Australian School of Abu Dhabi
Unit 2 Atomic Structure HL
Worksheets 3 Electrons in an Atom
The graph below shows a plot of first ionization energy against atomic number.
a) Explain how the graph provides evidence for:
i. The maximum number of electrons in the 2p sub-level is six.
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ii. The 3s sub-level is lower in energy than the 3p sub-level.
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iii. The 4s sub-level is lower in energy than the 3d sub-level.
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iv. Hund’s rule
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b) Describe how the graph of 2nd Ionization energy against atomic number would differ to the graph
above.
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2. The fourth ionization energy of vanadium is 4600 kJ mol-1, the fifth ionization energy of vanadium is
6280 kJ mol-1 and the sixth ionization energy of vanadium is 12400 kJ mol-1.
(a) Explain why the 5th ionization energy for vanadium is higher than the 4th ionization energy.
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(b) Explain why the difference between the 5th and 6th ionization energies is much larger than
the difference between the 4th and 5th ionization energies of vanadium.
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3. The lines in the ultraviolet emission spectrum of hydrogen gas converge at 9.12 x 10-8 m.
a) Use this value to calculate the ionization energy of hydrogen in kJ mol-1.
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b) Explain why the convergence line in the ultraviolet spectrum must be used to calculate the ionization
energy rather than the convergence line in the visible spectrum.
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Answers
1. (a) i. The second main energy level contains s and p sub-levels. The s sub-level is full after two
electrons have been added successively (Li & Be). The p sub-level is then successively filled
until six electrons have been added to give Ne. After Ne the third level starts to be filled (3s)
showing that the maximum number of electrons that can occupy the 2p sub-level is 6.
ii. It is easier to remove a p electron from Al ([Ne]3s23p1) than it is to remove an s electron from
Mg ([Ne]3s2).
iii. K has a very low IE with the configuration [Ar]4s1 so the 4s fills before the 3d sub-level.
iv. The regular increase from B to N then the drop between N and O shows that the three p
orbitals are filled singly before the electrons are paired up.
(b) The values will all be higher and the graph is shifted to the left by 1 atomic number so the peaks
will be given by Li+, Na+ and K+ but the basic shape will remain the same.
2. (a) For the 4th ionization energy the equation is V3+(g) → V4+(g) + e– and for the 5th ionization
energy the equation is V4+(g) → V5+(g) + e―. The V4+ ion will attract electrons more strongly
than the V3+ ion so the 5th electron will be harder to remove.
(b) Both the 4th and the 5th ionisation energies involve the loss of 3d electrons. The 5th
ionization gives the V5+ ion with an electronic configuration of 1s22s22p63s23p6 i.e. the same as
the noble gas argon. It will be much more difficult to remove the 6th electron as it involves
removal of a 3p electron and the 3p level is much more strongly attracted to the nucleus than a
3d electron.
3. (a) E = hv = hc/λ =(6.63 x 10-34 x 3.00 x 108)/9.12 x 10-8 = 2.18 x 10-18 J for one electron
= 2.181 x 10-16 x 6.02 x 1023 = 1.31 x 106 J mol-1 = 1310 kJ mol-1 (to 3 SF)
(b) The convergence line in the ultraviolet spectrum is due to the transition from n = ∞ to the
lowest level n = 1 which is the level occupied by the one hydrogen electron in the ground
state. The convergence line in the visible spectrum is due to the transition from n = ∞ to n = 2.
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