Animated Solution for Chemistry - Periodicity in Properties: The $1^{\text{st}}$, $2^{\text{nd}}$ and the $3^{\text{rd}}$ ionization enthalpies $I_1$, $I_2$ and $I_3$, of four atoms with atomic numbers $n$, $n+1$, $n+2$ and $n+3$, where $n<10$, are tabulated below. What is the value of $n$?

The Secret Language of Ionization Enthalpies

Imagine you are a detective, and the periodic table is your map. You are given a mysterious table of numbers—successive ionization enthalpies for four consecutive elements. At first glance, it just looks like a wall of data. But in chemistry, these numbers tell a thrilling story about the hidden architecture of atoms.

Whenever you are presented with successive ionization energies ($I_1,I_2,I_3,\dots$), your primary objective is to hunt for the sudden jump. This massive spike in energy is the atom's way of screaming, "Stop! You've reached my stable noble gas core!" Let's decode this table step by step.

Analyzing Element $(n+2)$

Let's direct our attention to the element with atomic number $(n+2)$.

Its first ionization enthalpy ($I_1$) is a relatively low $496\text{ kJ/mol}$. This means the first electron is loosely bound and easy to pluck away. However, look at what happens when we try to remove the second electron. The second ionization enthalpy ($I_2$) skyrockets to an astonishing $4562\text{ kJ/mol}$!

This is nearly a nine-fold increase. What does this physical reality tell us? It means that after losing just one electron, the atom achieved a highly stable, fully-filled noble gas configuration. Removing another electron from this deeply stable core requires a monumental amount of energy. Therefore, element $(n+2)$ has exactly one valence electron. It must be an Alkali Metal (Group 1).

Analyzing Element $(n+3)$

To ensure our logic is bulletproof, let's look at the very next element, $(n+3)$.

For this atom, the jump from $I_1$ ($738\text{ kJ/mol}$) to $I_2$ ($1451\text{ kJ/mol}$) is a normal, expected increase. But look at the transition to the third electron. The energy leaps from $1451\text{ kJ/mol}$ straight to $7733\text{ kJ/mol}$!

This massive jump indicates that the first two electrons were relatively easy to remove, but the third one is locked away inside a stable noble gas core. Thus, element $(n+3)$ has exactly two valence electrons, making it an Alkaline Earth Metal (Group 2). This perfectly confirms our previous deduction, as Group 2 naturally follows Group 1 in the periodic table.

Finding Element $n$

Now that we have anchored our position in the periodic table, we can simply walk backwards to find our mystery element $n$.

If $(n+2)$ is a Group 1 Alkali Metal, then the element immediately preceding it, $(n+1)$, must be the element that ends the previous period. It has to be a Noble Gas (Group 18).

Taking one more step back, the element before the noble gas is element $n$. This places element $n$ squarely in Group 17, meaning it is a Halogen.

We know that $n$ is a halogen. The halogens are Fluorine (Atomic No. 9), Chlorine (17), Bromine (35), and so on.

The problem provides a critical constraint:

Out of all the halogens, only Fluorine ($n=9$) satisfies this condition.

But wait, there is a catch here! Could $n$ be $1$ (Hydrogen)? Hydrogen is often placed above the halogens because it can also gain one electron. However, look closely at the data table. The table provides values for $I_2$ and $I_3$ for element $n$. Hydrogen only possesses a single electron! It is physically impossible for Hydrogen to have a second or third ionization enthalpy.

Therefore, the trap is avoided, and we can confidently conclude that the atomic number $n$ is exactly 9.