Question
Question: The five successive ionization energies of the element \(X\): \({{I}_{1}}\),\({{I}_{2}}\),\({{I}_{3}...
The five successive ionization energies of the element X: I1,I2,I3,I4, and I5 are 800, 2427, 3658, 25024, 32824 kJ mol−1 respectively, then the number of valence electron present of element X are?
A) 3
B) 5
C) 4
D) 2
Solution
Ionization energies are calculated for an isolated gaseous atom, when electrons are removed from its valence shell and converted it into a cation. Ionization energies are also called ionization enthalpies. when consecutive electrons are removed, then successive ionization energies are observed.
Complete answer:
We have given the successive ionization energies, I1,I2,I3,I4, and I5 for an element (X), which means after removal of first five electrons from the valence shell, the ionization energies are 800, 2427, 3658, 25024, 32824 kJ mol−1 respectively. We have to calculate the number of valence electrons in the element (X).
The ionization energies, to unpair a paired electron are high; also the ionization energy in removing electrons from the inner shell that has a fully filled configuration is much higher.
As we can see that I1 is lower than I2 and I3, this means that after I1, the remaining electrons are paired. But we can see an effective increase in the ionization energies after removal of 3 electrons, I1, I2, I3, and then for I4, and I5 the difference is not much. So, this means that I4, and I5 have been removed from a different shell, but I1,I2 and I3 are removed from the same shell, which is the valence shell. So, the element X has three electrons as after losing 3 electrons, it will acquire stable inert gas configuration.
Therefore, the number of electrons in the valence shell of element (X) is 3. So, option (A) is correct.
Note: The ionization energies of I4, and I5are higher as, the electrons in them are removed from the stable noble gas configuration. So, this also proves the presence of 3 electrons in the valence shell.