The ionic radius of this element is 1.95 times larger than its atomic radius. Its electormnegativity value is greater than 3.00 Paulings. Its 6th ionization energy is tremendously larger than its 5th ionization energy.
2006-10-29 04:04:55 · 3 answers · asked by Andrew 1 in Science & Mathematics ➔ Chemistry
The elements having electronegativies above 3 are: N, O, F and Cl.
Ionic radius of these elements are: 0.13, 1.40, 1.33, 1.81 (A)
atomic radius of these elements: 0.75, 0.65, 0.57, 0.97 (A)
Ratio is: 0.17, 2.15, 2.33, 1.87
6th / 5th ionization potential: 5.5, 1.21, 1.38, 1.43
Difficult to select one because the 1.95 figure in the text could not be validated. Nitrogen has the biggest jump in the ionization potential but the radius ratio is very low. If the radius ratio is to be understand as greater than 1.95 then the biggest jump in ionization ratio will be that of Chlorine.
2006-10-29 05:08:06 · answer #1 · answered by Dr. J. 6 · 0⤊ 0⤋
It must be fluorine (F).
It's ionic radius is greater than it's atomic radius, so it must be an anion.
It's electronegativity is great. F is the most electronegative of all the elements.
When it loses 5 electrons F(5+) has the same electronic configuration as the noble gas He. That's why it's 6th ionization energy is much larger than it's 5th
2006-10-29 04:12:15 · answer #2 · answered by Dimos F 4 · 2⤊ 1⤋
Fluorine.....
2006-10-29 04:19:41 · answer #3 · answered by jackwp2000 2 · 0⤊ 1⤋