Let’s start off with a question: Based on
information in the Periodic Table, which
do you think is bigger: Na or Mg?
Predict on paper (1) before continuing, and
explain why you picked the one you did.
11
12
Na
Mg
23.0
24.3
If you predicted that Mg is bigger, and you gave as
your reason that it has more protons and neutrons,
keep in mind that this might make the nucleus of
Mg bigger, but the nucleus is only a very, very tiny
portion of the whole atom.
11
12
Na
Mg
23.0
24.3
If you predicted that Mg is bigger because it has
more electrons, now you might be on to
something, but remember that those outermost
electrons in Na and Mg are being placed into the
same exact sublevel (the 3s).
11
12
Na
Mg
23.0
24.3
So perhaps this led you to predict that because
they both have the same number of occupied
electron energy levels, the two atoms should be
the same size.
11
12
Na
Mg
23.0
24.3
That may be closer to the truth, but remember
that each atom has its own unique spacing of the
energy levels...
So... let’s learn more about the forces at
play before we draw any conclusions...
11
12
Na
Mg
23.0
24.3
Let’s consider a lithium atom…
Since its atomic number is 3, we know it has three protons in
the nucleus. This means that its nucleus has a charge of 3+.
And Li’s electron configuration is 1s2 2s1
This means it has 2 electrons in the first level (1s2 2s1)
… and 1 electron in the second level (1s2 2s1)
So this is what Li’s Bohr diagram would look like:
Now let’s consider how much each electron is
attracted by the nucleus…
First, the two electrons on the lowest level… since they have no electrons
beneath them to shield them from the nucleus, they “feel” the full 3+ charge
pulling them in toward the nucleus. For these two electrons, the effective
nuclear charge (z*) is 3+.
Now, what about the one electron on the 2nd level… beneath it, there
are two electrons that act to shield it from the charge on the nucleus.
Thus, this electron only “feels” a charge of 1+ pulling it in toward the
nucleus. For this outermost electron, z* = 1+
z* = 1+
z* = 3+
In other words, these two electrons come
between the 3+ nucleus and the outermost
electron. These two electrons effectively
cancel out two of the protons. And this leaves
just one proton to attract the outermost
electron. So the effective nuclear charge “felt”
by the outermost electron is only 1+.
This blocking out of the nuclear charge by
lower level electrons is known as “shielding.”
Now let’s consider Bohr diagrams for the next two
elements on the periodic table: Be and B
How many protons would they each have in their nuclei? (Write down
your prediction before continuing
continuing.(2) Is this what you predicted?
And what about their electron configurations? (Write them down before
continuing
continuing. (3) Is this what you wrote down?
And so what would their Bohr diagrams look like? (Write them down
before continuing
continuing. (4) Is this what you were thinking?
Finally, what effective nuclear charges will be experienced by each of these
electrons? (Write them down before continuing (5) .) Is this what you had?
z* = 5+
z* = 4+
z* = 3+
Li
z* = 3+
z* = 2+
z* = 1+
Be
1s2 2s2
B
1s2 2s2 2p1
So what do these Bohr Diagrams and effective
nuclear charges tell us about these atoms?
First, compared to Li, Be’s electrons are being pulled in with a stronger
charge, both at the first level and the second. This stronger attractive
force would act to pull these levels in closer to the nucleus.
This means that Be is actually smaller than Li. One way to describe how
big atoms are is by the “atomic radius” -- the distance from the nucleus to
the outermost electrons. Li’s atomic radius is 145 pm while Be’s is only
105 pm. (A pm is a trillionth of a meter.) Now, how would B’s size
compare with the other two? Make a prediction (6) and explain why.
z* = 5+
z* = 4+
z* = 3+
145 pm
Li
z* = 3+
z* = 2+
z* = 1+
105 pm
Be
B
Is this what you predicted? And did you explain that B should be smaller
than either Li or Be because it has the greatest effective nuclear charge
(z*) pulling its electrons in? If Li’s atomic radius is 145 pm and Be’s is
105 pm, make a guess as to what B’s atomic radius might be (7).
Were you close?
z* = 1+
z* = 2+
z* = 3+
z* = 4+
145 pm
Li
z* = 3+
z* = 5+
105 pm
Be
85 pm
B
And this pattern of getting smaller and smaller as you move to the right
continues across the entire periodic table.
C
Li
Be
N
O
F
Ne
B
And this pattern of getting smaller and smaller as you move to the right
continues across the entire periodic table. But what happens after Ne???
Ne’s Bohr Diagram looks like this: With a 10+ charge on the nucleus and
effective nuclear charges of 10+ and 8+ pulling the electrons in very
close…
Li
Be
B
z* = 8+
z* = 10+
Ne
C
N
O
F
Ne
After Ne comes Na. Predict what Na’s Bohr diagram would look like (8).
It has an electron configuration: 1s2 2s2 2p6 3s1 (or [Ne] 3s1), which means
it has all the same electrons as Ne, plus one more electron in the 3rd level.
