Atomic structure - Electron Configurations and Periodic Table

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ATOMIC STRUCTURE Lesson by Dr.Chris UP, May 2014

description

Basic overview about structures of atoms and electron configuratoin which determines the chemical properties of elements

Transcript of Atomic structure - Electron Configurations and Periodic Table

Page 1: Atomic structure - Electron Configurations and Periodic Table

ATOMIC STRUCTURE

Lesson by Dr.ChrisUP, May 2014

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WHAT WE WILL LEARN … PART 1:

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STRUCTURE OF ATOMS

The space is made up of electrons The mass is in the core

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ATOMIC SCALES

1 mm

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ATOMIC MASS 1 atomic mass unit “amu”

= 1/12 of the mass of 12C atom

The carbon-12 atom has a mass of 12.000 u, and yet it contains 12 objects (6 protons and 6 neutrons) that each have a mass greater than 1.000 u, not to mention a small contribution from the 6 electrons.

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MASS DEFECT Mass of nucleus < sum of p + n + e !

Why ?

Some of the mass is converted to be the BINDING ENERGY of the nucleus that holds it together -> “mass defect”

For 12C:Mass defect = Dm = 6 * 1.008664 u + 6 * 1.007276 u + 6 * 0.00054858 u - 12.000 u = 0.098931 u

The binding energy in the carbon-12 atom is therefore 0.098931 u * 931.5 MeV/u = 92.15 MeV

(binding energies for electrons is just some eV !)

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PROTONS AND NEUTRONS ARE NOT FUNDAMENTAL PARTICLES

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ATOM SYMBOLS

Ne2010

Atomic number Z= no. of protons= no. of electrons

Mass number A= no. of protons + neutrons

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EXAMPLES

How many protons, electrons and neutrons are in:

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SOLUTION

Cl can have 18 or 20 neutrons 35.45 is a mix of 2/3 35Cl and 1/3 37Cl

Element number= no. of protons= no. of electrons

Mass number, not integer !=> mix of ISOTOPES withdifferent no. of neutrons !

Z

A

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ISOTOPES

Nearly all elements have isotopes,that means the same elements(no. of protons = Z) has different no. of neutrons, and therefore different mass

Example:Copper exists to 69.2% of 63Cu and the rest of 65Cu with masses 62.93 and 64.93

what is the atomic mass of the mixture ?

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ISOTOPES AND MS Isotope patterns are important to

identify fragments in MS

For example:

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***** BREAK *****

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WHERE ARE THE ELECTRONS ?

3 kinds of spectra:

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WHERE DO THE LINES COME FROM ?

Bohr (1913) emission spectra of hydrogen gas

Lines correspond to energies that are emitted by electrons:

emitted

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ELECTRONS ARE “FIXED” ON ORBITS ! Electrons can move between distinct

energy levels, they cannot exist just anywhere in the atom = quantum

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THE HYDROGEN EMISSION SPECTRUM

Electrical charged hydrogen gas will emit a blue color and also UV radiation

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QUESTION

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How many lines in the emission spectrum and at which energies (in cm-1) ?

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Solution: 3 levels 3 lines

Transition A:

∆E = E3 – E2 =-20’000 + 50’000 cm-1 =30’000 cm-1 =

λ = 1/30’000cm-1 * 107 nm/1 cm = 333 nm

We can express energy as wavenumber,because h and c are constant:

= const * 1/λ = const * ν

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RYDBERG EQUATION

From which energy level does an electron come to n=2 when visible light of 410 nm is emitted ?

What is the Ionization energy of hydrogen from this formula ?

-1

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ELECTRONS AS WAVES

Electrons can be regarded to be waves instead of particles.

DeBroglie:

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EXPLAINS WHY ELECTRONS CAN ONLY EXIST ON CERTAIN ORBITS

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ENERGY OF “WAVE-ELECTRONS”

We can model the behavior of an electron in a restricted area(“particle in a box”)

Quantization comes from the fact that a wave hasto “fit” into the boundaries

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Only wavelengths are allowed:

Use in DeBroglie:

Therefore the kinetic energy is:

n is the main quantum numberindicating the energy level

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3 QUANTUM NUMBERS

Because a particle is moving in 3D, we also get 3 quantum numbers

n: main quantum number (start with 1)

l : angular “ ( 0,1 .. n-1) m: magnetic “ ( -l … 0 … +l

)

Electrons can live only in these “orbitals” (spaces) defined by 3 quantum numbers

Up to 2 electrons can exist in one orbital

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WATCH A DEMO VIDEO

http://www.youtube.com/watch?v=Fw6dI7cguCg

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Main spectral lines = n

Fine structure = l

With magnetic field: Zeeman effectmagnetic quantum no. m

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QUESTIONS

How many orbitals are possible for the energy level n = 2 and how many electrons can live there maxium ?

n = 2 l = 0 and 1 (“s” and “p” level) m = 0 and -1, 0, +1 (px, y and z)

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Part 2:

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ELECTRONIC SHIELDINGAUFBAU PRINCIPLE (PERIODIC TABLE)

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ELECTRON SHIELDING The nuclear charge experienced by an

electron is reduced by shielding by other electrons.

Trends in effective nuclear charge can be used to rationalize the trends in many properties.

As a result of the combined effects of penetration and shielding, the order of energy levels in a shell of a many electron atom is s < p < d < f

(Atkin p.16)

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ELECTRON SHEILDING

Different from Hydrogen, there are electron-electron interactions in the other elements

Each electron experiences an effective nuclear charge: Zeff = Z - σ

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EXAMPLE LI (1S2 2S1)

The PAULI-EXCLUSION PRINCIPLE says that no 2 electrons can have the same 4 quantum numbers (n l m and s)=> the 1s orbital can not have 3 electrons, but max. 2 (n=1, l=0, m=0, s= +/- ½)

=> The 3rd electron goes to the next energy level n=2, but now, into s or p level ?

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LI ELECTRON CONFIGURATION

Because of the form of the function of 2sit has electron density closer to the nucleus than 2p

Þ less shielding by 1s electrons

Þ lower energy

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Z EFF FOR LI ELECTRONS

The 2 1s electrons influence each other and

shield the pos. charge

The 2s electron experiences the shielding by the 2 “inner” electrons very strongly, but not as

full 2 neg. charges

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CONSEQUENCE FOR ALL ELMENTS

As a result of penetration and shielding, the order of energies in many-electron atoms is typically

ns < np < nd < nf

because, in a given shell, s orbitals are the most penetrating and f orbitals are the least penetrating.

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The order of energy changes at

Ca – Sc !

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SLATER’S RULESZEFF = Z – S ESTIMATION OF S:

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EXAMPLE: K – WHERE IS THE 19TH ELECTRON ?

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HOMEWORK (PRESENT NEXT LESSON)Calculate the shielding for the valence electron(s) of:

Ca compare 4s2 <-> 3 d2Sc compare 3d1 <-> 4 p1Cu (1) compare 4s1 <-> 4 p1Cu (2) compare 3d10 4s1 <-> 3d9 4s2Mn compare 3d5 4s2 <-> 3d7Co (1) compare 3d7 4s2 <-> 3 d9Co (2) compare 3d7 4s2 <-> 3d8 4s1Cr (1) compare 3d5 4s1 <-> 3d4 4s2Cr (2) compare 3d5 4s1 <-> 3 d6

Questions: explain1. How shielding determines the AUFBAU

principle2. trend of atomic radius in PT (left to right)3. -”- ionization energies -- “ --4. - “ - electronegativities -- “ –