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 probably reason you don't have an answer is that an earlier post said the answer was B(carbon) and that makes no sense. The answer is Boron (A). Here is a link for you to look up the IP for boron.

The electron configuration for B is
1s2 2s2 2p1.
The first electron to be pulled off is the 2p1 and that is the easiest. The next two are approximately the same since both are 2s electrons but considerably higher than the first IP. The second one of any orbital is always higher because it is being pulled off against a more positive charge. The final two are in yet another closed shell and that makes them the highest of all.

In order to determine which element from the list could have the given successive ionization energies, we need to consider the trends of ionization energies across periods and groups in the periodic table.

The ionization energy is the energy required to remove an electron from an atom or ion in its gaseous state. Successive ionization energies refer to the energy required to remove each subsequent electron from an atom or ion.

Now, let's analyze the given data for the successive ionization energies: 0.807, 2.433, 3.666, 25.033, 32.834 MJ mol–1.

As we move across a period from left to right in the periodic table, the atomic number increases, indicating a greater number of protons in the nucleus. This results in stronger nuclear attraction for electrons, making it more difficult to remove them and leading to higher ionization energies.

Looking at the given data, we can see that the first ionization energy (0.807 MJ mol–1) is relatively low. This suggests that the element is in Group 1 or 2, as these groups tend to have low first ionization energies.

Now, let's analyze the trends in the successive ionization energies:

- The jump from the first ionization energy (0.807 MJ mol–1) to the second ionization energy (2.433 MJ mol–1) is significant. This indicates that the second electron is more tightly held and requires significantly more energy to remove.

- The second jump from the second ionization energy (2.433 MJ mol–1) to the third ionization energy (3.666 MJ mol–1) is relatively small. This suggests that the third electron is not as tightly held as the second electron.

- The third jump from the third ionization energy (3.666 MJ mol–1) to the fourth ionization energy (25.033 MJ mol–1) is large again. This indicates that the fourth electron is more strongly attracted and requires significantly more energy to remove.

- Finally, the jump from the fourth ionization energy (25.033 MJ mol–1) to the fifth ionization energy (32.834 MJ mol–1) is relatively small. This suggests that the fifth electron is not as tightly held as the fourth electron.

Based on these patterns, we can conclude that the element must have its configuration as follows:

1st electron - relatively low ionization energy
2nd electron - much higher ionization energy
3rd electron - relatively low ionization energy
4th electron - much higher ionization energy
5th electron - relatively low ionization energy

Among the given options, the only element that matches this pattern is nitrogen (N). Therefore, the correct answer is (C) N.