South Pasadena • Honors Chemistry
Name
9 • Atomic Structure
Period
9.3
PROBLEMS
1. Predict each of the following.
(a) Larger radius:
Co or Ni
(b) Higher ionization energy:
Sn or Sb
(c) Lower ionization energy:
Be+ or Be2+
(d) Lower electron affinity:
Ga or Ge
(e) Smaller radius:
I
(f) Larger radius:
Zn or Ag
(g) Lower electronegativity:
Sn or Pb
(h) Lower ionization energy:
K or Ca
(i) Smaller radius:
Cs+ or Ba2+
(j) Lower ionization energy:
Mg or K
(k) Smaller radius:
Ba or Ba2+
(l) Higher electronegativity:
Se or Te
(m) Lower ionization energy:
Ne or Cl
(n) Higher ionization energy:
N or Si
(o) Lower electronegativity:
Cs or Ba
(p) Lower ionization energy:
N or P
(q) Larger radius:
Rb or Cs
(r) Smaller radius:
Al or Si
or Br
2‒
(s) Smaller radius:
S
or S
(t) Larger radius:
Br‒ or Rb+
(u) Larger radius:
Se2‒ or Kr
(v) Higher ionization energy:
Ca or Sr
(w) Higher electronegativity:
Na or Mg
(x) Smaller radius:
I‒ or Ba2+
(y) Greater electron affinity:
Cl or Br
(z) Larger radius:
Cs or Cs+
(aa)
Larger radius:
Br or Br‒
(bb)
Smaller radius:
I
(cc)
Higher ionization energy:
Ca+ or Ca2+
(dd)
Higher second ionization energy:
K or Ca
or Po
–
Date
PERIODIC TRENDS
2. Provide a scientific explanation of the following
trends.
(a) The radius of the Br atom (0.111 nm) is less
than the radius of the Br– ion (0.196 nm).
Both Br and Br– have 35 protons. While Br
has 35 electrons, Br– has 36 electrons.
An atom with fewer electrons experiences
less electron-electron repulsions, so outer
electron is more attracted to the nucleus.
Because Br has fewer electrons than Br–, it
experiences less electron-electron repulsions
and has a smaller radius.
(b) A calcium atom (1.76 Å) is larger than a zinc
atom (1.22 Å)
The outer electron in Ca and Zn are both in
the 4s subshell. Ca has 20 protons, while Zn
has 30 protons.
The outer electron of an atom with more
protons is more attracted to the nucleus.
Because Ca has more protons, its outer
electron is more attracted to the nucleus and
it has a larger radius.
(c) The radius of the chlorine atom (0.99 Å) is
smaller than the radius of the chloride ion, Cl–
(1.81 Å).
Both Cl and Cl– have 17 protons. While Cl
has 17 electrons, Cl– has 18 electrons.
An atom with fewer electrons experiences
less electron-electron repulsions, so outer
electron is more attracted to the nucleus.
Because Cl has fewer electrons than Cl–, it
experiences less electron-electron repulsions
and has a smaller radius.
(d) The radius of Ca2+ (0.99 Å) is smaller than the
radius of Cl‒ (1.81 Å), even though they are
isoelectronic.
(g) The first ionization energy of selenium (941.0
kJ/mol) is less than bromine (1139.9 kJ/mol),
but greater than tellurium (869.3 kJ/mol).
Both Ca2+ and Cl– have 18 electrons.
However, Ca2+ has 20 protons and Cl– has 17
protons.
The outer electron in both Se and Br are in
the 4p orbital, but Se has 34 protons while
Br has 35 protons.
The outer electron of an atom with more
protons is more attracted to the nucleus.
The outer electron of an atom with more
protons is more attracted to the nucleus.
Because Ca2+ has more protons, its outer
electron is more attracted to the nucleus and
it has a smaller radius.
Because Se has fewer protons, its outer
electron is less attracted to the nucleus and
requires less electrons to remove, resulting
in a lower ionization energy.
(e) Potassium (418.8 kJ/mol) has a lower firstionization energy than sodium (495.8 kJ/mol).
The outer electron of K is on the 4s subshell,
while that of Na is on the 3s subshell.
The outer electron of Se is in the 4p orbital,
and that of Te is in the 5p orbital.
An electron on a higher subshell experiences
greater shielding by the core electrons, so it
is further away from and less attracted to
the nucleus.
An electron on a lower subshell experiences
less shielding by the core electrons, so it is
closer to and more attracted to the nucleus.
Because K’s outer electron is on a higher
subshell, it is less attracted to the nucleus
and requires less energy to remove, so it has
a lower first ionization energy.
(f) The first ionization energies of Si, P, and Cl are
786, 1012, and 1251 kJ/mol, respectively.
The outer electrons of Si, P, and Cl are in
the 3p subshell. Si, P, and Cl have 14, 15,
and 17 protons, respectively.
The outer electron of an atom with more
protons is more attracted to the nucleus.
Because Si has the fewest number of
protons, its outer electron is least attracted
to the nucleus and requires the least amount
of energy to remove, so Si has the lowest
ionization energy.
Because Cl has the most number of protons,
its outer electron is most attracted to the
nucleus and requires the most amount of
energy to remove, so Cl has the highest
ionization energy.
Because Se’s outer electron is on a lower
subshell, it is more attracted to the nucleus
and requires more energy to remove, so it
has a higher first ionization energy.
(h) The second ionization energy of K (3052
kJ/mol) is greater than the second ionization
energy of Ca (1145 kJ/mol).
The second ionization energy of K removes
an electron from the 3p subshell, while that
of Ca removes an electron from the 4s
subshell.
An electron on a lower subshell experiences
greater shielding by the core electrons, so
they are further away from and less
attracted to the nucleus.
Because the second ionization energy of K
removes an electron from a lower subshell,
that electron is more attracted to the nucleus
and requires more energy to remove,
resulting in a higher second ionization
energy.