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Answer :
Final answer:
The element in the second period that could have the given ionization energies (800, 2430, 3700, 25000, 33000 kJ/mol) is Beryllium (Be).
Explanation:
The ionization energy of an element refers to the energy required to remove an electron from an atom or ion in the gaseous state. Based on the provided ionization energy values, the element in the second period that could have those ionization energies is Beryllium (Be).
Beryllium has an atomic number of 4 and is located in the second period of the periodic table. It has 2 valence electrons in its outermost shell. The first ionization energy refers to the energy required to remove one of these valence electrons, which corresponds to the given value of 800 kJ/mol. The subsequent ionization energies (IE2, IE3, IE4, IE5) represent the energy required to remove the remaining electrons sequentially.
In summary, the element in the second period that could have the given ionization energies (800, 2430, 3700, 25000, 33000 kJ/mol) is Beryllium (Be).
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Final answer:
The pattern in the ionization energy values suggests that the element is oxygen (O), which resides in the second period and has six valence electrons. There's a large jump between the third and fourth ionization energy due to the removal of an electron from a filled or half-filled s orbital.
Explanation:
When we analyze the ionization energies (IE) provided, we observe a pattern where the first three ionization energies increase at a moderate rate, but the change from the third to the fourth ionization energy is very significant. This large jump typically indicates that the fourth electron is being removed from a newer, lower, and more stable energy level. Considering the values given, ie1 = 800, ie2 = 2430, ie3 = 3700, ie4 = 25000, ie5 = 33000 (all in kJ/mol), we are looking for an element in the second period of the periodic table as the question implies.
It is known that the elements in the second period have no more than eight electrons, and the large increase from ie3 to ie4 suggests that removing the fourth electron requires a much greater amount of energy, as it starts to remove electrons that are closest to the nucleus in a filled or half-filled s orbital. This pattern is consistent with the element oxygen (O), which has six valence electrons. The first two ionization energies would involve removing electrons from the p orbital, the third would remove the last p orbital electron, and the fourth would require removing one of the more strongly bound electrons in the s orbital.
