quantum mechanics n.
Skip this Video
Loading SlideShow in 5 Seconds..
QUANTUM MECHANICS PowerPoint Presentation
Download Presentation


100 Vues Download Presentation
Télécharger la présentation


- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript


  2. Electron clouds • Although we cannot know how the electron travels around the nucleus we can know where it spends the majority of its time (thus, we can know position but not trajectory). • The “probability” of finding an electron around a nucleus can be calculated. • Relative probability is indicated by a series of dots, indicating the “electron cloud”. • 90% electron probability/cloud for 1s orbital (notice higher probability toward the centre)

  3. 4 QUANTUM NUMBERS The shape, size, and energy of each orbital is a function of 4 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)

  4. 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

  5. How many electrons can an orbital hold? Schrodinger Wave Equation Y = fn(n, l, ml, ms) Shell – electrons with the same value of n Subshell – electrons with the same values of nandl Orbital – electrons with the same values of n, l, andml If n, l, and mlare 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

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

  7. Schrodinger Wave Equation Y = fn(n, l, ml, ms) spin quantum number ms ms = +½or -½ ms = +½ ms = -½ 7.6

  8. 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

  9. Properties of Waves Frequency (n) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = l x n 7.1

  10. Properties of Waves Wavelength (l) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough. 7.1

  11. 7.1

  12. Maxwell (1873), proposed that visible light consists of electromagnetic waves. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Speed of light (c) in vacuum = 3.00 x 108 m/s All electromagnetic radiation l x n = c 7.1

  13. 7.3

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

  15. Periodic table arrangement • the quantum theory helps to explain the structure of the periodic table. • n - 1 indicates that the d subshell in period 4 actually starts at 3 (4 - 1 = 3).

  16. Outermost subshell being filled with electrons 7.8

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

  18. What is the electron configuration of Mg? What are the possible quantum numbers for the last (outermost) electron in Cl? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons Abbreviated as [Ne]3s2 [Ne] 1s22s22p6 Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n = 3 l = 1 ml= -1, 0, or +1 ms = ½ or -½ 7.8

  19. Shorthand Notation • Step 1: Find the closest noble gas to the atom (or ion), WITHOUT GOING OVER the number of electrons in the atom (or ion). Write the noble gas in brackets [ ]. • Step 2: Find where to resume by finding the next energy level. • Step 3: Resume the configuration until it’s finished.

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

  21. Why are d and f orbitals always in lower energy levels? • d and forbitals require LARGE amounts of energy and have too many electrons to be valence (more than 8) • It’s better (lower in energy) to skip a sublevel that requires a large amount of energy (d and forbtials) 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!

  22. 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.

  23. 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.

  24. 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 outermost energy level (not the highest orbital in energy!)

  25. 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!

  26. Shapes and Orientations of Orbitals

  27. How many electrons can be in a sublevel? Remember: A maximum of two electrons can be placed in an orbital. s orbitals p orbitals d orbitals f orbitals Number of orbitals 1 3 5 7 Number of electrons 6 10 14 2

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

  29. p Orbitals 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

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

  31. Summary: p orbitals and d orbitals p orbitals look like a dumbell with 3 orientations: px, py, pz (“p sub z”). Four of the dorbitals resemble two dumbells in a clover shape. The last d orbital resembles a p orbital with a donut wrapped around the middle.

  32. Heisenberg’s uncertainty principle • Electrons are difficult to visualize. As a simplification we will picture them as tiny wave/particles around a nucleus. • The location of electrons is described by: n, l, ml • n = size, l = shape, ml = orientation • Heisenberg showed it is impossible to know both the position and velocity of an electron. • Think of measuring speed & position for a car. Slow Fast

  33. Heisenberg’s uncertainty principle • The distance between 2+ returning signals gives information on position and velocity. • A car is massive. The energy from the radar waves will not affect its path. However, because electrons are so small, anything that hits them will alter their course. • The first wave will knock the electron out of its normal path. • Thus, we cannot know both position and velocity because we cannot get 2 accurate signals to return.

  34. n l ml ms 3d 4s 3p 3s 2p 2s 1s 1 0(s) 0 ENERGY 2 0(s) 0 1(p) -1, 0, 1 3 0(s) 0 1(p) -1, 0, 1 2(d) -1, 0, 1, 2 -2, 4 0(s) 0 Movie: periodic table of the elements: t10-20