Atomic Wave Function Family Album Discussion

Discussion #11  Things you should know when you leave Discussion today   Atomic wave function family album a. Principle quantum number n=1,2,3,… specifies energy level n= j+l b. Number of nodal planes (l). c. Angular momentum quantum number (l=0, 1, 2,…n-1) defines the shape of the orbital. d. ml= -l,….0…l e. Number of loops j (j=n-l) f. S, P and D and F orbitals. Electron configuration a. Orbital, Shell, Subshell b. Shielding c. Pauli Exclusion Principle. Auf-bau. Hund’s rule Periodic Trends (I.E., E.A., Radius, Ionic Radius, Electronegativity) 1. The photoelectric effect threshold frequency of a metal is υ0 = 1*10 15 Hz. Gamma radiation of frequency 1* 10 Hz ejects electrons from the metal. Which of the following occurs when the intensity of the gamma radiation is reduced by 50 %? a. The velocity of the ejected electrons will be reduced by a factor of two. b. The kinetic energy of the ejected electrons will be reduced by a factor of two. c. The kinetic energy of the ejected electrons will be reduced by a factor of four. d. Number of ejected electrons will increase e. None of the above. 17 2. One molecule of substance A emits a photon of red (700nm) light and one molecule of substance B emits a photon of blue light(400nm). Give the letter of the following statements that is most accurate? a. Substance A releases more energy than substance B. b. Substance B releases more energy than substance A. c. Because each molecule releases one quantum of light, the amount of energy released by each substance is the same. d. There is no energy released because no heat is given off. 3. A red laser of 700nm and a violet laser of 350nm are both pointed at the same detector. The detector reads that the same number of red photons are absorbed per second as the number of violet photons absorbed per second. If the violet laser is 10 Watts, then what the intensity of the red laser? 10W # watts because detector reads that the same number of red photons  hc hc violet  red 10W # watts  hc hc 350nm 700nm ; 10W ·350nm=#watts· 700 nm #watts= 5W is the intensity of the red laser 4. Fill out the table: Name the orbital Draw the orbital n=2, l=0 j=2 2S n=3, l=2 j=1 3d n=2, l=1 j=1 2p n=2, l=2 j=0 Does not exist l=2, j=1 n=3 3d l=0, j=3 n=3 3s l=1, j=2 n=3 3p Rank in terms of Ionization energies. The same for the same n smaller for higher n 5. For the two hydrogen electron clouds below identify the quantum number “n”, the quantum number “l”, the number of radial loops “j”, and the specific name of the orbital (you must indicate orientation, e.g. 3dxy). Az z n =4 l=2 j=2 n=4 x l=1 j=3 x name:4px name: 4dzx 6. For the two hydrogen electron clouds above, which has the largest ionization energy? The same 7. Match each of the electron clouds above to one corresponding electron cloud cross section below. Indicate your answer by putting in the box next to the function. (x axis is distance from the nucleus, and y axis is a probability of finding an electron at the given distance) 4p 8. Below, draw (from left to right) a 3dxy, 3dx 2 2 -y 4d and 3dyz orbital. Be sure to correctly orient your orbitals in the x-y-z space. 9. Fill the table below Orbital l n=j+l j= n-l 1 2 3 etc 2 3 4 etc 1 2 3 j=n 1 2 3 j=n-1 s 0 p 1 d 2 3 4 etc 1 2 j=n-2 f 3 4 5 etc 1 2 j=n-3 Shape # e- that can fit Number of orbitals in a subshells(ml) 2 1 orientation in space Sphere 3 6 5 10 7 14 Dumbbell Cloverleaf 3+ 3+ 10. Write the ground state electron configuration for B, Al, Al , V, V B Al Al3+ V V3+ N3- = = = = = = 2 2 1 and N3-. 