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blogfast25
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It lacks a methyl on C No 5 for that. 
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aga
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That's a Deal !
I suppose the vast array of OC entities makes even the Naming pretty complex. Sure looks that way.
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aga
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Please Sir(s), what does a Hartree-Fock equation do ?
Or a 'Virtual Orbital' mean ?
[Edited on 15-11-2015 by aga]
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blogfast25
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Quote: Originally posted by aga  | Please Sir(s), what does a Hartree-Fock equation do ?
Or a 'Virtual Orbital' mean ?
[Edited on 15-11-2015 by aga] |
Hartree-Fock lies at the heart of several approximation methods for multi-atom multi-electron systems (usually called molecules) 
For multi electronic systems the trusted Schrödinger equation:
... has no analytical solutions because the inter-electron repulsive electrostatic forces cause Potential Energy (V) terms to occur in the
Hamiltonian Operator, that aren't encountered in simpler one electron systems.
To find numerical solutions for these cases a number of approximative/iterative techniques have been developed. Hartree Fock is an important
element in some of these methods.
Where did you find the term "Virtual Orbital"?
[Edited on 16-11-2015 by blogfast25]
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blogfast25
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QQuiz Questions for the week will come tomorrow. Been a bit busy today.
[Edited on 16-11-2015 by blogfast25]
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Metacelsus
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Basically, it ignores the instantaneous correlation of the wavefunctions, and instead treats wavefunctions like static charge distributions.
The Hamiltonian changes to an "effective Hamiltonian", which allows separation. The wavefunctions are solved for by an iterative process called the
self-consistent field method.
By "virtual orbitals", do you mean Slater orbitals?
[Edited on 11-16-2015 by Cheddite Cheese]
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aga
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| Quote: | | Where did you find the term "Virtual Orbital"? |
Same place i came across Hartree-Fock = York university Chem prospectus, QM module :-
http://www.york.ac.uk/chemistry/undergraduate/courses/option...
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blogfast25
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Question 9: (for 10 points)
Calculate the Lattice Enthalpy of NaF(s) from the following data (all in STP conditions):
Enthalpy of Formation of NaF = - 575 kJ/mol
Enthalpy of vapourisation of Na = + 97.42 kJ/mol
Ionisation energy of Na (Na == > Na<sup>+</sup> + e) = + 495.8 kJ/mol
Enthalpy of bond dissociation of F<sub>2</sub> = + 159 kJ/mol
Electron affinity of F ( F + e == > F<sup>-</sup> = - 328 kJ/mol
Question 10: (for 20 points)
Many d-block elements form colourful aqueous complexes with ammonia. Explain why the following are colourless:
a) Cu(NH<sub>3</sub><sub>4</sub><sup>+</sup>(aq)
b) Zn(NH<sub>3</sub><sub>4</sub><sup>2+</sup>(aq)
Oh, and these are the last questions of the current QQuiz!
[Edited on 17-11-2015 by blogfast25]
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blogfast25
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Answer to Question 7 (Question 8 has already been answered above):
The formation reaction equation of fluoromethane (methyl fluoride):
C(s) + 3/2 H<sub>2</sub>(g) + 1/2 F<sub>2</sub>(g) === > CH<sub>3</sub>F(g)
Enthalpies involved:
Enthalpy of atomisation of C = + 717 kJ
Breaking H2 bonds = 3/2 * 435 kJ
Breaking F2 bonds = 1/2 * 159 kJ
Three C-H bonds formed = 3 * (- 368) kJ
One C-F bond formed = - 536 kJ
Add it all up, gives - 191 kJ
No changes of state.
The estimated Enthalpy of fluoromethane in STP conditions is - 191 kJ/mol.
[Edited on 17-11-2015 by blogfast25]
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aga
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Woohoo ! I got that one ... hang on ... it says -191 ... DOH !
Q9 & 10 look hard. Real Hard.
Seeing as these are the last questions in this series, we need to organise the end-of-term party.
Who is bringing the booze & drugs this year ? Students or Masters ?
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Darkstar
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Why not just have students make it themselves for their final exam? It would test students on both theory as well as practice. Remember that reaction
between hydroiodic acid and benzyl alcohols you learned about? Time to put that knowledge to good use.
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aga
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Erm, do you not remember the Wilkinson Affair when the Students did the drugs a couple of years back ?
Still thinks he's The Hyper Hampster as far as i know.
Not visited him for a long time.
Do you a deal : Students make 100% ethanol in 12 days, age it in 2 hours to a 40% brandy worthy of consumption.
Masters bring the drugs.
Smithers the Lab Tech will arrange the chicks as usual.
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blogfast25
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Life on the edge of QM: Super-photons.
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blogfast25
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Found this excellent QM resource below:
Not a QM course but an alphabetical index to Quantum Mechanical concepts and problems, with beautifully rendered mathematics:
http://www.physicspages.com/index-physics-quantum-mechanics/
From the same site:
Index to Calculus:
http://www.physicspages.com/index-mathematics-calculus/
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Metacelsus
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Question:
How exactly does Russel-Saunders coupling work? Also, how does it give rise to Hund's Rule?
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aga
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At the 11th hour, submitted my answers to the last QM questions of this series.
Not sure i did too well with Q 9 or Q 10, at all.
I can see the answers in the text of this thread, yet somehow cannot connect the dots.
Everyone else has probably sent in their answers already, so D-Day looms Large ...
Edit:
Whatever the Outcome, this text has been Amazingly educational, and will continue to be so for quite a time.
