Final answer:
The period 3 element with the given ionization energies is likely phosphorus (P), as the significant jump between IE_5 and IE_6 implies that after the fifth electron is removed, the sixth electron is being removed from a core level, specifically from the neon core of phosphorus.
Step-by-step explanation:
To determine which period 3 element has the given ionization energies (IE_1 through IE_6), we need to analyze the pattern of the ionization energies. As we move across a period on the periodic table, ionization energies tend to increase. However, there's usually a significant jump in the energy required to remove an electron when an electron is being removed from a new, closer energy level to the nucleus (for example, moving from a valence electron to a core electron). This is because the core electrons are closer to the nucleus and more tightly bound due to a higher effective nuclear charge.
The provided ionization energies are:
- IE_1 = 1012 kJ/mol
- IE_2 = 1900 kJ/mol
- IE_3 = 2910 kJ/mol
- IE_4 = 4960 kJ/mol
- IE_5 = 6270 kJ/mol
- IE_6 = 22,200 kJ/mol
There is a significant jump between IE_5 and IE_6, suggesting that the first five electrons removed are valence electrons, and with the sixth electron, we are likely removing a core electron. Looking at the choices provided and considering the electron configurations of the period 3 elements, the element that has five valence electrons is phosphorus (P). Phosphorus has the configuration [Ne]3s23p3. Thus, the removal of the sixth electron would mean removing an electron from the more stable noble gas configuration of neon, hence the large increase in ionization energy.