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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 atom’s electron with emission spectrum electron can circle nucleus in certain paths, in which it has a certain amount of energy

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**n (principal quantum number) = 1,2,3,…**

Bohr’s Model of the Atom (1913) THIS CALCULATION HAS BEEN REMOVED e- can only have specific (quantized) energy values light is emitted as e- moves from one energy level to a lower energy level En = -RH ( ) 1 n2 n (principal quantum number) = 1,2,3,… RH (Rydberg constant) = 2.18 x 10-18J 7.3

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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 = hn E = hn 7.3

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**Why is e- energy quantized?**

De Broglie (1924) reasoned that e- is both particle and wave. 2pr = nl l = h/mu u = velocity of e- m = mass of e- THIS CALCULATION HAS BEEN REMOVED 7.4

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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 (Y) describes: . energy of e- with a given Y . 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

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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 ml (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)

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**1st Quantum Number Principal Quantum Number: n**

main energy level occupied by electron values are all positive integers (1,2,3,…) As n increases, the electron’s 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 n2

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**Schrodinger Wave Equation**

Y = fn(n, l, ml, ms) principal quantum number n n = 1, 2, 3, 4, …. distance of e- from the nucleus n=1 n=2 n=3 7.6

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1st Quantum Number Energy

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**2nd 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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**Schrodinger Wave Equation**

Y = fn(n, l, ml, ms) angular momentum quantum number l for a given value of n, l = 0, 1, 2, 3, … n-1 l = s orbital l = p orbital l = d orbital l = f orbital 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 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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2nd Quantum Number s orbitals: spherical l value of 0 Max 2 electronsd

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**2nd Quantum Number p orbitals: dumbbell-shaped l value of 1**

Max. 6 electrons

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**p Orbitals 3py orbital 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 3py orbital

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

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**2nd 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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**2nd 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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**3rd Quantum Number Magnetic Quantum Number: ml**

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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**Schrodinger Wave Equation**

Y = fn(n, l, ml, ms) magnetic quantum number ml for a given value of l ml = -l, …., 0, …. +l if l = 1 (p orbital), ml = -1, 0, or 1 if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2 orientation of the orbital in space 7.6

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**Total # of Orbitals in Level**

Energy Level (n) Sublevels in Level # Orbitals in Sublevel Total # of Orbitals in Level 1 l=0, s 2 l=0, s 4 l=1, p 3 9 l=2, d 5 16 l=3, f 7

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ml = -1 ml = 0 ml = 1 ml = -2 ml = -1 ml = 0 ml = 1 ml = 2 7.6

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**4th Quantum Number Spin Quantum Number: ms**

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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**Schrodinger Wave Equation**

Y = fn(n, l, ml, ms) spin quantum number ms ms = +½ or -½ ms = +½ ms = -½ 7.6

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**Schrodinger Wave Equation**

Y = fn(n, l, ml, ms) Existence (and energy) of electron in atom is described by its unique wave function Y. Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers. 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**

Y = fn(n, l, ml, ms) 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 ml How many electrons can an orbital hold? If n, l, and ml are fixed, then ms = ½ or - ½ Y = (n, l, ml, ½) or Y = (n, l, ml, -½) An orbital can hold 2 electrons 7.6

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**How many 2p orbitals are there in an atom?**

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

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**“Fill up” electrons in lowest energy orbitals (Aufbau principle)**

? ? Li 3 electrons Be 4 electrons C 6 electrons B 5 electrons B 1s22s22p1 Be 1s22s2 Li 1s22s1 H 1 electron He 2 electrons He 1s2 H 1s1 7.7

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The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule). F 9 electrons Ne 10 electrons C 6 electrons O 8 electrons N 7 electrons Ne 1s22s22p6 C 1s22s22p2 N 1s22s22p3 O 1s22s22p4 F 1s22s22p5 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 It’s 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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**in the orbital or subshell**

Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom. number of electrons in the orbital or subshell 1s1 principal quantum number n angular momentum quantum number l Orbital diagram 1s1 H 7.8

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**What is the electron configuration of Mg?**

Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s2 = 12 electrons Abbreviated as [Ne]3s2 [Ne] 1s22s22p6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s23p5 = 17 electrons Last electron added to 3p orbital n = 3 l = 1 ml = -1, 0, or +1 ms = ½ or -½ 7.8

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**Outermost subshell being filled with electrons**

7.8

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

7.8

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**Exceptions to the Aufbau Principle**

Remember d and f orbitals require LARGE amounts of energy If we can’t 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 d4 and d9 For the purposes of this class, we are going to assume that ALL atoms (or ions) that end in d4 or d9 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**

d4 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 d5 instead of d4. For example: Cr would be [Ar] 4s2 3d4, but since this ends exactly with a d4 it is an exception to the rule. Thus, Cr should be [Ar] 4s1 3d5. Procedure: Find the closest s orbital. Steal one electron from it, and add it to the d.

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**Write the shorthand notation for: Cu W Au**

Try These! Write the shorthand notation for: Cu W Au [Ar] 4s1 3d10 [Xe] 6s1 4f14 5d5 [Xe] 6s1 4f14 5d10

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**Exceptions to the Aufbau Principle**

The next most common are f1 and f8 The electron goes into the next d orbital Example: La [Xe]6s2 5d1 Gd [Xe]6s2 4f7 5d1

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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] 5s2 4d10 5p2 Sn+4 ion: [Kr] 4d10 Sn+2 ion: [Kr] 5s2 4d10 Note that the electrons came out of the highest energy level, not the highest energy orbital!

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