Animated Solution for Chemistry - Periodicity in Properties: The increasing order of the first ionisation enthalpies of the elements B, P, S and F (lowest first) is

The concept of Ionisation Enthalpy is one of the most fundamental and revealing properties in all of chemistry. It tells us exactly how tightly an atom holds onto its electrons. Imagine an atom as a fortress, and the electrons as its guards. The ionisation enthalpy is the amount of energy you need to supply to breach the fortress and pull away the outermost guard.

In this problem, we are tasked with ranking four specific elements—Boron ($B$), Phosphorus ($P$), Sulphur ($S$), and Fluorine ($F$)—in increasing order of their first ionisation enthalpies. To solve this, we cannot just rely on rote memorization. We must understand the intricate dance between the nucleus and the electron cloud.

The absolute first step in tackling any periodic property question is to visualize the battlefield. We must locate our contenders on the periodic table.

Boron ($B$) and Fluorine ($F$) reside in the second period. Boron is situated on the left side in Group 13, while Fluorine is positioned on the extreme right in Group 17.

Phosphorus ($P$) and Sulphur ($S$) are located directly below them in the third period. Phosphorus sits in Group 15, and Sulphur is right next to it in Group 16.

By mapping them out, we establish the horizontal and vertical relationships that will dictate our logic.

Before we dive into the specifics, let's establish the general trends that govern ionisation enthalpy.

As we move from left to right across a period, the effective nuclear charge ($Z_{eff}$) increases. The nucleus gains more protons, and because electrons are being added to the same principal shell, the shielding effect does not increase proportionally. The nucleus pulls the outermost electrons closer and holds them tighter. Consequently, the ionisation enthalpy increases.

Conversely, as we move down a group, new principal electron shells are added. The atomic radius increases significantly. The outermost electrons are now further away from the nucleus and are heavily shielded by the inner core electrons. This makes them easier to remove, meaning the ionisation enthalpy decreases down a group.

Let's apply these rules to our second-period elements, Boron and Fluorine.

Fluorine is just one electron shy of achieving a highly stable noble gas configuration. Its nucleus exerts a massive effective pull on its valence electrons. It holds onto them with incredible strength.

On the other hand, Boron is on the left side of the period. It has a single, relatively loosely bound electron in its $2p$ orbital ($1s^{2}2s^{2}2p^1$).

Therefore, it is abundantly clear that Fluorine has a significantly higher ionisation enthalpy than Boron. In fact, Fluorine has one of the highest ionisation enthalpies of all elements.

Now, we arrive at the most critical part of the problem—the comparison between Phosphorus and Sulphur in the third period.

If we blindly followed the left-to-right trend, we would assume that Sulphur, being to the right of Phosphorus, should have a higher ionisation enthalpy. However, nature loves symmetry, and this is where the trend breaks.

Let's examine their electronic configurations. Phosphorus has an atomic number of 15, giving it a valence configuration of $3s^{2}3p^3$. Notice that the $3p$ subshell is exactly half-filled. According to Hund's Rule, these three electrons occupy three separate $p$ orbitals with parallel spins. This symmetrical arrangement maximizes exchange energy, granting the atom an unexpected level of extra stability.

Sulphur, with an atomic number of 16, has a valence configuration of $3s^{2}3p^4$. The addition of that fourth electron forces pairing in one of the $p$ orbitals. This pairing introduces electron-electron repulsion, which slightly destabilizes the configuration.

Because of the profound stability of the half-filled $3p^3$ state, removing an electron from Phosphorus requires substantially more energy than removing one from Sulphur. Thus, $IE(P)>IE(S)$. This is a classic, high-yield exception that appears repeatedly in competitive exams.

We have established that $F>B$ and $P>S$. But how does Boron compare to Phosphorus and Sulphur?

A common trap is to assume that because Boron is in a higher period (Period 2), it must have a higher ionisation enthalpy than elements in Period 3. While the shell effect is important, we must also consider the group positions.

Boron is in Group 13. Its effective nuclear charge is quite low. Phosphorus and Sulphur are in Groups 15 and 16, respectively. As we move from Group 13 to Group 15/16, the nuclear pull increases dramatically.

In this specific comparison, the massive increase in effective nuclear charge for the Group 15 and 16 elements completely overpowers the fact that they are one shell lower than Boron. The nucleus of Phosphorus or Sulphur holds onto its valence electrons much more tightly than the nucleus of Boron holds onto its lone $2p$ electron.

Therefore, Boron actually has the lowest ionisation enthalpy among these three elements. $IE(B)<IE(S)<IE(P)$.

Let's bring all our logical deductions together to form the final sequence.

We know that Boron has the lowest value due to its low effective nuclear charge in Group 13.

Next comes Sulphur.

Then, stepping up, we have Phosphorus, which surpasses Sulphur purely because of its half-filled orbital stability.

Finally, sitting at the absolute top of the chart is Fluorine, with its immense nuclear pull and small atomic size.

Therefore, the correct increasing order of first ionisation enthalpies is:

Anticipating the Next Move

Now that we have mastered the first ionisation enthalpy, it is crucial to think like an examiner. What if the question asked for the second ionisation enthalpy?

Imagine we have already removed one electron from both Phosphorus and Sulphur. The $P^{+}$ ion now has a configuration of $3s^{2}3p^2$. The $S^{+}$ ion now has a configuration of $3s^{2}3p^3$.

Suddenly, the tables have turned! It is now Sulphur that possesses the highly stable, half-filled $p$ orbital configuration. Consequently, removing a second electron from Sulphur will require far more energy than removing a second electron from Phosphorus.

This means that while $I{E}_1(P)>I{E}_1(S)$, the trend reverses for the second electron: $I{E}_2(S)>I{E}_2(P)$. Always keep a close eye on how the electronic configuration evolves after each electron is removed. This dynamic thinking is what separates good students from great ones.