Download Chemistry Ms. Boon Unit 2: Standard 1c review – 08.30.12

Chemistry
Ms. Boon
Unit 2: Standard 1c review – 08.30.12
Ionization Energy Graphing Activity: Answers
1. In general, what happens to the 1st ionization energy as you go across a period?
Answer: The 1st ionization energy increases as you go across a period.
Explanation: To answer this question, you must remember that a period is a row on the periodic table. Then, look
at your data table or on your graph for the elements within a period. For example, the period 2 elements are Li, Be,
B, C, N, O, F, and Ne. The first ionization energy for Li is 0.5 and the values increase all the way to Ne with 2.1. You
can repeat this with the period 1 and 3 elements in other periods.
2. In general, what happens to the 1st ionization energy as you go down a group/family?
Answer: The 1st ionization energy decreases as you go down a group.
Explanation: To answer this question, you must remember that a group/family is a column on the periodic table.
Then, identify elements that are in the same group. For example, the group 1 elements or alkali metals represented
on your graph are Li, Na, and K. Compare their 1st ionization energies. Li is 0.5, Na is 0.5 and K is 0.4. This shows a
small decrease going down the group. Try another group. For example, group 2 or alkaline earth metals are Be,
Mg, and Ca. Be is 0.9, Mg is 0.7, and Ca is 0.6. Again, this shows a decrease.
3. List the elements for which the 2nd ionization energy is significantly higher than the 1st.
Answer: Li, Na, and K.
Explanation: Use your graph or your data table to find the difference between the 1st and 2nd ionization energies.
I just looked at the graph and found the elements with the lowest first ionization energies and noticed that their 2nd
ionization energies were really high. Here is how you would do the math:
Element
Li
Na
K
1st ionization energy
0.5
0.5
0.4
2nd ionization energy
7.3
4.6
3.1
Increase (difference)
6.8
4.1
2.7
4. Explain why the elements you listed in your answer to question three have such large 2nd ionization
energies.
Answer: These elements usually form ions with a +1 charge. That is why their first ionization energy is so low.
They usually lose one electron to meet the octet rule. The second ionization energy is much higher because this is
the energy to pull off another electron after the atom has already formed the stable ion with a full outer shell of
electrons. Alkali metals usually form +1 ions so trying to form a +2 ion uses way more energy.
Explanation: You need to notice that the elements you listed in #3 were the alkali metals. You need to remember
that they form +1 ions and they do that to get a full outer shell of electrons. This makes them stable and very
happy. Then, you have to understand that the 2nd ionization energy is the energy required to pull off another
electron from that same atom that already lost one. Finally, you need to infer that pulling off that second electron
takes a whole lot of energy because it is basically disrupting the stable octet of electrons that has formed.
5. List the elements for which the 3rd ionization energy is significantly higher than the 2nd.
Answer: Be, Mg, Ca
Explanation: Same process as above for #3 except compare the 2nd ionization energy to the 3rd.
6. Explain why the elements you listed in your answer to question three have such large 3rd ionization
energies.
Answer: These elements are alkaline earth metals that usually form ions with a 2+ charge. So, they have low 2nd
ionization energies because they want to lose 2 electrons. However, once they lose two, they are very happy with
that full outer shell of electrons and it takes a whole lot of energy to pull off a third electron.
Explanation: Pretty much the same as #4 but the elements form 2+ ions. After 2 electrons come off, the atom has
that stable octet, making the energy to pull off another electron very high.