Electronegativity and element identification

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plstevens
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As part of the exploration of one of the moons of Jupiter, a probe digs under the surface, where it finds a pocket of gas. It starts to do some on-board experimentation. The data stored in the probe's computer include the following table of Mulliken electronegativity values of gases.

Element Electronegativity,x
(KJ/mol)

He 1178
Ne 1030
Ar 746
O 728
N 701
Kr 659
Xe 568
Rn 500


After doing some analysis, the probe determines that the gas pocket contains an element that has an ionization energy of I=1527 KJ/mol and an electron affinity of EA=33KJ/mol. What is the identity of this element?
Express your answer using the symbol for the element (e.g., Hg).


The book I have doesn't offer any help with this problem, if someone could help me i would greatly appreciate it
 
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plstevens said:
As part of the exploration of one of the moons of Jupiter, a probe digs under the surface, where it finds a pocket of gas. It starts to do some on-board experimentation. The data stored in the probe's computer include the following table of Mulliken electronegativity values of gases.

Element Electronegativity,x
(KJ/mol)

He 1178
Ne 1030
Ar 746
O 728
N 701
Kr 659
Xe 568
Rn 500


After doing some analysis, the probe determines that the gas pocket contains an element that has an ionization energy of I=1527 KJ/mol and an electron affinity of EA=33KJ/mol. What is the identity of this element?
Express your answer using the symbol for the element (e.g., Hg).


The book I have doesn't offer any help with this problem, if someone could help me i would greatly appreciate it
This seems very much an homework.
 
plstevens said:
The book I have doesn't offer any help with this problem, if someone could help me i would greatly appreciate it
What are the 3 definitions you may need in order to solve this problem. If they are not in your book, can you find them on the internet?

PS: Next time round, use the Homework Forums (other sciences) for help with chemistry problems. This thread will be moved there by an Mentor.