Is this what you predicted? And now predict what the z* values would be
for each level (9). Is this what you were thinking?
Li
Be
B
C
N
z* = 1+
z* = 9+
z* = 11+
z* = 8+
z* = 10+
Ne
O
Na
F
And what would you predict for the size of an Na atom?
Its outermost electron is only being held in place with a very weak 1+
effective nuclear charge. So how far out do you think that valence
(outermost) level would be (10) ? Is this what you were picturing?
Li
Be
B
C
N
z* = 1+
z* = 9+
z* = 11+
z* = 8+
z* = 10+
Ne
O
180 pm
Na
F
Placing these back into the series… shows an obvious break in the trend
for this period. Thus it would seem to make more sense to have Na start
a new period. After all, Na does have just one valence electron (just like
Li), and it has a large radius (just like Li). In fact Na is bigger! Write
down your prediction as to why you think Na is bigger than Li (11) .
Li
Be
B
C
N
O
F
Mg
Al
Si
P
S
Cl
Ne
Na
Ar
Did you say that Na is bigger because it has a whole extra level
compared to Li? If so, good job. But what happens next?
After Na is Mg, which has two electrons in its valence level (just like Be),
each of which experiences a z* of 2+ (just like Be). This would make Mg
smaller than Na, just as Be was smaller than Li.
After that, the pattern continues for period 3 just as it did for period 2.
Ne
Na
Notice how as you go left to right across the periodic table, the atomic
radius steadily decreases. And as you move down the periodic table, the
atomic radius gradually increases.
Li
Be
B
C
N
O
F
Na
Mg
Al
Si
P
S
Cl
Ne
Ar
Using these trends, and the periodic table below…
Predict (12) which would be larger: C or Si? Ans: Si
Ans: As
(13) Which would be larger: As or Kr?
Ans: S
(14) Which would be larger: S or Ne?
As it turns out, size isn’t the only characteristic of the atom influenced by
the effective nuclear charge. “Ionization energy” is also affected.
Ionization energy is the energy required to remove an electron from an
atom. The stronger the force holding the electron in place, the more
energy it will take to remove it.
Try to remove Li’s outermost electron below (see arrow) by clicking on it
repeatedly. That didn’t take very much energy, did it? Now predict (15) how
hard it will be to try to remove one of Be’s outermost electrons (see arrow).
Click on it and see if you were right. That was a little harder, wasn’t it? How
about B? How will its ionization compare with the other two (16)?
z* = 1+
z* = 2+
z* = 3+
Li
z* = 3+
z* = 4+
Be
z* = 5+
B
Go ahead and click on the electron (see arrow) and find out.
Was your prediction correct? With an effective nuclear charge of 3+
acting on the valence electrons, they are especially difficult to remove.
So what do you think the general trend is for ionization energy as you go
from left to right across the periodic table: Increase or decrease? And
why (17) ? If you said increase, you were right! And if you also said that
it increases because there is an increase in the effective nuclear charge
holding these valence electrons in place, well then you are a true genius!
z* = 1+
z* = 2+
z* = 3+
Li
z* = 3+
z* = 4+
Be
z* = 5+
B
So, with an effective nuclear charge of 8+ holding it in place, how hard
would it be to remove a valence electron from an atom of Ne (18)?
Go ahead and click on the electron (see arrow) to see just how hard…
So… are you exhausted!!??
z* = 8+
z* = 10+
Ne
Li
Na
And how do you think Na’s ionization energy
z* = 1+
would compare with Li’s?
Na has a lot more protons than Li and therefore
a much greater positive charge on the nucleus
(11+ compared to only 3+), but all that really
matters is the charge “felt” by the valence
electron (the z*). What are the z*’s on the
valence electrons in these two atoms (19) ?
If you said z* = 1+ for both good job!
So how will their ionization energies compare?
z* = 1+ Write down your prediction before continuing (20).
Now click on Li’s valence electron (see arrow) to
see how much energy it takes to remove it
Pretty easy, huh? Now try Na’s (see arrow)
Wow, that was even easier! Why (21)?
Even though they both have the same z*, since
Na is bigger, the electron is a little farther away
from the nucleus, and that means the force
holding it in place is a little weaker. Since the
force is a little weaker, the ionization energy is a
little lower.
Fill in the following blanks (22 & 23):
As you go from left to right across the periodic table, ionization energy
effective nuclear charge (z*) tends to _______.
increase because the ______________
increase
generally tends to ______
decrease .
As you go down the per. table, ionization energy generally tends to _______
atomic radius tends to ________
The z* _____________,
but the ___________
and this
remains the same
increase
weaker
causes a ________
force of attraction.
Using these trends, and the periodic table below…
(24) Which would have a greater I.E.: C or Si?
Ans: C
(25) Which would have a greater I.E.: As or Kr?
Ans: Kr
(26) Which would have a greater IE? S or Ne
Ans: Ne
THE
END