Animated Solution for Chemistry - Periodicity in Properties: In which of the following arrangements the order is not according to the property indicated against it ?

When studying the periodic table, it is tempting to rely entirely on general trends. However, the true beauty of chemistry lies in the exceptions—the moments where quantum mechanics overrides simple left-to-right or top-to-bottom rules. Let's break down the four properties mentioned in this problem and see where the trap lies.

Metallic Radius

As we move down a group in the periodic table, such as Group 1 (Alkali Metals), a new principal quantum shell is added at each step. Even though the nuclear charge increases, the shielding effect of the inner electrons and the addition of a new shell cause the outermost electrons to be further from the nucleus.

Therefore, the metallic radius steadily increases. The order given, $\text{Li}<\text{Na}<\text{K}<\text{Rb}$, perfectly follows this general trend and is absolutely correct.

Electron Gain Enthalpy

Electron gain enthalpy is the energy released when an electron is added to a neutral gaseous atom. Generally, as we move down a group, the atomic size increases, and the added electron is further from the nucleus, resulting in a less negative electron gain enthalpy.

However, Group 17 (Halogens) presents a classic exception. You might expect Fluorine ($\text{F}$) to have the most negative electron gain enthalpy. But Fluorine is extremely small. When a new electron tries to enter its $2p$ orbital, it experiences severe inter-electronic repulsion from the existing electrons. Chlorine ($\text{Cl}$), being larger, accommodates the incoming electron much more easily.

Thus, the correct order is $\text{I}<\text{Br}<\text{F}<\text{Cl}$. The statement provided in the question is true.

Ionisation Enthalpy

First ionisation enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom. The general trend dictates that ionisation enthalpy increases across a period from left to right due to increasing effective nuclear charge.

If we blindly follow the trend, we would write $\text{B}<\text{C}<\text{N}<\text{O}$. But let's look at the electronic configurations: Nitrogen ($\text{N}$): $1s^{2}2s^{2}2p^3$ Oxygen ($\text{O}$): $1s^{2}2s^{2}2p^4$

Nitrogen possesses a exactly half-filled $p$-subshell ($p^3$), which grants it extra quantum mechanical exchange energy and symmetrical stability. Oxygen, conversely, has one extra electron ($p^4$) that experiences repulsion. Oxygen actually wants to lose that electron to achieve the stable $p^3$ configuration.

Because of this, it requires more energy to remove an electron from Nitrogen than from Oxygen. The correct order is $\text{B}<\text{C}<\text{O}<\text{N}$. The given order $\text{B}<\text{C}<\text{N}<\text{O}$ is incorrect, making it our answer!

Isoelectronic Species

Finally, let's examine the ions: ${\text{Al}}^{3+},{\text{Mg}}^{2+},{\text{Na}}^{+},{\text{F}}^{-}$. What do they have in common? They all have exactly 10 electrons. They are isoelectronic species.

When the number of electrons is constant, the size of the ion is dictated entirely by the nuclear charge (the number of protons). A higher number of protons exerts a stronger electrostatic pull on the electron cloud, shrinking the ion.

Comparing their atomic numbers ($Z$): $\text{Al}$ ($Z=13$) $\text{Mg}$ ($Z=12$) $\text{Na}$ ($Z=11$) $\text{F}$ ($Z=9$)

Since Aluminum has the most protons, ${\text{Al}}^{3+}$ is the smallest. Fluorine has the fewest protons, making ${\text{F}}^{-}$ the largest. The order ${\text{Al}}^{3+}<{\text{Mg}}^{2+}<{\text{Na}}^{+}<{\text{F}}^{-}$ is perfectly correct.