Quantum Theory and the Electronic Structure of Atoms Chapter 6

Do Electrons orbit the Nucleus? Do you Know that? Do you think that? Do you believe that?

7.3 Line Emission Spectrum of Hydrogen Atoms

Electrons are the key to chemistry Early experiments where elements were studied lead to an understanding that each element emitted a specific light pattern when excited. Hydrogen Linker Helium Linker Link to Flame Tests

7.3

“Fingerprint” Meaning Revealed quantizedThe energy of electrons in an atom is quantized –An electron in an atom can have only certain allowed energies. The fingerprint we see is an electron relaxing from an excited state to a lower excited state or the ground state thus giving off a specific amount of energy.

“Fingerprint” Meaning Revealed Neils Bohr - Interpreted fingerprint as transitions between allowed distances of an orbit from the nucleus. That si to say electrons orbit the nucleus! –Math works for 1 electron systems only Link to Bohr Model Explanation

“Fingerprint” Meaning Revealed Schrodinger – Interpreted fingerprint as transitions between allowed energies of waves. –Math works for all atoms spectra plus can be applied to molecules as well Electron energy descriptions are called electronic configurations. These come from the math of the “Schrodinger Equation.” Link to Idea … Play to give an idea of complexity.Link to Idea

1.e - can only have specific (quantized) energy values 2.light is emitted as e - moves from one energy level to a lower energy level Bohr’s Model of the Atom (1913) E n = -R H ( ) 1 n2n2 n (principal quantum number) = 1,2,3,… R H (Rydberg constant) = 2.18 x J 7.3

E = h 7.3

E photon =  E = E f - E i E f = -R H ( ) 1 n2n2 f E i = -R H ( ) 1 n2n2 i i f  E = R H ( ) 1 n2n2 1 n2n2 n f = 1 n i = 2 n f = 1 n i = 3 n f = 2 n i = 3 7.3

E photon = 2.18 x J x (1/25 - 1/9) E photon =  E = x J = 6.63 x (Js) x 3.00 x 10 8 (m/s)/1.55 x J = 1280 nm Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. E photon = h x c /  = h x c / E photon i f  E = R H ( ) 1 n2n2 1 n2n2 E photon = 7.3

De Broglie (1924) reasoned that e - is both particle and wave. 2  r = n = h/mu u = velocity of e - m = mass of e - Why is e - energy quantized? 7.4

Schrodinger Wave Equation In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e - Wave function (  ) describes: 1. energy of e - with a given  2. probability of finding e - in a volume of space Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems. 7.5

Schrodinger Wave Equation  fn(n, l, m l, m s ) principal quantum number n n = 1, 2, 3, 4, …. n=1 n=2 n=3 7.6 distance of e - from the nucleus

e - density (1s orbital) falls off rapidly as distance from nucleus increases Where 90% of the e - density is found for the 1s orbital 7.6

 = fn(n, l, m l, m s ) angular momentum quantum number l for a given value of n, l = 0, 1, 2, 3, … n-1 n = 1, l = 0 n = 2, l = 0 or 1 n = 3, l = 0, 1, or 2 Shape of the “volume” of space that the e - occupies l = 0 s orbital l = 1 p orbital l = 2 d orbital l = 3 f orbital Schrodinger Wave Equation 7.6

l = 0 (s orbitals) l = 1 (p orbitals) 7.6

l = 2 (d orbitals) 7.6

 = fn(n, l, m l, m s ) magnetic quantum number m l for a given value of l m l = -l, …., 0, …. +l orientation of the orbital in space if l = 1 (p orbital), m l = -1, 0, or 1 if l = 2 (d orbital), m l = -2, -1, 0, 1, or 2 Schrodinger Wave Equation 7.6

m l = -1m l = 0m l = 1 m l = -2m l = -1m l = 0m l = 1m l = 2 7.6

 = fn(n, l, m l, m s ) spin quantum number m s m s = +½ or -½ Schrodinger Wave Equation m s = -½m s = +½ 7.6

Existence (and energy) of electron in atom is described by its unique wave function . Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers. Schrodinger Wave Equation  = fn(n, l, m l, m s ) 7.6

Schrodinger Wave Equation  = fn(n, l, m l, m s ) Shell – electrons with the same value of n Subshell – electrons with the same values of n and l Orbital – electrons with the same values of n, l, and m l How many electrons can an orbital hold? If n, l, and m l are fixed, then m s = ½ or - ½  = (n, l, m l, ½ ) or  = (n, l, m l, - ½ ) An orbital can hold 2 electrons 7.6

How many 2p orbitals are there in an atom? 2p n=2 l = 1 If l = 1, then m l = -1, 0, or +1 3 orbitals How many electrons can be placed in the 3d subshell? 3d n=3 l = 2 If l = 2, then m l = -2, -1, 0, +1, or +2 5 orbitals which can hold a total of 10 e - 7.6

Energy of orbitals in a single electron atom Energy only depends on principal quantum number n E n = -R H ( ) 1 n2n2 n=1 n=2 n=3 7.7

Energy of orbitals in a multi-electron atom Energy depends on n and l n=1 l = 0 n=2 l = 0 n=2 l = 1 n=3 l = 0 n=3 l = 1 n=3 l = 2 7.7

“Fill up” electrons in lowest energy orbitals (Aufbau principle) H 1 electron H 1s 1 He 2 electronsHe 1s 2 Li 3 electrons Li 1s 2 2s 1 Be 4 electrons Be 1s 2 2s 2 B 5 electrons B 1s 2 2s 2 2p 1 C 6 electrons ?? 7.7

C 6 electrons The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule). C 1s 2 2s 2 2p 2 N 7 electrons N 1s 2 2s 2 2p 3 O 8 electrons O 1s 2 2s 2 2p 4 F 9 electrons F 1s 2 2s 2 2p 5 Ne 10 electrons Ne 1s 2 2s 2 2p 6 7.7

What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s = 12 electrons 7.7 Abbreviated as [Ne]3s 2 [Ne] 1s 2 2s 2 2p 6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 3p = 17 electrons Last electron added to 3p orbital n = 3l = 1m l = -1, 0, or +1m s = ½ or -½

Outermost subshell being filled with electrons 7.8

Paramagnetic unpaired electrons 2p Diamagnetic all electrons paired 2p 7.8