ap chemistry successive ionization energy question (you can do it!)the table above shows the first three ionization energies for atoms of four elements from the third period of the periodic table. the elements are numbered randomly. use the information in the table to answer the following questions.a. **identify element 3. explain your reasoning.b. **write the complete electron configuration for an atom of element 3.c. write the valence electron configuration for an atom of element 3. (note: valence configuration only includes configuration of valence electrons).d. what is the predicted oxidation state (charge) of the most common ion of element 2?e. *what is the symbol for element 2?f. **a neutral atom of which of the four elements will have the smallest radius?

Question

Accepted Answer

# A.
## Brief Explanations:
Element 3 has a large jump between the second and third ionization energies. This indicates that after losing two electrons, it is difficult to remove the third electron. In the third - period elements, magnesium ($Mg$) has an electron configuration of $[Ne]3s^{2}$. Losing two $3s$ electrons is relatively easy, but removing the third electron (from the $2p$ sub - shell, which is closer to the nucleus and more stable) requires much more energy.

# Answer:
Element 3 is magnesium ($Mg$).

# B.
## Brief Explanations:
The electron configuration of an atom is written based on the Aufbau principle (filling orbitals from the lowest energy to the highest). For magnesium ($Mg$, atomic number $12$):
- The first two electrons go into the $1s$ orbital ($1s^{2}$).
- The next two electrons go into the $2s$ orbital ($2s^{2}$).
- The next six electrons go into the $2p$ orbital ($2p^{6}$).
- The last two electrons go into the $3s$ orbital ($3s^{2}$).

# Answer:
$1s^{2}2s^{2}2p^{6}3s^{2}$

# C.
## Brief Explanations:
Valence electrons are the electrons in the outermost shell. For magnesium ($Mg$), the outermost shell is the third shell ($n = 3$), and the electrons in this shell are in the $3s$ orbital.

# Answer:
$3s^{2}$

# D.
## Brief Explanations:
Element 2 has a large jump between the first and second ionization energies. This means it is relatively easy to lose one electron. In the third - period elements, sodium ($Na$) has an electron configuration of $[Ne]3s^{1}$. Losing one $3s$ electron to form $Na^{+}$ (with a stable $[Ne]$ configuration) is favorable.

# Answer:
$+ 1$

# E.
## Brief Explanations:
As explained in part D, since element 2 has a large jump between the first and second ionization energies (similar to sodium's ionization energy trend, where $IE_{2}\gg IE_{1}$ because after losing one electron, it attains a noble - gas configuration), and sodium has the symbol $Na$.

# Answer:
$Na$

# F.
## Brief Explanations:
In the third period, as we move from left to right across the period, the atomic radius decreases. This is because the number of protons in the nucleus increases (increasing the nuclear charge), while the electrons are added to the same principal energy level. The increased nuclear charge pulls the electrons closer to the nucleus. Among the four elements (assuming they are $Na$, $Mg$, and two other third - period elements), the element with the highest first ionization energy (which is related to the effective nuclear charge and atomic size; higher ionization energy generally corresponds to a smaller atom due to a stronger hold on electrons) is likely to be the one on the right - hand side of the period. Element 1 has the highest first ionization energy among the four.

# Answer:
Element 1.

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Explanation:

A.

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B.