CHEMISTRY 161
HW CH#7B
72, 72, 74, 76, 80, 84, 86, 88, 90, 94, 102, 106, 114, 118, 122, 126.
7-72
The principal quantum number, n, gives the relative energy and size of the orbital. The angular
momentum quantum number, , gives the shape of the orbital. The magnetic quantum number, m ,
gives the orbital’s orientation in space.
7-74
To uniquely identify a particular electron, we need four quantum numbers: n to identify its shell,
to define its subshell (type of orbital), m to define its orbital’s orientation, and ms to define its spin
(+ 12 or  12 ) .
7-76
(a) When = 2, there are (2  2 + 1) = 5 orbitals
(b) When = 1, there are (2  1 + 1) = 3 orbitals
(c) Because only one m value is listed, this set of quantum numbers describes 1 orbital.
7-80
(a) 2p is the orbital for n = 2,
(b) 5f is the orbital for n = 5,
(c) 3d is the orbital for n = 3,
(d) 4f is the orbital for n = 4,
= 1.
= 3.
= 2.
= 3.
7-84
(a) For n = 3, the allowed values of are 2, 1, and 0 and for = 2 the allowed values of m are
–2, –1, 0, 1, and 2; this combination of n = 3, = 2, m = 0, ms = – 12 is allowed.
(b) For n = 5, the allowed values of are 4, 3, 2, 1, and 0 and for = 4 the allowed values of m are
–4, –3, –2, –1, 0, 1, 2, 3, and 4; this combination of n = 5, = 4, m = 4, ms = + 12 is allowed.
(c) For n = 3, the allowed values of
are 2, 1, and 0 and for
= 0 the allowed value of m is 0;
because m cannot equal 1 when = 0, this combination of n = 3, = 0, m = 1, ms = + 12 is not
allowed.
(d) For n = 4, the allowed values of
are 3, 2, 1, and 0; because
cannot equal 4 for n = 4, this
combination of n = 4, = 4, m = 1, ms = – 12 is not allowed.
7-86
The electron configuration of [core]ns2 that all the group 2 elements share indicates the same
reactivity with oxygen to form RO. The two ns2 electrons can be lost relatively easily to form R 2+,
which then combines with O2–.
7-88
We are to explain why the transition metals often form ions with a 2+ charge. The metals have an
electron configuration of [noble gas core](n – 1)d xns2.
7-90
The orbitals described are: (a) 2p, for n = 2, = 1; (b) 5f, for n = 5,
2; and (d) 4f, for n = 4, = 3.
In increasing order of energy: (a) 2p < (c) 3d < (d) 4f < (b) 5f.
= 3; (c) 3d, for n = 3,
=
7-94
All of these elements (the alkali metals) have an ns1 configuration after a noble gas core.
7-102
Ion’s electron configuration: [Kr]4d 105s2 = X+
Atom’s electron configuration: [Kr]4d 105s25p1 = X
This atom is indium and the monatomic cation is In+. The s and d orbitals are completely filled so
there are no unpaired electrons for In+ in its ground state.
7-106
(a) The order of filling of orbitals for atoms after neon is 3s < 3p. This atom has a total of 3
electrons in its outer shell: 2 fill the 3s orbital, and one is placed in a 3p orbital. This configuration,
[Ne]3s23p1, does not represent an excited state.
(b) Because the 4s orbital is lower in energy than the 3d orbital and the 3d orbital is lower in energy
than the 4p orbital, the lowest energy configuration for this atom is [Ar]3d 104s24p1, so the
configuration [Ar]3d 104s14p2 represents an excited state.
(c) Because the 5s orbital is lower in energy than the 4d orbital and the 4d orbital is lower in energy
than the 5p orbital, the lowest energy configuration for this atom is [Kr]4d 105s2, so the configuration
[Kr]4d 105s15p1 represents an excited state.
(d) The order of filling of orbitals for atoms after neon is 3s < 3p < 4s. This atom has a total of 9
electrons in its outer shell: 2 fill the 3s orbital, 6 fill the 3p orbitals and one is placed in the 4s orbital.
This configuration, [Ne]3s23p64s1, does not represent an excited state.
7-114
(a) He has a higher IE than Li because the effective nuclear charge on the electrons in the closed
shell (1s) is very high. The Li 2s electron, however, is shielded by the two 1s electrons, and therefore
feels less effective nuclear charge and is more easily ionized.
(b) Lithium has a lower IE than Be because its 2s electron feels a lower Zeff than the 2s electrons on
Be do. This is because of the increased nuclear charge on Be (4 protons) versus Li (3 protons).
(c) Beryllium has a higher ionization energy than B even though there are more protons in boron’s
nucleus, which would cause us to reason that the electron on B would feel a higher Zeff. However, the
electron that is ionized in B is in the p orbital, which has a higher orbital energy than the s orbital,
making it easier to remove.
(d) Nitrogen has a higher ionization energy than O because oxygen’s ionized electron comes from a p
orbital in which 2 electrons are paired. Placing two electrons in the same orbital destabilizes them,
making O easier to ionize.
7-118
The electron configurations for the elements Mg, Al, and Si are
Mg: [Ne]3s2
Al: [Ne]3s23p1
Si: [Ne]3s23p2
When aluminum is ionized, it loses the electron from the higher-lying p orbital, so its IE is slightly
lower than magnesium’s. Silicon, with its increased atomic number, increases the Zeff on the p
electron so it is harder to ionize than aluminum.
7-122
(a) Silver forms a 1+ ion through the loss of a high-lying 5s electron. Palladium ([Kr]4d 85s2) and
cadmium ([Kr]4d 105s2) each lose two 5s electrons to form 2+ cations.
(b) The heavier group 13 elements may form 1+ cations through the loss of the np electron and form
3+ cations through the loss of np electrons and the two ns electrons.
(c) The heavier group 14 elements may form 2+ cations through the loss of two np electrons
and form 4+ cations through the loss of both np electrons and the two ns electrons. The group 4
elements may lose the two ns electrons to form 2+ cations and may lose both the ns electrons and the two
(n – 1)d electrons to form 4+ cations.
7-126
(d)
(a) O5+: [He]2s1
(b) One 2s electron and all four 2p electrons are removed from oxygen to make O5+.
(c) As electrons are removed from the oxygen atom, the negative electron has to be separated from
increasingly positively charged ions. Separation of a 1– charge from a 5+ charge will require more
energy than separating a 1– charge from a 4+ or lower charge. Also, the small amount of electron
shielding from the other electrons in the same shell has decreased, contributing to higher and higher
successive ionization energies.
6.626  10 –34 J  s  3.00  108 m/s

 1.27  10 –8 m or 12.7 nm
1.559  10 –17 J