bonding in coordination complexes (part 1)
TRANSCRIPT
Bonding Models
for
Transition metal compounds
Part 1
1
Christoph Sontag PhD, Phayao University Aug. 2013 e-mail: [email protected] Office SCI 2201
2
Review: electronic structure and periodic table
VB Theory
Crystal Field Theory
Spectrochemical Series
UV/VIS spectroscopy
Ligand Field Theory
3
Where are metals and non-metals in the periodic table ? Where are the transition metals ? Which elements form cations and which anions ?
How atoms form bonds ?
Watch tutorial videos on my website
http://myphayao.com “Chemical Bonding”
Atoms use their VALENCE ELECTRONS to exchange or share electrons to form bonds !
Examples: how many VE are in these elements ?
Na Ca Cl Ar C N S P
Which rule can we find for the main group elements ?
4
Transition Metals have valence electrons
in d-orbitals !!
5
How many groups of transition metals exist ? (compare to the 8 MAIN GROUP elements) And why ?
Number of Valence Electrons:
The 2 s electrons in each row will stay in d-orbitals when we make a compound !!
Valence Bond (VB) Theory
Explain Metal-Ligand interaction:
• Each ligand gives 2 electrons to an empty metal orbital and forms a covalent bond
• The metal orbitals have to be hybridized in order to fit with the surrounding ligands
• We have to find out the right combination of metal orbitals (all empty !) to combine with a number of ligands (2, 4, 5 or 6)
Different idea from main group chemistry !
Here we assume that the central atom contributes 1 electron and the ligands also 1 each but in coordination chemistry, the electrons come all from the ligands !
“Strong” ligands as CN(-) cause single electrons in d-orbitals to combine together ! (forming “low-spin” compounds)
Similar to the Carbon sp3 in CH4
Important Hybrid orbitals
Exercises
What kind of complex will Ni(2+) form ? (a)With weak ligands like Cl(-) (b)With strong ligands like CN(-)
Possible answers:
(a) Consider [NiCl4]2+: Ni0 = 4s2 3d8 Ni2+ = 4s0 3d8
3d 4s 4p
sp3
(b) Consider [Ni(CN)4]2-: Ni0 = 4s2 3d8 Ni2+ = 4s0 3d8
3d 4s 4p
dsp2 4pz 3d
Strong ligands cause electron
pairing
Octahedral compounds
We need 6 empty hybrid orbitals on the metal -> for octahedral shape, we need: (a) d2sp3 (“inner shell”) or (b) sp3d2 (“outer shell”)
Example: [CoF6]3- (weak ligands)
How would [Co(NH3)6]3+ look like ?
Transition metal complexes
One interesting property of TM compounds is their colour in solution:
Show Cu(II) experiments on: http://www.youtube.com/watch?v=deNWxchzDRg
Energy Levels of d-Orbitals “Crystal Field Splitting”
20
In an atom, the d-orbitals have the same energy (“degenerated”) BUT: if molecules with high electron density approach the atom, then the energy levels for these 2 d-orbitals go up:
el
el
el
el
el
el
21
Look from the top down:
Energy Change in “crystal field”
22
el
el
el
el x2-y2
23
“Crystal Field Splitting Theory”
Combination of a metal ion (positive) and negative ligand molecules: 1. Electrostatic attraction 2. Metal-d electrons repulse ligand
electrons 3. Depending on the geometry of the
ligand sphere, metal d-orbitals are split in different energy levels
24
Colour of complexes
26
E = h * ν
Depends on the energy difference between the lower and higher metal d-orbital levels ! Visible light is absorbed and pushes electrons up => The higher Δ, the more “blue” is the light absorbed
27
Green colour => red is absorbed Yellow colour => blue is absorbed
Conclusion: The energy absorbed by the green complex is lower than by the yellow complex.
Calculate the absorbed light energy – example red light absorbed (700 nm wavelength) E = h * c / λ = 1240 eV * nm/ 700 nm = 1.77 eV = 2.82 * 10^(-19) J Blue light absorbed (400 nm wavelength) E =
Energy in electon volt eV the amount of energy gained (or lost) by the charge of a single electron moved across an electric potential difference of one volt. Thus it is 1 volt (1 joule per coulomb, 1 J/C) multiplied by the negative of the electron charge (−e, or −1.602176565(35)×10−19 C).
el
1 volt
Visible light photons have an energy of 1.5 – 3.5 eV
Energy in spectroscopy Energy of one photon: E = h * c / λ Expressed as eV: E = 1240 eV nm / λ Expressed in cm-1 E ~ 107 / λ Expressed in J: E = 1.988×10−16 J nm / λ Expressed in kJ/mol (for 1 mol of photons) E = 1.988×10−19 x 6.022x1023 / λ = = 1.2 x105 nm / λ
Calculate the energy of light of 500 nm in all 4 energies
Size of ∆
M (+)
M (n+)
low
high
L L
X(-) OH(-) H2O NH3 CN(-) CO
Estimate ∆
Co(3+) -> Ir(3+)
Fe(3+) -> Fe(2+)
Cr(F) 6 3- -> Cr(NH3)6 3+
Ni(NH3) 4 2+ -> Pd(NH3)4 2+
Spectrochemical Series
34
The splitting energy depends on: 1. metal ion:
high charge => high splitting : Ni(3+) > Ni(2+) high period => high splitting : Pd > Ni Pt4+ > Ir3+ > Rh3+ > Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+
2. ligands: empirical order by measurements = “spectrochemical series”
Energy Calculations
UV/VIS spectra: scala in cm-1 = 1/λ = ν “wavenumber”
BECAUSE: the wavenumber ~ energy (h* ν)
10.000 cm-1 ≈ 120 kJ/mol => wavelength λ = ..............
Example: Fe(H2O)6 2+ = d ?
Compare the energy of low and high spin ! (Δ = 10.400 cm-1 / P = 17.600 cm-1)
Is the complex dia- or paramagnetic ?
35
36
End of Part 1 Summary: • Transition metal ions have d-electrons which can interact with electron
rich small molecules like ammonia, water, chloride ... (we call them “ligands”)
• The electronic field of the ligands around the d-orbitals cause a splitting in their energy -> for ML6 the d-orbitals split into 3 low- and 2 high-energy orbitals
• The amount of splitting increases with higher charge on the metal and with higher electron density on the ligands
• Electrons in the lower d-orbitals can jump to the high energy levels by absorbing light energy => the complex has a colour !
THANKS FOR WATCHING … SEE YOU LATER !