Scps Chemistry Worksheet With Answers - Periodicity Page 8
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1
2
3
4
5
6
7
8
9
10
11Na
10
15
20
atomic number
The trend chart above is very similar to the 1
st
Ionization Energy trends located in your text book.
The trend shows the 1
st
IE for each element in the 3
rd
period of the PT. Starting the Na and
ending with Ar, the IE values in kJ/mol are, Na (495), Mg (737), Al (577), Si (786), P (1011), S
(999), Cl (1251), Ar (1520). As the nuclear charge increases moving across the periodic table,
the atomic radius decreases as valence e- are pulled closer to the nucleus. Moving across the
PT, there is an increase in valence e- but not an increase in inner core electrons. The shielding
effect of the inner core electrons is increasingly overcome by the increasing nuclear charge. Fore
these reasons, the energy required to remove the 1
st
electron increases as you move across the
PT. The above trend demonstrates this increase.
st
However, there are exceptions in that the 1
IE dips at elements Al and again at S. The reason
for this can be seen in the electron configurations. The electron configuration for Al shows 13
electrons, 1s
2
2s
2
2p
6
3s
2
3p
1
. The energy to remove the last e- from the 3p sublevel is less than the
energy to remove the electrons from the 3s sublevel because the 3s is a full sublevel and the
atom wants to hold onto the electrons in the full sublevels. Similar reasoning may be used when
explaining the dip between Phosphorus and Sulfur. P has an electron configuration that has a
half-filled 3p sublevel. The next element, S, has a 3p sublevel where one of the three orbitals
contains two e- while the other two orbitals only contain one e-. It is easier to remove the 4
th
e- in
Sulfur’s 3p sublevel than it is to remove the 3
rd
e- from Phosphorus 3p sublevel because half-filled
sublevels are more stable. Draw the orbital diagrams to prove this to yourself.
16. I am an element. I have a high electron affinity, (highly negative value), and my atomic
number is X. The element with atomic number X-1 has a lower ionization energy and a lower
electron affinity. The element with atomic number N+1 has a higher ionization energy and
basically no electron affinity (positive value). I am toxic in my elemental state, but I am very
commonly found in my nontoxic ionic form. Within my group, I have the second highest ionization
energy. Who am I? (support each step of your reasoning).
Chlorine
17. Write a chemical equation that shows the process or events in the formation of an anion.
A + e-
A-
18. What do transition metals have in common with respect to their electron configurations?
All transition metals are located in the d block. Their electron configurations begin with a filled s
sublevel of the energy level equal to the row number of the period they are located on. The s
sublevel is filled before the d sublevel ( s before d). Their valence shells all have 2 electrons in
the outermost shell.
SCPS Chemistry Worksheet – Periodicity -
page 8
19. Consider the table of the first four ionization energies for an element we will call A.
1
st
2
nd
3
rd
4
th
Ionization
energy in kJ/mol
578
1817
2745
11580
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