Bhavik Patel
Chemistry AP
Period: 3-4
Gangluff
Questions: 2
Explain each of the following observations using
principles of atomic structure and/or bonding
 a) Potassium has a lower first-ionization energy than
lithium
 b) The ionic radius of N3– is larger than that of O2–.
 c) A calcium atom is larger than a zinc atom.
 d) Boron has a lower first-ionization energy than
beryllium.
Answer: A
 Potassium has a lower first-ionization energy than lithium
 K (1s2)(2s2,2p6)(3s2,3p6) (3d0) (4s1) Z* = 19 – (10 · 1.00
+ 8 · 0.85) = 2.2
 Li (1s2)(2s1) Z* = 3 - (2 · 0.35) = 2.3
 Explanation not just in the difference in effective nuclear
charge. As one can there is not much. The primary reason is
related to the energy of a single s electron in the n = 4 level
and the n = 2 level. The electron in the n = 4 level is higher
in energy compared to an electron in n = 2 level and
therefore is easier to remove.
Answer: B
 The ionic radius of N3– is larger than that of O2–.
 N3– (1s2)(2s2,2p6) Z* = 7 – (2 · 0.85 + 7 · 0.35) =
2.85
 O2– (1s2)(2s2,2p6) Z* = 8 – (2 · 0.85 + 7 · 0.35) =
3.85
 The electrons in N3– experience a smaller effective
nuclear charge and are less bound to the nucleus
compared to the electrons in oxygen. The result is the
electrons in nitrogen will take up a occupy volume.
Answer: C
 A calcium atom is larger than a zinc atom.
 Ca (1s2)(2s2,2p6)(3s2,3p6) (3d0) (4s2) Z* = 20 – (10 · 1.00
+ 8 · 0.85 + 1 · 0.35) = 2.85
 Zn (1s2)(2s2,2p6)(3s2,3p6) (3d10) (4s2) Z* = 30 – (10 ·
1.00 + 18 · 0.85 + 1 · 0.35) = 4.35
 The electrons in the 4s level experience a greater effective
nuclear charge in Zn than in Ca. The electrons in Zn are
therefore attracted to the nucleus to a larger extent than the
electrons in calcium. So the atomic radius is smaller. In
general the Z* increases more slowly across a d block
compared to a p block. The reason is the electrons being added
in the dblock are inside the ns electrons, while for the p block
the electrons are not inside the ns electrons.
Answer: D
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Boron has a lower first-ionization energy than beryllium.
Be (1s2)(2s2) Z* = 4 – (2 · 0.85 + 1 · 0.35) = 1.95
B (1s2)(2s2,2p1) Z* = 8 – (2 · 0.85 + 2 · 0.35) = 2.6
We would expect from Z* arguments that it should require
more energy to remove the first electron in boron compared to
beryllium. However, the electron that is removed from boron
in in the 2p level which is higher in energy compared to the
2s level, therefore it is easier to remove. Recall that Z * for the
Group IA metals (after Li) are effectively the same value, yet
we all know the first ionization energy for the valence
electron in Cs is easier to remove than the valence electron in
sodium because the electron is located in a higher energy
level.