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 mikrometers 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. 93.341 kiloJoules per mole is the energy required to ionize the hydrogen atom.
        DE = 1.55 x 10-19 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.
      4. 6.4 Angstroms is our de Broglie wavelength. I got this by dividing Planck's constant by the momentum value which is merely the product of electron mass and the velocity specific to this numerical problem we've got here.
        Along the way, I determined that the kinetic energy of each photoelectron must amount to 5.2335 times ten to the nineteenth Joules

        Such kinetic energy value I considered to be the difference between hv and the photoelectric work function adjusted to give per photon Joules values rather than per mole.

        We must not mistake the Greek n for the Roman v. Although both quantities are taken over time, frequency has no consideration for the dimension L for length. In calculating for nde Broglie, however, we do quite need a length value, so as to complete our determination of quantitas motus
        ࢐ &#U+2192; &#rtarrow;
        p

        hn = K + E
        hn = Kinetic Energy plus Binding Energy
        hn = half the product of mass and Roman v for electron speed value squared, and all such added to the Binding Energy
        n = 1.25 x 1015 Hertz = 1.25 x 1015 second-1
        hn = (1.25 x 1015 Hertz = 1.25 x 1015 second-1) x Planck's constant =
        hn = 8.3 x 10-19 Joules
        hn = 8.285 x 10-19 Joules = K + B
      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.
        The promoted electron shirks its particular responsibility to go and help shield or "screen" the outside material micro-environment from Xenon's susbtantive full nuclear charge. Z stays the same when Zenon is excited. After all, the number of such element's protons remains as Z=54 = Zinitial = Zfinal (all not equal to , however, but greater than Zeff). In calculating our non-integral weighted average number of electrons, quantity-designated as S, such value must diminish so that the Z-to-S numerical gap increases, beseeming that of the next element in the period table--- Cesium, that is.
    1. The electron configuration of Oxygen atom is 1s22s22p4, or 1s22s22px22py12pz1
      As such, Oxygen has two unpaired electrons, distributed amongst the 2py and 2pz orbitals, we could say, although the x, y and z subscript designations are not as binding as s, p, d and f quantum letter designations.
      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

      Iodine has one unpaired electron in one of its three 5p orbitals (whether be it 5px, 5py or 5pz does not matter), and a configuration of [Kr]5s24d105p5

      For Iridium we have [Xe]6s24f145d106p3 as the configuration, with all three unpaired electrons being in 6p

      For Bismuth we have [Xe]6s24f145d106p3 as the configuration, with all three unpaired electrons being in 6p. Such configuration we get from Brown et al.'s Figure 6.28. The page before, we also have for Bismuth the long-hand confguration 1s22s22p63s23p63d104s24p64d104f145s25d105p66s26p3

      (Brown, et al., Sampble Exercise 6.9, pages 226-227).

      For Gadolinium [Xe]6s25p14f7
      There are how many unpaired electrons in Gadolinium? I'd say there would be eight. Most of these, seven to number, lay abouts in the 4f orbital. Gadolinium is paramagnetic. In fact, it and Iron are the only ferromagnetic elements in the entire universe.


    The website 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