The five successive ionization enthalpies of an element are 800, 2427, 3658, 35024 and 32824 kJ mol-1. The number of…

🧠 The big jump in successive IE values flags the moment you breach the noble-gas core — count the electrons removed BEFORE the jump Successive ionization enthalpies climb gradually while you are still removing valence electrons. The instant the next electron has to come from the inner noble-gas core, the energy needed jumps by a factor of 5-10. That jump tells you exactly how many valence electrons there were.

🗺️ Read the data $IE_1 = 800$, $IE_2 = 2427$, $IE_3 = 3658$, $IE_4 = 35024$, $IE_5 = 32824$.

Ratios: $IE_2/IE_1 = 3.0$ — gradual rise (still in valence shell). $IE_3/IE_2 = 1.5$ — gradual. $IE_4/IE_3 = 9.6$ — huge jump. The fourth electron broke into the core.

So the element had only 3 valence electrons. The fourth electron required core-breaking energy.

A Period 3 element with 3 valence electrons is Aluminium ($3s^2\,3p^1$). $IE_1$ for Al is around 577 — but the question's 800 may correspond to B (Group 13, $IE_1 = 801$). Either way, the count of valence electrons is 3.

⚠️ The trap Option (a) 5 is tempting because the question gives 5 IE values. But the count of values listed is irrelevant — what matters is the position of the BIG JUMP. The jump comes after $IE_3$, so 3 electrons were valence. Always look for the ratio jump, not the count of values.

$\boxed{\text{Answer: (c)}}$