxidized.
Week 1-10
Co2+, d7
ooooo o ooo ooooo 3d
4s
4p
4d
This does not occur with ferrous iron, why? Fe2+, d6
ooooo o ooo ooooo 3d
4s
4p
4d
This is the end of Valence bond theory that needs to be know for the exam.
Crystal Field Theory: (page 394-413) This theory is contemporary with valence bond theory but was developed by physicists. The biggest assumption is that the ligands are point charges. In the absence of a ligand field, all five d-orbitals are in equal energy. If the negative charge is brought in totally spherically then all five orbitals are increased in energy equally. _____ _____
Week 1-11
Note: Ligands are brought in as negative point charges so the five orbitals are no longer equally affected and so become non-degenerate.
The orbital energy is equally split around the center. 2 orbitals are increased by 6 Dq and 3 are decreased by 4 Dq. This is the crystal field splitting. The simplest example: [Ti(H2O)6]3+ This complex in water gives a purple color. The one d-electron is in the lower d orbitals.
Week 1-12
The absorption of energy is due to excitation of the electron into the upper orbitals.
Thus, the energy of the transition is equal to the splitting of the eg and t2g orbitals. t2g1 eg0 goes to t2g0 eg1 The Lambda max of [Ti(H2O)6]3+ = 492 nm (492 x 10-9 m) (wavelength) =1/(492x10-9)*100 = 20,300 cm-1 (frequency or inverse wavenumbers) where
Energy = Planck’s Const * frequency = h*c/lambda Note: This energy can be converted to KJ/mol
20,300 cm-1 * 1KJ/mol/83.6 cm-1 = 243 KJ/mol 1Kcal/mol = 4.18 KJ/mol so 10 Dq is 58 Kcal/mol Students are required to know how to do this conversion. Re(VI)F6 has a 10 Dq of 32,500 cm-1 = 388 KJ/mol = 92.9 Kcal/mol The 10 Dq has increased because Re(VI) has more charge and so greater splitting.
Week 1-1
Coordination Chemistry: (Chapter 11, pages 387-391) What is Inorganic Chemistry? Inorganic is everything other than carbon. You are prejudiced against inorganic chemistry due to a lack of knowledge but the majority of the chemistry in the world is inorganic. Just not as important to biology. Nevertheless, biology has critical uses of inorganic chemistry that are essential. Inorganic chemistry is made of coordinate bonds or “dative” bond These bonds can be increased in strength by using the chelate effect (connected ligands). Chela is greek for crab. Alfred Werner: First Nobel prize for inorganic chemistry, 1913
Werner used conductivity to figure out what was bound to the cobalt and what was a free ion.
These results also allowed Werner to postulate that the bonds were static and the complexes had fixed geometries.
Magnetism: (Chapter 11, pages 459-468) Measuring magnetism is critical to the inorganic chemist. (magnetic susceptibility,Chi, C) A diamagneitc substance placed in an external magnetic field has an induced circulation of electrons producing a net magnetic moment aligned in opposition to the field (repulsion). This diamagnetism is due to the paired electrons in the bond. The Chi value of a diamag. material is small, negative (≈ 1x10-6 cgs) and temperature independent.
Week 1-2
Week 1-3
Fig11.1 drago
Unpaired electrons produce paramagnetism. The Chi is positive and greater than diamagnetism (≈1x10-4 cgs). The material is attracted to the field and it is only induced when an external field is applied. Paramagnet in a Field: When the field is removed all electrons are randomized by thermal motion. This is why paramagnetism decreases as temperature increases. Curie Law: Chi(C) = Constant/T
or
1/Chi = T/Constant
Fig 11.57
Important point: the magnetism can be directly related to the number of unpaired electrons. We measure Chi (C m) yet m gives us information regarding the number of free electrons.
Xm = N2m2/3RT
or
m = (3RTXm/ N2)1/2 = 2.84(XmT)1/2
Where N is avogadros number, R ideal gas constant, T is temp, m is expressed in Bohr magnetons (BM) Paramagnetism originates with spins and orbital motion, spin-spin, orbital-orbital and spin-orbital interactions. However, the spin-spin interactions are the most prominent so we can derive the spin only equation. m = (4S(S+1))1/2 = (n(n+2))1/2 where n is the number of un-paired electrons. This is how experimentalists determine the number of free electrons. Magnetic Properties
Week 1-4
Structure of Inorganic Complexes: pg. 391-393 and 138-153 Week 1-5
Valence bond Theory Cobalt Hexamine Chloride- [Co(NH3)6]Cl3 Co has 27 electrons and Co3+ has 24 electrons
Pauling suggested the following method to understand the ligand bonding. Fill in 3 d-electrons per Hunds rule.
o 1s22s22p6 3s
ooo ooooo o ooo 3p
3d
4s
4p
Six NH3 approach the Co3+ core which will rearrange the Co+3 electrons.
o 1s22s22p6 3s
ooo ooooo o ooo 3p
3d
4s
4p
The six NH3 ligand fill up the outer 6 empty orbitals However, Pauling’s description did not account for the fact that all Co-NH3 bonds are equivalent. Hybridization was already applied to carbon where you get sp3 so Pauling applied this to cobalt hexamine. Hybridization of atomic orbitals for cobalt hexamine means we need six equal, empty orbitals, d2sp3.
