The five successive ionization enthalpies of an element are 800, 2427, 3658, 35024, 32824 kJ mol$$^{-1}$$. The number of valence electrons in the element is:

First of all, successive ionization enthalpies $$I_{1}, I_{2}, I_{3}, I_{4}, I_{5}, \ldots$$ represent the energies required to remove the first, second, third and so on electrons from an isolated gaseous atom of an element. In general, as more electrons are removed, the positive charge on the ion left behind increases, so the remaining electrons are held more strongly; therefore $$I_{2} > I_{1},\; I_{3} > I_{2}$$ and so forth.

However, when all the outer-shell (valence) electrons have been removed and we begin to take electrons out of the next inner shell, the required energy shows a very large sudden jump. This is because inner-shell electrons are much closer to the nucleus and feel a far stronger effective nuclear charge. Therefore, by locating the first big jump in the given data, we can identify how many electrons were present in the outermost shell, i.e. how many valence electrons the neutral atom had.

We are given the following five successive ionization enthalpies of the unknown element:

$$I_{1}=800\ \text{kJ mol}^{-1},$$ $$I_{2}=2427\ \text{kJ mol}^{-1},$$ $$I_{3}=3658\ \text{kJ mol}^{-1},$$ $$I_{4}=35024\ \text{kJ mol}^{-1},$$ $$I_{5}=32824\ \text{kJ mol}^{-1}.$$

Let us now examine the relative increases step by step.

First, we compare $$I_{2}$$ with $$I_{1}$$:

$$I_{2}-I_{1}=2427-800 = 1627\ \text{kJ mol}^{-1}.$$

Next, we compare $$I_{3}$$ with $$I_{2}$$:

$$I_{3}-I_{2}=3658-2427 = 1231\ \text{kJ mol}^{-1}.$$

Both of these increases (1627 and 1231) are of the same order of magnitude, showing the normal gradual rise expected while we are still removing electrons from the same outer shell.

Now we compare $$I_{4}$$ with $$I_{3}$$:

$$I_{4}-I_{3}=35024-3658 = 31366\ \text{kJ mol}^{-1}.$$

Here we see an enormous jump (over 30 000 kJ mol−1), far larger than the earlier differences. This indicates that after the removal of the first three electrons, we have exhausted the valence shell and are now attempting to ionize an inner-shell electron, which is much more tightly bound.

Therefore, the neutral atom must have had exactly three electrons in its outermost shell. In other words, the number of valence electrons is $$3$$.

Among the given options, “3” corresponds to Option C.

Hence, the correct answer is Option C.