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Institute of Robotics and Intelligent Systems Department of Mechanical and Process Engineering (DMAVT) ETH Zurich Exercise Session 2: Mini Lecture on Physical Chemistry NANOROBOTICS 2015 Assistants: Marcus Hoop Berna Özkale Dr. Salvador Pané Prof. Dr. Bradley J. Nelson

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Institute of Robotics and Intelligent Systems Department of Mechanical and Process Engineering (DMAVT) ETH Zurich

Exercise Session 2: Mini Lecture on Physical Chemistry NANOROBOTICS 2015 Assistants: Marcus Hoop

Berna Özkale Dr. Salvador Pané Prof. Dr. Bradley J. Nelson

2

A nanorobotic platform can be

-  Molecular Robot

-  A combination of nanostructures and functional molecules

The knowledge of molecular geometry is crucial if we want to design a nanorobot since togther with composition it will determine the future interactions and functionalities of the nanorobot.

Nanocar: an example of a molecular robot Y. Shirai et al., Nano Lett., 2005, 5 (11), pp 2330–2334

The role of physical chemistry

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•  A molecule consists of a discrete group of atoms that are bound via chemical bonds. •  Ionic bond: electrons are exchanged. •  Covalent bond: pairs of electrons are shared.

•  Electron configuration: the arrangement of electrons (e) of an atom or a molecule

1s2

2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 5f14 5g18 6s2 6p6 6d10 6f14 6g18 6h22

7s2 7p6 7d10 7f14 7g18 7h22 7i22

Orbitals are the physical region where the electron can be calculated to be, as defined by the particular mathematical form of the orbital.

1s

2s

3s

4s

5s

6s

7s

2p

3p

4p

5p

6p

3d

4d

5d4f

5fEnergy

Each orbital is described by the quantic numbers and can host only two electrons.

Construction of a molecule

4

Image taken from http://sciencenotes.org/

In a group, atomic number and

atomic radii increase, while electronegativity

decreases.

Elements and the periodic table

Atomic number Mass number

In a period, atomic radii decreases,

while electronegativity increases since elements have more protons and electrons.

Electronegativity (EN): the power of an atom to attract electrons to itself.

5

Let us consider the fluorine atom: F F has an atomic number A of 9. This means that F contains 9 protons (p). If the element has no charge, ne = np = 9. Then the electron configuration of F is : 1s2 2s2 2p5

Valence electrons

In usual conditions, F does not exist in the form of an atom but in the form of a diatomic molecule. Why?

2s2 2p5 One electron more would lead to a completely filled p orbitals!

Generally, atoms “will try to compensate incomplete configurations” towards “noble gas configurations”

In the case of F, two options are possible F “steals” an electron from another atom leading to the anionic form F-

Two F atoms share the electrons (covalent) leading to F2

An example: fluorine

http://www.rsc.org/periodic-table

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Let us consider the carbon dioxide molecule CO2 1.  Draw the atoms showing their valence electrons.

C: 1s2 2s2 2p2 O: 1s2 2s2 2p4

2.  Determine the central atom. It has the highest valence. That corresponds to the highest number of

single electrons in the valence shell. In case of atoms from the same column in the periodic table, the atom located lowest in the column is chosen.

3.  Pair up single electrons first to form bonds between the central atom and peripheral atoms. The most electronegative peripheral atoms are used first. If the central atom runs out of single electrons, its lone pairs may be used to form additional bonds. Single bonds are formed first. Multiple bonds may be formed as needed, leaving just enough single electrons to place the remaining peripheral atoms.

4.  Form anions by pairing up additional electrons to single electrons on the appropriate atoms. If no valence electrons are left on the central atom, add negatively charged peripheral atoms to account for the charge of polyatomic anions. Cations are formed by removing single electrons from the appropriate atoms.

B. B. Miburo, J. Chem. Ed., 1998, 75 (3), pp 317–320

C O O

CO O

CO O CO O CO O

Important! Elements of the second per iod cannot be surrounded by more than 8 e-! (Octet Rule)

Lewis representation of molecules

7

= Electron pair

AX3 trigonal planar AX2E bent

AX4 tetrahedral AX3E trigonal AX2E2 bent

AX4E butterfly AX3E2 T-shape AX2E3 linear

AX5E Square pyramidal AX4E2 square planar

AX2 linearAXEn linear (diatomic)

AX6 Octahedral

AX5 trigonal bipyramidal

Valence Shell Electron Pair Repulsion (VSEPR)

Carbon dioxide:

AX2 linear!

Formal charge φ of an atom i in a molecule:

ϕi = eV ,i − eNB ,i −eB ,i2

eV: Number of valence electrons eNB: Number of non-bonding electrons eB: Number of bonding electrons

8

Let us consider the following molecules: methane (CH4), ethane (C2H6), ethene or ethylene (C2H4) and ethyne or acetylene (C2H2):

C

H

H

H

H C

H

H

H

C

H

H

H C

H

H

C

H

H

CH C H

In 3D: H

CH

H

H

H

CC

HH

H

HH

Tetrahedron

C

H

H

C

H

H

Trigonal planar

CH C H

Linear

CH4 CH3CH3 CH2CH2 CHCH

Geometry of carbon molecules

9

In case of C, we usually do not draw the H attached to C and we represent C as a vertex or point

O

CC

HH

H

HH

H OH

Example for ethylene: CH2CH2 Example for acetylene: CH2CH2 Example for pentane: CH3CH2CH2CH2CH3

Example for ethanol: CH3CH2OH Example for benzene: C6H6

Geometry of carbon molecules

10

Question 1: Are the molecules below the same???

No. They are ISOMERS!

B is the specular image of A. When B is rotated 180 ° around the C-Br axe, you can see that they are not identical. In this molecule, carbon is said to be chiral!

A B

B

Br

CH

Cl

F

Br

CH

Cl

F

Br

CH Cl

F

Question 2: Given this empirical formula C6H6, which compound do we have?

In principle, we could have several options. Let us see two examples:

C

CC

C

CC

H

H

H

H

H

H

Benzene

C

C C

C

C C

HH

H H

HH

Prismane

They are also ISOMERS!

Geometry of carbon molecules

11

A polar molecule is a molecule with a permanent electric dipole moment. Let’s remember an example, hydrogen fluoride (HF)

EN(F) > EN(H)

δ+ δ−

Atkin’s Physical Chemistry, Oxford University Press (2010)

Electric dipole

Polarity of a molecule

Addition of dipole moments:

µ→

res = µ→

ii∑ = Qi ri

i∑

The permanent dipole moment stems from the partial charges on the atoms in the molecule that arise from differences in electronegativity or other features of bonding.

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Molecular geometry (symmetry) determines if the molecule is polar or not!!

Atkin’s Physical Chemistry, Oxford University Press (2010)

Carbon dioxide, CO2 (linear) Non-polar

Ozone, O3 (bent) Polar

Molecular interactions