The following 5 problems concern electron configuration, the shape of orbitals, and calculation of Bohr wavelengths. Please see the attached file for the fully formatted problems.
1. A hydrogen atom in a certain excited state has its electron in a 5f subshell. The electron drops down to the 3d subshell, releasing a photon in the process.
a) For each of these subshells. Give the n and 1 quantum numbers, and give the range of possible ml quantum numbers.
b) How many radial nodes and how many angular nodes does each of the orbitals in part a) have?
c) Determine the wavelength of light that would be emitted by this process.
d) The hydrogen atom now has a single electron in the 3d subshell. Calculate the energy (in kJ/mol) required to ionize this (excited state) hydrogen atom. This is the energy required to remove the electron completely from the atom.
a) Write the ground-state electron configuration and count the number of unpaired electrons in cesium.
b) The photoelectric binding energy of cesium is 183.7 kJ/mol. Light having a wavelength of 2.4 x iO in falls upon a cesium surface in an evacuated tube. Calculate the minimum deBroglie wavelength of the emitted photoelectrons.
3. For each of the following orbitals, provide a perspective sketch of the orbital in the Cartesian x, y, z coordinate system. Determine the number of angular nodes and the number of radial nodes in the orbital, and describe where those nodes fail. (For instance. the 2Px orbital has an angular node in the yz-plane.)
a) 1s b) 3px c) 3dx2-y2 d) 3dxz
4. Write the ground-state electron configuration and count the number of unpaired electrons in:
Si, Ni, Ir, and Gd
5. Identify the atom or ion corresponding to each of the following descriptions:
a) An atom with ground-state electron configuration [Kr]4d105s25p1
b) An atom with ground-state electron configuration [Xe]4f 145d66s2
c) An ion with charge ?2 and ground-state electron configuration [Ne]3s23p6
d) An ion with charge +4 and ground-state electron configuration [Ar]3d3
Please see the attached file for the complete solution.
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1.a. 5f principal quantum number (n) is 5
magnetic quantum number (l) is 3
(l always = 0,1,2,...n-1; and for s=0, p=1, d=2, f=3)
and the azimuthal (or angular momentum or "directional") quantum numbers (ml) can be -3,-2,-1,0,1,2,3 for the seven different f orbitals
(ml always = -l,-l+1...0...l-1,1)
Thus, for 3d, n=3, l=2, ml=-2,-1,0,1,2 (there are five d orbitals, whether they're 3d, 4d, or 5d)
1.b. For radial nodes, the formula is n-l-1, so for 5f, 5-3-1=2, and for 3d, 3-1-1=1
For angular nodes (sometimes "nodal planes," sometimes conical nodes), it's the same as l, so for 5f, there are 3, and for 3d, there are 2, per orbital.
FYI, there are some great pictures of orbitals at:
1.c. To calculate the wavelength of light emitted, you use the fact that the energy change is equal to the final energy minus the initial energy. The energy level is a function of the principal quantum number n.
The formula (the Bohr equation) ...
Five multi-part questions are fully solved and explained regarding ground-state electron configurations, excited-state electron configurations, calculation of a deBroglie wavelength for emission of a photoelectron, and the shape of orbitals.