Asked by dave
The first ionization potential of the elements B, C, and N (atomic numbers 5, 6, and 7) steadily increases, but that of O is less than that of N. The best interpretation of the lower
value for O is that:
1. the ionization potential of N is a maximum and the values decrease steadily for the elements O, F, and Ne.
2. there is more shielding of the nuclear charge in O than in B, C, or N.
3. the electron removed from O is farther from the nucleus and therefore less tightly bound than that in N.
4. the electron removed from O corresponds to a different value of the quantum number L than that of the electron removed from B, C,or N.
5. the half-filled set of p orbitals in N makes it more difficult to remove an electron from N than from O.
I think the answer is 1, but I'm not sure.
value for O is that:
1. the ionization potential of N is a maximum and the values decrease steadily for the elements O, F, and Ne.
2. there is more shielding of the nuclear charge in O than in B, C, or N.
3. the electron removed from O is farther from the nucleus and therefore less tightly bound than that in N.
4. the electron removed from O corresponds to a different value of the quantum number L than that of the electron removed from B, C,or N.
5. the half-filled set of p orbitals in N makes it more difficult to remove an electron from N than from O.
I think the answer is 1, but I'm not sure.
Answers
Answered by
DrBob222
#1 CAN'T be right.
I.P. N = 1402
I.P. O = 1314
I.P. F = 1681
I.P. Ne = 2080
You can look these up yourself at www.webelements.com as I did. The table shows I.P. does NOT decrease steadily.
I would look at #5 for the correct answer.
I.P. N = 1402
I.P. O = 1314
I.P. F = 1681
I.P. Ne = 2080
You can look these up yourself at www.webelements.com as I did. The table shows I.P. does NOT decrease steadily.
I would look at #5 for the correct answer.
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