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## Quantum Numbers

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**Quantum Numbers**Rules of Electron Location and Orbital Filling Order**Quantum NumbersAt the conclusion of our time together, you**should be able to: • List and define the 4 principles that are part of quantum numbers. • List and define each of the 4 quantum numbers. • Give the quantum number for every electron of every atom on the periodic table.**Principal Quantum Number - n**Symbol = n Represents the main energy level of the electron and its distance from the nucleus Equation: 2n2 - shows how many electrons can be in each energy level (e.g. 3rd energy level: 2(3)2 = 18 total possible e- in this energy level) Your turn: How many electrons in the 4th energy level? 32**Principal Quantum Number - n**• In an address analogy, this would be the state in which the electron would probably be found. Presently, we can find electrons in 7 states. • Values are 1-7 • Ex. = 1s1 (the electron configuration for H) Principal Quantum number = 1**The First of 4 Quantum Numbers for the One Electron of**Hydrogen**The Second Quantum Number - l**• This number describes sublevels, shapes of these sublevels and is the Angular momentum quantum number. The number of sublevels (shapes) in an energy level equals the value of n, the principle quantum number. Sublevels are named s, p, d, f. • Each sublevel has a unique shape:**Shape of the “s” Orbital**• s for "Sphere": the simplest shape, or shape of the simplest atoms like hydrogen and helium • Electrons don't interfere with, or block, each other from the pull of the nucleus - ball shape • Each energy level has an "s" orbital at the lowest energy within that level**Shapes of the “p” Orbitals**• p for "Peanut/Petal": a more complex shape that occurs at energy levels 2 and above**Shapes of the “d” Orbitals**• d for "Double Peanut/Petal": a complex shape occurring at energy levels 3 and above • The arrangement of these orbitals allows for "s" and "p" orbitals to fit closer to the middle/nucleus**Shapes of the “f” Orbitals**• f for "Flower": bizarre-shaped orbitals for electrons of very large atoms • electrons filling these orbitals are weakly attached to the atom because they are so far away from the pull of the nucleus**The Second Quantum Number - l**• In the address analogy, this would be the city in which the electron would probably be found. In the first state there would be 1 city, in the second state there would be 2 cities... • Values are 0 to (n-1) • Ex. = 1s1 , 0 to (1-1) = 0 Second Quantum number for Hydrogen = 0**The Second Quantum Number - l**• Values are 0 to (n-1) • Ex. = 1s1 , 0 to (1-1) = 0 • In the second energy level, 0 to (2-1) = 1 • In the third energy level, 0 to (3-1) = 2 • In the fourth energy level, 0 to (4-1) = 3 • Therefore: • For s, l = 0 • For p, l = 1 • For d, l = 2 • For f, l = 3**The Second of 4 Quantum Numbers for the One Electron of**Hydrogen**The Third Quantum Number - m**• This number describes orientations of the orbitals that each sublevel can have and is the Magnetic quantum number. s has one orientation, p has three orientations , d has five orientations and f has 7 orientations. In the address analogy, this would be the street on which the electron would probably be found. In the first city there would be one street, in the second city there would be one street and 3 more streets for a total of 4. In the 3rd city there would be the previous 4 streets plus 5 more streets...**The Third Quantum Number - m**• Values are –l to 0 to +l • We know what the shape looks like from the second Quantum Number, now we know how many orientations of these shapes each sublevel has!!**The Third Quantum Number - m**Values are –l to 0 to +l, therefore: • s = ___ • 0 Or just 1 orientation**The Third Quantum Number - m**Values are –l to 0 to +l, therefore: • p = ___ ___ ___ -1 0 +1 • Or 3 orientations**The Third Quantum Number - m**Values are –l to 0 to +l, therefore: • d = ___ ___ ___ ___ ___ • -2 -1 0 1 2 • or 5 orientations**The Third Quantum Number - m**Values are –l to 0 to +l, therefore: • f = ___ ___ ___ ___ ___ ___ ___ • -3 -2 -1 0 1 2 3 • or 7 orientations**The Third Quantum Number - m**Let’s Review the Values for m: • s = ___ p = ___ ___ ___ • 0 -1 0 +1 • d = ___ ___ ___ ___ ___ • -2 -1 0 1 2 • f = ___ ___ ___ ___ ___ ___ ___ • -3 -2 -1 0 1 2 3 • Ex. = 1s1 = 0 Third Quantum number from s = 0, or 1 shape**The