Final answer:
The Period 3 element corresponding to the given ionization energies is Phosphorus (P) as the significant jump in ionization energy between IE6 and IE7 indicates electrons are being removed from a deeper shell.
Step-by-step explanation:
The element from Period 3 that has the following ionization energies (IE1 = 1000, IE2 = 2250, IE3 = 3360, IE4 = 4560, IE5 = 7010, IE6 = 8500, IE7 = 27,100 kJ/mol) is Phosphorus (P). This can be determined by looking at the significant jump in ionization energy, which occurs when electrons are removed from a newly exposed energy shell. In this case, the jump between IE6 and IE7 indicates that the first six electrons are from the outer 3s and 3p subshells, while the seventh requires removing an electron from the more tightly bound 2p shell, characteristic of phosphorus with a configuration of [Ne]3s²3p³. Additionally, the provided ionization energies and trends of increasing ionization energy within a period supports this conclusion.