The lowest-lying energy levels for an atom with Z = 13 are shown. Note that thei
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Question
The lowest-lying energy levels for an atom with Z = 13 are shown. Note that their energies and quantum states are specified on the diagram, and that the energy scale is not linear. (i) Place the 13 electrons on the appropriate energy levels in the diagram for the ground state of the atom. (ii) List the possible quantum numbers for the electron (or electrons) in the highest energy level. (iii) What is the ionization energy of the atom? Use the periodic table to write down the full electron configurations of 6^C, and 32^Ge. What would their electron configurations be if it were not necessary to satisfy the Pauli exclusion principle?Explanation / Answer
Thus electron will be filled according to 1s2, 2s2, 2p6, 3s2,3p1 (this is the ground state configuration)
Principle quantum number, n = 3(energy level where it exists)
azimuthal quantum number, l (specific for each orbital) for p orbital, l = 1
(Note: when l = 0, it is an "s" orbital; when l = 2 it is a "d" orbital)
m = magnetic number, values range from -l to +l, so for a "p" orbital, could be -1, 0, or +1 and which means that it could be on the x, y, or z-axis
You may choose m = -1, m = 0, or m = +1 since there is no way of predicting which of the axes the electron will be on, pick any one from here.
s = spin number = +1/2 or -1/2
choose either value, since there is no way to predict the direction of spin.
Thus quantum numbers are
n = 3, l=1, m=-1, s = +1/2 for the highest occupied electron.
E = (Z2/n2) * 13.6 eV
Where z is the atomic number, and n is the principle quantum number.
Thus, ionization energy is 255.377 eV
Part B:
Electronic configuration of C is 1s2, 2s2, 2px1, 2py1,
Electronic configuration of Ge is 1s2, 2s2, 2p6, 3s2,3p6, 4s2, 3d10, 4px1, 4py1
If they were not obeying 2px or 4 px will be having two electrons of same spin (Pauli exclusion principle states that no two electrons in an atom can have identical quantum numbers)
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