1s 2s 2p 1s2 2s2 2p6 3s2 3p1 1s2 2s2 2p6 3s0 3p0 1s2 2s2 2p6 3s2 3p6 4s2 3d3 1s2 2s2 2p6 3s2 3p6 4s0 3d2 1s2 2s2 2p6 11. Draw the electron configuration diagram for C, but fill the diagram as if the C is in the ground state. Circle a subshell. Draw a dotted line around a shell. Draw a triangle around an orbital. 12. From the following neutral atom electron configurations, label the one with the highest third ionization energy, and lowest third ionization energy. Explain your answers. a. 1s2 2s2 2p6 3s2 3p6 4s2 highest b. 1s2 2s2 2p6 3s2 3p6 4s1 c.1s2 2s2 2p6 3s2 3p6 lowest 14 13. You have light with an intensity of 10 Watts and frequency 6·10 Hz. Calculate the change of energy of the light, in Joules, if the electron clouds of one trillion (1012) atoms are excited. (h=6.626x10-34 J·s). What was fraction of the light that was lost in 1s.? Energy change when one photon of energy is transferred: ΔΕlight=Ephoton= h νlight= 6.6·10-34·6·1014 s-1=39.6·10-20J/photon To jiggle 1012 atoms we need 1012 photons of energy transferred: ΔΕlight=Ephoton·1012= 39.6·10-20(J/photon)·1012(photons)= 39.6·10-8J=4·107 J per second is lost. ΔΕlight= -4·10-7J (light was emitted) Additional question: What was the fraction of the light energy that was lost in 1 s.?  light lost sec ond 10W  4  10 7 J / s  4  10 6 or 0.0004% (very small percentage) 10 J / s 14. A light beam of 10 Watts and 500nm illuminates a sample of matter. The matter is able to absorb 1.5x1019 photons of light every second at 500nm. What intensity of light (in Watts) is seen at the detector? 10 Watts Ephoton= hc  Matter ? Watts Detector  4 ·10-19J/photon matter is able to absorb 1.5·1019 photons of light every second hence: 1.5·1019photons/s x 4·10-19J/photon =6J/s = 6 Watts absorb by matter 10 Watts - 6 Watts=4Watts is left and will be detected 15. The work function (ionization energy), , of chromium metal is 7.2x10-19 J. What is the maximum kinetic energy of an electron, to one significant figure, if it is ejected from chromium metal by light of wavelength 250 nm? KE= ΔΕlight-IE( hc   7.2  10 19 J  7  10 18 J a. What will happen to the speed and quantity of the ejected electrons if the wavelength of the light is increased? Speed will be decrease but number of ejected electrons will be the same b. What will happen to the speed and quantity of the ejected electrons if the wavelength of the light is decreased? Speed will be increase but number of ejected electrons will be the same c. What will happen to the speed and quantity of electrons if the intensity of the light is increase? Speed will be the same but number of ejected electrons will increase Atom He Li Z Electron configuration B C N O F Ne Na Mg Al Zeff kJ mol n 2  IE 1312 1s2 2373 1s22s1 520 2 1.35 1.26 3 1s22s2 Be IE1 899 6 7 8 9 10 11 12 13 Ions (Ions Electron configuration) IE2 He+ 1s1 5248 a + Li 1s2 1757 + Be 1s22s1 a 1s 2 2s 2 2 p 1x 801 1s 2 2 s 2 2 p 1x 2 p 1y 1086 1s 2 2 s 2 2 p 1x 2 p 1y 2 p 1z 1s 2 2 s 2 2 p x2 2 p 1y 2 p 1z 1s 2 2 s 2 2 p x2 2 p y2 2 p 1z 1s 2 2 s 2 2 p x2 2 p y2 2 p z2 B 1s22s2 1.82 2350 + C 1s 2s 2 2 p 1x 2860 a 1314 + O 1s 2 s 2 p 1x 2 p 1y 2 p 1z 2 2 3370 + F 1s 2 s 2 p x2 2 p 1y 2 p 1z 2 2 3950 2.52 + Ne 1s 2 s 2 p x2 2 p y2 2 p 1z 2 2 4560 1.86 c + Na 1s 2 s 2 p x2 2 p y2 2 p z2 2 2 1450 2.25 a  [ Ne]3s 2 578 2 2.26 a 1s 2 2 s 2 2 p x2 2 p y2 2 p z2 3s 2 738 N 1s 2 s 2 p 1x 2 p 1y 2 3390 a 2080 + 2.00 b 1680 2 2.07  [ Ne]3s1 E n  13.6(eV) + a 1s 2 2 s 2 2 p x2 2 p y2 2 p z2 3s 1 496 [ Ne]3s 2 3 p 1x 2430 1.56 d 1400 kJ mol 7300 c 1.66 4 5 Trends in IE1 is Explained by: a. Z increases b. Electron –electron repulsion c. New shell d. l increases or Zeff decreases (shielding) + Mg [ Ne]3s 1 1820 1.99 d 2 2 2 Z eff Z eff  kJ  Z eff 18      2 . 18 10 (J) 1312   2 n2 n2  mol  n + Al [ Ne]3s 2