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Quantum Theory and the Electronic Structure of Atoms

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Bohr Model created by Niels Bohr (Danish physicist) in 1913 linked atoms electron with emission spectrum electron can circle nucleus in certain paths, in which it has a certain amount of energy

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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 Bohrs 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 THIS CALCULATION HAS BEEN REMOVED

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Bohr Model Can gain energy by moving to a higher rung on ladder Can lose energy by moving to lower rung on ladder Cannot gain or lose while on same rung of ladder

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Bohr Model a photon is released that has an energy equal to the difference between the initial and final energy orbits

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E = h 7.3

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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 THIS CALCULATION HAS BEEN REMOVED

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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 Schrodingers equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems. 7.5

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QUANTUM NUMBERS The shape, size, and energy of each orbital is a function of 3 quantum numbers which describe the location of an electron within an atom or ion n (principal) ---> energy level l (orbital) ---> shape of orbital m l (magnetic) ---> designates a particular suborbital The fourth quantum number is not derived from the wave function s(spin) ---> spin of the electron (clockwise or counterclockwise: ½ or – ½) s (spin) ---> spin of the electron (clockwise or counterclockwise: ½ or – ½)

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1 st Quantum Number Principal Quantum Number: n main energy level occupied by electron values are all positive integers (1,2,3,…) As n increases, the electrons energy and its average distance from the nucleus increase multiple electrons are in each level so have the same n value the total number of orbitals in a level is equal to n 2

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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

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1 st Quantum Number Energy

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2 nd Quantum Number Angular Momentum Quantum Number: l indicates the shape of the orbital (sublevel) the possible values of l are 0 to n-1 each atomic orbital is designated by the principal quantum number followed by the letter of the sublevel

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= 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

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Types of Orbitals ( l ) s orbital p orbital d orbital

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

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2 nd Quantum Number s orbitals: spherical l value of 0 Max 2 electronsd

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2 nd Quantum Number p orbitals: dumbbell-shaped l value of 1 Max. 6 electrons

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p Orbitals this is a p sublevel with 3 orbitals These are called x, y, and z this is a p sublevel with 3 orbitals These are called x, y, and z There is a PLANAR NODE thru the nucleus, which is an area of zero probability of finding an electron 3p y orbital

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p Orbitals The three p orbitals lie 90 o apart in spaceThe three p orbitals lie 90 o apart in space

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2 nd Quantum Number d orbitals: various shapes l value of 2 Max. 10 electrons

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l = 2 (d orbitals) 7.6

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2 nd Quantum Number f orbitals: various shapes l value of 3 Max. 14 electrons

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f Orbitals For l = 3, f sublevel with 7 orbitals

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3 rd Quantum Number Magnetic Quantum Number: m l indicates the orientation of an orbital around the nucleus has values from -l +l specifies the exact orbital that the electron is contained in each orbital holds maximum of 2 electrons

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= 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

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Energy Level (n) Sublevels in Level # Orbitals in Sublevel Total # of Orbitals in Level 1 l=0, s l=1, p 3 3 l=0, s 19 l=1, p 3 l=2, d 5 4 l=0, s 116 l=1, p 3 l=2, d 5 l=3, f 7

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

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4 th Quantum Number Spin Quantum Number: m s indicates the spin state of the electron only 2 possible directions only 2 possible values: -½ and +½ paired electrons must have opposite spins

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= 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

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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 ) Each seat is uniquely identified (E, R12, S8) Each seat can hold only one individual at a time 7.6

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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

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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

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Fill up electrons in lowest energy orbitals (Aufbau principle) H 1 electron H 1s 1 He 2 electrons He 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

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C 6 electrons The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hunds 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

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Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s 7.7

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Why are d and f orbitals always in lower energy levels? d and f orbitals require LARGE amounts of energy Its better (lower in energy) to skip a sublevel that requires a large amount of energy (d and f orbtials) for one in a higher level but lower energy This is the reason for the diagonal rule! BE SURE TO FOLLOW THE ARROWS IN ORDER!

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Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom. 1s 1 principal quantum number n angular momentum quantum number l number of electrons in the orbital or subshell Orbital diagram H 1s 1 7.8

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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 7.8 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

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Outermost subshell being filled with electrons 7.8

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

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Exceptions to the Aufbau Principle Remember d and f orbitals require LARGE amounts of energy If we cant fill these sublevels, then the next best thing is to be HALF full (one electron in each orbital in the sublevel) There are many exceptions, but the most common ones are d 4 and d 9 For the purposes of this class, we are going to assume that ALL atoms (or ions) that end in d 4 or d 9 are exceptions to the rule. This may or may not be true, it just depends on the atom.

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Exceptions to the Aufbau Principle d 4 is one electron short of being HALF full In order to become more stable (require less energy), one of the closest s electrons will actually go into the d, making it d 5 instead of d 4. For example: Cr would be [Ar] 4s 2 3d 4, but since this ends exactly with a d 4 it is an exception to the rule. Thus, Cr should be [Ar] 4s 1 3d 5. Procedure: Find the closest s orbital. Steal one electron from it, and add it to the d.

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Try These! Write the shorthand notation for: Cu W Au [Ar] 4s 1 3d 10 [Xe] 6s 1 4f 14 5d 5 [Xe] 6s 1 4f 14 5d 10

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Exceptions to the Aufbau Principle The next most common are f 1 and f 8 The electron goes into the next d orbital Example: –La [Xe]6s 2 5d 1 –Gd [Xe]6s 2 4f 7 5d 1

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Keep an Eye On Those Ions! Electrons are lost or gained like they always are with ions… negative ions have gained electrons, positive ions have lost electrons The electrons that are lost or gained should be added/removed from the highest energy level (not the highest orbital in energy!)

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Keep an Eye On Those Ions! Tin Atom: [Kr] 5s 2 4d 10 5p 2 Sn +4 ion: [Kr] 4d 10 Sn +2 ion: [Kr] 5s 2 4d 10 Note that the electrons came out of the highest energy level, not the highest energy orbital!

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