2.D
The fourth principal energy level has four sublevels: s, p, d,
and f. If the sublevel is completely filled, then s = 2 electrons,
p = 6 electrons, d = 10 electrons, and f = 14 electrons;
thus 2 + 6 + 10 + 14 = 32 total electrons for a full fourth principal energy
level.
3. B
The set of quantum numbers given was n = 3, l = 1, ml = 0, m
s = ±1/2. If n = 3,
this means that it’s a third energy level electron; if l = 1, then it’s a
p-sublevel electron; if ml
= 0, then it’s in the middle position of the set of three p orbitals. The
only tricky thing is that ms
= + or -?1/2. This means it’s either a p2 or a p5
electron. However, if it were p5, then one of the answer
choices would be argon (a noble gas), but it isn’t listed, so it must be the
p2, which makes silicon the correct answer.
4.A
The configuration given is 1s22s22p63s23p63d44s2.
The 3d4 is the important part—it means the element we desire
is in the first row of the d-block elements and is the fourth element in
that block, so it is Cr, or chromium.
5.C
The mass number and atomic number must be equal on both sides of the equation in
order for the equation to be properly balanced. We are looking for a component
that has a mass number of 4 and an atomic number of 2: helium. This is an
example of alpha decay, and the answer is
The complete equation is:
6.B
This problem is easily solved without a calculator, especially if you’ve been
practicing your math skills. The half-life is given as 9.98 minutes, which is
mighty close to 10 minutes. The total time is given as 60.0 minutes, so the
sample undergoes six half-lives. Start with this mass and keep cutting it in
half; each 10-minute half-life should be represented with an arrow, and you can
even put numbers under each arrow if you want, in order to keep track.
20.0 ? 10.0 ?5.0 ? 2.5 ? 1.25 ?0.625 ? 0.3125,
which is B.
7.E
This question asks about the second ionization energy. Remember that the
second ionization energy of any element is always larger than its first
ionization energy. The second ionization energy is significantly larger
if the second electron comes from a completed sublevel or principal energy
level. Na’s first electron removed is 3s1, while the second to
be removed comes from 2p6. There is a huge increase in
the amount of energy needed to remove that second electron because of the change
in principal energy levels.
8.T, T
(Do not fill in CE, for “correct explanation.”) Remember this type
of question? We told you about it in the first part of the book. Here we go:
statement I is true: hydrogen has a lower IE than helium. The electron removed
in each case is from the 1s sublevel, so Zeff becomes
very important: H has a Zeff of 1, while He has a Zeff
of 2. He attracts its electrons with more force; thus it requires more
energy to remove them. Statement II is also true: The halogens are highly
electronegative and form polar bonds with hydrogen. However, is statement II the
correct explanation for why statement I is true? No, so you would not fill in
the CE bubble.
9.E
Remember, atomic radii decrease moving across a period from left to
right. This is because protons are added to the nucleus and these protons
attract the electrons more strongly, pulling them in tighter and decreasing the
atomic radius. Atomic radius increases moving down a group or family,
however, because the value of n increases as you add another shell, the
electrons spend more time away from the nucleus, dissipating the attraction
between nuclear protons and themselves. Since fluorine and chlorine are both in
the same group but chlorine is below fluorine, we know that chlorine is bigger
than fluorine. Sulfur is to the left of both chlorine and fluorine, so it must
be the smallest of all. Arranged in order of increasing size, they are S, F, Cl,
or answer E.
10.T, T
(Do not fill in CE.) In general, the ionization energy increases with
increasing Zeff, and the same is true for second ionization
energies. B has a higher Zeff than does Be, and since the
second ionization energy would be needed to remove protons from the same
principal energy level, B’s higher Zeff means that its second
ionization energy would be higher than that of Be.
11.F, F
(Do not fill in CE.) While oxygen has a lower first ionization energy
than nitrogen due to the p4 anomaly, as you pass oxygen
(going from left to right across the period), the trend of increasing ionization
energy continues; the increased Zeff results in subsequent
elements in the period (such as fluorine) having a higher ionization energy than
oxygen. Oxygen does not have a higher Zeff than that of
fluorine, so the second statement is also false, and you would not fill in the
CE oval.
12.C
It simply requires the input of energy to remove an electron from any type of
atom. This is because energy must be put into the system in order to overcome
the favorable attraction between the electron and the positively charged atomic
nucleus. A reaction that must consume energy in order to proceed is known as
endothermic, while a reaction that gives off energy is exothermic; both of these
reactions are endothermic, and the answer is C.
13.C
Draw each structure using the rules for drawing Lewis dot structures. Remember
that resonance structures refer to two or more Lewis structures that are equally
good descriptions of a molecule. So, you’re looking for structures in which you
found yourself placing one double bond arbitrarily between a pair of atoms that,
elsewhere in the molecule, share only a single pair of atoms. You should also be
looking for a combination of single and multiple bonds occurring between the
central atom and a member of C, N, O, P, or S.
14.C
SF6 has six bonding sites as seen by its Lewis structure, drawn
below. It is octahedral. The four equatorial sites have F—S—F bond angles of
90°, while the two axial sites have F—S—F bond angles of 180°, so as you can
see, choice C is the best answer.
15.T, T, CE
The first statement is true: not all molecules that contain a polar bond are
themselves polar. And the second statement correctly explains why this is so: in
order for a molecule to be polar, it must contain at least one polar bond or
unshared pair of electrons that are not arranged symmetrically so as to
cancel each other out. If they are arranged so that they do cancel each other
out, the molecule will be nonpolar.
16.T, T
(Do not fill in CE.) While fluorine is indeed bigger than hydrogen, this
is not the reason NH3 is more polar than NF3. The true
reason for this is that the three N—F bonds are more polar than the N—H bonds:
the difference between the electronegativities of N and H are greater than that
between N and F, so in NH3, the nitrogen attracts electrons much more
strongly than do the hydrogens, and a significant dipole moment is created.
17.B
An octahedral geometry is produced from six bonding sites, sp3d2
hybridization, and an expanded octet. You know it cannot be BeCl2, BF3,
or CF4 since neither Be, B, nor C is from the third period or higher.
SeF6 and PF5 are both possibilities. The Lewis structures
for the remaining choices appear below:
You can see that PF5 has five bonding sites and is thus trigonal
bipyramidal. SeF6 has six bonding sites and is octahedral.
18.D
All of the Lewis structures appear below. Focus only on the positions of
the nuclei—you can’t see the lone pairs, but they determine the molecular shape
by their repulsions.
Next to display next topic in the chapter.
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