SPC Ramuel Mendoza Raagas for Teaching Fellow David Perlmutter

General Chemistry
Chemistry E-1a
Problem Set 6
Electron Configurations
Fall Term 2003
Submitted November 6, 2003, 5 p.m.
      1. for the 5f ("fundamental") subshell: n = principal quantum number = 5, l = 3; ml may be any integer ranging from negative three to positive three, inclusive
        for the 3d ("diffuse") subshell: n = 3, angular momentum quantum number = l = 2; ; ml may be any integer ranging from negative two to positive two, inclusive

        The way Logan phrased this first question in this sixth Problem Set of ours, I believe he used the old way of naming subshells, as opposed to the new way of naming subshells. If I were to name the subshells using the new system expediently presented in the website http://www26.brinkster.com/uptable/ups2.htm, I'd say that Logan's 5f subshell would be best called an 8f subshell under the new system, and that the 3d subshell, be called a 5d subshell. What is the old system's 5d subshell is nowadays a 7d subshell, and the new subshell naming system does not have anything by the name of 3d.

      2. 1.28 micrometers would be our wave length
        ...our wavelength being .2337 quadrillion Hertz. n = RH(1/n2 - 1/n2)
        What formula should we use? There is such a thing as c = nl as Logan told us in his October 30 lecture., but I guess we could focus on the quantum number n, particularly the two values for it that the question gives us. delta in energy equals Rydberg constant times the difference of the reciprocals of the initial and final principal quantum numbers' squares.
        How do we get the wavelength value that Logan wants? It would be the reciprocal of some interesting compound arithmetic calculation. That calculation is summed up as the product of the difference of the reciprocals of the initial and final principal quantum numbers's squares (5 and 3 in our case) with a number that itself is quite a cocktail of constants with the simple mass value. To wit, one of the juggernaut factors in our obtaining the wavelength's reciprocal is the quotient of mass and transcendental number e raised to the fourth power and (numerator follows as) eight times the vacuum permittivity constant's square times Planck's constant cubed and multiplied by the speed of light. Or we could just say heck that number must be like eleven million reciprocal meters.
      3. All we gotta do is multiplty our answer above by hc. c (the speed of light) must figure in because electromagnetic wave energy comes out.
      1. Zero radial nodes is what we get for the 1s orbital, considering the handy formula which expresses the number of radial nodes as the difference between the principal quantum number and the azimuthal quantum numberl. 1s also has got zero angular nodes.
      2. The 3pz orbital has got one radial node (a planar node). The xy plane is where such node exists for the 3pz orbital.
      3. As Israel and Kae showed on Sunday, 3dz2 has got two angular nodes. Kae said these two angular nodes may be projected as two cones one upon the other (sort of the spine we get for subsectioning conic sections in Analytical Geometry textbooks). I guess that the 3dz2 orbital has got no radial nodes.
      4. All original Ramuel Raagas chemical art using Microsoft paintbrush via Microsoft Word Nor would the 3dxy orbital give us any radial nodes. Does not sure matter whether our 3d orbital fetaure has a z2 or an xy orientation---- zero radial no density zones are all in store for it.
      1. [Kr]4d105s25p6 is Xenon's electron configuration.
      2. [Kr]4d105s25p56s1 is Xe*'s electron configuration.
      3. Zeff is the effective nuclear charge.
    1. The electron configuration of Oxygen atom is 1s22s22p4

      I could also say [He]2s22p4
      for Oxygen atom.
      Both first and second shell s orbitals have paired electrons so let's go look at O's 2p. Three orbitals are occupied, but I guess that two electrons are unpaired in the Oxygen atom, both these being 2p orbital loners.
      Silicon atom has got the configurtartion 1s22s22p63s23p2
      Silicon sure must got two lone electrons, unpaired they would be in their 3p orbitals.
      For Nickel I'd say the configuration is [Ar]4s23d8, or going long-hand, for Nickel we'd have 1s22s22p64s23d8
      For Gadolinium [Xe]6s25p14f7
      There are how many unpaired electrons in Gadolinium? I'd say there would be seven. All these seven unpaired electrons are in its 4f orbital. Gadolinium is paramagnetic. In fact, it and Iron are the only ferromagnetic elements in the entire universe.
    http://www.geocities.com/ultrastupidneal/Knowledge-Chemistry-Orbital.html
    http://www.chem.tue.nl/smo/MBE/gadolinium-DTPA.htm
    http://www.chemistry.uvic.ca/chem222/Notes/Lect2v3.PDF
    http://www.chemistry.uvic.ca/chem222/Tests/03MT1vak.PDF contains handy 3dxz illustration I will adapt for my question #2.d) drawing--- that is, for the 3dxy orbital