Tests are Good !
Looking back (to find answers) it becomes obvious which bits were not comprehended completely.
What do we want ?
More tests ! More QM ! More OC !
When do we want it ?
Now !
[Edited on 20-11-2015 by aga]
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aga
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Does this explain it ?
http://hyperphysics.phy-astr.gsu.edu/hbase/atomic/lcoup.html
Which one of Hund's 3 laws ?
The Multiplicity one ?
(yessir, i can google, i can google all night long ...)
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annaandherdad
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If you want to know exactly, it's a bit complicated. There is a very crude approximation to the structure of atoms, the central field approximation,
which is based on a self-consistent picture in which every electron moves in a central force field (a rotationally symmetric potential, including both
the "direct" and "exchange" potentials), which is determined by the states of all the electrons. The central field approximation is based on
Hartree-Fock theory, and is described in my notes:
http://bohr.physics.berkeley.edu/classes/221/1112/notes/hart...
The central field approximation is too crude to give even a qualitative understanding of the low lying energy levels of multielectron atoms, although
it does explain precisely the concept of the "electron configuration".
To get a qualitative and semi-quantitative understanding of the energy levels of atoms you need to do perturbation theory, based on the central field
Hamiltonian as a zeroth order approximation. I started some notes on this,
http://bohr.physics.berkeley.edu/classes/221/1112/notes/atom...
but these are incomplete. The topic is continued in some hand-written notes I have,
http://bohr.physics.berkeley.edu/classes/221/0708/lectures/L...
http://bohr.physics.berkeley.edu/classes/221/0708/lectures/L...
The problem is to construct wave functions that have the right exact quantum numbers L and S of the electrostatic approximation to the exact
Hamiltonian for the atom. This is carried out by some tabular methods that are explained in the notes I've linked. The main reference for my notes
is Intermediate Quantum Mechanics by Bethe and Jackiw.
The tabular methods give you the various LS "multiplets" that can be created out of a given electron configuration. This is called Russel-Saunders
coupling.
Hund's rules are semi-empirical rules that determine the order of the various LS multiplets within a given electron configuration.
In detail, it's a big topic, but if you want to know more detail about an aspect of it, let me know and I'll elaborate.
Any other SF Bay chemists?
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blogfast25
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Coming right up!
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blogfast25
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Oh bugger it: it's the weekend, end of term and blahblah, so I'll divulge the answers to 9 and 10 a bit prematurely.
| Quote: Originally posted by blogfast25 | Question 9: (for 10 points)
Calculate the Lattice Enthalpy of NaF(s) from the following data (all in STP conditions):
Enthalpy of Formation of NaF = - 575 kJ/mol
Enthalpy of vapourisation of Na = + 97.42 kJ/mol
Ionisation energy of Na (Na == > Na<sup>+</sup> + e) = + 495.8 kJ/mol
Enthalpy of bond dissociation of F<sub>2</sub> = + 159 kJ/mol
Electron affinity of F ( F + e == > F<sup>-</sup> = - 328 kJ/mol
Question 10: (for 20 points)
Many d-block elements form colourful aqueous complexes with ammonia. Explain why the following are colourless:
a) Cu(NH<sub>3</sub><sub>4</sub><sup>+</sup>(aq)
b) Zn(NH<sub>3</sub><sub>4</sub><sup>2+</sup>(aq)
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A. 9:
The knowledge needed lies in this part of the course:
http://www.sciencemadness.org/talk/viewthread.php?tid=62973&...
The equation for the Enthalpy of Formation of NaF (SRD) thus becomes:
+ 97.42 + 495.8 + 1/2 159 - 328 + L = - 575
Thus the Lattice Enthalpy of NaF is:
L = - 919.2 kJ/mol, a suitably high (but negative) value for an ionic compound made of relatively small ions packed closely together.
Remember that Lattice energy is mainly Coulombic potential energy.
A. 10:
The knowledge needed lies in this part of the course:
http://www.sciencemadness.org/talk/viewthread.php?tid=62973&...
For most colourful d-block element complexes the colour is explained by Ligand Field Theory. The ligands (ammonia molecules in this case) MO
electron repulsion field causes a difference in energy level between the e and t group d orbitals. Electrons can now absorb VIS photons to
transit between the two groups of d orbitals.
Now look at the specific cases of Cu(+1) and Zn(+2) complexes: from the electron configuration of both elements can be gleaned that
the 3d orbitals of the ions are in fact full: 3d<sup>10</sup>. Thus these electrons cannot transit because all 3d orbitals are full.
By contrast, Cu(+2) shows a 3d<sup>9</sup> configuration, so one 3d orbital is now half-full. Another 3d electron can now fill that gap.
Cu(+2) ammonia and aqua complexes are strongly blue.
[Edited on 21-11-2015 by blogfast25]
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Darkstar
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I'll post some more OC lessons soon. It's been a busy week for me!
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aga
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DOH! DOH! and DOH!
So the full d orbitals are the reason for the lack of colour.
DOH!
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aga
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You'd better, or we'll all dehydroxyfilicate, spontaneously.
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blogfast25
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Here's a quick one, off the cuff:
Explain why Yttrium (Y, Z = 39) compounds are colourless.
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blogfast25
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And one for the road.
Most alcohols react neutral with pure water. Solutions of phenol in water are slightly acidic.
Explain this difference from a structural point of view.
[Edited on 21-11-2015 by blogfast25]
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