Third of the 4 Quantum Numbers for the One Electron of**Hydrogen**The Fourth Quantum Number - s**• This number describes the spin of the electrons in an orbital. There can be two electrons in each orbital as long as they are spinning in opposite directions. • In the address analogy, this would be • the house number of the electron. • There can only be 2 houses on each street • Values are +1/2 = clockwise spin • -1/2 = counter clockwise spin**The Fourth Quantum Number - s**• Ex. = 1s1 , spin is up or +1/2 Fourth Quantum number from 1 = +1/2**The Fourth of the 4 Quantum Numbers for the One Electron of**Hydrogen**Quantum # Summary for Hydrogen**• 1s1 • Hydrogen has one electron spinning in a clockwise direction in the first energy level that has one orbital or orientation in its s sublevel shape which is spherical. • The 4 quantum numbers for H are: • 1, 0, 0, +1/2**Let’s Try Helium**• H is 1s1 • He has 2 electrons, can we add another electron spinning in the other direction in the first energy level of the s sublevel with its 1 spherical orbital? • Yes, He is 1s2 • He has 2 electrons spinning in opposite directions in the s sublevel with its spherical shape with 1 orientation.**Remember the Quantum # Summary for Hydrogen**• 1s1 • Hydrogen has one electron spinning in a clockwise direction in the first energy level that has one spherical orbital in its s sublevel. • The 4 quantum numbers for H are: • 1, 0, 0, +1/2**The Pauli Exclusion Principle**• No two electrons in an atom can have the same set of four quantum numbers. • Therefore, the second electron that He has must have a different set of 4 Quantum Numbers. • What would they be??**The 4 Quantum Numbers of Helium**• No two electrons in an atom can have the same set of four quantum numbers. • What principle?? • Pauli Exclusion Principle • Therefore, the second electron that He has must have a different set of 4 Quantum Numbers. • What would they be?? • 1, 0, 0, -1/2**Let’s Try Lithium**• He is 1s2 • He has 2 electrons, can we add another electron spinning in another direction in the first energy level of the s sublevel with its 1 spherical orbital? • No, the third electron must go to the 2nd energy level which has 2 sublevels, s and p, s with its one spherical orbital and p with its 3 orientations of its petal shaped orbitals.**Let’s Try Lithium**• So the address for all 3 electrons of Li is: • 1s2 2s1 • This is called the electron configuration for Li and basically says that the first two electrons of Li are in the s sublevel with its one spherical shape with the 3rd electron in the 2 energy level, s sublevel with its bigger spherical shape. • The Quantum #’s of the 3rd electron:**Now Beryllium:**• Add one more electron to Li : • 1s2 2s1? Where would it go?? • 1s2 2s2 • The 2s sublevel with its one spherical orbital can hold 2 electrons spinning in opposite directions. • The Quantum #’s of the 4th electron:**What About Boron:**• Add one more electron to Be: • 1s2 2s2? Where would it go?? • 1s2 2s3 ? • No • 1s2 2s2 2p1 • The 2s sublevel with its one spherical orbital is full. We must now start filling the 2p sublevel with its 3 orbitals • The Quantum #’s of the 5th electron:**The Electrons of the 2nd Energy Level for Boron**• s = ___ p = ___ ___ ___ • 0 -1 0 +1**Let’s Move to Carbon:**• Add one more electron to B: • 1s2 2s2 2p2 • The 2p sublevel has 3 orbitals. However, we can’t put another electron in the first orbital of p. Why? • Hund’s Rule: Orbitals of equal energy are each occupied by one electron before any orbital is occupied by a second electron.**The Electrons of the 2nd Energy Level for Carbon**• s = ___ p = ___ ___ ___ • 0 -1 0 +1**Nitrogen:**• Add one more electron to C: • 1s2 2s2 2p3 • The 3rd electron in the p sublevel must go into the 3rd orbital. Why? • Again Hund’s Rule: Orbitals of equal energy are each occupied by one electron before any orbital is occupied by a second electron.**The Electrons of the 2nd Energy Level for Carbon**• s = ___ p = ___ ___ ___ • 0 -1 0 +1**Oxygen:**• Add one more electron to N: • 1s2 2s2 2p4 • The 4th electron in the p sublevel must begin to double up the electrons of the three p orbitals.**The Electrons of the 2nd Energy Level for Carbon**• s = ___ p = ___ ___ ___ • 0 -1 0 +1**Fluorine:**• Add one more electron to O: • 1s2 2s2 2p5 • Where would the 5th electron for the 2p orbitals go?**The Electrons of the 2nd Energy Level for Fluorine**• s = ___ p = ___ ___ ___ • 0 -1 0 +1