The d orbitals chosen will be the orbitals along x,y,z axis (dz2 and dx2-y2)
Week 1-6
These 6 hybridized orbitals will house all 12 electrons from NH3 and lead to a perfect octahedron, 6 d2sp3 orbitals If this theory is to work it has to be applied to other compounds. Cobalt Hexaflouro Chloride The first observation is that this compound is paramagnetic so the electrons in the d-orbitals can not be paired.
ooooo o ooo ooooo 3d
4s
4p
4d
Hybridization is now sp3d2 which is the same as d2sp3. Measure ≈ 5 BM for magnetizm, this yields 4 unpaired electrons by the equation, m = (n(n+2))1/2. This theory then established two classifications of complexes. Innersphere complexes: [Co(NH3)6]2+ will hybridize the 3d orbitals Outersphere complexes: [Co(F6)]4- will hybridize the 4 d orbitals. Pauling explained this as weak ligands (F-) which stay in the outersphere and strong ligands (NH3) which interact with the innersphere.
Week 1-7
Paulings Theory now needed to explain other systems in which we know the magnetism. [Ni(CN)4]2- was known to exist and CN- was known to be a strong ligand. Ni2+ has 8 d orbital electrons
ooooo o ooo ooooo 3d
4s
4p
4d
Now rearrange d-electrons due to the strong ligands.
ooooo o ooo ooooo 3d
4s
4p
4d
Fill in 8 electrons for CN- so that have hybridization of dsp2 Pauling predicted square planar and diamagnetic with hybridization of the dx2-y2 orbital, which is correct. Note the empty p orbital could react with another approaching ligand.
Now Consider [NiCl4]2- where Cl- is a weak ligand.
Week 1-8
ooooo o ooo ooooo 3d
4s
4p
4d
Pauling predicts an sp3 hybridization which indicates tetrahedral geometry and two un-paired electrons. This is correct so theory is working. Consider [Ni(P(Et3))Cl2] This complex is diamagnetic and has a coordination of 4 so it can be either Tetrahedral
ooooo o ooo ooooo 3d
4s
4p
4d
or square planar
ooooo o ooo ooooo 3d
4s
4p
4d
This tells us the complex has to be square planar because its diamagnetic. Consider [Ni(NH3)6]2+ which is a paramagnetic compound, but why?
ooooo o ooo ooooo 3d
4s
4p
4d
Week 1-9
Ammonia acted as a stong ligand with Co3+ (sp3d2) but now with Ni2+, ammonia is acting as a weak ligand. To do this Pauling had to hybridize the d2sp3 orbitals and moved the two electrons in the 3d orbitals to the 4d orbitals.
ooooo o ooo ooooo 3d
4s
4p
4d
This was explained as a outersphere promoter. The strong ammonia ligands promote the two electrons to the outer sphere. Valence bond theory explains the magnetic moment and geometry but Pauling really found support for his theory when he was able to explain reactivites. For example: Co2+ = Co3+ + 1 electron.
-1.84V
Positive deltaG, Non-spontaneous reaction (deltaG = -nFEo) [Co(CN)6]4- = [Co(CN)6]3- + 1 electron
+0.83 V
Negative delta, this complex spontaneously oxidizes in solution. This is because the strong ligand, CN- promotes the electron to the 4d shell and thus is easily oxidized.
Week 1-10
Co2+, d7
ooooo o ooo ooooo 3d
4s
4p
4d
This does not occur with ferrous iron, why? Fe2+, d6
ooooo o ooo ooooo 3d
4s
4p
4d
This is the end of Valence bond theory that needs to be know for the exam.
Crystal Field Theory: (page 394-413) This theory is contemporary with valence bond theory but was developed by physicists. The biggest assumption is that the ligands are point charges. In the absence of a ligand field, all five d-orbitals are in equal energy. If the negative charge is brought in totally spherically then all five orbitals are increased in energy equally. _____ _____
Week 1-11
Note: Ligands are brought in as negative point charges so the five orbitals are no longer equally affected and so become non-degenerate.
The orbital energy is equally split around the center. 2 orbitals are increased by 6 Dq and 3 are decreased by 4 Dq. This is the crystal field splitting. The simplest example: [Ti(H2O)6]3+ This complex in water gives a purple color. The one d-electron is in the lower d orbitals.
Week 1-12
The absorption of energy is due to excitation of the electron into the upper orbitals.
Thus, the energy of the transition is equal to the splitting of the eg and t2g orbitals. t2g1 eg0 goes to t2g0 eg1 The Lambda max of [Ti(H2O)6]3+ = 492 nm (492 x 10-9 m) (wavelength) =1/(492x10-9)*100 = 20,300 cm-1 (frequency or inverse wavenumbers) where
Energy = Planck’s Const * frequency = h*c/lambda Note: This energy can be converted to KJ/mol
20,300 cm-1 * 1KJ/mol/83.6 cm-1 = 243 KJ/mol 1Kcal/mol = 4.18 KJ/mol so 10 Dq is 58 Kcal/mol Students are required to know how to do this conversion. Re(VI)F6 has a 10 Dq of 32,500 cm-1 = 388 KJ/mol = 92.9 Kcal/mol The 10 Dq has increased because Re(VI) has more charge and so greater splitting.
