26. The first five successive ionisation energies of an element are 0.807, 2.433, 3.666, 25.033,
32.834 MJ mol–1
. Which one element from those listed below could this element be?
•(A) B
(B) C
(C) N
(D) O
(E) F
the answer is B, but can you explain the process to answering it?
To determine which element from the given options could have the given ionization energies, we need to compare the pattern of ionization energies with the electron configuration of each element.
The ionization energy generally increases as we remove electrons from an atom. So, let's analyze the given values of ionization energies:
0.807, 2.433, 3.666, 25.033, 32.834 MJ mol–1.
In a periodic table, elements are arranged based on increasing atomic number. Going from left to right across a period, electrons are added to the same energy level (shell) but also successively filling subshells within each level. The first ionization energy corresponds to removing the outermost (valence) electron, and the successive ionization energies correspond to removing the inner electrons.
Based on this pattern, let's compare the given ionization energies to the electron configurations of the elements:
(A) B: 1s^2 2s^2 2p^1
(B) C: 1s^2 2s^2 2p^2
(C) N: 1s^2 2s^2 2p^3
(D) O: 1s^2 2s^2 2p^4
(E) F: 1s^2 2s^2 2p^5
Looking at the ionization energies and comparing them to the electron configurations, we see that the first ionization energy is relatively low for all elements because it corresponds to removing the outermost electron (valence electron). However, the second ionization energy shows a significant jump in energy for one of the elements.
The given ionization energies show a significant increase after the second ionization energy, suggesting that the element's electron configuration after losing two electrons corresponds to a stable electron configuration.
Based on this analysis, we can determine that the element from those listed above is Oxygen (O). The electron configuration of Oxygen is 1s² 2s² 2p⁴. After losing two electrons, the electron configuration becomes stable, and the remaining ionization energies increase significantly.
Therefore, the correct answer is (D) O.